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Paper IX: Arithmetic Packets, Dyadic Towers, and Local Polynomial Deformation

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1758 symbol spellings in 5516 mathematical passages across 48 selected source files. Subscripts and superscripts have their own entries alongside their base symbols.

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=Notation table · 2012 passages

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HNotation table · 231 passages

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jNotation table · 504 passages

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kNotation table · 473 passages

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  • Notation table · visual_body.tex, line 35

    k · number of differences · order k, independent of n

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    k
mNotation table · 440 passages

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  • Notation table · visual_body.tex, line 68

    llll Prime branch · Allowed m · q · Row step for the original shift

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    m
\midNotation table · 92 passages

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nNotation table · 722 passages

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pNotation table · 1233 passages

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p^{r-1}Notation table · 12 passages

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  • Notation table · visual_body.tex, line 72

    p odd · mmid p-1 · p-1 · dfracp-1mp^r-1

    Exact mathematical passage
    \dfrac{p-1}{m}p^{r-1}
qNotation table · 325 passages

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  • Notation table · visual_body.tex, line 68

    llll Prime branch · Allowed m · q · Row step for the original shift

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    q
rNotation table · 573 passages

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+Definition passage · 971 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 15

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    \label{eq:arith-euler-egf}
     \sum_{d\ge0}E_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
ADefinition passage · 215 passages

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  • Definition passage: Dyadic truncation · Read passage · 20_arithmetic.tex, line 880

    [Dyadic truncation] Define β_m by the formal unit-series identity ∑_m≥0β_m z^2m=(fraczsinh z)^2, β_0=1, where the quotient means the inverse of the unit series (sinh z/z)^2. Define A(n)=frac3^2n+2-14(2n+1)(2n+2), S_m(n)=β_m2^2mfrac(2n)!(2n-2m)!A(n-m) (0≤ m≤ n), and T_M(n)=∑_m=0^min(M,n)S_m(n) (M,n∈ N).

    Exact mathematical passage
    
     A(n)=\frac{3^{2n+2}-1}{4(2n+1)(2n+2)},\qquad
     S_m(n)=\beta_m2^{2m}\frac{(2n)!}{(2n-2m)!}A(n-m)
           \quad(0\le m\le n),
    
aDefinition passage · 541 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 757

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

    Exact mathematical passage
    
     S_d(p)=\sum_{a=0}^{p-1}(-1)^a(2a+1)^d.
    
bDefinition passage · 222 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 761

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

    Exact mathematical passage
    
     \operatorname{sech}^2t=\sum_{m\ge0}\gamma_m\frac{t^{2m}}{(2m)!},
     \qquad
     \gamma_m=\sum_{b=0}^{2m}\binom{2m}{b}E_bE_{2m-b},
    
CDefinition passage · 287 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 21

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    
     U_{N,j}(n)=E_{Nn}^{(N,j)},\qquad C_{N,j}(n)=\frac{U_{N,j}(n)}{j!}.
    
C_{N,j}Definition passage · 48 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 21

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    
     U_{N,j}(n)=E_{Nn}^{(N,j)},\qquad C_{N,j}(n)=\frac{U_{N,j}(n)}{j!}.
    
DDefinition passage · 135 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 48

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

    Exact mathematical passage
    
     D=\frac{d}{dt},\qquad L_\lambda=\frac{D^N-\lambda}{p}.
    
dDefinition passage · 398 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 15

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    \label{eq:arith-euler-egf}
     \sum_{d\ge0}E_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
D^NDefinition passage · 2 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 48

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

    Exact mathematical passage
    
     D=\frac{d}{dt},\qquad L_\lambda=\frac{D^N-\lambda}{p}.
    
EDefinition passage · 120 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 15

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    \label{eq:arith-euler-egf}
     \sum_{d\ge0}E_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
E_{2m-b}Definition passage · 1 passage

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 761

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

    Exact mathematical passage
    
     \operatorname{sech}^2t=\sum_{m\ge0}\gamma_m\frac{t^{2m}}{(2m)!},
     \qquad
     \gamma_m=\sum_{b=0}^{2m}\binom{2m}{b}E_bE_{2m-b},
    
E_bDefinition passage · 2 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 761

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

    Exact mathematical passage
    
     \operatorname{sech}^2t=\sum_{m\ge0}\gamma_m\frac{t^{2m}}{(2m)!},
     \qquad
     \gamma_m=\sum_{b=0}^{2m}\binom{2m}{b}E_bE_{2m-b},
    
E_d^{(N,j)}Definition passage · 2 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 15

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    \label{eq:arith-euler-egf}
     \sum_{d\ge0}E_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
E_{Nn}^{(N,j)}Definition passage · 10 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 21

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    
     U_{N,j}(n)=E_{Nn}^{(N,j)},\qquad C_{N,j}(n)=\frac{U_{N,j}(n)}{j!}.
    
fDefinition passage · 160 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 100

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

    Exact mathematical passage
    f
f_dDefinition passage · 1 passage

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 42

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

    Exact mathematical passage
    
     \mathcal H_p=\left\{\sum_{d\geq0}f_d\frac{t^d}{d!}:
                        f_d\in\mathbb Z_p\right\}
           \subset\mathbb Q_p[[t]].
    
\mathcal HDefinition passage · 22 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 42

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

    Exact mathematical passage
    
     \mathcal H_p=\left\{\sum_{d\geq0}f_d\frac{t^d}{d!}:
                        f_d\in\mathbb Z_p\right\}
           \subset\mathbb Q_p[[t]].
    
hDefinition passage · 142 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 100

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

    Exact mathematical passage
    h,k,n\in\N
\mathcal H_pDefinition passage · 12 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 42

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

    Exact mathematical passage
    
     \mathcal H_p=\left\{\sum_{d\geq0}f_d\frac{t^d}{d!}:
                        f_d\in\mathbb Z_p\right\}
           \subset\mathbb Q_p[[t]].
    
LDefinition passage · 388 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 107

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

    Exact mathematical passage
    v_p(x)\ge L
L_\lambdaDefinition passage · 3 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 48

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

    Exact mathematical passage
    
     D=\frac{d}{dt},\qquad L_\lambda=\frac{D^N-\lambda}{p}.
    
L_\lambda^kDefinition passage · 1 passage

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 51

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

    Exact mathematical passage
    L_\lambda^k A\in\mathcal H_p
MDefinition passage · 304 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 767

    rse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

    Exact mathematical passage
    
     T^{\mathrm{odd}}_{p,M}(n)=\sum_{m=0}^{\min(M,n)}
           \binom{2n}{2m}\gamma_m p^{2m}S_{2(n-m)}(p).
    
\minDefinition passage · 38 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 767

    rse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

    Exact mathematical passage
    
     T^{\mathrm{odd}}_{p,M}(n)=\sum_{m=0}^{\min(M,n)}
           \binom{2n}{2m}\gamma_m p^{2m}S_{2(n-m)}(p).
    
NDefinition passage · 412 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 14

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    N\ge1
\NDefinition passage · 50 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 100

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

    Exact mathematical passage
    h,k,n\in\N
\mathrm{odd}Definition passage · 2 passages

Search aliases: odd mathrm

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 767

    rse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

    Exact mathematical passage
    
     T^{\mathrm{odd}}_{p,M}(n)=\sum_{m=0}^{\min(M,n)}
           \binom{2n}{2m}\gamma_m p^{2m}S_{2(n-m)}(p).
    
p^{2m}Definition passage · 7 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 767

    rse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

    Exact mathematical passage
    
     T^{\mathrm{odd}}_{p,M}(n)=\sum_{m=0}^{\min(M,n)}
           \binom{2n}{2m}\gamma_m p^{2m}S_{2(n-m)}(p).
    
p^aDefinition passage · 34 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 47

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

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    N=p^a
p^LDefinition passage · 76 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 110

    put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0. This finite convention makes the displayed exponents meaningful at st…

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    x\in p^L\Z_p
\mathbb QDefinition passage · 16 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 42

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

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     \mathcal H_p=\left\{\sum_{d\geq0}f_d\frac{t^d}{d!}:
                        f_d\in\mathbb Z_p\right\}
           \subset\mathbb Q_p[[t]].
    
\mathbb Q_pDefinition passage · 9 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 42

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

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     \mathcal H_p=\left\{\sum_{d\geq0}f_d\frac{t^d}{d!}:
                        f_d\in\mathbb Z_p\right\}
           \subset\mathbb Q_p[[t]].
    
SDefinition passage · 95 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 757

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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     S_d(p)=\sum_{a=0}^{p-1}(-1)^a(2a+1)^d.
    
S_{2(n-m)}Definition passage · 2 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 767

    rse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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     T^{\mathrm{odd}}_{p,M}(n)=\sum_{m=0}^{\min(M,n)}
           \binom{2n}{2m}\gamma_m p^{2m}S_{2(n-m)}(p).
    
S_dDefinition passage · 1 passage

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 757

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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     S_d(p)=\sum_{a=0}^{p-1}(-1)^a(2a+1)^d.
    
S_mDefinition passage · 6 passages

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  • Definition passage: Dyadic truncation · Read passage · 20_arithmetic.tex, line 880

    [Dyadic truncation] Define β_m by the formal unit-series identity ∑_m≥0β_m z^2m=(fraczsinh z)^2, β_0=1, where the quotient means the inverse of the unit series (sinh z/z)^2. Define A(n)=frac3^2n+2-14(2n+1)(2n+2), S_m(n)=β_m2^2mfrac(2n)!(2n-2m)!A(n-m) (0≤ m≤ n), and T_M(n)=∑_m=0^min(M,n)S_m(n) (M,n∈ N).

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     A(n)=\frac{3^{2n+2}-1}{4(2n+1)(2n+2)},\qquad
     S_m(n)=\beta_m2^{2m}\frac{(2n)!}{(2n-2m)!}A(n-m)
           \quad(0\le m\le n),
    
\operatorname{sech}Definition passage · 32 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 760

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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    \operatorname{sech}^2
\operatorname{sech}^2Definition passage · 10 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 760

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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    \operatorname{sech}^2
\sinhDefinition passage · 32 passages

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  • Definition passage: Dyadic truncation · Read passage · 20_arithmetic.tex, line 874

    [Dyadic truncation] Define β_m by the formal unit-series identity ∑_m≥0β_m z^2m=(fraczsinh z)^2, β_0=1, where the quotient means the inverse of the unit series (sinh z/z)^2. Define A(n)=frac3^2n+2-14(2n+1)(2n+2), S_m(n)=β_m2^2mfrac(2n)!(2n-2m)!A(n-m) (0≤ m≤ n), and T_M(n)=∑_m=0^min(M,n)S_m(n) (M,n∈ N).

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     \sum_{m\ge0}\beta_m z^{2m}=\left(\frac{z}{\sinh z}\right)^2,
     \qquad \beta_0=1,
    
TDefinition passage · 182 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 767

    rse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

    Exact mathematical passage
    
     T^{\mathrm{odd}}_{p,M}(n)=\sum_{m=0}^{\min(M,n)}
           \binom{2n}{2m}\gamma_m p^{2m}S_{2(n-m)}(p).
    
tDefinition passage · 370 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 15

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    \label{eq:arith-euler-egf}
     \sum_{d\ge0}E_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
t^{2m}Definition passage · 3 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 761

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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     \operatorname{sech}^2t=\sum_{m\ge0}\gamma_m\frac{t^{2m}}{(2m)!},
     \qquad
     \gamma_m=\sum_{b=0}^{2m}\binom{2m}{b}E_bE_{2m-b},
    
t^dDefinition passage · 18 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 15

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    \label{eq:arith-euler-egf}
     \sum_{d\ge0}E_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
t^{Nn}Definition passage · 10 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 15

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    \label{eq:arith-euler-egf}
     \sum_{d\ge0}E_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
T^{\mathrm{odd}}_{p,M}Definition passage · 2 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 767

    rse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

    Exact mathematical passage
    
     T^{\mathrm{odd}}_{p,M}(n)=\sum_{m=0}^{\min(M,n)}
           \binom{2n}{2m}\gamma_m p^{2m}S_{2(n-m)}(p).
    
T_MDefinition passage · 12 passages

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  • Definition passage: Dyadic truncation · Read passage · 20_arithmetic.tex, line 886

    n] Define β_m by the formal unit-series identity ∑_m≥0β_m z^2m=(fraczsinh z)^2, β_0=1, where the quotient means the inverse of the unit series (sinh z/z)^2. Define A(n)=frac3^2n+2-14(2n+1)(2n+2), S_m(n)=β_m2^2mfrac(2n)!(2n-2m)!A(n-m) (0≤ m≤ n), and T_M(n)=∑_m=0^min(M,n)S_m(n) (M,n∈ N).

    Exact mathematical passage
    \label{eq:arith-tm}
     T_M(n)=\sum_{m=0}^{\min(M,n)}S_m(n)\qquad(M,n\in\N).
    
UDefinition passage · 133 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 21

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

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     U_{N,j}(n)=E_{Nn}^{(N,j)},\qquad C_{N,j}(n)=\frac{U_{N,j}(n)}{j!}.
    
uDefinition passage · 238 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 111

    d Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0. This finite convention makes the displayed exponents meaningful at step zero, where all positive-order differe…

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    \nu_p(u)=v_p(u)
U_{N,j}Definition passage · 8 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 21

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

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     U_{N,j}(n)=E_{Nn}^{(N,j)},\qquad C_{N,j}(n)=\frac{U_{N,j}(n)}{j!}.
    
vDefinition passage · 340 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 106

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

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    v_p(0)=+\infty
v_pDefinition passage · 42 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 106

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

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    v_p(0)=+\infty
xDefinition passage · 389 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 107

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

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    v_p(x)\ge L
\ZDefinition passage · 164 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 110

    put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0. This finite convention makes the displayed exponents meaningful at st…

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    x\in p^L\Z_p
\mathbb ZDefinition passage · 162 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 42

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

    Exact mathematical passage
    
     \mathcal H_p=\left\{\sum_{d\geq0}f_d\frac{t^d}{d!}:
                        f_d\in\mathbb Z_p\right\}
           \subset\mathbb Q_p[[t]].
    
zDefinition passage · 115 passages

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  • Definition passage: Dyadic truncation · Read passage · 20_arithmetic.tex, line 874

    [Dyadic truncation] Define β_m by the formal unit-series identity ∑_m≥0β_m z^2m=(fraczsinh z)^2, β_0=1, where the quotient means the inverse of the unit series (sinh z/z)^2. Define A(n)=frac3^2n+2-14(2n+1)(2n+2), S_m(n)=β_m2^2mfrac(2n)!(2n-2m)!A(n-m) (0≤ m≤ n), and T_M(n)=∑_m=0^min(M,n)S_m(n) (M,n∈ N).

    Exact mathematical passage
    
     \sum_{m\ge0}\beta_m z^{2m}=\left(\frac{z}{\sinh z}\right)^2,
     \qquad \beta_0=1,
    
z^{2m}Definition passage · 3 passages

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  • Definition passage: Dyadic truncation · Read passage · 20_arithmetic.tex, line 874

    [Dyadic truncation] Define β_m by the formal unit-series identity ∑_m≥0β_m z^2m=(fraczsinh z)^2, β_0=1, where the quotient means the inverse of the unit series (sinh z/z)^2. Define A(n)=frac3^2n+2-14(2n+1)(2n+2), S_m(n)=β_m2^2mfrac(2n)!(2n-2m)!A(n-m) (0≤ m≤ n), and T_M(n)=∑_m=0^min(M,n)S_m(n) (M,n∈ N).

    Exact mathematical passage
    
     \sum_{m\ge0}\beta_m z^{2m}=\left(\frac{z}{\sinh z}\right)^2,
     \qquad \beta_0=1,
    
\Z_pDefinition passage · 54 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 110

    put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0. This finite convention makes the displayed exponents meaningful at st…

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    x\in p^L\Z_p
\mathbb Z_pDefinition passage · 33 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 42

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

    Exact mathematical passage
    
     \mathcal H_p=\left\{\sum_{d\geq0}f_d\frac{t^d}{d!}:
                        f_d\in\mathbb Z_p\right\}
           \subset\mathbb Q_p[[t]].
    
\betaDefinition passage · 43 passages

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  • Definition passage: Dyadic truncation · Read passage · 20_arithmetic.tex, line 873

    [Dyadic truncation] Define β_m by the formal unit-series identity ∑_m≥0β_m z^2m=(fraczsinh z)^2, β_0=1, where the quotient means the inverse of the unit series (sinh z/z)^2. Define A(n)=frac3^2n+2-14(2n+1)(2n+2), S_m(n)=β_m2^2mfrac(2n)!(2n-2m)!A(n-m) (0≤ m≤ n), and T_M(n)=∑_m=0^min(M,n)S_m(n) (M,n∈ N).

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    \beta_m
\beta_0Definition passage · 4 passages

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  • Definition passage: Dyadic truncation · Read passage · 20_arithmetic.tex, line 874

    [Dyadic truncation] Define β_m by the formal unit-series identity ∑_m≥0β_m z^2m=(fraczsinh z)^2, β_0=1, where the quotient means the inverse of the unit series (sinh z/z)^2. Define A(n)=frac3^2n+2-14(2n+1)(2n+2), S_m(n)=β_m2^2mfrac(2n)!(2n-2m)!A(n-m) (0≤ m≤ n), and T_M(n)=∑_m=0^min(M,n)S_m(n) (M,n∈ N).

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     \sum_{m\ge0}\beta_m z^{2m}=\left(\frac{z}{\sinh z}\right)^2,
     \qquad \beta_0=1,
    
\beta_mDefinition passage · 17 passages

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  • Definition passage: Dyadic truncation · Read passage · 20_arithmetic.tex, line 873

    [Dyadic truncation] Define β_m by the formal unit-series identity ∑_m≥0β_m z^2m=(fraczsinh z)^2, β_0=1, where the quotient means the inverse of the unit series (sinh z/z)^2. Define A(n)=frac3^2n+2-14(2n+1)(2n+2), S_m(n)=β_m2^2mfrac(2n)!(2n-2m)!A(n-m) (0≤ m≤ n), and T_M(n)=∑_m=0^min(M,n)S_m(n) (M,n∈ N).

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    \beta_m
\gammaDefinition passage · 32 passages

Search aliases: γ gamma

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 760

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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    \gamma_m
\gamma_0Definition passage · 2 passages

Search aliases: γ_0 gamma

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 766

    -length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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    \gamma_0=1
\gamma_mDefinition passage · 11 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 760

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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    \gamma_m
\DeltaDefinition passage · 153 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 101

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

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     \Delta_h f(n)=f(n+h)-f(n),\qquad
     \Delta_h^0 f(n)=f(n),\qquad
     \Delta_h^{k+1}f(n)=\Delta_h^k f(n+h)-\Delta_h^k f(n).
    
\Delta_hDefinition passage · 7 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 101

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

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     \Delta_h f(n)=f(n+h)-f(n),\qquad
     \Delta_h^0 f(n)=f(n),\qquad
     \Delta_h^{k+1}f(n)=\Delta_h^k f(n+h)-\Delta_h^k f(n).
    
\Delta_h^0Definition passage · 2 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 101

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

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     \Delta_h f(n)=f(n+h)-f(n),\qquad
     \Delta_h^0 f(n)=f(n),\qquad
     \Delta_h^{k+1}f(n)=\Delta_h^k f(n+h)-\Delta_h^k f(n).
    
\Delta_h^kDefinition passage · 8 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 101

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

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     \Delta_h f(n)=f(n+h)-f(n),\qquad
     \Delta_h^0 f(n)=f(n),\qquad
     \Delta_h^{k+1}f(n)=\Delta_h^k f(n+h)-\Delta_h^k f(n).
    
\Delta_h^{k+1}Definition passage · 1 passage

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 101

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

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     \Delta_h f(n)=f(n+h)-f(n),\qquad
     \Delta_h^0 f(n)=f(n),\qquad
     \Delta_h^{k+1}f(n)=\Delta_h^k f(n+h)-\Delta_h^k f(n).
    
\lambdaDefinition passage · 27 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 47

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

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    \lambda\in\mathbb Z_p
\nuDefinition passage · 23 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 111

    d Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0. This finite convention makes the displayed exponents meaningful at step zero, where all positive-order differe…

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    \nu_p(u)=v_p(u)
\nu_pDefinition passage · 4 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 111

    d Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0. This finite convention makes the displayed exponents meaningful at step zero, where all positive-order differe…

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    \nu_p(u)=v_p(u)
\inDefinition passage · 452 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 100

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

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    h,k,n\in\N
\sumDefinition passage · 272 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 15

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    \label{eq:arith-euler-egf}
     \sum_{d\ge0}E_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
\sum_{a=0}^{p-1}Definition passage · 4 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 757

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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     S_d(p)=\sum_{a=0}^{p-1}(-1)^a(2a+1)^d.
    
\sum_{b=0}^{2m}Definition passage · 1 passage

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 761

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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     \operatorname{sech}^2t=\sum_{m\ge0}\gamma_m\frac{t^{2m}}{(2m)!},
     \qquad
     \gamma_m=\sum_{b=0}^{2m}\binom{2m}{b}E_bE_{2m-b},
    
\sum_{d\ge0}Definition passage · 8 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 15

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    \label{eq:arith-euler-egf}
     \sum_{d\ge0}E_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
\sum_{d\geq0}Definition passage · 3 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 42

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

    Exact mathematical passage
    
     \mathcal H_p=\left\{\sum_{d\geq0}f_d\frac{t^d}{d!}:
                        f_d\in\mathbb Z_p\right\}
           \subset\mathbb Q_p[[t]].
    
\sum_{m=0}^{\min(M,n)}Definition passage · 2 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 767

    rse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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     T^{\mathrm{odd}}_{p,M}(n)=\sum_{m=0}^{\min(M,n)}
           \binom{2n}{2m}\gamma_m p^{2m}S_{2(n-m)}(p).
    
\sum_{m\ge0}Definition passage · 4 passages

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  • Definition passage: Odd-length source and inverse coefficients · Read passage · 20_arithmetic.tex, line 761

    [Odd-length source and inverse coefficients] For a positive odd integer p, define S_d(p)=∑_a=0^p-1(-1)^a(2a+1)^d. Let γ_m be the integral Hurwitz coefficients of sech^2: sech^2t=∑_m≥0γ_mfract^2m(2m)!, γ_m=∑_b=0^2mbinom2mbE_bE_2m-b, where E_b=0 for odd b. Then γ_0=1. Define T^odd_p,M(n)=∑_m=0^min(M,n) binom2n2mγ_m p^2mS_2(n-m)(p). Here p need only be odd; primality is needed only to interpret the result as a prime-adic congruence ladder.

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     \operatorname{sech}^2t=\sum_{m\ge0}\gamma_m\frac{t^{2m}}{(2m)!},
     \qquad
     \gamma_m=\sum_{b=0}^{2m}\binom{2m}{b}E_bE_{2m-b},
    
\sum_{n\ge0}Definition passage · 20 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 15

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

    Exact mathematical passage
    \label{eq:arith-euler-egf}
     \sum_{d\ge0}E_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
\inftyDefinition passage · 22 passages

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  • Definition passage: Differences and valuation conventions · Read passage · 20_arithmetic.tex, line 106

    [Differences and valuation conventions] For a sequence f and h,k,n∈ N, put Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f(n)=f(n), Δ_h^k+1f(n)=Δ_h^k f(n+h)-Δ_h^k f(n). We use the extended valuation v_p(0)=+∞ in mathematical statements. Thus v_p(x)≥ L, including negative integers L, means x=0 or the ordinary valuation of the nonzero rational x is at least L. For integral coefficients and L≥0, it is equivalent to x∈ p^L Z_p. In expressions involving a step multiplier we write ν_p(u)=v_p(u) for u>0 and set ν_p(0)=0…

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    v_p(0)=+\infty
\neDefinition passage · 21 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 20

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

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    1/j!\ne0
\leDefinition passage · 91 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 14

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

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    0\le j<N
\geDefinition passage · 219 passages

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  • Definition passage: Congruential Euler coefficients · Read passage · 20_arithmetic.tex, line 14

    [Congruential Euler coefficients] For N≥1 and 0≤ j<N, define rational coefficients by the formal identity ∑_d≥0E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The reciprocal exists because its denominator has constant coefficient 1/j!≠0. Set U_N,j(n)=E_Nn^(N,j), C_N,j(n)=fracU_N,j(n)j!. Thus C_N,j(0)=1 and U_N,j(0)=j!. All coefficients outside the multiples of N vanish.

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    N\ge1
\geqDefinition passage · 229 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 42

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

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     \mathcal H_p=\left\{\sum_{d\geq0}f_d\frac{t^d}{d!}:
                        f_d\in\mathbb Z_p\right\}
           \subset\mathbb Q_p[[t]].
    
\subsetDefinition passage · 3 passages

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  • Definition passage: Hurwitz integral series · Read passage · strengthening.tex, line 42

    [Hurwitz integral series] For a prime p, let H_p={∑_d≥0f_dfract^dd!: f_d∈ Z_p} ⊂ Q_p[[t]]. For N=p^a and λ∈ Z_p, put D=fracddt, L_λ=fracD^N-λp. A series A is λ-flat if L_λ^k A∈ H_p for every k≥0.

    Exact mathematical passage
    
     \mathcal H_p=\left\{\sum_{d\geq0}f_d\frac{t^d}{d!}:
                        f_d\in\mathbb Z_p\right\}
           \subset\mathbb Q_p[[t]].
    
*Introduction passage · 1 passage

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  • Introduction passage · Read passage · 72_recovered.tex, line 2041

    Let c=a*b be the ordinary Cauchy convolution. Write A_r, B_r, and C_r for the phase ordinary generating functions.

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    c=a*b
\mathbf1Introduction passage · 10 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 129

    In characteristic p, factorial division is unavailable in general, but the coefficient operation remains meaningful over every commutative ring: (f⋆ g)_n=∑_r=0^nbinom nr f_r g_n-r, (D^d f)_n=f_n+d, 1_n=1, · n=0, 0, · n>0. Here ⋆ is Hurwitz multiplication. The symbol D^d denotes a coefficient shift, which agrees with formal differentiation after characteristic-zero encoding.

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    (f\star g)_n=\sum_{r=0}^{n}\binom nr f_r g_{n-r},\qquad
    (D^d f)_n=f_{n+d},\qquad \mathbf1_n=\begin{cases}1,&n=0,\\0,&n>0.\end{cases}
    
\mathbf1_nIntroduction passage · 2 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 129

    In characteristic p, factorial division is unavailable in general, but the coefficient operation remains meaningful over every commutative ring: (f⋆ g)_n=∑_r=0^nbinom nr f_r g_n-r, (D^d f)_n=f_n+d, 1_n=1, · n=0, 0, · n>0. Here ⋆ is Hurwitz multiplication. The symbol D^d denotes a coefficient shift, which agrees with formal differentiation after characteristic-zero encoding.

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    (f\star g)_n=\sum_{r=0}^{n}\binom nr f_r g_{n-r},\qquad
    (D^d f)_n=f_{n+d},\qquad \mathbf1_n=\begin{cases}1,&n=0,\\0,&n>0.\end{cases}
    
A_rIntroduction passage · 3 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2041

    Let c=a*b be the ordinary Cauchy convolution. Write A_r, B_r, and C_r for the phase ordinary generating functions.

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    A_r
A_xIntroduction passage · 25 passages

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  • Introduction passage · Read passage · continuation_01.tex, line 22

    Let F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, q≥1, L≥1, x∈ Z^d, F(x)≡0pmodq^L. Write R_q= Z/q Z, let ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

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    r_x=F(x)/q^L\in\mathbb Z^e,
    \qquad A_x=DF(x):\mathbb Z^d\longrightarrow\mathbb Z^e,
    \qquad J_x=\rho_q\circ A_x.
    
BIntroduction passage · 197 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2041

    Let c=a*b be the ordinary Cauchy convolution. Write A_r, B_r, and C_r for the phase ordinary generating functions.

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    B_r
B_1Introduction passage · 8 passages

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  • Introduction passage · Read passage · strengthening.tex, line 411

    Let B_m denote the Bernoulli numbers with B_1=-1/2. The identity obtained by substituting u=e^t-1 into log(1+u)/u is B_m=∑_r=0^m(-1)^rfracr!r+1 {m

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    B_1=-1/2
B_4^sIntroduction passage · 1 passage

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  • Introduction passage · Read passage · supplement_11.tex, line 78

    Write H_int for the subring of Q_2[[t]] whose factorial-scaled coefficients belong to Z_2. Flatness means B_4^sF∈ H_int for every s. The proof of multiplicative and reciprocal closure is induction on s, using the full product correction. It separately proves that the actual reciprocal has integral Hurwitz coefficients when F(0)=1.

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    B_4^sF\in\mathcal H_{\mathrm{int}}
B_jIntroduction passage · 16 passages

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  • Introduction passage · Read passage · strengthening.tex, line 1089

    [Exact image and residue enumeration] Let widetilde C_j denote the dyadic Newton interpolation. The map T_j(x)=fracwidetilde C_j(x)-12^b_j is a bijective isometry Z_2→ Z_2, and widetilde C_j( Z_2)=1+2^b_j Z_2, j!widetilde C_j( Z_2)=j!+2^B_j Z_2. For r>B_j, one least period modulo 2^r visits every residue in j!+2^B_j Z/2^r Z exactly once. For r≤ B_j, there is one residue and the least row period is one.

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     \widetilde C_j(\mathbb Z_2)=1+2^{b_j}\mathbb Z_2,
     \qquad j!\widetilde C_j(\mathbb Z_2)=j!+2^{B_j}\mathbb Z_2.
     \label{eq:four-interpolation-images-final}
    
b_jIntroduction passage · 5 passages

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  • Introduction passage · Read passage · strengthening.tex, line 1085

    [Exact image and residue enumeration] Let widetilde C_j denote the dyadic Newton interpolation. The map T_j(x)=fracwidetilde C_j(x)-12^b_j is a bijective isometry Z_2→ Z_2, and widetilde C_j( Z_2)=1+2^b_j Z_2, j!widetilde C_j( Z_2)=j!+2^B_j Z_2. For r>B_j, one least period modulo 2^r visits every residue in j!+2^B_j Z/2^r Z exactly once. For r≤ B_j, there is one residue and the least row period is one.

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     T_j(x)=\frac{\widetilde C_j(x)-1}{2^{b_j}}
    
B_mIntroduction passage · 3 passages

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  • Introduction passage · Read passage · strengthening.tex, line 411

    Let B_m denote the Bernoulli numbers with B_1=-1/2. The identity obtained by substituting u=e^t-1 into log(1+u)/u is B_m=∑_r=0^m(-1)^rfracr!r+1 {m

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    B_m
B_rIntroduction passage · 3 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2041

    Let c=a*b be the ordinary Cauchy convolution. Write A_r, B_r, and C_r for the phase ordinary generating functions.

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    B_r
cIntroduction passage · 225 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2041

    Let c=a*b be the ordinary Cauchy convolution. Write A_r, B_r, and C_r for the phase ordinary generating functions.

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    c=a*b
C^NIntroduction passage · 21 passages

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  • Introduction passage · Read passage · supplement_06.tex, line 69

    Principal targets. targetCompanion.orbit_of_recurrence targetCompanion.companion_jump, Companion.source_companion_jump targetCompanion.packet_step, Companion.packet_advance targetCompanion.packet_refinement, Companion.packet_zero_step pageheading2. From the characteristic polynomial to a recurrenceTheorem 9.1, continued Put D=C^N and let χ_D(X)=det(XI-D)=X^d+∑_a=0^d-1c_aX^a be its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands pol…

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    D=C^N
c_aIntroduction passage · 3 passages

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  • Introduction passage · Read passage · supplement_06.tex, line 70

    Principal targets. targetCompanion.orbit_of_recurrence targetCompanion.companion_jump, Companion.source_companion_jump targetCompanion.packet_step, Companion.packet_advance targetCompanion.packet_refinement, Companion.packet_zero_step pageheading2. From the characteristic polynomial to a recurrenceTheorem 9.1, continued Put D=C^N and let χ_D(X)=det(XI-D)=X^d+∑_a=0^d-1c_aX^a be its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands pol…

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     \chi_D(X)=\det(XI-D)=X^d+\sum_{a=0}^{d-1}c_aX^a
    
C_rIntroduction passage · 3 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2041

    Let c=a*b be the ordinary Cauchy convolution. Write A_r, B_r, and C_r for the phase ordinary generating functions.

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    C_r
\coshIntroduction passage · 29 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 66

    The central change in this edition is the treatment of the odd and dyadic Euler formulas. They are no longer presented as unrelated reconstruction devices. For each prime p we define an even source series H_p and an inverse kernel K_p satisfying K_p(t)H_p(t)=secht. At an odd prime, H_p(t)=cosh^2(pt)secht, K_p(t)=sech^2(pt). At the dyadic prime, H_2(t)=fracsinh(2t)sinh t2t^2, K_2(t)=frac4t^2sinh^2(2t). The coefficient convolution is identical in form, but the kernels have different integral weigh…

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    H_p(t)=\cosh^2(pt)\operatorname{sech}t,
    \qquad
    K_p(t)=\operatorname{sech}^2(pt).
    
\cosh^2Introduction passage · 8 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 66

    The central change in this edition is the treatment of the odd and dyadic Euler formulas. They are no longer presented as unrelated reconstruction devices. For each prime p we define an even source series H_p and an inverse kernel K_p satisfying K_p(t)H_p(t)=secht. At an odd prime, H_p(t)=cosh^2(pt)secht, K_p(t)=sech^2(pt). At the dyadic prime, H_2(t)=fracsinh(2t)sinh t2t^2, K_2(t)=frac4t^2sinh^2(2t). The coefficient convolution is identical in form, but the kernels have different integral weigh…

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    H_p(t)=\cosh^2(pt)\operatorname{sech}t,
    \qquad
    K_p(t)=\operatorname{sech}^2(pt).
    
\cosh_1Introduction passage · 1 passage

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  • Introduction passage · Read passage · supplement_02.tex, line 36

    r pageheading1. The actual odd-power generating functionLemma 4.1; manuscript page 10 Work in the rational formal power-series ring Q[[t]]. Define S_n(p)=∑_a=0^p-1(-1)^a(2a+1)^n∈ Z, F_p(t)=∑_n≥0S_n(p)fract^nn!. The formal exponential is Mathlib's mcodePowerSeries.exp; scalar substitution is its actual mcoderescale homomorphism. We define exp_a(t)=e^at, cosh_a(t)=frace^at+e^-at2, sech(t)=cosh_1(t)^-1. The constant term of cosh_a is proved to be one, so the reciprocal used here is a genuine unit i…

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     \exp_a(t)=e^{at},\qquad
     \cosh_a(t)=\frac{e^{at}+e^{-at}}2,\qquad
     \sech(t)=\cosh_1(t)^{-1}.
    
\cosh_aIntroduction passage · 3 passages

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  • Introduction passage · Read passage · supplement_02.tex, line 36

    r pageheading1. The actual odd-power generating functionLemma 4.1; manuscript page 10 Work in the rational formal power-series ring Q[[t]]. Define S_n(p)=∑_a=0^p-1(-1)^a(2a+1)^n∈ Z, F_p(t)=∑_n≥0S_n(p)fract^nn!. The formal exponential is Mathlib's mcodePowerSeries.exp; scalar substitution is its actual mcoderescale homomorphism. We define exp_a(t)=e^at, cosh_a(t)=frace^at+e^-at2, sech(t)=cosh_1(t)^-1. The constant term of cosh_a is proved to be one, so the reciprocal used here is a genuine unit i…

    Exact mathematical passage
    
     \exp_a(t)=e^{at},\qquad
     \cosh_a(t)=\frac{e^{at}+e^{-at}}2,\qquad
     \sech(t)=\cosh_1(t)^{-1}.
    
D^aIntroduction passage · 2 passages

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  • Introduction passage · Read passage · supplement_06.tex, line 74

    packet_refinement, Companion.packet_zero_step pageheading2. From the characteristic polynomial to a recurrenceTheorem 9.1, continued Put D=C^N and let χ_D(X)=det(XI-D)=X^d+∑_a=0^d-1c_aX^a be its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands polynomial evaluation into a finite sum and proves ∑_a=0^dχ_D[a]D^a=0, χ_D[d]=1. Here χ_D[a] denotes the coefficient at index a. The identities hold over any commutative coefficient ring; the …

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     \sum_{a=0}^{d}\chi_D[a]D^a=0,
     \qquad \chi_D[d]=1.
    
D^dIntroduction passage · 3 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 129

    In characteristic p, factorial division is unavailable in general, but the coefficient operation remains meaningful over every commutative ring: (f⋆ g)_n=∑_r=0^nbinom nr f_r g_n-r, (D^d f)_n=f_n+d, 1_n=1, · n=0, 0, · n>0. Here ⋆ is Hurwitz multiplication. The symbol D^d denotes a coefficient shift, which agrees with formal differentiation after characteristic-zero encoding.

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    (f\star g)_n=\sum_{r=0}^{n}\binom nr f_r g_{n-r},\qquad
    (D^d f)_n=f_{n+d},\qquad \mathbf1_n=\begin{cases}1,&n=0,\\0,&n>0.\end{cases}
    
D_TIntroduction passage · 11 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 17

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     J=D_YF(a)\in M_e(\mathbb K),\qquad
     K=D_TF(a)\in M_{e\times k}(\mathbb K).
     \label{eq:regularity-blocks}
    
D_YIntroduction passage · 13 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 17

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     J=D_YF(a)\in M_e(\mathbb K),\qquad
     K=D_TF(a)\in M_{e\times k}(\mathbb K).
     \label{eq:regularity-blocks}
    
\detIntroduction passage · 7 passages

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  • Introduction passage · Read passage · supplement_06.tex, line 70

    Principal targets. targetCompanion.orbit_of_recurrence targetCompanion.companion_jump, Companion.source_companion_jump targetCompanion.packet_step, Companion.packet_advance targetCompanion.packet_refinement, Companion.packet_zero_step pageheading2. From the characteristic polynomial to a recurrenceTheorem 9.1, continued Put D=C^N and let χ_D(X)=det(XI-D)=X^d+∑_a=0^d-1c_aX^a be its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands pol…

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     \chi_D(X)=\det(XI-D)=X^d+\sum_{a=0}^{d-1}c_aX^a
    
eIntroduction passage · 219 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 10

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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    e,k\in\N
E^{(N,j)}_{Nn}Introduction passage · 1 passage

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  • Introduction passage · Read passage · supplement_07.tex, line 184

    Normalization is also an explicit implication. Using the existing definition E^(N,j)_Nn=j!C_N,j(n), the source proves that multiplication by j! commutes with every iterated difference. If the unnormalized difference has precision A, the normalized difference has precision A-v_p(j!). Precision is defined as x=0 hboxor v_p(x)≥ A, so zero and negative factorial-loss thresholds are handled correctly. For j<p, the loss is zero.

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    E^{(N,j)}_{Nn}=j!C_{N,j}(n)
e^{-at}Introduction passage · 2 passages

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  • Introduction passage · Read passage · supplement_02.tex, line 36

    r pageheading1. The actual odd-power generating functionLemma 4.1; manuscript page 10 Work in the rational formal power-series ring Q[[t]]. Define S_n(p)=∑_a=0^p-1(-1)^a(2a+1)^n∈ Z, F_p(t)=∑_n≥0S_n(p)fract^nn!. The formal exponential is Mathlib's mcodePowerSeries.exp; scalar substitution is its actual mcoderescale homomorphism. We define exp_a(t)=e^at, cosh_a(t)=frace^at+e^-at2, sech(t)=cosh_1(t)^-1. The constant term of cosh_a is proved to be one, so the reciprocal used here is a genuine unit i…

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     \exp_a(t)=e^{at},\qquad
     \cosh_a(t)=\frac{e^{at}+e^{-at}}2,\qquad
     \sech(t)=\cosh_1(t)^{-1}.
    
e^{at}Introduction passage · 2 passages

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  • Introduction passage · Read passage · supplement_02.tex, line 36

    r pageheading1. The actual odd-power generating functionLemma 4.1; manuscript page 10 Work in the rational formal power-series ring Q[[t]]. Define S_n(p)=∑_a=0^p-1(-1)^a(2a+1)^n∈ Z, F_p(t)=∑_n≥0S_n(p)fract^nn!. The formal exponential is Mathlib's mcodePowerSeries.exp; scalar substitution is its actual mcoderescale homomorphism. We define exp_a(t)=e^at, cosh_a(t)=frace^at+e^-at2, sech(t)=cosh_1(t)^-1. The constant term of cosh_a is proved to be one, so the reciprocal used here is a genuine unit i…

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     \exp_a(t)=e^{at},\qquad
     \cosh_a(t)=\frac{e^{at}+e^{-at}}2,\qquad
     \sech(t)=\cosh_1(t)^{-1}.
    
e^tIntroduction passage · 7 passages

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  • Introduction passage · Read passage · strengthening.tex, line 412

    Let B_m denote the Bernoulli numbers with B_1=-1/2. The identity obtained by substituting u=e^t-1 into log(1+u)/u is B_m=∑_r=0^m(-1)^rfracr!r+1 {m

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    u=e^t-1
E_hIntroduction passage · 2 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 271

    Write E_h f(n)=f(n+h) for the shift operator. The unit shift is I+Δ_1. Expanding its nth power and evaluating at zero gives reconstruction. Composing difference operators gives the coefficient shift and applying reconstruction to Δ_1^s f gives the third identity. The binomial expansion of (E_h-I)^s gives the explicit stencil. Binomial coefficients vanish when their lower index exceeds n, which proves the interpolation claim. Equality of every point of the stencil proves equality of differences. …

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    E_h f(n)=f(n+h)
E_jIntroduction passage · 8 passages

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  • Introduction passage · Read passage · supplement_10.tex, line 5

    ntCheckpoint scope. This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution. Scope and final statement. Let p be a prime, let m>0 divide p-1, put N=mp and h=(p-1)/m, and let 0≤ j<N. The original rational packet is defined by ∑_q=0^nbinomNn+jNqC_N,j(q)=δ_n0, E_j(n)=j! C_N,j(n). For every n,k,u∈ N, the new formal result establishes E_j(n)∈ Z_p, Δ_uh^kE_j(n)∈ p^min(k,mn) Z_p. He…

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     \sum_{q=0}^{n}\binom{Nn+j}{Nq}C_{N,j}(q)=\delta_{n0},
     \qquad E_j(n)=j!\,C_{N,j}(n).
    
e_{n+h}^{(m,p)}Introduction passage · 3 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 55

    Fix an odd prime p, a divisor mmid p-1, and put h=fracp-1m. The reduced standard window inverse is defined by ∑_q=0^n binommnmqe_q^(m,p) = δ_n0 in F_p. We prove that e_n+h^(m,p)=e_n^(m,p) (n≥1). The result first follows from subgroup symmetry in the finite Hurwitz algebra. A second proof shows that the standard window operator preserves the finite space of sequences that are h periodic after their constant term. The two arguments expose different parts of the same finite character structure.

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    e_{n+h}^{(m,p)}=e_n^{(m,p)}
    \qquad(n\geq1).
    
e_{n+uh}^{(m,p)}Introduction passage · 1 passage

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  • Introduction passage · Read passage · supplement_07.tex, line 109

    The positive-tail period. Write h=(p-1)/m. For n≥1, evaluate U^pf_m=Uf_m at the nonnegative degree mn-1: f_m(mn+p-1)=f_m(mn). Since mn+p-1=m(n+h), the identified sequence satisfies e_n+h^(m,p)=e_n^(m,p) (n≥1). Induction gives e_n+uh^(m,p)=e_n^(m,p) for every u≥0. This proves a period, not necessarily the least positive period. The proof also includes p=2, where m=1 and h=1.

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    e_{n+uh}^{(m,p)}=e_n^{(m,p)}
e_n^{(m,p)}Introduction passage · 10 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 55

    Fix an odd prime p, a divisor mmid p-1, and put h=fracp-1m. The reduced standard window inverse is defined by ∑_q=0^n binommnmqe_q^(m,p) = δ_n0 in F_p. We prove that e_n+h^(m,p)=e_n^(m,p) (n≥1). The result first follows from subgroup symmetry in the finite Hurwitz algebra. A second proof shows that the standard window operator preserves the finite space of sequences that are h periodic after their constant term. The two arguments expose different parts of the same finite character structure.

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    e_{n+h}^{(m,p)}=e_n^{(m,p)}
    \qquad(n\geq1).
    
e_q^{(m,p)}Introduction passage · 2 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 47

    Fix an odd prime p, a divisor mmid p-1, and put h=fracp-1m. The reduced standard window inverse is defined by ∑_q=0^n binommnmqe_q^(m,p) = δ_n0 in F_p. We prove that e_n+h^(m,p)=e_n^(m,p) (n≥1). The result first follows from subgroup symmetry in the finite Hurwitz algebra. A second proof shows that the standard window operator preserves the finite space of sequences that are h periodic after their constant term. The two arguments expose different parts of the same finite character structure.

    Exact mathematical passage
    
    \sum_{q=0}^{n}
    \binom{mn}{mq}e_q^{(m,p)}
    =
    \delta_{n0}
    \quad\text{in }\FF_p.
    
\expIntroduction passage · 2 passages

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  • Introduction passage · Read passage · supplement_02.tex, line 36

    r pageheading1. The actual odd-power generating functionLemma 4.1; manuscript page 10 Work in the rational formal power-series ring Q[[t]]. Define S_n(p)=∑_a=0^p-1(-1)^a(2a+1)^n∈ Z, F_p(t)=∑_n≥0S_n(p)fract^nn!. The formal exponential is Mathlib's mcodePowerSeries.exp; scalar substitution is its actual mcoderescale homomorphism. We define exp_a(t)=e^at, cosh_a(t)=frace^at+e^-at2, sech(t)=cosh_1(t)^-1. The constant term of cosh_a is proved to be one, so the reciprocal used here is a genuine unit i…

    Exact mathematical passage
    
     \exp_a(t)=e^{at},\qquad
     \cosh_a(t)=\frac{e^{at}+e^{-at}}2,\qquad
     \sech(t)=\cosh_1(t)^{-1}.
    
\exp_aIntroduction passage · 1 passage

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  • Introduction passage · Read passage · supplement_02.tex, line 36

    r pageheading1. The actual odd-power generating functionLemma 4.1; manuscript page 10 Work in the rational formal power-series ring Q[[t]]. Define S_n(p)=∑_a=0^p-1(-1)^a(2a+1)^n∈ Z, F_p(t)=∑_n≥0S_n(p)fract^nn!. The formal exponential is Mathlib's mcodePowerSeries.exp; scalar substitution is its actual mcoderescale homomorphism. We define exp_a(t)=e^at, cosh_a(t)=frace^at+e^-at2, sech(t)=cosh_1(t)^-1. The constant term of cosh_a is proved to be one, so the reciprocal used here is a genuine unit i…

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     \exp_a(t)=e^{at},\qquad
     \cosh_a(t)=\frac{e^{at}+e^{-at}}2,\qquad
     \sech(t)=\cosh_1(t)^{-1}.
    
FIntroduction passage · 336 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 12

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     F\in\mathbb K[Y_1,\ldots,Y_e,T_1,\ldots,T_k]^e,
     \qquad a=(y_0,t_0),
    
F_0Introduction passage · 7 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2892

    [Fibonacci residue packets] Let F_0=0, F_1=1, and F_n+2=F_n+1+F_n. Let L_N denote the Lucas number. For every N≥1 and every address j, F_N(k+2)+j = L_NF_N(k+1)+j -(-1)^NF_Nk+j.

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    F_0=0
F_1Introduction passage · 9 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2638

    Let F=(F_1,…,F_r): Z_p^d⟶ Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    F=(F_1,\ldots,F_r):\Zp^d\longrightarrow\Zp^r
    
F_eIntroduction passage · 2 passages

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  • Introduction passage · Read passage · continuation_01.tex, line 16

    Let F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, q≥1, L≥1, x∈ Z^d, F(x)≡0pmodq^L. Write R_q= Z/q Z, let ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

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    F=(F_1,\ldots,F_e)\in\mathbb Z[X_1,\ldots,X_d]^e,
    \quad q\ge1,\quad L\ge1,
    \quad x\in\mathbb Z^d,
    \quad F(x)\equiv0\pmod{q^L}.
    
F_iIntroduction passage · 11 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2641

    Let F=(F_1,…,F_r): Z_p^d⟶ Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    F_i\in\Z_p[X_1,\ldots,X_d]
f_mIntroduction passage · 7 passages

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  • Introduction passage · Read passage · supplement_07.tex, line 101

    The positive-tail period. Write h=(p-1)/m. For n≥1, evaluate U^pf_m=Uf_m at the nonnegative degree mn-1: f_m(mn+p-1)=f_m(mn). Since mn+p-1=m(n+h), the identified sequence satisfies e_n+h^(m,p)=e_n^(m,p) (n≥1). Induction gives e_n+uh^(m,p)=e_n^(m,p) for every u≥0. This proves a period, not necessarily the least positive period. The proof also includes p=2, where m=1 and h=1.

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    U^pf_m=Uf_m
F_nIntroduction passage · 4 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2892

    [Fibonacci residue packets] Let F_0=0, F_1=1, and F_n+2=F_n+1+F_n. Let L_N denote the Lucas number. For every N≥1 and every address j, F_N(k+2)+j = L_NF_N(k+1)+j -(-1)^NF_Nk+j.

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    F_{n+2}=F_{n+1}+F_n
F_{N(k+1)+j}Introduction passage · 3 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2893

    [Fibonacci residue packets] Let F_0=0, F_1=1, and F_n+2=F_n+1+F_n. Let L_N denote the Lucas number. For every N≥1 and every address j, F_N(k+2)+j = L_NF_N(k+1)+j -(-1)^NF_Nk+j.

    Exact mathematical passage
    
    \boxed{
    F_{N(k+2)+j}
    =
    L_NF_{N(k+1)+j}
    -(-1)^NF_{Nk+j}.
    }
    
F_{N(k+2)+j}Introduction passage · 3 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2893

    [Fibonacci residue packets] Let F_0=0, F_1=1, and F_n+2=F_n+1+F_n. Let L_N denote the Lucas number. For every N≥1 and every address j, F_N(k+2)+j = L_NF_N(k+1)+j -(-1)^NF_Nk+j.

    Exact mathematical passage
    
    \boxed{
    F_{N(k+2)+j}
    =
    L_NF_{N(k+1)+j}
    -(-1)^NF_{Nk+j}.
    }
    
F_{n+1}Introduction passage · 2 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2892

    [Fibonacci residue packets] Let F_0=0, F_1=1, and F_n+2=F_n+1+F_n. Let L_N denote the Lucas number. For every N≥1 and every address j, F_N(k+2)+j = L_NF_N(k+1)+j -(-1)^NF_Nk+j.

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    F_{n+2}=F_{n+1}+F_n
F_{n+2}Introduction passage · 1 passage

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  • Introduction passage · Read passage · 72_recovered.tex, line 2892

    [Fibonacci residue packets] Let F_0=0, F_1=1, and F_n+2=F_n+1+F_n. Let L_N denote the Lucas number. For every N≥1 and every address j, F_N(k+2)+j = L_NF_N(k+1)+j -(-1)^NF_Nk+j.

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    F_{n+2}=F_{n+1}+F_n
f_{n+d}Introduction passage · 1 passage

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  • Introduction passage · Read passage · 10_foundations.tex, line 129

    In characteristic p, factorial division is unavailable in general, but the coefficient operation remains meaningful over every commutative ring: (f⋆ g)_n=∑_r=0^nbinom nr f_r g_n-r, (D^d f)_n=f_n+d, 1_n=1, · n=0, 0, · n>0. Here ⋆ is Hurwitz multiplication. The symbol D^d denotes a coefficient shift, which agrees with formal differentiation after characteristic-zero encoding.

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    (f\star g)_n=\sum_{r=0}^{n}\binom nr f_r g_{n-r},\qquad
    (D^d f)_n=f_{n+d},\qquad \mathbf1_n=\begin{cases}1,&n=0,\\0,&n>0.\end{cases}
    
F_{Nk+j}Introduction passage · 5 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2893

    [Fibonacci residue packets] Let F_0=0, F_1=1, and F_n+2=F_n+1+F_n. Let L_N denote the Lucas number. For every N≥1 and every address j, F_N(k+2)+j = L_NF_N(k+1)+j -(-1)^NF_Nk+j.

    Exact mathematical passage
    
    \boxed{
    F_{N(k+2)+j}
    =
    L_NF_{N(k+1)+j}
    -(-1)^NF_{Nk+j}.
    }
    
F_pIntroduction passage · 4 passages

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  • Introduction passage · Read passage · supplement_02.tex, line 31

    rence theorem remain pending. No continuum convergence or analytic summability theorem is inferred from the formal power-series equalities in this block. The historical numerical study is not being relabelled as a fresh experiment. pageheading1. The actual odd-power generating functionLemma 4.1; manuscript page 10 Work in the rational formal power-series ring Q[[t]]. Define S_n(p)=∑_a=0^p-1(-1)^a(2a+1)^n∈ Z, F_p(t)=∑_n≥0S_n(p)fract^nn!. The formal exponential is Mathlib's mcodePowerSeries.exp; s…

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     S_n(p)=\sum_{a=0}^{p-1}(-1)^a(2a+1)^n\in\Z,
     \qquad F_p(t)=\sum_{n\ge0}S_n(p)\frac{t^n}{n!}.
    
F_rIntroduction passage · 5 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2638

    Let F=(F_1,…,F_r): Z_p^d⟶ Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    F=(F_1,\ldots,F_r):\Zp^d\longrightarrow\Zp^r
    
f_rIntroduction passage · 1 passage

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  • Introduction passage · Read passage · 10_foundations.tex, line 129

    In characteristic p, factorial division is unavailable in general, but the coefficient operation remains meaningful over every commutative ring: (f⋆ g)_n=∑_r=0^nbinom nr f_r g_n-r, (D^d f)_n=f_n+d, 1_n=1, · n=0, 0, · n>0. Here ⋆ is Hurwitz multiplication. The symbol D^d denotes a coefficient shift, which agrees with formal differentiation after characteristic-zero encoding.

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    (f\star g)_n=\sum_{r=0}^{n}\binom nr f_r g_{n-r},\qquad
    (D^d f)_n=f_{n+d},\qquad \mathbf1_n=\begin{cases}1,&n=0,\\0,&n>0.\end{cases}
    
\FFIntroduction passage · 24 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 47

    Fix an odd prime p, a divisor mmid p-1, and put h=fracp-1m. The reduced standard window inverse is defined by ∑_q=0^n binommnmqe_q^(m,p) = δ_n0 in F_p. We prove that e_n+h^(m,p)=e_n^(m,p) (n≥1). The result first follows from subgroup symmetry in the finite Hurwitz algebra. A second proof shows that the standard window operator preserves the finite space of sequences that are h periodic after their constant term. The two arguments expose different parts of the same finite character structure.

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    \sum_{q=0}^{n}
    \binom{mn}{mq}e_q^{(m,p)}
    =
    \delta_{n0}
    \quad\text{in }\FF_p.
    
\FF_pIntroduction passage · 14 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 47

    Fix an odd prime p, a divisor mmid p-1, and put h=fracp-1m. The reduced standard window inverse is defined by ∑_q=0^n binommnmqe_q^(m,p) = δ_n0 in F_p. We prove that e_n+h^(m,p)=e_n^(m,p) (n≥1). The result first follows from subgroup symmetry in the finite Hurwitz algebra. A second proof shows that the standard window operator preserves the finite space of sequences that are h periodic after their constant term. The two arguments expose different parts of the same finite character structure.

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    \sum_{q=0}^{n}
    \binom{mn}{mq}e_q^{(m,p)}
    =
    \delta_{n0}
    \quad\text{in }\FF_p.
    
\FF_p^+Introduction passage · 4 passages

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\FF_p^rIntroduction passage · 3 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2522

    Let p be a prime. Let M∈ M_r,d( Z_p), b∈ Z_p^r. Suppose that x_L∈ Z_p^d satisfies Mx_L≡ bpmod p^L for some L≥1. Define the reduced residual ρ_L(x_L) = fracMx_L-bp^L pmod p ∈ F_p^r. Write overline M for the reduction of M modulo p.

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    \rho_L(x_L)
    =
    \frac{Mx_L-b}{p^L}
    \pmod p
    \in\FF_p^r.
    
gIntroduction passage · 113 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 129

    In characteristic p, factorial division is unavailable in general, but the coefficient operation remains meaningful over every commutative ring: (f⋆ g)_n=∑_r=0^nbinom nr f_r g_n-r, (D^d f)_n=f_n+d, 1_n=1, · n=0, 0, · n>0. Here ⋆ is Hurwitz multiplication. The symbol D^d denotes a coefficient shift, which agrees with formal differentiation after characteristic-zero encoding.

    Exact mathematical passage
    
    (f\star g)_n=\sum_{r=0}^{n}\binom nr f_r g_{n-r},\qquad
    (D^d f)_n=f_{n+d},\qquad \mathbf1_n=\begin{cases}1,&n=0,\\0,&n>0.\end{cases}
    
g_{n-r}Introduction passage · 2 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 129

    In characteristic p, factorial division is unavailable in general, but the coefficient operation remains meaningful over every commutative ring: (f⋆ g)_n=∑_r=0^nbinom nr f_r g_n-r, (D^d f)_n=f_n+d, 1_n=1, · n=0, 0, · n>0. Here ⋆ is Hurwitz multiplication. The symbol D^d denotes a coefficient shift, which agrees with formal differentiation after characteristic-zero encoding.

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    (f\star g)_n=\sum_{r=0}^{n}\binom nr f_r g_{n-r},\qquad
    (D^d f)_n=f_{n+d},\qquad \mathbf1_n=\begin{cases}1,&n=0,\\0,&n>0.\end{cases}
    
H_2Introduction passage · 8 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 72

    edition is the treatment of the odd and dyadic Euler formulas. They are no longer presented as unrelated reconstruction devices. For each prime p we define an even source series H_p and an inverse kernel K_p satisfying K_p(t)H_p(t)=secht. At an odd prime, H_p(t)=cosh^2(pt)secht, K_p(t)=sech^2(pt). At the dyadic prime, H_2(t)=fracsinh(2t)sinh t2t^2, K_2(t)=frac4t^2sinh^2(2t). The coefficient convolution is identical in form, but the kernels have different integral weights. This is the correct sen…

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    H_2(t)=\frac{\sinh(2t)\sinh t}{2t^2},
    \qquad
    K_2(t)=\frac{4t^2}{\sinh^2(2t)}.
    
\mathcal H_{\mathrm{int}}Introduction passage · 2 passages

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  • Introduction passage · Read passage · supplement_11.tex, line 78

    Write H_int for the subring of Q_2[[t]] whose factorial-scaled coefficients belong to Z_2. Flatness means B_4^sF∈ H_int for every s. The proof of multiplicative and reciprocal closure is induction on s, using the full product correction. It separately proves that the actual reciprocal has integral Hurwitz coefficients when F(0)=1.

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    \mathcal H_{\mathrm{int}}
H_pIntroduction passage · 10 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 61

    The central change in this edition is the treatment of the odd and dyadic Euler formulas. They are no longer presented as unrelated reconstruction devices. For each prime p we define an even source series H_p and an inverse kernel K_p satisfying K_p(t)H_p(t)=secht. At an odd prime, H_p(t)=cosh^2(pt)secht, K_p(t)=sech^2(pt). At the dyadic prime, H_2(t)=fracsinh(2t)sinh t2t^2, K_2(t)=frac4t^2sinh^2(2t). The coefficient convolution is identical in form, but the kernels have different integral weigh…

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    H_p
\hboxIntroduction passage · 2 passages

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  • Introduction passage · Read passage · supplement_07.tex, line 185

    Normalization is also an explicit implication. Using the existing definition E^(N,j)_Nn=j!C_N,j(n), the source proves that multiplication by j! commutes with every iterated difference. If the unnormalized difference has precision A, the normalized difference has precision A-v_p(j!). Precision is defined as x=0 hboxor v_p(x)≥ A, so zero and negative factorial-loss thresholds are handled correctly. For j<p, the loss is zero.

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     x=0\quad\hbox{or}\quad v_p(x)\ge A,
    
IIntroduction passage · 77 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 272

    Write E_h f(n)=f(n+h) for the shift operator. The unit shift is I+Δ_1. Expanding its nth power and evaluating at zero gives reconstruction. Composing difference operators gives the coefficient shift and applying reconstruction to Δ_1^s f gives the third identity. The binomial expansion of (E_h-I)^s gives the explicit stencil. Binomial coefficients vanish when their lower index exceeds n, which proves the interpolation claim. Equality of every point of the stencil proves equality of differences. …

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    I+\Delta_1
iIntroduction passage · 133 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2641

    Let F=(F_1,…,F_r): Z_p^d⟶ Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    F_i\in\Z_p[X_1,\ldots,X_d]
\imIntroduction passage · 14 passages

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  • Introduction passage · Read passage · 30_lifting.tex, line 170

    s over an arbitrary commutative ring. Let U,V,W be R-modules and let J:U→ V, K:U→ W be linear. Define C(J,K)=U/(ker J+ker K). An attainable value Ju has label [u]∈ C(J,K); this is independent of the choice of its preimage because ker J is killed in the quotient. Similarly, Kv has label [v]. Denote these maps by λ_J:im J→ C(J,K) and λ_K:im K→ C(J,K).

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    \lambda_J:\im J\to C(J,K)
\mathrm{int}Introduction passage · 2 passages

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  • Introduction passage · Read passage · supplement_11.tex, line 78

    Write H_int for the subring of Q_2[[t]] whose factorial-scaled coefficients belong to Z_2. Flatness means B_4^sF∈ H_int for every s. The proof of multiplicative and reciprocal closure is induction on s, using the full product correction. It separately proves that the actual reciprocal has integral Hurwitz coefficients when F(0)=1.

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    \mathcal H_{\mathrm{int}}
JIntroduction passage · 180 passages

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  • Introduction passage · Read passage · 30_lifting.tex, line 162

    The next obstruction concerns compatibility between individually attainable corrections. Its algebraic form works over an arbitrary commutative ring. Let U,V,W be R-modules and let J:U→ V, K:U→ W be linear. Define C(J,K)=U/(ker J+ker K). An attainable value Ju has label [u]∈ C(J,K); this is independent of the choice of its preimage because ker J is killed in the quotient. Similarly, Kv has label [v]. Denote these maps by λ_J:im J→ C(J,K) and λ_K:im K→ C(J,K).

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    J:U\to V
J^{-1}Introduction passage · 30 passages

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  • Introduction passage · Read passage · continuation_11.tex, line 36

    bseteq K^k, V⊆ K^e such that t_0∈ U, y_0∈ V, ψ(t_0)=y_0, ψ(U)⊆ V. The map is analytic at every point of U, and its graph describes all roots in the product neighborhood: ∀ t∈ U ∀ y∈ V, F(y,t)=0 Longleftrightarrow y=ψ(t). At the base parameter it has the strict Fréchet derivative Dψ(t_0)=-J^-1K. Consequently it is differentiable on U. The map is defined on the ambient parameter space, but its root, analyticity and uniqueness conclusions are restricted to the specified neighborhood. Uniqueness is …

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    \boxed{D\psi(t_0)=-J^{-1}K.}
    
J_FIntroduction passage · 3 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2652

    rrow Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    J_F(x_L)
J_xIntroduction passage · 40 passages

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  • Introduction passage · Read passage · continuation_01.tex, line 22

    Let F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, q≥1, L≥1, x∈ Z^d, F(x)≡0pmodq^L. Write R_q= Z/q Z, let ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

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    r_x=F(x)/q^L\in\mathbb Z^e,
    \qquad A_x=DF(x):\mathbb Z^d\longrightarrow\mathbb Z^e,
    \qquad J_x=\rho_q\circ A_x.
    
KIntroduction passage · 133 passages

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  • Introduction passage · Read passage · 30_lifting.tex, line 162

    The next obstruction concerns compatibility between individually attainable corrections. Its algebraic form works over an arbitrary commutative ring. Let U,V,W be R-modules and let J:U→ V, K:U→ W be linear. Define C(J,K)=U/(ker J+ker K). An attainable value Ju has label [u]∈ C(J,K); this is independent of the choice of its preimage because ker J is killed in the quotient. Similarly, Kv has label [v]. Denote these maps by λ_J:im J→ C(J,K) and λ_K:im K→ C(J,K).

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    K:U\to W
\mathbb KIntroduction passage · 23 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 10

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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    \mathbb K
\mathbb K^eIntroduction passage · 8 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 11

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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    (Y,T)\in\mathbb K^e\times\mathbb K^k
\mathbb K^kIntroduction passage · 9 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 11

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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    (Y,T)\in\mathbb K^e\times\mathbb K^k
K_2Introduction passage · 3 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 72

    edition is the treatment of the odd and dyadic Euler formulas. They are no longer presented as unrelated reconstruction devices. For each prime p we define an even source series H_p and an inverse kernel K_p satisfying K_p(t)H_p(t)=secht. At an odd prime, H_p(t)=cosh^2(pt)secht, K_p(t)=sech^2(pt). At the dyadic prime, H_2(t)=fracsinh(2t)sinh t2t^2, K_2(t)=frac4t^2sinh^2(2t). The coefficient convolution is identical in form, but the kernels have different integral weights. This is the correct sen…

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    H_2(t)=\frac{\sinh(2t)\sinh t}{2t^2},
    \qquad
    K_2(t)=\frac{4t^2}{\sinh^2(2t)}.
    
K_pIntroduction passage · 7 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 61

    The central change in this edition is the treatment of the odd and dyadic Euler formulas. They are no longer presented as unrelated reconstruction devices. For each prime p we define an even source series H_p and an inverse kernel K_p satisfying K_p(t)H_p(t)=secht. At an odd prime, H_p(t)=cosh^2(pt)secht, K_p(t)=sech^2(pt). At the dyadic prime, H_2(t)=fracsinh(2t)sinh t2t^2, K_2(t)=frac4t^2sinh^2(2t). The coefficient convolution is identical in form, but the kernels have different integral weigh…

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    K_p
\kerIntroduction passage · 29 passages

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  • Introduction passage · Read passage · 30_lifting.tex, line 164

    The next obstruction concerns compatibility between individually attainable corrections. Its algebraic form works over an arbitrary commutative ring. Let U,V,W be R-modules and let J:U→ V, K:U→ W be linear. Define C(J,K)=U/(ker J+ker K). An attainable value Ju has label [u]∈ C(J,K); this is independent of the choice of its preimage because ker J is killed in the quotient. Similarly, Kv has label [v]. Denote these maps by λ_J:im J→ C(J,K) and λ_K:im K→ C(J,K).

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    \label{eq:common-quotient}
     C(J,K)=U/(\ker J+\ker K).
    
\mathcal LIntroduction passage · 42 passages

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  • Introduction passage · Read passage · continuation_08.tex, line 49

    Every nonempty L_v has a checked free and transitive action of ker J. The action is defined without choosing a base lift. Choosing one valid solved digit then gives an explicit bijection L_v≅ker J. In particular, for every positive modulus, o_x(v)=0 Longrightarrow # L_v=#ker J. These results permit composite q, the degenerate modulus q=1, and zero dimensions. Neither J nor the full reduced Jacobian is assumed surjective.

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    \mathcal L_v
L_NIntroduction passage · 9 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2892

    [Fibonacci residue packets] Let F_0=0, F_1=1, and F_n+2=F_n+1+F_n. Let L_N denote the Lucas number. For every N≥1 and every address j, F_N(k+2)+j = L_NF_N(k+1)+j -(-1)^NF_Nk+j.

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    L_N
\mathcal L_vIntroduction passage · 9 passages

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  • Introduction passage · Read passage · continuation_08.tex, line 49

    Every nonempty L_v has a checked free and transitive action of ker J. The action is defined without choosing a base lift. Choosing one valid solved digit then gives an explicit bijection L_v≅ker J. In particular, for every positive modulus, o_x(v)=0 Longrightarrow # L_v=#ker J. These results permit composite q, the degenerate modulus q=1, and zero dimensions. Neither J nor the full reduced Jacobian is assumed surjective.

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    \mathcal L_v
\logIntroduction passage · 44 passages

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  • Introduction passage · Read passage · root_scope.tex, line 91

    Let Z_2⟨ x⟩ denote the restricted power series: their integral coefficients tend to zero dyadically. Extend A by mathscr A(x)=frac9^x+1-18(x+1)(2x+1), mathscr A(-1)=-fraclog98. The apparent singularity at x=-1 is removable. Since v_2(log9)=3, its convergent expansion is mathscr A(x)=fraclog9/81+2x ∑_h≥0frac(log9)^h(h+1)!(x+1)^h. It follows coefficientwise that mathscr A∈ Z_2⟨ x⟩^×, mathscr A≡1pmod2, and mathscr A(N)=A(N).

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     \mathscr A(x)=\frac{9^{x+1}-1}{8(x+1)(2x+1)},
     \qquad \mathscr A(-1)=-\frac{\log9}{8}.
    
\LongleftrightarrowIntroduction passage · 15 passages

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  • Introduction passage · Read passage · continuation_11.tex, line 33

    Local analytic graph theorem. There exist a map ψ: K^k⟶ K^e and open sets U⊆ K^k, V⊆ K^e such that t_0∈ U, y_0∈ V, ψ(t_0)=y_0, ψ(U)⊆ V. The map is analytic at every point of U, and its graph describes all roots in the product neighborhood: ∀ t∈ U ∀ y∈ V, F(y,t)=0 Longleftrightarrow y=ψ(t). At the base parameter it has the strict Fréchet derivative Dψ(t_0)=-J^-1K. Consequently it is differentiable on U. The map is defined on the ambient parameter space, but its root, analyticity and uniqueness co…

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    \forall t\in U\;\forall y\in V,
    \qquad F(y,t)=0\quad\Longleftrightarrow\quad y=\psi(t).
    
\LongrightarrowIntroduction passage · 7 passages

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  • Introduction passage · Read passage · continuation_08.tex, line 53

    Every nonempty L_v has a checked free and transitive action of ker J. The action is defined without choosing a base lift. Choosing one valid solved digit then gives an explicit bijection L_v≅ker J. In particular, for every positive modulus, o_x(v)=0 Longrightarrow # L_v=#ker J. These results permit composite q, the degenerate modulus q=1, and zero dimensions. Neither J nor the full reduced Jacobian is assumed surjective.

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    o_x(v)=0\quad\Longrightarrow\quad\#\mathcal L_v=\#\ker J.
    
M_eIntroduction passage · 7 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 17

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     J=D_YF(a)\in M_e(\mathbb K),\qquad
     K=D_TF(a)\in M_{e\times k}(\mathbb K).
     \label{eq:regularity-blocks}
    
M_{e\times k}Introduction passage · 4 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 17

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     J=D_YF(a)\in M_e(\mathbb K),\qquad
     K=D_TF(a)\in M_{e\times k}(\mathbb K).
     \label{eq:regularity-blocks}
    
M_{r,d}Introduction passage · 1 passage

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  • Introduction passage · Read passage · 72_recovered.tex, line 2512

    Let p be a prime. Let M∈ M_r,d( Z_p), b∈ Z_p^r. Suppose that x_L∈ Z_p^d satisfies Mx_L≡ bpmod p^L for some L≥1. Define the reduced residual ρ_L(x_L) = fracMx_L-bp^L pmod p ∈ F_p^r. Write overline M for the reduction of M modulo p.

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    M\in M_{r,d}(\Zp),
    \qquad
    b\in\Zp^r.
    
\mathscrIntroduction passage · 20 passages

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  • Introduction passage · Read passage · root_scope.tex, line 91

    Let Z_2⟨ x⟩ denote the restricted power series: their integral coefficients tend to zero dyadically. Extend A by mathscr A(x)=frac9^x+1-18(x+1)(2x+1), mathscr A(-1)=-fraclog98. The apparent singularity at x=-1 is removable. Since v_2(log9)=3, its convergent expansion is mathscr A(x)=fraclog9/81+2x ∑_h≥0frac(log9)^h(h+1)!(x+1)^h. It follows coefficientwise that mathscr A∈ Z_2⟨ x⟩^×, mathscr A≡1pmod2, and mathscr A(N)=A(N).

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     \mathscr A(x)=\frac{9^{x+1}-1}{8(x+1)(2x+1)},
     \qquad \mathscr A(-1)=-\frac{\log9}{8}.
    
oIntroduction passage · 22 passages

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  • Introduction passage · Read passage · continuation_08.tex, line 53

    Every nonempty L_v has a checked free and transitive action of ker J. The action is defined without choosing a base lift. Choosing one valid solved digit then gives an explicit bijection L_v≅ker J. In particular, for every positive modulus, o_x(v)=0 Longrightarrow # L_v=#ker J. These results permit composite q, the degenerate modulus q=1, and zero dimensions. Neither J nor the full reduced Jacobian is assumed surjective.

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    o_x(v)=0\quad\Longrightarrow\quad\#\mathcal L_v=\#\ker J.
    
o_xIntroduction passage · 16 passages

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  • Introduction passage · Read passage · continuation_08.tex, line 53

    Every nonempty L_v has a checked free and transitive action of ker J. The action is defined without choosing a base lift. Choosing one valid solved digit then gives an explicit bijection L_v≅ker J. In particular, for every positive modulus, o_x(v)=0 Longrightarrow # L_v=#ker J. These results permit composite q, the degenerate modulus q=1, and zero dimensions. Neither J nor the full reduced Jacobian is assumed surjective.

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    o_x(v)=0\quad\Longrightarrow\quad\#\mathcal L_v=\#\ker J.
    
\overline JIntroduction passage · 5 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2652

    tbfEditorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    \overline J_F(x_L)
\overline J_FIntroduction passage · 5 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2652

    tbfEditorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    \overline J_F(x_L)
\overline MIntroduction passage · 10 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2529

    Let p be a prime. Let M∈ M_r,d( Z_p), b∈ Z_p^r. Suppose that x_L∈ Z_p^d satisfies Mx_L≡ bpmod p^L for some L≥1. Define the reduced residual ρ_L(x_L) = fracMx_L-bp^L pmod p ∈ F_p^r. Write overline M for the reduction of M modulo p.

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    \overline M
p^{(p-1)x+j}Introduction passage · 2 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 1881

    The analytic function is defined by its unit source series. The expression p^(p-1)x+j in the interpolation identity is evaluated only at nonnegative integers. No analytic function x↦ p^x on all of Z_p is used.

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    p^{(p-1)x+j}
p^{\min(k,mn)}Introduction passage · 7 passages

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  • Introduction passage · Read passage · 20_arithmetic.tex, line 193

    [All-address differences on the divisor branch] Let p be prime, m>0, mmid p-1, N=mp, and 0≤ j<N. Write h=(p-1)/m. For every u,k,n∈ N, U_N,j(n) · ∈ Z_p, Δ_uh^kU_N,j(n) · ∈ p^min(k,mn) Z_p, v_p(Δ_uh^kC_N,j(n)) · ≥min(k,mn)-v_p(j!). The factorial correction is part of the normalized statement, including addresses j≥ p.

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     U_{N,j}(n)&\in\Z_p,\label{eq:arith-divisor-integral}\\
     \Delta_{uh}^{k}U_{N,j}(n)&\in p^{\min(k,mn)}\Z_p,\label{eq:arith-divisor-bound}\\
     v_p\bigl(\Delta_{uh}^{k}C_{N,j}(n)\bigr)
       &\ge\min(k,mn)-v_p(j!).\label{eq:arith-divisor-normalized}
    
p^xIntroduction passage · 1 passage

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  • Introduction passage · Read passage · 72_recovered.tex, line 1881

    The analytic function is defined by its unit source series. The expression p^(p-1)x+j in the interpolation identity is evaluated only at nonnegative integers. No analytic function x↦ p^x on all of Z_p is used.

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    x\mapsto p^x
\pmodIntroduction passage · 140 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2518

    Let p be a prime. Let M∈ M_r,d( Z_p), b∈ Z_p^r. Suppose that x_L∈ Z_p^d satisfies Mx_L≡ bpmod p^L for some L≥1. Define the reduced residual ρ_L(x_L) = fracMx_L-bp^L pmod p ∈ F_p^r. Write overline M for the reduction of M modulo p.

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    Mx_L\equiv b\pmod {p^L}
    
\QIntroduction passage · 35 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2641

    Let F=(F_1,…,F_r): Z_p^d⟶ Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    \Q_p
q^LIntroduction passage · 40 passages

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  • Introduction passage · Read passage · continuation_01.tex, line 16

    Let F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, q≥1, L≥1, x∈ Z^d, F(x)≡0pmodq^L. Write R_q= Z/q Z, let ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

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    F=(F_1,\ldots,F_e)\in\mathbb Z[X_1,\ldots,X_d]^e,
    \quad q\ge1,\quad L\ge1,
    \quad x\in\mathbb Z^d,
    \quad F(x)\equiv0\pmod{q^L}.
    
\Q_2Introduction passage · 5 passages

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  • Introduction passage · Read passage · supplement_11.tex, line 78

    Write H_int for the subring of Q_2[[t]] whose factorial-scaled coefficients belong to Z_2. Flatness means B_4^sF∈ H_int for every s. The proof of multiplicative and reciprocal closure is induction on s, using the full product correction. It separately proves that the actual reciprocal has integral Hurwitz coefficients when F(0)=1.

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    \Q_2[[t]]
\Q_pIntroduction passage · 19 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2641

    Let F=(F_1,…,F_r): Z_p^d⟶ Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    \Q_p
\QpIntroduction passage · 21 passages

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  • Introduction passage · Read passage · 00_scope.tex, line 19

    Throughout, N={0,1,2,…}. When p is prime, Z_p denotes the p-adic integers and Q_p their fraction field. We normalize v_p(p)=1. For a rational or p-adic quantity, a displayed lower valuation bound is understood with the mathematical convention v_p(0)=+∞. The corresponding formal predicates are zero-safe: the quantity is zero or its valuation meets the stated bound. An exact finite valuation assertion requires a nonzero quantity, and this is stated where used. Congruence in Z_p means ideal divisib…

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    \Qp
RIntroduction passage · 220 passages

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  • Introduction passage · Read passage · 30_lifting.tex, line 162

    The next obstruction concerns compatibility between individually attainable corrections. Its algebraic form works over an arbitrary commutative ring. Let U,V,W be R-modules and let J:U→ V, K:U→ W be linear. Define C(J,K)=U/(ker J+ker K). An attainable value Ju has label [u]∈ C(J,K); this is independent of the choice of its preimage because ker J is killed in the quotient. Similarly, Kv has label [v]. Denote these maps by λ_J:im J→ C(J,K) and λ_K:im K→ C(J,K).

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    R
R_qIntroduction passage · 2 passages

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  • Introduction passage · Read passage · continuation_01.tex, line 21

    Let F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, q≥1, L≥1, x∈ Z^d, F(x)≡0pmodq^L. Write R_q= Z/q Z, let ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

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    R_q=\mathbb Z/q\mathbb Z
R_q^dIntroduction passage · 1 passage

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  • Introduction passage · Read passage · continuation_01.tex, line 29

    et ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

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    R_q^d
R_q^eIntroduction passage · 3 passages

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  • Introduction passage · Read passage · continuation_01.tex, line 28

    F(x)≡0pmodq^L. Write R_q= Z/q Z, let ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

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    R_q^e
r_xIntroduction passage · 12 passages

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  • Introduction passage · Read passage · continuation_01.tex, line 22

    Let F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, q≥1, L≥1, x∈ Z^d, F(x)≡0pmodq^L. Write R_q= Z/q Z, let ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

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    r_x=F(x)/q^L\in\mathbb Z^e,
    \qquad A_x=DF(x):\mathbb Z^d\longrightarrow\mathbb Z^e,
    \qquad J_x=\rho_q\circ A_x.
    
sIntroduction passage · 339 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 274

    Write E_h f(n)=f(n+h) for the shift operator. The unit shift is I+Δ_1. Expanding its nth power and evaluating at zero gives reconstruction. Composing difference operators gives the coefficient shift and applying reconstruction to Δ_1^s f gives the third identity. The binomial expansion of (E_h-I)^s gives the explicit stencil. Binomial coefficients vanish when their lower index exceeds n, which proves the interpolation claim. Equality of every point of the stencil proves equality of differences. …

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    \Delta_1^s f
S_nIntroduction passage · 10 passages

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  • Introduction passage · Read passage · supplement_02.tex, line 31

    rence theorem remain pending. No continuum convergence or analytic summability theorem is inferred from the formal power-series equalities in this block. The historical numerical study is not being relabelled as a fresh experiment. pageheading1. The actual odd-power generating functionLemma 4.1; manuscript page 10 Work in the rational formal power-series ring Q[[t]]. Define S_n(p)=∑_a=0^p-1(-1)^a(2a+1)^n∈ Z, F_p(t)=∑_n≥0S_n(p)fract^nn!. The formal exponential is Mathlib's mcodePowerSeries.exp; s…

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     S_n(p)=\sum_{a=0}^{p-1}(-1)^a(2a+1)^n\in\Z,
     \qquad F_p(t)=\sum_{n\ge0}S_n(p)\frac{t^n}{n!}.
    
\sechIntroduction passage · 11 passages

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  • Introduction passage · Read passage · supplement_02.tex, line 36

    r pageheading1. The actual odd-power generating functionLemma 4.1; manuscript page 10 Work in the rational formal power-series ring Q[[t]]. Define S_n(p)=∑_a=0^p-1(-1)^a(2a+1)^n∈ Z, F_p(t)=∑_n≥0S_n(p)fract^nn!. The formal exponential is Mathlib's mcodePowerSeries.exp; scalar substitution is its actual mcoderescale homomorphism. We define exp_a(t)=e^at, cosh_a(t)=frace^at+e^-at2, sech(t)=cosh_1(t)^-1. The constant term of cosh_a is proved to be one, so the reciprocal used here is a genuine unit i…

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     \exp_a(t)=e^{at},\qquad
     \cosh_a(t)=\frac{e^{at}+e^{-at}}2,\qquad
     \sech(t)=\cosh_1(t)^{-1}.
    
\sinh^2Introduction passage · 11 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 72

    edition is the treatment of the odd and dyadic Euler formulas. They are no longer presented as unrelated reconstruction devices. For each prime p we define an even source series H_p and an inverse kernel K_p satisfying K_p(t)H_p(t)=secht. At an odd prime, H_p(t)=cosh^2(pt)secht, K_p(t)=sech^2(pt). At the dyadic prime, H_2(t)=fracsinh(2t)sinh t2t^2, K_2(t)=frac4t^2sinh^2(2t). The coefficient convolution is identical in form, but the kernels have different integral weights. This is the correct sen…

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    H_2(t)=\frac{\sinh(2t)\sinh t}{2t^2},
    \qquad
    K_2(t)=\frac{4t^2}{\sinh^2(2t)}.
    
t^2Introduction passage · 19 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 72

    edition is the treatment of the odd and dyadic Euler formulas. They are no longer presented as unrelated reconstruction devices. For each prime p we define an even source series H_p and an inverse kernel K_p satisfying K_p(t)H_p(t)=secht. At an odd prime, H_p(t)=cosh^2(pt)secht, K_p(t)=sech^2(pt). At the dyadic prime, H_2(t)=fracsinh(2t)sinh t2t^2, K_2(t)=frac4t^2sinh^2(2t). The coefficient convolution is identical in form, but the kernels have different integral weights. This is the correct sen…

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    H_2(t)=\frac{\sinh(2t)\sinh t}{2t^2},
    \qquad
    K_2(t)=\frac{4t^2}{\sinh^2(2t)}.
    
t^nIntroduction passage · 15 passages

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  • Introduction passage · Read passage · supplement_02.tex, line 31

    rence theorem remain pending. No continuum convergence or analytic summability theorem is inferred from the formal power-series equalities in this block. The historical numerical study is not being relabelled as a fresh experiment. pageheading1. The actual odd-power generating functionLemma 4.1; manuscript page 10 Work in the rational formal power-series ring Q[[t]]. Define S_n(p)=∑_a=0^p-1(-1)^a(2a+1)^n∈ Z, F_p(t)=∑_n≥0S_n(p)fract^nn!. The formal exponential is Mathlib's mcodePowerSeries.exp; s…

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     S_n(p)=\sum_{a=0}^{p-1}(-1)^a(2a+1)^n\in\Z,
     \qquad F_p(t)=\sum_{n\ge0}S_n(p)\frac{t^n}{n!}.
    
t_0Introduction passage · 55 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 12

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     F\in\mathbb K[Y_1,\ldots,Y_e,T_1,\ldots,T_k]^e,
     \qquad a=(y_0,t_0),
    
T_1Introduction passage · 7 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 12

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     F\in\mathbb K[Y_1,\ldots,Y_e,T_1,\ldots,T_k]^e,
     \qquad a=(y_0,t_0),
    
T_jIntroduction passage · 3 passages

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  • Introduction passage · Read passage · strengthening.tex, line 1085

    [Exact image and residue enumeration] Let widetilde C_j denote the dyadic Newton interpolation. The map T_j(x)=fracwidetilde C_j(x)-12^b_j is a bijective isometry Z_2→ Z_2, and widetilde C_j( Z_2)=1+2^b_j Z_2, j!widetilde C_j( Z_2)=j!+2^B_j Z_2. For r>B_j, one least period modulo 2^r visits every residue in j!+2^B_j Z/2^r Z exactly once. For r≤ B_j, there is one residue and the least row period is one.

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     T_j(x)=\frac{\widetilde C_j(x)-1}{2^{b_j}}
    
T_kIntroduction passage · 7 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 12

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     F\in\mathbb K[Y_1,\ldots,Y_e,T_1,\ldots,T_k]^e,
     \qquad a=(y_0,t_0),
    
U^pIntroduction passage · 9 passages

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  • Introduction passage · Read passage · supplement_07.tex, line 101

    The positive-tail period. Write h=(p-1)/m. For n≥1, evaluate U^pf_m=Uf_m at the nonnegative degree mn-1: f_m(mn+p-1)=f_m(mn). Since mn+p-1=m(n+h), the identified sequence satisfies e_n+h^(m,p)=e_n^(m,p) (n≥1). Induction gives e_n+uh^(m,p)=e_n^(m,p) for every u≥0. This proves a period, not necessarily the least positive period. The proof also includes p=2, where m=1 and h=1.

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    U^pf_m=Uf_m
VIntroduction passage · 59 passages

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  • Introduction passage · Read passage · 30_lifting.tex, line 162

    The next obstruction concerns compatibility between individually attainable corrections. Its algebraic form works over an arbitrary commutative ring. Let U,V,W be R-modules and let J:U→ V, K:U→ W be linear. Define C(J,K)=U/(ker J+ker K). An attainable value Ju has label [u]∈ C(J,K); this is independent of the choice of its preimage because ker J is killed in the quotient. Similarly, Kv has label [v]. Denote these maps by λ_J:im J→ C(J,K) and λ_K:im K→ C(J,K).

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    U,V,W
v_2Introduction passage · 158 passages

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  • Introduction passage · Read passage · root_scope.tex, line 96

    Let Z_2⟨ x⟩ denote the restricted power series: their integral coefficients tend to zero dyadically. Extend A by mathscr A(x)=frac9^x+1-18(x+1)(2x+1), mathscr A(-1)=-fraclog98. The apparent singularity at x=-1 is removable. Since v_2(log9)=3, its convergent expansion is mathscr A(x)=fraclog9/81+2x ∑_h≥0frac(log9)^h(h+1)!(x+1)^h. It follows coefficientwise that mathscr A∈ Z_2⟨ x⟩^×, mathscr A≡1pmod2, and mathscr A(N)=A(N).

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    v_2(\log9)=3
WIntroduction passage · 23 passages

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  • Introduction passage · Read passage · 30_lifting.tex, line 162

    The next obstruction concerns compatibility between individually attainable corrections. Its algebraic form works over an arbitrary commutative ring. Let U,V,W be R-modules and let J:U→ V, K:U→ W be linear. Define C(J,K)=U/(ker J+ker K). An attainable value Ju has label [u]∈ C(J,K); this is independent of the choice of its preimage because ker J is killed in the quotient. Similarly, Kv has label [v]. Denote these maps by λ_J:im J→ C(J,K) and λ_K:im K→ C(J,K).

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    U,V,W
\widetilde CIntroduction passage · 6 passages

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  • Introduction passage · Read passage · strengthening.tex, line 1083

    [Exact image and residue enumeration] Let widetilde C_j denote the dyadic Newton interpolation. The map T_j(x)=fracwidetilde C_j(x)-12^b_j is a bijective isometry Z_2→ Z_2, and widetilde C_j( Z_2)=1+2^b_j Z_2, j!widetilde C_j( Z_2)=j!+2^B_j Z_2. For r>B_j, one least period modulo 2^r visits every residue in j!+2^B_j Z/2^r Z exactly once. For r≤ B_j, there is one residue and the least row period is one.

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    \widetilde C_j
\widetilde C_jIntroduction passage · 4 passages

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  • Introduction passage · Read passage · strengthening.tex, line 1083

    [Exact image and residue enumeration] Let widetilde C_j denote the dyadic Newton interpolation. The map T_j(x)=fracwidetilde C_j(x)-12^b_j is a bijective isometry Z_2→ Z_2, and widetilde C_j( Z_2)=1+2^b_j Z_2, j!widetilde C_j( Z_2)=j!+2^B_j Z_2. For r>B_j, one least period modulo 2^r visits every residue in j!+2^B_j Z/2^r Z exactly once. For r≤ B_j, there is one residue and the least row period is one.

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    \widetilde C_j
XIntroduction passage · 103 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2641

    Let F=(F_1,…,F_r): Z_p^d⟶ Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    F_i\in\Z_p[X_1,\ldots,X_d]
X^aIntroduction passage · 1 passage

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  • Introduction passage · Read passage · supplement_06.tex, line 70

    Principal targets. targetCompanion.orbit_of_recurrence targetCompanion.companion_jump, Companion.source_companion_jump targetCompanion.packet_step, Companion.packet_advance targetCompanion.packet_refinement, Companion.packet_zero_step pageheading2. From the characteristic polynomial to a recurrenceTheorem 9.1, continued Put D=C^N and let χ_D(X)=det(XI-D)=X^d+∑_a=0^d-1c_aX^a be its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands pol…

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     \chi_D(X)=\det(XI-D)=X^d+\sum_{a=0}^{d-1}c_aX^a
    
X^dIntroduction passage · 2 passages

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  • Introduction passage · Read passage · supplement_06.tex, line 70

    Principal targets. targetCompanion.orbit_of_recurrence targetCompanion.companion_jump, Companion.source_companion_jump targetCompanion.packet_step, Companion.packet_advance targetCompanion.packet_refinement, Companion.packet_zero_step pageheading2. From the characteristic polynomial to a recurrenceTheorem 9.1, continued Put D=C^N and let χ_D(X)=det(XI-D)=X^d+∑_a=0^d-1c_aX^a be its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands pol…

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     \chi_D(X)=\det(XI-D)=X^d+\sum_{a=0}^{d-1}c_aX^a
    
X_1Introduction passage · 23 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2641

    Let F=(F_1,…,F_r): Z_p^d⟶ Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    F_i\in\Z_p[X_1,\ldots,X_d]
X_dIntroduction passage · 20 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2641

    Let F=(F_1,…,F_r): Z_p^d⟶ Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    F_i\in\Z_p[X_1,\ldots,X_d]
x_LIntroduction passage · 34 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2517

    Let p be a prime. Let M∈ M_r,d( Z_p), b∈ Z_p^r. Suppose that x_L∈ Z_p^d satisfies Mx_L≡ bpmod p^L for some L≥1. Define the reduced residual ρ_L(x_L) = fracMx_L-bp^L pmod p ∈ F_p^r. Write overline M for the reduction of M modulo p.

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    x_L\in\Zp^d
YIntroduction passage · 52 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 11

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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    (Y,T)\in\mathbb K^e\times\mathbb K^k
yIntroduction passage · 140 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 12

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     F\in\mathbb K[Y_1,\ldots,Y_e,T_1,\ldots,T_k]^e,
     \qquad a=(y_0,t_0),
    
y_0Introduction passage · 33 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 12

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     F\in\mathbb K[Y_1,\ldots,Y_e,T_1,\ldots,T_k]^e,
     \qquad a=(y_0,t_0),
    
Y_1Introduction passage · 9 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 12

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     F\in\mathbb K[Y_1,\ldots,Y_e,T_1,\ldots,T_k]^e,
     \qquad a=(y_0,t_0),
    
Y_eIntroduction passage · 4 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 12

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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     F\in\mathbb K[Y_1,\ldots,Y_e,T_1,\ldots,T_k]^e,
     \qquad a=(y_0,t_0),
    
\mathbb Z^dIntroduction passage · 13 passages

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  • Introduction passage · Read passage · continuation_01.tex, line 16

    Let F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, q≥1, L≥1, x∈ Z^d, F(x)≡0pmodq^L. Write R_q= Z/q Z, let ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

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    F=(F_1,\ldots,F_e)\in\mathbb Z[X_1,\ldots,X_d]^e,
    \quad q\ge1,\quad L\ge1,
    \quad x\in\mathbb Z^d,
    \quad F(x)\equiv0\pmod{q^L}.
    
\mathbb Z^eIntroduction passage · 4 passages

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  • Introduction passage · Read passage · continuation_01.tex, line 22

    Let F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, q≥1, L≥1, x∈ Z^d, F(x)≡0pmodq^L. Write R_q= Z/q Z, let ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

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    r_x=F(x)/q^L\in\mathbb Z^e,
    \qquad A_x=DF(x):\mathbb Z^d\longrightarrow\mathbb Z^e,
    \qquad J_x=\rho_q\circ A_x.
    
\Z_2Introduction passage · 28 passages

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  • Introduction passage · Read passage · supplement_11.tex, line 78

    Write H_int for the subring of Q_2[[t]] whose factorial-scaled coefficients belong to Z_2. Flatness means B_4^sF∈ H_int for every s. The proof of multiplicative and reciprocal closure is induction on s, using the full product correction. It separately proves that the actual reciprocal has integral Hurwitz coefficients when F(0)=1.

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    \Z_2
\mathbb Z_2Introduction passage · 61 passages

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  • Introduction passage · Read passage · root_scope.tex, line 89

    Let Z_2⟨ x⟩ denote the restricted power series: their integral coefficients tend to zero dyadically. Extend A by mathscr A(x)=frac9^x+1-18(x+1)(2x+1), mathscr A(-1)=-fraclog98. The apparent singularity at x=-1 is removable. Since v_2(log9)=3, its convergent expansion is mathscr A(x)=fraclog9/81+2x ∑_h≥0frac(log9)^h(h+1)!(x+1)^h. It follows coefficientwise that mathscr A∈ Z_2⟨ x⟩^×, mathscr A≡1pmod2, and mathscr A(N)=A(N).

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    \mathbb Z_2\langle x\rangle
\ZpIntroduction passage · 66 passages

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  • Introduction passage · Read passage · 00_scope.tex, line 18

    Throughout, N={0,1,2,…}. When p is prime, Z_p denotes the p-adic integers and Q_p their fraction field. We normalize v_p(p)=1. For a rational or p-adic quantity, a displayed lower valuation bound is understood with the mathematical convention v_p(0)=+∞. The corresponding formal predicates are zero-safe: the quantity is zero or its valuation meets the stated bound. An exact finite valuation assertion requires a nonzero quantity, and this is stated where used. Congruence in Z_p means ideal divisib…

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    \Zp
\Zp^dIntroduction passage · 11 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2517

    Let p be a prime. Let M∈ M_r,d( Z_p), b∈ Z_p^r. Suppose that x_L∈ Z_p^d satisfies Mx_L≡ bpmod p^L for some L≥1. Define the reduced residual ρ_L(x_L) = fracMx_L-bp^L pmod p ∈ F_p^r. Write overline M for the reduction of M modulo p.

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    x_L\in\Zp^d
\Zp^rIntroduction passage · 3 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2512

    Let p be a prime. Let M∈ M_r,d( Z_p), b∈ Z_p^r. Suppose that x_L∈ Z_p^d satisfies Mx_L≡ bpmod p^L for some L≥1. Define the reduced residual ρ_L(x_L) = fracMx_L-bp^L pmod p ∈ F_p^r. Write overline M for the reduction of M modulo p.

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    M\in M_{r,d}(\Zp),
    \qquad
    b\in\Zp^r.
    
\timesIntroduction passage · 45 passages

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  • Introduction passage · Read passage · 40_regularity.tex, line 11

    Let K be a complete nontrivially normed field. Let e,k∈ N, and write the variables as (Y,T)∈ K^e× K^k. For F∈ K[Y_1,…,Y_e,T_1,…,T_k]^e, a=(y_0,t_0), define the matrices of evaluated formal partial derivatives J=D_YF(a)∈ M_e( K), K=D_TF(a)∈ M_e× k( K). Rows index equations and columns index variables. We first establish that these formal derivatives are the derivatives of the evaluation map.

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    (Y,T)\in\mathbb K^e\times\mathbb K^k
\deltaIntroduction passage · 54 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 47

    Fix an odd prime p, a divisor mmid p-1, and put h=fracp-1m. The reduced standard window inverse is defined by ∑_q=0^n binommnmqe_q^(m,p) = δ_n0 in F_p. We prove that e_n+h^(m,p)=e_n^(m,p) (n≥1). The result first follows from subgroup symmetry in the finite Hurwitz algebra. A second proof shows that the standard window operator preserves the finite space of sequences that are h periodic after their constant term. The two arguments expose different parts of the same finite character structure.

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    \sum_{q=0}^{n}
    \binom{mn}{mq}e_q^{(m,p)}
    =
    \delta_{n0}
    \quad\text{in }\FF_p.
    
\Delta_1Introduction passage · 11 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 272

    Write E_h f(n)=f(n+h) for the shift operator. The unit shift is I+Δ_1. Expanding its nth power and evaluating at zero gives reconstruction. Composing difference operators gives the coefficient shift and applying reconstruction to Δ_1^s f gives the third identity. The binomial expansion of (E_h-I)^s gives the explicit stencil. Binomial coefficients vanish when their lower index exceeds n, which proves the interpolation claim. Equality of every point of the stencil proves equality of differences. …

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    I+\Delta_1
\Delta_1^sIntroduction passage · 15 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 274

    Write E_h f(n)=f(n+h) for the shift operator. The unit shift is I+Δ_1. Expanding its nth power and evaluating at zero gives reconstruction. Composing difference operators gives the coefficient shift and applying reconstruction to Δ_1^s f gives the third identity. The binomial expansion of (E_h-I)^s gives the explicit stencil. Binomial coefficients vanish when their lower index exceeds n, which proves the interpolation claim. Equality of every point of the stencil proves equality of differences. …

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    \Delta_1^s f
\delta_{n0}Introduction passage · 13 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 47

    Fix an odd prime p, a divisor mmid p-1, and put h=fracp-1m. The reduced standard window inverse is defined by ∑_q=0^n binommnmqe_q^(m,p) = δ_n0 in F_p. We prove that e_n+h^(m,p)=e_n^(m,p) (n≥1). The result first follows from subgroup symmetry in the finite Hurwitz algebra. A second proof shows that the standard window operator preserves the finite space of sequences that are h periodic after their constant term. The two arguments expose different parts of the same finite character structure.

    Exact mathematical passage
    
    \sum_{q=0}^{n}
    \binom{mn}{mq}e_q^{(m,p)}
    =
    \delta_{n0}
    \quad\text{in }\FF_p.
    
\Delta_uIntroduction passage · 1 passage

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  • Introduction passage · Read passage · 00_scope.tex, line 32

    Write Δ_u f(n)=f(n+u)-f(n) and use Δ_u^k for its k-fold iterate. Euler coefficients indexed by ordinary degree and by packet row are distinguished explicitly. A packet displacement by u rows is a degree displacement by Nu in a packet of width N.

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    \Delta_u f(n)=f(n+u)-f(n)
\Delta_u^kIntroduction passage · 3 passages

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  • Introduction passage · Read passage · 00_scope.tex, line 32

    Write Δ_u f(n)=f(n+u)-f(n) and use Δ_u^k for its k-fold iterate. Euler coefficients indexed by ordinary degree and by packet row are distinguished explicitly. A packet displacement by u rows is a degree displacement by Nu in a packet of width N.

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    \Delta_u^k
\Delta_{uh}^{k}Introduction passage · 8 passages

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  • Introduction passage · Read passage · 20_arithmetic.tex, line 193

    [All-address differences on the divisor branch] Let p be prime, m>0, mmid p-1, N=mp, and 0≤ j<N. Write h=(p-1)/m. For every u,k,n∈ N, U_N,j(n) · ∈ Z_p, Δ_uh^kU_N,j(n) · ∈ p^min(k,mn) Z_p, v_p(Δ_uh^kC_N,j(n)) · ≥min(k,mn)-v_p(j!). The factorial correction is part of the normalized statement, including addresses j≥ p.

    Exact mathematical passage
    
     U_{N,j}(n)&\in\Z_p,\label{eq:arith-divisor-integral}\\
     \Delta_{uh}^{k}U_{N,j}(n)&\in p^{\min(k,mn)}\Z_p,\label{eq:arith-divisor-bound}\\
     v_p\bigl(\Delta_{uh}^{k}C_{N,j}(n)\bigr)
       &\ge\min(k,mn)-v_p(j!).\label{eq:arith-divisor-normalized}
    
\lambda_JIntroduction passage · 1 passage

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  • Introduction passage · Read passage · 30_lifting.tex, line 170

    s over an arbitrary commutative ring. Let U,V,W be R-modules and let J:U→ V, K:U→ W be linear. Define C(J,K)=U/(ker J+ker K). An attainable value Ju has label [u]∈ C(J,K); this is independent of the choice of its preimage because ker J is killed in the quotient. Similarly, Kv has label [v]. Denote these maps by λ_J:im J→ C(J,K) and λ_K:im K→ C(J,K).

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    \lambda_J:\im J\to C(J,K)
\lambda_KIntroduction passage · 1 passage

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  • Introduction passage · Read passage · 30_lifting.tex, line 170

    ring. Let U,V,W be R-modules and let J:U→ V, K:U→ W be linear. Define C(J,K)=U/(ker J+ker K). An attainable value Ju has label [u]∈ C(J,K); this is independent of the choice of its preimage because ker J is killed in the quotient. Similarly, Kv has label [v]. Denote these maps by λ_J:im J→ C(J,K) and λ_K:im K→ C(J,K).

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    \lambda_K:\im K\to C(J,K)
\rhoIntroduction passage · 33 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2522

    Let p be a prime. Let M∈ M_r,d( Z_p), b∈ Z_p^r. Suppose that x_L∈ Z_p^d satisfies Mx_L≡ bpmod p^L for some L≥1. Define the reduced residual ρ_L(x_L) = fracMx_L-bp^L pmod p ∈ F_p^r. Write overline M for the reduction of M modulo p.

    Exact mathematical passage
    
    \rho_L(x_L)
    =
    \frac{Mx_L-b}{p^L}
    \pmod p
    \in\FF_p^r.
    
\rho_LIntroduction passage · 4 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2522

    Let p be a prime. Let M∈ M_r,d( Z_p), b∈ Z_p^r. Suppose that x_L∈ Z_p^d satisfies Mx_L≡ bpmod p^L for some L≥1. Define the reduced residual ρ_L(x_L) = fracMx_L-bp^L pmod p ∈ F_p^r. Write overline M for the reduction of M modulo p.

    Exact mathematical passage
    
    \rho_L(x_L)
    =
    \frac{Mx_L-b}{p^L}
    \pmod p
    \in\FF_p^r.
    
\rho_L^FIntroduction passage · 3 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2646

    Let F=(F_1,…,F_r): Z_p^d⟶ Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    \rho_L^F(x_L)
    =
    \frac{F(x_L)}{p^L}
    \pmod p.
    
\rho_qIntroduction passage · 7 passages

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  • Introduction passage · Read passage · continuation_01.tex, line 21

    Let F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, q≥1, L≥1, x∈ Z^d, F(x)≡0pmodq^L. Write R_q= Z/q Z, let ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

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    \rho_q
\chiIntroduction passage · 14 passages

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  • Introduction passage · Read passage · supplement_06.tex, line 70

    Principal targets. targetCompanion.orbit_of_recurrence targetCompanion.companion_jump, Companion.source_companion_jump targetCompanion.packet_step, Companion.packet_advance targetCompanion.packet_refinement, Companion.packet_zero_step pageheading2. From the characteristic polynomial to a recurrenceTheorem 9.1, continued Put D=C^N and let χ_D(X)=det(XI-D)=X^d+∑_a=0^d-1c_aX^a be its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands pol…

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     \chi_D(X)=\det(XI-D)=X^d+\sum_{a=0}^{d-1}c_aX^a
    
\chi_DIntroduction passage · 5 passages

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  • Introduction passage · Read passage · supplement_06.tex, line 70

    Principal targets. targetCompanion.orbit_of_recurrence targetCompanion.companion_jump, Companion.source_companion_jump targetCompanion.packet_step, Companion.packet_advance targetCompanion.packet_refinement, Companion.packet_zero_step pageheading2. From the characteristic polynomial to a recurrenceTheorem 9.1, continued Put D=C^N and let χ_D(X)=det(XI-D)=X^d+∑_a=0^d-1c_aX^a be its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands pol…

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     \chi_D(X)=\det(XI-D)=X^d+\sum_{a=0}^{d-1}c_aX^a
    
\psiIntroduction passage · 24 passages

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  • Introduction passage · Read passage · continuation_11.tex, line 26

    Local analytic graph theorem. There exist a map ψ: K^k⟶ K^e and open sets U⊆ K^k, V⊆ K^e such that t_0∈ U, y_0∈ V, ψ(t_0)=y_0, ψ(U)⊆ V. The map is analytic at every point of U, and its graph describes all roots in the product neighborhood: ∀ t∈ U ∀ y∈ V, F(y,t)=0 Longleftrightarrow y=ψ(t). At the base parameter it has the strict Fréchet derivative Dψ(t_0)=-J^-1K. Consequently it is differentiable on U. The map is defined on the ambient parameter space, but its root, analyticity and uniqueness co…

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    \psi:\mathbb K^k\longrightarrow\mathbb K^e
    
\ldotsIntroduction passage · 60 passages

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  • Introduction passage · Read passage · 00_scope.tex, line 18

    Throughout, N={0,1,2,…}. When p is prime, Z_p denotes the p-adic integers and Q_p their fraction field. We normalize v_p(p)=1. For a rational or p-adic quantity, a displayed lower valuation bound is understood with the mathematical convention v_p(0)=+∞. The corresponding formal predicates are zero-safe: the quantity is zero or its valuation meets the stated bound. An exact finite valuation assertion requires a nonzero quantity, and this is stated where used. Congruence in Z_p means ideal divisib…

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    \N=\{0,1,2,\ldots\}
\toIntroduction passage · 67 passages

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  • Introduction passage · Read passage · 30_lifting.tex, line 162

    The next obstruction concerns compatibility between individually attainable corrections. Its algebraic form works over an arbitrary commutative ring. Let U,V,W be R-modules and let J:U→ V, K:U→ W be linear. Define C(J,K)=U/(ker J+ker K). An attainable value Ju has label [u]∈ C(J,K); this is independent of the choice of its preimage because ker J is killed in the quotient. Similarly, Kv has label [v]. Denote these maps by λ_J:im J→ C(J,K) and λ_K:im K→ C(J,K).

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    J:U\to V
\mapstoIntroduction passage · 8 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 1881

    The analytic function is defined by its unit source series. The expression p^(p-1)x+j in the interpolation identity is evaluated only at nonnegative integers. No analytic function x↦ p^x on all of Z_p is used.

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    x\mapsto p^x
\forallIntroduction passage · 1 passage

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  • Introduction passage · Read passage · continuation_11.tex, line 33

    Local analytic graph theorem. There exist a map ψ: K^k⟶ K^e and open sets U⊆ K^k, V⊆ K^e such that t_0∈ U, y_0∈ V, ψ(t_0)=y_0, ψ(U)⊆ V. The map is analytic at every point of U, and its graph describes all roots in the product neighborhood: ∀ t∈ U ∀ y∈ V, F(y,t)=0 Longleftrightarrow y=ψ(t). At the base parameter it has the strict Fréchet derivative Dψ(t_0)=-J^-1K. Consequently it is differentiable on U. The map is defined on the ambient parameter space, but its root, analyticity and uniqueness co…

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    \forall t\in U\;\forall y\in V,
    \qquad F(y,t)=0\quad\Longleftrightarrow\quad y=\psi(t).
    
\sum_{a=0}^{d}Introduction passage · 2 passages

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  • Introduction passage · Read passage · supplement_06.tex, line 74

    packet_refinement, Companion.packet_zero_step pageheading2. From the characteristic polynomial to a recurrenceTheorem 9.1, continued Put D=C^N and let χ_D(X)=det(XI-D)=X^d+∑_a=0^d-1c_aX^a be its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands polynomial evaluation into a finite sum and proves ∑_a=0^dχ_D[a]D^a=0, χ_D[d]=1. Here χ_D[a] denotes the coefficient at index a. The identities hold over any commutative coefficient ring; the …

    Exact mathematical passage
    
     \sum_{a=0}^{d}\chi_D[a]D^a=0,
     \qquad \chi_D[d]=1.
    
\sum_{a=0}^{d-1}Introduction passage · 3 passages

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  • Introduction passage · Read passage · supplement_06.tex, line 70

    Principal targets. targetCompanion.orbit_of_recurrence targetCompanion.companion_jump, Companion.source_companion_jump targetCompanion.packet_step, Companion.packet_advance targetCompanion.packet_refinement, Companion.packet_zero_step pageheading2. From the characteristic polynomial to a recurrenceTheorem 9.1, continued Put D=C^N and let χ_D(X)=det(XI-D)=X^d+∑_a=0^d-1c_aX^a be its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands pol…

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     \chi_D(X)=\det(XI-D)=X^d+\sum_{a=0}^{d-1}c_aX^a
    
\sum_{h\ge0}Introduction passage · 1 passage

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  • Introduction passage · Read passage · root_scope.tex, line 97

    Let Z_2⟨ x⟩ denote the restricted power series: their integral coefficients tend to zero dyadically. Extend A by mathscr A(x)=frac9^x+1-18(x+1)(2x+1), mathscr A(-1)=-fraclog98. The apparent singularity at x=-1 is removable. Since v_2(log9)=3, its convergent expansion is mathscr A(x)=fraclog9/81+2x ∑_h≥0frac(log9)^h(h+1)!(x+1)^h. It follows coefficientwise that mathscr A∈ Z_2⟨ x⟩^×, mathscr A≡1pmod2, and mathscr A(N)=A(N).

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     \mathscr A(x)=\frac{\log9/8}{1+2x}
       \sum_{h\ge0}\frac{(\log9)^h}{(h+1)!}(x+1)^h.
     \label{eq:root:source}
    
\sum_{q=0}^{n}Introduction passage · 14 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 47

    Fix an odd prime p, a divisor mmid p-1, and put h=fracp-1m. The reduced standard window inverse is defined by ∑_q=0^n binommnmqe_q^(m,p) = δ_n0 in F_p. We prove that e_n+h^(m,p)=e_n^(m,p) (n≥1). The result first follows from subgroup symmetry in the finite Hurwitz algebra. A second proof shows that the standard window operator preserves the finite space of sequences that are h periodic after their constant term. The two arguments expose different parts of the same finite character structure.

    Exact mathematical passage
    
    \sum_{q=0}^{n}
    \binom{mn}{mq}e_q^{(m,p)}
    =
    \delta_{n0}
    \quad\text{in }\FF_p.
    
\sum_{r=0}^mIntroduction passage · 4 passages

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  • Introduction passage · Read passage · strengthening.tex, line 414

    Let B_m denote the Bernoulli numbers with B_1=-1/2. The identity obtained by substituting u=e^t-1 into log(1+u)/u is B_m=∑_r=0^m(-1)^rfracr!r+1 {m r}, where the braces are Stirling numbers of the second kind. Each 2r!/(r+1) is dyadically integral. Indeed, if s=v_2(r+1)>0, then r≥2^s-1 and v_2(r!)≥ v_2((2^s-1)!)=2^s-1-s≥ s-1. The last inequality is 2^s≥2s; the case s=0 is immediate. Consequently 2B_m∈ Z_2 (m≥0). Thus the transfer factors below require no appeal to the von Staudt--Clausen theorem.

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     B_m=\sum_{r=0}^m(-1)^r\frac{r!}{r+1}
                                      \left\{\begin{matrix}m\\r\end{matrix}\right\},
     \label{eq:bernoulli-stirling-denominator}
    
\sum_{r=0}^{n}Introduction passage · 15 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 129

    In characteristic p, factorial division is unavailable in general, but the coefficient operation remains meaningful over every commutative ring: (f⋆ g)_n=∑_r=0^nbinom nr f_r g_n-r, (D^d f)_n=f_n+d, 1_n=1, · n=0, 0, · n>0. Here ⋆ is Hurwitz multiplication. The symbol D^d denotes a coefficient shift, which agrees with formal differentiation after characteristic-zero encoding.

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    (f\star g)_n=\sum_{r=0}^{n}\binom nr f_r g_{n-r},\qquad
    (D^d f)_n=f_{n+d},\qquad \mathbf1_n=\begin{cases}1,&n=0,\\0,&n>0.\end{cases}
    
\circIntroduction passage · 5 passages

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  • Introduction passage · Read passage · continuation_01.tex, line 22

    Let F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, q≥1, L≥1, x∈ Z^d, F(x)≡0pmodq^L. Write R_q= Z/q Z, let ρ_q denote coordinatewise reduction, and put r_x=F(x)/q^L∈ Z^e, A_x=DF(x): Z^d⟶ Z^e, J_x=ρ_q∘ A_x. Here A_x is computed from the actual formal partial derivatives of the input polynomials. The Lean map J_x has domain Z^d and codomain R_q^e, viewed as integer modules. It factors through reduction to R_q^d; the cokernel below is presented directly using its integer-domain image.

    Exact mathematical passage
    
    r_x=F(x)/q^L\in\mathbb Z^e,
    \qquad A_x=DF(x):\mathbb Z^d\longrightarrow\mathbb Z^e,
    \qquad J_x=\rho_q\circ A_x.
    
\congIntroduction passage · 6 passages

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  • Introduction passage · Read passage · continuation_08.tex, line 52

    Every nonempty L_v has a checked free and transitive action of ker J. The action is defined without choosing a base lift. Choosing one valid solved digit then gives an explicit bijection L_v≅ker J. In particular, for every positive modulus, o_x(v)=0 Longrightarrow # L_v=#ker J. These results permit composite q, the degenerate modulus q=1, and zero dimensions. Neither J nor the full reduced Jacobian is assumed surjective.

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    \mathcal L_v\cong\ker J
\equivIntroduction passage · 135 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2518

    Let p be a prime. Let M∈ M_r,d( Z_p), b∈ Z_p^r. Suppose that x_L∈ Z_p^d satisfies Mx_L≡ bpmod p^L for some L≥1. Define the reduced residual ρ_L(x_L) = fracMx_L-bp^L pmod p ∈ F_p^r. Write overline M for the reduction of M modulo p.

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    Mx_L\equiv b\pmod {p^L}
    
\subseteqIntroduction passage · 23 passages

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  • Introduction passage · Read passage · continuation_11.tex, line 28

    Local analytic graph theorem. There exist a map ψ: K^k⟶ K^e and open sets U⊆ K^k, V⊆ K^e such that t_0∈ U, y_0∈ V, ψ(t_0)=y_0, ψ(U)⊆ V. The map is analytic at every point of U, and its graph describes all roots in the product neighborhood: ∀ t∈ U ∀ y∈ V, F(y,t)=0 Longleftrightarrow y=ψ(t). At the base parameter it has the strict Fréchet derivative Dψ(t_0)=-J^-1K. Consequently it is differentiable on U. The map is defined on the ambient parameter space, but its root, analyticity and uniqueness co…

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    U\subseteq\mathbb K^k
\starIntroduction passage · 15 passages

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  • Introduction passage · Read passage · 10_foundations.tex, line 129

    In characteristic p, factorial division is unavailable in general, but the coefficient operation remains meaningful over every commutative ring: (f⋆ g)_n=∑_r=0^nbinom nr f_r g_n-r, (D^d f)_n=f_n+d, 1_n=1, · n=0, 0, · n>0. Here ⋆ is Hurwitz multiplication. The symbol D^d denotes a coefficient shift, which agrees with formal differentiation after characteristic-zero encoding.

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    (f\star g)_n=\sum_{r=0}^{n}\binom nr f_r g_{n-r},\qquad
    (D^d f)_n=f_{n+d},\qquad \mathbf1_n=\begin{cases}1,&n=0,\\0,&n>0.\end{cases}
    
\langleIntroduction passage · 19 passages

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  • Introduction passage · Read passage · root_scope.tex, line 89

    Let Z_2⟨ x⟩ denote the restricted power series: their integral coefficients tend to zero dyadically. Extend A by mathscr A(x)=frac9^x+1-18(x+1)(2x+1), mathscr A(-1)=-fraclog98. The apparent singularity at x=-1 is removable. Since v_2(log9)=3, its convergent expansion is mathscr A(x)=fraclog9/81+2x ∑_h≥0frac(log9)^h(h+1)!(x+1)^h. It follows coefficientwise that mathscr A∈ Z_2⟨ x⟩^×, mathscr A≡1pmod2, and mathscr A(N)=A(N).

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    \mathbb Z_2\langle x\rangle
\rangleIntroduction passage · 19 passages

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  • Introduction passage · Read passage · root_scope.tex, line 89

    Let Z_2⟨ x⟩ denote the restricted power series: their integral coefficients tend to zero dyadically. Extend A by mathscr A(x)=frac9^x+1-18(x+1)(2x+1), mathscr A(-1)=-fraclog98. The apparent singularity at x=-1 is removable. Since v_2(log9)=3, its convergent expansion is mathscr A(x)=fraclog9/81+2x ∑_h≥0frac(log9)^h(h+1)!(x+1)^h. It follows coefficientwise that mathscr A∈ Z_2⟨ x⟩^×, mathscr A≡1pmod2, and mathscr A(N)=A(N).

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    \mathbb Z_2\langle x\rangle
\longrightarrowIntroduction passage · 24 passages

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  • Introduction passage · Read passage · 72_recovered.tex, line 2638

    Let F=(F_1,…,F_r): Z_p^d⟶ Z_p^r be a polynomial map. Editorial coefficient clarification. Here polynomial means F_i∈ Z_p[X_1,…,X_d], not merely an integer-valued polynomial with coefficients in Q_p. This makes the Jacobian and Taylor remainder integral and their reduction well typed. Suppose F(x_L)∈ p^L Z_p^r. Define ρ_L^F(x_L) = fracF(x_L)p^L pmod p. Let J_F(x_L) denote the Jacobian matrix and write overline J_F(x_L) for its reduction modulo p.

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    F=(F_1,\ldots,F_r):\Zp^d\longrightarrow\Zp^r
    
\mathbf1_{d\equiv r\ (N)}Use in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 83

    Modules. mcodeCyclotomicFiltration, mcodePrimeBlockPolynomial, and mcodeCyclicAmiceCoefficients. 2. Connection to the actual section. An integral coordinate estimate becomes relevant only after connecting it to the manuscript's generating functions. For a residue r∈ Z/N Z, define H_r(t)=∑_d≥01_d≡ r (N)fract^dd!, F_r(T)=H_r(log(1+T)). The implementation uses the actual formal logarithm and substitution in Q[[T]]. It proves H_r'=H_r-1, (1+T)F_r'=F_r-1, and hence the coefficient recurrence (d+1)coe…

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     H_r(t)=\sum_{d\ge0}\mathbf1_{d\equiv r\ (N)}\frac{t^d}{d!},
     \qquad F_r(T)=H_r(\log(1+T)).
    
\mathbf1_{r=0}Use in source · 2 passages

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  • Use in source · Read passage · supplement_09.tex, line 96

    For a residue r∈ Z/N Z, define H_r(t)=∑_d≥01_d≡ r (N)fract^dd!, F_r(T)=H_r(log(1+T)). The implementation uses the actual formal logarithm and substitution in Q[[T]]. It proves H_r'=H_r-1, (1+T)F_r'=F_r-1, and hence the coefficient recurrence (d+1)coefTd+1F_r =coefTdF_r-1-dcoefTdF_r. The initial values are F_r(0)=1_r=0.

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    F_r(0)=\mathbf1_{r=0}
\mathcal AUse in source · 12 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 927

    Its exponential generating function is A(t)=∑_n≥0A(n)fract^2n(2n)! =fraccosh(3t)-cosh t4t^2. The quotients by powers of t are well defined because the required initial coefficients vanish. The hyperbolic identity A(t)cosh t=(fracsinh(2t)2t)^2 then gives the exact formal-series product (∑_m≥0β_m(2t)^2m) A(t) =secht. Taking the factorial-scaled coefficient of t^2n produces the stated finite sum. Full depth leaves no omitted terms. The relevant declarations are leanKomorebiPaperIX.Dyadic.beta_integ…

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     \mathcal A(t)=\sum_{n\ge0}A(n)\frac{t^{2n}}{(2n)!}
                 =\frac{\cosh(3t)-\cosh t}{4t^2}.
    
a'Use in source · 7 passages

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  • Use in source · Read passage · 71_native.tex, line 238

    R(a,b), R(a',b), R(a',b') Longrightarrow R(a,b')

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    R(a,b),\ R(a',b),\ R(a',b')\quad\Longrightarrow\quad R(a,b')
a^{(m)}Use in source · 5 passages

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A^{-1}Use in source · 3 passages

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  • Use in source · Read passage · strengthening.tex, line 63

    Products preserve H_p because the coefficient at factorial degree d is (AB)_d=∑_i=0^dbinom di a_i b_d-i. The derivative D acts by the coefficient shift (DA)_d=a_d+1. Consequently every nonnegative power of D preserves the lattice. If A∈ H_p has unit constant term, its reciprocal belongs to the same lattice: with B=A^-1, b_0=a_0^-1, b_d=-a_0^-1∑_i=1^dbinom di a_i b_d-i. If A is supported at degrees divisible by N, this recurrence also proves that B has the same support.

    Exact mathematical passage
    B=A^{-1}
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  • Use in source · Read passage · strengthening.tex, line 730

    ties give sinh t-sin t =2asinh(at)cosh(bt)+2bcosh(at)sinh(bt), cosh t-cos t=2sinh(at)sinh(bt). Consequently D_3(t):=G_4,3(t)F_2(t) =frac32(fraccoth(at)at +fraccoth(bt)bt). This expression is a power series: its two polar terms cancel because a^-2+b^-2=0. Its constant is one, only degrees divisible by four survive, and the Bernoulli expansion gives D_3(t)=∑_i≥0d_ifract^4i(4i)!, d_i=3(-frac14)^i frac2^4i+2B_4i+2(4i+2)(4i+1). Again v_2(d_i)≥2i, and the inverse coefficients h_i have the same bound. …

    Exact mathematical passage
    a^{-2}+b^{-2}=0
A^2Use in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 467

    One sufficient new arithmetic input would be an odd-denominator projective quadratic A Z^2+2BZ+C≡ A W_r(Z)pmod2^P, A odd, 0<|B^2-AC|<2^P, at a proved precision for every admitted ordinary r. Then (B^2-AC)/A^2≡δ_rpmod2^P cannot vanish modulo 2^P. No depth-independent theorem supplying this inequality has been established. The tested recurrence and exterior-contact identities retain balancing terms and do not force a contradiction. Finite exponential models approximate the actual target at every d…

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    (B^2-AC)/A^2\equiv\delta_r\pmod{2^P}
a^2Use in source · 1 passage

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  • Use in source · Read passage · continuation_01.tex, line 51

    For every P∈ Z[X_1,…,X_d], integer a, and x,t∈ Z^d, Lean proves a^2mid P(x+at)-P(x)-a∑_i=1^d(∂_iP)(x)t_i. The proof uses polynomial induction: constants, addition, and multiplication by a variable. The product rule for the actual partial derivatives controls the multiplication step. It does not assume a Taylor identity or a liftability condition.

    Exact mathematical passage
    
    \boxed{
     a^2\mid P(x+at)-P(x)-a\sum_{i=1}^{d}(\partial_iP)(x)t_i.
    }
    
a^kUse in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 54

    Define a Kummer sequence over a commutative ring by P_a(f) Longleftrightarrow a^kmidΔ_1^k f(n) for all k,n. The exact discrete product rule Δ_1^s(fg)(n)=∑_i=0^sbinom si Δ_1^i f(n) Δ_1^s-ig(n+i) is proved by induction. It proves closure under products. Applied to fg=1, the same formula isolates Δ_1^s g(n)f(n+s); induction and multiplication by g(n+s) prove closure under actual pointwise inverses.

    Exact mathematical passage
    
    \PP_a(f)\quad\Longleftrightarrow\quad
     a^k\mid\Delta_1^k f(n)\ \text{for all }k,n.
    
a^{p^L}Use in source · 1 passage

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  • Use in source · Read passage · 10_foundations.tex, line 364

    [Finite unit coordinates] Let p be prime, L≥0, and a∈( Z/p^L+1 Z)^×. Define ω_L(a)=a^p^L, ⟨ a⟩_L=aω_L(a)^-1. Then a=ω_L(a)⟨ a⟩_L, ω_L(a)^p-1=1, ⟨ a⟩_L≡1pmod p. The two factors are uniquely determined by these properties and are compatible with reduction to lower precision. For a∈( Z/2^L+2 Z)^×, there is instead a unique decomposition a=ϵ(a)⟨ a⟩_4, ϵ(a)∈{1,-1}, ⟨ a⟩_4≡1pmod4.

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    \omega_L(a)=a^{p^L},\qquad \langle a\rangle_L=a\omega_L(a)^{-1}.
    
a^{p^{\ell-1}}Use in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 81

    ffPadic_coe, .standard_window_integral targetWindowIndependence.residue_independence, .reduced_unique targetWindowIndependence.two_window_addresses_congruent pageheading2. Explicit finite prime-power decompositionProposition 6.2; odd-prime branch Let p be prime, ℓ≥1, and a∈( Z/p^ℓ Z)^×. Define actual units in the same ring by ω_ℓ(a)=a^p^ℓ-1, ⟨ a⟩_ℓ=a ω_ℓ(a)^-1. The Lean source writes ℓ=L+1, so zero precision is excluded without truncated natural-number subtraction.

    Exact mathematical passage
    
     \boxed{\omega_\ell(a)=a^{p^{\ell-1}},\qquad
     \langle a\rangle_\ell=a\,\omega_\ell(a)^{-1}.}
    
a^{p^{\ell-1}(p-1)}Use in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 89

    The root equation and residue are derived. Euler's unit-group exponent gives ω_ℓ(a)^p-1 =a^p^ℓ-1(p-1)=a^φ(p^ℓ)=1. Reduction to the actual field F_p commutes with powers. Its Frobenius identity gives overlineω_ℓ(a) =bar a^p^ℓ-1=bar a. Consequently, a=ω_ℓ(a)⟨ a⟩_ℓ, ω_ℓ(a)^p-1=1, overline⟨ a⟩_ℓ=1. The manuscript's finite existence conclusion is obtained directly. The proof does not assume a Hensel lift or an already supplied root representative.

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     \omega_\ell(a)^{p-1}
     =a^{p^{\ell-1}(p-1)}=a^{\varphi(p^\ell)}=1.
    
a^{\varphi(p^\ell)}Use in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 89

    The root equation and residue are derived. Euler's unit-group exponent gives ω_ℓ(a)^p-1 =a^p^ℓ-1(p-1)=a^φ(p^ℓ)=1. Reduction to the actual field F_p commutes with powers. Its Frobenius identity gives overlineω_ℓ(a) =bar a^p^ℓ-1=bar a. Consequently, a=ω_ℓ(a)⟨ a⟩_ℓ, ω_ℓ(a)^p-1=1, overline⟨ a⟩_ℓ=1. The manuscript's finite existence conclusion is obtained directly. The proof does not assume a Hensel lift or an already supplied root representative.

    Exact mathematical passage
    
     \omega_\ell(a)^{p-1}
     =a^{p^{\ell-1}(p-1)}=a^{\varphi(p^\ell)}=1.
    
A_{-c}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 525

    d product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting from the independently established integrality of the reciprocal G, proves that inversion negates the charge. Finally A_c+2=A_c-1, so increasing …

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     A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').
    
A_0Use in source · 4 passages

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  • Use in source · Read passage · edition_summary.tex, line 16

    The strongest common-step conclusion in width four is particularly compact. For n≥0 and k,H≥1, ∑_i=0^k(-1)^k-ibinom ki E_4(n+iH)^(4,j) ≡0pmod2^ v_2(k!)+kv_2(H)+A_j(k), where A_0(k)=A_1(k)=k, A_2(k)=max{k+4, 2k+lfloor k/2rfloor+1}, A_3(k)=k+2. There is no starting-row threshold. The exponent is exact in addresses zero and one for every order and positive step. It is exact in address two at every power-of-two order. The proof chapters also give mixed-step forms, the cancellation criterion in addre…

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     A_0(k)=A_1(k)=k,\qquad
     A_2(k)=\max\{k+4,\,2k+\lfloor k/2\rfloor+1\},\qquad
     A_3(k)=k+2.
    
\mathcal A_0Use in source · 5 passages

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  • Use in source · Read passage · strengthening.tex, line 339

    Let Γ_R(P,Q) be the R-subalgebra generated by all γ_m(P) and γ_r(Q). Adjoin E, F with E^2=1+2P, F^2=1+2Q, and call the resulting algebra A_0. Its elements are finite expressions; no completion is being presumed. It is generated as a Γ_R(P,Q)-module by 1, E, F, E F. The functional maps this module to R. For an even basis term this is [source reference] and independence. For a basis term with an odd Euler factor the moment is zero, since P and Q are even polynomials in their respective variables.

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    \mathcal A_0
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  • Use in source · Read passage · 10_foundations.tex, line 12

    The original arithmetic seed notation S=(A,a_0, O, R,{P_N,j},w,τ_p) records a coefficient ring or module, a seed, allowed operators, relations, residue selectors, a depth weight, and a topology. This notation describes the ingredients of the construction; a category of seed presentations and a notion of equivalence still require separate definitions. We therefore state each theorem for a concrete object. Section~[source reference] discusses the additional structure needed for the original native…

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    \mathcal S=(A,a_0,\mathcal O,\mathcal R,\{P_{N,j}\},w,\tau_p)
    
a_{0,1}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 672

    The first Newton coefficients, derived from [source reference], are a_0,1=-2, a_1,1=-frac65, a_2,1=-frac1615, a_3,1=-frac3635. Their exact valuations are c_j. Theorem~[source reference] gives v_2(a_j,k)≥ k for all orders. To obtain the stronger logarithmic envelope required by Lemma~[source reference], only a finite initial segment needs improvement. The recurrence proves the following exact valuations: cll Address j · Further orders k · v_2(a_j,k) in that order

    Exact mathematical passage
    
     a_{0,1}=-2,\quad a_{1,1}=-\frac65,\quad
     a_{2,1}=-\frac{16}{15},\quad a_{3,1}=-\frac{36}{35}.
    
a_0^{-1}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 64

    Products preserve H_p because the coefficient at factorial degree d is (AB)_d=∑_i=0^dbinom di a_i b_d-i. The derivative D acts by the coefficient shift (DA)_d=a_d+1. Consequently every nonnegative power of D preserves the lattice. If A∈ H_p has unit constant term, its reciprocal belongs to the same lattice: with B=A^-1, b_0=a_0^-1, b_d=-a_0^-1∑_i=1^dbinom di a_i b_d-i. If A is supported at degrees divisible by N, this recurrence also proves that B has the same support.

    Exact mathematical passage
    
     b_0=a_0^{-1},\qquad
     b_d=-a_0^{-1}\sum_{i=1}^d\binom di a_i b_{d-i}.
     \label{eq:hurwitz-reciprocal-recurrence}
    
a_0^{[M]}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1977

    For a sequence a=(a_n)_n≥0 and M≥2, define a_r^[M](q)=a_Mq+r (0≤ r<M). The analysis map is P_Ma = (a_0^[M],…,a_M-1^[M]). Synthesis interleaves the branches by the rule a_Mq+r=a_r^[M](q).

    Exact mathematical passage
    
    \mathcal P_Ma
    =
    \left(a_0^{[M]},\ldots,a_{M-1}^{[M]}\right).
    
A_1Use in source · 4 passages

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  • Use in source · Read passage · edition_summary.tex, line 16

    The strongest common-step conclusion in width four is particularly compact. For n≥0 and k,H≥1, ∑_i=0^k(-1)^k-ibinom ki E_4(n+iH)^(4,j) ≡0pmod2^ v_2(k!)+kv_2(H)+A_j(k), where A_0(k)=A_1(k)=k, A_2(k)=max{k+4, 2k+lfloor k/2rfloor+1}, A_3(k)=k+2. There is no starting-row threshold. The exponent is exact in addresses zero and one for every order and positive step. It is exact in address two at every power-of-two order. The proof chapters also give mixed-step forms, the cancellation criterion in addre…

    Exact mathematical passage
    
     A_0(k)=A_1(k)=k,\qquad
     A_2(k)=\max\{k+4,\,2k+\lfloor k/2\rfloor+1\},\qquad
     A_3(k)=k+2.
    
a_1Use in source · 14 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 615

    [Finite Newton dominance] Let f: N→ Q and write a_k=Δ_1^kf(0). Suppose that c∈ N, a_1≠0, v_2(a_1)=c, and v_2(a_k)≥ c+1+lfloorlog_2krfloor (k≥2). Then for every n≥0 and T>0, f(n+T)-f(n)≠0, v_2(f(n+T)-f(n))=v_2(T)+c.

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    a_1\ne0
a_{1,1}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 672

    The first Newton coefficients, derived from [source reference], are a_0,1=-2, a_1,1=-frac65, a_2,1=-frac1615, a_3,1=-frac3635. Their exact valuations are c_j. Theorem~[source reference] gives v_2(a_j,k)≥ k for all orders. To obtain the stronger logarithmic envelope required by Lemma~[source reference], only a finite initial segment needs improvement. The recurrence proves the following exact valuations: cll Address j · Further orders k · v_2(a_j,k) in that order

    Exact mathematical passage
    
     a_{0,1}=-2,\quad a_{1,1}=-\frac65,\quad
     a_{2,1}=-\frac{16}{15},\quad a_{3,1}=-\frac{36}{35}.
    
A_1^sUse in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 541

    Packing g_j into degrees divisible by four gives K_4,j. Its source property is precisely charge-one flatness. Inversion gives charge -1, and the charge shift restores charge one for the reciprocal P_4,j. If dec(F)(d)=d![t^d]F, the exact coefficient identity is dec(A_1^sF)(d) =2^-sΔ_4^sdec(F)(d). At d=4n, packing turns the right side into 2^-sΔ_1^sC_j(n). This extraction requires no negative degree and therefore holds at every n, including zero. It proves the unit-step bound leanKomorebiPaperIX.D…

    Exact mathematical passage
    
     \operatorname{dec}(A_1^sF)(d)
           =2^{-s}\Delta_4^s\operatorname{dec}(F)(d).
    
A_2Use in source · 4 passages

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  • Use in source · Read passage · edition_summary.tex, line 16

    The strongest common-step conclusion in width four is particularly compact. For n≥0 and k,H≥1, ∑_i=0^k(-1)^k-ibinom ki E_4(n+iH)^(4,j) ≡0pmod2^ v_2(k!)+kv_2(H)+A_j(k), where A_0(k)=A_1(k)=k, A_2(k)=max{k+4, 2k+lfloor k/2rfloor+1}, A_3(k)=k+2. There is no starting-row threshold. The exponent is exact in addresses zero and one for every order and positive step. It is exact in address two at every power-of-two order. The proof chapters also give mixed-step forms, the cancellation criterion in addre…

    Exact mathematical passage
    
     A_0(k)=A_1(k)=k,\qquad
     A_2(k)=\max\{k+4,\,2k+\lfloor k/2\rfloor+1\},\qquad
     A_3(k)=k+2.
    
\mathcal A_2Use in source · 4 passages

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  • Use in source · Read passage · strengthening.tex, line 640

    Let A_2 be the divided-power algebra over R generated by P,Q, adjoining B, D with B^2=1+4P, D^2=1+4Q. The functional is integral on its module basis 1, B, D, B D by the Bernoulli lemma, independence, and vanishing of odd moments. Every component in [source reference] has integral divided powers, by the same finite identities as before; P^2 is even easier than P^2/2. Consequently L((U^4-1)^m)∈4^m m! Z_2, v_2(Δ_1^kC_2(n))≥2k+v_2(k!). The starting-row conclusion follows from the finite base expansi…

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    \mathcal A_2
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  • Use in source · Read passage · strengthening.tex, line 659

    The low Newton coefficients of the actual reciprocal are C_2(0)=1, C_2(1)=-frac115, C_2(2)=frac1345, a_1=-frac1615, a_2=frac6445, where a_m=Δ_1^m C_2(0). Thus v_2(a_1)=4 and v_2(a_2)=6. For m≥3, [source reference] bounds a_m by 2m+v_2(m!). The finite shifted Newton formula Δ_1^kC_2(n)=∑_m=k^n+ka_mbinom nm-k therefore gives v_2(Δ_1^kC_2(n))≥ k+v_2(k!)+3 for every k≥1. For k≥3 this is 2k≥ k+3; for k=1,2 it uses the exact initial coefficients and the higher tail bound. The finite step expansion pro…

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     C_2(0)=1,\quad C_2(1)=-\frac1{15},\quad C_2(2)=\frac{13}{45},
     \qquad a_1=-\frac{16}{15},\quad a_2=\frac{64}{45},
     \label{eq:address-two-initial-newton}
    
a_{2,1}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 672

    The first Newton coefficients, derived from [source reference], are a_0,1=-2, a_1,1=-frac65, a_2,1=-frac1615, a_3,1=-frac3635. Their exact valuations are c_j. Theorem~[source reference] gives v_2(a_j,k)≥ k for all orders. To obtain the stronger logarithmic envelope required by Lemma~[source reference], only a finite initial segment needs improvement. The recurrence proves the following exact valuations: cll Address j · Further orders k · v_2(a_j,k) in that order

    Exact mathematical passage
    
     a_{0,1}=-2,\quad a_{1,1}=-\frac65,\quad
     a_{2,1}=-\frac{16}{15},\quad a_{3,1}=-\frac{36}{35}.
    
a_{2,d}Use in source · 1 passage

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  • Use in source · visual_body.tex, line 456

    At every center x_0∈ Z_2, the interpolations admit ordinary Taylor series C_j(x_0+z)=∑_d≥0a_j,d(x_0)z^d. For positive degrees, v_2(a_j,d(x_0)) · =d · · (j=0,1), v_2(a_2,d(x_0)) · ≥max{2d+lfloor d/2rfloor, d+3}, v_2(a_3,d(x_0)) · ≥ d. At address two equality with 5d/2 holds along every power-of-two degree d≥4.

    Exact mathematical passage
    
    \begin{aligned}
     v_2(a_{j,d}(x_0))&=d &&(j=0,1),\\
     v_2(a_{2,d}(x_0))&\ge\max\{2d+\lfloor d/2\rfloor,\ d+3\},\\
     v_2(a_{3,d}(x_0))&\ge d.
    \end{aligned}
    
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  • Use in source · Read passage · edition_summary.tex, line 16

    The strongest common-step conclusion in width four is particularly compact. For n≥0 and k,H≥1, ∑_i=0^k(-1)^k-ibinom ki E_4(n+iH)^(4,j) ≡0pmod2^ v_2(k!)+kv_2(H)+A_j(k), where A_0(k)=A_1(k)=k, A_2(k)=max{k+4, 2k+lfloor k/2rfloor+1}, A_3(k)=k+2. There is no starting-row threshold. The exponent is exact in addresses zero and one for every order and positive step. It is exact in address two at every power-of-two order. The proof chapters also give mixed-step forms, the cancellation criterion in addre…

    Exact mathematical passage
    
     A_0(k)=A_1(k)=k,\qquad
     A_2(k)=\max\{k+4,\,2k+\lfloor k/2\rfloor+1\},\qquad
     A_3(k)=k+2.
    
A_{3,1}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1277

    [Exact change between local source charts] Editorial range correction. This comparison holds for the even source subsequences, equivalently the evenized generating functions H_p. It does not identify their full odd-degree source values. For p=3, the original odd source has A_3,1=S_1(3)=3, whereas every odd dyadic source coefficient is zero; the even kernel L_3K_2 cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity H_p=L_pK_ℓ H_ℓ…

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    A_{3,1}=S_1(3)=3
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  • Use in source · Read passage · 20_arithmetic.tex, line 672

    The first Newton coefficients, derived from [source reference], are a_0,1=-2, a_1,1=-frac65, a_2,1=-frac1615, a_3,1=-frac3635. Their exact valuations are c_j. Theorem~[source reference] gives v_2(a_j,k)≥ k for all orders. To obtain the stronger logarithmic envelope required by Lemma~[source reference], only a finite initial segment needs improvement. The recurrence proves the following exact valuations: cll Address j · Further orders k · v_2(a_j,k) in that order

    Exact mathematical passage
    
     a_{0,1}=-2,\quad a_{1,1}=-\frac65,\quad
     a_{2,1}=-\frac{16}{15},\quad a_{3,1}=-\frac{36}{35}.
    
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  • Use in source · visual_body.tex, line 456

    At every center x_0∈ Z_2, the interpolations admit ordinary Taylor series C_j(x_0+z)=∑_d≥0a_j,d(x_0)z^d. For positive degrees, v_2(a_j,d(x_0)) · =d · · (j=0,1), v_2(a_2,d(x_0)) · ≥max{2d+lfloor d/2rfloor, d+3}, v_2(a_3,d(x_0)) · ≥ d. At address two equality with 5d/2 holds along every power-of-two degree d≥4.

    Exact mathematical passage
    
    \begin{aligned}
     v_2(a_{j,d}(x_0))&=d &&(j=0,1),\\
     v_2(a_{2,d}(x_0))&\ge\max\{2d+\lfloor d/2\rfloor,\ d+3\},\\
     v_2(a_{3,d}(x_0))&\ge d.
    \end{aligned}
    
a_{B^s f}Use in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 91

    The earlier coefficient criterion is recovered exactly. Define the literal coefficient defect (δ a)(d)=a(d+p^2)-a(d+p). The new source proves, for every s,d, a_B^s f(d)=p^-s(δ^s a_f)(d). For an integral coefficient x, it proves the bridge p^-sx∈ Z_p Longleftrightarrow p^smid x in Z_p. Thus the differential definition is equivalent to the existing coefficient-level Dwork criterion. Division occurs in Q_p, while the assertion after division is membership in Z_p. It is not divisibility in the field…

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     a_{B^s f}(d)=p^{-s}(\delta^s a_f)(d).
    
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  • Use in source · Read passage · 20_arithmetic.tex, line 216

    Set H_c(t)=∑_n≥0fract^Nn+c(Nn+c)!, A_c(t)=∑_n≥0fract^Nn(Nn+c)!, R_c(t)=fracA_c-1(t)A_c(t) (1≤ c<N). The denominator of this last quotient has nonzero constant coefficient. The section identities give R_cH_c=tH_c-1, R_c(0)=c, j! P_N,j= P_N,0∏_c=1^jR_c. Cancellation of consecutive source sections proves the product identity and ties the ratios directly to the family under study.

    Exact mathematical passage
    
     H_c(t)=\sum_{n\ge0}\frac{t^{Nn+c}}{(Nn+c)!},\qquad
     A_c(t)=\sum_{n\ge0}\frac{t^{Nn}}{(Nn+c)!},\qquad
     R_c(t)=\frac{A_{c-1}(t)}{A_c(t)}\quad(1\le c<N).
    
A_{c+2}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 530

    atives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting from the independently established integrality of the reciprocal G, proves that inversion negates the charge. Finally A_c+2=A_c-1, so increasing a charge by two preserves flatness. These are leanKomorebiPaperIX.ChargedDworkClosure.B_mul, leanKomorebiPaperIX.Cha…

    Exact mathematical passage
    A_{c+2}=A_c-1
A_{c+d}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 517

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

    Exact mathematical passage
    \label{eq:arith-charged-product}
     A_{c+d}(FG)=(A_cF)G+F(A_dG)
           +2F'G'''+3F''G''+2F'''G'.
    
A_{c-1}Use in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 216

    Set H_c(t)=∑_n≥0fract^Nn+c(Nn+c)!, A_c(t)=∑_n≥0fract^Nn(Nn+c)!, R_c(t)=fracA_c-1(t)A_c(t) (1≤ c<N). The denominator of this last quotient has nonzero constant coefficient. The section identities give R_cH_c=tH_c-1, R_c(0)=c, j! P_N,j= P_N,0∏_c=1^jR_c. Cancellation of consecutive source sections proves the product identity and ties the ratios directly to the family under study.

    Exact mathematical passage
    
     H_c(t)=\sum_{n\ge0}\frac{t^{Nn+c}}{(Nn+c)!},\qquad
     A_c(t)=\sum_{n\ge0}\frac{t^{Nn}}{(Nn+c)!},\qquad
     R_c(t)=\frac{A_{c-1}(t)}{A_c(t)}\quad(1\le c<N).
    
A_c^sUse in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 515

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

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    A_c^s
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  • Use in source · Read passage · 20_arithmetic.tex, line 517

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

    Exact mathematical passage
    \label{eq:arith-charged-product}
     A_{c+d}(FG)=(A_cF)G+F(A_dG)
           +2F'G'''+3F''G''+2F'''G'.
    
a_dUse in source · 9 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 231

    At the singular addresses c=ap, 1≤ a<m, use the weighted ring W_p={∑_d≥0a_dT^d: p^lfloor d/prfloora_d∈ Z_p}. The key estimate is R_ap(log(1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!)…

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     \mathcal W_p=\left\{\sum_{d\ge0}a_dT^d:
              p^{\lfloor d/p\rfloor}a_d\in\Z_p\right\}.
    
a_{d+1}Use in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 272

    To deduce flatness from the weighted estimate, use Θ=(1+T)d/dT. Both Θ and (Θ^p^2-Θ^p)/p preserve W_p. For the first operator, its coefficient formula is (d+1)a_d+1+da_d; whenever the floor weight rises, the factor d+1 provides the missing p. For the second operator, the proof uses its full Leibniz correction and the scalar congruence p^2mid b^p^2-b^p. Weighted monomials factor as a_dT^d=(p^lfloor d/prfloora_d) (T^p/p)^lfloor d/prfloorT^dbmod p. The operator preserves these monomials and hence f…

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    (d+1)a_{d+1}+da_d
a_{d+pk}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 333

    Here is the coefficient extraction explicitly. If a_d=d![t^d]F, then d![t^d]B_p^kF=p^-kΔ_p(p-1)^ka_d+pk. For F=j! P_N,j and k≤ mn, take d=mpn-pk. Support in multiples of mp turns this into p^-kΔ_h^kU_N,j(n), so flatness gives divisibility by p^k. For k>mn, expand Δ_h^k-mn applied to Δ_h^mnU_N,j(n); every shifted row is at least n, so each summand is divisible by p^mn. Finally Δ_uh=((I+Δ_h)^u-I), whose kth power for u>0 is an integral combination of powers of Δ_h of order at least k. The case u=0…

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     d![t^d]B_p^kF=p^{-k}\Delta_{p(p-1)}^ka_{d+pk}.
    
a_fUse in source · 4 passages

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  • Use in source · Read passage · supplement_08.tex, line 66

    pageheading2. The actual integral Hurwitz latticeConnecting differential expressions to coefficient divisibility The specialization works inside Q_p[[t]] and uses factorial-scaled coefficients a_f(d)=d![t^d]f, I_p={f∈ Q_p[[t]]:a_f(d)∈ Z_p for every d}. The source constructs I_p as a subring. Actual exponential-series multiplication gives a_fg(d)=∑_r=0^dbinom dr a_f(r)a_g(d-r), a_Uf(d)=a_f(d+1). These identities prove multiplication and derivative stability of the lattice. Membership does not req…

    Exact mathematical passage
    
     a_f(d)=d![t^d]f,\qquad
     I_p=\{f\in\Q_p[[t]]:a_f(d)\in\Z_p\text{ for every }d\}.
    
a_{f^{-1}}Use in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 79

    The reciprocal is genuinely integral. For f∈ I_p with constant coefficient one, the usual power-series reciprocal is used. Its inverse equation yields the exact coefficient recurrence a_f^-1(d)=δ_d0-∑_r=0^d-1 binom dr a_f^-1(r)a_f(d-r). Strong induction proves that all of these coefficients are integral. Combining this construction with the preceding all-order theorem proves flatness of the actual reciprocal. No flatness or integrality of that reciprocal is assumed in the normalized-series speci…

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     a_{f^{-1}}(d)=\delta_{d0}-\sum_{r=0}^{d-1}
         \binom dr a_{f^{-1}}(r)a_f(d-r).
    
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  • Use in source · Read passage · supplement_08.tex, line 71

    pageheading2. The actual integral Hurwitz latticeConnecting differential expressions to coefficient divisibility The specialization works inside Q_p[[t]] and uses factorial-scaled coefficients a_f(d)=d![t^d]f, I_p={f∈ Q_p[[t]]:a_f(d)∈ Z_p for every d}. The source constructs I_p as a subring. Actual exponential-series multiplication gives a_fg(d)=∑_r=0^dbinom dr a_f(r)a_g(d-r), a_Uf(d)=a_f(d+1). These identities prove multiplication and derivative stability of the lattice. Membership does not req…

    Exact mathematical passage
    
     a_{fg}(d)=\sum_{r=0}^{d}\binom dr a_f(r)a_g(d-r),
     \qquad a_{Uf}(d)=a_f(d+1).
    
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  • Use in source · Read passage · supplement_08.tex, line 71

    pageheading2. The actual integral Hurwitz latticeConnecting differential expressions to coefficient divisibility The specialization works inside Q_p[[t]] and uses factorial-scaled coefficients a_f(d)=d![t^d]f, I_p={f∈ Q_p[[t]]:a_f(d)∈ Z_p for every d}. The source constructs I_p as a subring. Actual exponential-series multiplication gives a_fg(d)=∑_r=0^dbinom dr a_f(r)a_g(d-r), a_Uf(d)=a_f(d+1). These identities prove multiplication and derivative stability of the lattice. Membership does not req…

    Exact mathematical passage
    
     a_{fg}(d)=\sum_{r=0}^{d}\binom dr a_f(r)a_g(d-r),
     \qquad a_{Uf}(d)=a_f(d+1).
    
a_{H_ax+j}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2393

    , v_0,w∈ Z_p^d, H>0, and a,j,n,r∈ N. These are the hypotheses of the two corrected matrix statements used by the retained proof. Without integral observation vectors or positive L, the preceding counterexamples invalidate the claimed precision. Assume C^H=I+p^LB. Put H_a=Hp^a. Then C^H_a=I+p^L+aB_a for an integral matrix B_a. Every refined packet b_a,j(x)=a_H_ax+j has Mahler coefficients satisfying v_p(c_a,j,r)≥(L+a)r. Equivalently, Δ_H_a^ ra_n ≡0 pmod p^(L+a)r.

    Exact mathematical passage
    
    b_{a,j}(x)=a_{H_ax+j}
    
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  • Use in source · Read passage · 72_recovered.tex, line 1681

    means the matrix-binomial interpolation. With p=3, C=4, L=1, w=1 and v_0=1/3, one has Δ_1a_0=1, which is not in 3 Z_3. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses. Assume C^H=I+p^LB with B∈ M_d( Z_p). For every address j and every x∈ Z_p, a_Hx+j = ∑_r≥0 p^Lrbinom xr w^TC^jB^rv_0. The series converges on Z_p, and Δ_H^ra_n ≡0 pmod p^Lr.

    Exact mathematical passage
    
    \boxed{
    a_{Hx+j}
    =
    \sum_{r\geq0}
    p^{Lr}\binom xr
    w^TC^jB^rv_0.
    }
    
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  • Use in source · Read passage · 10_foundations.tex, line 337

    [Differences of unit modes] Over every commutative ring, Δ_h^s(u^n)=u^n(u^h-1)^s. For any prime p, units u_i∈ Z_p^×, coefficients a_i∈ Z_p, and a finite index set I, the sequence f(n)=∑_i∈ Ia_i u_i^n satisfies Δ_p-1^s f(n)∈ p^s Z_p (n,s∈ N).

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    a_i\in\Zp
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  • Use in source · Read passage · edition_summary.tex, line 11

    The strongest common-step conclusion in width four is particularly compact. For n≥0 and k,H≥1, ∑_i=0^k(-1)^k-ibinom ki E_4(n+iH)^(4,j) ≡0pmod2^ v_2(k!)+kv_2(H)+A_j(k), where A_0(k)=A_1(k)=k, A_2(k)=max{k+4, 2k+lfloor k/2rfloor+1}, A_3(k)=k+2. There is no starting-row threshold. The exponent is exact in addresses zero and one for every order and positive step. It is exact in address two at every power-of-two order. The proof chapters also give mixed-step forms, the cancellation criterion in addre…

    Exact mathematical passage
    
     \sum_{i=0}^k(-1)^{k-i}\binom ki E_{4(n+iH)}^{(4,j)}
     \equiv0\pmod{2^{\,v_2(k!)+kv_2(H)+A_j(k)}},
    
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  • Use in source · Read passage · 40_regularity.tex, line 40

    ts strict derivative is the continuous linear map represented by the matrix of evaluated formal partial derivatives. For the derivative statement, use polynomial induction on constants, addition, and multiplication by a variable. Constants have zero derivative. The sum rule agrees with the additivity of formal partial differentiation. For PX_j, the product rule gives d(PX_j)_a(v)=a_j dP_a(v)+P(a)v_j, which agrees with the evaluated formal product rule for every coordinate. The finite product rul…

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     d(PX_j)_a(v)=a_j\,dP_a(v)+P(a)v_j,
    
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  • Use in source · visual_body.tex, line 455

    At every center x_0∈ Z_2, the interpolations admit ordinary Taylor series C_j(x_0+z)=∑_d≥0a_j,d(x_0)z^d. For positive degrees, v_2(a_j,d(x_0)) · =d · · (j=0,1),

    Exact mathematical passage
    C_j(x_0+z)=\sum_{d\ge0}a_{j,d}(x_0)z^d
a_{j,k}Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 677

    The first Newton coefficients, derived from [source reference], are a_0,1=-2, a_1,1=-frac65, a_2,1=-frac1615, a_3,1=-frac3635. Their exact valuations are c_j. Theorem~[source reference] gives v_2(a_j,k)≥ k for all orders. To obtain the stronger logarithmic envelope required by Lemma~[source reference], only a finite initial segment needs improvement. The recurrence proves the following exact valuations: cll Address j · Further orders k · v_2(a_j,k) in that order

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    v_2(a_{j,k})\ge k
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  • Use in source · Read passage · 20_arithmetic.tex, line 614

    [Finite Newton dominance] Let f: N→ Q and write a_k=Δ_1^kf(0). Suppose that c∈ N, a_1≠0, v_2(a_1)=c, and v_2(a_k)≥ c+1+lfloorlog_2krfloor (k≥2). Then for every n≥0 and T>0, f(n+T)-f(n)≠0, v_2(f(n+T)-f(n))=v_2(T)+c.

    Exact mathematical passage
    a_k=\Delta_1^kf(0)
a_{k+i}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 628

    Finite Newton reconstruction at an arbitrary row gives Δ_1^kf(n)=∑_i=0^nbinom ni a_k+i. The monotonicity of lfloorlog_2krfloor therefore preserves the higher-coefficient bound after translation. In addition, Δ_1f(n)-a_1∈2^c+1 Z_2, so the first difference is nonzero of exact valuation c.

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     \Delta_1^kf(n)=\sum_{i=0}^{n}\binom ni a_{k+i}.
    
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  • Use in source · Read passage · strengthening.tex, line 664

    The low Newton coefficients of the actual reciprocal are C_2(0)=1, C_2(1)=-frac115, C_2(2)=frac1345, a_1=-frac1615, a_2=frac6445, where a_m=Δ_1^m C_2(0). Thus v_2(a_1)=4 and v_2(a_2)=6. For m≥3, [source reference] bounds a_m by 2m+v_2(m!). The finite shifted Newton formula Δ_1^kC_2(n)=∑_m=k^n+ka_mbinom nm-k therefore gives v_2(Δ_1^kC_2(n))≥ k+v_2(k!)+3 for every k≥1. For k≥3 this is 2k≥ k+3; for k=1,2 it uses the exact initial coefficients and the higher tail bound. The finite step expansion pro…

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    a_m=\Delta_1^m C_2(0)
a_{M(Nq+s)+r}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2031

    Direct substitution gives (a_r^[M])_s^[N](q) = a_M(Nq+s)+r. Iterate this identity along the mixed radix path.

    Exact mathematical passage
    
    \left(a_r^{[M]}\right)_s^{[N]}(q)
    =
    a_{M(Nq+s)+r}.
    
a_{M(q+1)}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2094

    For r<M-1, (Ea)_Mq+r=a_Mq+r+1=a_r+1^[M](q). For r=M-1, (Ea)_Mq+M-1=a_M(q+1)=E_qa_0^[M](q). After M shifts every phase has advanced by one coarse index.

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    (Ea)_{Mq+M-1}=a_{M(q+1)}=E_qa_0^{[M]}(q).
    
a_{M-1}^{[M]}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1977

    For a sequence a=(a_n)_n≥0 and M≥2, define a_r^[M](q)=a_Mq+r (0≤ r<M). The analysis map is P_Ma = (a_0^[M],…,a_M-1^[M]). Synthesis interleaves the branches by the rule a_Mq+r=a_r^[M](q).

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    \mathcal P_Ma
    =
    \left(a_0^{[M]},\ldots,a_{M-1}^{[M]}\right).
    
a_{Mq+r}Use in source · 5 passages

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  • Use in source · Read passage · 10_foundations.tex, line 485

    The recovered Annals text develops a broader polyphase vocabulary. We explain it through elementary identities with explicit algebraic derivations. These identities have not been separately formalized in Lean for this consolidation. For a sequence a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplicati…

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    a_r^{[M]}(q)=a_{Mq+r}
a_{Mq+r+1}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2090

    For r<M-1, (Ea)_Mq+r=a_Mq+r+1=a_r+1^[M](q). For r=M-1, (Ea)_Mq+M-1=a_M(q+1)=E_qa_0^[M](q). After M shifts every phase has advanced by one coarse index.

    Exact mathematical passage
    
    (Ea)_{Mq+r}=a_{Mq+r+1}=a_{r+1}^{[M]}(q).
    
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  • Use in source · Read passage · 72_recovered.tex, line 484

    [Exact coefficient transport] Let A(t)=∑_n≥0A_nfract^nn!, S(t)=H(t)A(t). Then S_n = ∑_r=0^n binom nr h_rA_n-r. The inverse coordinate formula is A_n = ∑_r=0^n binom nr k_rS_n-r. Both transforms are lower triangular with diagonal one.

    Exact mathematical passage
    
    A(t)=\sum_{n\geq0}A_n\frac{t^n}{n!},
    \qquad
    S(t)=H(t)A(t).
    
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  • Use in source · Read passage · 10_foundations.tex, line 78

    Over a characteristic-zero field K, put E(a)=∑_n≥0fraca_nn!t^n, D(A)_n=n![t^n]A. These inverse maps between sequences and K[[t]] provide formal exponential coordinates. No analytic convergence is needed for the coefficient identities below.

    Exact mathematical passage
    
    \mathcal E(a)=\sum_{n\geq0}\frac{a_n}{n!}t^n,
    \qquad \mathcal D(A)_n=n![t^n]A.
    
a_{n+1}Use in source · 2 passages

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  • Use in source · Read passage · 71_native.tex, line 161

    Consider the recurrence a_n+1=a_n with integer initial value. Its global parameter space is X= Z. Its local parameter spaces are X_p= Z_p, and every local realization has the same prime-independent recurrence. Give the local factors their usual p-adic topologies and the product its product topology. For these integral local factors, the restricted product relative to all Z_p is the ordinary product. Write

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    a_{n+1}=a_n
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  • Use in source · Read passage · strengthening.tex, line 174

    [All-row prime-power transfer] Let p be prime, a≥1, N=p^a, and 0≤ j<N. The coefficients C_N,j(n) belong to Z_p for all n≥0. Set A_N,j(n)=(-1)^n C_N,j(n) and T_H=S^H-(-1)^H. For k≥1 and positive H_1,…,H_k, v_p(Δ_H_1⋯Δ_H_kA_N,j(n)) · ≥ k+∑_ν=1^k v_p(H_ν),

    Exact mathematical passage
    A_{N,j}(n)=(-1)^n C_{N,j}(n)
A_{n-r}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 490

    [Exact coefficient transport] Let A(t)=∑_n≥0A_nfract^nn!, S(t)=H(t)A(t). Then S_n = ∑_r=0^n binom nr h_rA_n-r. The inverse coordinate formula is A_n = ∑_r=0^n binom nr k_rS_n-r. Both transforms are lower triangular with diagonal one.

    Exact mathematical passage
    
    \boxed{
    S_n
    =
    \sum_{r=0}^{n}
    \binom nr h_rA_{n-r}.
    }
    
a_{n-r}Use in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 95

    [Constructed unit-triangular coordinates] Let h: N→ K satisfy h_0=1, write H= E(h) and H^-1= E(k), and define U_Ha= D(H E(a)). Then U_H is an equivalence, with inverse U_H^-1a= D(H^-1 E(a)). Its coefficient formulas are (U_Ha)_n=∑_r=0^nbinom nr h_r a_n-r, (U_H^-1a)_n=∑_r=0^nbinom nr k_r a_n-r. The resulting triangular matrix has diagonal entries 1. The transported selectors widehat P_N,j=U_HP_N,jU_H^-1 are idempotent and pairwise orthogonal, and their sum is the identity.

    Exact mathematical passage
    
    (U_Ha)_n=\sum_{r=0}^{n}\binom nr h_r a_{n-r},\qquad
    (U_H^{-1}a)_n=\sum_{r=0}^{n}\binom nr k_r a_{n-r}.
    
a_n^{(m)}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 260

    [Multiple selector] Define a^(m)∈ H( F_p) by a_n^(m) = 1, · mmid n, 0, · mnmid n.

    Exact mathematical passage
    
    a_n^{(m)}
    =
    \begin{cases}
    1,&m\mid n,\\
    0,&m\nmid n.
    \end{cases}
    
A_{p,2m}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1188

    [Unified local Euler transfer] For every prime p and every even n≥0, E_n = ∑_r=0^n/2 binom n2r κ_p,r A_p,n-2r. If H_p(t) = ∑_m≥0A_p,2mfract^2m(2m)! and K_p(t) = ∑_r≥0κ_p,rfract^2r(2r)!, then K_p(t)H_p(t)=secht. For odd p, H_p(t)=cosh^2(pt)secht, K_p(t)=sech^2(pt). For p=2, H_2(t)=fracsinh(2t)sinh t2t^2, K_2(t)=frac4t^2sinh^2(2t).

    Exact mathematical passage
    
    H_p(t)
    =
    \sum_{m\geq0}A_{p,2m}\frac{t^{2m}}{(2m)!}
    
A_{p,n}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1156

    For a prime p, define A_p,n = S_n(p), · p odd, R_n, · p=2 and n even, 0, · p=2 and n odd, and κ_p,r = (-1)^rZ_2r+1p^2r, · p odd,

    Exact mathematical passage
    
    A_{p,n}
    =
    \begin{cases}
    S_n(p),&p\text{ odd},\\
    R_n,&p=2\text{ and }n\text{ even},\\
    0,&p=2\text{ and }n\text{ odd},
    \end{cases}
    
A_{p,n-2r}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1177

    [Unified local Euler transfer] For every prime p and every even n≥0, E_n = ∑_r=0^n/2 binom n2r κ_p,r A_p,n-2r. If H_p(t) = ∑_m≥0A_p,2mfract^2m(2m)! and K_p(t) = ∑_r≥0κ_p,rfract^2r(2r)!, then K_p(t)H_p(t)=secht. For odd p, H_p(t)=cosh^2(pt)secht, K_p(t)=sech^2(pt). For p=2, H_2(t)=fracsinh(2t)sinh t2t^2, K_2(t)=frac4t^2sinh^2(2t).

    Exact mathematical passage
    
    \boxed{
    E_n
    =
    \sum_{r=0}^{n/2}
    \binom n{2r}
    \kappa_{p,r}
    A_{p,n-2r}.
    }
    
a_qUse in source · 4 passages

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  • Use in source · Read passage · 10_foundations.tex, line 113

    The factorials are nonzero in K, so encoding and decoding cancel coefficientwise. The constant coefficient of H is 1; hence its formal power-series inverse exists. Multiplication by H and H^-1 gives the two inverse laws. The ordinary product coefficient formula, multiplied by n!, gives the displayed binomial sums. In row n, the coefficient of a_n is h_0=1 and coefficients of a_q with q>n vanish. Conjugating the identities of Theorem~[source reference] by the constructed equivalence proves the tr…

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    a_q
a_rUse in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 34

    This article begins with the Hurwitz product (amathbin⋆ b)_n = ∑_r=0^n binom nr a_rb_n-r. Over a field of characteristic zero it is multiplication of exponential generating functions. Over a finite field it realizes the additive group algebra through the exponential modes ϵ_u=(u^n)_n≥0. This finite algebra gives an exact proof of the standard window period. Hurwitz series rings are treated in [source reference]. The finite difference and umbral background may be found in [source reference].

    Exact mathematical passage
    
    (a\starh b)_n
    =
    \sum_{r=0}^{n}
    \binom nr a_rb_{n-r}.
    
a_{r+1}^{[M]}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2090

    For r<M-1, (Ea)_Mq+r=a_Mq+r+1=a_r+1^[M](q). For r=M-1, (Ea)_Mq+M-1=a_M(q+1)=E_qa_0^[M](q). After M shifts every phase has advanced by one coarse index.

    Exact mathematical passage
    
    (Ea)_{Mq+r}=a_{Mq+r+1}=a_{r+1}^{[M]}(q).
    
a_{r+Ms}^{[MN]}Use in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 488

    polyphase vocabulary. We explain it through elementary identities with explicit algebraic derivations. These identities have not been separately formalized in Lean for this consolidation. For a sequence a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplication, the phase-r product has a term without a …

    Exact mathematical passage
    
    A(z)=\sum_{r=0}^{M-1}z^rA_r(z^M),\qquad
    \bigl(a_r^{[M]}\bigr)_s^{[N]}=a_{r+Ms}^{[MN]}.
    
A_r^{[M]}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1993

    [Exact discrete polyphase reconstruction] The analysis map P_M is a bijection between sequences and ordered families of M sequences. If A(z)=∑_n≥0a_nz^n and A_r^[M](w)=∑_q≥0a_Mq+rw^q, then A(z)=∑_r=0^M-1z^rA_r^[M](z^M).

    Exact mathematical passage
    
    A_r^{[M]}(w)=\sum_{q\geq0}a_{Mq+r}w^q,
    
a_r^{[M]}Use in source · 6 passages

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  • Use in source · Read passage · 10_foundations.tex, line 485

    The recovered Annals text develops a broader polyphase vocabulary. We explain it through elementary identities with explicit algebraic derivations. These identities have not been separately formalized in Lean for this consolidation. For a sequence a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplicati…

    Exact mathematical passage
    a_r^{[M]}(q)=a_{Mq+r}
A_uUse in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 494

    a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplication, the phase-r product has a term without a carry and a term with one carry: C_r(w)=∑_u+v=rA_u(w)B_v(w) +w∑_u+v=r+MA_u(w)B_v(w), 0≤ u,v<M. Adding the two phase addresses gives the formula. It describes ordinary Cauchy multiplication; Hurwitz convo…

    Exact mathematical passage
    
    C_r(w)=\sum_{u+v=r}A_u(w)B_v(w)
    +w\sum_{u+v=r+M}A_u(w)B_v(w),\qquad 0\leq u,v<M.
    
a_{Uf}Use in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 71

    pageheading2. The actual integral Hurwitz latticeConnecting differential expressions to coefficient divisibility The specialization works inside Q_p[[t]] and uses factorial-scaled coefficients a_f(d)=d![t^d]f, I_p={f∈ Q_p[[t]]:a_f(d)∈ Z_p for every d}. The source constructs I_p as a subring. Actual exponential-series multiplication gives a_fg(d)=∑_r=0^dbinom dr a_f(r)a_g(d-r), a_Uf(d)=a_f(d+1). These identities prove multiplication and derivative stability of the lattice. Membership does not req…

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     a_{fg}(d)=\sum_{r=0}^{d}\binom dr a_f(r)a_g(d-r),
     \qquad a_{Uf}(d)=a_f(d+1).
    
A_{z_n}Use in source · 2 passages

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  • Use in source · Read passage · continuation_03.tex, line 49

    The one-step Taylor theorem already supplies the exact correction equation b_n+A_z_nt=0, b_n=F(z_n)/p^L+nbmod p. Surjectivity supplies a correcting digit for every residual b_n. The chosen correction gives an actual new integer point z_n+1, using the proved nonlinear Taylor remainder.

    Exact mathematical passage
    
    b_n+A_{z_n}t=0,\qquad
    b_n=F(z_n)/p^{L+n}\bmod p.
    
A_{z_{n+1}}Use in source · 1 passage

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  • Use in source · Read passage · continuation_03.tex, line 58

    Every coordinate change is divisible by p^L+n, hence by p. Each formal partial derivative is itself an integer polynomial, so its value is unchanged modulo p. Therefore A_z_n+1=A_z_n=A_x. The proof assumes no separate surjectivity or solvability statement at later levels. A dependent recursion chooses each next point from the proved one-step existence theorem. It produces one compatible sequence, rather than unrelated witnesses at different precisions.

    Exact mathematical passage
    
    A_{z_{n+1}}=A_{z_n}=A_x.
    
A_{\ell,n}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1278

    . For p=3, the original odd source has A_3,1=S_1(3)=3, whereas every odd dyadic source coefficient is zero; the even kernel L_3K_2 cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity H_p=L_pK_ℓ H_ℓ and its proof remain valid. For any two primes p and ℓ, the source sequences A_p,n and A_ℓ,n are related by an exact unit lower triangular transformation obtained by composing the inverse kernel L_p with the reconstruction kernel K_ℓ.

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    A_{\ell,n}
\mathrm{act}Use in source · 6 passages

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  • Use in source · Read passage · root_scope.tex, line 374

    ion K_r(Z)=2^-178G_r(10202+2^15Z). The complete coefficient theorem gives K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z), a_r,b_r∈ Z_2, u_r∈ Z_2^×, R_r∈ Z_2⟨ Z⟩. Its half derivative is a dyadic isometry. It therefore has a unique critical point c_r∈ Z_2. Set Φ_r^act=K_r(c_r). The earlier independent-undepletion notation for this same scalar is Φ_r(2^2r-25). Excluding a multiple root on the surviving chart is exactly the assertion Φ_r^act≠0.

    Exact mathematical passage
    \Phi_r^{\mathrm{act}}=K_r(c_r)
\operatorname{aug}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 210

    [Augmentation criterion] Let x=∑_u∈ F_pc_u[u] ∈ F_p[ F_p^+]. Then x is a unit if and only if aug(x) = ∑_uc_u is nonzero. If J is the augmentation ideal, then J^p=0.

    Exact mathematical passage
    
    \operatorname{aug}(x)
    =
    \sum_uc_u
    
\mathcal BUse in source · 10 passages

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  • Use in source · Read passage · supplement_10.tex, line 58

    The normalized defect is not assumed to be a derivation. Its proved product rule contains the two middle Leibniz sums: B_Θ(FG)=(B_Θ F)G+F(B_Θ G)+ B_p(F,G). The coefficients of B_p are the actual integral quotients binomp^2a/p and binom pa/p. Derivative stability and this correction formula show that the elements F∈ W_p satisfying B_Θ F∈ W_p are closed under the finite ring operations.

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     B_\Theta(FG)=(B_\Theta F)G+F(B_\Theta G)+\mathcal B_p(F,G).
    
b'Use in source · 5 passages

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  • Use in source · Read passage · 71_native.tex, line 238

    R(a,b), R(a',b), R(a',b') Longrightarrow R(a,b')

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    R(a,b),\ R(a',b),\ R(a',b')\quad\Longrightarrow\quad R(a,b')
b^{(m,p)}Use in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 300

    Since the subgroup average has augmentation one, it has a unique Hurwitz inverse. Write b^(m,p) = (a^(m))^-1.

    Exact mathematical passage
    
    b^{(m,p)}
    =
    (a^{(m)})^{-1}.
    
B^{-1}Use in source · 1 passage

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  • Use in source · Read passage · continuation_10.tex, line 76

    0 · cI_k=cI_e+k. The second equality uses JA=cI_e. The generic block proof works over any commutative ring. The p-adic specialization also explicitly proves that B is invertible over Q and that p^sB^-1 is integral. There is no denominator loss from appending the parameter equations.

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    p^sB^{-1}
b^{-2}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 730

    ties give sinh t-sin t =2asinh(at)cosh(bt)+2bcosh(at)sinh(bt), cosh t-cos t=2sinh(at)sinh(bt). Consequently D_3(t):=G_4,3(t)F_2(t) =frac32(fraccoth(at)at +fraccoth(bt)bt). This expression is a power series: its two polar terms cancel because a^-2+b^-2=0. Its constant is one, only degrees divisible by four survive, and the Bernoulli expansion gives D_3(t)=∑_i≥0d_ifract^4i(4i)!, d_i=3(-frac14)^i frac2^4i+2B_4i+2(4i+2)(4i+1). Again v_2(d_i)≥2i, and the inverse coefficients h_i have the same bound. …

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    a^{-2}+b^{-2}=0
B^2Use in source · 2 passages

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  • Use in source · Read passage · root_scope.tex, line 462

    One sufficient new arithmetic input would be an odd-denominator projective quadratic A Z^2+2BZ+C≡ A W_r(Z)pmod2^P, A odd, 0<|B^2-AC|<2^P, at a proved precision for every admitted ordinary r. Then (B^2-AC)/A^2≡δ_rpmod2^P cannot vanish modulo 2^P. No depth-independent theorem supplying this inequality has been established. The tested recurrence and exterior-contact identities retain balancing terms and do not force a contradiction. Finite exponential models approximate the actual target at every d…

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     A Z^2+2BZ+C\equiv A\mathcal W_r(Z)\pmod{2^P},\qquad A\text{ odd},
     \qquad 0<|B^2-AC|<2^P,
    
\mathcal B^2Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 624

    The variable B=2Z+1 has transform e^tfrac2te^2t-1=fractsinh t. Its odd moments vanish, and P=( B^2-1)/4=Z(Z+1) has the divided-power moments of Lemma~[source reference]. Introduce an independent copy D and Q=( D^2-1)/4. The exact factorization F_2(t)=fract^2cosh t-cos t =fracatsinh(at)fracbtsinh(bt) shows that U=a B+b D satisfies L(U^4n)=C_2(n). Direct expansion gives H_2:=fracU^4-14 =P+Q-P^2-Q^2+6PQ+i B D(P-Q).

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    P=(\mathcal B^2-1)/4=Z(Z+1)
b^2Use in source · 1 passage

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  • Use in source · Read passage · continuation_07.tex, line 88

    For p=2, take F(X)=X+X^2, H(X)=1, a=0, s=0, M=1. There is a unique root b∈2 Z_2 of X+X^2+2. Its equation gives b+2=-b^2. It also forces b≡2pmod4. Therefore v_2(b+2)=2=2M-3s. The accompanying Lean controls show that the error is divisible by four and is not divisible by eight. This refutes a uniform replacement of 2M-3s by 2M-3s+1, even for a one-variable unit Jacobian.

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    b+2=-b^2
b^pUse in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 275

    To deduce flatness from the weighted estimate, use Θ=(1+T)d/dT. Both Θ and (Θ^p^2-Θ^p)/p preserve W_p. For the first operator, its coefficient formula is (d+1)a_d+1+da_d; whenever the floor weight rises, the factor d+1 provides the missing p. For the second operator, the proof uses its full Leibniz correction and the scalar congruence p^2mid b^p^2-b^p. Weighted monomials factor as a_dT^d=(p^lfloor d/prfloora_d) (T^p/p)^lfloor d/prfloorT^dbmod p. The operator preserves these monomials and hence f…

    Exact mathematical passage
    p^2\mid b^{p^2}-b^p
b^{p^2}Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 275

    To deduce flatness from the weighted estimate, use Θ=(1+T)d/dT. Both Θ and (Θ^p^2-Θ^p)/p preserve W_p. For the first operator, its coefficient formula is (d+1)a_d+1+da_d; whenever the floor weight rises, the factor d+1 provides the missing p. For the second operator, the proof uses its full Leibniz correction and the scalar congruence p^2mid b^p^2-b^p. Weighted monomials factor as a_dT^d=(p^lfloor d/prfloora_d) (T^p/p)^lfloor d/prfloorT^dbmod p. The operator preserves these monomials and hence f…

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    p^2\mid b^{p^2}-b^p
B^rUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1681

    means the matrix-binomial interpolation. With p=3, C=4, L=1, w=1 and v_0=1/3, one has Δ_1a_0=1, which is not in 3 Z_3. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses. Assume C^H=I+p^LB with B∈ M_d( Z_p). For every address j and every x∈ Z_p, a_Hx+j = ∑_r≥0 p^Lrbinom xr w^TC^jB^rv_0. The series converges on Z_p, and Δ_H^ra_n ≡0 pmod p^Lr.

    Exact mathematical passage
    
    \boxed{
    a_{Hx+j}
    =
    \sum_{r\geq0}
    p^{Lr}\binom xr
    w^TC^jB^rv_0.
    }
    
B^sUse in source · 2 passages

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  • Use in source · Read passage · supplement_08.tex, line 46

    All orders, without a circular closure assumption. Let I⊂ A be a subring stable under U. Define Bounded_r(f)iff B^s f∈ I (0≤ s≤ r), Flat(f)iff B^s f∈ I (s≥0). The definitions contain no product or inverse closure requirement. Since B commutes with U, derivatives of a bounded element remain bounded. Induction on r in the displayed product rule then proves closure under multiplication. The finite middle sums use the lower-order induction hypothesis, not the desired next-order conclusion.

    Exact mathematical passage
    
     \operatorname{Bounded}_r(f)\iff B^s f\in I\ (0\le s\le r),
     \qquad \operatorname{Flat}(f)\iff B^s f\in I\ (s\ge0).
    
B_0Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 80

    [The Bernoulli boundary address] At N=2, the address j=1 gives a second classical family. Define the Bernoulli numbers by B(z)=fracze^z-1=∑_r≥0B_rfracz^rr!, B_0=1, B_1=-frac12. All quotients here are regular formal series. The identity 2B(z)-B(2z)=fraczsinh z =∑_n≥0C_2,1(n)fracz^2n(2n)! follows by putting the two fractions on a common denominator and using [source reference]. Comparing even coefficients yields C_2,1(n)=(2-2^2n)B_2n (n∈ N). This classical derivation restores the boundary example …

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     B(z)=\frac{z}{e^z-1}=\sum_{r\ge0}B_r\frac{z^r}{r!},
     \qquad B_0=1,\quad B_1=-\frac12.
    
b_0Use in source · 7 passages

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  • Use in source · worked_newton_values.tex, line 1

    (b_0,b_1,b_2,b_3,b_4)=(1,-2,72,-33872,60511872).

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    (b_0,b_1,b_2,b_3,b_4)=(1,-2,72,-33872,60511872).
    
b_{0,r}Use in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 99

    Independently, let b_d,r be the residue-r coordinate of (x)_d in R_N. Actual group-ring multiplication gives b_0,r=1_r=0, b_d+1,r=b_d,r-1-d b_d,r. Induction comparing these two recurrences proves d!coefTdH_r(log(1+T))=b_d,r. Thus the cyclic coefficient is not an unrelated stand-in for a section coefficient. For the canonical label 0≤ c<N, the formal section is also proved equal to the original mcodeExponentialSection.allowbreakmcodeoriginalSection used in the existing Euler project.

    Exact mathematical passage
    
     b_{0,r}=\mathbf1_{r=0},\qquad
     b_{d+1,r}=b_{d,r-1}-d\,b_{d,r}.
    
b_1Use in source · 10 passages

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  • Use in source · worked_newton_values.tex, line 1

    (b_0,b_1,b_2,b_3,b_4)=(1,-2,72,-33872,60511872).

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    (b_0,b_1,b_2,b_3,b_4)=(1,-2,72,-33872,60511872).
    
B_2Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 1013

    [Exact four-address first differences] For every n≥0, H>0, and j∈{0,1,2,3}, v_2(C_j(n+H)-C_j(n)) · =v_2(H)+b_j, · (b_0,b_1,b_2,b_3) · =(1,1,4,2), v_2(E_4(n+H)^(4,j)-E_4n^(4,j)) · =v_2(H)+B_j, · (B_0,B_1,B_2,B_3) · =(1,1,5,3). Consequently, the least positive row period modulo 2^r is exactly P_j(r)=2^max(0,r-B_j) (r≥1). The corresponding displacement in the original Euler subscript is 4P_j(r).

    Exact mathematical passage
    
     v_2(C_j(n+H)-C_j(n))&=v_2(H)+b_j,
          &(b_0,b_1,b_2,b_3)&=(1,1,4,2),\label{eq:rank-four-normalized-isometry}\\
     v_2(E_{4(n+H)}^{(4,j)}-E_{4n}^{(4,j)})&=v_2(H)+B_j,
          &(B_0,B_1,B_2,B_3)&=(1,1,5,3).
     \label{eq:rank-four-original-isometry}
    
b_2Use in source · 8 passages

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  • Use in source · worked_newton_values.tex, line 1

    (b_0,b_1,b_2,b_3,b_4)=(1,-2,72,-33872,60511872).

    Exact mathematical passage
    
    (b_0,b_1,b_2,b_3,b_4)=(1,-2,72,-33872,60511872).
    
B_{2m}Use in source · 8 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 952

    [Exact weighted inverse coefficient] Let κ_m=β_m2^2m(2m)!. For every m≥1, κ_m=-2^4m(2m-1)B_2m≠0, v_2(κ_m)=4m-1, where B_2m is the usual even Bernoulli number.

    Exact mathematical passage
    
     \kappa_m=-2^{4m}(2m-1)B_{2m}\ne0,\qquad
     v_2(\kappa_m)=4m-1,
    
B_{2n}Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 91

    oulli numbers by B(z)=fracze^z-1=∑_r≥0B_rfracz^rr!, B_0=1, B_1=-frac12. All quotients here are regular formal series. The identity 2B(z)-B(2z)=fraczsinh z =∑_n≥0C_2,1(n)fracz^2n(2n)! follows by putting the two fractions on a common denominator and using [source reference]. Comparing even coefficients yields C_2,1(n)=(2-2^2n)B_2n (n∈ N). This classical derivation restores the boundary example from the original manuscript. No dedicated Lean bridge for this Bernoulli specialization was identified i…

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     C_{2,1}(n)=(2-2^{2n})B_{2n}\qquad(n\in\N).
    
B_{2r}Use in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1082

    [Dyadic inverse coefficients] For r≥1, β_r = - frac2^2r(2r-1)B_2r(2r)!. Consequently, v_2(β_r)=s_2(r)-1. Every R_n is a dyadic unit.

    Exact mathematical passage
    
    \boxed{
    \beta_r
    =
    -
    \frac{2^{2r}(2r-1)B_{2r}}{(2r)!}.
    }
    
B_3Use in source · 2 passages

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  • Use in source · Read passage · supplement_04.tex, line 173

    A sharp binomial-weight improvement. Let F(n)=81binom n1+3binom n2+3binom n3+81binom n4. At n=3 and s=1, the four coefficient valuations are (4,1,1,4), while the binomial valuations are (0,1,1,0). Thus the refined bound is B_3(w;3,1)=min(4,2,2,4)=2, stronger than the coefficient-only lower bound one. The source checks both the minimum and the actual result: Δ F(3)=180, ord_3(180)=2. The valuation equality is proved in the actual completed ring Z_3, with an independent exact integer divisibility …

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     B_3(w;3,1)=\min(4,2,2,4)=2,
    
b_3Use in source · 4 passages

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  • Use in source · worked_newton_values.tex, line 1

    (b_0,b_1,b_2,b_3,b_4)=(1,-2,72,-33872,60511872).

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    (b_0,b_1,b_2,b_3,b_4)=(1,-2,72,-33872,60511872).
    
B_4Use in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 73

    3. The dyadic differential operator and reciprocity. The operator required here is B_4=fracD^8-D^42, D=fracddt. It is not the preceding p=2 specialization (D^4-D^2)/2. It is also not an ordinary derivation. The new algebra module proves the actual product formula, including the middle derivative sums whose coefficients are binom 8i/2 and binom 4i/2.

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    \boxed{B_4=\frac{D^8-D^4}{2},\qquad D=\frac{d}{dt}.}
    
b_4Use in source · 3 passages

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  • Use in source · worked_newton_values.tex, line 1

    (b_0,b_1,b_2,b_3,b_4)=(1,-2,72,-33872,60511872).

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    (b_0,b_1,b_2,b_3,b_4)=(1,-2,72,-33872,60511872).
    
B_{4i+2}Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 452

    G_4,1(t)F_0(t) =frac12(fractanh(at)at +fractanh(bt)bt). For example, its numerator follows from sinh t+sin t =2bsinh(at)cosh(bt)+2acosh(at)sinh(bt), while cosh t+cos t=2cosh(at)cosh(bt). Only degrees divisible by four survive. Its actual Hurwitz packets are D_1(t)=∑_i≥0d_ifract^4i(4i)!, d_i=(-frac14)^i frac2^4i+2(2^4i+2-1)B_4i+2(4i+2)(4i+1). The constant is one. Since v_2(4i+2)=1, the elementary Bernoulli bound shows v_2(d_i)≥2i. Hurwitz inversion therefore gives D_1(t)^-1=∑_i≥0h_ifract^4i(4i)!,…

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     D_1(t)=\sum_{i\geq0}d_i\frac{t^{4i}}{(4i)!},\qquad
     d_i=\left(-\frac14\right)^i
     \frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.
     \label{eq:address-one-packets}
    
B_aUse in source · 7 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2372

    atrix_power_amplification Editorial hypothesis correction. Require L≥1. At L=0 the standalone claim is false: for p=3, d=1 and X=2=1+B, one has X^3-1=7∉3 Z_3. With L≥1, every higher binomial term in the retained proof has the required valuation, including p=2. If X=I+p^LB with B∈ M_d( Z_p), then for every a≥0 there is B_a∈ M_d( Z_p) such that X^p^a=I+p^L+aB_a.

    Exact mathematical passage
    B_a\in M_d(\Zp)
b_{a,j}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2393

    , v_0,w∈ Z_p^d, H>0, and a,j,n,r∈ N. These are the hypotheses of the two corrected matrix statements used by the retained proof. Without integral observation vectors or positive L, the preceding counterexamples invalidate the claimed precision. Assume C^H=I+p^LB. Put H_a=Hp^a. Then C^H_a=I+p^L+aB_a for an integral matrix B_a. Every refined packet b_a,j(x)=a_H_ax+j has Mahler coefficients satisfying v_p(c_a,j,r)≥(L+a)r. Equivalently, Δ_H_a^ ra_n ≡0 pmod p^(L+a)r.

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    b_{a,j}(x)=a_{H_ax+j}
    
B_{a-1}Use in source · 3 passages

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  • Use in source · Read passage · strengthening.tex, line 1143

    ents from the reciprocal recurrence are a_1=-g(1)-1, a_2=binom2NNg(1)^2-g(2)+2g(1)+1. The first has valuation exactly one. Also binom2NN≡6pmod8 for every N=2^a≥2. To see this, write B_a=binom2^a+12^a =2∏_h=1^2^a-1(1+frac2^ah). The even-h factors give B_a-1; the odd-h factors are one modulo four for a≥2. Since B_a-1 has valuation one, B_a≡ B_a-1pmod8, and B_1=6. Therefore a_2≡6-1+2+1=0pmod8. The global Mahler bound covers all k≥3, completing the criterion with b=1. The period and image deductions…

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    B_{a-1}
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  • Use in source · Read passage · supplement_10.tex, line 82

    p^2. Replacing F by this finite prefix proves the result for the entire formal series, without an analytic convergence assumption: B_Θ W_p⊆ W_p, B_Θ^s W_p⊆ W_p (s≥0). Modules. mcodeWeightedAmiceRing, mcodeFrobeniusSquare, mcodeThetaWeighted. 2. Actual substitution and singular higher differences. Let E(F)(t)=F(e^t-1), D=fracddt, B_D=fracD^p^2-D^pp. All substitutions use Mathlib's actual formal exponential, logarithm and substitution maps. The chain rule gives D E(F)= E(Θ F), B_D^s E(F)= E(B_Θ^sF…

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     \mathcal E(F)(t)=F(e^t-1),\qquad D=\frac{d}{dt},\qquad
     B_D=\frac{D^{p^2}-D^p}{p}.
    
b_dUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 64

    Products preserve H_p because the coefficient at factorial degree d is (AB)_d=∑_i=0^dbinom di a_i b_d-i. The derivative D acts by the coefficient shift (DA)_d=a_d+1. Consequently every nonnegative power of D preserves the lattice. If A∈ H_p has unit constant term, its reciprocal belongs to the same lattice: with B=A^-1, b_0=a_0^-1, b_d=-a_0^-1∑_i=1^dbinom di a_i b_d-i. If A is supported at degrees divisible by N, this recurrence also proves that B has the same support.

    Exact mathematical passage
    
     b_0=a_0^{-1},\qquad
     b_d=-a_0^{-1}\sum_{i=1}^d\binom di a_i b_{d-i}.
     \label{eq:hurwitz-reciprocal-recurrence}
    
b_{d+1,r}Use in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 99

    Independently, let b_d,r be the residue-r coordinate of (x)_d in R_N. Actual group-ring multiplication gives b_0,r=1_r=0, b_d+1,r=b_d,r-1-d b_d,r. Induction comparing these two recurrences proves d!coefTdH_r(log(1+T))=b_d,r. Thus the cyclic coefficient is not an unrelated stand-in for a section coefficient. For the canonical label 0≤ c<N, the formal section is also proved equal to the original mcodeExponentialSection.allowbreakmcodeoriginalSection used in the existing Euler project.

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     b_{0,r}=\mathbf1_{r=0},\qquad
     b_{d+1,r}=b_{d,r-1}-d\,b_{d,r}.
    
b_{d,r}Use in source · 3 passages

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  • Use in source · Read passage · supplement_09.tex, line 98

    Independently, let b_d,r be the residue-r coordinate of (x)_d in R_N. Actual group-ring multiplication gives b_0,r=1_r=0, b_d+1,r=b_d,r-1-d b_d,r. Induction comparing these two recurrences proves d!coefTdH_r(log(1+T))=b_d,r. Thus the cyclic coefficient is not an unrelated stand-in for a section coefficient. For the canonical label 0≤ c<N, the formal section is also proved equal to the original mcodeExponentialSection.allowbreakmcodeoriginalSection used in the existing Euler project.

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    b_{d,r}
b_{d,r-1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 99

    Independently, let b_d,r be the residue-r coordinate of (x)_d in R_N. Actual group-ring multiplication gives b_0,r=1_r=0, b_d+1,r=b_d,r-1-d b_d,r. Induction comparing these two recurrences proves d!coefTdH_r(log(1+T))=b_d,r. Thus the cyclic coefficient is not an unrelated stand-in for a section coefficient. For the canonical label 0≤ c<N, the formal section is also proved equal to the original mcodeExponentialSection.allowbreakmcodeoriginalSection used in the existing Euler project.

    Exact mathematical passage
    
     b_{0,r}=\mathbf1_{r=0},\qquad
     b_{d+1,r}=b_{d,r-1}-d\,b_{d,r}.
    
b_{d-i}Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 57

    Products preserve H_p because the coefficient at factorial degree d is (AB)_d=∑_i=0^dbinom di a_i b_d-i. The derivative D acts by the coefficient shift (DA)_d=a_d+1. Consequently every nonnegative power of D preserves the lattice. If A∈ H_p has unit constant term, its reciprocal belongs to the same lattice: with B=A^-1, b_0=a_0^-1, b_d=-a_0^-1∑_i=1^dbinom di a_i b_d-i. If A is supported at degrees divisible by N, this recurrence also proves that B has the same support.

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     (AB)_d=\sum_{i=0}^d\binom di a_i b_{d-i}.
    
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  • Use in source · Read passage · supplement_10.tex, line 87

    (s≥0). Modules. mcodeWeightedAmiceRing, mcodeFrobeniusSquare, mcodeThetaWeighted. 2. Actual substitution and singular higher differences. Let E(F)(t)=F(e^t-1), D=fracddt, B_D=fracD^p^2-D^pp. All substitutions use Mathlib's actual formal exponential, logarithm and substitution maps. The chain rule gives D E(F)= E(Θ F), B_D^s E(F)= E(B_Θ^sF). The implementation also proves the inverse-substitution identity E(f(log(1+T)))=f(t).

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     D\mathcal E(F)=\mathcal E(\Theta F),\qquad
     B_D^s\mathcal E(F)=\mathcal E(B_\Theta^sF).
    
B_iUse in source · 5 passages

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  • Use in source · Read passage · root_scope.tex, line 223

    For 0≤ i<j, let d_i=v_2(i!)+v_2((j-1-i)!), k_i=1+lfloorlog_2max(i,j-1-i)rfloor, B_i=i+2^k_i Z_2, t_i=d_i+k_i. At j=1 use d_0=k_0=t_0=0. The balls B_i partition Z_2. Indeed a residue represented twice among 0,…,j-1 splits into its two children; every other represented residue remains one ball. On each such ball, P_j(i+2^k_iy)=2^t_iyQ_i(y), Q_i∈ Z_2⟨ y⟩^×, Q_i≡1pmod2.

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     B_i=i+2^{k_i}\mathbb Z_2,\qquad t_i=d_i+k_i.
    
b_kUse in source · 7 passages

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  • Use in source · Read passage · 10_foundations.tex, line 411

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

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    b_k=x_{Nk+j}
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  • Use in source · Read passage · 10_foundations.tex, line 412

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

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    b_k=C^j(C^N)^kx_0,\qquad b_{k+1}=C^Nb_k.
    
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  • Use in source · Read passage · 10_foundations.tex, line 416

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

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    b_{k+d}=-\sum_{r=0}^{d-1}q_r b_{k+r}.
    
b_{k+r}Use in source · 1 passage

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  • Use in source · Read passage · 10_foundations.tex, line 416

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

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    b_{k+d}=-\sum_{r=0}^{d-1}q_r b_{k+r}.
    
B_LUse in source · 16 passages

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  • Use in source · Read passage · 30_lifting.tex, line 16

    Unless a different coefficient ring is specified, a polynomial system is a finite list of integer polynomials. All divisibility conditions on vectors are coordinatewise. For a positive integer q, write S_q= Z/q Z and use an overline for reduction modulo q. For a prime p, write F_p=S_p, normalize |p|_p=p^-1, and put B_L(x)=x+p^L Z_p^d. An integer center is included in this notation by its natural image in Z_p^d. All dimensions below are finite and may be zero. A statement over S_q allows composit…

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     B_L(x)=x+p^L\Zp^d.
    
b_mUse in source · 5 passages

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  • Use in source · Read passage · supplement_02.tex, line 86

    The inverse coefficients are integral by construction. Define the integer binomial convolution B_n=∑_r=0^nbinom nr E_rE_n-r, b_m=B_2m. The product formula for actual EGFs gives ∑_n≥0B_nfract^nn!=sech^2 t. Odd coefficients vanish by parity. In particular, every b_m is an integer; this is the integrality required by the later congruence.

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     B_n=\sum_{r=0}^{n}\binom nr E_rE_{n-r},\qquad b_m=B_{2m}.
    
b_{mn}^{(m,p)}Use in source · 1 passage

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B_nUse in source · 3 passages

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  • Use in source · Read passage · supplement_02.tex, line 86

    The inverse coefficients are integral by construction. Define the integer binomial convolution B_n=∑_r=0^nbinom nr E_rE_n-r, b_m=B_2m. The product formula for actual EGFs gives ∑_n≥0B_nfract^nn!=sech^2 t. Odd coefficients vanish by parity. In particular, every b_m is an integer; this is the integrality required by the later congruence.

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     B_n=\sum_{r=0}^{n}\binom nr E_rE_{n-r},\qquad b_m=B_{2m}.
    
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  • Use in source · Read passage · 10_foundations.tex, line 156

    Pascal's identity proves the first product rule, and induction proves the higher rule. The intermediate binomial coefficients binom pr vanish in characteristic p. For the inverse, solve row n using the coefficient of b_n, which is g_0=1: b_n=1_n-∑_r<nbinom nr b_r g_n-r. Subtract the first and pth product rules for b⋆ g=1. The assumed identity for g leaves (D^p b-Db)⋆ g=0. A second triangular induction shows that convolution with a sequence having constant coefficient 1 has trivial kernel.

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    b_n
b_{n+p-1}^{(m,p)}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 311

    [Support and positive degree period] The inverse b^(m,p) has the following properties. item b_n^(m,p)=0 unless mmid n. item For every positive n, b_n+p-1^(m,p) = b_n^(m,p).

    Exact mathematical passage
    
    \boxed{
    b_{n+p-1}^{(m,p)}
    =
    b_n^{(m,p)}.
    }
    
B_{N,j}Use in source · 3 passages

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  • Use in source · Read passage · strengthening.tex, line 1163

    [Small-rank complete isometry table] Every address at N∈{2,4,8} satisfies v_2(C_N,j(n+H)-C_N,j(n))=b_N,j+v_2(H). The following entries are ordered by j=0,…,N-1: c l l N · normalized shifts b_N,j · original shifts B_N,j=b_N,j+v_2(j!)

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    B_{N,j}=b_{N,j}+v_2(j!)
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  • Use in source · Read passage · strengthening.tex, line 1157

    [Small-rank complete isometry table] Every address at N∈{2,4,8} satisfies v_2(C_N,j(n+H)-C_N,j(n))=b_N,j+v_2(H). The following entries are ordered by j=0,…,N-1: c l l N · normalized shifts b_N,j · original shifts B_N,j=b_N,j+v_2(j!)

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     v_2(C_{N,j}(n+H)-C_{N,j}(n))=b_{N,j}+v_2(H).
    
b_{n-r}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 34

    This article begins with the Hurwitz product (amathbin⋆ b)_n = ∑_r=0^n binom nr a_rb_n-r. Over a field of characteristic zero it is multiplication of exponential generating functions. Over a finite field it realizes the additive group algebra through the exponential modes ϵ_u=(u^n)_n≥0. This finite algebra gives an exact proof of the standard window period. Hurwitz series rings are treated in [source reference]. The finite difference and umbral background may be found in [source reference].

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    (a\starh b)_n
    =
    \sum_{r=0}^{n}
    \binom nr a_rb_{n-r}.
    
b_n^{(m,p)}Use in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 309

    [Support and positive degree period] The inverse b^(m,p) has the following properties. item b_n^(m,p)=0 unless mmid n. item For every positive n, b_n+p-1^(m,p) = b_n^(m,p).

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    b_n^{(m,p)}=0
B_pUse in source · 5 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 209

    Work in the Hurwitz integral ring of formal series H_p={F∈ Q_p[[t]]: d![t^d]F∈ Z_p for every d}. For D=d/dt, let B_p=(D^p^2-D^p)/p. Call F flat if B_p^sF∈ H_p for every s≥0. We will prove that j! P_N,j is flat at every address. The main difficulty is to include the addresses at which the usual unit-pivot argument fails.

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    B_p=(D^{p^2}-D^p)/p
\mathcal B_pUse in source · 2 passages

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  • Use in source · Read passage · supplement_10.tex, line 58

    The normalized defect is not assumed to be a derivation. Its proved product rule contains the two middle Leibniz sums: B_Θ(FG)=(B_Θ F)G+F(B_Θ G)+ B_p(F,G). The coefficients of B_p are the actual integral quotients binomp^2a/p and binom pa/p. Derivative stability and this correction formula show that the elements F∈ W_p satisfying B_Θ F∈ W_p are closed under the finite ring operations.

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     B_\Theta(FG)=(B_\Theta F)G+F(B_\Theta G)+\mathcal B_p(F,G).
    
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  • Use in source · Read passage · 20_arithmetic.tex, line 333

    Here is the coefficient extraction explicitly. If a_d=d![t^d]F, then d![t^d]B_p^kF=p^-kΔ_p(p-1)^ka_d+pk. For F=j! P_N,j and k≤ mn, take d=mpn-pk. Support in multiples of mp turns this into p^-kΔ_h^kU_N,j(n), so flatness gives divisibility by p^k. For k>mn, expand Δ_h^k-mn applied to Δ_h^mnU_N,j(n); every shifted row is at least n, so each summand is divisible by p^mn. Finally Δ_uh=((I+Δ_h)^u-I), whose kth power for u>0 is an integral combination of powers of Δ_h of order at least k. The case u=0…

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     d![t^d]B_p^kF=p^{-k}\Delta_{p(p-1)}^ka_{d+pk}.
    
B_p^rUse in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 307

    Writing q=R_c-1-(c-1) and y=R_c-c, this becomes (tD+c)y=cq+yq-y^2. On degrees divisible by p, the inverse of tD+c divides the degree-d coefficient by the unit d+c. Starting with y_0=0, form y_s+1=(tD+c)^-1(cq+y_sq-y_s^2). Flatness and support are preserved. Indeed, put E_c=tD+c and C_p=pD^p^2-D^p. The commutator is [B_p,E_c]=C_p, and iteration gives B_p^rE_cy=E_cB_p^ry+rC_pB_p^r-1y. For the inverse equation E_cy=f, induction on r makes the correction term Hurwitz-integral. Both operators preserv…

    Exact mathematical passage
    
     B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.
    
B_p^{r-1}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 307

    Writing q=R_c-1-(c-1) and y=R_c-c, this becomes (tD+c)y=cq+yq-y^2. On degrees divisible by p, the inverse of tD+c divides the degree-d coefficient by the unit d+c. Starting with y_0=0, form y_s+1=(tD+c)^-1(cq+y_sq-y_s^2). Flatness and support are preserved. Indeed, put E_c=tD+c and C_p=pD^p^2-D^p. The commutator is [B_p,E_c]=C_p, and iteration gives B_p^rE_cy=E_cB_p^ry+rC_pB_p^r-1y. For the inverse equation E_cy=f, induction on r makes the correction term Hurwitz-integral. Both operators preserv…

    Exact mathematical passage
    
     B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.
    
B_p^sUse in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 210

    Work in the Hurwitz integral ring of formal series H_p={F∈ Q_p[[t]]: d![t^d]F∈ Z_p for every d}. For D=d/dt, let B_p=(D^p^2-D^p)/p. Call F flat if B_p^sF∈ H_p for every s≥0. We will prove that j! P_N,j is flat at every address. The main difficulty is to include the addresses at which the usual unit-pivot argument fails.

    Exact mathematical passage
    B_p^sF\in\mathcal H_p
b_rUse in source · 3 passages

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  • Use in source · Read passage · 10_foundations.tex, line 157

    Pascal's identity proves the first product rule, and induction proves the higher rule. The intermediate binomial coefficients binom pr vanish in characteristic p. For the inverse, solve row n using the coefficient of b_n, which is g_0=1: b_n=1_n-∑_r<nbinom nr b_r g_n-r. Subtract the first and pth product rules for b⋆ g=1. The assumed identity for g leaves (D^p b-Db)⋆ g=0. A second triangular induction shows that convolution with a sequence having constant coefficient 1 has trivial kernel.

    Exact mathematical passage
    
    b_n=\mathbf1_n-\sum_{r<n}\binom nr b_r g_{n-r}.
    
b_{r-q}Use in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 142

    The row action is connected to the matrix entries. Reversing the finite sum and using binomial symmetry gives the actual source-index row action E_2r=∑_q=0^rU_r,q(p)S_2q(p), where U_r,q(p)= displaystylebinom2r2qb_r-qp^2(r-q), · q≤ r, 0, · q>r. The row-action equality, the diagonal identity U_r,r=1, and vanishing above the diagonal are all checked. The arithmetic weight is also explicit: p^2(r-q)mid U_r,q(p) in Z, 0≤ q≤ r. No inverse-coordinate identity, unit diagonal or weight bound is assumed f…

    Exact mathematical passage
    
     U_{r,q}(p)=
     \begin{cases}
     \displaystyle\binom{2r}{2q}b_{r-q}p^{2(r-q)},&q\le r,\\
     0,&q>r.
     \end{cases}
    
b_TUse in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 867

    Theorem~[source reference] gives a unique root b of F+p^M H in a+p^s+1R^e+k, with displacement depth M-s. Its last equations say exactly b_T=t_0+p^Mv=τ. Its first equations are G(b)=0. The integral first-order expression uses CbinomH(a)-v =binomA(H(a)+Kv)-p^sv. Projection onto the solved coordinates gives b_Y-y_0+p^M-sA(H(a)+Kv)∈ p^2M-3sR^e. The same inclusion identity as before converts this to [source reference].

    Exact mathematical passage
    b_T=t_0+p^Mv=\tau
B_vUse in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 494

    a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplication, the phase-r product has a term without a carry and a term with one carry: C_r(w)=∑_u+v=rA_u(w)B_v(w) +w∑_u+v=r+MA_u(w)B_v(w), 0≤ u,v<M. Adding the two phase addresses gives the formula. It describes ordinary Cauchy multiplication; Hurwitz convo…

    Exact mathematical passage
    
    C_r(w)=\sum_{u+v=r}A_u(w)B_v(w)
    +w\sum_{u+v=r+M}A_u(w)B_v(w),\qquad 0\leq u,v<M.
    
b_xUse in source · 4 passages

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  • Use in source · Read passage · continuation_02.tex, line 19

    Let p be prime, L≥1, and F∈ Z[X_1,…,X_d]^e, x∈ Z^d, F(x)≡0pmodp^L. Write r_x=F(x)/p^L, b_x=r_xbmod p, A_x=DF(x)bmod p: F_p^d⟶ F_p^e, ρ_x=rank_ F_pA_x. The actual lift set is L_x,L= {y∈( Z/p^L+1 Z)^d: y≡ xpmodp^L, F(y)=0}, where polynomial evaluation in the displayed set is modulo p^L+1. Define O_x= F_p^e/imA_x, o_x=[b_x]. Lean proves # L_x,L= p^d-ρ_x, · o_x=0,

    Exact mathematical passage
    
    r_x=F(x)/p^L,\qquad b_x=r_x\bmod p,
    \qquad A_x=DF(x)\bmod p:\mathbb F_p^d\longrightarrow\mathbb F_p^e,
    \qquad \rho_x=\operatorname{rank}_{\mathbb F_p}A_x.
    
b_YUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 874

    Theorem~[source reference] gives a unique root b of F+p^M H in a+p^s+1R^e+k, with displacement depth M-s. Its last equations say exactly b_T=t_0+p^Mv=τ. Its first equations are G(b)=0. The integral first-order expression uses CbinomH(a)-v =binomA(H(a)+Kv)-p^sv. Projection onto the solved coordinates gives b_Y-y_0+p^M-sA(H(a)+Kv)∈ p^2M-3sR^e. The same inclusion identity as before converts this to [source reference].

    Exact mathematical passage
    
     b_Y-y_0+p^{M-s}A(H(a)+Kv)\in p^{2M-3s}R^e.
    
B_\ThetaUse in source · 5 passages

Search aliases: B_Θ Theta

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  • Use in source · Read passage · supplement_10.tex, line 46

    Put Θ=(1+T)fracddT, B_Θ=fracΘ^p^2-Θ^pp. The coefficient formula [T^d]Θ F=(d+1)a_d+1+da_d proves Θ W_p⊆ W_p. If the floor weight increases between d and d+1, then pmid d+1, which compensates for that increase.

    Exact mathematical passage
    
     \Theta=(1+T)\frac{d}{dT},\qquad
     B_\Theta=\frac{\Theta^{p^2}-\Theta^p}{p}.
    
B_\Theta^sUse in source · 3 passages

Search aliases: B_Θ^s Theta

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  • Use in source · Read passage · supplement_10.tex, line 75

    Finally, [T^d]B_Θ F depends only on coefficients through d+p^2. Replacing F by this finite prefix proves the result for the entire formal series, without an analytic convergence assumption: B_Θ W_p⊆ W_p, B_Θ^s W_p⊆ W_p (s≥0). Modules. mcodeWeightedAmiceRing, mcodeFrobeniusSquare, mcodeThetaWeighted. 2. Actual substitution and singular higher differences. Let E(F)(t)=F(e^t-1), D=fracddt, B_D=fracD^p^2-D^pp. All substitutions use Mathlib's actual formal exponential, logarithm and substitution maps…

    Exact mathematical passage
    
     \boxed{B_\Theta\Wp\subseteq\Wp,\qquad B_\Theta^s\Wp\subseteq\Wp\ (s\ge0).}
    
\bar aUse in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 94

    The root equation and residue are derived. Euler's unit-group exponent gives ω_ℓ(a)^p-1 =a^p^ℓ-1(p-1)=a^φ(p^ℓ)=1. Reduction to the actual field F_p commutes with powers. Its Frobenius identity gives overlineω_ℓ(a) =bar a^p^ℓ-1=bar a. Consequently, a=ω_ℓ(a)⟨ a⟩_ℓ, ω_ℓ(a)^p-1=1, overline⟨ a⟩_ℓ=1. The manuscript's finite existence conclusion is obtained directly. The proof does not assume a Hensel lift or an already supplied root representative.

    Exact mathematical passage
    
     \overline{\omega_\ell(a)}
     =\bar a^{p^{\ell-1}}=\bar a.
    
\bar a^{p^{\ell-1}}Use in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 94

    The root equation and residue are derived. Euler's unit-group exponent gives ω_ℓ(a)^p-1 =a^p^ℓ-1(p-1)=a^φ(p^ℓ)=1. Reduction to the actual field F_p commutes with powers. Its Frobenius identity gives overlineω_ℓ(a) =bar a^p^ℓ-1=bar a. Consequently, a=ω_ℓ(a)⟨ a⟩_ℓ, ω_ℓ(a)^p-1=1, overline⟨ a⟩_ℓ=1. The manuscript's finite existence conclusion is obtained directly. The proof does not assume a Hensel lift or an already supplied root representative.

    Exact mathematical passage
    
     \overline{\omega_\ell(a)}
     =\bar a^{p^{\ell-1}}=\bar a.
    
\bar rUse in source · 7 passages

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  • Use in source · Read passage · continuation_08.tex, line 16

    Let q>0, L≥1, and m,k,e≥0. Write S= Z/q Z. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0)∈ Z^m+k, F(x)≡0pmodq^L. The actual lower root supplies the unique integral residual r=q^-LF(x)∈ Z^e, bar r=rbmod q∈ S^e. Let the actual reduced Jacobian blocks be J=(D_YF)(x)bmod q:S^m→ S^e, K=(D_TF)(x)bmod q:S^k→ S^e. For a prescribed parameter digit v∈ S^k, choose its canonical integral representative widehat v. Define L_v to be the set of actual residue vectors z∈( Z/q^L+1 Z)^m+k satisfying z≡ xpmodq^L, F(z…

    Exact mathematical passage
    
    r=q^{-L}F(x)\in\mathbb Z^e,\qquad \bar r=r\bmod q\in S^e.
    
\bar wUse in source · 2 passages

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  • Use in source · Read passage · continuation_09.tex, line 59

    Exact branching. At any stage n, put N=L+n. For every parameter digit v∈ F_p^k, let L_n,v={bar w∈( Z/p^N+1 Z)^m+k: l bar w≡ z(n)pmodp^N, F(bar w)=0, free(bar w)=t(n)+p^Nvpmodp^N+1 }. Here p^Nv means any integer representative of v, multiplied by p^N; the residue class does not depend on the representative. Then # L_n,v=p^m-e for every n and every v. Surjectivity implies e≤ m. This is a count of distinct actual next-level residue classes. It does not count integer representatives or entire infini…

    Exact mathematical passage
    
    \mathcal L_{n,v}=\left\{\bar w\in(\mathbb Z/p^{N+1}\mathbb Z)^{m+k}:
    \begin{array}{l}
    \bar w\equiv z(n)\pmod{p^N},\\
    F(\bar w)=0,\\
    \operatorname{free}(\bar w)=t(n)+p^Nv\pmod{p^{N+1}}
    \end{array}\right\}.
    
\bar xUse in source · 9 passages

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  • Use in source · Read passage · 50_open_questions.tex, line 20

    For a fixed polynomial system F∈ Z[X_1,…,X_d]^e, a precision L≥1, and a root bar x modulo p^L, let L_F,L(bar x) ={x'∈( Z/p^L+1 Z)^d: F(x')=0, x'≡bar xpmodp^L}. The finite obstruction theorem decides whether this set is empty from the residual digit and reduced Jacobian. When it is nonempty, the theorem also determines its cardinality and affine kernel structure. The conclusion concerns one level: vanishing of the obstruction permits the next step, but does not guarantee that subsequent steps are…

    Exact mathematical passage
    \bar x
\bigmUse in source · 1 passage

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  • Use in source · Read passage · supplement_01.tex, line 135

    0, · w=∞. The infinity convention expresses a vanishing coefficient, not an infinite exponent operation. For any finite index set, every term with weight at least L is divisible by p^L. A finite-sum proof yields p^L bigm| ∑_r=0^nbinom nr c_rp^[w(r)]S_n-r -∑_substack0≤ r≤ n w(r)<L binom nr c_rp^[w(r)]S_n-r. The result is specialized to actual p-adic integers R= Z_p for prime p. Divisibility there is precisely the asserted membership in p^L Z_p.

    Exact mathematical passage
    
     \boxed{p^L\ \bigm|\ 
     \sum_{r=0}^{n}\binom nr c_rp^{[w(r)]}S_{n-r}
     -\sum_{\substack{0\le r\le n\\w(r)<L}}
         \binom nr c_rp^{[w(r)]}S_{n-r}.}
    
\bmodUse in source · 31 passages

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  • Use in source · Read passage · 10_foundations.tex, line 57

    At each index n, exactly one standard residue label equals nbmod N. Consequently a repeated selector changes nothing, two different selectors have zero composite, and the sum retains a(n) exactly once. If Nmid M, reducing nbmod M modulo N gives nbmod N. This proves the refinement identity at every coefficient. Equality of all coefficients proves the formal-series versions.

    Exact mathematical passage
    n\bmod N
\bottomruleUse in source · 1 passage

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  • Use in source · Read passage · supplement_03.tex, line 178

    [3pt] 1 · -91019/45 · 4 [3pt] 2 · 398197/45 · 6 [3pt] 3 · -208601/15 · 8 [3pt] 4 · 1385 · 10 The checked nonzero error valuations are E_8-T_0(8)=frac5494445, v_2(E_8-T_0(8))=5, E_8-T_3(8)=frac22937615, v_2(E_8-T_3(8))=15. They exceed the uniform guarantees; these two examples do not establish a universally stronger exponent.

    Exact mathematical passage
    3pt]
    1 & $-91019/45$ & 4\\[3pt]
    2 & $398197/45$ & 6\\[3pt]
    3 & $-208601/15$ & 8\\[3pt]
    4 & $1385$ & 10\\\bottomrule
    \end{tabular}
    \end{center}
    The checked nonzero error valuations are
    \[
     E_8-T_0(8)=\frac{54944}{45},\qquad v_2(E_8-T_0(8))=5,
    
\operatorname{Bounded}Use in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 46

    All orders, without a circular closure assumption. Let I⊂ A be a subring stable under U. Define Bounded_r(f)iff B^s f∈ I (0≤ s≤ r), Flat(f)iff B^s f∈ I (s≥0). The definitions contain no product or inverse closure requirement. Since B commutes with U, derivatives of a bounded element remain bounded. Induction on r in the displayed product rule then proves closure under multiplication. The finite middle sums use the lower-order induction hypothesis, not the desired next-order conclusion.

    Exact mathematical passage
    
     \operatorname{Bounded}_r(f)\iff B^s f\in I\ (0\le s\le r),
     \qquad \operatorname{Flat}(f)\iff B^s f\in I\ (s\ge0).
    
\operatorname{Bounded}_rUse in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 46

    All orders, without a circular closure assumption. Let I⊂ A be a subring stable under U. Define Bounded_r(f)iff B^s f∈ I (0≤ s≤ r), Flat(f)iff B^s f∈ I (s≥0). The definitions contain no product or inverse closure requirement. Since B commutes with U, derivatives of a bounded element remain bounded. Induction on r in the displayed product rule then proves closure under multiplication. The finite middle sums use the lower-order induction hypothesis, not the desired next-order conclusion.

    Exact mathematical passage
    
     \operatorname{Bounded}_r(f)\iff B^s f\in I\ (0\le s\le r),
     \qquad \operatorname{Flat}(f)\iff B^s f\in I\ (s\ge0).
    
\bulletUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 847

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    
    T_{m,p}(\chi^\bullet)_n
    =
    \frac1m
    \sum_{\zeta\in\mu_m}
    (1+\zeta y)^{mn}.
    
\mathcal CUse in source · 4 passages

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  • Use in source · Read passage · continuation_01.tex, line 104

    Let G∈ Z[X_1,…,X_d]^f be a second fixed system with the same variables and the same lower root x. Let K_x=ρ_q∘ DG(x), and define C_x= Z^d/(ker J_x+ker K_x). This quotient is finite: it is a quotient of the finite image of J_x, or equivalently of the finite image of K_x.

    Exact mathematical passage
    
    \mathcal C_x=\mathbb Z^d/(\ker J_x+\ker K_x).
    
c^{(p)}_{m,j}Use in source · 4 passages

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  • Use in source · Read passage · supplement_05.tex, line 49

    Specializing to R= Z_p constructs c^(p)_m,j(n). Mapping to Q_p and comparing with the original rational forward substitution gives c^(p)_m,j(n)=C_mp,j(n) in Q_p. Thus every original standard-window coefficient is p-integral, for every row. Integrality is a conclusion, not an additional hypothesis on the coefficients.

    Exact mathematical passage
    c^{(p)}_{m,j}(n)
c^{(p)}_{m,k}Use in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 65

    ply the residue map Independently define e^(m,p) over F_p by the unit-diagonal system ∑_q=0^nbinommnmqe^(m,p)_q=δ_n0. Its recurrence and uniqueness are checked. Applying the actual homomorphism π: Z_p→ F_p to the integral row, and then Lucas reduction, proves π(c^(p)_m,j(n))=e^(m,p)_n. For j,k<p, a separate ideal statement gives pmid c^(p)_m,j(n)-c^(p)_m,k(n) inside Z_p. The formal solver also covers m=0; the manuscript uses m≥1. No global finite-difference theorem or period assertion enters thi…

    Exact mathematical passage
    
     p\mid c^{(p)}_{m,j}(n)-c^{(p)}_{m,k}(n)\quad\text{inside }\Z_p.
    
C^{12}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2424

    1 · 0 , one has C^12 =I+8 29 · 18 18 · 11 . Therefore Δ_12·2^a^ rF_n ≡0 pmod 2^(a+3)r. Every dyadic refinement doubles the step and adds one guaranteed binary digit at each Mahler depth.

    Exact mathematical passage
    
    C^{12}
    =I+8
    \begin{pmatrix}
    29&18\\
    18&11
    \end{pmatrix}.
    
C^2Use in source · 2 passages

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c^2Use in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 40

    [Polynomial Taylor remainder] Let R be a commutative ring, P∈ R[X_1,…,X_d], a∈ R^d, and c∈ R. There is a polynomial Q∈ R[U_1,…,U_d], with Q(0)=0, such that for every u∈ R^d, P(a+cu)=P(a)+c∑_j=1^d(∂_jP)(a)u_j+c^2Q(u). In particular, when R= Z and a,u,c are integral, subtracting the first two terms on the right from P(a+cu) gives a multiple of c^2.

    Exact mathematical passage
    \label{eq:polynomial-taylor}
     P(a+cu)=P(a)+c\sum_{j=1}^d(\partial_jP)(a)u_j+c^2Q(u).
    
C^{24}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1749

    1 · 0 . The matrix C is invertible modulo twenty four, and C^24-I = 24 3126 · 1932 1932 · 1194 . For every scalar coordinate of the state sequence C^nv_0, Δ_24^ ra_n ≡0 pmod 24^r. One integral jump therefore gives simultaneous congruence ladders at the primes two and three.

    Exact mathematical passage
    
    C^{24}-I
    =
    24
    \begin{pmatrix}
    3126&1932\\
    1932&1194
    \end{pmatrix}.
    
C^3Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2810

    [t] centering includegraphics[width=.95linewidth,height=3.15in,keepaspectratio]../../figures/reconstructed/recurrence_packets.pdf captionNew reconstruction. Fibonacci residues F_3k+j, with F_0=0 and F_1=1. Each packet evolves by the same jump matrix C^3 and has a different initial address.

    Exact mathematical passage
    C^3
C^HUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1677

    0,w∈ Z_p^d, L≥1, and H,j,n∈ N with H>0; the value at Hx+j for x∈ Z_p means the matrix-binomial interpolation. With p=3, C=4, L=1, w=1 and v_0=1/3, one has Δ_1a_0=1, which is not in 3 Z_3. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses. Assume C^H=I+p^LB with B∈ M_d( Z_p). For every address j and every x∈ Z_p, a_Hx+j = ∑_r≥0 p^Lrbinom xr w^TC^jB^rv_0. The series converges on Z_p, and Δ_H^ra_n …

    Exact mathematical passage
    
    C^H=I+p^LB
    
C^{H_a}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2389

    dentEditorial hypothesis correction. Require L≥1, C,B∈ M_d( Z_p), v_0,w∈ Z_p^d, H>0, and a,j,n,r∈ N. These are the hypotheses of the two corrected matrix statements used by the retained proof. Without integral observation vectors or positive L, the preceding counterexamples invalidate the claimed precision. Assume C^H=I+p^LB. Put H_a=Hp^a. Then C^H_a=I+p^L+aB_a for an integral matrix B_a. Every refined packet b_a,j(x)=a_H_ax+j has Mahler coefficients satisfying v_p(c_a,j,r)≥(L+a)r. Equivalently,…

    Exact mathematical passage
    
    C^{H_a}=I+p^{L+a}B_a
    
C^{Hx+j}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1702

    Use C^Hx+j=C^j(I+p^LB)^x and the matrix binomial series. One difference in the coarse variable multiplies the series by p^LB.

    Exact mathematical passage
    
    C^{Hx+j}=C^j(I+p^LB)^x
    
C^jUse in source · 6 passages

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  • Use in source · Read passage · 10_foundations.tex, line 412

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

    Exact mathematical passage
    
    b_k=C^j(C^N)^kx_0,\qquad b_{k+1}=C^Nb_k.
    
C^nUse in source · 10 passages

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  • Use in source · Read passage · 10_foundations.tex, line 411

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

    Exact mathematical passage
    x_n=C^n x_0
C^{n+1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 121

    An independent Lucas sequence. Define L_0=2, L_1=1, L_n+2=L_n+L_n+1. Direct finite matrix arithmetic proves C^2=C+I. Multiplication by C^n gives C^n+2=C^n+C^n+1. Taking traces shows that tr(C^n) satisfies the same recurrence and initial values as L_n. Induction yields tr(C^N)=L_N. The implementation separately proves det C=-1, det(C^N)=(-1)^N, L_N=2F_N+1-F_N. No Binet roots or diagonalizability argument is needed.

    Exact mathematical passage
    C^{n+2}=C^n+C^{n+1}
C^{n+2}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 121

    An independent Lucas sequence. Define L_0=2, L_1=1, L_n+2=L_n+L_n+1. Direct finite matrix arithmetic proves C^2=C+I. Multiplication by C^n gives C^n+2=C^n+C^n+1. Taking traces shows that tr(C^n) satisfies the same recurrence and initial values as L_n. Induction yields tr(C^N)=L_N. The implementation separately proves det C=-1, det(C^N)=(-1)^N, L_N=2F_N+1-F_N. No Binet roots or diagonalizability argument is needed.

    Exact mathematical passage
    C^{n+2}=C^n+C^{n+1}
C^{Nk+j}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 31

    pageheading1. Actual matrix orbits and arithmetic jumpsTheorem 9.1; manuscript page 22 Let R be a commutative ring, I a finite coordinate type and d=|I|. For C∈ M_I(R) and v_0∈ R^I, define orbit(C,v_0,n)=C^nv_0, u_k=packet(C,v_0,N,j,k)=C^Nk+jv_0. The implementation uses actual matrix powers and matrix--vector multiplication. A separately supplied sequence v with v_n+1=Cv_n is proved by induction to equal this power orbit. Thus the source-sequence formulation is connected to the construction.

    Exact mathematical passage
    
     \operatorname{orbit}(C,v_0,n)=C^nv_0,
     \qquad u_k=\operatorname{packet}(C,v_0,N,j,k)=C^{Nk+j}v_0.
    
C_0Use in source · 12 passages

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  • Use in source · Read passage · strengthening.tex, line 323

    Put K= Q_2(i), R= Z_2[i], a=(1+i)/2, and b=(1-i)/2. The exact identity F_0(t)=frac2cosh t+cos t =sech(at)sech(bt) gives, for U=a E+b F, L(U^4n)=C_0(n). Although a and b have negative valuation, their denominators disappear from the following exact polynomial identities: U^2 · = E F+i(P-Q),nonumber

    Exact mathematical passage
    \mathcal L(U^{4n})=C_0(n)
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  • Use in source · Read passage · 10_foundations.tex, line 266

    Δ_h^s f(n) · =∑_j=0^s(-1)^s-jbinom sj f(n+jh). The finite interpolant I_Nf(n)=∑_r=0^Nc_r(f)binom nr agrees with f for n≤ N. It has the same order-s unit-step difference at n whenever n+s≤ N. For arbitrary coefficients c_0,…,c_N, the corresponding binomial interpolant has exactly those coefficients and zero coefficients above N.

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    c_0,\ldots,c_N
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  • Use in source · Read passage · strengthening.tex, line 468

    the elementary Bernoulli bound shows v_2(d_i)≥2i. Hurwitz inversion therefore gives D_1(t)^-1=∑_i≥0h_ifract^4i(4i)!, h_0=1, v_2(h_i)≥2i. This follows by induction from the inverse recurrence: for packet index i, every product d_rh_i-r has valuation at least 2i, and its binomial multiplier is integral. Because F_1=F_0D_1^-1, C_1(n)-C_0(n) =∑_i≥1h_ibinom4n4iC_0(n-i). At a natural n this is a finite sum, with i≤ n.

    Exact mathematical passage
    
     C_1(n)-C_0(n)
         =\sum_{i\geq1}h_i\binom{4n}{4i}C_0(n-i).
     \label{eq:address-one-exact-transfer}
    
c_1Use in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 656

    [Exact four-packet displacement law] Set (c_0,c_1,c_2,c_3)=(1,1,4,2), (d_0,d_1,d_2,d_3)=(1,1,5,3). For 0≤ j<4, n≥0, and every positive integer T, C_j(n+T)-C_j(n) · ≠0, · v_2(C_j(n+T)-C_j(n)) · =v_2(T)+c_j,

    Exact mathematical passage
    
     (c_0,c_1,c_2,c_3)=(1,1,4,2),\qquad
     (d_0,d_1,d_2,d_3)=(1,1,5,3).
    
C_2Use in source · 15 passages

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  • Use in source · Read passage · strengthening.tex, line 267

    egintheorem[Four-address factorial bound] For j∈{0,1,2,3}, n≥0, k≥1, and positive mixed steps, v_2(Δ_H_1⋯Δ_H_kC_j(n)) ≥ k+v_2(k!)+∑_ν=1^k v_2(H_ν)+β_j, (β_0,β_1,β_2,β_3)=(0,0,3,1). Address two also satisfies v_2(Δ_H_1⋯Δ_H_kC_2(n)) ≥ 2k+v_2(k!)+lfloorfrack2rfloor +∑_ν=1^k v_2(H_ν). For the original coefficients, the offsets in [source reference] are (0,0,4,2). No positive-order offset is asserted for the empty difference.

    Exact mathematical passage
    
     v_2(\Delta_{H_1}\cdots\Delta_{H_k}C_2(n))
     \geq 2k+v_2(k!)+\left\lfloor\frac{k}{2}\right\rfloor
                 +\sum_{\nu=1}^k v_2(H_\nu).
     \label{eq:factorial-four-j2-filtration}
    
c_2Use in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 656

    [Exact four-packet displacement law] Set (c_0,c_1,c_2,c_3)=(1,1,4,2), (d_0,d_1,d_2,d_3)=(1,1,5,3). For 0≤ j<4, n≥0, and every positive integer T, C_j(n+T)-C_j(n) · ≠0, · v_2(C_j(n+T)-C_j(n)) · =v_2(T)+c_j,

    Exact mathematical passage
    
     (c_0,c_1,c_2,c_3)=(1,1,4,2),\qquad
     (d_0,d_1,d_2,d_3)=(1,1,5,3).
    
C_{2,0}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 668

    [Euler and Bernoulli boundaries] For (N,j)=(2,0), C_2,0(t)=secht, so C_2,0(n)=E_2n. For (N,j)=(2,1), C_2,1(t)=fractsinh t, and C_2,1(n) = 2(1-2^2n-1)B_2n. The usual convention B_1=-1/2 is used. Euler and Bernoulli number conventions are reviewed in [source reference].

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    C_{2,0}(n)=E_{2n}.
    
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  • Use in source · Read passage · 20_arithmetic.tex, line 85

    [The Bernoulli boundary address] At N=2, the address j=1 gives a second classical family. Define the Bernoulli numbers by B(z)=fracze^z-1=∑_r≥0B_rfracz^rr!, B_0=1, B_1=-frac12. All quotients here are regular formal series. The identity 2B(z)-B(2z)=fraczsinh z =∑_n≥0C_2,1(n)fracz^2n(2n)! follows by putting the two fractions on a common denominator and using [source reference]. Comparing even coefficients yields C_2,1(n)=(2-2^2n)B_2n (n∈ N). This classical derivation restores the boundary example …

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     2B(z)-B(2z)=\frac{z}{\sinh z}
                =\sum_{n\ge0}C_{2,1}(n)\frac{z^{2n}}{(2n)!}
    
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  • Use in source · Read passage · strengthening.tex, line 740

    is one, only degrees divisible by four survive, and the Bernoulli expansion gives D_3(t)=∑_i≥0d_ifract^4i(4i)!, d_i=3(-frac14)^i frac2^4i+2B_4i+2(4i+2)(4i+1). Again v_2(d_i)≥2i, and the inverse coefficients h_i have the same bound. Since F_3=F_2D_3^-1, C_3(n)-C_2(n)=∑_i≥1h_ibinom4n4iC_2(n-i). The scaled-binomial argument [source reference] applies unchanged. The basic bound for C_2 includes order zero; its continuous extension is available from the threshold-free theorem. Thus the shifted sums c…

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     C_3(n)-C_2(n)=\sum_{i\geq1}h_i\binom{4n}{4i}C_2(n-i).
     \label{eq:address-three-exact-transfer}
    
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  • Use in source · Read passage · 20_arithmetic.tex, line 656

    [Exact four-packet displacement law] Set (c_0,c_1,c_2,c_3)=(1,1,4,2), (d_0,d_1,d_2,d_3)=(1,1,5,3). For 0≤ j<4, n≥0, and every positive integer T, C_j(n+T)-C_j(n) · ≠0, · v_2(C_j(n+T)-C_j(n)) · =v_2(T)+c_j,

    Exact mathematical passage
    
     (c_0,c_1,c_2,c_3)=(1,1,4,2),\qquad
     (d_0,d_1,d_2,d_3)=(1,1,5,3).
    
C_{3,0}Use in source · 1 passage

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  • Use in source · worked_atlas.tex, line 46

    At N=p=3 and j=0, put A(n)=(-1)^nC_3,0(n). The first three rows are AtlasImageodd_prime_sign The ordinary first difference at row zero is -1-1=-2. It is a 3-adic unit, so an asserted lower bound of one for that difference would be false. The twisted first difference is zero, whose mathematical valuation is +∞. At the next row, the twisted difference is 19-1=18, of valuation two.

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    A(n)=(-1)^nC_{3,0}(n)
C_{4,0}Use in source · 5 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 870

    Return now to the classical signed Euler numbers secht=frac1cosh t =∑_n≥0E_2nfract^2n(2n)!. Thus E_2n=E_2n^(2,0) in the convention [source reference]. In particular E_0=1, E_2=-1, and E_4=5. The following construction concerns these N=2 coefficients, not C_4,0.

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    C_{4,0}
C_{4,1}Use in source · 6 passages

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  • Use in source · Read passage · supplement_11.tex, line 137

    An exact arithmetic check, not an optimality claim. The independent rational recurrence gives C_4,1(1)=-1/5 and C_4,1(2)=121/45. Their difference is 26/9, of 2-adic valuation one. This agrees with the new bound at n=k=u=1. The source also contains universal instantiations of the degree-indexed theorem at all four dyadic addresses and at an odd singular address.

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    C_{4,1}(1)=-1/5
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  • Use in source · Read passage · 20_arithmetic.tex, line 454

    For an odd prime, 2mid p-1, so the option m=2 introduces no further case. The sole remaining branch is p=2,m=2,N=4. The all-row theorem proved next gives 2^kmidΔ_1^kC_4,j(n) at every row and address. After multiplying by j!, Lemma~[source reference] gives packet period 2^r-1 and degree displacement 4·2^r-1=2^r+1=qp^r. It also gives integrality. Hence the complete statement follows. The formal degree conversion is leanKomorebiPaperIX.NishibuchiCompletion.original_degree_shift; the all-parameter d…

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    2^k\mid\Delta_1^kC_{4,j}(n)
C_{6,0}Use in source · 1 passage

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  • Use in source · Read passage · supplement_07.tex, line 142

    No automatic extra power. The original rational triangular coefficients satisfy C_6,0(1)=-1, C_6,0(2)=923, v_3(C_6,0(2)-C_6,0(1))=v_3(924)=1. A period modulo 3 therefore cannot simply be reported as a period modulo 9. This is consistent with the global theorem's proposed first-difference bound; it is not a counterexample to Theorem 7.5.

    Exact mathematical passage
    
     C_{6,0}(1)=-1,\quad C_{6,0}(2)=923,\quad
     v_3\bigl(C_{6,0}(2)-C_{6,0}(1)\bigr)=v_3(924)=1.
    
C_{6,2}Use in source · 2 passages

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  • Use in source · Read passage · supplement_08.tex, line 184

    pageheading5. Examples, verification and the remaining gapThe coverage count is unchanged; the unresolved theorem now has a stronger proved part The original rational recurrence gives C_6,2(1)=-frac128, C_6,2(2)=frac425364, C_6,2(3)=-frac20397592660. Consequently, the new source proves Δ_1^2 C_6,2(1)=-frac9499591235, v_3 (Δ_1^2 C_6,2(1))=2. This attains the bound min(2,2·1)=2. A claim of a third power is rejected. The separate cubic example proves that the normalized differential defect is not a…

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     C_{6,2}(1)=-\frac1{28},\quad C_{6,2}(2)=\frac{425}{364},\quad
     C_{6,2}(3)=-\frac{2039759}{2660}.
    
C_{6,3}Use in source · 12 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 357

    [Why the factorial loss is necessary] Take p=3, m=2, N=6, and j=3. The recurrence gives C_6,3(1)=-frac184, C_6,3(2)=frac7035460, C_6,3(2)-C_6,3(1)=frac64455. The last difference has 3-adic valuation zero. Here h=1, so at n=1 the order-one bound without factorial loss would require valuation at least one. The corrected normalized bound is sharp, since v_3(C_6,3(2)-C_6,3(1))=0=1-v_3(3!). Multiplication by 3! gives U_6,3(2)-U_6,3(1)=frac384455, v_3(U_6,3(2)-U_6,3(1))=1, consistent with the unnormal…

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     C_{6,3}(1)=-\frac1{84},\qquad
     C_{6,3}(2)=\frac{703}{5460},\qquad
     C_{6,3}(2)-C_{6,3}(1)=\frac{64}{455}.
    
C_{6,4}Use in source · 1 passage

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  • Use in source · Read passage · supplement_10.tex, line 202

    w For (p,m,j)=(3,2,3), the source derives C_6,3(1)=-frac184, C_6,3(2)=frac7035460, C_6,3(3)=-frac2231997980. Consequently, Δ_1^2(3!C_6,3)(1)=-frac14647688645, v_3 (-frac14647688645)=2. This attains min(2,2·1)=2. At the regular address j=4 beyond the singular boundary, the source likewise proves Δ_1^2(4!C_6,4)(1)=-frac1027455122377375, and applies the universal exponent-two bound to this actual coefficient function.

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     \Delta_1^2\bigl(4!C_{6,4}\bigr)(1)=-\frac{102745512}{2377375},
    
c_{a,j,r}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2397

    ypotheses of the two corrected matrix statements used by the retained proof. Without integral observation vectors or positive L, the preceding counterexamples invalidate the claimed precision. Assume C^H=I+p^LB. Put H_a=Hp^a. Then C^H_a=I+p^L+aB_a for an integral matrix B_a. Every refined packet b_a,j(x)=a_H_ax+j has Mahler coefficients satisfying v_p(c_a,j,r)≥(L+a)r. Equivalently, Δ_H_a^ ra_n ≡0 pmod p^(L+a)r.

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    \boxed{\vp(c_{a,j,r})\geq(L+a)r.}
    
c_iUse in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 75

    [Translated product formula] For N=p^a, a≥1, and λ,μ∈ Z_p, L_λ+μ(AB) =(L_λ A)B+A(L_μ B) +∑_i=1^N-1c_i(D^iA)(D^N-iB), c_i=frac1pbinom Ni∈ Z. Flatness types add under multiplication.

    Exact mathematical passage
    
     L_{\lambda+\mu}(AB)
      =(L_\lambda A)B+A(L_\mu B)
           +\sum_{i=1}^{N-1}c_i(D^iA)(D^{N-i}B),
     \qquad c_i=\frac1p\binom Ni\in\mathbb Z.
     \label{eq:translated-product}
    
c_i^{e_i}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 101

    e with unspecified remainders. On the tensor of the two factors use the mutually commuting operators L_λ⊗1, 1⊗ L_μ, c_iD^i⊗ D^N-i (1≤ i<N). Applying multiplication after the kth power of their sum gives exactly L_λ+μ^k(AB). Expanding it yields terms indexed by a,b,e_1,…,e_N-1≥0, with a+b+∑_i e_i=k: frack!a!b!∏_i e_i!∏_i c_i^e_i (D^α L_λ^a A)(D^β L_μ^b B), α=∑_i i e_i, β=∑_i(N-i)e_i. Every scalar is integral. If A and B have the stated flatness types, both factors in every term belong to H_p, as …

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     \frac{k!}{a!b!\prod_i e_i!}\prod_i c_i^{e_i}
     (D^\alpha L_\lambda^a A)(D^\beta L_\mu^b B),
     \quad
     \alpha=\sum_i i e_i,\quad
     \beta=\sum_i(N-i)e_i.
     \label{eq:translated-multinomial}
    
C_jUse in source · 16 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 467

    For the remainder of the four-packet discussion abbreviate C_j=C_4,j and U_j=U_4,j. Removing the original row cap requires an unshifted coefficient operator.

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    C_j=C_{4,j}
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  • Use in source · Read passage · 20_arithmetic.tex, line 661

    [Exact four-packet displacement law] Set (c_0,c_1,c_2,c_3)=(1,1,4,2), (d_0,d_1,d_2,d_3)=(1,1,5,3). For 0≤ j<4, n≥0, and every positive integer T, C_j(n+T)-C_j(n) · ≠0, · v_2(C_j(n+T)-C_j(n)) · =v_2(T)+c_j, U_j(n+T)-U_j(n) · ≠0, · v_2(U_j(n+T)-U_j(n)) · =v_2(T)+d_j.

    Exact mathematical passage
    
     C_j(n+T)-C_j(n)&\ne0,&
     v_2\bigl(C_j(n+T)-C_j(n)\bigr)&=v_2(T)+c_j,\\
     U_j(n+T)-U_j(n)&\ne0,&
     v_2\bigl(U_j(n+T)-U_j(n)\bigr)&=v_2(T)+d_j.
    
C_{J,K}Use in source · 2 passages

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  • Use in source · Read passage · 50_open_questions.tex, line 107

    The linear compatibility theorem provides one concrete part of this program. If two local observations J:U→ V and K:U→ W are linear maps from the same correction module, define C_J,K=U/(ker J+ker K). The attainable observations im J and im K have canonical labels in this quotient. Equality of labels is equivalent to simultaneous realization by a single correction. The joint image is exactly the fiber product over C_J,K, with the proved linear universal property. The declarations are begingroupsm…

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    C_{J,K}=U/(\ker J+\ker K).
    
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  • Use in source · Read passage · 72_recovered.tex, line 1635

    [Polynomial exponential Mahler transfer] Let f(n)=P(n)u^n with P∈ Z_p[n] of degree d. Let H=Λ_p,L, fix an address j, and write Q_j(x)=P(Hx+j), y=u^H, q=y-1. Then f(Hx+j)=∑_r≥0c_j,rbinom xr, where c_j,r = u^j ∑_k=0^min(d,r) binom rk Δ^kQ_j(0)y^kq^r-k. Consequently, v_p(c_j,r)≥ Lmax(r-d,0). For R≥ d, truncation at depth R is valid modulo p^L(R+1-d) uniformly on Z_p.

    Exact mathematical passage
    
    \boxed{
    f(Hx+j)=\sum_{r\geq0}c_{j,r}\binom xr,
    }
    
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  • Use in source · Read passage · 10_foundations.tex, line 509

    For M≥1 and s,r,k∈ N, define the affine Mahler kernel by Θ_M,s(r,k)=[z^r](1+z)^s((1+z)^M-1)^k =∑_j=0^k(-1)^k-jbinom kjbinomMj+sr. It vanishes unless k≤ r≤ Mk+s. For a finite binomial expansion f(n)=∑_r=0^Dc_rbinom nr, the kth Newton coefficient of g(q)=f(Mq+s) is therefore c_k(g)=∑_r=k^min(D,Mk+s)c_rΘ_M,s(r,k). Indeed, insert the finite stencil for Δ_1^k g(0) and interchange its two finite sums. Coefficient extraction of the binomial expansion of ((1+z)^M-1)^k gives the kernel. In the binary zer…

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    c_k(g)=\sum_{r=k}^{\min(D,Mk+s)}c_r\Theta_{M,s}(r,k).
    
C_{mp,j}Use in source · 17 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 134

    [Lucas reduction and address independence] Let p be prime, m>0, and N=mp. For 0≤ j<p and n,q∈ N, binommpn+jmpq≡binommnmqpmod p. Every C_mp,j(n) belongs to Z_p. Its reduction is the unique sequence overline C_m: N→ F_p satisfying ∑_q=0^nbinommnmqoverline C_m(q)=δ_n0 in F_p. Consequently, for any two addresses 0≤ j,k<p, C_mp,j(n)≡ C_mp,k(n)pmod p (n∈ N).

    Exact mathematical passage
    C_{mp,j}(n)
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  • Use in source · Read passage · 20_arithmetic.tex, line 141

    prop:arith-lucas-window Let p be prime, m>0, and N=mp. For 0≤ j<p and n,q∈ N, binommpn+jmpq≡binommnmqpmod p. Every C_mp,j(n) belongs to Z_p. Its reduction is the unique sequence overline C_m: N→ F_p satisfying ∑_q=0^nbinommnmqoverline C_m(q)=δ_n0 in F_p. Consequently, for any two addresses 0≤ j,k<p, C_mp,j(n)≡ C_mp,k(n)pmod p (n∈ N).

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     C_{mp,j}(n)\equiv C_{mp,k}(n)\pmod p\qquad(n\in\N).
    
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  • Use in source · Read passage · 10_foundations.tex, line 266

    Δ_h^s f(n) · =∑_j=0^s(-1)^s-jbinom sj f(n+jh). The finite interpolant I_Nf(n)=∑_r=0^Nc_r(f)binom nr agrees with f for n≤ N. It has the same order-s unit-step difference at n whenever n+s≤ N. For arbitrary coefficients c_0,…,c_N, the corresponding binomial interpolant has exactly those coefficients and zero coefficients above N.

    Exact mathematical passage
    c_0,\ldots,c_N
C_{N,0}Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 242

    [Necessary boundaries] At an odd prime the ordinary step-one difference cannot replace the sign-twisted operator in the general theorem. For j=0, C_N,0(0)=1 and C_N,0(1)=-1, so their ordinary difference is -2, a unit at every odd prime. The proof also does not extend unchanged to non-prime-power N, whose interior binomial coefficients need not be divisible by the chosen prime. Neither observation is a counterexample to Nishibuchi's original eventual congruence.

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    C_{N,0}(0)=1
C_pUse in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 305

    fferentiation of the section identities gives tR_c'=R_c(1+R_c-1-R_c). Writing q=R_c-1-(c-1) and y=R_c-c, this becomes (tD+c)y=cq+yq-y^2. On degrees divisible by p, the inverse of tD+c divides the degree-d coefficient by the unit d+c. Starting with y_0=0, form y_s+1=(tD+c)^-1(cq+y_sq-y_s^2). Flatness and support are preserved. Indeed, put E_c=tD+c and C_p=pD^p^2-D^p. The commutator is [B_p,E_c]=C_p, and iteration gives B_p^rE_cy=E_cB_p^ry+rC_pB_p^r-1y. For the inverse equation E_cy=f, induction o…

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    C_p=pD^{p^2}-D^p
c_{p^ak}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2306

    (q)=q has d_1=1; thus d_1not≡ c_2pmod2. The coefficient consequence is read with integral original coefficients. Let M=p^a. Then Θ_p^a,s(r,k) ≡ binomsr-p^ak pmod p, where the binomial coefficient is zero outside its natural range. Consequently, d_s,k ≡ ∑_u=0^s binom su c_p^ak+u pmod p. In particular, d_0,k≡ c_p^akpmod p.

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    \boxed{d_{0,k}\equiv c_{p^ak}\pmod p.}
    
c_{p^ak+u}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2295

    insufficient. For p=2, M=2, s=0 and f(x)=x/2, one has c_1=1/2, c_2=0, while g(q)=q has d_1=1; thus d_1not≡ c_2pmod2. The coefficient consequence is read with integral original coefficients. Let M=p^a. Then Θ_p^a,s(r,k) ≡ binomsr-p^ak pmod p, where the binomial coefficient is zero outside its natural range. Consequently, d_s,k ≡ ∑_u=0^s binom su c_p^ak+u pmod p. In particular, d_0,k≡ c_p^akpmod p.

    Exact mathematical passage
    
    \boxed{
    d_{s,k}
    \equiv
    \sum_{u=0}^{s}
    \binom su
    c_{p^ak+u}
    \pmod p.
    }
    
C_RUse in source · 4 passages

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  • Use in source · Read passage · 71_native.tex, line 248

    for maps from A and B to a common set if and only if R is rectangular. In that case a canonical finite choice of common set is the connected-component set C_R of the bipartite graph with vertex set Asqcup B and edges given by R.

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    C_R
c_rUse in source · 40 passages

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  • Use in source · Read passage · 10_foundations.tex, line 215

    [Filtered finite reconstruction] For every n,L∈ N and sequences c,S in that ring, · ∑_r=0^nbinom nr c_rp^[w(r)]S_n-r · ≡ ∑_substack0≤ r≤ n w(r)<L binom nr c_rp^[w(r)]S_n-rpmodp^L. Here congruence means divisibility of the difference by p^L in the coefficient ring. In Z_p, it means membership in p^L Z_p.

    Exact mathematical passage
    
    \begin{split}
    &\sum_{r=0}^{n}\binom nr c_rp^{[w(r)]}S_{n-r}\\
    &\hspace{8mm}\equiv
    \sum_{\substack{0\leq r\leq n\\w(r)<L}}
    \binom nr c_rp^{[w(r)]}S_{n-r}\pmod{p^L}.
    \end{split}
    
c_sUse in source · 1 passage

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  • Use in source · Read passage · supplement_04.tex, line 54

    An independently defined interpolant. For a finite coefficient list, define I_N(c)(n)=∑_r=0^Nc_rbinom nr. The source proves exact coefficient extraction: Δ_1^s I_N(c)(0)= c_s, · s≤ N, 0, · s>N. It follows that I_N(c) has the stated shifted-coefficient formula at every order, with zero coefficients beyond N. Every difference of order greater than N vanishes. Its coefficients are uniquely determined by its values at 0,…,N.

    Exact mathematical passage
    
     \Delta_1^s I_N(c)(0)=
     \begin{cases}c_s,&s\leq N,\\0,&s>N.\end{cases}
    
c_{s+j}Use in source · 5 passages

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  • Use in source · Read passage · 10_foundations.tex, line 255

    [Finite Newton--Mahler identities] For every n,s,h∈ N, f(n) · =∑_r=0^nc_r(f)binom nr, c_r(Δ_1^s f) · =c_s+r(f), Δ_1^s f(n) · =∑_j=0^nc_s+j(f)binom nj, Δ_h^s f(n) · =∑_j=0^s(-1)^s-jbinom sj f(n+jh). The finite interpolant I_Nf(n)=∑_r=0^Nc_r(f)binom nr agrees with f for n≤ N. It has the same order-s unit-step difference at n whenever n+s≤ N. For arbitrary coefficients c_0,…,c_N, the corresponding binomial interpolant has exactly those coefficients and zero coefficients above N.

    Exact mathematical passage
    
    \begin{aligned}
    f(n)&=\sum_{r=0}^{n}c_r(f)\binom nr,\\
    c_r(\Delta_1^s f)&=c_{s+r}(f),\\
    \Delta_1^s f(n)&=\sum_{j=0}^{n}c_{s+j}(f)\binom nj,\\
    \Delta_h^s f(n)&=\sum_{j=0}^{s}(-1)^{s-j}\binom sj f(n+jh).
    \end{aligned}
    
c_{s+r}Use in source · 1 passage

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  • Use in source · Read passage · 10_foundations.tex, line 255

    [Finite Newton--Mahler identities] For every n,s,h∈ N, f(n) · =∑_r=0^nc_r(f)binom nr, c_r(Δ_1^s f) · =c_s+r(f), Δ_1^s f(n) · =∑_j=0^nc_s+j(f)binom nj, Δ_h^s f(n) · =∑_j=0^s(-1)^s-jbinom sj f(n+jh). The finite interpolant I_Nf(n)=∑_r=0^Nc_r(f)binom nr agrees with f for n≤ N. It has the same order-s unit-step difference at n whenever n+s≤ N. For arbitrary coefficients c_0,…,c_N, the corresponding binomial interpolant has exactly those coefficients and zero coefficients above N.

    Exact mathematical passage
    
    \begin{aligned}
    f(n)&=\sum_{r=0}^{n}c_r(f)\binom nr,\\
    c_r(\Delta_1^s f)&=c_{s+r}(f),\\
    \Delta_1^s f(n)&=\sum_{j=0}^{n}c_{s+j}(f)\binom nj,\\
    \Delta_h^s f(n)&=\sum_{j=0}^{s}(-1)^{s-j}\binom sj f(n+jh).
    \end{aligned}
    
c_uUse in source · 8 passages

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  • Use in source · Read passage · 72_recovered.tex, line 192

    The product law follows from [source reference]. To prove injectivity, suppose ∑_u∈ F_pc_uϵ_u=0. Looking at the coordinates n=0,1,…,p-1 gives a Vandermonde system with matrix (u^n)_0≤ n≤ p-1, u∈ F_p. Its determinant is the product of all differences between distinct field elements, so it is nonzero. Hence every c_u vanishes.

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    \sum_{u\in\FF_p}c_u\epsilon_u=0.
    
\mathcal C_xUse in source · 4 passages

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  • Use in source · Read passage · continuation_01.tex, line 104

    Let G∈ Z[X_1,…,X_d]^f be a second fixed system with the same variables and the same lower root x. Let K_x=ρ_q∘ DG(x), and define C_x= Z^d/(ker J_x+ker K_x). This quotient is finite: it is a quotient of the finite image of J_x, or equivalently of the finite image of K_x.

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    \mathcal C_x=\mathbb Z^d/(\ker J_x+\ker K_x).
    
c_{\eta u}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 334

    Write b^(m,p) = ∑_u∈ F_pc_uϵ_u. Invariance gives c_η u=c_u. After the change of variable u↦η u, b_n^(m,p) = η^n b_n^(m,p) for every η∈ H_m. If mnmid n, some η satisfies η^n≠1, so the coefficient vanishes.

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    c_{\eta u}=c_u
c_\chiUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 843

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

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    d_q=\sum_{\chi\in\mu_h}c_\chi\chi^q.
    
\CcalUse in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 586

    [Congruential reciprocal packet] Define C_N,j(t) = fract^j/j! E_N,j(t) = ∑_n≥0 C_N,j(n) fract^Nn(Nn)!.

    Exact mathematical passage
    
    \boxed{
    \Ccal_{N,j}(t)
    =
    \frac{t^j/j!}{\Ecal_{N,j}(t)}
    =
    \sum_{n\geq0}
    C_{N,j}(n)
    \frac{t^{Nn}}{(Nn)!}.
    }
    
\Ccal_{2,0}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 664

    [Euler and Bernoulli boundaries] For (N,j)=(2,0), C_2,0(t)=secht, so C_2,0(n)=E_2n. For (N,j)=(2,1), C_2,1(t)=fractsinh t, and C_2,1(n) = 2(1-2^2n-1)B_2n. The usual convention B_1=-1/2 is used. Euler and Bernoulli number conventions are reviewed in [source reference].

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    \Ccal_{2,0}(t)=\operatorname{sech}t,
    
\Ccal_{2,1}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 672

    [Euler and Bernoulli boundaries] For (N,j)=(2,0), C_2,0(t)=secht, so C_2,0(n)=E_2n. For (N,j)=(2,1), C_2,1(t)=fractsinh t, and C_2,1(n) = 2(1-2^2n-1)B_2n. The usual convention B_1=-1/2 is used. Euler and Bernoulli number conventions are reviewed in [source reference].

    Exact mathematical passage
    
    \Ccal_{2,1}(t)=\frac{t}{\sinh t},
    
\Ccal_{N,j}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 586

    [Congruential reciprocal packet] Define C_N,j(t) = fract^j/j! E_N,j(t) = ∑_n≥0 C_N,j(n) fract^Nn(Nn)!.

    Exact mathematical passage
    
    \boxed{
    \Ccal_{N,j}(t)
    =
    \frac{t^j/j!}{\Ecal_{N,j}(t)}
    =
    \sum_{n\geq0}
    C_{N,j}(n)
    \frac{t^{Nn}}{(Nn)!}.
    }
    
\operatorname{coe}Use in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 96

    Identity with the original rational coefficients. The integral lift is constructed from the decoded coefficients of the actual reciprocal. The code proves coe(e_j(n))=C_4,j(n) in Q_2, and proves that the coefficient map commutes with every difference stencil. Multiplying by j! yields the unnormalized family. The valuation result is not divisibility in a field.

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     \operatorname{coe}(e_j(n))=C_{4,j}(n)\quad\text{in }\Q_2,
    
\coefUse in source · 6 passages

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  • Use in source · Read passage · supplement_09.tex, line 92

    mate becomes relevant only after connecting it to the manuscript's generating functions. For a residue r∈ Z/N Z, define H_r(t)=∑_d≥01_d≡ r (N)fract^dd!, F_r(T)=H_r(log(1+T)). The implementation uses the actual formal logarithm and substitution in Q[[T]]. It proves H_r'=H_r-1, (1+T)F_r'=F_r-1, and hence the coefficient recurrence (d+1)coefTd+1F_r =coefTdF_r-1-dcoefTdF_r. The initial values are F_r(0)=1_r=0.

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     (d+1)\coef{T}{d+1}F_r
     =\coef{T}{d}F_{r-1}-d\coef{T}{d}F_r.
    
\cokerUse in source · 4 passages

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  • Use in source · Read passage · 50_open_questions.tex, line 53

    The original conjecture asks for a canonical finite obstruction packet whose vanishing is necessary and sufficient for a one-level lift in a finitely presented arithmetic seed system. For a fixed finite polynomial presentation, the paper now supplies the precise obstruction o_F,bar x=[r]∈ coker (J_F,bar x: F_p^d⟶ F_p^e), where r is the residual digit. We have proved independence from the chosen integer representative. Invariance under a change of equations or presentation requires more.

    Exact mathematical passage
    
    o_{F,\bar x}=[r]\in
    \coker\!\left(J_{F,\bar x}:\F_p^d\longrightarrow\F_p^e\right),
    
\operatorname{coker}Use in source · 1 passage

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  • Use in source · Read passage · 71_native.tex, line 134

    coker(Hom_B(Ω_P/ Z⊗_P B, F_p)⟶Hom_B(I/I^2, F_p)).

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    \operatorname{coker}\left(\operatorname{Hom}_B(\Omega_{P/\mathbb Z}\otimes_P B,\mathbb F_p)\longrightarrow\operatorname{Hom}_B(I/I^2,\mathbb F_p)\right).
\cosUse in source · 6 passages

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  • Use in source · Read passage · supplement_02.tex, line 97

    Connecting to the paper's Z_n convention. The paper defines Z_n through sec t+tan t. The Lean construction uses Mathlib's actual formal sine and cosine series and defines sec t=(cos t)^-1, tan t=(sin t)sec t, Z_n=n![t^n](sec t+tan t). It proves the sine and cosine derivative identities, sin^2t+cos^2t=1, and (tan t)'=sec^2t. Sign substitution in the squared variable carries the hyperbolic cosine kernel to the actual cosine kernel. The same homomorphism carries their inverses and their squares. Ex…

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     \sec t=(\cos t)^{-1},\quad \tan t=(\sin t)\sec t,
     \quad Z_n=n![t^n](\sec t+\tan t).
    
\cos^2Use in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 101

    Connecting to the paper's Z_n convention. The paper defines Z_n through sec t+tan t. The Lean construction uses Mathlib's actual formal sine and cosine series and defines sec t=(cos t)^-1, tan t=(sin t)sec t, Z_n=n![t^n](sec t+tan t). It proves the sine and cosine derivative identities, sin^2t+cos^2t=1, and (tan t)'=sec^2t. Sign substitution in the squared variable carries the hyperbolic cosine kernel to the actual cosine kernel. The same homomorphism carries their inverses and their squares. Ex…

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    \sin^2t+\cos^2t=1
\cothUse in source · 5 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 960

    Let Q(z)=zcoth z, the regular formal series with constant coefficient one. The elementary differential identity (frac zsinh z)^2=Q(z)-zQ'(z) and the Bernoulli expansion Q(z)=1+∑_m≥1frac2^2mB_2m(2m)!z^2m give β_m=-2^2m(2m-1)B_2m/(2m)!. Multiplication by 2^2m(2m)! proves the identity for κ_m. The Bernoulli denominator theorem gives v_2(B_2m)=-1 for m>0: its denominator contains each prime q with (q-1)mid 2m exactly once, including 2[source reference]. Since 2m-1 is odd, the valuation is exactly 4m…

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    Q(z)=z\coth z
\mathcal DUse in source · 16 passages

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  • Use in source · Read passage · 10_foundations.tex, line 78

    Over a characteristic-zero field K, put E(a)=∑_n≥0fraca_nn!t^n, D(A)_n=n![t^n]A. These inverse maps between sequences and K[[t]] provide formal exponential coordinates. No analytic convergence is needed for the coefficient identities below.

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    \mathcal E(a)=\sum_{n\geq0}\frac{a_n}{n!}t^n,
    \qquad \mathcal D(A)_n=n![t^n]A.
    
D^2Use in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 76

    3. The dyadic differential operator and reciprocity. The operator required here is B_4=fracD^8-D^42, D=fracddt. It is not the preceding p=2 specialization (D^4-D^2)/2. It is also not an ordinary derivation. The new algebra module proves the actual product formula, including the middle derivative sums whose coefficients are binom 8i/2 and binom 4i/2.

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    (D^4-D^2)/2
\mathcal D^2Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 626

    The variable B=2Z+1 has transform e^tfrac2te^2t-1=fractsinh t. Its odd moments vanish, and P=( B^2-1)/4=Z(Z+1) has the divided-power moments of Lemma~[source reference]. Introduce an independent copy D and Q=( D^2-1)/4. The exact factorization F_2(t)=fract^2cosh t-cos t =fracatsinh(at)fracbtsinh(bt) shows that U=a B+b D satisfies L(U^4n)=C_2(n). Direct expansion gives H_2:=fracU^4-14 =P+Q-P^2-Q^2+6PQ+i B D(P-Q).

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    Q=(\mathcal D^2-1)/4
D^4Use in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 512

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

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     A_c=\frac{D^4-c}{2}.
    
D^8Use in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 73

    3. The dyadic differential operator and reciprocity. The operator required here is B_4=fracD^8-D^42, D=fracddt. It is not the preceding p=2 specialization (D^4-D^2)/2. It is also not an ordinary derivation. The new algebra module proves the actual product formula, including the middle derivative sums whose coefficients are binom 8i/2 and binom 4i/2.

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    \boxed{B_4=\frac{D^8-D^4}{2},\qquad D=\frac{d}{dt}.}
    
D^{d-r}Use in source · 1 passage

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  • Use in source · Read passage · 10_foundations.tex, line 140

    [Hurwitz product rules] For d≥0, D^d(f⋆ g)=∑_r=0^dbinom dr(D^r f)⋆(D^d-rg). If R has prime characteristic p, then D^p(f⋆ g)=f⋆ D^p g+D^p f⋆ g. Over a field, if g_0=1, a sequence b is constructed recursively with b⋆ g=1. If also D^p g=Dg in characteristic p, then D^p b=Db.

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    D^d(f\star g)=\sum_{r=0}^{d}\binom dr(D^r f)\star(D^{d-r}g).
    
D^iUse in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 75

    [Translated product formula] For N=p^a, a≥1, and λ,μ∈ Z_p, L_λ+μ(AB) =(L_λ A)B+A(L_μ B) +∑_i=1^N-1c_i(D^iA)(D^N-iB), c_i=frac1pbinom Ni∈ Z. Flatness types add under multiplication.

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     L_{\lambda+\mu}(AB)
      =(L_\lambda A)B+A(L_\mu B)
           +\sum_{i=1}^{N-1}c_i(D^iA)(D^{N-i}B),
     \qquad c_i=\frac1p\binom Ni\in\mathbb Z.
     \label{eq:translated-product}
    
D^{N-i}Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 75

    [Translated product formula] For N=p^a, a≥1, and λ,μ∈ Z_p, L_λ+μ(AB) =(L_λ A)B+A(L_μ B) +∑_i=1^N-1c_i(D^iA)(D^N-iB), c_i=frac1pbinom Ni∈ Z. Flatness types add under multiplication.

    Exact mathematical passage
    
     L_{\lambda+\mu}(AB)
      =(L_\lambda A)B+A(L_\mu B)
           +\sum_{i=1}^{N-1}c_i(D^iA)(D^{N-i}B),
     \qquad c_i=\frac1p\binom Ni\in\mathbb Z.
     \label{eq:translated-product}
    
D^pUse in source · 7 passages

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  • Use in source · Read passage · 10_foundations.tex, line 144

    [Hurwitz product rules] For d≥0, D^d(f⋆ g)=∑_r=0^dbinom dr(D^r f)⋆(D^d-rg). If R has prime characteristic p, then D^p(f⋆ g)=f⋆ D^p g+D^p f⋆ g. Over a field, if g_0=1, a sequence b is constructed recursively with b⋆ g=1. If also D^p g=Dg in characteristic p, then D^p b=Db.

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    D^p(f\star g)=f\star D^p g+D^p f\star g.
    
D^{p^2}Use in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 209

    Work in the Hurwitz integral ring of formal series H_p={F∈ Q_p[[t]]: d![t^d]F∈ Z_p for every d}. For D=d/dt, let B_p=(D^p^2-D^p)/p. Call F flat if B_p^sF∈ H_p for every s≥0. We will prove that j! P_N,j is flat at every address. The main difficulty is to include the addresses at which the usual unit-pivot argument fails.

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    B_p=(D^{p^2}-D^p)/p
D^rUse in source · 1 passage

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  • Use in source · Read passage · 10_foundations.tex, line 140

    [Hurwitz product rules] For d≥0, D^d(f⋆ g)=∑_r=0^dbinom dr(D^r f)⋆(D^d-rg). If R has prime characteristic p, then D^p(f⋆ g)=f⋆ D^p g+D^p f⋆ g. Over a field, if g_0=1, a sequence b is constructed recursively with b⋆ g=1. If also D^p g=Dg in characteristic p, then D^p b=Db.

    Exact mathematical passage
    
    D^d(f\star g)=\sum_{r=0}^{d}\binom dr(D^r f)\star(D^{d-r}g).
    
D^\alphaUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 101

    e with unspecified remainders. On the tensor of the two factors use the mutually commuting operators L_λ⊗1, 1⊗ L_μ, c_iD^i⊗ D^N-i (1≤ i<N). Applying multiplication after the kth power of their sum gives exactly L_λ+μ^k(AB). Expanding it yields terms indexed by a,b,e_1,…,e_N-1≥0, with a+b+∑_i e_i=k: frack!a!b!∏_i e_i!∏_i c_i^e_i (D^α L_λ^a A)(D^β L_μ^b B), α=∑_i i e_i, β=∑_i(N-i)e_i. Every scalar is integral. If A and B have the stated flatness types, both factors in every term belong to H_p, as …

    Exact mathematical passage
    
     \frac{k!}{a!b!\prod_i e_i!}\prod_i c_i^{e_i}
     (D^\alpha L_\lambda^a A)(D^\beta L_\mu^b B),
     \quad
     \alpha=\sum_i i e_i,\quad
     \beta=\sum_i(N-i)e_i.
     \label{eq:translated-multinomial}
    
D^\betaUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 101

    e with unspecified remainders. On the tensor of the two factors use the mutually commuting operators L_λ⊗1, 1⊗ L_μ, c_iD^i⊗ D^N-i (1≤ i<N). Applying multiplication after the kth power of their sum gives exactly L_λ+μ^k(AB). Expanding it yields terms indexed by a,b,e_1,…,e_N-1≥0, with a+b+∑_i e_i=k: frack!a!b!∏_i e_i!∏_i c_i^e_i (D^α L_λ^a A)(D^β L_μ^b B), α=∑_i i e_i, β=∑_i(N-i)e_i. Every scalar is integral. If A and B have the stated flatness types, both factors in every term belong to H_p, as …

    Exact mathematical passage
    
     \frac{k!}{a!b!\prod_i e_i!}\prod_i c_i^{e_i}
     (D^\alpha L_\lambda^a A)(D^\beta L_\mu^b B),
     \quad
     \alpha=\sum_i i e_i,\quad
     \beta=\sum_i(N-i)e_i.
     \label{eq:translated-multinomial}
    
d_0Use in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 656

    [Exact four-packet displacement law] Set (c_0,c_1,c_2,c_3)=(1,1,4,2), (d_0,d_1,d_2,d_3)=(1,1,5,3). For 0≤ j<4, n≥0, and every positive integer T, C_j(n+T)-C_j(n) · ≠0, · v_2(C_j(n+T)-C_j(n)) · =v_2(T)+c_j,

    Exact mathematical passage
    
     (c_0,c_1,c_2,c_3)=(1,1,4,2),\qquad
     (d_0,d_1,d_2,d_3)=(1,1,5,3).
    
d_{0,k}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2306

    (q)=q has d_1=1; thus d_1not≡ c_2pmod2. The coefficient consequence is read with integral original coefficients. Let M=p^a. Then Θ_p^a,s(r,k) ≡ binomsr-p^ak pmod p, where the binomial coefficient is zero outside its natural range. Consequently, d_s,k ≡ ∑_u=0^s binom su c_p^ak+u pmod p. In particular, d_0,k≡ c_p^akpmod p.

    Exact mathematical passage
    
    \boxed{d_{0,k}\equiv c_{p^ak}\pmod p.}
    
D_1Use in source · 3 passages

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  • Use in source · Read passage · strengthening.tex, line 439

    The address-one source is G_4,1(t)=(sinh t+sin t)/(2t). The identities a+b=1, a-b=i, and ab=1/2 give D_1(t):=G_4,1(t)F_0(t) =frac12(fractanh(at)at +fractanh(bt)bt). For example, its numerator follows from sinh t+sin t =2bsinh(at)cosh(bt)+2acosh(at)sinh(bt), while cosh t+cos t=2cosh(at)cosh(bt). Only degrees divisible by four survive. Its actual Hurwitz packets are D_1(t)=∑_i≥0d_ifract^4i(4i)!, d_i=(-frac14)^i frac2^4i+2(2^4i+2-1)B_4i+2(4i+2)(4i+1). The constant is one. Since v_2(4i+2)=1, the ele…

    Exact mathematical passage
    
     D_1(t):=G_{4,1}(t)F_0(t)
     =\frac12\left(\frac{\tanh(at)}{at}
                             +\frac{\tanh(bt)}{bt}\right).
     \label{eq:address-one-tangent-factor}
    
d_1Use in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 656

    [Exact four-packet displacement law] Set (c_0,c_1,c_2,c_3)=(1,1,4,2), (d_0,d_1,d_2,d_3)=(1,1,5,3). For 0≤ j<4, n≥0, and every positive integer T, C_j(n+T)-C_j(n) · ≠0, · v_2(C_j(n+T)-C_j(n)) · =v_2(T)+c_j,

    Exact mathematical passage
    
     (c_0,c_1,c_2,c_3)=(1,1,4,2),\qquad
     (d_0,d_1,d_2,d_3)=(1,1,5,3).
    
D_1^{-1}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 467

    stant is one. Since v_2(4i+2)=1, the elementary Bernoulli bound shows v_2(d_i)≥2i. Hurwitz inversion therefore gives D_1(t)^-1=∑_i≥0h_ifract^4i(4i)!, h_0=1, v_2(h_i)≥2i. This follows by induction from the inverse recurrence: for packet index i, every product d_rh_i-r has valuation at least 2i, and its binomial multiplier is integral. Because F_1=F_0D_1^-1, C_1(n)-C_0(n) =∑_i≥1h_ibinom4n4iC_0(n-i). At a natural n this is a finite sum, with i≤ n.

    Exact mathematical passage
    F_1=F_0D_1^{-1}
d_2Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 656

    [Exact four-packet displacement law] Set (c_0,c_1,c_2,c_3)=(1,1,4,2), (d_0,d_1,d_2,d_3)=(1,1,5,3). For 0≤ j<4, n≥0, and every positive integer T, C_j(n+T)-C_j(n) · ≠0, · v_2(C_j(n+T)-C_j(n)) · =v_2(T)+c_j,

    Exact mathematical passage
    
     (c_0,c_1,c_2,c_3)=(1,1,4,2),\qquad
     (d_0,d_1,d_2,d_3)=(1,1,5,3).
    
d_{2,\ell}Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 1257

    =0,1,2,3, the dyadic interpolation has an ordinary power series widetilde C_j(x)=1+∑_ℓ≥1d_j,ℓx^ℓ, v_2(d_j,ℓ)≥ℓ+β_j. Its Taylor series converges for |x|_2<2. At addresses zero and one, v_2(d_j,ℓ)=ℓ (ℓ≥1), R_0=R_1=2. At address two, v_2(d_2,ℓ)≥ max(2ℓ+lfloorfracℓ2rfloor, ℓ+3), R_2=2^5/2. The radius at address three is at least two; its exact radius is not asserted here. The stated exact radii at addresses zero, one, and two hold at every center in Z_2.

    Exact mathematical passage
    
     v_2(d_{2,\ell})\geq
           \max\left(2\ell+\left\lfloor\frac\ell2\right\rfloor,
                              \ell+3\right),
     \qquad R_2=2^{5/2}.
     \label{eq:two-exact-taylor-radius}
    
D_3Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 723

    The address-three source is G_4,3(t)=3(sinh t-sin t)/t^3. Addition identities give sinh t-sin t =2asinh(at)cosh(bt)+2bcosh(at)sinh(bt), cosh t-cos t=2sinh(at)sinh(bt). Consequently D_3(t):=G_4,3(t)F_2(t) =frac32(fraccoth(at)at +fraccoth(bt)bt). This expression is a power series: its two polar terms cancel because a^-2+b^-2=0. Its constant is one, only degrees divisible by four survive, and the Bernoulli expansion gives D_3(t)=∑_i≥0d_ifract^4i(4i)!, d_i=3(-frac14)^i frac2^4i+2B_4i+2(4i+2)(4i+1). …

    Exact mathematical passage
    
     D_3(t):=G_{4,3}(t)F_2(t)
     =\frac32\left(\frac{\coth(at)}{at}
                                +\frac{\coth(bt)}{bt}\right).
     \label{eq:address-three-cotangent-factor}
    
d_3Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 656

    [Exact four-packet displacement law] Set (c_0,c_1,c_2,c_3)=(1,1,4,2), (d_0,d_1,d_2,d_3)=(1,1,5,3). For 0≤ j<4, n≥0, and every positive integer T, C_j(n+T)-C_j(n) · ≠0, · v_2(C_j(n+T)-C_j(n)) · =v_2(T)+c_j,

    Exact mathematical passage
    
     (c_0,c_1,c_2,c_3)=(1,1,4,2),\qquad
     (d_0,d_1,d_2,d_3)=(1,1,5,3).
    
D_3^{-1}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 739

    e a^-2+b^-2=0. Its constant is one, only degrees divisible by four survive, and the Bernoulli expansion gives D_3(t)=∑_i≥0d_ifract^4i(4i)!, d_i=3(-frac14)^i frac2^4i+2B_4i+2(4i+2)(4i+1). Again v_2(d_i)≥2i, and the inverse coefficients h_i have the same bound. Since F_3=F_2D_3^-1, C_3(n)-C_2(n)=∑_i≥1h_ibinom4n4iC_2(n-i). The scaled-binomial argument [source reference] applies unchanged. The basic bound for C_2 includes order zero; its continuous extension is available from the threshold-free theo…

    Exact mathematical passage
    F_3=F_2D_3^{-1}
D_aUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2754

    [Packet decomposition of a lift obstruction] Suppose F_p^d = ⊕_a∈ AU_a, F_p^r = ⊕_a∈ AV_a, and suppose a linear correction map D: F_p^d→ F_p^r satisfies D(U_a)⊆ V_a for every a. Write D_a=D|_U_a. Then coker D ≅ ⊕_a∈ Acoker D_a. A residual lifts if and only if every packet obstruction vanishes.

    Exact mathematical passage
    D_a=D|_{U_a}
d_iUse in source · 8 passages

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  • Use in source · Read passage · root_scope.tex, line 219

    For 0≤ i<j, let d_i=v_2(i!)+v_2((j-1-i)!), k_i=1+lfloorlog_2max(i,j-1-i)rfloor, B_i=i+2^k_i Z_2, t_i=d_i+k_i. At j=1 use d_0=k_0=t_0=0. The balls B_i partition Z_2. Indeed a residue represented twice among 0,…,j-1 splits into its two children; every other represented residue remains one ball. On each such ball, P_j(i+2^k_iy)=2^t_iyQ_i(y), Q_i∈ Z_2⟨ y⟩^×, Q_i≡1pmod2.

    Exact mathematical passage
    
     d_i=v_2(i!)+v_2((j-1-i)!),\qquad
     k_i=1+\left\lfloor\log_2\max(i,j-1-i)\right\rfloor,
    
d_jUse in source · 11 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 661

    [Exact four-packet displacement law] Set (c_0,c_1,c_2,c_3)=(1,1,4,2), (d_0,d_1,d_2,d_3)=(1,1,5,3). For 0≤ j<4, n≥0, and every positive integer T, C_j(n+T)-C_j(n) · ≠0, · v_2(C_j(n+T)-C_j(n)) · =v_2(T)+c_j, U_j(n+T)-U_j(n) · ≠0, · v_2(U_j(n+T)-U_j(n)) · =v_2(T)+d_j.

    Exact mathematical passage
    
     C_j(n+T)-C_j(n)&\ne0,&
     v_2\bigl(C_j(n+T)-C_j(n)\bigr)&=v_2(T)+c_j,\\
     U_j(n+T)-U_j(n)&\ne0,&
     v_2\bigl(U_j(n+T)-U_j(n)\bigr)&=v_2(T)+d_j.
    
d_{j,m}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 1290

    For j=0,1, the term m=ℓ in [source reference] has exact valuation ℓ, since s(ℓ,ℓ)=1 and the Newton valuation is exact. Every later term has valuation at least ℓ+1. Thus v_2(d_j,ℓ)=ℓ, and the coefficient radius is exactly two. At a center x_0∈ Z_2, the coefficient of y^ℓ in the translated series is ∑_m≥ℓd_j,mbinom mℓ x_0^m-ℓ. Its initial term again has exact valuation ℓ, and all later terms have at least one more bit. The exact radius is therefore two at every integral center.

    Exact mathematical passage
    
     \sum_{m\geq\ell}d_{j,m}\binom m\ell x_0^{m-\ell}.
     \label{eq:taylor-translation-coefficient}
    
d_{j,\ell}Use in source · 4 passages

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  • Use in source · Read passage · strengthening.tex, line 1245

    [Taylor coefficient bounds and radii] For j=0,1,2,3, the dyadic interpolation has an ordinary power series widetilde C_j(x)=1+∑_ℓ≥1d_j,ℓx^ℓ, v_2(d_j,ℓ)≥ℓ+β_j. Its Taylor series converges for |x|_2<2. At addresses zero and one, v_2(d_j,ℓ)=ℓ (ℓ≥1), R_0=R_1=2. At address two, v_2(d_2,ℓ)≥ max(2ℓ+lfloorfracℓ2rfloor, ℓ+3), R_2=2^5/2. The radius at address three is at least two; its exact radius is not asserted here. The stated exact radii at addresses zero, one, and two hold at every center in Z_2.

    Exact mathematical passage
    
     \widetilde C_j(x)=1+\sum_{\ell\geq1}d_{j,\ell}x^\ell,
     \qquad v_2(d_{j,\ell})\geq\ell+\beta_j.
    
d_nUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 842

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    d_n=f_n
D_{N,j}Use in source · 12 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 59

    [A denominator bound for every address] For N≥1, 0≤ j<N, and n∈ N, set D_N,j(n)=∏_r=1^nbinomNr+jNr, D_N,j(0)=1. Then D_N,j(n)C_N,j(n)∈ Z.

    Exact mathematical passage
    
     D_{N,j}(n)=\prod_{r=1}^{n}\binom{Nr+j}{Nr},\qquad D_{N,j}(0)=1.
    
d_n^{-1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 44

    First construct the integral solution. For a ring R, a triangular row system T, right-hand side b, and proved unit pivots d_n=T_nn, define s_n=d_n^-1(b_n-∑_q<nT_nqs_q). The new ring-level solver uses well-founded recursion. Its row identities and unique-solution property are proved. The source also proves compatibility of this solver with ring homomorphisms into fields.

    Exact mathematical passage
    
     s_n=d_n^{-1}\left(b_n-\sum_{q<n}T_{nq}s_q\right).
    
D_PUse in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 51

    Apply polynomial induction using constants, addition, and multiplication by a variable. Constants have zero remainder, and remainders add. For the multiplication step, suppose the assertion is known for P and write D_P(U)=∑_j(∂_jP)(a)U_j. Multiplication of its expansion by a_i+cU_i gives constant term P(a)a_i, linear term c(a_iD_P(U)+P(a)U_i), and remainder c^2(Q(U)(a_i+cU_i)+D_P(U)U_i). The product rule identifies the linear term with the derivative of PX_i. The displayed remainder vanishes at …

    Exact mathematical passage
    D_P(U)=\sum_j(\partial_jP)(a)U_j
d_qUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 843

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    
    d_q=\sum_{\chi\in\mu_h}c_\chi\chi^q.
    
d_rUse in source · 5 passages

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  • Use in source · Read passage · root_scope.tex, line 159

    [Exact classical tail valuation] For every r≥1 and N≥ r, v_2( T_r-1(N)) =4r-1+v_2binom Nr+μ_r+v_2G_r(N-r). There are unique factors G_r(X)=W_r(X)U_r(X), where W_r∈ Z_2[X] is monic of degree d_r=max J_r and U_r∈ Z_2⟨ X⟩^× satisfies U_r≡1pmod2. Consequently v_2G_r(x)=v_2W_r(x) for every x∈ Z_2, and deg W_r≤lfloorfracs_2(r)-13rfloor. Equation~[source reference], the exact weight in [source reference], and the digit-sum identity v_2binom2N2r=v_2binom Nr prove [source reference]. The binomial coeffic…

    Exact mathematical passage
    d_r=\max J_r
d_{s,k}Use in source · 6 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2212

    [Mahler polyphase transfer] Let f(x)=∑_r≥0c_rbinom xr be a convergent Mahler expansion. For g_s(q)=f(Mq+s), one has g_s(q)=∑_k≥0d_s,kbinom qk, where d_s,k = ∑_r=k^Mk+s c_rΘ_M,s(r,k). Every refined coefficient uses a finite window of the original Mahler coefficients.

    Exact mathematical passage
    
    \boxed{
    g_s(q)=\sum_{k\geq0}d_{s,k}\binom qk,}
    
d_vUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 428

    [Exponential form and minimal period] Let G_m=( F_p^×)^m. This cyclic group has order h. There are uniquely determined coefficients d_v∈ F_p, indexed by v∈ G_m, such that e_n^(m,p) = ∑_v∈ G_md_vv^n (n≥1). The minimal positive period is the least common multiple of the orders of those v for which d_v≠0. In particular, it divides h.

    Exact mathematical passage
    d_v\in\FF_p
\ddotsUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2071

    [Polyphase shift matrix] Let E be the scalar shift (Ea)_n=a_n+1, and let E_q shift the coarse index. Then P_M(Ea) = S_M(E_q) P_Ma, where S_M(E_q) = 0 · 1 · 0 · ⋯ · 0 0 · 0 · 1 · ⋯ · 0 vdots · · · ddots · vdots 0 · 0 · 0 · ⋯ · 1 E_q · 0 · 0 · ⋯ · 0 . It also satisfies S_M(E_q)^M=E_qI_M.

    Exact mathematical passage
    
    \mathsf S_M(E_q)
    =
    \begin{pmatrix}
    0&1&0&\cdots&0\\
    0&0&1&\cdots&0\\
    \vdots&&&\ddots&\vdots\\
    0&0&0&\cdots&1\\
    E_q&0&0&\cdots&0
    \end{pmatrix}.
    
\operatorname{dec}Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 539

    Packing g_j into degrees divisible by four gives K_4,j. Its source property is precisely charge-one flatness. Inversion gives charge -1, and the charge shift restores charge one for the reciprocal P_4,j. If dec(F)(d)=d![t^d]F, the exact coefficient identity is dec(A_1^sF)(d) =2^-sΔ_4^sdec(F)(d). At d=4n, packing turns the right side into 2^-sΔ_1^sC_j(n). This extraction requires no negative degree and therefore holds at every n, including zero. It proves the unit-step bound leanKomorebiPaperIX.D…

    Exact mathematical passage
    \operatorname{dec}(F)(d)=d![t^d]F
\degUse in source · 2 passages

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  • Use in source · Read passage · root_scope.tex, line 162

    eginequation v_2( T_r-1(N)) =4r-1+v_2binom Nr+μ_r+v_2G_r(N-r). There are unique factors G_r(X)=W_r(X)U_r(X), where W_r∈ Z_2[X] is monic of degree d_r=max J_r and U_r∈ Z_2⟨ X⟩^× satisfies U_r≡1pmod2. Consequently v_2G_r(x)=v_2W_r(x) for every x∈ Z_2, and deg W_r≤lfloorfracs_2(r)-13rfloor. Equation~[source reference], the exact weight in [source reference], and the digit-sum identity v_2binom2N2r=v_2binom Nr prove [source reference]. The binomial coefficients themselves need not be equal.

    Exact mathematical passage
    
     \deg W_r\le\left\lfloor\frac{s_2(r)-1}{3}\right\rfloor.
    
\dimUse in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 313

    For prime p, rank-nullity gives dimker J_x=d-ρ_x and dim( F_p^e/im J_x)=e-ρ_x. The cardinality of an n-dimensional F_p-space is p^n. Combine these identities with the obstruction criterion and the residue equivalence.

    Exact mathematical passage
    \dim\ker J_x=d-\rho_x
\eUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 918

    [Odd source generating function] For every positive odd q, ∑_n≥0S_n(q)fract^nn! = e^qtcosh(qt)secht. Its even part is ∑_n≥0S_2n(q)fract^2n(2n)! = cosh^2(qt)secht.

    Exact mathematical passage
    
    \boxed{
    \sum_{n\geq0}S_n(q)\frac{t^n}{n!}
    =
    \e^{qt}\cosh(qt)\operatorname{sech}t.
    }
    
\mathcal EUse in source · 29 passages

Search aliases: E mathcal

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  • Use in source · Read passage · 10_foundations.tex, line 78

    Over a characteristic-zero field K, put E(a)=∑_n≥0fraca_nn!t^n, D(A)_n=n![t^n]A. These inverse maps between sequences and K[[t]] provide formal exponential coordinates. No analytic convergence is needed for the coefficient identities below.

    Exact mathematical passage
    
    \mathcal E(a)=\sum_{n\geq0}\frac{a_n}{n!}t^n,
    \qquad \mathcal D(A)_n=n![t^n]A.
    
e'Use in source · 1 passage

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  • Use in source · Read passage · 71_native.tex, line 58

    Another representative has the form x'=x+p^Ls. Since L≥1, its Jacobian is the same modulo p. Dividing (1) by p^L shows that its residual digit is e'=e+Js, so it gives exactly the same quotient class. If t_0 solves (2), all its solutions are t_0+ker J, and no others. The finite cardinalities follow from rank-nullity. This also covers no variables, no equations, zero Jacobian and singular solutions. square

    Exact mathematical passage
    e'=e+Js
\e^{(2a+1)t}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 937

    The finite geometric sum gives ∑_a=0^q-1(-1)^a e^(2a+1)t = e^tfrac1+ e^2qt1+ e^2t = e^qtcosh(qt)secht. Averaging at t and -t gives the even part.

    Exact mathematical passage
    
    \sum_{a=0}^{q-1}(-1)^a\e^{(2a+1)t}
    =
    \e^t\frac{1+\e^{2qt}}{1+\e^{2t}}
    =
    \e^{qt}\cosh(qt)\operatorname{sech}t.
    
e^{(2a+1)t}Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 814

    The exponential generating series of the source is the finite sum S_p(t)=∑_a=0^p-1(-1)^a e^(2a+1)t. Multiplication by cosh t telescopes. Since p is odd, S_p(t)=frac1+e^2pt2secht. Taking its even part gives ∑_n≥0S_2n(p)fract^2n(2n)! =cosh^2(pt)secht. Multiplication by the reciprocal sech^2(pt) yields secht. The EGF product rule now gives the reconstruction coefficient by coefficient. Replacing m by n-q produces the displayed lower triangular row. Its diagonal and weights follow from γ_0=1 and the…

    Exact mathematical passage
    
     \mathcal S_p(t)=\sum_{a=0}^{p-1}(-1)^a e^{(2a+1)t}.
    
E^{(4,j)}Use in source · 4 passages

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  • Use in source · visual_body.tex, line 310

    In dyadic distance, |x|_2=2^-v_2(x). After continuous interpolation to Z_2, the law reads |E^(4,j)(x)-E^(4,j)(y)|_2 =2^-B_j|x-y|_2. It says exactly how much the map shrinks distances. It also prevents any hidden collision at finer precision: distinct inputs remain distinct after the known scale is removed.

    Exact mathematical passage
    
     |E^{(4,j)}(x)-E^{(4,j)}(y)|_2
     =2^{-B_j}|x-y|_2.
    
e^{(m,p)}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 865

    [A second proof of the standard window period] The reduced inverse e^(m,p) and every standard address satisfy the period in [source reference].

    Exact mathematical passage
    e^{(m,p)}
e^{(m,p)}_nUse in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 61

    Then apply the residue map. Independently define e^(m,p) over F_p by the unit-diagonal system ∑_q=0^nbinommnmqe^(m,p)_q=δ_n0. Its recurrence and uniqueness are checked. Applying the actual homomorphism π: Z_p→ F_p to the integral row, and then Lucas reduction, proves π(c^(p)_m,j(n))=e^(m,p)_n. For j,k<p, a separate ideal statement gives pmid c^(p)_m,j(n)-c^(p)_m,k(n) inside Z_p. The formal solver also covers m=0; the manuscript uses m≥1. No global finite-difference theorem or period assertion en…

    Exact mathematical passage
    
     \boxed{\pi(c^{(p)}_{m,j}(n))=e^{(m,p)}_n.}
    
e^{(m,p)}_qUse in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 57

    Then apply the residue map. Independently define e^(m,p) over F_p by the unit-diagonal system ∑_q=0^nbinommnmqe^(m,p)_q=δ_n0. Its recurrence and uniqueness are checked. Applying the actual homomorphism π: Z_p→ F_p to the integral row, and then Lucas reduction, proves π(c^(p)_m,j(n))=e^(m,p)_n. For j,k<p, a separate ideal statement gives pmid c^(p)_m,j(n)-c^(p)_m,k(n) inside Z_p. The formal solver also covers m=0; the manuscript uses m≥1. No global finite-difference theorem or period assertion en…

    Exact mathematical passage
    
     \sum_{q=0}^{n}\binom{mn}{mq}e^{(m,p)}_q=\delta_{n0}.
    
E^{(N,j)}Use in source · 2 passages

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  • Use in source · visual_body.tex, line 160

    There is a prime-power extension. For N=p^a, a≥1, the sign-corrected operator is T_H=S^H-(-1)^H, v_p (T_H_1⋯ T_H_kE^(N,j)(n)) ≥ v_p(j!)+∑_i=1^k(1+v_p(H_i)). Here E^(N,j)(n)=E_Nn^(N,j). For an odd prime, an ordinary first difference need not have positive valuation: at address zero its first two normalized values are 1 and -1, with difference -2. The sign correction is therefore essential to this prime-power theorem.

    Exact mathematical passage
    
     T_H=S^H-(-1)^H,
     \qquad
     v_p\!\left(T_{H_1}\cdots T_{H_k}E^{(N,j)}(n)\right)
     \ge v_p(j!)+\sum_{i=1}^k\bigl(1+v_p(H_i)\bigr).
    
e^{-2pt}Use in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 60

    Even projection is identified coefficientwise. For any formal series f, let EvenPart(f)=(f(t)+f(-t))/2. Its nth coefficient is proved to be the coefficient of f when n is even and zero otherwise. Rescaling the inverse law proves sech(-t)=sech(t). The exponential multiplication identity then gives cosh^2(pt)=frac2+e^2pt+e^-2pt4, and hence ∑_r≥0S_2r(p)fract^2r(2r)! =cosh^2(pt) sech t. Both the full and even EGF formulas are derived from the original integer sum. Neither is supplied as a hypothesis…

    Exact mathematical passage
    
     \cosh^2(pt)=\frac{2+e^{2pt}+e^{-2pt}}4,
    
e^{-t}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 200

    j(n)∈ p^k Z_p. Moreover, Δ_1^k A_N,j(n)=(-1)^n+k(S+1)^k C_N,j(n), so A_N,j has the ordinary p^k difference property. At p=2, L_1=L_-1-1; its integral binomial expansion on F_N,j gives Δ_1^k C_N,j(n)∈2^k Z_2 directly. For N=1 the interior sum in the product formula is empty, so the same argument applies without change; equivalently, F_1,0(t)=e^-t.

    Exact mathematical passage
    F_{1,0}(t)=e^{-t}
\mathcal E^2Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 305

    dependent variables mean that moments of polynomials in separate variables factor. Odd moments vanish. Frobenius's congruence is v_2(∑_r=0^m(-1)^m-rbinom mr Euler_2n+2r) =m+v_2(m!) (n,m≥0). For m=0, the assertion is that every even Euler number is odd. With P=( E^2-1)/2 and γ_m(P)=P^m/m!, the case n=0 says L(γ_m(P))∈ Z_2^× (m≥0). Introduce an independent Euler variable F and Q=( F^2-1)/2.

    Exact mathematical passage
    P=(\mathcal E^2-1)/2
e^{2pt}Use in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 818

    The exponential generating series of the source is the finite sum S_p(t)=∑_a=0^p-1(-1)^a e^(2a+1)t. Multiplication by cosh t telescopes. Since p is odd, S_p(t)=frac1+e^2pt2secht. Taking its even part gives ∑_n≥0S_2n(p)fract^2n(2n)! =cosh^2(pt)secht. Multiplication by the reciprocal sech^2(pt) yields secht. The EGF product rule now gives the reconstruction coefficient by coefficient. Replacing m by n-q produces the displayed lower triangular row. Its diagonal and weights follow from γ_0=1 and the…

    Exact mathematical passage
    
     \mathcal S_p(t)=\frac{1+e^{2pt}}{2}\operatorname{sech}t.
    
\e^{2qt}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 937

    The finite geometric sum gives ∑_a=0^q-1(-1)^a e^(2a+1)t = e^tfrac1+ e^2qt1+ e^2t = e^qtcosh(qt)secht. Averaging at t and -t gives the even part.

    Exact mathematical passage
    
    \sum_{a=0}^{q-1}(-1)^a\e^{(2a+1)t}
    =
    \e^t\frac{1+\e^{2qt}}{1+\e^{2t}}
    =
    \e^{qt}\cosh(qt)\operatorname{sech}t.
    
\e^{2t}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 937

    The finite geometric sum gives ∑_a=0^q-1(-1)^a e^(2a+1)t = e^tfrac1+ e^2qt1+ e^2t = e^qtcosh(qt)secht. Averaging at t and -t gives the even part.

    Exact mathematical passage
    
    \sum_{a=0}^{q-1}(-1)^a\e^{(2a+1)t}
    =
    \e^t\frac{1+\e^{2qt}}{1+\e^{2t}}
    =
    \e^{qt}\cosh(qt)\operatorname{sech}t.
    
e^{2t}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 620

    The variable B=2Z+1 has transform e^tfrac2te^2t-1=fractsinh t. Its odd moments vanish, and P=( B^2-1)/4=Z(Z+1) has the divided-power moments of Lemma~[source reference]. Introduce an independent copy D and Q=( D^2-1)/4. The exact factorization F_2(t)=fract^2cosh t-cos t =fracatsinh(at)fracbtsinh(bt) shows that U=a B+b D satisfies L(U^4n)=C_2(n). Direct expansion gives H_2:=fracU^4-14 =P+Q-P^2-Q^2+6PQ+i B D(P-Q).

    Exact mathematical passage
    
     e^t\frac{2t}{e^{2t}-1}=\frac{t}{\sinh t}.
    
\mathcal E^dUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 290

    Let L be the ordinary Euler evaluation functional on a variable E, defined by L( E^d)=Euler_d, ∑_d≥0Euler_dfract^dd! =secht. The functional is linear; it is not a multiplicative evaluation. Independent variables mean that moments of polynomials in separate variables factor. Odd moments vanish. Frobenius's congruence is v_2(∑_r=0^m(-1)^m-rbinom mr Euler_2n+2r) =m+v_2(m!) (n,m≥0). For m=0, the assertion is that every even Euler number is odd. With P=( E^2-1)/2 and γ_m(P)=P^m/m!, the case n=0 says …

    Exact mathematical passage
    
     \mathcal L(\mathcal E^d)=\operatorname{Euler}_d,
     \qquad \sum_{d\geq0}\operatorname{Euler}_d\frac{t^d}{d!}
                      =\operatorname{sech}t.
    
\e^{qt}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 918

    [Odd source generating function] For every positive odd q, ∑_n≥0S_n(q)fract^nn! = e^qtcosh(qt)secht. Its even part is ∑_n≥0S_2n(q)fract^2n(2n)! = cosh^2(qt)secht.

    Exact mathematical passage
    
    \boxed{
    \sum_{n\geq0}S_n(q)\frac{t^n}{n!}
    =
    \e^{qt}\cosh(qt)\operatorname{sech}t.
    }
    
\e^tUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 937

    The finite geometric sum gives ∑_a=0^q-1(-1)^a e^(2a+1)t = e^tfrac1+ e^2qt1+ e^2t = e^qtcosh(qt)secht. Averaging at t and -t gives the even part.

    Exact mathematical passage
    
    \sum_{a=0}^{q-1}(-1)^a\e^{(2a+1)t}
    =
    \e^t\frac{1+\e^{2qt}}{1+\e^{2t}}
    =
    \e^{qt}\cosh(qt)\operatorname{sech}t.
    
e^zUse in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 80

    [The Bernoulli boundary address] At N=2, the address j=1 gives a second classical family. Define the Bernoulli numbers by B(z)=fracze^z-1=∑_r≥0B_rfracz^rr!, B_0=1, B_1=-frac12. All quotients here are regular formal series. The identity 2B(z)-B(2z)=fraczsinh z =∑_n≥0C_2,1(n)fracz^2n(2n)! follows by putting the two fractions on a common denominator and using [source reference]. Comparing even coefficients yields C_2,1(n)=(2-2^2n)B_2n (n∈ N). This classical derivation restores the boundary example …

    Exact mathematical passage
    
     B(z)=\frac{z}{e^z-1}=\sum_{r\ge0}B_r\frac{z^r}{r!},
     \qquad B_0=1,\quad B_1=-\frac12.
    
E_{(p-1)x+j}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1858

    [Odd prime analytic Euler branch] The series defining mathscr E_p,j^× converges in the Tate algebra on Z_p. For every nonnegative integer x, mathscr E_p,j^×(x) = (1-σ_pp^(p-1)x+j) E_(p-1)x+j.

    Exact mathematical passage
    
    \boxed{
    \mathscr E_{p,j}^\times(x)
    =
    \left(1-\sigma_pp^{(p-1)x+j}\right)
    E_{(p-1)x+j}.
    }
    
E_0Use in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 869

    Return now to the classical signed Euler numbers secht=frac1cosh t =∑_n≥0E_2nfract^2n(2n)!. Thus E_2n=E_2n^(2,0) in the convention [source reference]. In particular E_0=1, E_2=-1, and E_4=5. The following construction concerns these N=2 coefficients, not C_4,0.

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    E_0=1
e_0Use in source · 2 passages

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  • Use in source · Read passage · supplement_07.tex, line 133

    Three checked boundaries. The initial row. For (p,m)=(3,2), the same reduced recurrence gives e_0=1 and e_1=-1=2 in F_3. Its proposed period is one, so the positive-tail result cannot in general include row zero.

    Exact mathematical passage
    e_0=1
e_0^{(m,p)}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 374

    [Compressed inverse recurrence] The compressed coefficients satisfy ∑_q=0^n binommnmq e_q^(m,p) = δ_n0. In particular, e_0^(m,p)=1, and the positive coefficients are determined recursively because the diagonal is one.

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    e_0^{(m,p)}=1
E_0^{(N,j)}Use in source · 1 passage

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  • Use in source · visual_body.tex, line 18

    There are two coefficient conventions, and they must stay distinct: C_N,j(0)=1, E_0^(N,j)=j!, v_2(E_Nn^(N,j))=v_2(C_N,j(n))+v_2(j!). The last equality uses v_2(0)=+∞. Congruences of dyadic rational numbers mean membership in the corresponding ideal of Z_2; a displayed rational denominator is invertible modulo 2^r when it is odd.

    Exact mathematical passage
    
     C_{N,j}(0)=1,
     \qquad E_0^{(N,j)}=j!,
     \qquad
     v_2(E_{Nn}^{(N,j)})=v_2(C_{N,j}(n))+v_2(j!).
    
e_1Use in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 447

    [A compressed inverse modulo seven] Editorial numerical correction. The positive-index standard-window tail printed below is erroneous. The defining recurrence gives e_1=-1≡6, e_2=5 and e_3=-61≡2pmod7, so the correct positive tail is 6,5,2,6,5,2,…. The period-three conclusion remains correct. The original line is retained only as a corrected historical record. Take p=7 and m=2, so h=3. The recurrence gives e_n^(2,7) = 1,6,4,1,6,4,… (n≥1). The positive tail has exact period three.

    Exact mathematical passage
    e_1=-1\equiv6
E_2Use in source · 8 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 869

    Return now to the classical signed Euler numbers secht=frac1cosh t =∑_n≥0E_2nfract^2n(2n)!. Thus E_2n=E_2n^(2,0) in the convention [source reference]. In particular E_0=1, E_2=-1, and E_4=5. The following construction concerns these N=2 coefficients, not C_4,0.

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    E_2=-1
e_2Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 447

    [A compressed inverse modulo seven] Editorial numerical correction. The positive-index standard-window tail printed below is erroneous. The defining recurrence gives e_1=-1≡6, e_2=5 and e_3=-61≡2pmod7, so the correct positive tail is 6,5,2,6,5,2,…. The period-three conclusion remains correct. The original line is retained only as a corrected historical record. Take p=7 and m=2, so h=3. The recurrence gives e_n^(2,7) = 1,6,4,1,6,4,… (n≥1). The positive tail has exact period three.

    Exact mathematical passage
    e_2=5
E_{2M+2}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 988

    [Optimal uniform dyadic tower] For all M,n∈ N, E_2n-T_M(n)∈2^4M+3 Z_2. At the first omitted row n=M+1, the error is nonzero and v_2(E_2M+2-T_M(M+1))=4M+3. Thus the exponent 4M+3 cannot be increased uniformly in the Euler row.

    Exact mathematical passage
    
     v_2\bigl(E_{2M+2}-T_M(M+1)\bigr)=4M+3.
    
E_{2n}Use in source · 15 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 749

    The classical signed Euler numbers have generating function secht=∑_n≥0E_2nt^2n/(2n)!. They are the N=2, j=0 member of [source reference]. The coordinate theorem reconstructs them from finite alternating sums of odd powers.

    Exact mathematical passage
    \operatorname{sech}t=\sum_{n\ge0}E_{2n}t^{2n}/(2n)!
\mathcal E_{2N}Use in source · 2 passages

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  • Use in source · Read passage · root_scope.tex, line 56

    The classical dyadic Bernoulli denominator law v_2(B_2m)=-1 and Legendre's formula yield v_2(β_m)=s_2(m)-1, v_2(κ_m)=4m-1. The source identity ∑_N≥0A(N)fract^2N(2N)! =fracsinh(2t)sinh t2t^2 implies, after multiplication by 4t^2/sinh^2(2t), E_2N=∑_m=0^N κ_mbinom2N2mA(N-m). For a retained depth M≥0, put T_M(N)= E_2N -∑_m=0^min(M,N)κ_mbinom2N2mA(N-m). An empty tail, N≤ M, is exactly zero.

    Exact mathematical passage
    
     \mathcal E_{2N}=\sum_{m=0}^N
        \kappa_m\binom{2N}{2m}A(N-m).
     \label{eq:root:packet}
    
E_{2n}^{(2,0)}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 868

    Return now to the classical signed Euler numbers secht=frac1cosh t =∑_n≥0E_2nfract^2n(2n)!. Thus E_2n=E_2n^(2,0) in the convention [source reference]. In particular E_0=1, E_2=-1, and E_4=5. The following construction concerns these N=2 coefficients, not C_4,0.

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    E_{2n}=E_{2n}^{(2,0)}
E_{2r}Use in source · 8 passages

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  • Use in source · Read passage · supplement_02.tex, line 78

    The source defines E_2r as the integer numerator of that rational coefficient and proves that its rational cast equals C_2,0(r). Set E_2r+1=0. The actual exponential-section identity from the first block proves ∑_n≥0E_nfract^nn!=sech t. Thus the integer sequence is connected to the intended signed Euler family, rather than merely assigned the name Euler.

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    E_{2r}
E_{2r+1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 78

    The source defines E_2r as the integer numerator of that rational coefficient and proves that its rational cast equals C_2,0(r). Set E_2r+1=0. The actual exponential-section identity from the first block proves ∑_n≥0E_nfract^nn!=sech t. Thus the integer sequence is connected to the intended signed Euler family, rather than merely assigned the name Euler.

    Exact mathematical passage
    E_{2r+1}=0
e_3Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 447

    [A compressed inverse modulo seven] Editorial numerical correction. The positive-index standard-window tail printed below is erroneous. The defining recurrence gives e_1=-1≡6, e_2=5 and e_3=-61≡2pmod7, so the correct positive tail is 6,5,2,6,5,2,…. The period-three conclusion remains correct. The original line is retained only as a corrected historical record. Take p=7 and m=2, so h=3. The recurrence gives e_n^(2,7) = 1,6,4,1,6,4,… (n≥1). The positive tail has exact period three.

    Exact mathematical passage
    e_3=-61\equiv2\pmod7
E_4Use in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 869

    Return now to the classical signed Euler numbers secht=frac1cosh t =∑_n≥0E_2nfract^2n(2n)!. Thus E_2n=E_2n^(2,0) in the convention [source reference]. In particular E_0=1, E_2=-1, and E_4=5. The following construction concerns these N=2 coefficients, not C_4,0.

    Exact mathematical passage
    E_4=5
E_{4(n+H)}^{(4,j)}Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 1013

    [Exact four-address first differences] For every n≥0, H>0, and j∈{0,1,2,3}, v_2(C_j(n+H)-C_j(n)) · =v_2(H)+b_j, · (b_0,b_1,b_2,b_3) · =(1,1,4,2), v_2(E_4(n+H)^(4,j)-E_4n^(4,j)) · =v_2(H)+B_j, · (B_0,B_1,B_2,B_3) · =(1,1,5,3). Consequently, the least positive row period modulo 2^r is exactly P_j(r)=2^max(0,r-B_j) (r≥1). The corresponding displacement in the original Euler subscript is 4P_j(r).

    Exact mathematical passage
    
     v_2(C_j(n+H)-C_j(n))&=v_2(H)+b_j,
          &(b_0,b_1,b_2,b_3)&=(1,1,4,2),\label{eq:rank-four-normalized-isometry}\\
     v_2(E_{4(n+H)}^{(4,j)}-E_{4n}^{(4,j)})&=v_2(H)+B_j,
          &(B_0,B_1,B_2,B_3)&=(1,1,5,3).
     \label{eq:rank-four-original-isometry}
    
E_{4(n+iH)}^{(4,j)}Use in source · 3 passages

Search aliases: E_4(n+iH)^(4,j)

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  • Use in source · Read passage · edition_summary.tex, line 11

    The strongest common-step conclusion in width four is particularly compact. For n≥0 and k,H≥1, ∑_i=0^k(-1)^k-ibinom ki E_4(n+iH)^(4,j) ≡0pmod2^ v_2(k!)+kv_2(H)+A_j(k), where A_0(k)=A_1(k)=k, A_2(k)=max{k+4, 2k+lfloor k/2rfloor+1}, A_3(k)=k+2. There is no starting-row threshold. The exponent is exact in addresses zero and one for every order and positive step. It is exact in address two at every power-of-two order. The proof chapters also give mixed-step forms, the cancellation criterion in addre…

    Exact mathematical passage
    
     \sum_{i=0}^k(-1)^{k-i}\binom ki E_{4(n+iH)}^{(4,j)}
     \equiv0\pmod{2^{\,v_2(k!)+kv_2(H)+A_j(k)}},
    
E_{4n}^{(4,j)}Use in source · 3 passages

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  • Use in source · Read passage · strengthening.tex, line 1013

    [Exact four-address first differences] For every n≥0, H>0, and j∈{0,1,2,3}, v_2(C_j(n+H)-C_j(n)) · =v_2(H)+b_j, · (b_0,b_1,b_2,b_3) · =(1,1,4,2), v_2(E_4(n+H)^(4,j)-E_4n^(4,j)) · =v_2(H)+B_j, · (B_0,B_1,B_2,B_3) · =(1,1,5,3). Consequently, the least positive row period modulo 2^r is exactly P_j(r)=2^max(0,r-B_j) (r≥1). The corresponding displacement in the original Euler subscript is 4P_j(r).

    Exact mathematical passage
    
     v_2(C_j(n+H)-C_j(n))&=v_2(H)+b_j,
          &(b_0,b_1,b_2,b_3)&=(1,1,4,2),\label{eq:rank-four-normalized-isometry}\\
     v_2(E_{4(n+H)}^{(4,j)}-E_{4n}^{(4,j)})&=v_2(H)+B_j,
          &(B_0,B_1,B_2,B_3)&=(1,1,5,3).
     \label{eq:rank-four-original-isometry}
    
E_6Use in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 174

    The displayed manuscript row is kernel-checked. The scalar anchors are derived from the same general coefficient recurrence: (E_0,E_2,E_4,E_6,E_8)=(1,-1,5,-61,1385), (b_0,b_1,b_2,b_3,b_4)=(1,-2,16,-272,7936). The literal source sums at p=5 give (S_0,S_2,S_4,S_6,S_8)=(1,49,4705,428689,37665985). Consequently, T_2(5,8)=2 731 001 385, E_8-T_2(5,8)=-2 731 000 000. The source proves both 5^6mid E_8-T_2(5,8), 5^7nmid E_8-T_2(5,8). Thus the guaranteed power is attained in this example. No stronger unif…

    Exact mathematical passage
    
     (E_0,E_2,E_4,E_6,E_8)=(1,-1,5,-61,1385),
    
E_8Use in source · 7 passages

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  • Use in source · Read passage · supplement_02.tex, line 174

    The displayed manuscript row is kernel-checked. The scalar anchors are derived from the same general coefficient recurrence: (E_0,E_2,E_4,E_6,E_8)=(1,-1,5,-61,1385), (b_0,b_1,b_2,b_3,b_4)=(1,-2,16,-272,7936). The literal source sums at p=5 give (S_0,S_2,S_4,S_6,S_8)=(1,49,4705,428689,37665985). Consequently, T_2(5,8)=2 731 001 385, E_8-T_2(5,8)=-2 731 000 000. The source proves both 5^6mid E_8-T_2(5,8), 5^7nmid E_8-T_2(5,8). Thus the guaranteed power is attained in this example. No stronger unif…

    Exact mathematical passage
    
     (E_0,E_2,E_4,E_6,E_8)=(1,-1,5,-61,1385),
    
E_cUse in source · 5 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 305

    pnmid c, differentiation of the section identities gives tR_c'=R_c(1+R_c-1-R_c). Writing q=R_c-1-(c-1) and y=R_c-c, this becomes (tD+c)y=cq+yq-y^2. On degrees divisible by p, the inverse of tD+c divides the degree-d coefficient by the unit d+c. Starting with y_0=0, form y_s+1=(tD+c)^-1(cq+y_sq-y_s^2). Flatness and support are preserved. Indeed, put E_c=tD+c and C_p=pD^p^2-D^p. The commutator is [B_p,E_c]=C_p, and iteration gives B_p^rE_cy=E_cB_p^ry+rC_pB_p^r-1y. For the inverse equation E_cy=f, …

    Exact mathematical passage
    E_c=tD+c
E_FUse in source · 4 passages

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  • Use in source · Read passage · 30_lifting.tex, line 798

    For the error estimate, let δ=b-a and D=M-s. Taylor's theorem for the preconditioned polynomial gives E_F=(AF)(b)-(AF)(a)-p^sδ∈ p^2DR^d. Polynomial congruence gives E_H=(AH)(b)-(AH)(a)∈ p^DR^d. The equations F(a)=0 and G(b)=0 imply p^sδ+p^MAH(a)=-E_F-p^ME_H∈ p^2DR^d, because M+D≥2D. Cancel p^s in the integral domain R to obtain δ+p^M-sAH(a)∈ p^2M-3sR^d. This integral vector is ε. After inclusion into Q_p, p^M-sA=p^MJ^-1 gives [source reference].

    Exact mathematical passage
    
     E_F=(AF)(b)-(AF)(a)-p^s\delta\in p^{2D}R^d.
    
E_HUse in source · 4 passages

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  • Use in source · Read passage · 30_lifting.tex, line 802

    For the error estimate, let δ=b-a and D=M-s. Taylor's theorem for the preconditioned polynomial gives E_F=(AF)(b)-(AF)(a)-p^sδ∈ p^2DR^d. Polynomial congruence gives E_H=(AH)(b)-(AH)(a)∈ p^DR^d. The equations F(a)=0 and G(b)=0 imply p^sδ+p^MAH(a)=-E_F-p^ME_H∈ p^2DR^d, because M+D≥2D. Cancel p^s in the integral domain R to obtain δ+p^M-sAH(a)∈ p^2M-3sR^d. This integral vector is ε. After inclusion into Q_p, p^M-sA=p^MJ^-1 gives [source reference].

    Exact mathematical passage
    E_H=(AH)(b)-(AH)(a)\in p^DR^d
e_iUse in source · 4 passages

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  • Use in source · Read passage · 71_native.tex, line 20

    e_i=F_i(x)/p^Lpmod p, J_ij=frac∂ F_i∂ X_j(x)pmod p.

    Exact mathematical passage
    e_i=F_i(x)/p^L\pmod p,\qquad J_{ij}=\frac{\partial F_i}{\partial X_j}(x)\pmod p.
e_jUse in source · 3 passages

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  • Use in source · Read passage · supplement_11.tex, line 96

    Identity with the original rational coefficients. The integral lift is constructed from the decoded coefficients of the actual reciprocal. The code proves coe(e_j(n))=C_4,j(n) in Q_2, and proves that the coefficient map commutes with every difference stencil. Multiplying by j! yields the unnormalized family. The valuation result is not divisibility in a field.

    Exact mathematical passage
    
     \operatorname{coe}(e_j(n))=C_{4,j}(n)\quad\text{in }\Q_2,
    
E_{mpn+qp^r}^{(mp,j)}Use in source · 2 passages

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  • Use in source · Read passage · edition_summary.tex, line 4

    nal edition establishes addcontentslinetocsectionWhat this final edition establishes The original external problem is Nishibuchi's Conjecture 3.9 for generalized Euler numbers. Its full parameter range is resolved. If p is prime, m is positive with mmid p-1 or m=2, 0≤ j<mp, and q=lcm(2,p-1), then the original Euler coefficients are p-adically integral and E_mpn+qp^r^(mp,j)≡ E_mpn^(mp,j)pmodp^r (r≥1, n≥ r). The paper states the generating function and the conversion between ordinary degree and pa…

    Exact mathematical passage
    
     E_{mpn+qp^r}^{(mp,j)}\equiv E_{mpn}^{(mp,j)}\pmod{p^r}
     \qquad (r\ge1,\ n\ge r).
    
E_{mpn}^{(mp,j)}Use in source · 2 passages

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  • Use in source · Read passage · edition_summary.tex, line 4

    nal edition establishes addcontentslinetocsectionWhat this final edition establishes The original external problem is Nishibuchi's Conjecture 3.9 for generalized Euler numbers. Its full parameter range is resolved. If p is prime, m is positive with mmid p-1 or m=2, 0≤ j<mp, and q=lcm(2,p-1), then the original Euler coefficients are p-adically integral and E_mpn+qp^r^(mp,j)≡ E_mpn^(mp,j)pmodp^r (r≥1, n≥ r). The paper states the generating function and the conversion between ordinary degree and pa…

    Exact mathematical passage
    
     E_{mpn+qp^r}^{(mp,j)}\equiv E_{mpn}^{(mp,j)}\pmod{p^r}
     \qquad (r\ge1,\ n\ge r).
    
E_nUse in source · 13 passages

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  • Use in source · Read passage · 72_recovered.tex, line 904

    Let q be a positive odd integer. Define S_n(q) = ∑_a=0^q-1 (-1)^a(2a+1)^n. Let the signed Euler numbers and Euler zigzag numbers be defined by secht = ∑_n≥0E_nfract^nn! and sec t+tan t = ∑_n≥0Z_nfract^nn!.

    Exact mathematical passage
    
    \operatorname{sech}t
    =
    \sum_{n\geq0}E_n\frac{t^n}{n!}
    
\mathcal E_nUse in source · 1 passage

Search aliases: E_n mathcal

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  • Use in source · Read passage · root_scope.tex, line 24

    To distinguish ordinary signed Euler numbers from the generalized numbers of the main text, write secht=∑_n≥0 E_nfract^nn!, fracte^t-1=∑_n≥0B_nfract^nn!, B_1=-frac12. Throughout this appendix, v_2(2)=1 and v_2(0)=∞. Let s_2(n) be the number of ones in the binary expansion of the nonnegative integer n. Define A(N)=frac3^2N+2-14(2N+1)(2N+2), (frac zsinh z)^2=∑_m≥0β_m z^2m, κ_m=2^2m(2m)!β_m. The initial values are β_0=κ_0=1. For m≥1, differentiation of zcoth z gives β_m=-frac2^2m(2m-1)B_2m(2m)!, κ_…

    Exact mathematical passage
    
     \operatorname{sech}t=\sum_{n\ge0}\mathcal E_n\frac{t^n}{n!},
     \qquad \frac{t}{e^t-1}=\sum_{n\ge0}B_n\frac{t^n}{n!},
     \qquad B_1=-\frac12.
    
E_{N,j}Use in source · 2 passages

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  • Use in source · Read passage · supplement_01.tex, line 111

    For N>0 and 0≤ j<N, the independently defined original exponential section is E_N,j(t)=∑_d≡ j (rm mod N)fract^dd!. Using the actual power-series expansion homomorphism z↦ t^N, the code proves E_N,j(t)=fract^jj!K_N,j(t^N), F_N,j(t^N)E_N,j(t)=fract^jj!. Thus the recurrence is connected back to the original normalized EGF definition. The section vanishes at zero when j>0; no inverse of that nonunit section is assumed.

    Exact mathematical passage
    
     E_{N,j}(t)=\sum_{d\equiv j\ ({\rm mod}\ N)}\frac{t^d}{d!}.
    
e_{N-1}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 100

    d formula is an integral multinomial formula, rather than a product rule with unspecified remainders. On the tensor of the two factors use the mutually commuting operators L_λ⊗1, 1⊗ L_μ, c_iD^i⊗ D^N-i (1≤ i<N). Applying multiplication after the kth power of their sum gives exactly L_λ+μ^k(AB). Expanding it yields terms indexed by a,b,e_1,…,e_N-1≥0, with a+b+∑_i e_i=k: frack!a!b!∏_i e_i!∏_i c_i^e_i (D^α L_λ^a A)(D^β L_μ^b B), α=∑_i i e_i, β=∑_i(N-i)e_i. Every scalar is integral. If A and B have t…

    Exact mathematical passage
    a,b,e_1,\ldots,e_{N-1}\geq0
E_{n-2r}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 976

    [Odd modulus Euler coordinate pair] For every positive odd q and every even n≥0, E_n = ∑_r=0^n/2 (-1)^rZ_2r+1 binom n2r q^2rS_n-2r(q). The inverse transform is S_n(q) = E_n + ∑_r=1^n/2 2^2r-1 binom n2r q^2rE_n-2r. Both transforms are lower triangular with diagonal one.

    Exact mathematical passage
    
    \boxed{
    S_n(q)
    =
    E_n
    +
    \sum_{r=1}^{n/2}
    2^{2r-1}
    \binom n{2r}
    q^{2r}E_{n-2r}.
    }
    
E_{n-r}Use in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 86

    The inverse coefficients are integral by construction. Define the integer binomial convolution B_n=∑_r=0^nbinom nr E_rE_n-r, b_m=B_2m. The product formula for actual EGFs gives ∑_n≥0B_nfract^nn!=sech^2 t. Odd coefficients vanish by parity. In particular, every b_m is an integer; this is the integrality required by the later congruence.

    Exact mathematical passage
    
     B_n=\sum_{r=0}^{n}\binom nr E_rE_{n-r},\qquad b_m=B_{2m}.
    
e_n^{(2,7)}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 449

    ction. The positive-index standard-window tail printed below is erroneous. The defining recurrence gives e_1=-1≡6, e_2=5 and e_3=-61≡2pmod7, so the correct positive tail is 6,5,2,6,5,2,…. The period-three conclusion remains correct. The original line is retained only as a corrected historical record. Take p=7 and m=2, so h=3. The recurrence gives e_n^(2,7) = 1,6,4,1,6,4,… (n≥1). The positive tail has exact period three.

    Exact mathematical passage
    
    e_n^{(2,7)}
    =
    1,6,4,1,6,4,\ldots
    \qquad(n\geq1).
    
E_{Nn+qp^r}^{(N,j)}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 170

    [Nishibuchi congruence] Let p be prime, let m>0, and suppose that mmid p-1 or m=2. Put N=mp and q=lcm(2,p-1). For every 0≤ j<N, all U_N,j(n) belong to Z_p. For r≥1 and n≥ r, E_Nn+qp^r^(N,j)≡ E_Nn^(N,j)pmodp^r. Thus the positive sufficient starting index n_0=r works uniformly. If mmid p-1, the stronger sufficient condition r≤ mn is valid.

    Exact mathematical passage
    \label{eq:arith-nishibuchi}
     E_{Nn+qp^r}^{(N,j)}\equiv E_{Nn}^{(N,j)}\pmod{p^r}.
    
E_{p,j}^\timesUse in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1841

    Fix an even address j∈{0,2,…,p-3}. For x∈ Z_p, define mathscr E_p,j^×(x) = ∑_r≥0 · (-1)^rZ_2r+1p^2r binom(p-1)x+j2r · × ∑_substack0≤ a<p 2a+1≠ p (-1)^a (2a+1)^j-2r ((2a+1)^p-1)^x.

    Exact mathematical passage
    
    \begin{aligned}
    \mathscr E_{p,j}^\times(x)
    =
    \sum_{r\geq0}
    &(-1)^rZ_{2r+1}p^{2r}
    \binom{(p-1)x+j}{2r}\\
    &\times
    \sum_{\substack{0\leq a<p\\2a+1\neq p}}
    (-1)^a
    (2a+1)^{j-2r}
    \left((2a+1)^{p-1}\right)^x.
    \end{aligned}
    
E_qUse in source · 5 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2062

    [Polyphase shift matrix] Let E be the scalar shift (Ea)_n=a_n+1, and let E_q shift the coarse index. Then P_M(Ea) = S_M(E_q) P_Ma, where S_M(E_q) = 0 · 1 · 0 · ⋯ · 0

    Exact mathematical passage
    E_q
E_rUse in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 86

    The inverse coefficients are integral by construction. Define the integer binomial convolution B_n=∑_r=0^nbinom nr E_rE_n-r, b_m=B_2m. The product formula for actual EGFs gives ∑_n≥0B_nfract^nn!=sech^2 t. Odd coefficients vanish by parity. In particular, every b_m is an integer; this is the integrality required by the later congruence.

    Exact mathematical passage
    
     B_n=\sum_{r=0}^{n}\binom nr E_rE_{n-r},\qquad b_m=B_{2m}.
    
e_{r,0}Use in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 428

    The latest continuation retains the exact Bernoulli undepletion factor e_r,q=frac1-2^2r-11-2^2r-14^q, q≥0. It is a dyadic unit, including e_r,0=1. Multiplication by this factor preserves the all-q error of the shifted Jacobi moment model. Consequently the actual model critical values satisfy, for every m≥25, v_2(Φ_r,m^act-Φ_r^act) ≥ L_m=8m-4s_2(m)-178, L_m⟶∞. The estimate includes every packet and controls motion of the critical point. There is no ceiling at the undepletion exponent 2r-25. For e…

    Exact mathematical passage
    e_{r,0}=1
e_{r,q}Use in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 425

    The latest continuation retains the exact Bernoulli undepletion factor e_r,q=frac1-2^2r-11-2^2r-14^q, q≥0. It is a dyadic unit, including e_r,0=1. Multiplication by this factor preserves the all-q error of the shifted Jacobi moment model. Consequently the actual model critical values satisfy, for every m≥25, v_2(Φ_r,m^act-Φ_r^act) ≥ L_m=8m-4s_2(m)-178, L_m⟶∞. The estimate includes every packet and controls motion of the critical point. There is no ceiling at the undepletion exponent 2r-25. For e…

    Exact mathematical passage
    
     e_{r,q}=\frac{1-2^{2r-1}}{1-2^{2r-1}4^q},\qquad q\ge0.
    
e_r^{(m,p)}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 407

    [Rational ordinary generating function] In F_p[[z]], ∑_n≥0e_n^(m,p)z^n = 1+ frac ∑_r=1^he_r^(m,p)z^r 1-z^h.

    Exact mathematical passage
    
    \boxed{
    \sum_{n\geq0}e_n^{(m,p)}z^n
    =
    1+
    \frac{
    \sum_{r=1}^{h}e_r^{(m,p)}z^r
    }{1-z^h}.
    }
    
\EcalUse in source · 3 passages

Search aliases: Ecal E mathcal

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  • Use in source · Read passage · 72_recovered.tex, line 576

    For integers N≥1 and 0≤ j<N, define E_N,j(t) = ∑_n≥0 fract^Nn+j(Nn+j)!. The forced factor t^j may be removed without changing the residue address.

    Exact mathematical passage
    
    \Ecal_{N,j}(t)
    =
    \sum_{n\geq0}
    \frac{t^{Nn+j}}{(Nn+j)!}.
    
\Ecal_{N,j}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 576

    For integers N≥1 and 0≤ j<N, define E_N,j(t) = ∑_n≥0 fract^Nn+j(Nn+j)!. The forced factor t^j may be removed without changing the residue address.

    Exact mathematical passage
    
    \Ecal_{N,j}(t)
    =
    \sum_{n\geq0}
    \frac{t^{Nn+j}}{(Nn+j)!}.
    
\EEUse in source · 4 passages

Search aliases: EE E mathcal

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  • Use in source · Read passage · supplement_11.tex, line 5

    Nishibuchi conjecture completion Checkpoint scope. This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution. 1. The complete external statement. For N>0 and 0≤ j<N, use the original unnormalized convention ∑_d≥0 E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The existing packet convention satisfies E_Nn^(N,j)=j!C_N,j(n). The formalization defines the degree-indexed coefficient…

    Exact mathematical passage
    
     \sum_{d\ge0}\EE_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
\EE_d^{(N,j)}Use in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 5

    Nishibuchi conjecture completion Checkpoint scope. This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution. 1. The complete external statement. For N>0 and 0≤ j<N, use the original unnormalized convention ∑_d≥0 E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The existing packet convention satisfies E_Nn^(N,j)=j!C_N,j(n). The formalization defines the degree-indexed coefficient…

    Exact mathematical passage
    
     \sum_{d\ge0}\EE_d^{(N,j)}\frac{t^d}{d!}
     =\left(\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!}\right)^{-1}.
    
\EE_{Nn+qp^r}^{(N,j)}Use in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 14

    Let p be prime, m>0, and assume mmid p-1 or m=2. Put N=mp and q=lcm(2,p-1). Every E_Nn^(N,j) is p-adically integral. For every r≥1, E_Nn+qp^r^(N,j)≡ E_Nn^(N,j)pmodp^r (0≤ j<N, n≥ r). In particular, the positive sufficient threshold n_0=r works uniformly in all allowed parameters.

    Exact mathematical passage
    
    \boxed{\EE_{Nn+qp^r}^{(N,j)}\equiv\EE_{Nn}^{(N,j)}\pmod{p^r}
           \qquad(0\le j<N,\ n\ge r).}
    
\EE_{Nn}^{(N,j)}Use in source · 3 passages

Search aliases: EE_Nn^(N,j) E_Nn^(N,j) EE mathcal

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  • Use in source · Read passage · supplement_11.tex, line 10

    ed original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution. 1. The complete external statement. For N>0 and 0≤ j<N, use the original unnormalized convention ∑_d≥0 E_d^(N,j)fract^dd! =(∑_n≥0fract^Nn(Nn+j)!)^-1. The existing packet convention satisfies E_Nn^(N,j)=j!C_N,j(n). The formalization defines the degree-indexed coefficients through the actual factorial-scaled series coefficients, proves their packet identification, and checks …

    Exact mathematical passage
    \EE_{Nn}^{(N,j)}=j!C_{N,j}(n)
\operatorname{Euler}Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 290

    Let L be the ordinary Euler evaluation functional on a variable E, defined by L( E^d)=Euler_d, ∑_d≥0Euler_dfract^dd! =secht. The functional is linear; it is not a multiplicative evaluation. Independent variables mean that moments of polynomials in separate variables factor. Odd moments vanish. Frobenius's congruence is v_2(∑_r=0^m(-1)^m-rbinom mr Euler_2n+2r) =m+v_2(m!) (n,m≥0). For m=0, the assertion is that every even Euler number is odd. With P=( E^2-1)/2 and γ_m(P)=P^m/m!, the case n=0 says …

    Exact mathematical passage
    
     \mathcal L(\mathcal E^d)=\operatorname{Euler}_d,
     \qquad \sum_{d\geq0}\operatorname{Euler}_d\frac{t^d}{d!}
                      =\operatorname{sech}t.
    
\operatorname{Euler}_{2n+2r}Use in source · 1 passage

Search aliases: Euler_2n+2r operatorname

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  • Use in source · Read passage · strengthening.tex, line 298

    ariable E, defined by L( E^d)=Euler_d, ∑_d≥0Euler_dfract^dd! =secht. The functional is linear; it is not a multiplicative evaluation. Independent variables mean that moments of polynomials in separate variables factor. Odd moments vanish. Frobenius's congruence is v_2(∑_r=0^m(-1)^m-rbinom mr Euler_2n+2r) =m+v_2(m!) (n,m≥0). For m=0, the assertion is that every even Euler number is odd. With P=( E^2-1)/2 and γ_m(P)=P^m/m!, the case n=0 says L(γ_m(P))∈ Z_2^× (m≥0). Introduce an independent Euler v…

    Exact mathematical passage
    
     v_2\left(\sum_{r=0}^m(-1)^{m-r}\binom mr
                     \operatorname{Euler}_{2n+2r}\right)
           =m+v_2(m!)\qquad(n,m\geq0).
     \label{eq:classical-frobenius-exact}
    
\operatorname{Euler}_dUse in source · 1 passage

Search aliases: Euler_d operatorname

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  • Use in source · Read passage · strengthening.tex, line 290

    Let L be the ordinary Euler evaluation functional on a variable E, defined by L( E^d)=Euler_d, ∑_d≥0Euler_dfract^dd! =secht. The functional is linear; it is not a multiplicative evaluation. Independent variables mean that moments of polynomials in separate variables factor. Odd moments vanish. Frobenius's congruence is v_2(∑_r=0^m(-1)^m-rbinom mr Euler_2n+2r) =m+v_2(m!) (n,m≥0). For m=0, the assertion is that every even Euler number is odd. With P=( E^2-1)/2 and γ_m(P)=P^m/m!, the case n=0 says …

    Exact mathematical passage
    
     \mathcal L(\mathcal E^d)=\operatorname{Euler}_d,
     \qquad \sum_{d\geq0}\operatorname{Euler}_d\frac{t^d}{d!}
                      =\operatorname{sech}t.
    
\operatorname{Even}Use in source · 1 passage

Search aliases: Even operatorname

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  • Use in source · Read passage · supplement_03.tex, line 129

    Both row indexing by r and the literal even-degree version with hypothesis Even(n) are checked. All rows include degree zero. The exact identity is a rational equality with the Euler coefficient's integer-to-rational cast made explicit.

    Exact mathematical passage
    \operatorname{Even}(n)
\operatorname{EvenPart}Use in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 59

    Even projection is identified coefficientwise. For any formal series f, let EvenPart(f)=(f(t)+f(-t))/2. Its nth coefficient is proved to be the coefficient of f when n is even and zero otherwise. Rescaling the inverse law proves sech(-t)=sech(t). The exponential multiplication identity then gives cosh^2(pt)=frac2+e^2pt+e^-2pt4, and hence ∑_r≥0S_2r(p)fract^2r(2r)! =cosh^2(pt) sech t. Both the full and even EGF formulas are derived from the original integer sum. Neither is supplied as a hypothesis…

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    \operatorname{EvenPart}(f)=(f(t)+f(-t))/2
\operatorname{Ext}Use in source · 2 passages

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  • Use in source · Read passage · 50_open_questions.tex, line 158

    The accompanying nolinkurlnative/Native_Conjectures_Progress.md records further results in restricted settings: an intrinsic finite Ext^1 obstruction for finitely presented commutative integer algebras, finite CRT reconstruction for polynomial solution sets, and a characterization of rectangular finite relations by equality of common labels. The obstruction result is a stated specialization of a classical external theorem; the CRT and rectangularity results have prose proofs in the note. These a…

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    \operatorname{Ext}^1
\operatorname{Ext}^1Use in source · 1 passage

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  • Use in source · Read passage · 50_open_questions.tex, line 158

    The accompanying nolinkurlnative/Native_Conjectures_Progress.md records further results in restricted settings: an intrinsic finite Ext^1 obstruction for finitely presented commutative integer algebras, finite CRT reconstruction for polynomial solution sets, and a characterization of rectangular finite relations by equality of common labels. The obstruction result is a stated specialization of a classical external theorem; the CRT and rectangularity results have prose proofs in the note. These a…

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    \operatorname{Ext}^1
\operatorname{Ext}^1_BUse in source · 1 passage

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  • Use in source · Read passage · 71_native.tex, line 110

    o(f)∈Ext^1_B(NL_B/ Z, F_p)

    Exact mathematical passage
    o(f)\in\operatorname{Ext}^1_B(\mathrm{NL}_{B/\mathbb Z},\mathbb F_p)
\FUse in source · 24 passages

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  • Use in source · Read passage · 10_foundations.tex, line 173

    begingroupsmallraggedright Formal correspondence. leanHurwitz.shift_convolution, leanHurwitz.prime_shift_convolution, leanHurwitz.convolution_left_eq_zero, leanHurwitzInverse.inverse_convolution, and leanHurwitzInverse.inverse_prime_difference. Working directly with coefficient sequences avoids the undefined expression 1/n! in F_p when pmid n!. endgroup

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    \F_p
\mathbb FUse in source · 32 passages

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  • Use in source · Read passage · 71_native.tex, line 23

    The divisions are exact integer divisions. Regard J as a linear map F_p^d→ F_p^r. Define

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    \mathbb F_p^d\to\mathbb F_p^r
\mathcal FUse in source · 17 passages

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  • Use in source · Read passage · 30_lifting.tex, line 848

    Use the augmented polynomial systems F(Y,T)=(F(Y,T),T-t_0), H(Y,T)=(H(Y,T),-v). Then F(a)=0, and its derivative is B=D F(a)=J · K

    Exact mathematical passage
    
     \mathcal F(Y,T)=(F(Y,T),T-t_0),\qquad
     \mathcal H(Y,T)=(H(Y,T),-v).
    
F'Use in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 517

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

    Exact mathematical passage
    \label{eq:arith-charged-product}
     A_{c+d}(FG)=(A_cF)G+F(A_dG)
           +2F'G'''+3F''G''+2F'''G'.
    
F''Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 517

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

    Exact mathematical passage
    \label{eq:arith-charged-product}
     A_{c+d}(FG)=(A_cF)G+F(A_dG)
           +2F'G'''+3F''G''+2F'''G'.
    
F'''Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 517

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

    Exact mathematical passage
    \label{eq:arith-charged-product}
     A_{c+d}(FG)=(A_cF)G+F(A_dG)
           +2F'G'''+3F''G''+2F'''G'.
    
f^{-1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 79

    The reciprocal is genuinely integral. For f∈ I_p with constant coefficient one, the usual power-series reciprocal is used. Its inverse equation yields the exact coefficient recurrence a_f^-1(d)=δ_d0-∑_r=0^d-1 binom dr a_f^-1(r)a_f(d-r). Strong induction proves that all of these coefficients are integral. Combining this construction with the preceding all-order theorem proves flatness of the actual reciprocal. No flatness or integrality of that reciprocal is assumed in the normalized-series speci…

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     a_{f^{-1}}(d)=\delta_{d0}-\sum_{r=0}^{d-1}
         \binom dr a_{f^{-1}}(r)a_f(d-r).
    
\mathcal F^2Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 313

    0). For m=0, the assertion is that every even Euler number is odd. With P=( E^2-1)/2 and γ_m(P)=P^m/m!, the case n=0 says L(γ_m(P))∈ Z_2^× (m≥0). Introduce an independent Euler variable F and Q=( F^2-1)/2.

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    Q=(\mathcal F^2-1)/2
f_0Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 842

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

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    c=f_0-f_h
f_1Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 861

    ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h. It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by f_0,f_1,…,f_h is injective and hence bijective.

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    f_0,f_1,\ldots,f_h
F_{1,0}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 200

    j(n)∈ p^k Z_p. Moreover, Δ_1^k A_N,j(n)=(-1)^n+k(S+1)^k C_N,j(n), so A_N,j has the ordinary p^k difference property. At p=2, L_1=L_-1-1; its integral binomial expansion on F_N,j gives Δ_1^k C_N,j(n)∈2^k Z_2 directly. For N=1 the interior sum in the product formula is empty, so the same argument applies without change; equivalently, F_1,0(t)=e^-t.

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    F_{1,0}(t)=e^{-t}
F_{11}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 159

    ndary checksExample 9.3 and supporting results The independently defined Lucas recurrence gives (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11). For N=3, the general theorem therefore specializes to u_k+2=4u_k+1+u_k. The implementation proves this for every offset j and every k, not only the displayed sample. At j=2, direct evaluation of the actual Fibonacci sequence gives (F_2,F_5,F_8,F_11,F_14,F_17)=(1,5,21,89,377,1597). This is the exact manuscript example.

    Exact mathematical passage
    
     (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).
    
F_{14}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 159

    ndary checksExample 9.3 and supporting results The independently defined Lucas recurrence gives (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11). For N=3, the general theorem therefore specializes to u_k+2=4u_k+1+u_k. The implementation proves this for every offset j and every k, not only the displayed sample. At j=2, direct evaluation of the actual Fibonacci sequence gives (F_2,F_5,F_8,F_11,F_14,F_17)=(1,5,21,89,377,1597). This is the exact manuscript example.

    Exact mathematical passage
    
     (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).
    
F_{17}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 159

    ndary checksExample 9.3 and supporting results The independently defined Lucas recurrence gives (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11). For N=3, the general theorem therefore specializes to u_k+2=4u_k+1+u_k. The implementation proves this for every offset j and every k, not only the displayed sample. At j=2, direct evaluation of the actual Fibonacci sequence gives (F_2,F_5,F_8,F_11,F_14,F_17)=(1,5,21,89,377,1597). This is the exact manuscript example.

    Exact mathematical passage
    
     (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).
    
F_2Use in source · 4 passages

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  • Use in source · Read passage · supplement_06.tex, line 159

    ndary checksExample 9.3 and supporting results The independently defined Lucas recurrence gives (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11). For N=3, the general theorem therefore specializes to u_k+2=4u_k+1+u_k. The implementation proves this for every offset j and every k, not only the displayed sample. At j=2, direct evaluation of the actual Fibonacci sequence gives (F_2,F_5,F_8,F_11,F_14,F_17)=(1,5,21,89,377,1597). This is the exact manuscript example.

    Exact mathematical passage
    
     (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).
    
\mathbb F_2Use in source · 4 passages

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  • Use in source · Read passage · root_scope.tex, line 140

    ad J_r={q:λ_r(q)=μ_r}, G_r=2^-μ_rH_r. Since λ_r(0)=0 and λ_r(q)≥3q+1-s_2(r), every minimizing address satisfies 0≤ q≤lfloorfracs_2(r)-13rfloor. In particular μ_r≤0, and the minimum is found by a finite carry calculation. Reduction of the whole series gives overline G_r(X)=∑_q∈ J_rX^underline q in F_2[X]. Its highest degree is max J_r, with leading coefficient one. Thus the normalization is the actual Gauss normalization.

    Exact mathematical passage
    
     \overline G_r(X)=\sum_{q\in J_r}X^{\underline q}
           \quad\text{in }\mathbb F_2[X].
     \label{eq:root:reduction}
    
F_3Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 739

    e a^-2+b^-2=0. Its constant is one, only degrees divisible by four survive, and the Bernoulli expansion gives D_3(t)=∑_i≥0d_ifract^4i(4i)!, d_i=3(-frac14)^i frac2^4i+2B_4i+2(4i+2)(4i+1). Again v_2(d_i)≥2i, and the inverse coefficients h_i have the same bound. Since F_3=F_2D_3^-1, C_3(n)-C_2(n)=∑_i≥1h_ibinom4n4iC_2(n-i). The scaled-binomial argument [source reference] applies unchanged. The basic bound for C_2 includes order zero; its continuous extension is available from the threshold-free theo…

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    F_3=F_2D_3^{-1}
\F_3Use in source · 1 passage

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  • Use in source · Read passage · supplement_07.tex, line 133

    Three checked boundaries. The initial row. For (p,m)=(3,2), the same reduced recurrence gives e_0=1 and e_1=-1=2 in F_3. Its proposed period is one, so the positive-tail result cannot in general include row zero.

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    \F_3
\mathbb F_3Use in source · 1 passage

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  • Use in source · Read passage · 71_native.tex, line 78

    setlengthitemsep0ptsetlengthparskip0pt item A failed singular lift: p=3, L=1, F(X)=X^2-3, x=0. Here e=-1≠0 in F_3 and J=0. There are no lifts of zero modulo 3 to a root modulo 9. item Representative independence is a quotient assertion: for F(X)=X^2-1, p=3, L=1, representatives 1 and 4 have residual digits 0 and 2. Their common Jacobian is 2, with full image, and both give zero obstruction. Their corrected representatives 1 and 10 are the same root modulo 9. item A successful step need not give …

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    \mathbb F_3
F_{3k+j}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2810

    [t] centering includegraphics[width=.95linewidth,height=3.15in,keepaspectratio]../../figures/reconstructed/recurrence_packets.pdf captionNew reconstruction. Fibonacci residues F_3k+j, with F_0=0 and F_1=1. Each packet evolves by the same jump matrix C^3 and has a different initial address.

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    F_{3k+j}
F_{4,j}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 252

    Write C_j=C_4,j and F_j=F_4,j. The all-row theorem removes the threshold, but at N=4 it leaves a further factorial gain unused. The exact algebra below proves that gain for all four addresses and identifies several infinite families where it is sharp.

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    F_j=F_{4,j}
F_5Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 159

    ndary checksExample 9.3 and supporting results The independently defined Lucas recurrence gives (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11). For N=3, the general theorem therefore specializes to u_k+2=4u_k+1+u_k. The implementation proves this for every offset j and every k, not only the displayed sample. At j=2, direct evaluation of the actual Fibonacci sequence gives (F_2,F_5,F_8,F_11,F_14,F_17)=(1,5,21,89,377,1597). This is the exact manuscript example.

    Exact mathematical passage
    
     (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).
    
F_8Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 159

    ndary checksExample 9.3 and supporting results The independently defined Lucas recurrence gives (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11). For N=3, the general theorem therefore specializes to u_k+2=4u_k+1+u_k. The implementation proves this for every offset j and every k, not only the displayed sample. At j=2, direct evaluation of the actual Fibonacci sequence gives (F_2,F_5,F_8,F_11,F_14,F_17)=(1,5,21,89,377,1597). This is the exact manuscript example.

    Exact mathematical passage
    
     (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).
    
f_{a+b,s+p^at}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2115

    [Exact local branch reconstruction] The family {f_a,s}_0≤ s<p^a determines f uniquely on Z_p. More precisely, if y∈ s+p^a Z_p, then f(y)=f_a,s(fracy-sp^a). For a,b≥1, (f_a,s)_b,t = f_a+b,s+p^at. Thus local refinement is exactly the p adic digit tree.

    Exact mathematical passage
    
    \boxed{
    (f_{a,s})_{b,t}
    =
    f_{a+b,s+p^at}.
    }
    
f_{a,s}Use in source · 4 passages

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F_dUse in source · 1 passage

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  • Use in source · Read passage · continuation_06.tex, line 12

    Let p be prime, R= Z_p, K= Q_p, and d,s,r∈ N. Let F=(F_1,…,F_d)∈ R[X_1,…,X_d]^d, a∈ R^d, and let J=DF(a)=(∂_jF_i(a))_i,j. Assume J∈GL_d(K), p^sJ^-1∈ M_d(R), F(a)∈ p^rR^d, r>2s. Then ∃! b∈ a+p^s+1R^d: F(b)=0, b-a∈ p^r-sR^d. All vector divisibility assertions are coordinatewise. The uniqueness predicate is the entire ball of depth s+1: competitors need not be assumed to have depth r-s. The theorem permits arbitrary integral p-adic coefficients and centers, every prime including two, and dimension …

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    F=(F_1,\ldots,F_d)\in R[X_1,\ldots,X_d]^d
f_hUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 842

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

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    d_0=f_h
F_jUse in source · 2 passages

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  • Use in source · Read passage · supplement_06.tex, line 174

    Uniqueness from the first two packet values. For a fixed N,j, any integer sequence satisfying the proved sampled recurrence and starting at F_j,F_N+j agrees with F_Nk+j for all k. This is proved by two-step induction. It is a consequence of the established coefficients and recurrence, not a premise used to identify Fibonacci values initially.

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    F_j,F_{N+j}
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  • Use in source · Read passage · 20_arithmetic.tex, line 1045

    [Isometric packet bijections] For 0≤ j<4, define on natural numbers f_j(n)=fracU_j(n)-j!2^d_j, (d_0,d_1,d_2,d_3)=(1,1,5,3). There is a unique continuous map Φ_j: Z_2→ Z_2 whose value at every n∈ N equals the embedded rational f_j(n). It is an isometric bijection and Φ_j(0)=0: |Φ_j(x)-Φ_j(y)|_2=|x-y|_2 (x,y∈ Z_2). In particular, every element of Z_2 has exactly one preimage.

    Exact mathematical passage
    
     f_j(n)=\frac{U_j(n)-j!}{2^{d_j}},\qquad (d_0,d_1,d_2,d_3)=(1,1,5,3).
    
F_NUse in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 126

    n independent Lucas sequence Define L_0=2, L_1=1, L_n+2=L_n+L_n+1. Direct finite matrix arithmetic proves C^2=C+I. Multiplication by C^n gives C^n+2=C^n+C^n+1. Taking traces shows that tr(C^n) satisfies the same recurrence and initial values as L_n. Induction yields tr(C^N)=L_N. The implementation separately proves det C=-1, det(C^N)=(-1)^N, L_N=2F_N+1-F_N. No Binet roots or diagonalizability argument is needed.

    Exact mathematical passage
    
     \det C=-1,\qquad \det(C^N)=(-1)^N,
     \qquad L_N=2F_{N+1}-F_N.
    
f_nUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 830

    [Fourier invariance of the standard window operator] Let p be odd, let mmid p-1, and let h=(p-1)/m. Define (T_m,pf)_n = ∑_q=0^nbinommnmqf_q in F_p. Let V_h be the vector space of sequences satisfying f_n+h=f_n (n≥1). Then T_m,p(V_h)=V_h. The restriction of T_m,p to V_h is an automorphism.

    Exact mathematical passage
    
    f_{n+h}=f_n
    \qquad(n\geq1).
    
F_{N+1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 126

    n independent Lucas sequence Define L_0=2, L_1=1, L_n+2=L_n+L_n+1. Direct finite matrix arithmetic proves C^2=C+I. Multiplication by C^n gives C^n+2=C^n+C^n+1. Taking traces shows that tr(C^n) satisfies the same recurrence and initial values as L_n. Induction yields tr(C^N)=L_N. The implementation separately proves det C=-1, det(C^N)=(-1)^N, L_N=2F_N+1-F_N. No Binet roots or diagonalizability argument is needed.

    Exact mathematical passage
    
     \det C=-1,\qquad \det(C^N)=(-1)^N,
     \qquad L_N=2F_{N+1}-F_N.
    
f_{n+h}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 830

    [Fourier invariance of the standard window operator] Let p be odd, let mmid p-1, and let h=(p-1)/m. Define (T_m,pf)_n = ∑_q=0^nbinommnmqf_q in F_p. Let V_h be the vector space of sequences satisfying f_n+h=f_n (n≥1). Then T_m,p(V_h)=V_h. The restriction of T_m,p to V_h is an automorphism.

    Exact mathematical passage
    
    f_{n+h}=f_n
    \qquad(n\geq1).
    
F_{N+j}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 174

    Uniqueness from the first two packet values. For a fixed N,j, any integer sequence satisfying the proved sampled recurrence and starting at F_j,F_N+j agrees with F_Nk+j for all k. This is proved by two-step induction. It is a consequence of the established coefficients and recurrence, not a premise used to identify Fibonacci values initially.

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    F_j,F_{N+j}
F_{N,0}Use in source · 2 passages

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  • Use in source · Read passage · supplement_10.tex, line 171

    Modules. mcodeEulerSeriesInverse, mcodeRiccatiFlatPropagation, mcodeRatioRiccati. 4. All-residue assembly and original coefficients. The standard-window source and inverse results supply R_1 and F_N,0. Induction over 1≤ c<N now has two proved cases. At pmid c, the singular weighted estimate and the new substitution theorem give flatness. At pnmid c, the preceding Riccati propagation applies. Thus every actual ratio R_c is flat.

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    F_{N,0}
F_{N,j}Use in source · 10 passages

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  • Use in source · Read passage · supplement_01.tex, line 104

    The formalization does not stop at a recurrence. In the variable z=t^N, construct K_N,j(z)=∑_n≥0fracj!(Nn+j)!z^n, F_N,j(z)=∑_n≥0fracC_N,j(n)(Nn)!z^n. It proves the factorial coefficient identity, then F_N,jK_N,j=1 and therefore F_N,j=K_N,j^-1.

    Exact mathematical passage
    
     K_{N,j}(z)=\sum_{n\ge0}\frac{j!}{(Nn+j)!}z^n,\qquad
     F_{N,j}(z)=\sum_{n\ge0}\frac{C_{N,j}(n)}{(Nn)!}z^n.
    
F_{N,j+1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_10.tex, line 178

    Let K_j(t) be the normalized section kernel with constant coefficient one. The constructed unnormalized Euler series is F_N,j(t)=j! K_j(t)^-1. The source/inverse identities and the factorial successor identity prove F_N,j+1=F_N,jR_j+1, F_N,j=F_N,0∏_c=1^jR_c. The product formula is not an assembly hypothesis. Proved flatness of all factors and the original product-closure theorem establish flatness of the full family.

    Exact mathematical passage
    
     F_{N,j+1}=F_{N,j}R_{j+1},\qquad
     F_{N,j}=F_{N,0}\prod_{c=1}^{j}R_c.
    
\F_pUse in source · 18 passages

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  • Use in source · Read passage · 10_foundations.tex, line 173

    begingroupsmallraggedright Formal correspondence. leanHurwitz.shift_convolution, leanHurwitz.prime_shift_convolution, leanHurwitz.convolution_left_eq_zero, leanHurwitzInverse.inverse_convolution, and leanHurwitzInverse.inverse_prime_difference. Working directly with coefficient sequences avoids the undefined expression 1/n! in F_p when pmid n!. endgroup

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    \F_p
\mathbb F_pUse in source · 12 passages

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  • Use in source · Read passage · 71_native.tex, line 95

    odulo 2 has zero obstruction and lifts to 1,3 modulo 4. Both have nonzero obstruction at the next step. There is no root modulo 8 and no compatible infinite lift. item The raw Jacobian cokernel need not be presentation independent: the presentations F=(X) and G=(X,0) define the same solution system. Their raw obstruction groups at zero are respectively zero and F_p. Both actual distinguished obstructions are zero. Thus one cannot claim a canonical isomorphism of these two raw groups merely becau…

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    \mathbb F_p
\F_p^2Use in source · 1 passage

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  • Use in source · Read passage · 50_open_questions.tex, line 64

    The raw cokernel already reveals the difficulty. The presentations F=(X) and G=(X,0) describe the same zero locus over every coefficient ring. At zero their Jacobians are, respectively, F_pxrightarrow 1 F_p, F_pxrightarrow (1,0) F_p^2. Their raw cokernels are 0 and F_p. Both distinguished obstruction classes vanish, but the ambient obstruction groups are not isomorphic. The example rules out the raw cokernel as a presentation-independent object while leaving open the possibility of an intrinsic …

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    \F_p\xrightarrow{\,1\,}\F_p,
    \qquad
    \F_p\xrightarrow{\,(1,0)\,}\F_p^2.
    
\F_p^dUse in source · 1 passage

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  • Use in source · Read passage · 50_open_questions.tex, line 53

    The original conjecture asks for a canonical finite obstruction packet whose vanishing is necessary and sufficient for a one-level lift in a finitely presented arithmetic seed system. For a fixed finite polynomial presentation, the paper now supplies the precise obstruction o_F,bar x=[r]∈ coker (J_F,bar x: F_p^d⟶ F_p^e), where r is the residual digit. We have proved independence from the chosen integer representative. Invariance under a change of equations or presentation requires more.

    Exact mathematical passage
    
    o_{F,\bar x}=[r]\in
    \coker\!\left(J_{F,\bar x}:\F_p^d\longrightarrow\F_p^e\right),
    
\mathbb F_p^dUse in source · 7 passages

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  • Use in source · Read passage · 71_native.tex, line 23

    The divisions are exact integer divisions. Regard J as a linear map F_p^d→ F_p^r. Define

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    \mathbb F_p^d\to\mathbb F_p^r
\F_p^eUse in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 314

    For prime p, rank-nullity gives dimker J_x=d-ρ_x and dim( F_p^e/im J_x)=e-ρ_x. The cardinality of an n-dimensional F_p-space is p^n. Combine these identities with the obstruction criterion and the residue equivalence.

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    \dim(\F_p^e/\im J_x)=e-\rho_x
\mathbb F_p^eUse in source · 8 passages

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  • Use in source · Read passage · continuation_02.tex, line 19

    Let p be prime, L≥1, and F∈ Z[X_1,…,X_d]^e, x∈ Z^d, F(x)≡0pmodp^L. Write r_x=F(x)/p^L, b_x=r_xbmod p, A_x=DF(x)bmod p: F_p^d⟶ F_p^e, ρ_x=rank_ F_pA_x. The actual lift set is L_x,L= {y∈( Z/p^L+1 Z)^d: y≡ xpmodp^L, F(y)=0}, where polynomial evaluation in the displayed set is modulo p^L+1. Define O_x= F_p^e/imA_x, o_x=[b_x]. Lean proves # L_x,L= p^d-ρ_x, · o_x=0,

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    r_x=F(x)/p^L,\qquad b_x=r_x\bmod p,
    \qquad A_x=DF(x)\bmod p:\mathbb F_p^d\longrightarrow\mathbb F_p^e,
    \qquad \rho_x=\operatorname{rank}_{\mathbb F_p}A_x.
    
\mathbb F_p^{e+q}Use in source · 1 passage

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  • Use in source · Read passage · continuation_05.tex, line 16

    Let p be prime, L≥1, e,q≥0, and F∈ Z[X_1,…,X_e+q]^e, x∈ Z^e+q, F(x)≡0pmodp^L. Assume the actual reduced Jacobian A_x=DF(x)bmod p: F_p^e+q⟶ F_p^e is surjective. All vector congruences are coordinatewise.

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    A_x=DF(x)\bmod p:\mathbb F_p^{e+q}\longrightarrow\mathbb F_p^e
    
\F_p^kUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 635

    ints] Let q=p be prime in Theorem~[source reference]. For every τ∈ t_0+p^L Z_p^k, there are an integer tower z_n and b∈ Z_p^m+k such that F(b)=0, free(b)=τ, z_n⟶ b, b-z_n∈ p^L+n Z_p^m+k. The initial value is z_0=x. At every stage and for every proposed parameter digit v∈ F_p^k, the corresponding next residue fiber has exactly p^m-e elements. If m=e, then for each τ there is exactly one such root in B_L(x), independently of all choices of integer approximants and correcting digits.

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    v\in\F_p^k
\mathbb F_p^kUse in source · 1 passage

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  • Use in source · Read passage · continuation_09.tex, line 59

    Exact branching. At any stage n, put N=L+n. For every parameter digit v∈ F_p^k, let L_n,v={bar w∈( Z/p^N+1 Z)^m+k: l bar w≡ z(n)pmodp^N,

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    v\in\mathbb F_p^k
\mathbb F_p^rUse in source · 5 passages

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  • Use in source · Read passage · 71_native.tex, line 23

    The divisions are exact integer divisions. Regard J as a linear map F_p^d→ F_p^r. Define

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    \mathbb F_p^d\to\mathbb F_p^r
\F_p^\timesUse in source · 1 passage

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  • Use in source · Read passage · 10_foundations.tex, line 346

    Induction gives the geometric identity. Reduction of a unit to F_p^× gives u_i^p-1=1 modulo p. Taking s powers supplies p^s in each summand. The difference operator is linear over finite sums.

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    \F_p^\times
f_qUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 823

    [Fourier invariance of the standard window operator] Let p be odd, let mmid p-1, and let h=(p-1)/m. Define (T_m,pf)_n = ∑_q=0^nbinommnmqf_q in F_p. Let V_h be the vector space of sequences satisfying f_n+h=f_n (n≥1). Then T_m,p(V_h)=V_h. The restriction of T_m,p to V_h is an automorphism.

    Exact mathematical passage
    
    (T_{m,p}f)_n
    =
    \sum_{q=0}^{n}\binom{mn}{mq}f_q
    \quad\text{in }\FF_p.
    
f_RUse in source · 2 passages

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  • Use in source · Read passage · 71_native.tex, line 251

    The component maps f_R:A→ C_R and g_R:B→ C_R have the following universal property. For every set Y and maps h:A→ Y, k:B→ Y satisfying h(a)=k(b) whenever R(a,b), there is a unique map u:C_R→ Y with u∘ f_R=h and u∘ g_R=k.

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    f_R:A\to C_R
F_r'Use in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 88

    ip 2. Connection to the actual section. An integral coordinate estimate becomes relevant only after connecting it to the manuscript's generating functions. For a residue r∈ Z/N Z, define H_r(t)=∑_d≥01_d≡ r (N)fract^dd!, F_r(T)=H_r(log(1+T)). The implementation uses the actual formal logarithm and substitution in Q[[T]]. It proves H_r'=H_r-1, (1+T)F_r'=F_r-1, and hence the coefficient recurrence (d+1)coefTd+1F_r =coefTdF_r-1-dcoefTdF_r. The initial values are F_r(0)=1_r=0.

    Exact mathematical passage
    
     H_r'=H_{r-1},\qquad (1+T)F_r'=F_{r-1},
    
F_{r-1}Use in source · 2 passages

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  • Use in source · Read passage · supplement_09.tex, line 88

    ip 2. Connection to the actual section. An integral coordinate estimate becomes relevant only after connecting it to the manuscript's generating functions. For a residue r∈ Z/N Z, define H_r(t)=∑_d≥01_d≡ r (N)fract^dd!, F_r(T)=H_r(log(1+T)). The implementation uses the actual formal logarithm and substitution in Q[[T]]. It proves H_r'=H_r-1, (1+T)F_r'=F_r-1, and hence the coefficient recurrence (d+1)coefTd+1F_r =coefTdF_r-1-dcoefTdF_r. The initial values are F_r(0)=1_r=0.

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     H_r'=H_{r-1},\qquad (1+T)F_r'=F_{r-1},
    
\fallUse in source · 5 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1113

    [Dyadic Euler coordinate pair] For every even n≥0, E_n = ∑_r=0^n/2 β_r2^2r fall n2r R_n-2r. The inverse transform is R_n = ∑_r=0^n/2 frac2^4r+1(2r+2)! fall n2r E_n-2r.

    Exact mathematical passage
    
    \boxed{
    E_n
    =
    \sum_{r=0}^{n/2}
    \beta_r2^{2r}
    \fall n{2r}
    R_{n-2r}.
    }
    
\FF_3Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2724

    [A singular polynomial obstruction] Let p=3 and consider the class x_1=0 modulo three. For F_0(x)=x^2, the residual and the reduced derivative both vanish. Every correction y∈ F_3 gives F_0(3y)≡0pmod9, so the class has three children. For F_1(x)=x^2+3, the reduced derivative is still zero, but the reduced residual is one. Its class in the cokernel is nonzero, so no lift modulo nine exists.

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    y\in\FF_3
\FF_p^dUse in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2551

    [One level linear lifting] A class represented by x_L lifts to a solution modulo p^L+1 if and only if Ob^lin_L(x_L)=0. More precisely, a correction y∈ F_p^d gives a lift x_L+1=x_L+p^Ly pmod p^L+1 if and only if overline My=-ρ_L(x_L). When one correction exists, the set of all correction classes is a coset of keroverline M.

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    y\in\FF_p^d
\FF_p^\timesUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 251

    Fix a divisor m of p-1. Let H_m = {ζ∈ F_p^×:ζ^m=1}. This group has order m.

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    H_m
    =
    \{\zeta\in\FF_p^\times:\zeta^m=1\}.
    
\operatorname{Flat}Use in source · 2 passages

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  • Use in source · Read passage · supplement_08.tex, line 46

    All orders, without a circular closure assumption. Let I⊂ A be a subring stable under U. Define Bounded_r(f)iff B^s f∈ I (0≤ s≤ r), Flat(f)iff B^s f∈ I (s≥0). The definitions contain no product or inverse closure requirement. Since B commutes with U, derivatives of a bounded element remain bounded. Induction on r in the displayed product rule then proves closure under multiplication. The finite middle sums use the lower-order induction hypothesis, not the desired next-order conclusion.

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     \operatorname{Bounded}_r(f)\iff B^s f\in I\ (0\le s\le r),
     \qquad \operatorname{Flat}(f)\iff B^s f\in I\ (s\ge0).
    
\operatorname{free}Use in source · 11 passages

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  • Use in source · Read passage · 30_lifting.tex, line 461

    ∈ p^L Z^e. Suppose the full reduced Jacobian is surjective. There is a permutation σ of the coordinates such that, writing σ x=(y_0,t_0) with e solved coordinates and k parameter coordinates, the parameter ball T=t_0+p^L Z_p^k is homeomorphic to the entire root locus R={z∈ B_L(x):F(z)=0}. The inverse homeomorphism is the actual free-coordinate projection z↦free(σ z). If h:T→ R is the forward map, then for every N≥0, h(t)≡ h(u)pmodp^N Longleftrightarrow t≡ upmodp^N. Equivalently, σ h(t)=(ϕ(t),t) …

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    z\mapsto\operatorname{free}(\sigma z)
GUse in source · 92 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 517

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

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    \label{eq:arith-charged-product}
     A_{c+d}(FG)=(A_cF)G+F(A_dG)
           +2F'G'''+3F''G''+2F'''G'.
    
G'Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 517

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

    Exact mathematical passage
    \label{eq:arith-charged-product}
     A_{c+d}(FG)=(A_cF)G+F(A_dG)
           +2F'G'''+3F''G''+2F'''G'.
    
G''Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 517

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

    Exact mathematical passage
    \label{eq:arith-charged-product}
     A_{c+d}(FG)=(A_cF)G+F(A_dG)
           +2F'G'''+3F''G''+2F'''G'.
    
G'''Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 517

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

    Exact mathematical passage
    \label{eq:arith-charged-product}
     A_{c+d}(FG)=(A_cF)G+F(A_dG)
           +2F'G'''+3F''G''+2F'''G'.
    
G^2Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 525

    d product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting from the independently established integrality of the reciprocal G, proves that inversion negates the charge. Finally A_c+2=A_c-1, so increasing …

    Exact mathematical passage
    
     A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').
    
g^2Use in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 53

    For an actual inverse fg=1, applying the product rule and multiplying by g gives B(g)=-g^2B(f)-gM_p^2(f,g)+gM_p(f,g). If g∈ I, induction on r proves boundedness of g from boundedness of f. At each successor stage the preceding stage supplies boundedness of g needed on the right. Passing to all finite orders proves reciprocal closure for an integral inverse.

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     B(g)=-g^2B(f)-gM_{p^2}(f,g)+gM_p(f,g).
    
g^{b-a}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 1074

    To prove surjectivity, use compactness. For an isometry g of a compact metric space into itself, the sequence of iterates g^n(x) has a convergent subsequence. Two sufficiently close iterates, with exponents a<b, yield d(x,g^b-a(x)) equally small after removing a isometric iterations. Since b-a>0, the latter point is in the image of g. Thus the image is dense. It is also compact and hence closed, so it is the whole space. Apply this to Φ_j on Z_2.

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    d(x,g^{b-a}(x))
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  • Use in source · Read passage · 20_arithmetic.tex, line 1072

    To prove surjectivity, use compactness. For an isometry g of a compact metric space into itself, the sequence of iterates g^n(x) has a convergent subsequence. Two sufficiently close iterates, with exponents a<b, yield d(x,g^b-a(x)) equally small after removing a isometric iterations. Since b-a>0, the latter point is in the image of g. Thus the image is dense. It is also compact and hence closed, so it is the whole space. Apply this to Φ_j on Z_2.

    Exact mathematical passage
    g^n(x)
g_0Use in source · 7 passages

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  • Use in source · Read passage · 10_foundations.tex, line 147

    [Hurwitz product rules] For d≥0, D^d(f⋆ g)=∑_r=0^dbinom dr(D^r f)⋆(D^d-rg). If R has prime characteristic p, then D^p(f⋆ g)=f⋆ D^p g+D^p f⋆ g. Over a field, if g_0=1, a sequence b is constructed recursively with b⋆ g=1. If also D^p g=Dg in characteristic p, then D^p b=Db.

    Exact mathematical passage
    g_0=1
g_1Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 488

    The Hurwitz coefficients of the normalized source K_4,j at degrees 4n are g_j(n)=binom4n+j4n^-1. Their explicit expressions are g_0(n) · =1, · g_1(n) · =frac14n+1, g_2(n) · =frac1(4n+1)(2n+1), · g_3(n) · =frac3(4n+1)(2n+1)(4n+3). Every denominator is odd. An affine sequence with even slope satisfies 2^kmidΔ_1^k f(n) for every k,n. The exact product identity Δ_1^s(fg)(n)=∑_i=0^sbinom si Δ_1^if(n)Δ_1^s-ig(n+i) proves closure of this property under products. Applying the formula with the factors in…

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     \begin{aligned}
     g_0(n)&=1,&
     g_1(n)&=\frac1{4n+1},\\
     g_2(n)&=\frac1{(4n+1)(2n+1)},&
     g_3(n)&=\frac3{(4n+1)(2n+1)(4n+3)}.
     \end{aligned}
    
g_2Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 488

    The Hurwitz coefficients of the normalized source K_4,j at degrees 4n are g_j(n)=binom4n+j4n^-1. Their explicit expressions are g_0(n) · =1, · g_1(n) · =frac14n+1, g_2(n) · =frac1(4n+1)(2n+1), · g_3(n) · =frac3(4n+1)(2n+1)(4n+3). Every denominator is odd. An affine sequence with even slope satisfies 2^kmidΔ_1^k f(n) for every k,n. The exact product identity Δ_1^s(fg)(n)=∑_i=0^sbinom si Δ_1^if(n)Δ_1^s-ig(n+i) proves closure of this property under products. Applying the formula with the factors in…

    Exact mathematical passage
    
     \begin{aligned}
     g_0(n)&=1,&
     g_1(n)&=\frac1{4n+1},\\
     g_2(n)&=\frac1{(4n+1)(2n+1)},&
     g_3(n)&=\frac3{(4n+1)(2n+1)(4n+3)}.
     \end{aligned}
    
g_3Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 488

    The Hurwitz coefficients of the normalized source K_4,j at degrees 4n are g_j(n)=binom4n+j4n^-1. Their explicit expressions are g_0(n) · =1, · g_1(n) · =frac14n+1, g_2(n) · =frac1(4n+1)(2n+1), · g_3(n) · =frac3(4n+1)(2n+1)(4n+3). Every denominator is odd. An affine sequence with even slope satisfies 2^kmidΔ_1^k f(n) for every k,n. The exact product identity Δ_1^s(fg)(n)=∑_i=0^sbinom si Δ_1^if(n)Δ_1^s-ig(n+i) proves closure of this property under products. Applying the formula with the factors in…

    Exact mathematical passage
    
     \begin{aligned}
     g_0(n)&=1,&
     g_1(n)&=\frac1{4n+1},\\
     g_2(n)&=\frac1{(4n+1)(2n+1)},&
     g_3(n)&=\frac3{(4n+1)(2n+1)(4n+3)}.
     \end{aligned}
    
G_{4,1}Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 437

    The address-one source is G_4,1(t)=(sinh t+sin t)/(2t). The identities a+b=1, a-b=i, and ab=1/2 give D_1(t):=G_4,1(t)F_0(t) =frac12(fractanh(at)at +fractanh(bt)bt). For example, its numerator follows from sinh t+sin t =2bsinh(at)cosh(bt)+2acosh(at)sinh(bt), while cosh t+cos t=2cosh(at)cosh(bt). Only degrees divisible by four survive. Its actual Hurwitz packets are D_1(t)=∑_i≥0d_ifract^4i(4i)!, d_i=(-frac14)^i frac2^4i+2(2^4i+2-1)B_4i+2(4i+2)(4i+1). The constant is one. Since v_2(4i+2)=1

    Exact mathematical passage
    G_{4,1}(t)=(\sinh t+\sin t)/(2t)
G_{4,3}Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 716

    The address-three source is G_4,3(t)=3(sinh t-sin t)/t^3. Addition identities give sinh t-sin t =2asinh(at)cosh(bt)+2bcosh(at)sinh(bt), cosh t-cos t=2sinh(at)sinh(bt). Consequently D_3(t):=G_4,3(t)F_2(t) =frac32(fraccoth(at)at +fraccoth(bt)bt). This expression is a power series: its two polar terms cancel because a^-2+b^-2=0. Its constant is one, only degrees divisible by four survive, and the Bernoulli expansion gives D_3(t)=∑_i≥0d_ifract^4i(4i)!, d_i=3(-frac14)^i frac2^4i+2B_4i+2(4i+2)(4i+1). …

    Exact mathematical passage
    G_{4,3}(t)=3(\sinh t-\sin t)/t^3
G_AUse in source · 1 passage

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  • Use in source · worked_atlas.tex, line 99

    AtlasHeadA derivative weight can cost one digit of root precisionatlas:weighted Compare the linear equations F(x)=3x=0 and G_A(x)=3x+3^A=0, where A≥2. Their coefficient difference has 3-adic valuation A, but their roots differ by x_A-0=-3^A-1, v_3(x_A)=A-1. AtlasImageweighted_motion The derivative has valuation one. Solving the correction equation divides by three, so one digit of coefficient precision is lost in root displacement. The calculation is exact for every displayed A and also follows …

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    G_A(x)=3x+3^A=0
G_iUse in source · 1 passage

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  • Use in source · Read passage · continuation_04.tex, line 61

    If p^Lmid c, the actual identities give G(x)=F(x)+cH(x), DG(x)≡ DF(x)pmod p. Thus if F(x)≡0pmodp^L and A_x is surjective, then ∃ b∈ B_L(x):G(b)=0. This theorem allows rectangular systems. The evaluation, derivative and Jacobian identities are proved from the polynomial definition G_i=F_i+C(c)H_i, rather than supplied as hypotheses. Algebraic congruence helpers also permit composite moduli; exact p-adic root existence uses prime p.

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    G_i=F_i+C(c)H_i
G_jUse in source · 2 passages

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  • Use in source · Read passage · supplement_11.tex, line 81

    Pack g_j into degrees divisible by four and form its actual EGF. The formal coefficient identity is G_j(t)=∑_n≥0g_j(n)fract^4n(4n)! =j!∑_n≥0fract^4n(4n+j)!. Its inverse is therefore the original normalized packet EGF. The source Kummer property gives flatness of G_j. The proved reciprocal closure gives flatness of G_j^-1.

    Exact mathematical passage
    
    G_j(t)=\sum_{n\ge0}g_j(n)\frac{t^{4n}}{(4n)!}
           =j!\sum_{n\ge0}\frac{t^{4n}}{(4n+j)!}.
    
g_jUse in source · 10 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 484

    The Hurwitz coefficients of the normalized source K_4,j at degrees 4n are g_j(n)=binom4n+j4n^-1. Their explicit expressions are g_0(n) · =1, · g_1(n) · =frac14n+1,

    Exact mathematical passage
    
     g_j(n)=\binom{4n+j}{4n}^{-1}.
    
g_{j+1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 137

    For j=0, g_0 is the literal N-divisibility indicator. Because Nmid p(p-1), the two shifts in δ g_0 agree and δ g_0=0. It is flat at every order. The actual binomial identity gives Λ_j+1g_j+1=(j+1)g_j. As 1≤ j+1<p, the nonsingular reflection theorem inductively proves flatness of every g_j in the standard window. There is no uniform-convergence assumption in this argument.

    Exact mathematical passage
    
     \Lambda_{j+1}g_{j+1}=(j+1)g_j.
    
G_j^{-1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 85

    Pack g_j into degrees divisible by four and form its actual EGF. The formal coefficient identity is G_j(t)=∑_n≥0g_j(n)fract^4n(4n)! =j!∑_n≥0fract^4n(4n+j)!. Its inverse is therefore the original normalized packet EGF. The source Kummer property gives flatness of G_j. The proved reciprocal closure gives flatness of G_j^-1.

    Exact mathematical passage
    G_j^{-1}
G_mUse in source · 5 passages

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  • Use in source · Read passage · 72_recovered.tex, line 425

    [Exponential form and minimal period] Let G_m=( F_p^×)^m. This cyclic group has order h. There are uniquely determined coefficients d_v∈ F_p, indexed by v∈ G_m, such that e_n^(m,p) = ∑_v∈ G_md_vv^n (n≥1). The minimal positive period is the least common multiple of the orders of those v for which d_v≠0. In particular, it divides h.

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    G_m=(\FF_p^\times)^m.
    
g_mUse in source · 4 passages

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  • Use in source · Read passage · supplement_07.tex, line 78

    pageheading2. Identification with the reduced Euler sequenceCorollary 8.5; manuscript page 21 Work over F_p= Z/p Z. For positive mmid p-1, take the literal section g_m(d)=1, · mmid d, 0, · mnmid d. It satisfies g_m(0)=1. Since p≡1pmod m, the conditions mmid d+p and mmid d+1 are equivalent. The source therefore proves U^pg_m=Ug_m.

    Exact mathematical passage
    
     g_m(d)=\begin{cases}1,&m\mid d,\\0,&m\nmid d.\end{cases}
    
G_{N,j}Use in source · 4 passages

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  • Use in source · Read passage · strengthening.tex, line 14

    Throughout these sections, the normalized source and its reciprocal are G_N,j(t)=j!∑_n≥0fract^Nn(Nn+j)!, F_N,j(t)=G_N,j(t)^-1 =∑_n≥0C_N,j(n)fract^Nn(Nn)!, where N≥1 and 0≤ j<N. In particular, C_N,j(0)=1. The coefficients in the original convention satisfy E_Nn^(N,j)=j!C_N,j(n), E_d^(N,j)=0 (Nnmid d). Thus every valuation of a nonzero difference gains v_p(j!) on returning to the original coefficients. This factor must be retained when stating optimal periods or the final congruence.

    Exact mathematical passage
    
     G_{N,j}(t)=j!\sum_{n\geq0}\frac{t^{Nn}}{(Nn+j)!},\qquad
     F_{N,j}(t)=G_{N,j}(t)^{-1}
       =\sum_{n\geq0}C_{N,j}(n)\frac{t^{Nn}}{(Nn)!},
     \label{eq:final-normalization}
    
g_{N,j}Use in source · 4 passages

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  • Use in source · Read passage · strengthening.tex, line 140

    The packed factorial coefficient of G_N,j is g_N,j(n)=fracj!(Nn)!(Nn+j)! =∏_i=1^j(1+frac Ni n)^-1. The empty product is one. If N=p^a and 1≤ i<N, then v_p(i)<a, so N/i∈ p Z_p. Every affine denominator in [source reference] is therefore a unit on Z_p.

    Exact mathematical passage
    
     g_{N,j}(n)=\frac{j!(Nn)!}{(Nn+j)!}
       =\prod_{i=1}^j\left(1+\frac Ni n\right)^{-1}.
     \label{eq:actual-prime-power-kernel}
    
G_rUse in source · 12 passages

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  • Use in source · Read passage · root_scope.tex, line 128

    The scalar packet weights have exact valuations λ_r(q)=3q+s_2(r+q)-s_2(r). Put μ_r=min_q≥0λ_r(q), J_r={q:λ_r(q)=μ_r}, G_r=2^-μ_rH_r. Since λ_r(0)=0 and λ_r(q)≥3q+1-s_2(r), every minimizing address satisfies 0≤ q≤lfloorfracs_2(r)-13rfloor. In particular μ_r≤0, and the minimum is found by a finite carry calculation. Reduction of the whole series gives overline G_r(X)=∑_q∈ J_rX^underline q in F_2[X]. Its highest degree is max J_r, with leading coefficient one. Thus the normalization is the actual G…

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     \mu_r=\min_{q\ge0}\lambda_r(q),\qquad
     J_r=\{q:\lambda_r(q)=\mu_r\},\qquad G_r=2^{-\mu_r}H_r.
    
g_RUse in source · 2 passages

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  • Use in source · Read passage · 71_native.tex, line 251

    The component maps f_R:A→ C_R and g_R:B→ C_R have the following universal property. For every set Y and maps h:A→ Y, k:B→ Y satisfying h(a)=k(b) whenever R(a,b), there is a unique map u:C_R→ Y with u∘ f_R=h and u∘ g_R=k.

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    g_R:B\to C_R
G_r'Use in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 294

    For a general depth r, call j_r=max J_r its true terminal. This definition depends on the actual carry minimum; rewriting the same integer as another wall expression does not change j_r. The universal simplicity target is for every integer r≥1 and x∈ Z_2, G_r(x)=0 Longrightarrow G_r'(x)≠0. The following are completed restricted results of the later continuations. Their family statements include a proof that the indicated label is the true terminal throughout the displayed ranges.

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     \text{for every integer }r\ge1\text{ and }x\in\mathbb Z_2,
     \qquad G_r(x)=0\ \Longrightarrow\ G_r'(x)\ne0.
     \label{eq:root:simplicitytarget}
    
g_sUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2208

    [Mahler polyphase transfer] Let f(x)=∑_r≥0c_rbinom xr be a convergent Mahler expansion. For g_s(q)=f(Mq+s), one has g_s(q)=∑_k≥0d_s,kbinom qk, where d_s,k = ∑_r=k^Mk+s c_rΘ_M,s(r,k). Every refined coefficient uses a finite window of the original Mahler coefficients.

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    g_s(q)=f(Mq+s),
    
\gcdUse in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 179

    Principal targets. targetUnitRefinement.omega_compatible, .principal_compatible, .principal_power targetDyadicUnits.dyadic_unit_decomposition, .decomposition_unique pageheading4. CRT addresses and the actual packet selectorsTheorem 10.1; manuscript page 22 Let a≥0, let M be odd, and put N=2^aM. Coprimality is derived from oddness: gcd(2^a,M)=1. The construction uses the actual Chinese-remainder ring equivalence Z/N Z ≃ Z/2^a Z× Z/M Z. Its bijectivity and component formula are checked. On a natur…

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     \gcd(2^a,M)=1.
    
\operatorname{GL}Use in source · 14 passages

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  • Use in source · Read passage · 30_lifting.tex, line 686

    [Weighted nonlinear lifting] Let p be prime, d,s,r∈ N, F∈ R[X_1,…,X_d]^d, and a∈ R^d. Set J=DF(a). Suppose J∈GL_d( Q_p), p^sJ^-1∈ M_d(R), F(a)∈ p^rR^d, r>2s. Then there is exactly one root b of F in a+p^s+1R^d, and it satisfies b-a∈ p^r-sR^d. Uniqueness holds throughout the full ball of depth s+1.

    Exact mathematical passage
    \label{eq:weighted-hypotheses}
     J\in\operatorname{GL}_d(\Qp),\qquad
     p^sJ^{-1}\in M_d(R),\qquad F(a)\in p^rR^d,\qquad r>2s.
    
\operatorname{GL}_dUse in source · 5 passages

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  • Use in source · Read passage · 30_lifting.tex, line 686

    [Weighted nonlinear lifting] Let p be prime, d,s,r∈ N, F∈ R[X_1,…,X_d]^d, and a∈ R^d. Set J=DF(a). Suppose J∈GL_d( Q_p), p^sJ^-1∈ M_d(R), F(a)∈ p^rR^d, r>2s. Then there is exactly one root b of F in a+p^s+1R^d, and it satisfies b-a∈ p^r-sR^d. Uniqueness holds throughout the full ball of depth s+1.

    Exact mathematical passage
    \label{eq:weighted-hypotheses}
     J\in\operatorname{GL}_d(\Qp),\qquad
     p^sJ^{-1}\in M_d(R),\qquad F(a)\in p^rR^d,\qquad r>2s.
    
\operatorname{GL}_eUse in source · 9 passages

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  • Use in source · Read passage · 30_lifting.tex, line 828

    [Joint coefficient and parameter deformation] leavevmode Let F,H∈ R[Y_1,…,Y_e,T_1,…,T_k]^e and a=(y_0,t_0)∈ R^e+k satisfy F(a)=0. Define the actual blocks J=D_YF(a)∈ M_e(R), K=D_TF(a)∈ M_e× k(R). Suppose J∈GL_e( Q_p), p^sJ^-1∈ M_e(R), and M>2s. For every v∈ R^k, set τ=t_0+p^Mv, G=F+p^MH. There is exactly one y∈ y_0+p^s+1R^e with G(y,τ)=0. It satisfies y-y_0 · ∈ p^M-sR^e,

    Exact mathematical passage
    J\in\operatorname{GL}_e(\Qp)
H^{-1}Use in source · 5 passages

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  • Use in source · Read passage · 10_foundations.tex, line 89

    [Constructed unit-triangular coordinates] Let h: N→ K satisfy h_0=1, write H= E(h) and H^-1= E(k), and define U_Ha= D(H E(a)). Then U_H is an equivalence, with inverse U_H^-1a= D(H^-1 E(a)). Its coefficient formulas are (U_Ha)_n=∑_r=0^nbinom nr h_r a_n-r, (U_H^-1a)_n=∑_r=0^nbinom nr k_r a_n-r. The resulting triangular matrix has diagonal entries 1. The transported selectors widehat P_N,j=U_HP_N,jU_H^-1 are idempotent and pairwise orthogonal, and their sum is the identity.

    Exact mathematical passage
    H^{-1}=\mathcal E(k)
h^{-1}Use in source · 1 passage

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  • Use in source · Read passage · continuation_05.tex, line 27

    There is a coordinate permutation σ, with (σ z)_i=z_σ(i). Write σ z=(y,t), using e solved coordinates and q free coordinates. Define (y_0,t_0)=σ x, B=x+p^L Z_p^e+q, T=t_0+p^L Z_p^q, R={z∈ B:F(z)=0}. There is a homeomorphism h:Txrightarrow ∼ R, h^-1(z)=free(σ z). Equivalently, a continuous function ϕ:T→ y_0+p^L Z_p^e gives the entire root locus through σ h(t)=(ϕ(t),t). Every parameter has exactly one corresponding root in B. For every N≥0, h(t)≡ h(u)pmodp^Niff t≡ upmodp^N.

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    h:T\xrightarrow{\ \sim\ }\mathcal R,\qquad
    h^{-1}(z)=\operatorname{free}(\sigma z).
    
H_0Use in source · 9 passages

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  • Use in source · Read passage · strengthening.tex, line 326

    , and b=(1-i)/2. The exact identity F_0(t)=frac2cosh t+cos t =sech(at)sech(bt) gives, for U=a E+b F, L(U^4n)=C_0(n). Although a and b have negative valuation, their denominators disappear from the following exact polynomial identities: U^2 · = E F+i(P-Q),nonumber H_0:=fracU^4-12 · =P+Q-fracP^22-fracQ^22 +3PQ+i E F(P-Q).

    Exact mathematical passage
    
     U^2&=\mathcal E\mathcal F+i(P-Q),\nonumber\\
     H_0:=\frac{U^4-1}{2}
       &=P+Q-\frac{P^2}{2}-\frac{Q^2}{2}
                     +3PQ+i\mathcal E\mathcal F(P-Q).
     \label{eq:four-euler-H}
    
h_0Use in source · 5 passages

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  • Use in source · Read passage · 10_foundations.tex, line 88

    [Constructed unit-triangular coordinates] Let h: N→ K satisfy h_0=1, write H= E(h) and H^-1= E(k), and define U_Ha= D(H E(a)). Then U_H is an equivalence, with inverse U_H^-1a= D(H^-1 E(a)). Its coefficient formulas are (U_Ha)_n=∑_r=0^nbinom nr h_r a_n-r, (U_H^-1a)_n=∑_r=0^nbinom nr k_r a_n-r. The resulting triangular matrix has diagonal entries 1. The transported selectors widehat P_N,j=U_HP_N,jU_H^-1 are idempotent and pairwise orthogonal, and their sum is the identity.

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    h_0=1
H_1Use in source · 15 passages

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  • Use in source · Read passage · strengthening.tex, line 32

    Write S f(n)=f(n+1) and Δ_H=S^H-1. A mixed difference Δ_H_1⋯Δ_H_k acts on the row variable n, rather than on the full Euler subscript. We use v_p(0)=+∞. Orders and starting rows are nonnegative integers; steps are positive integers unless stated otherwise. An empty mixed difference is the original value, and a zero ordinary step makes every nonempty product containing it zero.

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    \Delta_{H_1}\cdots\Delta_{H_k}
\mathcal H_2Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 516

    For an integer charge c, put A_c=fracD^4-c2. A series is flat of charge c when applying any power A_c^s to it gives an element of H_2. The required product formula is A_c+d(FG)=(A_cF)G+F(A_dG) +2F'G'''+3F''G''+2F'''G'. All middle coefficients are integral. Derivatives preserve H_2 and commute with A_c. Induction on the number of iterations in [source reference] proves that multiplication adds charges. If FG=1, the same formula yields A_-cG=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G'). Induction, starting…

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    \mathcal H_2
H_{3,0}Use in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 202

    It also derives coefT3H_3,0(log(1+T))=1/6 from the actual differential recurrence and checks its valuation v_3(1/6)=-1. Thus weighted membership does not mean ordinary coefficient integrality. This value attains the allowed floor weight at degree three.

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    \coef{T}{3}H_{3,0}(\log(1+T))=1/6
H_aUse in source · 10 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2330

    [Scalar polyphase precision amplification] Let u∈ Z_p^×, let a,r≥0, and put H_a= (p-1)p^a, · p odd, 2^a+1, · p=2. Then Δ_H_a^ ru^n = u^n(u^H_a-1)^r. For odd p, v_p(Δ_H_a^ ru^n) ≥(a+1)r. For p=2 and odd u, v_2(Δ_H_a^ ru^n) ≥(a+3)r. Equivalently, every additional prime power digit increases the certified Mahler slope by one.

    Exact mathematical passage
    
    H_a=
    \begin{cases}
    (p-1)p^a,&p\text{ odd},\\
    2^{a+1},&p=2.
    \end{cases}
    
H_cUse in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 216

    Set H_c(t)=∑_n≥0fract^Nn+c(Nn+c)!, A_c(t)=∑_n≥0fract^Nn(Nn+c)!, R_c(t)=fracA_c-1(t)A_c(t) (1≤ c<N). The denominator of this last quotient has nonzero constant coefficient. The section identities give R_cH_c=tH_c-1, R_c(0)=c, j! P_N,j= P_N,0∏_c=1^jR_c. Cancellation of consecutive source sections proves the product identity and ties the ratios directly to the family under study.

    Exact mathematical passage
    
     H_c(t)=\sum_{n\ge0}\frac{t^{Nn+c}}{(Nn+c)!},\qquad
     A_c(t)=\sum_{n\ge0}\frac{t^{Nn}}{(Nn+c)!},\qquad
     R_c(t)=\frac{A_{c-1}(t)}{A_c(t)}\quad(1\le c<N).
    
H_{c-1}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 223

    Set H_c(t)=∑_n≥0fract^Nn+c(Nn+c)!, A_c(t)=∑_n≥0fract^Nn(Nn+c)!, R_c(t)=fracA_c-1(t)A_c(t) (1≤ c<N). The denominator of this last quotient has nonzero constant coefficient. The section identities give R_cH_c=tH_c-1, R_c(0)=c, j! P_N,j= P_N,0∏_c=1^jR_c. Cancellation of consecutive source sections proves the product identity and ties the ratios directly to the family under study.

    Exact mathematical passage
    
     R_cH_c=tH_{c-1},\qquad R_c(0)=c,\qquad
     j!\mathcal P_{N,j}=\mathcal P_{N,0}\prod_{c=1}^{j}R_c.
    
h_dUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 606

    If m is not a power of two, s_2(m)≥2 makes μ_m even. If m is a positive power of two, the only central indices which can contribute an odd residue are r=1 and r=3, since r-2s_2(r)=-1 only at these two positive integers. The r=1 term is odd: T(m,1)=2^m-1 and v_2(m!)=m-1. For m≥4 a power of two, the r=3 term is even. More explicitly, fracT(m,3)2^m-3=h_m-3(1,3,6), ∑_d≥0h_d(1,3,6)t^d =frac1(1-t)(1-3t)(1-6t). Reduction modulo two gives (1-t)^-2, whose degree m-3 coefficient has parity m-2=0. At m=1,2…

    Exact mathematical passage
    
     \frac{T(m,3)}{2^{m-3}}=h_{m-3}(1,3,6),
     \qquad
     \sum_{d\geq0}h_d(1,3,6)t^d
           =\frac1{(1-t)(1-3t)(1-6t)}.
    
H_iUse in source · 7 passages

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  • Use in source · Read passage · continuation_04.tex, line 61

    If p^Lmid c, the actual identities give G(x)=F(x)+cH(x), DG(x)≡ DF(x)pmod p. Thus if F(x)≡0pmodp^L and A_x is surjective, then ∃ b∈ B_L(x):G(b)=0. This theorem allows rectangular systems. The evaluation, derivative and Jacobian identities are proved from the polynomial definition G_i=F_i+C(c)H_i, rather than supplied as hypotheses. Algebraic congruence helpers also permit composite moduli; exact p-adic root existence uses prime p.

    Exact mathematical passage
    G_i=F_i+C(c)H_i
h_iUse in source · 6 passages

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  • Use in source · Read passage · strengthening.tex, line 460

    nly degrees divisible by four survive. Its actual Hurwitz packets are D_1(t)=∑_i≥0d_ifract^4i(4i)!, d_i=(-frac14)^i frac2^4i+2(2^4i+2-1)B_4i+2(4i+2)(4i+1). The constant is one. Since v_2(4i+2)=1, the elementary Bernoulli bound shows v_2(d_i)≥2i. Hurwitz inversion therefore gives D_1(t)^-1=∑_i≥0h_ifract^4i(4i)!, h_0=1, v_2(h_i)≥2i. This follows by induction from the inverse recurrence: for packet index i, every product d_rh_i-r has valuation at least 2i, and its binomial multiplier is integral. B…

    Exact mathematical passage
    
     D_1(t)^{-1}=\sum_{i\geq0}h_i\frac{t^{4i}}{(4i)!},
     \qquad h_0=1,\quad v_2(h_i)\geq2i.
    
h_{i-r}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 465

    (2^4i+2-1)B_4i+2(4i+2)(4i+1). The constant is one. Since v_2(4i+2)=1, the elementary Bernoulli bound shows v_2(d_i)≥2i. Hurwitz inversion therefore gives D_1(t)^-1=∑_i≥0h_ifract^4i(4i)!, h_0=1, v_2(h_i)≥2i. This follows by induction from the inverse recurrence: for packet index i, every product d_rh_i-r has valuation at least 2i, and its binomial multiplier is integral. Because F_1=F_0D_1^-1, C_1(n)-C_0(n) =∑_i≥1h_ibinom4n4iC_0(n-i). At a natural n this is a finite sum, with i≤ n.

    Exact mathematical passage
    d_rh_{i-r}
H_kUse in source · 14 passages

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  • Use in source · Read passage · strengthening.tex, line 32

    Write S f(n)=f(n+1) and Δ_H=S^H-1. A mixed difference Δ_H_1⋯Δ_H_k acts on the row variable n, rather than on the full Euler subscript. We use v_p(0)=+∞. Orders and starting rows are nonnegative integers; steps are positive integers unless stated otherwise. An empty mixed difference is the original value, and a zero ordinary step makes every nonempty product containing it zero.

    Exact mathematical passage
    \Delta_{H_1}\cdots\Delta_{H_k}
H_mUse in source · 6 passages

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  • Use in source · Read passage · 72_recovered.tex, line 251

    Fix a divisor m of p-1. Let H_m = {ζ∈ F_p^×:ζ^m=1}. This group has order m.

    Exact mathematical passage
    
    H_m
    =
    \{\zeta\in\FF_p^\times:\zeta^m=1\}.
    
h_{m-3}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 606

    If m is not a power of two, s_2(m)≥2 makes μ_m even. If m is a positive power of two, the only central indices which can contribute an odd residue are r=1 and r=3, since r-2s_2(r)=-1 only at these two positive integers. The r=1 term is odd: T(m,1)=2^m-1 and v_2(m!)=m-1. For m≥4 a power of two, the r=3 term is even. More explicitly, fracT(m,3)2^m-3=h_m-3(1,3,6), ∑_d≥0h_d(1,3,6)t^d =frac1(1-t)(1-3t)(1-6t). Reduction modulo two gives (1-t)^-2, whose degree m-3 coefficient has parity m-2=0. At m=1,2…

    Exact mathematical passage
    
     \frac{T(m,3)}{2^{m-3}}=h_{m-3}(1,3,6),
     \qquad
     \sum_{d\geq0}h_d(1,3,6)t^d
           =\frac1{(1-t)(1-3t)(1-6t)}.
    
H_{N,c}Use in source · 9 passages

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  • Use in source · Read passage · supplement_09.tex, line 5

    Checkpoint scope. This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution. Scope and result. Let p be prime, let mmid p-1, and put N=mp. For 1≤ a<m, put c=ap and define the original exponential section and ratio by H_N,c(t)=∑_n≥0fract^Nn+c(Nn+c)!, R_c(t)=fractH_N,c-1(t)H_N,c(t). The quotient is interpreted after removing its forced zero. This block proves, for the actual rati…

    Exact mathematical passage
    
     H_{N,c}(t)=\sum_{n\ge0}\frac{t^{Nn+c}}{(Nn+c)!},\qquad
     R_c(t)=\frac{tH_{N,c-1}(t)}{H_{N,c}(t)}.
    
H_{N,c}'Use in source · 1 passage

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  • Use in source · Read passage · supplement_10.tex, line 130

    Modules. mcodeWeightedDworkTransfer, mcodeSeriesDifferenceBridge, mcodeRatioSeriesSupport. 3. Nonsingular propagation without an assumed limit. For consecutive original sections, the already defined ratios satisfy R_cH_N,c=tH_N,c-1, H_N,c'=H_N,c-1. Differentiating the first relation, using the relation for R_c-1, and cancelling a section whose leading coefficient is proved nonzero gives the actual Riccati equation tR_c'=R_c(1+R_c-1-R_c), 2≤ c<N. It is not an equation imposed on a replacement rat…

    Exact mathematical passage
    
     R_cH_{N,c}=tH_{N,c-1},\qquad H_{N,c}'=H_{N,c-1}.
    
H_{N,c-1}Use in source · 3 passages

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  • Use in source · Read passage · supplement_09.tex, line 5

    Checkpoint scope. This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution. Scope and result. Let p be prime, let mmid p-1, and put N=mp. For 1≤ a<m, put c=ap and define the original exponential section and ratio by H_N,c(t)=∑_n≥0fract^Nn+c(Nn+c)!, R_c(t)=fractH_N,c-1(t)H_N,c(t). The quotient is interpreted after removing its forced zero. This block proves, for the actual rati…

    Exact mathematical passage
    
     H_{N,c}(t)=\sum_{n\ge0}\frac{t^{Nn+c}}{(Nn+c)!},\qquad
     R_c(t)=\frac{tH_{N,c-1}(t)}{H_{N,c}(t)}.
    
H_rUse in source · 6 passages

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  • Use in source · Read passage · supplement_09.tex, line 83

    Modules. mcodeCyclotomicFiltration, mcodePrimeBlockPolynomial, and mcodeCyclicAmiceCoefficients. 2. Connection to the actual section. An integral coordinate estimate becomes relevant only after connecting it to the manuscript's generating functions. For a residue r∈ Z/N Z, define H_r(t)=∑_d≥01_d≡ r (N)fract^dd!, F_r(T)=H_r(log(1+T)). The implementation uses the actual formal logarithm and substitution in Q[[T]]. It proves H_r'=H_r-1, (1+T)F_r'=F_r-1, and hence the coefficient recurrence (d+1)coe…

    Exact mathematical passage
    
     H_r(t)=\sum_{d\ge0}\mathbf1_{d\equiv r\ (N)}\frac{t^d}{d!},
     \qquad F_r(T)=H_r(\log(1+T)).
    
h_rUse in source · 4 passages

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  • Use in source · Read passage · 10_foundations.tex, line 95

    [Constructed unit-triangular coordinates] Let h: N→ K satisfy h_0=1, write H= E(h) and H^-1= E(k), and define U_Ha= D(H E(a)). Then U_H is an equivalence, with inverse U_H^-1a= D(H^-1 E(a)). Its coefficient formulas are (U_Ha)_n=∑_r=0^nbinom nr h_r a_n-r, (U_H^-1a)_n=∑_r=0^nbinom nr k_r a_n-r. The resulting triangular matrix has diagonal entries 1. The transported selectors widehat P_N,j=U_HP_N,jU_H^-1 are idempotent and pairwise orthogonal, and their sum is the identity.

    Exact mathematical passage
    
    (U_Ha)_n=\sum_{r=0}^{n}\binom nr h_r a_{n-r},\qquad
    (U_H^{-1}a)_n=\sum_{r=0}^{n}\binom nr k_r a_{n-r}.
    
H_r'Use in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 88

    ip 2. Connection to the actual section. An integral coordinate estimate becomes relevant only after connecting it to the manuscript's generating functions. For a residue r∈ Z/N Z, define H_r(t)=∑_d≥01_d≡ r (N)fract^dd!, F_r(T)=H_r(log(1+T)). The implementation uses the actual formal logarithm and substitution in Q[[T]]. It proves H_r'=H_r-1, (1+T)F_r'=F_r-1, and hence the coefficient recurrence (d+1)coefTd+1F_r =coefTdF_r-1-dcoefTdF_r. The initial values are F_r(0)=1_r=0.

    Exact mathematical passage
    
     H_r'=H_{r-1},\qquad (1+T)F_r'=F_{r-1},
    
H_{r-1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 88

    ip 2. Connection to the actual section. An integral coordinate estimate becomes relevant only after connecting it to the manuscript's generating functions. For a residue r∈ Z/N Z, define H_r(t)=∑_d≥01_d≡ r (N)fract^dd!, F_r(T)=H_r(log(1+T)). The implementation uses the actual formal logarithm and substitution in Q[[T]]. It proves H_r'=H_r-1, (1+T)F_r'=F_r-1, and hence the coefficient recurrence (d+1)coefTd+1F_r =coefTdF_r-1-dcoefTdF_r. The initial values are F_r(0)=1_r=0.

    Exact mathematical passage
    
     H_r'=H_{r-1},\qquad (1+T)F_r'=F_{r-1},
    
H_\nuUse in source · 11 passages

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  • Use in source · Read passage · strengthening.tex, line 176

    [All-row prime-power transfer] Let p be prime, a≥1, N=p^a, and 0≤ j<N. The coefficients C_N,j(n) belong to Z_p for all n≥0. Set A_N,j(n)=(-1)^n C_N,j(n) and T_H=S^H-(-1)^H. For k≥1 and positive H_1,…,H_k, v_p(Δ_H_1⋯Δ_H_kA_N,j(n)) · ≥ k+∑_ν=1^k v_p(H_ν), v_p(T_H_1⋯ T_H_kE_Nn^(N,j)) · ≥ v_p(j!)+k+∑_ν=1^k v_p(H_ν). At p=2, the ordinary differences of C_N,j and of the original coefficients satisfy the same bounds. These dyadic statements also hold for N=1.

    Exact mathematical passage
    
     v_p(\Delta_{H_1}\cdots\Delta_{H_k}A_{N,j}(n))
      &\geq k+\sum_{\nu=1}^k v_p(H_\nu),
     \label{eq:prime-power-signed-bound}\\
     v_p(T_{H_1}\cdots T_{H_k}E_{Nn}^{(N,j)})
      &\geq v_p(j!)+k+\sum_{\nu=1}^k v_p(H_\nu).
     \label{eq:prime-power-original-twist}
    
H_\ellUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1277

    the even source subsequences, equivalently the evenized generating functions H_p. It does not identify their full odd-degree source values. For p=3, the original odd source has A_3,1=S_1(3)=3, whereas every odd dyadic source coefficient is zero; the even kernel L_3K_2 cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity H_p=L_pK_ℓ H_ℓ and its proof remain valid. For any two primes p and ℓ, the source sequences A_p,n and A_ℓ,n are…

    Exact mathematical passage
    H_p=L_pK_\ell H_\ell
\HcalUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 124

    [Hurwitz series ring] Let R be a commutative ring with identity. The Hurwitz series ring H(R) is the set R^ N with componentwise addition and product (amathbin⋆ b)_n = ∑_r=0^n binom nr a_r b_n-r. Its identity is δ=(1,0,0,…).

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    \Hcal(R)
\hlineUse in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 717

    [All positive periods and their least element] For r∈ N and 0≤ j<4, a positive integer T is a packet period of U_j modulo 2^r if and only if 2^max(0,r-d_j)mid T. Here a packet period means U_j(n+T)≡ U_j(n)pmod2^r for every n≥0. Consequently the least positive packet periods are c|cccc j · 0 · 1 · 2 · 3 least period · 2^max(0,r-1) · 2^max(0,r-1) · 2^max(0,r-5) · 2^max(0,r-3).

    Exact mathematical passage
    
     \begin{array}{c|cccc}
     j&0&1&2&3\\\hline
     \text{least period}&2^{\max(0,r-1)}&2^{\max(0,r-1)}&
                     2^{\max(0,r-5)}&2^{\max(0,r-3)}.
     \end{array}
    
\operatorname{Hom}Use in source · 1 passage

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  • Use in source · Read passage · 71_native.tex, line 134

    coker(Hom_B(Ω_P/ Z⊗_P B, F_p)⟶Hom_B(I/I^2, F_p)).

    Exact mathematical passage
    \operatorname{coker}\left(\operatorname{Hom}_B(\Omega_{P/\mathbb Z}\otimes_P B,\mathbb F_p)\longrightarrow\operatorname{Hom}_B(I/I^2,\mathbb F_p)\right).
\operatorname{Hom}_BUse in source · 1 passage

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  • Use in source · Read passage · 71_native.tex, line 134

    coker(Hom_B(Ω_P/ Z⊗_P B, F_p)⟶Hom_B(I/I^2, F_p)).

    Exact mathematical passage
    \operatorname{coker}\left(\operatorname{Hom}_B(\Omega_{P/\mathbb Z}\otimes_P B,\mathbb F_p)\longrightarrow\operatorname{Hom}_B(I/I^2,\mathbb F_p)\right).
\HpUse in source · 3 passages

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  • Use in source · Read passage · supplement_10.tex, line 97

    The relevant integral subring is the Hurwitz subring H( Z_p)={f∈ Q_p[[t]]: n![t^n]f∈ Z_p for every n}. It is not coefficientwise integrality of ordinary power-series coefficients. The precise connection is n![t^n] E(F)=[T^0]Θ^nF. Since every iterate of Θ preserves W_p, the right side is integral. Therefore E( W_p)⊆ H( Z_p). Applying the same statement to every B_Θ^sF proves f(log(1+T))∈ W_p Longrightarrow B_D^sf∈ H( Z_p) for every s. This is the proved transfer that was missing in the earlier at…

    Exact mathematical passage
    
     \Hp=\{f\in\Q_p[[t]]:\ n![t^n]f\in\Z_p\ \text{for every }n\}.
    
I^2Use in source · 2 passages

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  • Use in source · Read passage · 71_native.tex, line 134

    coker(Hom_B(Ω_P/ Z⊗_P B, F_p)⟶Hom_B(I/I^2, F_p)).

    Exact mathematical passage
    \operatorname{coker}\left(\operatorname{Hom}_B(\Omega_{P/\mathbb Z}\otimes_P B,\mathbb F_p)\longrightarrow\operatorname{Hom}_B(I/I^2,\mathbb F_p)\right).
I^bUse in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 48

    ficient in prime row p gives (z-1)^p∈ PI. Every other degree-p product of the two generators of I also belongs to PI. The formal ideal proof therefore establishes I^p⊆ PI. For q=plfloor q/prfloor+(qbmod p), strong induction and the factorial valuation identity give I^q⊆ P^v_p(q!). In detail, write b=lfloor q/prfloor. Then I^q⊆ P^b I^b⊆ P^b+v_p(b!)=P^v_p(q!). No integral-domain hypothesis or division in a field is used here.

    Exact mathematical passage
    
     I^q\subseteq P^b I^b\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.
    
I^{\lfloor d/p\rfloor}Use in source · 1 passage

Search aliases: I^lfloor d/prfloor lfloor rfloor

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  • Use in source · Read passage · supplement_09.tex, line 68

    For the falling factorial (x)_d=∏_i=0^d-1(x-i), grouping into consecutive blocks gives (x)_d∈ I^lfloor d/prfloor ⊆(p^v_p(lfloor d/prfloor!)). Every integral residue coordinate therefore satisfies p^v_p(lfloor d/prfloor!)mid [(r)](x)_d in Z. Dividing a coordinate by d! is done only afterward, using v_p(d!)=lfloor d/prfloor+v_p(lfloor d/prfloor!). This establishes the weighted denominator estimate before any field-valued divisibility could make it vacuous.

    Exact mathematical passage
    
     (x)_d\in I^{\lfloor d/p\rfloor}
     \subseteq\bigl(p^{v_p(\lfloor d/p\rfloor!)}\bigr).
    
I^pUse in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 238

    At the singular addresses c=ap, 1≤ a<m, use the weighted ring W_p={∑_d≥0a_dT^d: p^lfloor d/prfloora_d∈ Z_p}. The key estimate is R_ap(log(1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!)…

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    I^p\subseteq(p)I
I^qUse in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 239

    sses c=ap, 1≤ a<m, use the weighted ring W_p={∑_d≥0a_dT^d: p^lfloor d/prfloora_d∈ Z_p}. The key estimate is R_ap(log(1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!). The identity v_p(d!…

    Exact mathematical passage
    I^q\subseteq(p^{v_p(q!)})
I_2Use in source · 2 passages

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  • Use in source · Read passage · supplement_04.tex, line 165

    ar targetUnitModes.unit_mode_valuation_gain targetUnitModes.finite_unit_modes_valuation_gain pageheading4. Exact examples and adversarial checksThe examples use the same definitions as the general results A polynomial anchor. The actual difference table for f(n)=n^2 yields (c_0,c_1,c_2,c_3)=(0,1,2,0). The finite interpolant therefore reconstructs I_2(c)(7)=49. This checks the sign and indexing conventions, not just the names of the defined objects.

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    I_2(c)(7)=49
I_dUse in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 701

    Let A=p^sJ^-1∈ M_d(R). Inclusion into Q_p and its injectivity give the two integral identities AJ=JA=p^sI_d. These identities are also an equivalent integral certificate for the first two hypotheses of [source reference]. In particular, A is injective: Aw=0 implies p^sw=JAw=0, hence w=0.

    Exact mathematical passage
    
     AJ=JA=p^sI_d.
    
I_eUse in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 860

    0 · cI_k. The identities AJ=JA=cI_e give CB=BC=cI_e+k. In the second product, the upper right block is -JAK+cK=0; in the first, it is AK-AK=0. Thus augmentation preserves the original inverse bound s.

    Exact mathematical passage
    AJ=JA=cI_e
I_{e+k}Use in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 860

    0 · cI_k. The identities AJ=JA=cI_e give CB=BC=cI_e+k. In the second product, the upper right block is -JAK+cK=0; in the first, it is AK-AK=0. Thus augmentation preserves the original inverse bound s.

    Exact mathematical passage
    CB=BC=cI_{e+k}
I_kUse in source · 11 passages

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  • Use in source · Read passage · 30_lifting.tex, line 506

    Now suppose a chosen set of e columns of DF(x)bmod p is independent, and put those coordinates first. Apply the square result to the augmented polynomial map Ψ(Y,T)=(F(Y,T),T). Its reduced Jacobian has upper blocks (D_YF,D_TF) and lower blocks (0,I_k), so it is invertible. For t∈ T, the target (0,t) belongs to Ψ(x)+p^L Z_p^e+k because F(x)≡0pmodp^L. The inverse of Ψ at that target gives the unique root with parameters t. The inverse congruence property gives one direction of [source reference]; …

    Exact mathematical passage
    (0,I_k)
I_MUse in source · 1 passage

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I_NUse in source · 5 passages

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  • Use in source · Read passage · 10_foundations.tex, line 264

    Δ_h^s f(n) · =∑_j=0^s(-1)^s-jbinom sj f(n+jh). The finite interpolant I_Nf(n)=∑_r=0^Nc_r(f)binom nr agrees with f for n≤ N. It has the same order-s unit-step difference at n whenever n+s≤ N. For arbitrary coefficients c_0,…,c_N, the corresponding binomial interpolant has exactly those coefficients and zero coefficients above N.

    Exact mathematical passage
    I_Nf(n)=\sum_{r=0}^{N}c_r(f)\binom nr
I_pUse in source · 3 passages

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  • Use in source · Read passage · supplement_08.tex, line 66

    pageheading2. The actual integral Hurwitz latticeConnecting differential expressions to coefficient divisibility The specialization works inside Q_p[[t]] and uses factorial-scaled coefficients a_f(d)=d![t^d]f, I_p={f∈ Q_p[[t]]:a_f(d)∈ Z_p for every d}. The source constructs I_p as a subring. Actual exponential-series multiplication gives a_fg(d)=∑_r=0^dbinom dr a_f(r)a_g(d-r), a_Uf(d)=a_f(d+1). These identities prove multiplication and derivative stability of the lattice. Membership does not req…

    Exact mathematical passage
    
     a_f(d)=d![t^d]f,\qquad
     I_p=\{f\in\Q_p[[t]]:a_f(d)\in\Z_p\text{ for every }d\}.
    
\iffUse in source · 8 passages

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  • Use in source · Read passage · continuation_01.tex, line 35

    Define the finite obstruction module and its distinguished class by O_x=R_q^e/imJ_x, o_x=[ρ_q(r_x)]∈ O_x. The checked criterion is o_x=0 iff ∃ y∈ Z^d: y≡ xpmodq^L, F(y)≡0pmodq^L+1. The residual exists by the lower congruence and is unique because q>0. The quotient is finite since its ambient module R_q^e is finite. The theorem permits zero variables or zero equations as well.

    Exact mathematical passage
    
    \boxed{
     o_x=0
     \iff
     \exists y\in\mathbb Z^d:\quad
     y\equiv x\pmod{q^L},\qquad F(y)\equiv0\pmod{q^{L+1}}.
    }
    
\operatorname{im}Use in source · 7 passages

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  • Use in source · Read passage · 71_native.tex, line 25

    O_F,bar x= F_p^r/imJ, o_F,bar x=[e].

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    O_{F,\bar x}=\mathbb F_p^r/\operatorname{im}J,\qquad o_{F,\bar x}=[e].
\infUse in source · 1 passage

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  • Use in source · Read passage · supplement_04.tex, line 104

    For a uniform coefficient hypothesis, the simpler conclusion follows: inf_j≥0 w(s+j)≤ord_p(Δ_1^s f(n)). For each finite L≤ B_p(w;n,s), a separate theorem yields p^LmidΔ_1^s f(n) inside Z_p. Another specialization handles arbitrary finite interpolants with infinite weights beyond their degree.

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     \boxed{\inf_{j\geq0} w(s+j)\leq\ord_p(\Delta_1^s f(n)).}
    
\inf_{j\geq0}Use in source · 1 passage

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  • Use in source · Read passage · supplement_04.tex, line 104

    For a uniform coefficient hypothesis, the simpler conclusion follows: inf_j≥0 w(s+j)≤ord_p(Δ_1^s f(n)). For each finite L≤ B_p(w;n,s), a separate theorem yields p^LmidΔ_1^s f(n) inside Z_p. Another specialization handles arbitrary finite interpolants with infinite weights beyond their degree.

    Exact mathematical passage
    
     \boxed{\inf_{j\geq0} w(s+j)\leq\ord_p(\Delta_1^s f(n)).}
    
J^pUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 216

    [Augmentation criterion] Let x=∑_u∈ F_pc_u[u] ∈ F_p[ F_p^+]. Then x is a unit if and only if aug(x) = ∑_uc_u is nonzero. If J is the augmentation ideal, then J^p=0.

    Exact mathematical passage
    
    J^p=0.
    
J_aUse in source · 3 passages

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  • Use in source · Read passage · continuation_09.tex, line 89

    At depth N≥1, suppose F(a)=q^Nr and a desired next integer parameter target satisfies t'-free(a)=q^Nw, with w∈ Z^k. Surjectivity supplies a solved digit u∈ S_q^m satisfying J_a u=-bar r-K_abar w. Choose an integer representative widetilde u, and set a'=a+q^N(widetilde u,w). The parameter quotient is the supplied integer w. Therefore free(a')=t' exactly. The checked polynomial Taylor identity gives F(a')≡ F(a)+q^NDF(a)(widetilde u,w) ≡0pmodq^N+1. Its quadratic remainder is divisible by q^2N, and …

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    J_a u=-\bar r-K_a\bar w.
    
J_{F,\bar x}Use in source · 1 passage

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  • Use in source · Read passage · 50_open_questions.tex, line 53

    The original conjecture asks for a canonical finite obstruction packet whose vanishing is necessary and sufficient for a one-level lift in a finitely presented arithmetic seed system. For a fixed finite polynomial presentation, the paper now supplies the precise obstruction o_F,bar x=[r]∈ coker (J_F,bar x: F_p^d⟶ F_p^e), where r is the residual digit. We have proved independence from the chosen integer representative. Invariance under a change of equations or presentation requires more.

    Exact mathematical passage
    
    o_{F,\bar x}=[r]\in
    \coker\!\left(J_{F,\bar x}:\F_p^d\longrightarrow\F_p^e\right),
    
J_{ij}Use in source · 1 passage

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  • Use in source · Read passage · 71_native.tex, line 20

    e_i=F_i(x)/p^Lpmod p, J_ij=frac∂ F_i∂ X_j(x)pmod p.

    Exact mathematical passage
    e_i=F_i(x)/p^L\pmod p,\qquad J_{ij}=\frac{\partial F_i}{\partial X_j}(x)\pmod p.
J_rUse in source · 6 passages

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  • Use in source · Read passage · root_scope.tex, line 128

    The scalar packet weights have exact valuations λ_r(q)=3q+s_2(r+q)-s_2(r). Put μ_r=min_q≥0λ_r(q), J_r={q:λ_r(q)=μ_r}, G_r=2^-μ_rH_r. Since λ_r(0)=0 and λ_r(q)≥3q+1-s_2(r), every minimizing address satisfies 0≤ q≤lfloorfracs_2(r)-13rfloor. In particular μ_r≤0, and the minimum is found by a finite carry calculation. Reduction of the whole series gives overline G_r(X)=∑_q∈ J_rX^underline q in F_2[X]. Its highest degree is max J_r, with leading coefficient one. Thus the normalization is the actual G…

    Exact mathematical passage
    
     \mu_r=\min_{q\ge0}\lambda_r(q),\qquad
     J_r=\{q:\lambda_r(q)=\mu_r\},\qquad G_r=2^{-\mu_r}H_r.
    
j_rUse in source · 4 passages

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  • Use in source · Read passage · root_scope.tex, line 290

    For a general depth r, call j_r=max J_r its true terminal. This definition depends on the actual carry minimum; rewriting the same integer as another wall expression does not change j_r. The universal simplicity target is for every integer r≥1 and x∈ Z_2, G_r(x)=0 Longrightarrow G_r'(x)≠0. The following are completed restricted results of the later continuations. Their family statements include a proof that the indicated label is the true terminal throughout the displayed ranges.

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    j_r=\max J_r
J_tUse in source · 2 passages

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  • Use in source · Read passage · 40_regularity.tex, line 174

    [Regularity throughout an open parameter set] Let U⊆ K^k be open, and let h be continuous on U. Suppose that for every t∈ U, F(h(t),t)=0, J_t=D_YF(h(t),t)∈GL_e( K). Then h is analytic at every point of U and its strict derivative is Dh(t)=-J_t^-1D_TF(h(t),t) (t∈ U). At each t∈ U, openness turns continuity on U into continuity at t and makes the root equation hold on a neighborhood of t. Apply Theorem~[source reference] at (h(t),t). This is leanKomorebi.PolynomialAnalyticSelection.continuous_sele…

    Exact mathematical passage
    
     F(h(t),t)=0,
     \qquad J_t=D_YF(h(t),t)\in\operatorname{GL}_e(\mathbb K).
    
J_t^{-1}Use in source · 2 passages

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  • Use in source · Read passage · 40_regularity.tex, line 179

    [Regularity throughout an open parameter set] Let U⊆ K^k be open, and let h be continuous on U. Suppose that for every t∈ U, F(h(t),t)=0, J_t=D_YF(h(t),t)∈GL_e( K). Then h is analytic at every point of U and its strict derivative is Dh(t)=-J_t^-1D_TF(h(t),t) (t∈ U). At each t∈ U, openness turns continuity on U into continuity at t and makes the root equation hold on a neighborhood of t. Apply Theorem~[source reference] at (h(t),t). This is leanKomorebi.PolynomialAnalyticSelection.continuous_sele…

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     Dh(t)=-J_t^{-1}D_TF(h(t),t)\qquad(t\in U).
    
J_{x'}Use in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 134

    [Independence of the integer representative] Under the hypotheses of Theorem~[source reference], let x'=x+q^Ls and write F(x')=q^Lr_x'. Then J_x'=J_x, overliner_x'=overliner_x+J_xoverline s. Consequently the two residual classes coincide in the common quotient S_q^e/im J_x.

    Exact mathematical passage
    
     J_{x'}=J_x,\qquad
     \overline{r_{x'}}=\overline{r_x}+J_x\overline s.
    
J_{z(n)}Use in source · 1 passage

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  • Use in source · Read passage · continuation_09.tex, line 40

    _0, t(n+1)-t(n)∈ q^L+n Z^k, there is a single tower z: N→ Z^m+k with z(0)=x, free(z(n))=t(n), F(z(n))∈ q^L+n Z^e, z(n+1)-z(n)∈ q^L+n Z^m+k. The parameter equality is exact in integers. For a≤ b, z(b)-z(a)∈ q^L+a Z^m+k, DF(z(n))bmod q=DF(x)bmod q. Consequently J_z(n)=J_x at every level. No later surjectivity or per-step solvability is assumed. This theorem includes positive composite bases and q=1.

    Exact mathematical passage
    J_{z(n)}=J_x
J_{z_n}Use in source · 4 passages

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  • Use in source · Read passage · 30_lifting.tex, line 338

    [Compatible polynomial tower] Let q≥1, L≥1, F∈ Z[X_1,…,X_d]^e, and x∈ Z^d satisfy F(x)∈ q^L Z^e. Suppose J_x:S_q^d→ S_q^e is surjective. There is a single sequence z_n∈ Z^d such that z_0 · =x, · F(z_n) · ∈ q^L+n Z^e, z_n+1-z_n · ∈ q^L+n Z^d, · J_z_n · =J_x. For a≤ b, it also satisfies z_b-z_a∈ q^L+a Z^d.

    Exact mathematical passage
    
     z_0&=x,& F(z_n)&\in q^{L+n}\Z^e,\\
     z_{n+1}-z_n&\in q^{L+n}\Z^d,& J_{z_n}&=J_x.
    
J_{z_{n+1}}Use in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 350

    At level N=L+n, divide the residual by q^N. Surjectivity supplies t∈ S_q^d with J_z_nt=-overlineF(z_n)/q^N. Choose an integer representative and apply [source reference] to obtain z_n+1. The new point is congruent to z_n modulo q. Evaluation of each formal partial derivative therefore gives J_z_n+1=J_z_n.

    Exact mathematical passage
    J_{z_{n+1}}=J_{z_n}
K^dUse in source · 1 passage

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  • Use in source · Read passage · continuation_07.tex, line 19

    mathbb Q_p, and d,s,M∈ N. Let F,H∈ R[X_1,…,X_d]^d, let a∈ R^d satisfy F(a)=0, and set J=DF(a), G=F+p^M H. Assume J∈GL_d(K), p^sJ^-1∈ M_d(R), M>2s. Then there is exactly one root b of G in a+p^s+1R^d. It satisfies b-a∈ p^M-sR^d, and there is ε∈ p^2M-3sR^d such that, in K^d, b-a=-p^M J^-1H(a)+ε. The inverse in this formula belongs to the original Jacobian DF(a). The theorem constructs the perturbed root from these hypotheses and proves uniqueness throughout the full ball of depth s+1. It does not …

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    K^d
\mathbb K^nUse in source · 1 passage

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  • Use in source · Read passage · 40_regularity.tex, line 28

    [Derivative of polynomial evaluation] For P∈ K[X_1,…,X_n], polynomial evaluation is analytic at every point. At a∈ K^n its strict Fr echet derivative is v⟼∑_j=1^n (∂_jP)(a)v_j. For a finite system, its strict derivative is the continuous linear map represented by the matrix of evaluated formal partial derivatives. For the derivative statement, use polynomial induction on constants, addition, and multiplication by a variable. Constants have zero derivative. The sum rule agrees with the additivity…

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    a\in\mathbb K^n
k_0Use in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 226

    For 0≤ i<j, let d_i=v_2(i!)+v_2((j-1-i)!), k_i=1+lfloorlog_2max(i,j-1-i)rfloor, B_i=i+2^k_i Z_2, t_i=d_i+k_i. At j=1 use d_0=k_0=t_0=0. The balls B_i partition Z_2. Indeed a residue represented twice among 0,…,j-1 splits into its two children; every other represented residue remains one ball. On each such ball, P_j(i+2^k_iy)=2^t_iyQ_i(y), Q_i∈ Z_2⟨ y⟩^×, Q_i≡1pmod2.

    Exact mathematical passage
    d_0=k_0=t_0=0
K_{4,j}Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 482

    The Hurwitz coefficients of the normalized source K_4,j at degrees 4n are g_j(n)=binom4n+j4n^-1. Their explicit expressions are g_0(n) · =1, · g_1(n) · =frac14n+1,

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    K_{4,j}
K_aUse in source · 2 passages

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  • Use in source · Read passage · continuation_09.tex, line 89

    At depth N≥1, suppose F(a)=q^Nr and a desired next integer parameter target satisfies t'-free(a)=q^Nw, with w∈ Z^k. Surjectivity supplies a solved digit u∈ S_q^m satisfying J_a u=-bar r-K_abar w. Choose an integer representative widetilde u, and set a'=a+q^N(widetilde u,w). The parameter quotient is the supplied integer w. Therefore free(a')=t' exactly. The checked polynomial Taylor identity gives F(a')≡ F(a)+q^NDF(a)(widetilde u,w) ≡0pmodq^N+1. Its quadratic remainder is divisible by q^2N, and …

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    J_a u=-\bar r-K_a\bar w.
    
K_cUse in source · 3 passages

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  • Use in source · Read passage · supplement_09.tex, line 151

    Removing a singular forced zero. Put c=ap<N and define the actual normalized kernel K_c(t)=c!∑_n≥0fract^Nn(Nn+c)!, K_c(0)=1. Let ℓ(T)=log(1+T)/T, defined by shifting logarithm coefficients, not by inverting T. Set U(T)=ℓ(T)^cK_c(log(1+T)). The formal proof establishes U(0)=1 and the exact factorization H_N,c(log(1+T))=fracT^cc!U(T). It follows coefficientwise that coefTdU=c!coefTd+cH_N,c(log(1+T)). Because p^amid(ap)! and lfloor(d+ap)/prfloor=a+lfloor d/prfloor, the already proved section estima…

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     K_c(t)=c!\sum_{n\ge0}\frac{t^{Nn}}{(Nn+c)!},\qquad K_c(0)=1.
    
K_{c-1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 172

    Boundary. The original positive-address section has zero constant coefficient. It is not the series to which unit inverse closure is applied. 4. The singular ratio estimate. For a positive address c, the rational ratio is independently constructed from the original kernels: R_c(t)=c K_c-1(t)K_c(t)^-1. The scalar factor c is necessary. The source proves the cleared identity R_c(t)H_N,c(t)=tH_N,c-1(t). This connects the constructed object to the original unnormalized ratio A_c-1/A_c. It does not a…

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     R_c(t)=c\,K_{c-1}(t)K_c(t)^{-1}.
    
k_iUse in source · 3 passages

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  • Use in source · Read passage · root_scope.tex, line 219

    For 0≤ i<j, let d_i=v_2(i!)+v_2((j-1-i)!), k_i=1+lfloorlog_2max(i,j-1-i)rfloor, B_i=i+2^k_i Z_2, t_i=d_i+k_i. At j=1 use d_0=k_0=t_0=0. The balls B_i partition Z_2. Indeed a residue represented twice among 0,…,j-1 splits into its two children; every other represented residue remains one ball. On each such ball, P_j(i+2^k_iy)=2^t_iyQ_i(y), Q_i∈ Z_2⟨ y⟩^×, Q_i≡1pmod2.

    Exact mathematical passage
    
     d_i=v_2(i!)+v_2((j-1-i)!),\qquad
     k_i=1+\left\lfloor\log_2\max(i,j-1-i)\right\rfloor,
    
K_jUse in source · 2 passages

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  • Use in source · Read passage · supplement_10.tex, line 173

    Let K_j(t) be the normalized section kernel with constant coefficient one. The constructed unnormalized Euler series is F_N,j(t)=j! K_j(t)^-1. The source/inverse identities and the factorial successor identity prove F_N,j+1=F_N,jR_j+1, F_N,j=F_N,0∏_c=1^jR_c. The product formula is not an assembly hypothesis. Proved flatness of all factors and the original product-closure theorem establish flatness of the full family.

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    K_j(t)
k_mUse in source · 3 passages

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  • Use in source · Read passage · supplement_03.tex, line 37

    The unit kernel and its reciprocal. Use a squared formal variable u and define k_m=frac2^2m+1(2m+2)!, K(u)=∑_m≥0k_mu^m, β_m=[u^m]K(u)^-1. The constant term is proved to be one. The factorial valuation bound gives v_2(k_m)=2m+1-v_2((2m+2)!)≥0. Thus every k_m belongs to O_2. A general coefficientwise subring theorem is proved by strong induction using the actual unit-inverse coefficient formula. Applied to K, it establishes β_m∈ O_2 for every m≥0. The reciprocal's coefficients are therefore proved…

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     k_m=\frac{2^{2m+1}}{(2m+2)!},\qquad
     K(u)=\sum_{m\geq0}k_mu^m,\qquad
     \beta_m=[u^m]K(u)^{-1}.
    
K_{N,j}Use in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 29

    The normalized source and its inverse are K_N,j(t)=j!∑_n≥0fract^Nn(Nn+j)!, P_N,j(t)=∑_n≥0C_N,j(n)fract^Nn(Nn)!. Their product is one. Clearing factorials in its coefficient of t^Nn gives the triangular system ∑_q=0^nbinomNn+jNqC_N,j(q)=δ_n0. In particular, for n>0, C_N,j(n)=-binomNn+jNn^-1 ∑_q=0^n-1binomNn+jNqC_N,j(q). The diagonal binomial coefficient is positive, so this recurrence determines the sequence uniquely. The Lean construction begins with this recurrence and derives the reciprocal id…

    Exact mathematical passage
    \label{eq:arith-kernel}
     K_{N,j}(t)=j!\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!},\qquad
     \mathcal P_{N,j}(t)=\sum_{n\ge0}C_{N,j}(n)\frac{t^{Nn}}{(Nn)!}.
    
K_{N,j}^{-1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_01.tex, line 108

    The formalization does not stop at a recurrence. In the variable z=t^N, construct K_N,j(z)=∑_n≥0fracj!(Nn+j)!z^n, F_N,j(z)=∑_n≥0fracC_N,j(n)(Nn)!z^n. It proves the factorial coefficient identity, then F_N,jK_N,j=1 and therefore F_N,j=K_N,j^-1.

    Exact mathematical passage
    F_{N,j}=K_{N,j}^{-1}
K_rUse in source · 4 passages

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  • Use in source · Read passage · root_scope.tex, line 361

    The current focused obstruction is the admitted ordinary-index family r=(17659+2^18T)2^640-161, T∈ Z_≥0, j_r=161. On its surviving chart define the actual, undepleted function K_r(Z)=2^-178G_r(10202+2^15Z). The complete coefficient theorem gives K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z), a_r,b_r∈ Z_2, u_r∈ Z_2^×, R_r∈ Z_2⟨ Z⟩. Its half derivative is a dyadic isometry. It therefore has a unique critical point c_r∈ Z_2. Set Φ_r^act=K_r(c_r). The earlier independent-undepletion notation for this same sc…

    Exact mathematical passage
    
     K_r(Z)=2^{-178}G_r(10202+2^{15}Z).
    
k_rUse in source · 3 passages

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  • Use in source · Read passage · 10_foundations.tex, line 95

    [Constructed unit-triangular coordinates] Let h: N→ K satisfy h_0=1, write H= E(h) and H^-1= E(k), and define U_Ha= D(H E(a)). Then U_H is an equivalence, with inverse U_H^-1a= D(H^-1 E(a)). Its coefficient formulas are (U_Ha)_n=∑_r=0^nbinom nr h_r a_n-r, (U_H^-1a)_n=∑_r=0^nbinom nr k_r a_n-r. The resulting triangular matrix has diagonal entries 1. The transported selectors widehat P_N,j=U_HP_N,jU_H^-1 are idempotent and pairwise orthogonal, and their sum is the identity.

    Exact mathematical passage
    
    (U_Ha)_n=\sum_{r=0}^{n}\binom nr h_r a_{n-r},\qquad
    (U_H^{-1}a)_n=\sum_{r=0}^{n}\binom nr k_r a_{n-r}.
    
K_tUse in source · 2 passages

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  • Use in source · Read passage · continuation_11.tex, line 97

    There is also an explicit open-set version. Let U be open, let g be continuous on U, and suppose for every t∈ U that F(g(t),t)=0, J_t:=D_YF(g(t),t)∈GL_e( K). Writing K_t=D_TF(g(t),t), the theorem mcodecontinuous_selection_analyticOnNhd proves g is analytic at every t∈ U, Dg(t)=-J_t^-1K_t (t∈ U), with a strict Fréchet derivative at each point. The per-point invertibility hypothesis is part of this theorem. This is a transfer statement for continuous local root selections satisfying its hypotheses…

    Exact mathematical passage
    K_t=D_TF(g(t),t)
K_xUse in source · 8 passages

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  • Use in source · Read passage · 30_lifting.tex, line 539

    Suppose now that the split into solved variables and parameter variables is fixed in advance. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0), and assume F(x)=q^Lr_x with q≥1 and L≥1. Write the reduced Jacobian as DF(x)bmod q=(J_x K_x), J_x:S_q^m→ S_q^e, K_x:S_q^k→ S_q^e. For a prescribed parameter digit v∈ S_q^k, let L_x,L,v be the set of next residue roots over x whose parameter coordinates are t_0+q^Lwidetilde v modulo q^L+1.

    Exact mathematical passage
    
     DF(x)\bmod q=(J_x\ K_x),\qquad
     J_x:S_q^m\to S_q^e,\quad K_x:S_q^k\to S_q^e.
    
K_{z_n}Use in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 610

    At level N=L+n, write the supplied parameter increment exactly as t_n+1-t_n=q^Nw with w∈ Z^k. Surjectivity of the solved block gives a digit u satisfying J_z_nu=-overlineF(z_n)/q^N-K_z_noverline w. Choose an integer representative widetilde u and set z_n+1=z_n+q^N(widetilde u,w). Its parameter coordinates equal the supplied target exactly. Taylor expansion proves the next root condition, while congruence modulo q preserves the full reduced Jacobian. Construct the sequence recursively, preserving…

    Exact mathematical passage
    J_{z_n}u=-\overline{F(z_n)/q^N}-K_{z_n}\overline w
K_\ellUse in source · 3 passages

Search aliases: K_ℓ ell

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  • Use in source · Read passage · 72_recovered.tex, line 1277

    the even source subsequences, equivalently the evenized generating functions H_p. It does not identify their full odd-degree source values. For p=3, the original odd source has A_3,1=S_1(3)=3, whereas every odd dyadic source coefficient is zero; the even kernel L_3K_2 cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity H_p=L_pK_ℓ H_ℓ and its proof remain valid. For any two primes p and ℓ, the source sequences A_p,n and A_ℓ,n are…

    Exact mathematical passage
    H_p=L_pK_\ell H_\ell
lUse in source · 2 passages

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  • Use in source · Read passage · continuation_09.tex, line 59

    Exact branching. At any stage n, put N=L+n. For every parameter digit v∈ F_p^k, let L_n,v={bar w∈( Z/p^N+1 Z)^m+k: l bar w≡ z(n)pmodp^N, F(bar w)=0, free(bar w)=t(n)+p^Nvpmodp^N+1 }. Here p^Nv means any integer representative of v, multiplied by p^N; the residue class does not depend on the representative. Then # L_n,v=p^m-e for every n and every v. Surjectivity implies e≤ m. This is a count of distinct actual next-level residue classes. It does not count integer representatives or entire infini…

    Exact mathematical passage
    
    \mathcal L_{n,v}=\left\{\bar w\in(\mathbb Z/p^{N+1}\mathbb Z)^{m+k}:
    \begin{array}{l}
    \bar w\equiv z(n)\pmod{p^N},\\
    F(\bar w)=0,\\
    \operatorname{free}(\bar w)=t(n)+p^Nv\pmod{p^{N+1}}
    \end{array}\right\}.
    
L_{-1}Use in source · 1 passage

Search aliases: L_-1

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  • Use in source · Read passage · strengthening.tex, line 197

    Reciprocal transfer gives L_-1^kF_N,j∈ H_p. Extraction of the packed coefficient gives (S+1)^k C_N,j(n)∈ p^k Z_p. Moreover, Δ_1^k A_N,j(n)=(-1)^n+k(S+1)^k C_N,j(n), so A_N,j has the ordinary p^k difference property. At p=2, L_1=L_-1-1; its integral binomial expansion on F_N,j gives Δ_1^k C_N,j(n)∈2^k Z_2 directly. For N=1 the interior sum in the product formula is empty, so the same argument applies without change; equivalently, F_1,0(t)=e^-t.

    Exact mathematical passage
    L_1=L_{-1}-1
L_{-1}^kUse in source · 1 passage

Search aliases: L_-1^k

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  • Use in source · Read passage · strengthening.tex, line 190

    Reciprocal transfer gives L_-1^kF_N,j∈ H_p. Extraction of the packed coefficient gives (S+1)^k C_N,j(n)∈ p^k Z_p. Moreover, Δ_1^k A_N,j(n)=(-1)^n+k(S+1)^k C_N,j(n), so A_N,j has the ordinary p^k difference property. At p=2, L_1=L_-1-1; its integral binomial expansion on F_N,j gives Δ_1^k C_N,j(n)∈2^k Z_2 directly. For N=1 the interior sum in the product formula is empty, so the same argument applies without change; equivalently, F_1,0(t)=e^-t.

    Exact mathematical passage
    L_{-1}^kF_{N,j}\in\mathcal H_p
L_{-\lambda}^bUse in source · 1 passage

Search aliases: L_-λ^b lambda

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  • Use in source · Read passage · strengthening.tex, line 122

    Set B=A^-1. Its order-zero integrality was established in [source reference]. Assume by induction that L_-λ^bB∈ H_p for b<k. Since AB=1, L_0^k(AB)=(fracD^Np)^k1=0 (k>0). Use [source reference] with μ=-λ. Exactly one term has inverse order b=k: it has a=0 and all e_i=0, and is A L_-λ^k B. In every other term b<k, so its second factor is integral by induction and derivative-shift integrality. Its first factor is integral by the flatness of A, and its scalar is integral. Hence A L_-λ^k B∈ H_p. Mult…

    Exact mathematical passage
    L_{-\lambda}^bB\in\mathcal H_p
L_{-\lambda}^kUse in source · 3 passages

Search aliases: L_-λ^k lambda

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  • Use in source · Read passage · strengthening.tex, line 128

    beginproof Set B=A^-1. Its order-zero integrality was established in [source reference]. Assume by induction that L_-λ^bB∈ H_p for b<k. Since AB=1, L_0^k(AB)=(fracD^Np)^k1=0 (k>0). Use [source reference] with μ=-λ. Exactly one term has inverse order b=k: it has a=0 and all e_i=0, and is A L_-λ^k B. In every other term b<k, so its second factor is integral by induction and derivative-shift integrality. Its first factor is integral by the flatness of A, and its scalar is integral. Hence A L_-λ^k B…

    Exact mathematical passage
    A L_{-\lambda}^k B
L_0Use in source · 3 passages

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  • Use in source · Read passage · supplement_06.tex, line 118

    An independent Lucas sequence. Define L_0=2, L_1=1, L_n+2=L_n+L_n+1. Direct finite matrix arithmetic proves C^2=C+I. Multiplication by C^n gives C^n+2=C^n+C^n+1. Taking traces shows that tr(C^n) satisfies the same recurrence and initial values as L_n. Induction yields tr(C^N)=L_N. The implementation separately proves det C=-1, det(C^N)=(-1)^N, L_N=2F_N+1-F_N. No Binet roots or diagonalizability argument is needed.

    Exact mathematical passage
    
     L_0=2,\quad L_1=1,\quad L_{n+2}=L_n+L_{n+1}.
    
L_0^kUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 123

    Set B=A^-1. Its order-zero integrality was established in [source reference]. Assume by induction that L_-λ^bB∈ H_p for b<k. Since AB=1, L_0^k(AB)=(fracD^Np)^k1=0 (k>0). Use [source reference] with μ=-λ. Exactly one term has inverse order b=k: it has a=0 and all e_i=0, and is A L_-λ^k B. In every other term b<k, so its second factor is integral by induction and derivative-shift integrality. Its first factor is integral by the flatness of A, and its scalar is integral. Hence A L_-λ^k B∈ H_p. Mult…

    Exact mathematical passage
    
     L_0^k(AB)=\left(\frac{D^N}{p}\right)^k1=0\qquad(k>0).
    
L_1Use in source · 3 passages

Search aliases: L_1

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  • Use in source · Read passage · supplement_06.tex, line 118

    An independent Lucas sequence. Define L_0=2, L_1=1, L_n+2=L_n+L_n+1. Direct finite matrix arithmetic proves C^2=C+I. Multiplication by C^n gives C^n+2=C^n+C^n+1. Taking traces shows that tr(C^n) satisfies the same recurrence and initial values as L_n. Induction yields tr(C^N)=L_N. The implementation separately proves det C=-1, det(C^N)=(-1)^N, L_N=2F_N+1-F_N. No Binet roots or diagonalizability argument is needed.

    Exact mathematical passage
    
     L_0=2,\quad L_1=1,\quad L_{n+2}=L_n+L_{n+1}.
    
L_1^kUse in source · 1 passage

Search aliases: L_1^k

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  • Use in source · Read passage · strengthening.tex, line 166

    n adds the factor -(k+1)s in the numerator. Its denominator is a unit, so it belongs to p^k Z_p. The discrete Leibniz formula Δ_1^k(fg)(n) =∑_r=0^kbinom kr (Δ_1^r f)(n)(Δ_1^k-rg)(n+r) then proves Δ_1^k g_N,j(n)∈ p^k Z_p for every k,n. The coefficient at degree Nn of L_1^kG_N,j is precisely p^-kΔ_1^k g_N,j(n), and the other residue classes vanish. Thus the actual source is 1-flat.

    Exact mathematical passage
    L_1^kG_{N,j}
L_2Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1256

    dfrac2^4r+1(2r+1)(2r+2), · p=2. Then for every prime p and every even n≥0, A_p,n = ∑_r=0^n/2 binom n2r λ_p,r E_n-2r. Equivalently, if L_p(t)=K_p(t)^-1, then H_p(t)=L_p(t)secht, where L_p(t)=cosh^2(pt) for odd p, and L_2(t)=(fracsinh(2t)2t)^2 for p=2.

    Exact mathematical passage
    
    L_2(t)=\left(\frac{\sinh(2t)}{2t}\right)^2
    
L_3Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1277

    l source charts] Editorial range correction. This comparison holds for the even source subsequences, equivalently the evenized generating functions H_p. It does not identify their full odd-degree source values. For p=3, the original odd source has A_3,1=S_1(3)=3, whereas every odd dyadic source coefficient is zero; the even kernel L_3K_2 cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity H_p=L_pK_ℓ H_ℓ and its proof remain vali…

    Exact mathematical passage
    L_3K_2
L_4Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 151

    s.state_is_orbit, FibonacciPackets.packet_is_fibonacci targetFibonacciPackets.trace_power_is_lucas, FibonacciPackets.det_power targetFibonacciPackets.jump_charpoly targetFibonacciPackets.fibonacci_packet_recurrence pageheading4. Exact examples and meaningful boundary checksExample 9.3 and supporting results The independently defined Lucas recurrence gives (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11). For N=3, the general theorem therefore specializes to u_k+2=4u_k+1+u_k. The implementation proves th…

    Exact mathematical passage
    
     (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11).
    
L_5Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 151

    s.state_is_orbit, FibonacciPackets.packet_is_fibonacci targetFibonacciPackets.trace_power_is_lucas, FibonacciPackets.det_power targetFibonacciPackets.jump_charpoly targetFibonacciPackets.fibonacci_packet_recurrence pageheading4. Exact examples and meaningful boundary checksExample 9.3 and supporting results The independently defined Lucas recurrence gives (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11). For N=3, the general theorem therefore specializes to u_k+2=4u_k+1+u_k. The implementation proves th…

    Exact mathematical passage
    
     (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11).
    
\mathcal L_{F,L}Use in source · 1 passage

Search aliases: L_F,L mathcal

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  • Use in source · Read passage · 50_open_questions.tex, line 21

    For a fixed polynomial system F∈ Z[X_1,…,X_d]^e, a precision L≥1, and a root bar x modulo p^L, let L_F,L(bar x) ={x'∈( Z/p^L+1 Z)^d: F(x')=0, x'≡bar xpmodp^L}. The finite obstruction theorem decides whether this set is empty from the residual digit and reduced Jacobian. When it is nonempty, the theorem also determines its cardinality and affine kernel structure. The conclusion concerns one level: vanishing of the obstruction permits the next step, but does not guarantee that subsequent steps are…

    Exact mathematical passage
    
    \mathcal L_{F,L}(\bar x)
    =\{x'\in(\Z/p^{L+1}\Z)^d:
    F(x')=0,\ x'\equiv\bar x\pmod{p^L}\}.
    
L_jUse in source · 3 passages

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  • Use in source · Read passage · root_scope.tex, line 204

    For a wall label j≥1 and 0≤ q<j, define e_j(q)=3q+s_2(j-1)-s_2(j-q-1), E_j=min_q<je_j(q), Θ(j)=3j+1+s_2(j-1), L_j=Θ(j)-E_j. Here E_j is a carry minimum, not an Euler number. A wall label is regular when e_j(q)>0 for 1≤ q<j. For r=2^L-j with L>L_j, put δ=L-L_j. The carry identities give μ_r=E_j-δ, J_r={j}, G_r(x)=P_j(x)U(x-j)+2^δ V(x), where U≡1pmod2. For a regular label, also E_j=0 and V≡1pmod2.

    Exact mathematical passage
    
     \Theta(j)=3j+1+s_2(j-1),\qquad L_j=\Theta(j)-E_j.
    
L_mUse in source · 3 passages

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  • Use in source · Read passage · root_scope.tex, line 431

    The latest continuation retains the exact Bernoulli undepletion factor e_r,q=frac1-2^2r-11-2^2r-14^q, q≥0. It is a dyadic unit, including e_r,0=1. Multiplication by this factor preserves the all-q error of the shifted Jacobi moment model. Consequently the actual model critical values satisfy, for every m≥25, v_2(Φ_r,m^act-Φ_r^act) ≥ L_m=8m-4s_2(m)-178, L_m⟶∞. The estimate includes every packet and controls motion of the critical point. There is no ceiling at the undepletion exponent 2r-25. For e…

    Exact mathematical passage
    
     v_2\bigl(\Phi_{r,m}^{\mathrm{act}}-\Phi_r^{\mathrm{act}}\bigr)
     \ge L_m=8m-4s_2(m)-178,
     \qquad L_m\longrightarrow\infty.
     \label{eq:root:actualapprox}
    
L_nUse in source · 2 passages

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  • Use in source · Read passage · supplement_06.tex, line 118

    An independent Lucas sequence. Define L_0=2, L_1=1, L_n+2=L_n+L_n+1. Direct finite matrix arithmetic proves C^2=C+I. Multiplication by C^n gives C^n+2=C^n+C^n+1. Taking traces shows that tr(C^n) satisfies the same recurrence and initial values as L_n. Induction yields tr(C^N)=L_N. The implementation separately proves det C=-1, det(C^N)=(-1)^N, L_N=2F_N+1-F_N. No Binet roots or diagonalizability argument is needed.

    Exact mathematical passage
    
     L_0=2,\quad L_1=1,\quad L_{n+2}=L_n+L_{n+1}.
    
L_{n+1}Use in source · 1 passage

Search aliases: L_n+1

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  • Use in source · Read passage · supplement_06.tex, line 118

    An independent Lucas sequence. Define L_0=2, L_1=1, L_n+2=L_n+L_n+1. Direct finite matrix arithmetic proves C^2=C+I. Multiplication by C^n gives C^n+2=C^n+C^n+1. Taking traces shows that tr(C^n) satisfies the same recurrence and initial values as L_n. Induction yields tr(C^N)=L_N. The implementation separately proves det C=-1, det(C^N)=(-1)^N, L_N=2F_N+1-F_N. No Binet roots or diagonalizability argument is needed.

    Exact mathematical passage
    
     L_0=2,\quad L_1=1,\quad L_{n+2}=L_n+L_{n+1}.
    
L_{n+2}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 118

    An independent Lucas sequence. Define L_0=2, L_1=1, L_n+2=L_n+L_n+1. Direct finite matrix arithmetic proves C^2=C+I. Multiplication by C^n gives C^n+2=C^n+C^n+1. Taking traces shows that tr(C^n) satisfies the same recurrence and initial values as L_n. Induction yields tr(C^N)=L_N. The implementation separately proves det C=-1, det(C^N)=(-1)^N, L_N=2F_N+1-F_N. No Binet roots or diagonalizability argument is needed.

    Exact mathematical passage
    
     L_0=2,\quad L_1=1,\quad L_{n+2}=L_n+L_{n+1}.
    
\mathcal L_{n,v}Use in source · 2 passages

Search aliases: L_n,v mathcal

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  • Use in source · Read passage · continuation_09.tex, line 59

    Exact branching. At any stage n, put N=L+n. For every parameter digit v∈ F_p^k, let L_n,v={bar w∈( Z/p^N+1 Z)^m+k: l bar w≡ z(n)pmodp^N, F(bar w)=0, free(bar w)=t(n)+p^Nvpmodp^N+1 }. Here p^Nv means any integer representative of v, multiplied by p^N; the residue class does not depend on the representative. Then # L_n,v=p^m-e for every n and every v. Surjectivity implies e≤ m. This is a count of distinct actual next-level residue classes. It does not count integer representatives or entire infini…

    Exact mathematical passage
    
    \mathcal L_{n,v}=\left\{\bar w\in(\mathbb Z/p^{N+1}\mathbb Z)^{m+k}:
    \begin{array}{l}
    \bar w\equiv z(n)\pmod{p^N},\\
    F(\bar w)=0,\\
    \operatorname{free}(\bar w)=t(n)+p^Nv\pmod{p^{N+1}}
    \end{array}\right\}.
    
L_pUse in source · 11 passages

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  • Use in source · Read passage · 71_native.tex, line 207

    For a finite set S of primes, positive precisions L_p, and a fixed integer polynomial system F, put

    Exact mathematical passage
    L_p
L_TUse in source · 5 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 258

    To pass from the section to the singular ratio, factor out its initial zero: for c=ap, the section has the form H_c(log(1+T))=fracT^cc! V(T), V(0)=1, V∈ W_p. The weight of c! compensates for the coefficient shift by c. The weighted ring is closed under inversion of series with constant coefficient one and under L_T=log(1+T)(1+T)d/dT. With ℓ(T)=log(1+T)/T, interpreted by coefficient shifting, differentiation of the displayed factorization gives R_c(log(1+T))=c(1+T)ℓ(T)+(L_TV)V^-1. Since pℓ(T)∈ W_…

    Exact mathematical passage
    L_T=\log(1+T)(1+T)d/dT
\mathcal L_{x,L}Use in source · 6 passages

Search aliases: L_x,L mathcal

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  • Use in source · Read passage · 30_lifting.tex, line 281

    At positive polynomial depth, restrict it to the correction fiber T_x={t∈ S_q^d:overliner_x+J_xt=0} and the lift set L_x,L= {y∈( Z/q^L+1 Z)^d:y≡ xpmodq^L, F(y)=0}. The polynomial predicate in the second display is evaluation in the residue ring. Evaluation on canonical integer representatives gives the same predicate, because integer polynomial evaluation respects congruence.

    Exact mathematical passage
    
     \mathcal L_{x,L}=
     \{y\in(\Z/q^{L+1}\Z)^d:y\equiv x\pmod{q^L},\ F(y)=0\}.
    
\mathcal L_{x,L,v}Use in source · 3 passages

Search aliases: L_x,L,v mathcal

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  • Use in source · Read passage · 30_lifting.tex, line 543

    Suppose now that the split into solved variables and parameter variables is fixed in advance. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0), and assume F(x)=q^Lr_x with q≥1 and L≥1. Write the reduced Jacobian as DF(x)bmod q=(J_x K_x), J_x:S_q^m→ S_q^e, K_x:S_q^k→ S_q^e. For a prescribed parameter digit v∈ S_q^k, let L_x,L,v be the set of next residue roots over x whose parameter coordinates are t_0+q^Lwidetilde v modulo q^L+1.

    Exact mathematical passage
    \mathcal L_{x,L,v}
\mathcal L_{z_n,L+n}Use in source · 1 passage

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  • Use in source · Read passage · continuation_03.tex, line 100

    Surjectivity implies rank_ F_pA_x=e, e≤ d. Combining this with the preceding exact residue-lift count gives # L_z_n,L+n=p^d-e for every n. These are distinct solutions modulo p^L+n+1 lying above the selected point modulo p^L+n. The original finite obstruction group at each smooth point has cardinality one.

    Exact mathematical passage
    
    \boxed{\#\mathcal L_{z_n,L+n}=p^{d-e}\quad\text{for every }n.}
    
L_{\lambda+\mu}Use in source · 1 passage

Search aliases: L_λ+μ lambda mu

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  • Use in source · Read passage · strengthening.tex, line 75

    [Translated product formula] For N=p^a, a≥1, and λ,μ∈ Z_p, L_λ+μ(AB) =(L_λ A)B+A(L_μ B) +∑_i=1^N-1c_i(D^iA)(D^N-iB), c_i=frac1pbinom Ni∈ Z. Flatness types add under multiplication.

    Exact mathematical passage
    
     L_{\lambda+\mu}(AB)
      =(L_\lambda A)B+A(L_\mu B)
           +\sum_{i=1}^{N-1}c_i(D^iA)(D^{N-i}B),
     \qquad c_i=\frac1p\binom Ni\in\mathbb Z.
     \label{eq:translated-product}
    
L_{\lambda+\mu}^kUse in source · 1 passage

Search aliases: L_λ+μ^k lambda mu

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  • Use in source · Read passage · strengthening.tex, line 99

    For completeness, the iterated formula is an integral multinomial formula, rather than a product rule with unspecified remainders. On the tensor of the two factors use the mutually commuting operators L_λ⊗1, 1⊗ L_μ, c_iD^i⊗ D^N-i (1≤ i<N). Applying multiplication after the kth power of their sum gives exactly L_λ+μ^k(AB). Expanding it yields terms indexed by a,b,e_1,…,e_N-1≥0, with a+b+∑_i e_i=k: frack!a!b!∏_i e_i!∏_i c_i^e_i (D^α L_λ^a A)(D^β L_μ^b B), α=∑_i i e_i, β=∑_i(N-i)e_i. Every scalar i…

    Exact mathematical passage
    L_{\lambda+\mu}^k(AB)
L_\lambda^aUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 101

    e with unspecified remainders. On the tensor of the two factors use the mutually commuting operators L_λ⊗1, 1⊗ L_μ, c_iD^i⊗ D^N-i (1≤ i<N). Applying multiplication after the kth power of their sum gives exactly L_λ+μ^k(AB). Expanding it yields terms indexed by a,b,e_1,…,e_N-1≥0, with a+b+∑_i e_i=k: frack!a!b!∏_i e_i!∏_i c_i^e_i (D^α L_λ^a A)(D^β L_μ^b B), α=∑_i i e_i, β=∑_i(N-i)e_i. Every scalar is integral. If A and B have the stated flatness types, both factors in every term belong to H_p, as …

    Exact mathematical passage
    
     \frac{k!}{a!b!\prod_i e_i!}\prod_i c_i^{e_i}
     (D^\alpha L_\lambda^a A)(D^\beta L_\mu^b B),
     \quad
     \alpha=\sum_i i e_i,\quad
     \beta=\sum_i(N-i)e_i.
     \label{eq:translated-multinomial}
    
L_\muUse in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 75

    [Translated product formula] For N=p^a, a≥1, and λ,μ∈ Z_p, L_λ+μ(AB) =(L_λ A)B+A(L_μ B) +∑_i=1^N-1c_i(D^iA)(D^N-iB), c_i=frac1pbinom Ni∈ Z. Flatness types add under multiplication.

    Exact mathematical passage
    
     L_{\lambda+\mu}(AB)
      =(L_\lambda A)B+A(L_\mu B)
           +\sum_{i=1}^{N-1}c_i(D^iA)(D^{N-i}B),
     \qquad c_i=\frac1p\binom Ni\in\mathbb Z.
     \label{eq:translated-product}
    
L_\mu^bUse in source · 1 passage

Search aliases: L_μ^b mu

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  • Use in source · Read passage · strengthening.tex, line 101

    e with unspecified remainders. On the tensor of the two factors use the mutually commuting operators L_λ⊗1, 1⊗ L_μ, c_iD^i⊗ D^N-i (1≤ i<N). Applying multiplication after the kth power of their sum gives exactly L_λ+μ^k(AB). Expanding it yields terms indexed by a,b,e_1,…,e_N-1≥0, with a+b+∑_i e_i=k: frack!a!b!∏_i e_i!∏_i c_i^e_i (D^α L_λ^a A)(D^β L_μ^b B), α=∑_i i e_i, β=∑_i(N-i)e_i. Every scalar is integral. If A and B have the stated flatness types, both factors in every term belong to H_p, as …

    Exact mathematical passage
    
     \frac{k!}{a!b!\prod_i e_i!}\prod_i c_i^{e_i}
     (D^\alpha L_\lambda^a A)(D^\beta L_\mu^b B),
     \quad
     \alpha=\sum_i i e_i,\quad
     \beta=\sum_i(N-i)e_i.
     \label{eq:translated-multinomial}
    
\lceilUse in source · 1 passage

Search aliases: lceil

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  • Use in source · Read passage · root_scope.tex, line 268

    The largest emergence threshold is max_i t_i=v_2((j-1)!)+lceillog_2 jrceil. For j=2^a, all thresholds equal j-1. There are no roots before that distance and exactly j simple roots at and after it, within the positive-distance range. The equality and pre-wall cases are also resolved: regular labels j≥2 have no roots there; j=1 has one simple odd root at equality.

    Exact mathematical passage
    
     \max_i t_i=v_2((j-1)!)+\lceil\log_2 j\rceil.
    
\operatorname{lcm}Use in source · 5 passages

Search aliases: lcm operatorname

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  • Use in source · Read passage · edition_summary.tex, line 3

    What this final edition establishes. addcontentslinetocsectionWhat this final edition establishes The original external problem is Nishibuchi's Conjecture 3.9 for generalized Euler numbers. Its full parameter range is resolved. If p is prime, m is positive with mmid p-1 or m=2, 0≤ j<mp, and q=lcm(2,p-1), then the original Euler coefficients are p-adically integral and E_mpn+qp^r^(mp,j)≡ E_mpn^(mp,j)pmodp^r (r≥1, n≥ r). The paper states the generating function and the conversion between ordinary …

    Exact mathematical passage
    q=\operatorname{lcm}(2,p-1)
\lfloorUse in source · 45 passages

Search aliases: lfloor

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  • Use in source · Read passage · edition_summary.tex, line 16

    The strongest common-step conclusion in width four is particularly compact. For n≥0 and k,H≥1, ∑_i=0^k(-1)^k-ibinom ki E_4(n+iH)^(4,j) ≡0pmod2^ v_2(k!)+kv_2(H)+A_j(k), where A_0(k)=A_1(k)=k, A_2(k)=max{k+4, 2k+lfloor k/2rfloor+1}, A_3(k)=k+2. There is no starting-row threshold. The exponent is exact in addresses zero and one for every order and positive step. It is exact in address two at every power-of-two order. The proof chapters also give mixed-step forms, the cancellation criterion in addre…

    Exact mathematical passage
    
     A_0(k)=A_1(k)=k,\qquad
     A_2(k)=\max\{k+4,\,2k+\lfloor k/2\rfloor+1\},\qquad
     A_3(k)=k+2.
    
\liminfUse in source · 1 passage

Search aliases: liminf

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  • Use in source · visual_body.tex, line 474

    The radius is 2^liminf_d v_2(a_d)/d in the normalized dyadic absolute value. Hence R_0=R_1=2, R_2=2^5/2, R_3≥2. The exact radius of address three is not asserted. For address two, the lower bound gives convergence inside the claimed radius and the unbounded equality family prevents a larger one.

    Exact mathematical passage
    2^{\liminf_d v_2(a_d)/d}
\liminf_dUse in source · 1 passage

Search aliases: liminf_d liminf

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  • Use in source · visual_body.tex, line 474

    The radius is 2^liminf_d v_2(a_d)/d in the normalized dyadic absolute value. Hence R_0=R_1=2, R_2=2^5/2, R_3≥2. The exact radius of address three is not asserted. For address two, the lower bound gives convergence inside the claimed radius and the unbounded equality family prevents a larger one.

    Exact mathematical passage
    2^{\liminf_d v_2(a_d)/d}
\mathrm{lin}Use in source · 3 passages

Search aliases: lin mathrm

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  • Use in source · Read passage · 72_recovered.tex, line 2533

    [Linear lift obstruction] The linear obstruction at depth L is Ob^lin_L(x_L) = [ρ_L(x_L)] ∈ coker(overline M).

    Exact mathematical passage
    
    \boxed{
    \operatorname{Ob}^{\mathrm{lin}}_L(x_L)
    =
    [\rho_L(x_L)]
    \in
    \coker(\overline M).
    }
    
\log_2Use in source · 11 passages

Search aliases: log_2 log

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  • Use in source · Read passage · 20_arithmetic.tex, line 616

    [Finite Newton dominance] Let f: N→ Q and write a_k=Δ_1^kf(0). Suppose that c∈ N, a_1≠0, v_2(a_1)=c, and v_2(a_k)≥ c+1+lfloorlog_2krfloor (k≥2). Then for every n≥0 and T>0, f(n+T)-f(n)≠0, v_2(f(n+T)-f(n))=v_2(T)+c.

    Exact mathematical passage
    
     v_2(a_k)\ge c+1+\lfloor\log_2k\rfloor\qquad(k\ge2).
    
M^kUse in source · 1 passage

Search aliases: M^k

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  • Use in source · Read passage · 72_recovered.tex, line 2164

    [Edge coefficients of the affine kernel] For every M≥1, s≥0, and k≥0, Θ_M,s(k,k)=M^k, Θ_M,s(Mk+s,k)=1. For binary radix and zero address, Θ_2,0(k+j,k) = 2^k-jbinom kj (0≤ j≤ k). Hence v_2(Θ_2,0(k+j,k)) = k-j+v_2binom kj.

    Exact mathematical passage
    
    \boxed{
    \Theta_{M,s}(k,k)=M^k,
    \qquad
    \Theta_{M,s}(Mk+s,k)=1.
    }
    
M_1Use in source · 2 passages

Search aliases: M_1

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  • Use in source · Read passage · 72_recovered.tex, line 2017

    [Nested discrete refinement] For positive integers M,N, (a_r^[M])_s^[N] = a_r+Ms^[MN]. For a mixed radix path (M_1,…,M_ℓ) with digits 0≤ s_i<M_i, the final address is r( s) = s_1+M_1s_2+⋯+ (∏_i=1^ℓ-1M_i)s_ℓ. Only the product radix and accumulated address determine the final branch.

    Exact mathematical passage
    (M_1,\ldots,M_\ell)
m_1Use in source · 1 passage

Search aliases: m_1

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  • Use in source · Read passage · strengthening.tex, line 394

    The conversion from unit steps to arbitrary mixed steps is used several times, so it is useful to make it explicit. Write Δ=Δ_1. For H>0, Δ_H=∑_m=1^Hbinom HmΔ^m. A mixed term with local orders m_1,…,m_k≥1 and M=∑_ν m_ν therefore has valuation at least M+v_2(M!)+β +∑_ν(v_2(H_ν)-v_2(m_ν)) if the unit-order estimate has offset β for every positive order. The inequalities m-1≥ v_2(m), M≥ k, v_2(M!)≥ v_2(k!) make this at least k+v_2(k!)+β+∑_ν v_2(H_ν). Every expansion is finite. This proves the desir…

    Exact mathematical passage
    m_1,\ldots,m_k\geq1
M_dUse in source · 17 passages

Search aliases: M_d

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  • Use in source · Read passage · 30_lifting.tex, line 686

    [Weighted nonlinear lifting] Let p be prime, d,s,r∈ N, F∈ R[X_1,…,X_d]^d, and a∈ R^d. Set J=DF(a). Suppose J∈GL_d( Q_p), p^sJ^-1∈ M_d(R), F(a)∈ p^rR^d, r>2s. Then there is exactly one root b of F in a+p^s+1R^d, and it satisfies b-a∈ p^r-sR^d. Uniqueness holds throughout the full ball of depth s+1.

    Exact mathematical passage
    \label{eq:weighted-hypotheses}
     J\in\operatorname{GL}_d(\Qp),\qquad
     p^sJ^{-1}\in M_d(R),\qquad F(a)\in p^rR^d,\qquad r>2s.
    
M_IUse in source · 1 passage

Search aliases: M_I

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  • Use in source · Read passage · supplement_06.tex, line 30

    pageheading1. Actual matrix orbits and arithmetic jumpsTheorem 9.1; manuscript page 22 Let R be a commutative ring, I a finite coordinate type and d=|I|. For C∈ M_I(R) and v_0∈ R^I, define orbit(C,v_0,n)=C^nv_0, u_k=packet(C,v_0,N,j,k)=C^Nk+jv_0. The implementation uses actual matrix powers and matrix--vector multiplication. A separately supplied sequence v with v_n+1=Cv_n is proved by induction to equal this power orbit. Thus the source-sequence formulation is connected to the construction.

    Exact mathematical passage
    C\in M_I(R)
M_iUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2017

    [Nested discrete refinement] For positive integers M,N, (a_r^[M])_s^[N] = a_r+Ms^[MN]. For a mixed radix path (M_1,…,M_ℓ) with digits 0≤ s_i<M_i, the final address is r( s) = s_1+M_1s_2+⋯+ (∏_i=1^ℓ-1M_i)s_ℓ. Only the product radix and accumulated address determine the final branch.

    Exact mathematical passage
    0\leq s_i<M_i
m_kUse in source · 1 passage

Search aliases: m_k

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  • Use in source · Read passage · strengthening.tex, line 394

    The conversion from unit steps to arbitrary mixed steps is used several times, so it is useful to make it explicit. Write Δ=Δ_1. For H>0, Δ_H=∑_m=1^Hbinom HmΔ^m. A mixed term with local orders m_1,…,m_k≥1 and M=∑_ν m_ν therefore has valuation at least M+v_2(M!)+β +∑_ν(v_2(H_ν)-v_2(m_ν)) if the unit-order estimate has offset β for every positive order. The inequalities m-1≥ v_2(m), M≥ k, v_2(M!)≥ v_2(k!) make this at least k+v_2(k!)+β+∑_ν v_2(H_ν). Every expansion is finite. This proves the desir…

    Exact mathematical passage
    m_1,\ldots,m_k\geq1
M_pUse in source · 2 passages

Search aliases: M_p

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  • Use in source · Read passage · supplement_08.tex, line 39

    ∑_a=1^q-1frac1pbinom qa (U^a f)(U^q-ag), q=p,p^2. The coefficients of each M_q are proved integral: the middle entries of a prime-power binomial row are divisible by p. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence B(fg)=B(f)g+fB(g)+M_p^2(f,g)-M_p(f,g). The middle terms are essential. At p=3, f=t and g=t^2, the constant coefficient of B(fg) is -2, wher…

    Exact mathematical passage
    
    \boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}
    
M_{p^2}Use in source · 2 passages

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  • Use in source · Read passage · supplement_08.tex, line 39

    ∑_a=1^q-1frac1pbinom qa (U^a f)(U^q-ag), q=p,p^2. The coefficients of each M_q are proved integral: the middle entries of a prime-power binomial row are divisible by p. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence B(fg)=B(f)g+fB(g)+M_p^2(f,g)-M_p(f,g). The middle terms are essential. At p=3, f=t and g=t^2, the constant coefficient of B(fg) is -2, wher…

    Exact mathematical passage
    
    \boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}
    
M_qUse in source · 2 passages

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  • Use in source · Read passage · supplement_08.tex, line 33

    pageheading1. The characteristic-zero product ruleThe normalized defect is not an ordinary derivation Let A be a commutative algebra over a characteristic-zero field K, and let U be an actual K-derivation. The source defines B=fracU^p^2-U^pp, M_q(f,g)=∑_a=1^q-1frac1pbinom qa (U^a f)(U^q-ag), q=p,p^2. The coefficients of each M_q are proved integral: the middle entries of a prime-power binomial row are divisible by p. In Lean they are implemented as natural quotients, with exact division proved b…

    Exact mathematical passage
    
     B=\frac{U^{p^2}-U^p}{p},\qquad
     M_q(f,g)=\sum_{a=1}^{q-1}\frac{1}{p}\binom qa
              (U^a f)(U^{q-a}g),\quad q=p,p^2.
    
m_\nuUse in source · 8 passages

Search aliases: m_ν nu

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  • Use in source · Read passage · strengthening.tex, line 395

    The conversion from unit steps to arbitrary mixed steps is used several times, so it is useful to make it explicit. Write Δ=Δ_1. For H>0, Δ_H=∑_m=1^Hbinom HmΔ^m. A mixed term with local orders m_1,…,m_k≥1 and M=∑_ν m_ν therefore has valuation at least M+v_2(M!)+β +∑_ν(v_2(H_ν)-v_2(m_ν)) if the unit-order estimate has offset β for every positive order. The inequalities m-1≥ v_2(m), M≥ k, v_2(M!)≥ v_2(k!) make this at least k+v_2(k!)+β+∑_ν v_2(H_ν). Every expansion is finite. This proves the desir…

    Exact mathematical passage
    M=\sum_\nu m_\nu
M_\ellUse in source · 1 passage

Search aliases: M_ℓ ell

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  • Use in source · Read passage · 72_recovered.tex, line 2017

    [Nested discrete refinement] For positive integers M,N, (a_r^[M])_s^[N] = a_r+Ms^[MN]. For a mixed radix path (M_1,…,M_ℓ) with digits 0≤ s_i<M_i, the final address is r( s) = s_1+M_1s_2+⋯+ (∏_i=1^ℓ-1M_i)s_ℓ. Only the product radix and accumulated address determine the final branch.

    Exact mathematical passage
    (M_1,\ldots,M_\ell)
\mathbinUse in source · 1 passage

Search aliases: mathbin

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  • Use in source · Read passage · 30_lifting.tex, line 181

    [Common correction and fiber product] For every u,v∈ U, [u]=[v] in C(J,K) Longleftrightarrow ∃ w∈ U: Jw=Ju, Kw=Kv. Both label maps are surjective, and the image of the joint map is exactly the fiber product: im(U→im J×im K, u↦(Ju,Ku)) =im Jmathbin×_C(J,K)im K. If either V or W is finite, then C(J,K) is finite.

    Exact mathematical passage
    \label{eq:joint-fiber-product}
     \im\bigl(U\to\im J\times\im K,\ u\mapsto(Ju,Ku)\bigr)
     =\im J\mathbin{\times}_{C(J,K)}\im K.
    
\maxUse in source · 23 passages

Search aliases: max

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  • Use in source · Read passage · edition_summary.tex, line 16

    The strongest common-step conclusion in width four is particularly compact. For n≥0 and k,H≥1, ∑_i=0^k(-1)^k-ibinom ki E_4(n+iH)^(4,j) ≡0pmod2^ v_2(k!)+kv_2(H)+A_j(k), where A_0(k)=A_1(k)=k, A_2(k)=max{k+4, 2k+lfloor k/2rfloor+1}, A_3(k)=k+2. There is no starting-row threshold. The exponent is exact in addresses zero and one for every order and positive step. It is exact in address two at every power-of-two order. The proof chapters also give mixed-step forms, the cancellation criterion in addre…

    Exact mathematical passage
    
     A_0(k)=A_1(k)=k,\qquad
     A_2(k)=\max\{k+4,\,2k+\lfloor k/2\rfloor+1\},\qquad
     A_3(k)=k+2.
    
\max_iUse in source · 1 passage

Search aliases: max_i max

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  • Use in source · Read passage · root_scope.tex, line 268

    The largest emergence threshold is max_i t_i=v_2((j-1)!)+lceillog_2 jrceil. For j=2^a, all thresholds equal j-1. There are no roots before that distance and exactly j simple roots at and after it, within the positive-distance range. The equality and pre-wall cases are also resolved: regular labels j≥2 have no roots there; j=1 has one simple odd root at equality.

    Exact mathematical passage
    
     \max_i t_i=v_2((j-1)!)+\lceil\log_2 j\rceil.
    
\mcodeUse in source · 1 passage

Search aliases: mcode

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  • Use in source · Read passage · supplement_05.tex, line 221

    _bijective, .address_components, .residue_equivalence targetCRTPackets.packet_factorization, .series_packet_factorization targetFiniteArithmeticExamples.outside_window_not_integral pageheading5. Verification, regressions and remaining workCumulative release evidence The fresh build started with an empty output directory under Lean 4.34.0-rc2 and Mathlib revision mcodeaeb72f1e9981b6d35b426b16dca7759be3e7ba06. All nine dependency revisions matched the runtime manifest and repository heads.

    Exact mathematical passage
    
     \mcode{aeb72f1e9981b6d35b426b16dca7759be3e7ba06}.
    
\min_{0\leq j\leq n}Use in source · 2 passages

Search aliases: min_0≤ j≤ n min leq leq

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  • Use in source · Read passage · 10_foundations.tex, line 306

    [Weighted difference bound] Let f: N→ Z_p and let w: N→ N∪{∞} satisfy w(s+j)≤ord_p(c_s+j(f)) for 0≤ j≤ n. Then ord_p(Δ_1^s f(n))≥ min_0≤ j≤ n {w(s+j)+v_pbinom nj}.

    Exact mathematical passage
    
    \operatorname{ord}_p(\Delta_1^s f(n))\geq
    \min_{0\leq j\leq n}
    \left\{w(s+j)+v_p\binom nj\right\}.
    
\min_{k\leq r\leq Mk+s}Use in source · 1 passage

Search aliases: min_k≤ r≤ Mk+s min leq leq

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  • Use in source · Read passage · 72_recovered.tex, line 2262

    [Polyphase valuation ledger] For every refined coefficient, v_p(d_s,k) ≥ min_k≤ r≤ Mk+s ( v_p(c_r)+v_p(Θ_M,s(r,k)) ). If v_p(c_r)≥ w(r) for a nondecreasing function w, then v_p(d_s,k)≥ w(k). Thus every monotone Mahler precision ladder survives arbitrary polyphase refinement.

    Exact mathematical passage
    
    \boxed{
    \vp(d_{s,k})
    \geq
    \min_{k\leq r\leq Mk+s}
    \left(
    \vp(c_r)+\vp(\Theta_{M,s}(r,k))
    \right).
    }
    
\min_{M+1\leq r\leq n/2}Use in source · 1 passage

Search aliases: min_M+1≤ r≤ n/2 min leq leq

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  • Use in source · Read passage · 72_recovered.tex, line 1409

    [Optimal dyadic truncation] Let T_2,M(n) = ∑_r=0^min(M,n/2) κ_2,rbinom n2rR_n-2r. Then E_n≡ T_2,M(n)pmod 2^4M+3. More precisely, v_2(E_n-T_2,M(n)) ≥ min_M+1≤ r≤ n/2 ( 4r-1+v_2binom n2r ). The uniform exponent 4M+3 is best possible.

    Exact mathematical passage
    
    \boxed{
    \vTwo\bigl(E_n-T_{2,M}(n)\bigr)
    \geq
    \min_{M+1\leq r\leq n/2}
    \left(
    4r-1+\vTwo\binom n{2r}
    \right).
    }
    
\min_{q<j}Use in source · 1 passage

Search aliases: min_q<j min

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  • Use in source · Read passage · root_scope.tex, line 200

    For a wall label j≥1 and 0≤ q<j, define e_j(q)=3q+s_2(j-1)-s_2(j-q-1), E_j=min_q<je_j(q), Θ(j)=3j+1+s_2(j-1), L_j=Θ(j)-E_j. Here E_j is a carry minimum, not an Euler number. A wall label is regular when e_j(q)>0 for 1≤ q<j. For r=2^L-j with L>L_j, put δ=L-L_j. The carry identities give μ_r=E_j-δ, J_r={j}, G_r(x)=P_j(x)U(x-j)+2^δ V(x), where U≡1pmod2. For a regular label, also E_j=0 and V≡1pmod2.

    Exact mathematical passage
    
     e_j(q)=3q+s_2(j-1)-s_2(j-q-1),\qquad
     E_j=\min_{q<j}e_j(q),
    
\min_{q\ge0}Use in source · 1 passage

Search aliases: min_q≥0 min ge

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  • Use in source · Read passage · root_scope.tex, line 128

    The scalar packet weights have exact valuations λ_r(q)=3q+s_2(r+q)-s_2(r). Put μ_r=min_q≥0λ_r(q), J_r={q:λ_r(q)=μ_r}, G_r=2^-μ_rH_r. Since λ_r(0)=0 and λ_r(q)≥3q+1-s_2(r), every minimizing address satisfies 0≤ q≤lfloorfracs_2(r)-13rfloor. In particular μ_r≤0, and the minimum is found by a finite carry calculation. Reduction of the whole series gives overline G_r(X)=∑_q∈ J_rX^underline q in F_2[X]. Its highest degree is max J_r, with leading coefficient one. Thus the normalization is the actual G…

    Exact mathematical passage
    
     \mu_r=\min_{q\ge0}\lambda_r(q),\qquad
     J_r=\{q:\lambda_r(q)=\mu_r\},\qquad G_r=2^{-\mu_r}H_r.
    
\min_{r\geq s}Use in source · 1 passage

Search aliases: min_r≥ s min geq

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  • Use in source · Read passage · 72_recovered.tex, line 1497

    [Mahler valuation ledger] If c_r∈ Z_p and v_p(c_r)≥ w(r), then v_p(Δ^sf(n)) ≥ min_r≥ s ( w(r)+v_pbinom nr-s ).

    Exact mathematical passage
    
    \boxed{
    \vp(\Delta^sf(n))
    \geq
    \min_{r\geq s}
    \left(
    w(r)+\vp\binom n{r-s}
    \right).
    }
    
\min_{\substack{0\leq r\leq n\\w(r)\geq L}}Use in source · 1 passage

Search aliases: min_substack0≤ r≤ n w(r)≥ L min substack leq leq w geq

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  • Use in source · Read passage · 72_recovered.tex, line 553

    w(r)<L binom nr c_rp^w(r)S_n-r pmod p^L. When the omitted set is nonempty, the tail has valuation at least min_substack0≤ r≤ n w(r)≥ L ( w(r)+v_pbinom nr+v_p(S_n-r) ).

    Exact mathematical passage
    
    \boxed{
    \min_{\substack{0\leq r\leq n\\w(r)\geq L}}
    \left(
    w(r)+\vp\binom nr+\vp(S_{n-r})
    \right).
    }
    
\mathbb NUse in source · 5 passages

Search aliases: N mathbb

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  • Use in source · Read passage · continuation_06.tex, line 11

    Let p be prime, R= Z_p, K= Q_p, and d,s,r∈ N. Let F=(F_1,…,F_d)∈ R[X_1,…,X_d]^d, a∈ R^d, and let J=DF(a)=(∂_jF_i(a))_i,j. Assume J∈GL_d(K), p^sJ^-1∈ M_d(R), F(a)∈ p^rR^d, r>2s. Then ∃! b∈ a+p^s+1R^d: F(b)=0, b-a∈ p^r-sR^d. All vector divisibility assertions are coordinatewise. The uniqueness predicate is the entire ball of depth s+1: competitors need not be assumed to have depth r-s. The theorem permits arbitrary integral p-adic coefficients and centers, every prime including two, and dimension …

    Exact mathematical passage
    d,s,r\in\mathbb N
n^2Use in source · 1 passage

Search aliases: n^2

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  • Use in source · Read passage · supplement_04.tex, line 161

    rincipal targets. targetUnitModes.geometric_difference, UnitModes.unit_fermat_divisibility targetUnitModes.unit_mode_valuation_gain targetUnitModes.finite_unit_modes_valuation_gain pageheading4. Exact examples and adversarial checksThe examples use the same definitions as the general results A polynomial anchor. The actual difference table for f(n)=n^2 yields (c_0,c_1,c_2,c_3)=(0,1,2,0). The finite interpolant therefore reconstructs I_2(c)(7)=49. This checks the sign and indexing conventions, no…

    Exact mathematical passage
    f(n)=n^2
n^pUse in source · 1 passage

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n_0Use in source · 3 passages

Search aliases: n_0

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  • Use in source · Read passage · 20_arithmetic.tex, line 173

    chi congruence] Let p be prime, let m>0, and suppose that mmid p-1 or m=2. Put N=mp and q=lcm(2,p-1). For every 0≤ j<N, all U_N,j(n) belong to Z_p. For r≥1 and n≥ r, E_Nn+qp^r^(N,j)≡ E_Nn^(N,j)pmodp^r. Thus the positive sufficient starting index n_0=r works uniformly. If mmid p-1, the stronger sufficient condition r≤ mn is valid.

    Exact mathematical passage
    n_0=r
N_1Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 3026

    [Several coprime moduli] Let N_1,…,N_s be pairwise coprime and put N=∏_j=1^sN_j. For compatible addresses r_j, P_N,r = ∏_j=1^sP_N_j,r_j, where r is the unique combined address modulo N.

    Exact mathematical passage
    N_1,\ldots,N_s
N_jUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 3027

    [Several coprime moduli] Let N_1,…,N_s be pairwise coprime and put N=∏_j=1^sN_j. For compatible addresses r_j, P_N,r = ∏_j=1^sP_N_j,r_j, where r is the unique combined address modulo N.

    Exact mathematical passage
    
    N=\prod_{j=1}^{s}N_j.
    
N_sUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 3026

    [Several coprime moduli] Let N_1,…,N_s be pairwise coprime and put N=∏_j=1^sN_j. For compatible addresses r_j, P_N,r = ∏_j=1^sP_N_j,r_j, where r is the unique combined address modulo N.

    Exact mathematical passage
    N_1,\ldots,N_s
\mathrm{NL}Use in source · 1 passage

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  • Use in source · Read passage · 71_native.tex, line 110

    o(f)∈Ext^1_B(NL_B/ Z, F_p)

    Exact mathematical passage
    o(f)\in\operatorname{Ext}^1_B(\mathrm{NL}_{B/\mathbb Z},\mathbb F_p)
\mathrm{NL}_{B/\mathbb Z}Use in source · 1 passage

Search aliases: NL_B/ Z mathrm mathbb

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  • Use in source · Read passage · 71_native.tex, line 110

    o(f)∈Ext^1_B(NL_B/ Z, F_p)

    Exact mathematical passage
    o(f)\in\operatorname{Ext}^1_B(\mathrm{NL}_{B/\mathbb Z},\mathbb F_p)
\nmidUse in source · 22 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 291

    At a nonsingular address 2≤ c<N with pnmid c, differentiation of the section identities gives tR_c'=R_c(1+R_c-1-R_c). Writing q=R_c-1-(c-1) and y=R_c-c, this becomes (tD+c)y=cq+yq-y^2. On degrees divisible by p, the inverse of tD+c divides the degree-d coefficient by the unit d+c. Starting with y_0=0, form y_s+1=(tD+c)^-1(cq+y_sq-y_s^2). Flatness and support are preserved. Indeed, put E_c=tD+c and C_p=pD^p^2-D^p. The commutator is [B_p,E_c]=C_p, and iteration gives B_p^rE_cy=E_cB_p^ry+rC_pB_p^r-…

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    p\nmid c
\NNUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 124

    [Hurwitz series ring] Let R be a commutative ring with identity. The Hurwitz series ring H(R) is the set R^ N with componentwise addition and product (amathbin⋆ b)_n = ∑_r=0^n binom nr a_r b_n-r. Its identity is δ=(1,0,0,…).

    Exact mathematical passage
    R^{\NN}
\notUse in source · 3 passages

Search aliases: not

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  • Use in source · Read passage · 72_recovered.tex, line 2284

    rime_power_sparsification Editorial hypothesis correction. The affine-kernel congruence below needs no change, but its consequence for transformed Mahler coefficients requires c_r∈ Z_p for every r. Convergence of a Q_p-valued Mahler expansion alone is insufficient. For p=2, M=2, s=0 and f(x)=x/2, one has c_1=1/2, c_2=0, while g(q)=q has d_1=1; thus d_1not≡ c_2pmod2. The coefficient consequence is read with integral original coefficients. Let M=p^a. Then Θ_p^a,s(r,k) ≡ binomsr-p^ak pmod p, where …

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    d_1\not\equiv c_2\pmod2
OUse in source · 1 passage

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  • Use in source · Read passage · 71_native.tex, line 25

    O_F,bar x= F_p^r/imJ, o_F,bar x=[e].

    Exact mathematical passage
    O_{F,\bar x}=\mathbb F_p^r/\operatorname{im}J,\qquad o_{F,\bar x}=[e].
\mathcal OUse in source · 13 passages

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  • Use in source · Read passage · 10_foundations.tex, line 12

    The original arithmetic seed notation S=(A,a_0, O, R,{P_N,j},w,τ_p) records a coefficient ring or module, a seed, allowed operators, relations, residue selectors, a depth weight, and a topology. This notation describes the ingredients of the construction; a category of seed presentations and a notion of equivalence still require separate definitions. We therefore state each theorem for a concrete object. Section~[source reference] discusses the additional structure needed for the original native…

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    \mathcal S=(A,a_0,\mathcal O,\mathcal R,\{P_{N,j}\},w,\tau_p)
    
\mathcal O_2Use in source · 5 passages

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  • Use in source · Read passage · supplement_03.tex, line 30

    cients is not universally formalized here. Formal-series identities do not establish analytic convergence. The corrected global odd-prime finite-difference theorem remains pending. Historical numerical evidence is not presented as a fresh reproduction. pageheading1. Integral coefficients and an exact unit sourceLemmas 5.1 and 5.2; manuscript page 14 Define the rational subring O_2={q∈ Q:|q|_2≤1}. Its ring laws follow from the actual nonarchimedean norm. For every q∈ O_2, the code constructs its …

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     \mathcal O_2=\{q\in\Q:|q|_2\leq1\}.
    
O_{F,\bar x}Use in source · 1 passage

Search aliases: O_F,bar x bar

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  • Use in source · Read passage · 71_native.tex, line 25

    O_F,bar x= F_p^r/imJ, o_F,bar x=[e].

    Exact mathematical passage
    O_{F,\bar x}=\mathbb F_p^r/\operatorname{im}J,\qquad o_{F,\bar x}=[e].
o_{F,\bar x}Use in source · 2 passages

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  • Use in source · Read passage · 50_open_questions.tex, line 53

    The original conjecture asks for a canonical finite obstruction packet whose vanishing is necessary and sufficient for a one-level lift in a finitely presented arithmetic seed system. For a fixed finite polynomial presentation, the paper now supplies the precise obstruction o_F,bar x=[r]∈ coker (J_F,bar x: F_p^d⟶ F_p^e), where r is the residual digit. We have proved independence from the chosen integer representative. Invariance under a change of equations or presentation requires more.

    Exact mathematical passage
    
    o_{F,\bar x}=[r]\in
    \coker\!\left(J_{F,\bar x}:\F_p^d\longrightarrow\F_p^e\right),
    
\mathcal O_xUse in source · 7 passages

Search aliases: O_x mathcal

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  • Use in source · Read passage · 30_lifting.tex, line 72

    [Finite polynomial obstruction] Let q≥1, L≥1, x∈ Z^d, and suppose F(x)∈ q^L Z^e. Define the exact integer residual r_x by F(x)=q^Lr_x, and put O_x=S_q^e/im J_x, o_x=[overliner_x]. Then O_x is finite and o_x=0 Longleftrightarrow ∃ y∈ Z^d: y≡ xpmodq^L, F(y)≡0pmodq^L+1. No surjectivity or nonsingularity assumption is required.

    Exact mathematical passage
    
     \mathcal O_x=S_q^e/\im J_x,\qquad o_x=[\overline{r_x}].
    
\operatorname{Ob}Use in source · 6 passages

Search aliases: Ob operatorname

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  • Use in source · Read passage · 72_recovered.tex, line 2533

    [Linear lift obstruction] The linear obstruction at depth L is Ob^lin_L(x_L) = [ρ_L(x_L)] ∈ coker(overline M).

    Exact mathematical passage
    
    \boxed{
    \operatorname{Ob}^{\mathrm{lin}}_L(x_L)
    =
    [\rho_L(x_L)]
    \in
    \coker(\overline M).
    }
    
\operatorname{Ob}^{F}_LUse in source · 3 passages

Search aliases: Ob^F_L operatorname

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  • Use in source · Read passage · 72_recovered.tex, line 2656

    [Polynomial lift obstruction] The polynomial obstruction at depth L is Ob^F_L(x_L) = [ρ_L^F(x_L)] ∈ coker(overline J_F(x_L)).

    Exact mathematical passage
    
    \boxed{
    \operatorname{Ob}^{F}_L(x_L)
    =
    [\rho_L^F(x_L)]
    \in
    \coker\bigl(\overline J_F(x_L)\bigr).
    }
    
\operatorname{Ob}^{\mathrm{lin}}_LUse in source · 3 passages

Search aliases: Ob^lin_L operatorname mathrm

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  • Use in source · Read passage · 72_recovered.tex, line 2533

    [Linear lift obstruction] The linear obstruction at depth L is Ob^lin_L(x_L) = [ρ_L(x_L)] ∈ coker(overline M).

    Exact mathematical passage
    
    \boxed{
    \operatorname{Ob}^{\mathrm{lin}}_L(x_L)
    =
    [\rho_L(x_L)]
    \in
    \coker(\overline M).
    }
    
\operatorname{orbit}Use in source · 1 passage

Search aliases: orbit operatorname

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  • Use in source · Read passage · supplement_06.tex, line 31

    pageheading1. Actual matrix orbits and arithmetic jumpsTheorem 9.1; manuscript page 22 Let R be a commutative ring, I a finite coordinate type and d=|I|. For C∈ M_I(R) and v_0∈ R^I, define orbit(C,v_0,n)=C^nv_0, u_k=packet(C,v_0,N,j,k)=C^Nk+jv_0. The implementation uses actual matrix powers and matrix--vector multiplication. A separately supplied sequence v with v_n+1=Cv_n is proved by induction to equal this power orbit. Thus the source-sequence formulation is connected to the construction.

    Exact mathematical passage
    
     \operatorname{orbit}(C,v_0,n)=C^nv_0,
     \qquad u_k=\operatorname{packet}(C,v_0,N,j,k)=C^{Nk+j}v_0.
    
\operatorname{ord}Use in source · 4 passages

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  • Use in source · Read passage · 10_foundations.tex, line 295

    For x∈ Z_p, extend the valuation to zero by setting ord_p(x)= v_p(x), · x≠0, ∞, · x=0. Thus L≤ord_p(x) is equivalent to x∈ p^L Z_p.

    Exact mathematical passage
    
    \operatorname{ord}_p(x)=
    \begin{cases}v_p(x),&x\ne0,\\\infty,&x=0.\end{cases}
    
\ordUse in source · 8 passages

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  • Use in source · Read passage · supplement_04.tex, line 24

    The additive valuation is extended to ord_p(0)=∞. This avoids incorrectly applying Mathlib's finite-valued default at zero to an inequality that must also cover cancellation and identically zero differences.

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    \ord_p(0)=\infty
\ord_3Use in source · 1 passage

Search aliases: ord_3 ord operatorname

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  • Use in source · Read passage · supplement_04.tex, line 177

    ubsection*A sharp binomial-weight improvement Let F(n)=81binom n1+3binom n2+3binom n3+81binom n4. At n=3 and s=1, the four coefficient valuations are (4,1,1,4), while the binomial valuations are (0,1,1,0). Thus the refined bound is B_3(w;3,1)=min(4,2,2,4)=2, stronger than the coefficient-only lower bound one. The source checks both the minimum and the actual result: Δ F(3)=180, ord_3(180)=2. The valuation equality is proved in the actual completed ring Z_3, with an independent exact integer divi…

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     \boxed{\Delta F(3)=180,\qquad\ord_3(180)=2.}
    
\operatorname{ord}_pUse in source · 4 passages

Search aliases: ord_p operatorname

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  • Use in source · Read passage · 10_foundations.tex, line 295

    For x∈ Z_p, extend the valuation to zero by setting ord_p(x)= v_p(x), · x≠0, ∞, · x=0. Thus L≤ord_p(x) is equivalent to x∈ p^L Z_p.

    Exact mathematical passage
    
    \operatorname{ord}_p(x)=
    \begin{cases}v_p(x),&x\ne0,\\\infty,&x=0.\end{cases}
    
\ord_pUse in source · 7 passages

Search aliases: ord_p ord operatorname

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  • Use in source · Read passage · supplement_04.tex, line 24

    The additive valuation is extended to ord_p(0)=∞. This avoids incorrectly applying Mathlib's finite-valued default at zero to an inequality that must also cover cancellation and identically zero differences.

    Exact mathematical passage
    \ord_p(0)=\infty
\overline CUse in source · 4 passages

Search aliases: overline C overline

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  • Use in source · Read passage · 20_arithmetic.tex, line 135

    [Lucas reduction and address independence] Let p be prime, m>0, and N=mp. For 0≤ j<p and n,q∈ N, binommpn+jmpq≡binommnmqpmod p. Every C_mp,j(n) belongs to Z_p. Its reduction is the unique sequence overline C_m: N→ F_p satisfying ∑_q=0^nbinommnmqoverline C_m(q)=δ_n0 in F_p. Consequently, for any two addresses 0≤ j,k<p, C_mp,j(n)≡ C_mp,k(n)pmod p (n∈ N).

    Exact mathematical passage
    \overline C_m:\N\to\F_p
\overline C_mUse in source · 4 passages

Search aliases: overline C_m overline

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  • Use in source · Read passage · 20_arithmetic.tex, line 135

    [Lucas reduction and address independence] Let p be prime, m>0, and N=mp. For 0≤ j<p and n,q∈ N, binommpn+jmpq≡binommnmqpmod p. Every C_mp,j(n) belongs to Z_p. Its reduction is the unique sequence overline C_m: N→ F_p satisfying ∑_q=0^nbinommnmqoverline C_m(q)=δ_n0 in F_p. Consequently, for any two addresses 0≤ j,k<p, C_mp,j(n)≡ C_mp,k(n)pmod p (n∈ N).

    Exact mathematical passage
    \overline C_m:\N\to\F_p
\overline GUse in source · 2 passages

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  • Use in source · Read passage · root_scope.tex, line 140

    ad J_r={q:λ_r(q)=μ_r}, G_r=2^-μ_rH_r. Since λ_r(0)=0 and λ_r(q)≥3q+1-s_2(r), every minimizing address satisfies 0≤ q≤lfloorfracs_2(r)-13rfloor. In particular μ_r≤0, and the minimum is found by a finite carry calculation. Reduction of the whole series gives overline G_r(X)=∑_q∈ J_rX^underline q in F_2[X]. Its highest degree is max J_r, with leading coefficient one. Thus the normalization is the actual Gauss normalization.

    Exact mathematical passage
    
     \overline G_r(X)=\sum_{q\in J_r}X^{\underline q}
           \quad\text{in }\mathbb F_2[X].
     \label{eq:root:reduction}
    
\overline G_rUse in source · 2 passages

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  • Use in source · Read passage · root_scope.tex, line 140

    ad J_r={q:λ_r(q)=μ_r}, G_r=2^-μ_rH_r. Since λ_r(0)=0 and λ_r(q)≥3q+1-s_2(r), every minimizing address satisfies 0≤ q≤lfloorfracs_2(r)-13rfloor. In particular μ_r≤0, and the minimum is found by a finite carry calculation. Reduction of the whole series gives overline G_r(X)=∑_q∈ J_rX^underline q in F_2[X]. Its highest degree is max J_r, with leading coefficient one. Thus the normalization is the actual Gauss normalization.

    Exact mathematical passage
    
     \overline G_r(X)=\sum_{q\in J_r}X^{\underline q}
           \quad\text{in }\mathbb F_2[X].
     \label{eq:root:reduction}
    
\overline rUse in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 657

    The solved Jacobian remains surjective, so every shifted obstruction [overline r+K_z_nv] vanishes. Formula~[source reference] therefore gives p^m-e choices for every parameter digit at every stage. When m=e, the fixed-coordinate chart in the proof of Theorem~[source reference] contains every root in the initial ball with the specified parameters. Its uniqueness proves the endpoint assertion. This is uniqueness of the actual endpoint, not of the integer representatives used to approach it.

    Exact mathematical passage
    [\overline r+K_{z_n}v]
\overline sUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 134

    [Independence of the integer representative] Under the hypotheses of Theorem~[source reference], let x'=x+q^Ls and write F(x')=q^Lr_x'. Then J_x'=J_x, overliner_x'=overliner_x+J_xoverline s. Consequently the two residual classes coincide in the common quotient S_q^e/im J_x.

    Exact mathematical passage
    
     J_{x'}=J_x,\qquad
     \overline{r_{x'}}=\overline{r_x}+J_x\overline s.
    
\overline tUse in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 92

    Every point in the lower residue class has the form y=x+q^Lt with t∈ Z^d. By Lemma~[source reference] and 2L≥ L+1, F(x+q^Lt)≡ q^L(r_x+DF(x)t)pmodq^L+1. The integer q^L is nonzero. Cancelling it in the corresponding divisibility condition proves F(x+q^Lt)≡0pmodq^L+1 Longleftrightarrow overliner_x+J_xoverline t=0. Integer vectors reduce onto S_q^d. The equation on the right is solvable exactly when overliner_x belongs to im J_x, since that image is closed under negation. This is precisely o_x=0. F…

    Exact mathematical passage
    \label{eq:digit-correction}
     F(x+q^Lt)\equiv0\pmod{q^{L+1}}
     \quad\Longleftrightarrow\quad
     \overline{r_x}+J_x\overline t=0.
    
\overline wUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 610

    At level N=L+n, write the supplied parameter increment exactly as t_n+1-t_n=q^Nw with w∈ Z^k. Surjectivity of the solved block gives a digit u satisfying J_z_nu=-overlineF(z_n)/q^N-K_z_noverline w. Choose an integer representative widetilde u and set z_n+1=z_n+q^N(widetilde u,w). Its parameter coordinates equal the supplied target exactly. Taylor expansion proves the next root condition, while congruence modulo q preserves the full reduced Jacobian. Construct the sequence recursively, preserving…

    Exact mathematical passage
    J_{z_n}u=-\overline{F(z_n)/q^N}-K_{z_n}\overline w
\overline{(G_r-W_kU_k)/2^k}Use in source · 1 passage

Search aliases: overline(G_r-W_kU_k)/2^k overline

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  • Use in source · Read passage · root_scope.tex, line 179

    For preparation, start from the monic polynomial g=overline G_r and its zero-one lift W_1, with U_1=1. Suppose G_r-W_kU_k is coefficientwise divisible by 2^k. Its divided reduction is a finite polynomial because G_r is restricted. Divide it by g in F_2[X]: overline(G_r-W_kU_k)/2^k=gq+s, deg s<d_r. With zero-one coefficient lifts, set W_k+1=W_k+2^ks and U_k+1=U_k+2^kq. The error is now divisible by 2^k+1, the polynomial remains monic of degree d_r, and the unit reduces to one. The sequences conve…

    Exact mathematical passage
    
     \overline{(G_r-W_kU_k)/2^k}=gq+s,\qquad\deg s<d_r.
    
\overline{F(z_n)/q^N}Use in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 347

    At level N=L+n, divide the residual by q^N. Surjectivity supplies t∈ S_q^d with J_z_nt=-overlineF(z_n)/q^N. Choose an integer representative and apply [source reference] to obtain z_n+1. The new point is congruent to z_n modulo q. Evaluation of each formal partial derivative therefore gives J_z_n+1=J_z_n.

    Exact mathematical passage
    J_{z_n}t=-\overline{F(z_n)/q^N}
\overline{r_x}Use in source · 7 passages

Search aliases: overliner_x overline

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  • Use in source · Read passage · 30_lifting.tex, line 72

    [Finite polynomial obstruction] Let q≥1, L≥1, x∈ Z^d, and suppose F(x)∈ q^L Z^e. Define the exact integer residual r_x by F(x)=q^Lr_x, and put O_x=S_q^e/im J_x, o_x=[overliner_x]. Then O_x is finite and o_x=0 Longleftrightarrow ∃ y∈ Z^d: y≡ xpmodq^L, F(y)≡0pmodq^L+1. No surjectivity or nonsingularity assumption is required.

    Exact mathematical passage
    
     \mathcal O_x=S_q^e/\im J_x,\qquad o_x=[\overline{r_x}].
    
\overline{r_{x'}}Use in source · 1 passage

Search aliases: overliner_x' overline

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  • Use in source · Read passage · 30_lifting.tex, line 134

    [Independence of the integer representative] Under the hypotheses of Theorem~[source reference], let x'=x+q^Ls and write F(x')=q^Lr_x'. Then J_x'=J_x, overliner_x'=overliner_x+J_xoverline s. Consequently the two residual classes coincide in the common quotient S_q^e/im J_x.

    Exact mathematical passage
    
     J_{x'}=J_x,\qquad
     \overline{r_{x'}}=\overline{r_x}+J_x\overline s.
    
\overline{\omega_\ell(a)}Use in source · 1 passage

Search aliases: overlineω_ℓ(a) overline omega ell

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  • Use in source · Read passage · supplement_05.tex, line 94

    The root equation and residue are derived. Euler's unit-group exponent gives ω_ℓ(a)^p-1 =a^p^ℓ-1(p-1)=a^φ(p^ℓ)=1. Reduction to the actual field F_p commutes with powers. Its Frobenius identity gives overlineω_ℓ(a) =bar a^p^ℓ-1=bar a. Consequently, a=ω_ℓ(a)⟨ a⟩_ℓ, ω_ℓ(a)^p-1=1, overline⟨ a⟩_ℓ=1. The manuscript's finite existence conclusion is obtained directly. The proof does not assume a Hensel lift or an already supplied root representative.

    Exact mathematical passage
    
     \overline{\omega_\ell(a)}
     =\bar a^{p^{\ell-1}}=\bar a.
    
\overline{\langle a\rangle_\ell}Use in source · 1 passage

Search aliases: overline⟨ a⟩_ℓ overline langle rangle ell

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  • Use in source · Read passage · supplement_05.tex, line 99

    The root equation and residue are derived. Euler's unit-group exponent gives ω_ℓ(a)^p-1 =a^p^ℓ-1(p-1)=a^φ(p^ℓ)=1. Reduction to the actual field F_p commutes with powers. Its Frobenius identity gives overlineω_ℓ(a) =bar a^p^ℓ-1=bar a. Consequently, a=ω_ℓ(a)⟨ a⟩_ℓ, ω_ℓ(a)^p-1=1, overline⟨ a⟩_ℓ=1. The manuscript's finite existence conclusion is obtained directly. The proof does not assume a Hensel lift or an already supplied root representative.

    Exact mathematical passage
    
     \boxed{a=\omega_\ell(a)\langle a\rangle_\ell,\quad
     \omega_\ell(a)^{p-1}=1,\quad
     \overline{\langle a\rangle_\ell}=1.}
    
PUse in source · 113 passages

Search aliases: P

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  • Use in source · Read passage · 10_foundations.tex, line 12

    The original arithmetic seed notation S=(A,a_0, O, R,{P_N,j},w,τ_p) records a coefficient ring or module, a seed, allowed operators, relations, residue selectors, a depth weight, and a topology. This notation describes the ingredients of the construction; a category of seed presentations and a notion of equivalence still require separate definitions. We therefore state each theorem for a concrete object. Section~[source reference] discusses the additional structure needed for the original native…

    Exact mathematical passage
    
    \mathcal S=(A,a_0,\mathcal O,\mathcal R,\{P_{N,j}\},w,\tau_p)
    
\mathcal PUse in source · 9 passages

Search aliases: P mathcal

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  • Use in source · Read passage · 20_arithmetic.tex, line 29

    The normalized source and its inverse are K_N,j(t)=j!∑_n≥0fract^Nn(Nn+j)!, P_N,j(t)=∑_n≥0C_N,j(n)fract^Nn(Nn)!. Their product is one. Clearing factorials in its coefficient of t^Nn gives the triangular system ∑_q=0^nbinomNn+jNqC_N,j(q)=δ_n0. In particular, for n>0, C_N,j(n)=-binomNn+jNn^-1 ∑_q=0^n-1binomNn+jNqC_N,j(q). The diagonal binomial coefficient is positive, so this recurrence determines the sequence uniquely. The Lean construction begins with this recurrence and derives the reciprocal id…

    Exact mathematical passage
    \label{eq:arith-kernel}
     K_{N,j}(t)=j!\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!},\qquad
     \mathcal P_{N,j}(t)=\sum_{n\ge0}C_{N,j}(n)\frac{t^{Nn}}{(Nn)!}.
    
p^{(L+a)r}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2401

    by the retained proof. Without integral observation vectors or positive L, the preceding counterexamples invalidate the claimed precision. Assume C^H=I+p^LB. Put H_a=Hp^a. Then C^H_a=I+p^L+aB_a for an integral matrix B_a. Every refined packet b_a,j(x)=a_H_ax+j has Mahler coefficients satisfying v_p(c_a,j,r)≥(L+a)r. Equivalently, Δ_H_a^ ra_n ≡0 pmod p^(L+a)r.

    Exact mathematical passage
    
    \boxed{
    \Delta_{H_a}^{\,r}a_n
    \equiv0
    \pmod {p^{(L+a)r}}.
    }
    
p^{-(L+n)}Use in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 378

    Cross-level compatibility bounds the supremum distance between later terms by p^-(L+n). Completeness gives a limit, and closed norm bounds retain the congruence to every z_n. View the coefficients of the original integer polynomials in Z_p. Their evaluations satisfy both F_i(z_n)→ F_i(a), by continuity, and F_i(z_n)→0, by their growing divisibility. Uniqueness of limits gives exact vanishing.

    Exact mathematical passage
    p^{-(L+n)}
p^{-1}Use in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 15

    Unless a different coefficient ring is specified, a polynomial system is a finite list of integer polynomials. All divisibility conditions on vectors are coordinatewise. For a positive integer q, write S_q= Z/q Z and use an overline for reduction modulo q. For a prime p, write F_p=S_p, normalize |p|_p=p^-1, and put B_L(x)=x+p^L Z_p^d. An integer center is included in this notation by its natural image in Z_p^d. All dimensions below are finite and may be zero. A statement over S_q allows composit…

    Exact mathematical passage
    |p|_p=p^{-1}
p^{-2r+\vp((2r)!)}Use in source · 1 passage

Search aliases: p^-2r+vp((2r)!) p^-2r+v_p((2r)!) vp

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  • Use in source · Read passage · 72_recovered.tex, line 1874

    Every retained base is a unit, and its (p-1)st power lies in 1+p Z_p. The polynomial p^2rbinom(p-1)x+j2r has Gauss norm at most p^-2r+v_p((2r)!). This tends to zero because p is odd. The remaining factors have bounded Gauss norm. Thus the series converges in the Tate algebra. At a nonnegative integer x, the binomial factor vanishes for large r, and [source reference] gives the value.

    Exact mathematical passage
    
    p^{-2r+\vp((2r)!)}.
    
p^{-k}Use in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 333

    Here is the coefficient extraction explicitly. If a_d=d![t^d]F, then d![t^d]B_p^kF=p^-kΔ_p(p-1)^ka_d+pk. For F=j! P_N,j and k≤ mn, take d=mpn-pk. Support in multiples of mp turns this into p^-kΔ_h^kU_N,j(n), so flatness gives divisibility by p^k. For k>mn, expand Δ_h^k-mn applied to Δ_h^mnU_N,j(n); every shifted row is at least n, so each summand is divisible by p^mn. Finally Δ_uh=((I+Δ_h)^u-I), whose kth power for u>0 is an integral combination of powers of Δ_h of order at least k. The case u=0…

    Exact mathematical passage
    
     d![t^d]B_p^kF=p^{-k}\Delta_{p(p-1)}^ka_{d+pk}.
    
p^{-s}Use in source · 2 passages

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  • Use in source · Read passage · supplement_08.tex, line 91

    The earlier coefficient criterion is recovered exactly. Define the literal coefficient defect (δ a)(d)=a(d+p^2)-a(d+p). The new source proves, for every s,d, a_B^s f(d)=p^-s(δ^s a_f)(d). For an integral coefficient x, it proves the bridge p^-sx∈ Z_p Longleftrightarrow p^smid x in Z_p. Thus the differential definition is equivalent to the existing coefficient-level Dwork criterion. Division occurs in Q_p, while the assertion after division is membership in Z_p. It is not divisibility in the field…

    Exact mathematical passage
    
     a_{B^s f}(d)=p^{-s}(\delta^s a_f)(d).
    
P^2Use in source · 5 passages

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  • Use in source · Read passage · strengthening.tex, line 326

    , and b=(1-i)/2. The exact identity F_0(t)=frac2cosh t+cos t =sech(at)sech(bt) gives, for U=a E+b F, L(U^4n)=C_0(n). Although a and b have negative valuation, their denominators disappear from the following exact polynomial identities: U^2 · = E F+i(P-Q),nonumber H_0:=fracU^4-12 · =P+Q-fracP^22-fracQ^22 +3PQ+i E F(P-Q).

    Exact mathematical passage
    
     U^2&=\mathcal E\mathcal F+i(P-Q),\nonumber\\
     H_0:=\frac{U^4-1}{2}
       &=P+Q-\frac{P^2}{2}-\frac{Q^2}{2}
                     +3PQ+i\mathcal E\mathcal F(P-Q).
     \label{eq:four-euler-H}
    
p^2Use in source · 16 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 209

    Work in the Hurwitz integral ring of formal series H_p={F∈ Q_p[[t]]: d![t^d]F∈ Z_p for every d}. For D=d/dt, let B_p=(D^p^2-D^p)/p. Call F flat if B_p^sF∈ H_p for every s≥0. We will prove that j! P_N,j is flat at every address. The main difficulty is to include the addresses at which the usual unit-pivot argument fails.

    Exact mathematical passage
    B_p=(D^{p^2}-D^p)/p
p^{2(n-q)}Use in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 805

    [Classical odd-length reconstruction] For every positive odd integer p and every n∈ N, E_2n=∑_m=0^nbinom2n2mγ_m p^2mS_2(n-m)(p). Equivalently, E_2n=∑_q=0^n V_nqS_2q(p) for the lower triangular matrix V_nq=binom2n2qγ_n-qp^2(n-q) (q≤ n), V_nq=0 (q>n). It has V_nn=1 and p^2(n-q)mid V_nq when q≤ n.

    Exact mathematical passage
    
     V_{nq}=\binom{2n}{2q}\gamma_{n-q}p^{2(n-q)}\quad(q\le n),
     \qquad V_{nq}=0\quad(q>n).
    
p^{2(r-q)}Use in source · 2 passages

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  • Use in source · Read passage · supplement_02.tex, line 142

    The row action is connected to the matrix entries. Reversing the finite sum and using binomial symmetry gives the actual source-index row action E_2r=∑_q=0^rU_r,q(p)S_2q(p), where U_r,q(p)= displaystylebinom2r2qb_r-qp^2(r-q), · q≤ r, 0, · q>r. The row-action equality, the diagonal identity U_r,r=1, and vanishing above the diagonal are all checked. The arithmetic weight is also explicit: p^2(r-q)mid U_r,q(p) in Z, 0≤ q≤ r. No inverse-coordinate identity, unit diagonal or weight bound is assumed f…

    Exact mathematical passage
    
     U_{r,q}(p)=
     \begin{cases}
     \displaystyle\binom{2r}{2q}b_{r-q}p^{2(r-q)},&q\le r,\\
     0,&q>r.
     \end{cases}
    
p^{2a(M+1)}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1020

    [Odd modulus truncation] For every M≥0, every positive odd q, and every even n, E_n ≡ ∑_r=0^min(M,n/2) (-1)^rZ_2r+1 binom n2rq^2rS_n-2r(q) pmod q^2M+2. If pmid q and a=v_p(q), the same truncation is valid modulo p^2a(M+1).

    Exact mathematical passage
    p^{2a(M+1)}
p^{2D}Use in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 798

    For the error estimate, let δ=b-a and D=M-s. Taylor's theorem for the preconditioned polynomial gives E_F=(AF)(b)-(AF)(a)-p^sδ∈ p^2DR^d. Polynomial congruence gives E_H=(AH)(b)-(AH)(a)∈ p^DR^d. The equations F(a)=0 and G(b)=0 imply p^sδ+p^MAH(a)=-E_F-p^ME_H∈ p^2DR^d, because M+D≥2D. Cancel p^s in the integral domain R to obtain δ+p^M-sAH(a)∈ p^2M-3sR^d. This integral vector is ε. After inclusion into Q_p, p^M-sA=p^MJ^-1 gives [source reference].

    Exact mathematical passage
    
     E_F=(AF)(b)-(AF)(a)-p^s\delta\in p^{2D}R^d.
    
p^{2L}Use in source · 4 passages

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  • Use in source · Read passage · 71_native.tex, line 41

    Indeed, each term involving at least two correction factors is divisible by p^2L, and 2L≥ L+1. Constant and linear correction terms give the two displayed summands. Adding the monomial expansions proves (1) for every polynomial, with no denominator or analytic convergence issue.

    Exact mathematical passage
    p^{2L}
p^{2M+2}Use in source · 6 passages

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  • Use in source · Read passage · 10_foundations.tex, line 237

    begingroupsmallraggedright Formal correspondence. leanFiltered.filtered_triangular_reconstruction and leanFiltered.padic_filtered_reconstruction. For weights 2r, retaining r≤ M gives the bound p^2M+2 through leanFiltered.even_weight_tail. A sharper kernel weight can improve this bound, as in the classical dyadic reconstruction later in the paper. Each coefficient row is finite, so no hypothesis w(r)→∞ is needed here. endgroup

    Exact mathematical passage
    p^{2M+2}
p^{2M-3s}Use in source · 9 passages

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  • Use in source · Read passage · 30_lifting.tex, line 767

    [Weighted coefficient deformation] Let F,H∈ R[X_1,…,X_d]^d, F(a)=0, and J=DF(a). Suppose J∈GL_d( Q_p), p^sJ^-1∈ M_d(R), and M>2s. For G=F+p^MH, there is exactly one root b∈ a+p^s+1R^d. It satisfies b-a∈ p^M-sR^d and, in Q_p^d, b-a=-p^MJ^-1H(a)+ε, ε∈ p^2M-3sR^d. Only the original Jacobian DF(a) is subject to an inverse hypothesis.

    Exact mathematical passage
    \label{eq:coefficient-first-order}
     b-a=-p^MJ^{-1}H(a)+\varepsilon,
     \qquad\varepsilon\in p^{2M-3s}R^d.
    
p^{2r}Use in source · 5 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1166

    0, · p=2 and n odd, and κ_p,r = (-1)^rZ_2r+1p^2r, · p odd, β_r2^2r(2r)!, · p=2.

    Exact mathematical passage
    
    \kappa_{p,r}
    =
    \begin{cases}
    (-1)^rZ_{2r+1}p^{2r},&p\text{ odd},\\
    \beta_r2^{2r}(2r)!,&p=2.
    \end{cases}
    
p^{2s+1}Use in source · 11 passages

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  • Use in source · Read passage · 30_lifting.tex, line 711

    Precondition the polynomial system by A and set c=p^s+1. Its derivative at a is AJ=p^sI_d. Lemma~[source reference] gives a polynomial vector Q with Q(0)=0 and (AF)(a+cu)=(AF)(a)+p^2s+1u+p^2s+2Q(u). Since A is integral, (AF)(a)∈ p^rR^d. The strict integer inequality r>2s permits k=r-(2s+1)≥0 and a vector C∈ p^kR^d such that (AF)(a)=p^2s+1C. The normalized root equation is therefore u=-C-pQ(u).

    Exact mathematical passage
    
     (AF)(a+cu)=(AF)(a)+p^{2s+1}u+p^{2s+2}Q(u).
    
p^{2s+2}Use in source · 4 passages

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  • Use in source · Read passage · 30_lifting.tex, line 711

    Precondition the polynomial system by A and set c=p^s+1. Its derivative at a is AJ=p^sI_d. Lemma~[source reference] gives a polynomial vector Q with Q(0)=0 and (AF)(a+cu)=(AF)(a)+p^2s+1u+p^2s+2Q(u). Since A is integral, (AF)(a)∈ p^rR^d. The strict integer inequality r>2s permits k=r-(2s+1)≥0 and a vector C∈ p^kR^d such that (AF)(a)=p^2s+1C. The normalized root equation is therefore u=-C-pQ(u).

    Exact mathematical passage
    
     (AF)(a+cu)=(AF)(a)+p^{2s+1}u+p^{2s+2}Q(u).
    
p^{[r]}Use in source · 1 passage

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  • Use in source · Read passage · 10_foundations.tex, line 209

    To isolate the filtration argument, let p be any element of a commutative ring and let w: N→ N∪{∞}. Set p^[r]=p^r for finite r and p^[∞]=0. The latter notation means the coefficient vanishes.

    Exact mathematical passage
    p^{[r]}=p^r
p^{[w(r)]}Use in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 215

    [Filtered finite reconstruction] For every n,L∈ N and sequences c,S in that ring, · ∑_r=0^nbinom nr c_rp^[w(r)]S_n-r · ≡ ∑_substack0≤ r≤ n w(r)<L binom nr c_rp^[w(r)]S_n-rpmodp^L. Here congruence means divisibility of the difference by p^L in the coefficient ring. In Z_p, it means membership in p^L Z_p.

    Exact mathematical passage
    
    \begin{split}
    &\sum_{r=0}^{n}\binom nr c_rp^{[w(r)]}S_{n-r}\\
    &\hspace{8mm}\equiv
    \sum_{\substack{0\leq r\leq n\\w(r)<L}}
    \binom nr c_rp^{[w(r)]}S_{n-r}\pmod{p^L}.
    \end{split}
    
p^{[w]}Use in source · 1 passage

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  • Use in source · Read passage · supplement_01.tex, line 131

    ach earlier clearing denominator to the common product, then cancels the new nonzero diagonal. This is a denominator bound, not a claim that every normalized coefficient is integral. pageheading3. Filtered tails and the standard windowSource items 2.2 and 8.2 Filtration at a finite coefficient row. Let R be any commutative ring. For w∈ N∪{∞} define p^[w]=p^w, · w<∞, 0, · w=∞. The infinity convention expresses a vanishing coefficient, not an infinite exponent operation. For any finite index set, …

    Exact mathematical passage
    
     p^{[w]}=\begin{cases}p^w,&w<\infty,\\0,&w=\infty.\end{cases}
    
p^{[\infty]}Use in source · 1 passage

Search aliases: p^[∞] infty

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  • Use in source · Read passage · 10_foundations.tex, line 210

    To isolate the filtration argument, let p be any element of a commutative ring and let w: N→ N∪{∞}. Set p^[r]=p^r for finite r and p^[∞]=0. The latter notation means the coefficient vanishes.

    Exact mathematical passage
    p^{[\infty]}=0
p^{a+b}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2127

    The cosets s+p^a Z_p form a disjoint partition of Z_p. The first formula is the inverse affine change of variables on one coset. The second follows from p^a(p^bx+t)+s = p^a+bx+p^at+s.

    Exact mathematical passage
    
    p^a(p^bx+t)+s
    =
    p^{a+b}x+p^at+s.
    
P^bUse in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 48

    ficient in prime row p gives (z-1)^p∈ PI. Every other degree-p product of the two generators of I also belongs to PI. The formal ideal proof therefore establishes I^p⊆ PI. For q=plfloor q/prfloor+(qbmod p), strong induction and the factorial valuation identity give I^q⊆ P^v_p(q!). In detail, write b=lfloor q/prfloor. Then I^q⊆ P^b I^b⊆ P^b+v_p(b!)=P^v_p(q!). No integral-domain hypothesis or division in a field is used here.

    Exact mathematical passage
    
     I^q\subseteq P^b I^b\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.
    
p^bUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2127

    The cosets s+p^a Z_p form a disjoint partition of Z_p. The first formula is the inverse affine change of variables on one coset. The second follows from p^a(p^bx+t)+s = p^a+bx+p^at+s.

    Exact mathematical passage
    
    p^a(p^bx+t)+s
    =
    p^{a+b}x+p^at+s.
    
P^{b+v_p(b!)}Use in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 48

    ficient in prime row p gives (z-1)^p∈ PI. Every other degree-p product of the two generators of I also belongs to PI. The formal ideal proof therefore establishes I^p⊆ PI. For q=plfloor q/prfloor+(qbmod p), strong induction and the factorial valuation identity give I^q⊆ P^v_p(q!). In detail, write b=lfloor q/prfloor. Then I^q⊆ P^b I^b⊆ P^b+v_p(b!)=P^v_p(q!). No integral-domain hypothesis or division in a field is used here.

    Exact mathematical passage
    
     I^q\subseteq P^b I^b\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.
    
p^DUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 802

    For the error estimate, let δ=b-a and D=M-s. Taylor's theorem for the preconditioned polynomial gives E_F=(AF)(b)-(AF)(a)-p^sδ∈ p^2DR^d. Polynomial congruence gives E_H=(AH)(b)-(AH)(a)∈ p^DR^d. The equations F(a)=0 and G(b)=0 imply p^sδ+p^MAH(a)=-E_F-p^ME_H∈ p^2DR^d, because M+D≥2D. Cancel p^s in the integral domain R to obtain δ+p^M-sAH(a)∈ p^2M-3sR^d. This integral vector is ε. After inclusion into Q_p, p^M-sA=p^MJ^-1 gives [source reference].

    Exact mathematical passage
    E_H=(AH)(b)-(AH)(a)\in p^DR^d
p^{d-d}Use in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 392

    For square uniqueness, let a,b∈ B_L(x) be arbitrary exact roots. Their canonical integer truncations are genuine roots in the respective finite residue rings, since evaluation commutes with reduction. They agree at depth L. If they agree at depth N≥ L, both next residues belong to the same lift set, which has cardinality p^d-d=1. Induction gives agreement at every depth. The residue maps separate p-adic integers, so a=b. Existence was already supplied by the tower.

    Exact mathematical passage
    p^{d-d}=1
p^{d-e}Use in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 372

    [Completion and smooth branching] In Theorem~[source reference], let q=p be prime. The tower can be chosen with a limit a∈ Z_p^d such that F(a)=0, z_n⟶ a, a-z_n∈ p^L+n Z_p^d (n≥0). At each stage, the number of distinct next-level residue roots over z_n is exactly p^d-e, and the finite obstruction module has one element. If d=e, there is exactly one root of F in B_L(x).

    Exact mathematical passage
    p^{d-e}
p^{d-\operatorname{rank}J}Use in source · 1 passage

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  • Use in source · Read passage · 71_native.tex, line 33

    Theorem. For this fixed polynomial presentation, the group and distinguished class are independent of the chosen integer representative. The class vanishes if and only if bar x lifts to a solution modulo p^L+1. The obstruction group has p^r-rankJ elements. When lifts exist, their set is a torsor for ker J and contains exactly p^d-rankJ elements.

    Exact mathematical passage
    p^{d-\operatorname{rank}J}
p^{d-\rho}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2589

    [Exact number of linear children] If ρ=rank_ F_poverline M, then a solvable depth L class has exactly p^d-ρ children modulo p^L+1.

    Exact mathematical passage
    
    \boxed{
    p^{d-\rho}
    }
    
p^{d-\rho_x}Use in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 295

    [Torsor and lift cardinalities] Under the hypotheses of Theorem~[source reference], there is an equivalence T_x≃ L_x,L. Every nonempty fiber is a torsor for ker J_x under addition of correction digits. If q=p is prime and ρ_x=rank_ F_pJ_x, then # L_x,L= p^d-ρ_x, · o_x=0, 0, · o_x≠0, # O_x=p^e-ρ_x.

    Exact mathematical passage
    \label{eq:lift-count}
     \#\mathcal L_{x,L}=
     \begin{cases}p^{d-\rho_x},&o_x=0,\\0,&o_x\ne0,\end{cases}
     \qquad
     \#\mathcal O_x=p^{e-\rho_x}.
    
p^{e-\rho}Use in source · 1 passage

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  • Use in source · Read passage · continuation_08.tex, line 64

    When q=p is prime, put ρ=rank_ F_pJ. Then # L_v= p^m-ρ, · o_x(v)=0, 0, · o_x(v)≠0, # C_x=p^e-ρ. The count is of distinct actual residue classes at level L+1. The solved dimension m, rather than the total number m+k of variables, appears in the exponent because the parameter digit is fixed. These rank formulas are asserted only over prime fields.

    Exact mathematical passage
    
    \boxed{\#\mathcal L_v=
    \begin{cases}
    p^{m-\rho},&o_x(v)=0,\\
    0,&o_x(v)\ne0,
    \end{cases}}
    \qquad
    \boxed{\#\mathcal C_x=p^{e-\rho}.}
    
p^{e-\rho_x}Use in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 295

    [Torsor and lift cardinalities] Under the hypotheses of Theorem~[source reference], there is an equivalence T_x≃ L_x,L. Every nonempty fiber is a torsor for ker J_x under addition of correction digits. If q=p is prime and ρ_x=rank_ F_pJ_x, then # L_x,L= p^d-ρ_x, · o_x=0, 0, · o_x≠0, # O_x=p^e-ρ_x.

    Exact mathematical passage
    \label{eq:lift-count}
     \#\mathcal L_{x,L}=
     \begin{cases}p^{d-\rho_x},&o_x=0,\\0,&o_x\ne0,\end{cases}
     \qquad
     \#\mathcal O_x=p^{e-\rho_x}.
    
p^iUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 209

    To retain the valuation of a step, let B be any sequence satisfying Δ_1^m B∈ p^m Z_p at every order and row. Use the finite operator identity Δ_H=∑_i=1^Hbinom HiΔ_1^i. The relation ibinom Hi=HbinomH-1i-1 implies v_p(p^ibinom Hi) ≥ i+v_p(H)-v_p(i)≥1+v_p(H). Multiplying these identities for the k steps, every term of total order M has its scalar multiplied by p^M, and is divisible by p^k+∑_ν v_p(H_ν). This proves the mixed-step bound. Finally, Δ_H A_N,j(n)=(-1)^n+HT_H C_N,j(n). Iteration contribut…

    Exact mathematical passage
    
     v_p\left(p^i\binom Hi\right)
     \geq i+v_p(H)-v_p(i)\geq1+v_p(H).
    
p^kUse in source · 15 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 338

    Here is the coefficient extraction explicitly. If a_d=d![t^d]F, then d![t^d]B_p^kF=p^-kΔ_p(p-1)^ka_d+pk. For F=j! P_N,j and k≤ mn, take d=mpn-pk. Support in multiples of mp turns this into p^-kΔ_h^kU_N,j(n), so flatness gives divisibility by p^k. For k>mn, expand Δ_h^k-mn applied to Δ_h^mnU_N,j(n); every shifted row is at least n, so each summand is divisible by p^mn. Finally Δ_uh=((I+Δ_h)^u-I), whose kth power for u>0 is an integral combination of powers of Δ_h of order at least k. The case u=0…

    Exact mathematical passage
    p^k
p^{k(1+v_p(u))}Use in source · 1 passage

Search aliases: p^k(1+v_p(u))

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  • Use in source · Read passage · 20_arithmetic.tex, line 557

    To incorporate v_2(u), expand a displacement using finite Newton coefficients. For any prime p, u>0, and 1≤ a≤ u, p^v_p(u)+1midbinom ua p^a, as follows from abinom ua=ubinomu-1a-1 and v_p(a)<a. Repeated displacement, retaining the remaining base-difference order at each induction step, gives a factor p^k(1+v_p(u)) from k displacements. This induction is formalized in leanKomorebiPaperIX.StepAmplification.repeated_precision. Applying it with p=2 proves [source reference]; multiplication by j! pro…

    Exact mathematical passage
    p^{k(1+v_p(u))}
p^{k+\sum_\nu v_p(H_\nu)}Use in source · 1 passage

Search aliases: p^k+∑_ν v_p(H_ν) sum nu nu

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  • Use in source · Read passage · strengthening.tex, line 215

    ng Δ_1^m B∈ p^m Z_p at every order and row. Use the finite operator identity Δ_H=∑_i=1^Hbinom HiΔ_1^i. The relation ibinom Hi=HbinomH-1i-1 implies v_p(p^ibinom Hi) ≥ i+v_p(H)-v_p(i)≥1+v_p(H). Multiplying these identities for the k steps, every term of total order M has its scalar multiplied by p^M, and is divisible by p^k+∑_ν v_p(H_ν). This proves the mixed-step bound. Finally, Δ_H A_N,j(n)=(-1)^n+HT_H C_N,j(n). Iteration contributes only a global sign. Multiplying by j! proves [source reference…

    Exact mathematical passage
    p^{k+\sum_\nu v_p(H_\nu)}
p^{k-1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 145

    c branch For 1≤ k≤ℓ, let ρ_ℓ,k: Z/p^ℓ Z⟶ Z/p^k Z be the actual reduction homomorphism. It induces a map on units. The explicit construction satisfies ρ_ℓ,k(ω_ℓ(a)) =ω_k(ρ_ℓ,k(a)), ρ_ℓ,k(⟨ a⟩_ℓ) =⟨ρ_ℓ,k(a)⟩_k. The proof uses the exponent factorization p^ℓ-1=p^k-1p^ℓ-k and the already proved Frobenius invariance of a lifted root. Principal factors also satisfy ⟨ a⟩_ℓ^ p^ℓ-1=1. At precision one the principal factor is exactly one. These are coherent finite certificates; no infinite inverse limit or…

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    p^{\ell-1}=p^{k-1}p^{\ell-k}
p^{ks}Use in source · 1 passage

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  • Use in source · Read passage · supplement_04.tex, line 120

    -field reductionTheorem 7.4; manuscript page 18 For u in any commutative ring, and every natural step h, iteration proves Δ_h^s(u^n)=u^n(u^h-1)^s. The proof unfolds the actual forward-difference operator and uses the power multiplication law. No valuation premise or geometric-difference identity is assumed. More generally, a proved divisibility p^kmid u^h-1 implies p^ksmidΔ_h^s(u^n).

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     p^{ks}\mid\Delta_h^s(u^n).
    
p^{L(R+1-d)}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1657

    f degree d. Let H=Λ_p,L, fix an address j, and write Q_j(x)=P(Hx+j), y=u^H, q=y-1. Then f(Hx+j)=∑_r≥0c_j,rbinom xr, where c_j,r = u^j ∑_k=0^min(d,r) binom rk Δ^kQ_j(0)y^kq^r-k. Consequently, v_p(c_j,r)≥ Lmax(r-d,0). For R≥ d, truncation at depth R is valid modulo p^L(R+1-d) uniformly on Z_p.

    Exact mathematical passage
    p^{L(R+1-d)}
p^{L+1}Use in source · 19 passages

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  • Use in source · Read passage · 10_foundations.tex, line 363

    [Finite unit coordinates] Let p be prime, L≥0, and a∈( Z/p^L+1 Z)^×. Define ω_L(a)=a^p^L, ⟨ a⟩_L=aω_L(a)^-1. Then a=ω_L(a)⟨ a⟩_L, ω_L(a)^p-1=1, ⟨ a⟩_L≡1pmod p. The two factors are uniquely determined by these properties and are compatible with reduction to lower precision. For a∈( Z/2^L+2 Z)^×, there is instead a unique decomposition a=ϵ(a)⟨ a⟩_4, ϵ(a)∈{1,-1}, ⟨ a⟩_4≡1pmod4.

    Exact mathematical passage
    a\in(\Z/p^{L+1}\Z)^\times
p^{L+a}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2373

    r Editorial hypothesis correction. Require L≥1. At L=0 the standalone claim is false: for p=3, d=1 and X=2=1+B, one has X^3-1=7∉3 Z_3. With L≥1, every higher binomial term in the retained proof has the required valuation, including p=2. If X=I+p^LB with B∈ M_d( Z_p), then for every a≥0 there is B_a∈ M_d( Z_p) such that X^p^a=I+p^L+aB_a.

    Exact mathematical passage
    
    \boxed{X^{p^a}=I+p^{L+a}B_a.}
    
p^{L+n}Use in source · 13 passages

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  • Use in source · Read passage · 30_lifting.tex, line 367

    [Completion and smooth branching] In Theorem~[source reference], let q=p be prime. The tower can be chosen with a limit a∈ Z_p^d such that F(a)=0, z_n⟶ a, a-z_n∈ p^L+n Z_p^d (n≥0). At each stage, the number of distinct next-level residue roots over z_n is exactly p^d-e, and the finite obstruction module has one element. If d=e, there is exactly one root of F in B_L(x).

    Exact mathematical passage
    
     F(a)=0,\qquad z_n\longrightarrow a,\qquad
     a-z_n\in p^{L+n}\Zp^d\quad(n\ge0).
    
p^{L+n+1}Use in source · 1 passage

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  • Use in source · Read passage · continuation_03.tex, line 102

    Surjectivity implies rank_ F_pA_x=e, e≤ d. Combining this with the preceding exact residue-lift count gives # L_z_n,L+n=p^d-e for every n. These are distinct solutions modulo p^L+n+1 lying above the selected point modulo p^L+n. The original finite obstruction group at each smooth point has cardinality one.

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    p^{L+n+1}
p^{L-1}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1538

    For odd p and L≥1, put Λ_p,L=(p-1)p^L-1. For p=2 and L≥3, put Λ_2,L=2^L-2. These are exponents of the unit groups modulo the declared prime powers.

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    \Lambda_{p,L}=(p-1)p^{L-1}.
    
p^{L_p}Use in source · 3 passages

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  • Use in source · Read passage · 71_native.tex, line 210

    N=∏_p∈ Sp^L_p, Sol_F(R)={x∈ R^d:F(x)=0}.

    Exact mathematical passage
    N=\prod_{p\in S}p^{L_p},\qquad\operatorname{Sol}_F(R)=\{x\in R^d:F(x)=0\}.
p^{\lfloor d/p\rfloor}Use in source · 5 passages

Search aliases: p^lfloor d/prfloor lfloor rfloor

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  • Use in source · Read passage · 20_arithmetic.tex, line 231

    At the singular addresses c=ap, 1≤ a<m, use the weighted ring W_p={∑_d≥0a_dT^d: p^lfloor d/prfloora_d∈ Z_p}. The key estimate is R_ap(log(1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!)…

    Exact mathematical passage
    
     \mathcal W_p=\left\{\sum_{d\ge0}a_dT^d:
              p^{\lfloor d/p\rfloor}a_d\in\Z_p\right\}.
    
p^{Lr}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1681

    means the matrix-binomial interpolation. With p=3, C=4, L=1, w=1 and v_0=1/3, one has Δ_1a_0=1, which is not in 3 Z_3. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses. Assume C^H=I+p^LB with B∈ M_d( Z_p). For every address j and every x∈ Z_p, a_Hx+j = ∑_r≥0 p^Lrbinom xr w^TC^jB^rv_0. The series converges on Z_p, and Δ_H^ra_n ≡0 pmod p^Lr.

    Exact mathematical passage
    
    \boxed{
    a_{Hx+j}
    =
    \sum_{r\geq0}
    p^{Lr}\binom xr
    w^TC^jB^rv_0.
    }
    
P^mUse in source · 3 passages

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  • Use in source · Read passage · strengthening.tex, line 306

    at moments of polynomials in separate variables factor. Odd moments vanish. Frobenius's congruence is v_2(∑_r=0^m(-1)^m-rbinom mr Euler_2n+2r) =m+v_2(m!) (n,m≥0). For m=0, the assertion is that every even Euler number is odd. With P=( E^2-1)/2 and γ_m(P)=P^m/m!, the case n=0 says L(γ_m(P))∈ Z_2^× (m≥0). Introduce an independent Euler variable F and Q=( F^2-1)/2.

    Exact mathematical passage
    \gamma_m(P)=P^m/m!
p^MUse in source · 31 passages

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  • Use in source · Read passage · 30_lifting.tex, line 418

    [Smooth coefficient stability] Let p be prime, M≥1, F,H∈ Z[X_1,…,X_d]^e, and c∈ p^M Z. Suppose DF(x)bmod p is surjective for an integer x∈ Z^d. If a∈ Z_p^d satisfies F(a)=0 and a≡ xpmod p, then there is a root b of G=F+cH with b-a∈ p^M Z_p^d.

    Exact mathematical passage
    c\in p^M\Z
p^mUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 203

    To retain the valuation of a step, let B be any sequence satisfying Δ_1^m B∈ p^m Z_p at every order and row. Use the finite operator identity Δ_H=∑_i=1^Hbinom HiΔ_1^i. The relation ibinom Hi=HbinomH-1i-1 implies v_p(p^ibinom Hi) ≥ i+v_p(H)-v_p(i)≥1+v_p(H). Multiplying these identities for the k steps, every term of total order M has its scalar multiplied by p^M, and is divisible by p^k+∑_ν v_p(H_ν). This proves the mixed-step bound. Finally, Δ_H A_N,j(n)=(-1)^n+HT_H C_N,j(n). Iteration contribut…

    Exact mathematical passage
    \Delta_1^m B\in p^m\mathbb Z_p
p^{M-(2s+1)}Use in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 787

    Retain the original certificate A=p^sJ^-1 and put c=p^s+1. Taylor expansion and a polynomial difference identity yield polynomial vectors Q_F,Q_H, both zero at the origin, such that (AF)(a+cu)=p^2s+1u+p^2s+2Q_F(u), (AH)(a+cu)=AH(a)+cQ_H(u). Consequently (AG)(a+cu)=p^2s+1(C+u+pQ(u)), where C=p^M-(2s+1)AH(a), Q=Q_F+p^M-s-1Q_H, Q(0)=0. All exponents are nonnegative. The same contraction and divisibility argument as in Theorem~[source reference] gives the root, the displacement p^M-s, and full-ball …

    Exact mathematical passage
    
     C=p^{M-(2s+1)}AH(a),\qquad
     Q=Q_F+p^{M-s-1}Q_H,\qquad Q(0)=0.
    
p^{m-e}Use in source · 4 passages

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  • Use in source · Read passage · 30_lifting.tex, line 636

    Theorem~[source reference]. For every τ∈ t_0+p^L Z_p^k, there are an integer tower z_n and b∈ Z_p^m+k such that F(b)=0, free(b)=τ, z_n⟶ b, b-z_n∈ p^L+n Z_p^m+k. The initial value is z_0=x. At every stage and for every proposed parameter digit v∈ F_p^k, the corresponding next residue fiber has exactly p^m-e elements. If m=e, then for each τ there is exactly one such root in B_L(x), independently of all choices of integer approximants and correcting digits.

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    p^{m-e}
p^{M-s}Use in source · 15 passages

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  • Use in source · Read passage · 30_lifting.tex, line 766

    [Weighted coefficient deformation] Let F,H∈ R[X_1,…,X_d]^d, F(a)=0, and J=DF(a). Suppose J∈GL_d( Q_p), p^sJ^-1∈ M_d(R), and M>2s. For G=F+p^MH, there is exactly one root b∈ a+p^s+1R^d. It satisfies b-a∈ p^M-sR^d and, in Q_p^d, b-a=-p^MJ^-1H(a)+ε, ε∈ p^2M-3sR^d. Only the original Jacobian DF(a) is subject to an inverse hypothesis.

    Exact mathematical passage
    b-a\in p^{M-s}R^d
p^{M-s-1}Use in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 787

    Retain the original certificate A=p^sJ^-1 and put c=p^s+1. Taylor expansion and a polynomial difference identity yield polynomial vectors Q_F,Q_H, both zero at the origin, such that (AF)(a+cu)=p^2s+1u+p^2s+2Q_F(u), (AH)(a+cu)=AH(a)+cQ_H(u). Consequently (AG)(a+cu)=p^2s+1(C+u+pQ(u)), where C=p^M-(2s+1)AH(a), Q=Q_F+p^M-s-1Q_H, Q(0)=0. All exponents are nonnegative. The same contraction and divisibility argument as in Theorem~[source reference] gives the root, the displacement p^M-s, and full-ball …

    Exact mathematical passage
    
     C=p^{M-(2s+1)}AH(a),\qquad
     Q=Q_F+p^{M-s-1}Q_H,\qquad Q(0)=0.
    
p^{m-\rho}Use in source · 1 passage

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  • Use in source · Read passage · continuation_08.tex, line 64

    When q=p is prime, put ρ=rank_ F_pJ. Then # L_v= p^m-ρ, · o_x(v)=0, 0, · o_x(v)≠0, # C_x=p^e-ρ. The count is of distinct actual residue classes at level L+1. The solved dimension m, rather than the total number m+k of variables, appears in the exponent because the parameter digit is fixed. These rank formulas are asserted only over prime fields.

    Exact mathematical passage
    
    \boxed{\#\mathcal L_v=
    \begin{cases}
    p^{m-\rho},&o_x(v)=0,\\
    0,&o_x(v)\ne0,
    \end{cases}}
    \qquad
    \boxed{\#\mathcal C_x=p^{e-\rho}.}
    
p^{m-\rho_x}Use in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 556

    [Prescribed-parameter obstruction and count] Define o_x(v)=[overliner_x+K_xv]∈ S_q^e/im J_x. Then L_x,L,v is nonempty if and only if o_x(v)=0. When nonempty, it is a torsor for ker J_x. If q=p is prime and ρ_x=rank_ F_pJ_x, then # L_x,L,v =p^m-ρ_x, · o_x(v)=0, 0, · o_x(v)≠0, #( F_p^e/im J_x)=p^e-ρ_x. The criterion holds uniformly for every prescribed v without a surjectivity hypothesis.

    Exact mathematical passage
    \label{eq:prescribed-count}
     \#\mathcal L_{x,L,v}
     =\begin{cases}p^{m-\rho_x},&o_x(v)=0,\\0,&o_x(v)\ne0,\end{cases}
     \qquad \#(\F_p^e/\im J_x)=p^{e-\rho_x}.
    
p^{\min(k(1+\nu_p(u)),mn)}Use in source · 1 passage

Search aliases: p^min(k(1+ν_p(u)),mn) min nu

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  • Use in source · Read passage · 20_arithmetic.tex, line 571

    [Step-sensitive divisor-branch bound] Under the hypotheses of Theorem~[source reference], for all u,k,n∈ N, Δ_u(p-1)/m^kU_mp,j(n) ∈ p^min(k(1+ν_p(u)),mn) Z_p. For C_mp,j the corresponding valuation lower bound is min(k(1+ν_p(u)),mn)-v_p(j!).

    Exact mathematical passage
    
     \Delta_{u(p-1)/m}^{k}U_{mp,j}(n)
           \in p^{\min(k(1+\nu_p(u)),mn)}\Z_p.
    
p^{\min(k,r)}Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 416

    [Prime-power period from finite differences] Let R be a commutative ring, p a prime, f: N→ R, h,n∈ N, and r≥1. If p^min(k,r)midΔ_h^k f(n) for every k∈ N, then p^rmid f(n+p^r-1h)-f(n).

    Exact mathematical passage
    
     p^{\min(k,r)}\mid\Delta_h^k f(n)\quad\text{for every }k\in\N,
    
p^{mn}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 341

    ly. If a_d=d![t^d]F, then d![t^d]B_p^kF=p^-kΔ_p(p-1)^ka_d+pk. For F=j! P_N,j and k≤ mn, take d=mpn-pk. Support in multiples of mp turns this into p^-kΔ_h^kU_N,j(n), so flatness gives divisibility by p^k. For k>mn, expand Δ_h^k-mn applied to Δ_h^mnU_N,j(n); every shifted row is at least n, so each summand is divisible by p^mn. Finally Δ_uh=((I+Δ_h)^u-I), whose kth power for u>0 is an integral combination of powers of Δ_h of order at least k. The case u=0 gives zero for k>0, and order zero is the …

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    p^{mn}
p^NUse in source · 8 passages

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  • Use in source · Read passage · 30_lifting.tex, line 463

    t, writing σ x=(y_0,t_0) with e solved coordinates and k parameter coordinates, the parameter ball T=t_0+p^L Z_p^k is homeomorphic to the entire root locus R={z∈ B_L(x):F(z)=0}. The inverse homeomorphism is the actual free-coordinate projection z↦free(σ z). If h:T→ R is the forward map, then for every N≥0, h(t)≡ h(u)pmodp^N Longleftrightarrow t≡ upmodp^N. Equivalently, σ h(t)=(ϕ(t),t) for a continuous ϕ:T→ y_0+p^L Z_p^e whose graph is the entire reordered root locus.

    Exact mathematical passage
    \label{eq:chart-congruence}
     h(t)\equiv h(u)\pmod{p^N}
     \quad\Longleftrightarrow\quad t\equiv u\pmod{p^N}.
    
p^nUse in source · 9 passages

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  • Use in source · Read passage · 30_lifting.tex, line 315

    For prime p, rank-nullity gives dimker J_x=d-ρ_x and dim( F_p^e/im J_x)=e-ρ_x. The cardinality of an n-dimensional F_p-space is p^n. Combine these identities with the obstruction criterion and the residue equivalence.

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    p^n
p^{N+1}Use in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 475

    First consider a square integer polynomial map P whose reduced Jacobian at x is surjective. Its Jacobian is invertible over F_p. A digit comparison using Taylor expansion shows that equal outputs modulo p^N+1 force equal inputs modulo p^N+1 whenever the inputs already agree modulo p^N and belong to B_1(x). Induction and integer truncation give, for arbitrary a,b∈ B_1(x), P(a)≡ P(b)pmodp^N Longleftrightarrow a≡ bpmodp^N.

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    p^{N+1}
p^{n+1}Use in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 730

    To obtain the sharper displacement, begin with u∈ R^d. If u∈ p^nR^d, then Q(0)=0 and polynomial congruence give Q(u)∈ p^nR^d. For n<k, both C and pQ(u) are divisible by p^n+1. Equation~[source reference] therefore improves the divisibility of u by one digit. Induction gives u∈ p^kR^d. Thus b=a+p^s+1u satisfies [source reference], because s+1+k=r-s. Its normalized preconditioned value vanishes, and injectivity of A gives F(b)=0.

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    p^{n+1}
p^qUse in source · 2 passages

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  • Use in source · Read passage · supplement_10.tex, line 69

    The scalar input is proved for every natural b, including nonunits: p^2mid b^p^2-b^p in Z. Because Θ(1+T)^b=b(1+T)^b, the binomial expansion of T^p=((1+T)-1)^p then proves stability of the divided generator T^p/p. For d=pq+r, 0≤ r<p, every weighted monomial has the exact factorization a_dT^d=(p^qa_d)(fracT^pp)^qT^r. The scalar p^qa_d is integral by the actual weight condition. Thus all weighted monomials, and all finite weighted polynomials, satisfy the defect condition.

    Exact mathematical passage
    
     a_dT^d=(p^qa_d)\left(\frac{T^p}{p}\right)^qT^r.
    
p^rUse in source · 21 passages

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  • Use in source · Read passage · edition_summary.tex, line 4

    nal edition establishes addcontentslinetocsectionWhat this final edition establishes The original external problem is Nishibuchi's Conjecture 3.9 for generalized Euler numbers. Its full parameter range is resolved. If p is prime, m is positive with mmid p-1 or m=2, 0≤ j<mp, and q=lcm(2,p-1), then the original Euler coefficients are p-adically integral and E_mpn+qp^r^(mp,j)≡ E_mpn^(mp,j)pmodp^r (r≥1, n≥ r). The paper states the generating function and the conversion between ordinary degree and pa…

    Exact mathematical passage
    
     E_{mpn+qp^r}^{(mp,j)}\equiv E_{mpn}^{(mp,j)}\pmod{p^r}
     \qquad (r\ge1,\ n\ge r).
    
p^{r-\operatorname{rank}J}Use in source · 1 passage

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  • Use in source · Read passage · 71_native.tex, line 31

    Theorem. For this fixed polynomial presentation, the group and distinguished class are independent of the chosen integer representative. The class vanishes if and only if bar x lifts to a solution modulo p^L+1. The obstruction group has p^r-rankJ elements. When lifts exist, their set is a torsor for ker J and contains exactly p^d-rankJ elements.

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    p^{r-\operatorname{rank}J}
p^{r-s}Use in source · 5 passages

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  • Use in source · Read passage · 30_lifting.tex, line 692

    hted nonlinear lifting] Let p be prime, d,s,r∈ N, F∈ R[X_1,…,X_d]^d, and a∈ R^d. Set J=DF(a). Suppose J∈GL_d( Q_p), p^sJ^-1∈ M_d(R), F(a)∈ p^rR^d, r>2s. Then there is exactly one root b of F in a+p^s+1R^d, and it satisfies b-a∈ p^r-sR^d. Uniqueness holds throughout the full ball of depth s+1.

    Exact mathematical passage
    \label{eq:weighted-displacement}
     b-a\in p^{r-s}R^d.
    
p^{r-s+1}Use in source · 2 passages

Search aliases: p^r-s+1

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  • Use in source · Read passage · 30_lifting.tex, line 757

    and displacement boundary] For p=2, F(X)=X^2+3, and a=1, one has F(a)=4 and DF(a)=2. Thus s=1 and r=2 satisfy every displayed hypothesis except the strict inequality. No root exists in Z_2, since none exists modulo eight. Hence r≥2s cannot replace r>2s uniformly. Also, F(X)=p^sX+p^r has root -p^r-s, whose displacement is not divisible by p^r-s+1. The guaranteed displacement cannot uniformly gain a digit.

    Exact mathematical passage
    p^{r-s+1}
p^sUse in source · 41 passages

Search aliases: p^s

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  • Use in source · Read passage · 10_foundations.tex, line 339

    [Differences of unit modes] Over every commutative ring, Δ_h^s(u^n)=u^n(u^h-1)^s. For any prime p, units u_i∈ Z_p^×, coefficients a_i∈ Z_p, and a finite index set I, the sequence f(n)=∑_i∈ Ia_i u_i^n satisfies Δ_p-1^s f(n)∈ p^s Z_p (n,s∈ N).

    Exact mathematical passage
    
    \Delta_{p-1}^s f(n)\in p^s\Zp
    \qquad(n,s\in\N).
    
p^{s+1}Use in source · 20 passages

Search aliases: p^s+1

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  • Use in source · Read passage · 30_lifting.tex, line 690

    [Weighted nonlinear lifting] Let p be prime, d,s,r∈ N, F∈ R[X_1,…,X_d]^d, and a∈ R^d. Set J=DF(a). Suppose J∈GL_d( Q_p), p^sJ^-1∈ M_d(R), F(a)∈ p^rR^d, r>2s. Then there is exactly one root b of F in a+p^s+1R^d, and it satisfies b-a∈ p^r-sR^d. Uniqueness holds throughout the full ball of depth s+1.

    Exact mathematical passage
    a+p^{s+1}R^d
p^{s+1+k}Use in source · 1 passage

Search aliases: p^s+1+k

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  • Use in source · Read passage · continuation_06.tex, line 61

    For completeness, the sharp precision is additional to this fixed-point existence. Inductively, if u∈ p^nR^d, then polynomial congruence and Q(0)=0 imply Q(u)∈ p^nR^d. For n<k, the equation u=-C-pQ(u) therefore gives u∈ p^n+1R^d. Starting at n=0, this proves u∈ p^kR^d. Hence b=a+tu satisfies b-a∈ p^s+1+kR^d=p^r-sR^d. The normalization and injectivity of A give F(b)=0.

    Exact mathematical passage
    
    b-a\in p^{s+1+k}R^d=p^{r-s}R^d.
    
p^{v_p(\lfloor d/p\rfloor!)}Use in source · 3 passages

Search aliases: p^v_p(lfloor d/prfloor!) lfloor rfloor

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  • Use in source · Read passage · 20_arithmetic.tex, line 242

    1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!). The identity v_p(d!)=lfloor d/prfloor+v_p(lfloor d/prfloor!) accounts for precisely the weight in W_p after dividing by d!. Comparison o…

    Exact mathematical passage
    p^{v_p(\lfloor d/p\rfloor!)}
P^{v_p(q!)}Use in source · 2 passages

Search aliases: P^v_p(q!)

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  • Use in source · Read passage · supplement_09.tex, line 44

    p). Expanding ((z-1)+1)^p=1 and using divisibility of every interior binomial coefficient in prime row p gives (z-1)^p∈ PI. Every other degree-p product of the two generators of I also belongs to PI. The formal ideal proof therefore establishes I^p⊆ PI. For q=plfloor q/prfloor+(qbmod p), strong induction and the factorial valuation identity give I^q⊆ P^v_p(q!). In detail, write b=lfloor q/prfloor. Then I^q⊆ P^b I^b⊆ P^b+v_p(b!)=P^v_p(q!). No integral-domain hypothesis or division in a field is u…

    Exact mathematical passage
    
     \boxed{I^q\subseteq P^{v_p(q!)}}.
    
p^{v_p(q!)}Use in source · 1 passage

Search aliases: p^v_p(q!)

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  • Use in source · Read passage · 20_arithmetic.tex, line 239

    sses c=ap, 1≤ a<m, use the weighted ring W_p={∑_d≥0a_dT^d: p^lfloor d/prfloora_d∈ Z_p}. The key estimate is R_ap(log(1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!). The identity v_p(d!…

    Exact mathematical passage
    I^q\subseteq(p^{v_p(q!)})
p^{v_p(u)+1}Use in source · 1 passage

Search aliases: p^v_p(u)+1

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  • Use in source · Read passage · 20_arithmetic.tex, line 552

    To incorporate v_2(u), expand a displacement using finite Newton coefficients. For any prime p, u>0, and 1≤ a≤ u, p^v_p(u)+1midbinom ua p^a, as follows from abinom ua=ubinomu-1a-1 and v_p(a)<a. Repeated displacement, retaining the remaining base-difference order at each induction step, gives a factor p^k(1+v_p(u)) from k displacements. This induction is formalized in leanKomorebiPaperIX.StepAmplification.repeated_precision. Applying it with p=2 proves [source reference]; multiplication by j! pro…

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     p^{v_p(u)+1}\mid\binom ua p^a,
    
p^wUse in source · 1 passage

Search aliases: p^w

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  • Use in source · Read passage · supplement_01.tex, line 131

    ach earlier clearing denominator to the common product, then cancels the new nonzero diagonal. This is a denominator bound, not a claim that every normalized coefficient is integral. pageheading3. Filtered tails and the standard windowSource items 2.2 and 8.2 Filtration at a finite coefficient row. Let R be any commutative ring. For w∈ N∪{∞} define p^[w]=p^w, · w<∞, 0, · w=∞. The infinity convention expresses a vanishing coefficient, not an infinite exponent operation. For any finite index set, …

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     p^{[w]}=\begin{cases}p^w,&w<\infty,\\0,&w=\infty.\end{cases}
    
p^{w(r)}Use in source · 2 passages

Search aliases: p^w(r)

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  • Use in source · Read passage · 72_recovered.tex, line 530

    [Filtered triangular reconstruction] Let p be prime. Suppose A_n = ∑_r=0^n binom nr c_rp^w(r)S_n-r, where c_r∈ Z_p, S_n∈ Z_p, and w(r)∈ N. For every precision L≥1, A_n ≡ ∑_substack0≤ r≤ n

    Exact mathematical passage
    
    A_n
    =
    \sum_{r=0}^{n}
    \binom nr
    c_rp^{w(r)}S_{n-r},
    
p^\ellUse in source · 5 passages

Search aliases: p^ℓ ell

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  • Use in source · Read passage · supplement_05.tex, line 25

    The specification matters. Rational coefficients are not reduced through an imaginary map Q→ F_p. The proof first constructs their integral lifts in Z_p and establishes equality in Q_p. Root representatives are actual units at the requested precision, not unchanged residues that need not satisfy the root equation modulo p^ℓ.

    Exact mathematical passage
    p^\ell
p^{\ell-1}Use in source · 6 passages

Search aliases: p^ℓ-1 ell

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  • Use in source · Read passage · supplement_05.tex, line 81

    ffPadic_coe, .standard_window_integral targetWindowIndependence.residue_independence, .reduced_unique targetWindowIndependence.two_window_addresses_congruent pageheading2. Explicit finite prime-power decompositionProposition 6.2; odd-prime branch Let p be prime, ℓ≥1, and a∈( Z/p^ℓ Z)^×. Define actual units in the same ring by ω_ℓ(a)=a^p^ℓ-1, ⟨ a⟩_ℓ=a ω_ℓ(a)^-1. The Lean source writes ℓ=L+1, so zero precision is excluded without truncated natural-number subtraction.

    Exact mathematical passage
    
     \boxed{\omega_\ell(a)=a^{p^{\ell-1}},\qquad
     \langle a\rangle_\ell=a\,\omega_\ell(a)^{-1}.}
    
p^{\ell-k}Use in source · 1 passage

Search aliases: p^ℓ-k ell

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  • Use in source · Read passage · supplement_05.tex, line 145

    c branch For 1≤ k≤ℓ, let ρ_ℓ,k: Z/p^ℓ Z⟶ Z/p^k Z be the actual reduction homomorphism. It induces a map on units. The explicit construction satisfies ρ_ℓ,k(ω_ℓ(a)) =ω_k(ρ_ℓ,k(a)), ρ_ℓ,k(⟨ a⟩_ℓ) =⟨ρ_ℓ,k(a)⟩_k. The proof uses the exponent factorization p^ℓ-1=p^k-1p^ℓ-k and the already proved Frobenius invariance of a lifted root. Principal factors also satisfy ⟨ a⟩_ℓ^ p^ℓ-1=1. At precision one the principal factor is exactly one. These are coherent finite certificates; no infinite inverse limit or…

    Exact mathematical passage
    p^{\ell-1}=p^{k-1}p^{\ell-k}
P_0Use in source · 2 passages

Search aliases: P_0

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  • Use in source · Read passage · root_scope.tex, line 107

    Set r=M+1≥1 and define P_q(x)=∏_i=0^q-1(x-i)(2x-(2i+1)), P_0=1, H_r(x)=∑_q≥08^qfracβ_r+qβ_r P_q(x)mathscr A(x-q). The coefficient valuations tend to infinity. At any nonnegative integer x, terms with q>x vanish. Cancelling the factorials in [source reference] therefore gives the exact identity T_r-1(N)=κ_rbinom2N2rH_r(N-r) (N≥ r). This is a finite rational identity at every original Euler row.

    Exact mathematical passage
    
     P_q(x)=\prod_{i=0}^{q-1}(x-i)(2x-(2i+1)),\qquad P_0=1,
    
P_1Use in source · 1 passage

Search aliases: P_1

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  • Use in source · visual_body.tex, line 340

    Modulo 256=2^8, the least periods are (P_0(8),P_1(8),P_2(8),P_3(8))=(128,128,8,32). In actual Euler degree, multiply by four: (512,512,32,128). For address two, precision at most 2^5 already has row period one. Its Euler coefficients all lie in 2+32 Z_2; further precision resolves additional dyadic digits within that residue class.

    Exact mathematical passage
    
     (P_0(8),P_1(8),P_2(8),P_3(8))=(128,128,8,32).
    
P_2Use in source · 1 passage

Search aliases: P_2

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  • Use in source · visual_body.tex, line 340

    Modulo 256=2^8, the least periods are (P_0(8),P_1(8),P_2(8),P_3(8))=(128,128,8,32). In actual Euler degree, multiply by four: (512,512,32,128). For address two, precision at most 2^5 already has row period one. Its Euler coefficients all lie in 2+32 Z_2; further precision resolves additional dyadic digits within that residue class.

    Exact mathematical passage
    
     (P_0(8),P_1(8),P_2(8),P_3(8))=(128,128,8,32).
    
P_{2^a,\,r\bmod2^a}Use in source · 1 passage

Search aliases: P_2^a, rbmod2^a bmod

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  • Use in source · Read passage · supplement_05.tex, line 201

    0, · otherwise, take the canonical label 0≤ r<N. The logical equivalence yields the actual equality of sequence operators P_N,rf= P_2^a, rbmod2^a(P_M, rbmod Mf). This statement holds for coefficients in any type with zero. A separate theorem lifts the equality coefficientwise to the existing formal power-series selectors over any semiring.

    Exact mathematical passage
    
     \boxed{P_{N,r}f=
     P_{2^a,\,r\bmod2^a}\bigl(P_{M,\,r\bmod M}f\bigr).}
    
P_{2^a,r\bmod 2^a}Use in source · 1 passage

Search aliases: P_2^a,rbmod 2^a bmod

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  • Use in source · Read passage · 10_foundations.tex, line 456

    [Dyadic--odd packet factorization] Let a≥0 and let M be a positive odd integer. The Chinese remainder map is a ring isomorphism Z/(2^aM) Z≃ Z/2^a Z× Z/M Z. For 0≤ r<2^aM and any coefficient sequence f, P_2^aM,rf =P_2^a,rbmod 2^a P_M,rbmod Mf. The same identity holds for formal-series coefficient selectors.

    Exact mathematical passage
    
    P_{2^aM,r}f
    =P_{2^a,r\bmod 2^a}\,P_{M,r\bmod M}f.
    
P_{2^aM,r}Use in source · 1 passage

Search aliases: P_2^aM,r

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  • Use in source · Read passage · 10_foundations.tex, line 456

    [Dyadic--odd packet factorization] Let a≥0 and let M be a positive odd integer. The Chinese remainder map is a ring isomorphism Z/(2^aM) Z≃ Z/2^a Z× Z/M Z. For 0≤ r<2^aM and any coefficient sequence f, P_2^aM,rf =P_2^a,rbmod 2^a P_M,rbmod Mf. The same identity holds for formal-series coefficient selectors.

    Exact mathematical passage
    
    P_{2^aM,r}f
    =P_{2^a,r\bmod 2^a}\,P_{M,r\bmod M}f.
    
P_3Use in source · 1 passage

Search aliases: P_3

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  • Use in source · visual_body.tex, line 340

    Modulo 256=2^8, the least periods are (P_0(8),P_1(8),P_2(8),P_3(8))=(128,128,8,32). In actual Euler degree, multiply by four: (512,512,32,128). For address two, precision at most 2^5 already has row period one. Its Euler coefficients all lie in 2+32 Z_2; further precision resolves additional dyadic digits within that residue class.

    Exact mathematical passage
    
     (P_0(8),P_1(8),P_2(8),P_3(8))=(128,128,8,32).
    
\mathcal P_{4,j}Use in source · 1 passage

Search aliases: P_4,j mathcal

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  • Use in source · Read passage · 20_arithmetic.tex, line 539

    Packing g_j into degrees divisible by four gives K_4,j. Its source property is precisely charge-one flatness. Inversion gives charge -1, and the charge shift restores charge one for the reciprocal P_4,j. If dec(F)(d)=d![t^d]F, the exact coefficient identity is dec(A_1^sF)(d) =2^-sΔ_4^sdec(F)(d). At d=4n, packing turns the right side into 2^-sΔ_1^sC_j(n). This extraction requires no negative degree and therefore holds at every n, including zero. It proves the unit-step bound leanKomorebiPaperIX.D…

    Exact mathematical passage
    \mathcal P_{4,j}
P_aUse in source · 1 passage

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  • Use in source · Read passage · 40_regularity.tex, line 40

    ts strict derivative is the continuous linear map represented by the matrix of evaluated formal partial derivatives. For the derivative statement, use polynomial induction on constants, addition, and multiplication by a variable. Constants have zero derivative. The sum rule agrees with the additivity of formal partial differentiation. For PX_j, the product rule gives d(PX_j)_a(v)=a_j dP_a(v)+P(a)v_j, which agrees with the evaluated formal product rule for every coordinate. The finite product rul…

    Exact mathematical passage
    
     d(PX_j)_a(v)=a_j\,dP_a(v)+P(a)v_j,
    
P_{A,r_A}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 3010

    [Coprime packet factorization] Let A and B be coprime positive integers. Let r_A be an address modulo A and r_B an address modulo B. Let r be the unique address modulo AB satisfying r≡ r_Apmod A, r≡ r_Bpmod B. Then P_AB,r = P_A,r_AP_B,r_B = P_B,r_BP_A,r_A.

    Exact mathematical passage
    
    \boxed{
    P_{AB,r}
    =
    P_{A,r_A}P_{B,r_B}
    =
    P_{B,r_B}P_{A,r_A}.
    }
    
P_{AB,r}Use in source · 1 passage

Search aliases: P_AB,r

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  • Use in source · Read passage · 72_recovered.tex, line 3010

    [Coprime packet factorization] Let A and B be coprime positive integers. Let r_A be an address modulo A and r_B an address modulo B. Let r be the unique address modulo AB satisfying r≡ r_Apmod A, r≡ r_Bpmod B. Then P_AB,r = P_A,r_AP_B,r_B = P_B,r_BP_A,r_A.

    Exact mathematical passage
    
    \boxed{
    P_{AB,r}
    =
    P_{A,r_A}P_{B,r_B}
    =
    P_{B,r_B}P_{A,r_A}.
    }
    
P_{B,r_B}Use in source · 1 passage

Search aliases: P_B,r_B

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  • Use in source · Read passage · 72_recovered.tex, line 3010

    [Coprime packet factorization] Let A and B be coprime positive integers. Let r_A be an address modulo A and r_B an address modulo B. Let r be the unique address modulo AB satisfying r≡ r_Apmod A, r≡ r_Bpmod B. Then P_AB,r = P_A,r_AP_B,r_B = P_B,r_BP_A,r_A.

    Exact mathematical passage
    
    \boxed{
    P_{AB,r}
    =
    P_{A,r_A}P_{B,r_B}
    =
    P_{B,r_B}P_{A,r_A}.
    }
    
P_iUse in source · 5 passages

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  • Use in source · Read passage · strengthening.tex, line 476

    Here is the polynomial estimate that supplies one additional bit. For i≥1, put P_i(x)=4^ibinom4x4i. Its constant is zero, and its degree-d coefficient is [x^d]P_i(x)=frac4^i4^d s(4i,d)(4i)!, where s(m,d) is a signed Stirling number of the first kind. In each product contributing to s(4i,d), at most d-1 of the 2i-1 even integers in 1,…,4i-1 can be omitted. Thus v_2(s(4i,d))≥max(0,2i-d). Writing s_2(i) for the number of ones in the binary expansion of i, and using v_2((4i)!)=4i-s_2(i), gives v_2([…

    Exact mathematical passage
    P_i(x)=4^i\binom{4x}{4i}
P_jUse in source · 6 passages

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  • Use in source · Read passage · root_scope.tex, line 211

    For a wall label j≥1 and 0≤ q<j, define e_j(q)=3q+s_2(j-1)-s_2(j-q-1), E_j=min_q<je_j(q), Θ(j)=3j+1+s_2(j-1), L_j=Θ(j)-E_j. Here E_j is a carry minimum, not an Euler number. A wall label is regular when e_j(q)>0 for 1≤ q<j. For r=2^L-j with L>L_j, put δ=L-L_j. The carry identities give μ_r=E_j-δ, J_r={j}, G_r(x)=P_j(x)U(x-j)+2^δ V(x), where U≡1pmod2. For a regular label, also E_j=0 and V≡1pmod2.

    Exact mathematical passage
    
     \mu_r=E_j-\delta,\qquad J_r=\{j\},\qquad
     G_r(x)=P_j(x)U(x-j)+2^\delta V(x),
    
p_j^{a_j}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 3054

    [Finite local reconstruction] Let N=∏_j=1^sp_j^a_j be the prime power factorization of a positive integer. The map Z/N Z ⟶ ∏_j=1^s Z/p_j^a_j Z is a ring isomorphism. Thus a compatible collection of prime power residues reconstructs one residue modulo N [source reference].

    Exact mathematical passage
    
    N=\prod_{j=1}^{s}p_j^{a_j}
    
\mathcal P_MUse in source · 3 passages

Search aliases: P_M mathcal

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  • Use in source · Read passage · 72_recovered.tex, line 1977

    For a sequence a=(a_n)_n≥0 and M≥2, define a_r^[M](q)=a_Mq+r (0≤ r<M). The analysis map is P_Ma = (a_0^[M],…,a_M-1^[M]). Synthesis interleaves the branches by the rule a_Mq+r=a_r^[M](q).

    Exact mathematical passage
    
    \mathcal P_Ma
    =
    \left(a_0^{[M]},\ldots,a_{M-1}^{[M]}\right).
    
P_{M,\,r\bmod M}Use in source · 1 passage

Search aliases: P_M, rbmod M bmod

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  • Use in source · Read passage · supplement_05.tex, line 201

    0, · otherwise, take the canonical label 0≤ r<N. The logical equivalence yields the actual equality of sequence operators P_N,rf= P_2^a, rbmod2^a(P_M, rbmod Mf). This statement holds for coefficients in any type with zero. A separate theorem lifts the equality coefficientwise to the existing formal power-series selectors over any semiring.

    Exact mathematical passage
    
     \boxed{P_{N,r}f=
     P_{2^a,\,r\bmod2^a}\bigl(P_{M,\,r\bmod M}f\bigr).}
    
P_{M,r}Use in source · 2 passages

Search aliases: P_M,r

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  • Use in source · Read passage · 10_foundations.tex, line 50

    [Residue reconstruction] The selectors satisfy P_N,j^2=P_N,j, P_N,iP_N,j=0 (i≠ j), ∑_j=0^N-1P_N,j=I. If Nmid M and M>0, then P_N,ja=∑_substack0≤ r<M r≡ jpmod NP_M,ra. The same identities hold coefficientwise for formal power series.

    Exact mathematical passage
    
    P_{N,j}a=\sum_{\substack{0\leq r<M\\r\equiv j\pmod N}}P_{M,r}a.
    
P_{M,r\bmod M}Use in source · 1 passage

Search aliases: P_M,rbmod M bmod

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  • Use in source · Read passage · 10_foundations.tex, line 456

    [Dyadic--odd packet factorization] Let a≥0 and let M be a positive odd integer. The Chinese remainder map is a ring isomorphism Z/(2^aM) Z≃ Z/2^a Z× Z/M Z. For 0≤ r<2^aM and any coefficient sequence f, P_2^aM,rf =P_2^a,rbmod 2^a P_M,rbmod Mf. The same identity holds for formal-series coefficient selectors.

    Exact mathematical passage
    
    P_{2^aM,r}f
    =P_{2^a,r\bmod 2^a}\,P_{M,r\bmod M}f.
    
\mathcal P_{N,0}Use in source · 1 passage

Search aliases: P_N,0 mathcal

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  • Use in source · Read passage · 20_arithmetic.tex, line 223

    Set H_c(t)=∑_n≥0fract^Nn+c(Nn+c)!, A_c(t)=∑_n≥0fract^Nn(Nn+c)!, R_c(t)=fracA_c-1(t)A_c(t) (1≤ c<N). The denominator of this last quotient has nonzero constant coefficient. The section identities give R_cH_c=tH_c-1, R_c(0)=c, j! P_N,j= P_N,0∏_c=1^jR_c. Cancellation of consecutive source sections proves the product identity and ties the ratios directly to the family under study.

    Exact mathematical passage
    
     R_cH_c=tH_{c-1},\qquad R_c(0)=c,\qquad
     j!\mathcal P_{N,j}=\mathcal P_{N,0}\prod_{c=1}^{j}R_c.
    
P_{N,i}Use in source · 1 passage

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  • Use in source · Read passage · 10_foundations.tex, line 45

    [Residue reconstruction] The selectors satisfy P_N,j^2=P_N,j, P_N,iP_N,j=0 (i≠ j), ∑_j=0^N-1P_N,j=I. If Nmid M and M>0, then P_N,ja=∑_substack0≤ r<M

    Exact mathematical passage
    
    P_{N,j}^2=P_{N,j},\qquad P_{N,i}P_{N,j}=0\quad(i\ne j),
    \qquad \sum_{j=0}^{N-1}P_{N,j}=I.
    
P_{N,j}Use in source · 9 passages

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  • Use in source · Read passage · 10_foundations.tex, line 12

    The original arithmetic seed notation S=(A,a_0, O, R,{P_N,j},w,τ_p) records a coefficient ring or module, a seed, allowed operators, relations, residue selectors, a depth weight, and a topology. This notation describes the ingredients of the construction; a category of seed presentations and a notion of equivalence still require separate definitions. We therefore state each theorem for a concrete object. Section~[source reference] discusses the additional structure needed for the original native…

    Exact mathematical passage
    
    \mathcal S=(A,a_0,\mathcal O,\mathcal R,\{P_{N,j}\},w,\tau_p)
    
\mathcal P_{N,j}Use in source · 5 passages

Search aliases: P_N,j mathcal

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  • Use in source · Read passage · 20_arithmetic.tex, line 29

    The normalized source and its inverse are K_N,j(t)=j!∑_n≥0fract^Nn(Nn+j)!, P_N,j(t)=∑_n≥0C_N,j(n)fract^Nn(Nn)!. Their product is one. Clearing factorials in its coefficient of t^Nn gives the triangular system ∑_q=0^nbinomNn+jNqC_N,j(q)=δ_n0. In particular, for n>0, C_N,j(n)=-binomNn+jNn^-1 ∑_q=0^n-1binomNn+jNqC_N,j(q). The diagonal binomial coefficient is positive, so this recurrence determines the sequence uniquely. The Lean construction begins with this recurrence and derives the reciprocal id…

    Exact mathematical passage
    \label{eq:arith-kernel}
     K_{N,j}(t)=j!\sum_{n\ge0}\frac{t^{Nn}}{(Nn+j)!},\qquad
     \mathcal P_{N,j}(t)=\sum_{n\ge0}C_{N,j}(n)\frac{t^{Nn}}{(Nn)!}.
    
P_{N,j}^2Use in source · 1 passage

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  • Use in source · Read passage · 10_foundations.tex, line 45

    [Residue reconstruction] The selectors satisfy P_N,j^2=P_N,j, P_N,iP_N,j=0 (i≠ j), ∑_j=0^N-1P_N,j=I. If Nmid M and M>0, then P_N,ja=∑_substack0≤ r<M

    Exact mathematical passage
    
    P_{N,j}^2=P_{N,j},\qquad P_{N,i}P_{N,j}=0\quad(i\ne j),
    \qquad \sum_{j=0}^{N-1}P_{N,j}=I.
    
P_{N,r}Use in source · 7 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2973

    [Residue projector] For a sequence a=(a_n)_n≥0, a positive integer N, and an address r modulo N, define (P_N,ra)_n = a_n, · n≡ rpmod N, 0, · nnot≡ rpmod N.

    Exact mathematical passage
    
    (P_{N,r}a)_n
    =
    \begin{cases}
    a_n,&n\equiv r\pmod N,\\
    0,&n\not\equiv r\pmod N.
    \end{cases}
    
P_{N,s}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2985

    [Complete residue packet system] For fixed N, the projectors P_N,r satisfy P_N,rP_N,s = δ_rsP_N,r and ∑_rbmod NP_N,r=I.

    Exact mathematical passage
    
    P_{N,r}P_{N,s}
    =
    \delta_{rs}P_{N,r}
    
P_{N_j,r_j}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 3031

    [Several coprime moduli] Let N_1,…,N_s be pairwise coprime and put N=∏_j=1^sN_j. For compatible addresses r_j, P_N,r = ∏_j=1^sP_N_j,r_j, where r is the unique combined address modulo N.

    Exact mathematical passage
    
    \boxed{
    P_{N,r}
    =
    \prod_{j=1}^{s}P_{N_j,r_j},
    }
    
P_qUse in source · 2 passages

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  • Use in source · Read passage · root_scope.tex, line 107

    Set r=M+1≥1 and define P_q(x)=∏_i=0^q-1(x-i)(2x-(2i+1)), P_0=1, H_r(x)=∑_q≥08^qfracβ_r+qβ_r P_q(x)mathscr A(x-q). The coefficient valuations tend to infinity. At any nonnegative integer x, terms with q>x vanish. Cancelling the factorials in [source reference] therefore gives the exact identity T_r-1(N)=κ_rbinom2N2rH_r(N-r) (N≥ r). This is a finite rational identity at every original Euler row.

    Exact mathematical passage
    
     P_q(x)=\prod_{i=0}^{q-1}(x-i)(2x-(2i+1)),\qquad P_0=1,
    
P_rUse in source · 5 passages

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  • Use in source · Read passage · strengthening.tex, line 552

    Define the central-factorial basis polynomials P_r(X)=∏_a=0^r-1(X-a(a+1)). At X=Z(Z+1), they have the exact factorization P_r(Z(Z+1)) =∏_a=0^r-1(Z-a)(Z+a+1) =(2r)!binomZ+r2r. The Bernoulli transform gives L((1+T)^Z)=log(1+T)/T, whence L(binom Zj)=(-1)^j/(j+1). Finite Vandermonde expansion therefore gives L(binomZ+r2r) · =(-1)^r∑_q=0^r(-1)^qbinom rqfrac1r+q+1nonumber

    Exact mathematical passage
    
     P_r(X)=\prod_{a=0}^{r-1}(X-a(a+1)).
    
P_{r+1}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 581

    Expand X^m=∑_r=0^mT(m,r)P_r(X). Since XP_r(X)=P_r+1(X)+r(r+1)P_r(X), these integer coefficients satisfy T(m+1,r)=T(m,r-1)+r(r+1)T(m,r), T(0,0)=1. Every r(r+1) is even. Induction, including the boundary indices, proves 2^m-rmid T(m,r) for 0≤ r≤ m. For m>0 the r=0 term vanishes. Each remaining summand of L(P^m/m!) has valuation at least m-r+2v_2(r!)-v_2(m!) =r+s_2(m)-2s_2(r)≥ s_2(m)-1≥0. Here r≥2s_2(r)-1 follows from r≥2^s_2(r)-1 and 2^u≥2u for u≥1. This proves integrality and the stronger bound.

    Exact mathematical passage
    XP_r(X)=P_{r+1}(X)+r(r+1)P_r(X)
\operatorname{pack}Use in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 88

    For a packed coefficient sequence e, the exact extraction formula is (U^8-U^4)^kpack_4(e)(4n-4k) =Δ_1^k e(n), k≤ n. Here U is the coefficient shift. The nonnegative extraction index is why the first estimate covers k≤ n. The existing saturation lemma extends it to all k with exponent min(k,n), and the step-multiple lemma extends step one to every natural u.

    Exact mathematical passage
    
    \bigl(U^8-U^4\bigr)^k\operatorname{pack}_4(e)(4n-4k)
     =\Delta_1^k e(n),\qquad k\le n.
    
\operatorname{pack}_4Use in source · 1 passage

Search aliases: pack_4 operatorname

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  • Use in source · Read passage · supplement_11.tex, line 88

    For a packed coefficient sequence e, the exact extraction formula is (U^8-U^4)^kpack_4(e)(4n-4k) =Δ_1^k e(n), k≤ n. Here U is the coefficient shift. The nonnegative extraction index is why the first estimate covers k≤ n. The existing saturation lemma extends it to all k with exponent min(k,n), and the step-multiple lemma extends step one to every natural u.

    Exact mathematical passage
    
    \bigl(U^8-U^4\bigr)^k\operatorname{pack}_4(e)(4n-4k)
     =\Delta_1^k e(n),\qquad k\le n.
    
\operatorname{packet}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 31

    pageheading1. Actual matrix orbits and arithmetic jumpsTheorem 9.1; manuscript page 22 Let R be a commutative ring, I a finite coordinate type and d=|I|. For C∈ M_I(R) and v_0∈ R^I, define orbit(C,v_0,n)=C^nv_0, u_k=packet(C,v_0,N,j,k)=C^Nk+jv_0. The implementation uses actual matrix powers and matrix--vector multiplication. A separately supplied sequence v with v_n+1=Cv_n is proved by induction to equal this power orbit. Thus the source-sequence formulation is connected to the construction.

    Exact mathematical passage
    
     \operatorname{orbit}(C,v_0,n)=C^nv_0,
     \qquad u_k=\operatorname{packet}(C,v_0,N,j,k)=C^{Nk+j}v_0.
    
\PPUse in source · 3 passages

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  • Use in source · Read passage · supplement_11.tex, line 54

    Define a Kummer sequence over a commutative ring by P_a(f) Longleftrightarrow a^kmidΔ_1^k f(n) for all k,n. The exact discrete product rule Δ_1^s(fg)(n)=∑_i=0^sbinom si Δ_1^i f(n) Δ_1^s-ig(n+i) is proved by induction. It proves closure under products. Applied to fg=1, the same formula isolates Δ_1^s g(n)f(n+s); induction and multiplication by g(n+s) prove closure under actual pointwise inverses.

    Exact mathematical passage
    
    \PP_a(f)\quad\Longleftrightarrow\quad
     a^k\mid\Delta_1^k f(n)\ \text{for all }k,n.
    
\PP_2Use in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 66

    An affine sequence whose slope is divisible by a satisfies P_a. Consequently the four displayed source sequences satisfy P_2. No numerical interpolation or finite set of samples is used in that deduction.

    Exact mathematical passage
    \PP_2
\PP_aUse in source · 2 passages

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  • Use in source · Read passage · supplement_11.tex, line 54

    Define a Kummer sequence over a commutative ring by P_a(f) Longleftrightarrow a^kmidΔ_1^k f(n) for all k,n. The exact discrete product rule Δ_1^s(fg)(n)=∑_i=0^sbinom si Δ_1^i f(n) Δ_1^s-ig(n+i) is proved by induction. It proves closure under products. Applied to fg=1, the same formula isolates Δ_1^s g(n)f(n+s); induction and multiplication by g(n+s) prove closure under actual pointwise inverses.

    Exact mathematical passage
    
    \PP_a(f)\quad\Longleftrightarrow\quad
     a^k\mid\Delta_1^k f(n)\ \text{for all }k,n.
    
QUse in source · 74 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 960

    Let Q(z)=zcoth z, the regular formal series with constant coefficient one. The elementary differential identity (frac zsinh z)^2=Q(z)-zQ'(z) and the Bernoulli expansion Q(z)=1+∑_m≥1frac2^2mB_2m(2m)!z^2m give β_m=-2^2m(2m-1)B_2m/(2m)!. Multiplication by 2^2m(2m)! proves the identity for κ_m. The Bernoulli denominator theorem gives v_2(B_2m)=-1 for m>0: its denominator contains each prime q with (q-1)mid 2m exactly once, including 2[source reference]. Since 2m-1 is odd, the valuation is exactly 4m…

    Exact mathematical passage
    Q(z)=z\coth z
Q'Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 962

    Let Q(z)=zcoth z, the regular formal series with constant coefficient one. The elementary differential identity (frac zsinh z)^2=Q(z)-zQ'(z) and the Bernoulli expansion Q(z)=1+∑_m≥1frac2^2mB_2m(2m)!z^2m give β_m=-2^2m(2m-1)B_2m/(2m)!. Multiplication by 2^2m(2m)! proves the identity for κ_m. The Bernoulli denominator theorem gives v_2(B_2m)=-1 for m>0: its denominator contains each prime q with (q-1)mid 2m exactly once, including 2[source reference]. Since 2m-1 is odd, the valuation is exactly 4m…

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     \left(\frac z{\sinh z}\right)^2=Q(z)-zQ'(z)
    
q'Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2058

    Write the input indices as Mq+u and Mq'+v. Since 0≤ u,v<M, their sum has residue u+v when u+v<M, and residue u+v-M with one additional coarse index when u+v≥ M.

    Exact mathematical passage
    Mq'+v
q^{-L}Use in source · 1 passage

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  • Use in source · Read passage · continuation_08.tex, line 16

    Let q>0, L≥1, and m,k,e≥0. Write S= Z/q Z. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0)∈ Z^m+k, F(x)≡0pmodq^L. The actual lower root supplies the unique integral residual r=q^-LF(x)∈ Z^e, bar r=rbmod q∈ S^e. Let the actual reduced Jacobian blocks be J=(D_YF)(x)bmod q:S^m→ S^e, K=(D_TF)(x)bmod q:S^k→ S^e. For a prescribed parameter digit v∈ S^k, choose its canonical integral representative widehat v. Define L_v to be the set of actual residue vectors z∈( Z/q^L+1 Z)^m+k satisfying z≡ xpmodq^L, F(z…

    Exact mathematical passage
    
    r=q^{-L}F(x)\in\mathbb Z^e,\qquad \bar r=r\bmod q\in S^e.
    
Q^2Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 326

    , and b=(1-i)/2. The exact identity F_0(t)=frac2cosh t+cos t =sech(at)sech(bt) gives, for U=a E+b F, L(U^4n)=C_0(n). Although a and b have negative valuation, their denominators disappear from the following exact polynomial identities: U^2 · = E F+i(P-Q),nonumber H_0:=fracU^4-12 · =P+Q-fracP^22-fracQ^22 +3PQ+i E F(P-Q).

    Exact mathematical passage
    
     U^2&=\mathcal E\mathcal F+i(P-Q),\nonumber\\
     H_0:=\frac{U^4-1}{2}
       &=P+Q-\frac{P^2}{2}-\frac{Q^2}{2}
                     +3PQ+i\mathcal E\mathcal F(P-Q).
     \label{eq:four-euler-H}
    
q^{2M+2}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1010

    [Odd modulus truncation] For every M≥0, every positive odd q, and every even n, E_n ≡ ∑_r=0^min(M,n/2) (-1)^rZ_2r+1 binom n2rq^2rS_n-2r(q) pmod q^2M+2. If pmid q and a=v_p(q), the same truncation is valid modulo p^2a(M+1).

    Exact mathematical passage
    
    \boxed{
    E_n
    \equiv
    \sum_{r=0}^{\min(M,n/2)}
    (-1)^rZ_{2r+1}
    \binom n{2r}q^{2r}S_{n-2r}(q)
    \pmod {q^{2M+2}}.
    }
    
q^{2N}Use in source · 1 passage

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  • Use in source · Read passage · continuation_09.tex, line 98

    ^m satisfying J_a u=-bar r-K_abar w. Choose an integer representative widetilde u, and set a'=a+q^N(widetilde u,w). The parameter quotient is the supplied integer w. Therefore free(a')=t' exactly. The checked polynomial Taylor identity gives F(a')≡ F(a)+q^NDF(a)(widetilde u,w) ≡0pmodq^N+1. Its quadratic remainder is divisible by q^2N, and 2N≥ N+1. Also a'≡ apmod q, so evaluation of every polynomial partial derivative gives the same reduced Jacobian. Induction on the level, carrying the root, exa…

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    q^{2N}
q^{2r}Use in source · 5 passages

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  • Use in source · Read passage · 72_recovered.tex, line 965

    [Odd modulus Euler coordinate pair] For every positive odd q and every even n≥0, E_n = ∑_r=0^n/2 (-1)^rZ_2r+1 binom n2r q^2rS_n-2r(q). The inverse transform is S_n(q) = E_n + ∑_r=1^n/2 2^2r-1 binom n2r q^2rE_n-2r. Both transforms are lower triangular with diagonal one.

    Exact mathematical passage
    
    \boxed{
    E_n
    =
    \sum_{r=0}^{n/2}
    (-1)^rZ_{2r+1}
    \binom n{2r}
    q^{2r}S_{n-2r}(q).
    }
    
q^dUse in source · 1 passage

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  • Use in source · Read passage · continuation_02.tex, line 57

    For every positive integer q, including composite moduli and modulus one, and every L≥0, the checked map Φ_x,L:( Z/q Z)^d⟶ {ybmod q^L+1:y≡ xpmodq^L}, t⟼ x+q^Lwidetilde t is a bijection. Here widetilde t is the canonical integer representative of the digit vector. The entire residue fiber therefore has q^d elements. Injectivity is proved by cancelling q^L in the corresponding integer divisibility statement; surjectivity comes from the lower-residue congruence.

    Exact mathematical passage
    q^d
Q^eUse in source · 1 passage

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  • Use in source · Read passage · continuation_10.tex, line 33

    )∈ M_e× k(R). Suppose, for natural numbers s,M, that J∈GL_e(Q), p^sJ^-1∈ M_e(R), M>2s. For every v∈ R^k, set τ=t_0+p^Mv, G=F+p^MH. Then there is exactly one y∈ y_0+p^s+1R^e with G(y,τ)=0. It satisfies y-y_0∈ p^M-sR^e and there is an integral vector ε∈ p^2M-3sR^e such that, in Q^e, displaystyle y-y_0=-p^MJ^-1(H(a)+Kv)+ε. The same original exponent s appears throughout. No inverse or rank assumption on the perturbed Jacobian is imposed. The statement includes arbitrary integral p-adic coefficients…

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    Q^e
q^{L+1}Use in source · 18 passages

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  • Use in source · Read passage · 30_lifting.tex, line 76

    [Finite polynomial obstruction] Let q≥1, L≥1, x∈ Z^d, and suppose F(x)∈ q^L Z^e. Define the exact integer residual r_x by F(x)=q^Lr_x, and put O_x=S_q^e/im J_x, o_x=[overliner_x]. Then O_x is finite and o_x=0 Longleftrightarrow ∃ y∈ Z^d: y≡ xpmodq^L, F(y)≡0pmodq^L+1. No surjectivity or nonsingularity assumption is required.

    Exact mathematical passage
    \label{eq:finite-obstruction}
     o_x=0\quad\Longleftrightarrow\quad
     \exists y\in\Z^d:\quad
     y\equiv x\pmod{q^L},\qquad F(y)\equiv0\pmod{q^{L+1}}.
    
q^{L+a}Use in source · 4 passages

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  • Use in source · Read passage · 30_lifting.tex, line 342

    z_n+1-z_n · ∈ q^L+n Z^d, · J_z_n · =J_x. For a≤ b, it also satisfies z_b-z_a∈ q^L+a Z^d.

    Exact mathematical passage
    z_b-z_a\in q^{L+a}\Z^d
q^{L+n}Use in source · 5 passages

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  • Use in source · Read passage · 30_lifting.tex, line 338

    [Compatible polynomial tower] Let q≥1, L≥1, F∈ Z[X_1,…,X_d]^e, and x∈ Z^d satisfy F(x)∈ q^L Z^e. Suppose J_x:S_q^d→ S_q^e is surjective. There is a single sequence z_n∈ Z^d such that z_0 · =x, · F(z_n) · ∈ q^L+n Z^e, z_n+1-z_n · ∈ q^L+n Z^d, · J_z_n · =J_x. For a≤ b, it also satisfies z_b-z_a∈ q^L+a Z^d.

    Exact mathematical passage
    
     z_0&=x,& F(z_n)&\in q^{L+n}\Z^e,\\
     z_{n+1}-z_n&\in q^{L+n}\Z^d,& J_{z_n}&=J_x.
    
Q^NUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2869

    [Residue subsequences of a second order recurrence] Suppose x_n+2=Px_n+1-Qx_n over a commutative ring. Define V_0=2, V_1=P, and V_n+2=PV_n+1-QV_n. For every positive integer N and every address j, the subsequence u_k=x_Nk+j satisfies u_k+2=V_Nu_k+1-Q^Nu_k.

    Exact mathematical passage
    
    \boxed{
    u_{k+2}=V_Nu_{k+1}-Q^Nu_k.}
    
q^NUse in source · 7 passages

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  • Use in source · Read passage · 30_lifting.tex, line 346

    At level N=L+n, divide the residual by q^N. Surjectivity supplies t∈ S_q^d with J_z_nt=-overlineF(z_n)/q^N. Choose an integer representative and apply [source reference] to obtain z_n+1. The new point is congruent to z_n modulo q. Evaluation of each formal partial derivative therefore gives J_z_n+1=J_z_n.

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    q^N
q^{N+1}Use in source · 1 passage

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  • Use in source · Read passage · continuation_09.tex, line 95

    s t'-free(a)=q^Nw, with w∈ Z^k. Surjectivity supplies a solved digit u∈ S_q^m satisfying J_a u=-bar r-K_abar w. Choose an integer representative widetilde u, and set a'=a+q^N(widetilde u,w). The parameter quotient is the supplied integer w. Therefore free(a')=t' exactly. The checked polynomial Taylor identity gives F(a')≡ F(a)+q^NDF(a)(widetilde u,w) ≡0pmodq^N+1. Its quadratic remainder is divisible by q^2N, and 2N≥ N+1. Also a'≡ apmod q, so evaluation of every polynomial partial derivative give…

    Exact mathematical passage
    
    F(a')\equiv F(a)+q^NDF(a)(\widetilde u,w)
    \equiv0\pmod{q^{N+1}}.
    
Q^rUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1719

    /3 in the one-prime example above loses the advertised digit; invertibility of C alone does not restore it. The original proof is retained under these hypotheses. Let Q=∏_p∈ Sp^L_p for a finite set of primes. Let C∈ M_d( Z) be invertible modulo Q, and choose H such that C^H=I+QB with B∈ M_d( Z). Then for every nonnegative integer x, a_Hx+j = ∑_r=0^xQ^rbinom xr w^TC^jB^rv_0. For every r, Δ_H^ra_n≡0pmod Q^r.

    Exact mathematical passage
    
    \boxed{
    a_{Hx+j}
    =
    \sum_{r=0}^{x}Q^r\binom xr
    w^TC^jB^rv_0.
    }
    
q^{r-d}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1661

    On the polynomial factor the difference operator acts as q+yΔ. The two terms commute. Expand (q+yΔ)^r and evaluate at zero. Since Δ^kQ_j=0 for k>d, every surviving term contains at least q^r-d.

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    q^{r-d}
q^{r-k}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1641

    [Polynomial exponential Mahler transfer] Let f(n)=P(n)u^n with P∈ Z_p[n] of degree d. Let H=Λ_p,L, fix an address j, and write Q_j(x)=P(Hx+j), y=u^H, q=y-1. Then f(Hx+j)=∑_r≥0c_j,rbinom xr, where c_j,r = u^j ∑_k=0^min(d,r) binom rk Δ^kQ_j(0)y^kq^r-k. Consequently, v_p(c_j,r)≥ Lmax(r-d,0). For R≥ d, truncation at depth R is valid modulo p^L(R+1-d) uniformly on Z_p.

    Exact mathematical passage
    
    \boxed{
    c_{j,r}
    =
    u^j
    \sum_{k=0}^{\min(d,r)}
    \binom rk
    \Delta^kQ_j(0)y^kq^{r-k}.
    }
    
\mathbb Q_2Use in source · 4 passages

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  • Use in source · Read passage · strengthening.tex, line 315

    Put K= Q_2(i), R= Z_2[i], a=(1+i)/2, and b=(1-i)/2. The exact identity F_0(t)=frac2cosh t+cos t =sech(at)sech(bt) gives, for U=a E+b F, L(U^4n)=C_0(n). Although a and b have negative valuation, their denominators disappear from the following exact polynomial identities: U^2 · = E F+i(P-Q),nonumber

    Exact mathematical passage
    K=\mathbb Q_2(i)
Q_FUse in source · 6 passages

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  • Use in source · Read passage · 30_lifting.tex, line 777

    Retain the original certificate A=p^sJ^-1 and put c=p^s+1. Taylor expansion and a polynomial difference identity yield polynomial vectors Q_F,Q_H, both zero at the origin, such that (AF)(a+cu)=p^2s+1u+p^2s+2Q_F(u), (AH)(a+cu)=AH(a)+cQ_H(u). Consequently (AG)(a+cu)=p^2s+1(C+u+pQ(u)), where C=p^M-(2s+1)AH(a), Q=Q_F+p^M-s-1Q_H, Q(0)=0. All exponents are nonnegative. The same contraction and divisibility argument as in Theorem~[source reference] gives the root, the displacement p^M-s, and full-ball …

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    Q_F,Q_H
Q_HUse in source · 6 passages

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  • Use in source · Read passage · 30_lifting.tex, line 777

    Retain the original certificate A=p^sJ^-1 and put c=p^s+1. Taylor expansion and a polynomial difference identity yield polynomial vectors Q_F,Q_H, both zero at the origin, such that (AF)(a+cu)=p^2s+1u+p^2s+2Q_F(u), (AH)(a+cu)=AH(a)+cQ_H(u). Consequently (AG)(a+cu)=p^2s+1(C+u+pQ(u)), where C=p^M-(2s+1)AH(a), Q=Q_F+p^M-s-1Q_H, Q(0)=0. All exponents are nonnegative. The same contraction and divisibility argument as in Theorem~[source reference] gives the root, the displacement p^M-s, and full-ball …

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    Q_F,Q_H
Q_iUse in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 230

    For 0≤ i<j, let d_i=v_2(i!)+v_2((j-1-i)!), k_i=1+lfloorlog_2max(i,j-1-i)rfloor, B_i=i+2^k_i Z_2, t_i=d_i+k_i. At j=1 use d_0=k_0=t_0=0. The balls B_i partition Z_2. Indeed a residue represented twice among 0,…,j-1 splits into its two children; every other represented residue remains one ball. On each such ball, P_j(i+2^k_iy)=2^t_iyQ_i(y), Q_i∈ Z_2⟨ y⟩^×, Q_i≡1pmod2.

    Exact mathematical passage
    
     P_j(i+2^{k_i}y)=2^{t_i}yQ_i(y),\qquad
     Q_i\in\mathbb Z_2\langle y\rangle^\times,
     \qquad Q_i\equiv1\pmod2.
    
Q_jUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1628

    [Polynomial exponential Mahler transfer] Let f(n)=P(n)u^n with P∈ Z_p[n] of degree d. Let H=Λ_p,L, fix an address j, and write Q_j(x)=P(Hx+j), y=u^H, q=y-1. Then f(Hx+j)=∑_r≥0c_j,rbinom xr, where c_j,r = u^j ∑_k=0^min(d,r) binom rk Δ^kQ_j(0)y^kq^r-k. Consequently, v_p(c_j,r)≥ Lmax(r-d,0). For R≥ d, truncation at depth R is valid modulo p^L(R+1-d) uniformly on Z_p.

    Exact mathematical passage
    
    Q_j(x)=P(Hx+j),
    \qquad
    y=u^H,
    \qquad q=y-1.
    
\mathbb Q_p^eUse in source · 1 passage

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  • Use in source · Read passage · continuation_11.tex, line 118

    hbb Z_p^e+k, F(a)=0. Let ι: Z_p↪ Q_p be the coefficient inclusion, applied entrywise to points and matrices. Assume only that ι(D_YF(a))∈GL_e( Q_p). The theorem mcodePolynomialRegularityIntegral.exists_analytic_integral_implicit constructs the preceding analytic graph in the ambient spaces Q_p^k and Q_p^e. Its root equation evaluates the original integral polynomial coefficients using ι, and its base derivative is -ι(D_YF(a))^-1 ι(D_TF(a)). The source proves that polynomial evaluation, formal pa…

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    \mathbb Q_p^e
\mathbb Q_p^kUse in source · 1 passage

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  • Use in source · Read passage · continuation_11.tex, line 118

    ,T]^e, a∈ Z_p^e+k, F(a)=0. Let ι: Z_p↪ Q_p be the coefficient inclusion, applied entrywise to points and matrices. Assume only that ι(D_YF(a))∈GL_e( Q_p). The theorem mcodePolynomialRegularityIntegral.exists_analytic_integral_implicit constructs the preceding analytic graph in the ambient spaces Q_p^k and Q_p^e. Its root equation evaluates the original integral polynomial coefficients using ι, and its base derivative is -ι(D_YF(a))^-1 ι(D_TF(a)). The source proves that polynomial evaluation, for…

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    \mathbb Q_p^k
q_rUse in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 415

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

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    \chi_{C^N}(X)=X^d+\sum_{r<d}q_rX^r
q_{u,L}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1549

    [One unit mode] Let u∈ Z_p^×, let H=Λ_p,L, and put q_u,L=u^H-1∈ p^L Z_p. For every integer address j and every x∈ Z_p, u^Hx+j = u^j∑_r≥0q_u,L^ rbinom xr. The rth Mahler coefficient has valuation at least Lr.

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    q_{u,L}=u^H-1\in p^L\Zp.
    
q_{u,L}^{\,r}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1553

    [One unit mode] Let u∈ Z_p^×, let H=Λ_p,L, and put q_u,L=u^H-1∈ p^L Z_p. For every integer address j and every x∈ Z_p, u^Hx+j = u^j∑_r≥0q_u,L^ rbinom xr. The rth Mahler coefficient has valuation at least Lr.

    Exact mathematical passage
    
    \boxed{
    u^{Hx+j}
    =
    u^j\sum_{r\geq0}q_{u,L}^{\,r}\binom xr.
    }
    
\Qp^dUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 766

    [Weighted coefficient deformation] Let F,H∈ R[X_1,…,X_d]^d, F(a)=0, and J=DF(a). Suppose J∈GL_d( Q_p), p^sJ^-1∈ M_d(R), and M>2s. For G=F+p^MH, there is exactly one root b∈ a+p^s+1R^d. It satisfies b-a∈ p^M-sR^d and, in Q_p^d, b-a=-p^MJ^-1H(a)+ε, ε∈ p^2M-3sR^d. Only the original Jacobian DF(a) is subject to an inverse hypothesis.

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    \Qp^d
\Qp^eUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 842

    y-y_0 · =-p^MJ^-1(H(a)+Kv)+ε, ε∈ p^2M-3sR^e. The equality involving J^-1 is in Q_p^e, with an integral error vector. The parameter equality is exact.

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    \Qp^e
\QQUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 468

    The examples later in the paper use different kernels, but their coefficient algebra has one common form. This section records that form once. Let R be a commutative algebra over Q, and take all coefficients in R.

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    \QQ
\mathcal RUse in source · 5 passages

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  • Use in source · Read passage · 10_foundations.tex, line 12

    The original arithmetic seed notation S=(A,a_0, O, R,{P_N,j},w,τ_p) records a coefficient ring or module, a seed, allowed operators, relations, residue selectors, a depth weight, and a topology. This notation describes the ingredients of the construction; a category of seed presentations and a notion of equivalence still require separate definitions. We therefore state each theorem for a concrete object. Section~[source reference] discusses the additional structure needed for the original native…

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    \mathcal S=(A,a_0,\mathcal O,\mathcal R,\{P_{N,j}\},w,\tau_p)
    
R^dUse in source · 44 passages

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  • Use in source · Read passage · 30_lifting.tex, line 37

    [Polynomial Taylor remainder] Let R be a commutative ring, P∈ R[X_1,…,X_d], a∈ R^d, and c∈ R. There is a polynomial Q∈ R[U_1,…,U_d], with Q(0)=0, such that for every u∈ R^d, P(a+cu)=P(a)+c∑_j=1^d(∂_jP)(a)u_j+c^2Q(u). In particular, when R= Z and a,u,c are integral, subtracting the first two terms on the right from P(a+cu) gives a multiple of c^2.

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    a\in R^d
R^eUse in source · 9 passages

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  • Use in source · Read passage · 30_lifting.tex, line 833

    rmation] leavevmode Let F,H∈ R[Y_1,…,Y_e,T_1,…,T_k]^e and a=(y_0,t_0)∈ R^e+k satisfy F(a)=0. Define the actual blocks J=D_YF(a)∈ M_e(R), K=D_TF(a)∈ M_e× k(R). Suppose J∈GL_e( Q_p), p^sJ^-1∈ M_e(R), and M>2s. For every v∈ R^k, set τ=t_0+p^Mv, G=F+p^MH. There is exactly one y∈ y_0+p^s+1R^e with G(y,τ)=0. It satisfies y-y_0 · ∈ p^M-sR^e,

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    y\in y_0+p^{s+1}R^e
R^{e+k}Use in source · 6 passages

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  • Use in source · Read passage · 30_lifting.tex, line 824

    [Joint coefficient and parameter deformation] leavevmode Let F,H∈ R[Y_1,…,Y_e,T_1,…,T_k]^e and a=(y_0,t_0)∈ R^e+k satisfy F(a)=0. Define the actual blocks J=D_YF(a)∈ M_e(R), K=D_TF(a)∈ M_e× k(R). Suppose J∈GL_e( Q_p), p^sJ^-1∈ M_e(R), and M>2s. For every v∈ R^k, set τ=t_0+p^Mv, G=F+p^MH. There is exactly one y∈ y_0+p^s+1R^e with G(y,τ)=0. It satisfies y-y_0 · ∈ p^M-sR^e,

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    a=(y_0,t_0)\in R^{e+k}
R^IUse in source · 2 passages

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  • Use in source · Read passage · supplement_06.tex, line 30

    pageheading1. Actual matrix orbits and arithmetic jumpsTheorem 9.1; manuscript page 22 Let R be a commutative ring, I a finite coordinate type and d=|I|. For C∈ M_I(R) and v_0∈ R^I, define orbit(C,v_0,n)=C^nv_0, u_k=packet(C,v_0,N,j,k)=C^Nk+jv_0. The implementation uses actual matrix powers and matrix--vector multiplication. A separately supplied sequence v with v_n+1=Cv_n is proved by induction to equal this power orbit. Thus the source-sequence formulation is connected to the construction.

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    v_0\in R^I
R^kUse in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 829

    [Joint coefficient and parameter deformation] leavevmode Let F,H∈ R[Y_1,…,Y_e,T_1,…,T_k]^e and a=(y_0,t_0)∈ R^e+k satisfy F(a)=0. Define the actual blocks J=D_YF(a)∈ M_e(R), K=D_TF(a)∈ M_e× k(R). Suppose J∈GL_e( Q_p), p^sJ^-1∈ M_e(R), and M>2s. For every v∈ R^k, set τ=t_0+p^Mv, G=F+p^MH. There is exactly one y∈ y_0+p^s+1R^e with G(y,τ)=0. It satisfies y-y_0 · ∈ p^M-sR^e,

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    v\in R^k
R^{\NN}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 124

    [Hurwitz series ring] Let R be a commutative ring with identity. The Hurwitz series ring H(R) is the set R^ N with componentwise addition and product (amathbin⋆ b)_n = ∑_r=0^n binom nr a_r b_n-r. Its identity is δ=(1,0,0,…).

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    R^{\NN}
R_0Use in source · 5 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1324

    [The first nontrivial dyadic row] For n=4, R_4=frac9115, R_2=frac53, R_0=1. The exact kernel coefficients are κ_2,1=-frac83, κ_2,2=frac1285. Therefore E_4 = frac9115 -frac83binom42frac53 +frac1285 =5. The rational source values are dyadic units, while the transfer depth is carried by the kernel.

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    R_4=\frac{91}{15},
    \qquad
    R_2=\frac53,
    \qquad
    R_0=1.
    
R_1Use in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 324

    The standard window supplies the initial ratio R_1. The weighted estimate handles pmid c, and Riccati propagation handles pnmid c. Flatness is closed under products, so the exact factorization establishes flatness of j! P_N,j. Extracting its supported Hurwitz coefficients and applying the proved saturation and step-multiple lemmas yields [source reference]--[source reference].

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    R_1
R_2Use in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1324

    [The first nontrivial dyadic row] For n=4, R_4=frac9115, R_2=frac53, R_0=1. The exact kernel coefficients are κ_2,1=-frac83, κ_2,2=frac1285. Therefore E_4 = frac9115 -frac83binom42frac53 +frac1285 =5. The rational source values are dyadic units, while the transfer depth is carried by the kernel.

    Exact mathematical passage
    
    R_4=\frac{91}{15},
    \qquad
    R_2=\frac53,
    \qquad
    R_0=1.
    
R_{2(r-m)}Use in source · 2 passages

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  • Use in source · Read passage · supplement_03.tex, line 122

    rgetDyadicSeries.dyadic_reconstruction_series pageheading3. Exact reconstruction and genuine dyadic congruenceTheorem 5.3 and Corollary 5.4; manuscript pages 14--15 The reciprocal coefficients β_m use an ordinary power expansion. In contrast, the source and Euler series use exponential coefficients. Extracting degree 2r from B(2t)F(t)=sech t therefore gives E_2r=∑_m=0^rβ_m2^2mfall2r2mR_2(r-m), fall2r2m=frac(2r)!(2r-2m)!. The formal proof uses the actual natural-number descending factorial and pr…

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     \boxed{E_{2r}=\sum_{m=0}^{r}\beta_m2^{2m}\fall{2r}{2m}R_{2(r-m)},}
     \qquad
     \fall{2r}{2m}=\frac{(2r)!}{(2r-2m)!}.
    
R_{2n}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1898

    [Analytic dyadic source] The function mathscr R belongs to the dyadic Tate algebra and takes values in Z_2. For every nonnegative integer n, mathscr R(n)=R_2n.

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    \boxed{\mathscr R(n)=R_{2n}.}
    
R_{2r}Use in source · 4 passages

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  • Use in source · Read passage · supplement_03.tex, line 54

    The source has valuation exactly zero. At even degree 2r, define the literal rational source R_2r=frac3^2r+2-14(2r+1)(2r+2). Both numerator and denominator are nonzero. The pinned Mathlib lifting-the-exponent theorem gives v_2(3^2r+2-1)=2+v_2(2r+2). Because 2r+1 is odd, the denominator has the same valuation. Consequently, v_2(R_2r)=0, |R_2r|_2=1. The code proves actual mcodeIsUnit status for the constructed Z_2 element. This strengthens the manuscript's integrality conclusion without changing i…

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     R_{2r}=\frac{3^{2r+2}-1}{4(2r+1)(2r+2)}.
    
R_3Use in source · 1 passage

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  • Use in source · visual_body.tex, line 476

    The radius is 2^liminf_d v_2(a_d)/d in the normalized dyadic absolute value. Hence R_0=R_1=2, R_2=2^5/2, R_3≥2. The exact radius of address three is not asserted. For address two, the lower bound gives convergence inside the claimed radius and the unbounded equality family prevents a larger one.

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     R_0=R_1=2,\qquad R_2=2^{5/2},\qquad R_3\ge2.
    
R_4Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1324

    [The first nontrivial dyadic row] For n=4, R_4=frac9115, R_2=frac53, R_0=1. The exact kernel coefficients are κ_2,1=-frac83, κ_2,2=frac1285. Therefore E_4 = frac9115 -frac83binom42frac53 +frac1285 =5. The rational source values are dyadic units, while the transfer depth is carried by the kernel.

    Exact mathematical passage
    
    R_4=\frac{91}{15},
    \qquad
    R_2=\frac53,
    \qquad
    R_0=1.
    
R_6Use in source · 1 passage

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  • Use in source · Read passage · supplement_03.tex, line 163

    al checksExample 5.5 and controls against incorrect normalizations The actual reciprocal recurrence gives (β_0,β_1,β_2,β_3,β_4) =(1,-frac13,frac115,-frac2189,frac1675). These values are obtained from the general formal-series definition. They are not substituted as independent input data for the reconstruction theorem. The source formula gives (R_0,R_2,R_4,R_6,R_8) =(1,frac53,frac9115,frac2057,frac738145). Using the retained Euler anchor E_8=1385, Lean checks the literal decomposition E_8=frac73…

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     (R_0,R_2,R_4,R_6,R_8)
     =\left(1,\frac53,\frac{91}{15},\frac{205}{7},\frac{7381}{45}\right).
    
R_8Use in source · 1 passage

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  • Use in source · Read passage · supplement_03.tex, line 163

    al checksExample 5.5 and controls against incorrect normalizations The actual reciprocal recurrence gives (β_0,β_1,β_2,β_3,β_4) =(1,-frac13,frac115,-frac2189,frac1675). These values are obtained from the general formal-series definition. They are not substituted as independent input data for the reconstruction theorem. The source formula gives (R_0,R_2,R_4,R_6,R_8) =(1,frac53,frac9115,frac2057,frac738145). Using the retained Euler anchor E_8=1385, Lean checks the literal decomposition E_8=frac73…

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     (R_0,R_2,R_4,R_6,R_8)
     =\left(1,\frac53,\frac{91}{15},\frac{205}{7},\frac{7381}{45}\right).
    
r_AUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 3003

    [Coprime packet factorization] Let A and B be coprime positive integers. Let r_A be an address modulo A and r_B an address modulo B. Let r be the unique address modulo AB satisfying r≡ r_Apmod A, r≡ r_Bpmod B. Then P_AB,r = P_A,r_AP_B,r_B = P_B,r_BP_A,r_A.

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    r_A
r_aUse in source · 2 passages

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  • Use in source · Read passage · supplement_10.tex, line 120

    The actual ratio is also proved to be an expansion in t^N, so its coefficients vanish outside multiples of N. Constructed Z_p lifts and the established coefficient criterion now give, for its literal Hurwitz coefficients r_a(n), Δ_uh^kr_a(n)∈ p^min(k,mn) Z_p. The separate rational precision theorem is obtained through the coefficient-lift equality. Divisibility in the field Q_p is never substituted for this assertion.

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    r_a(n)
R_{ap}Use in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 235

    At the singular addresses c=ap, 1≤ a<m, use the weighted ring W_p={∑_d≥0a_dT^d: p^lfloor d/prfloora_d∈ Z_p}. The key estimate is R_ap(log(1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!)…

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    R_{ap}(\log(1+T))\in\mathcal W_p
r_BUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 3003

    [Coprime packet factorization] Let A and B be coprime positive integers. Let r_A be an address modulo A and r_B an address modulo B. Let r be the unique address modulo AB satisfying r≡ r_Apmod A, r≡ r_Bpmod B. Then P_AB,r = P_A,r_AP_B,r_B = P_B,r_BP_A,r_A.

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    r_B
R_cUse in source · 17 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 216

    Set H_c(t)=∑_n≥0fract^Nn+c(Nn+c)!, A_c(t)=∑_n≥0fract^Nn(Nn+c)!, R_c(t)=fracA_c-1(t)A_c(t) (1≤ c<N). The denominator of this last quotient has nonzero constant coefficient. The section identities give R_cH_c=tH_c-1, R_c(0)=c, j! P_N,j= P_N,0∏_c=1^jR_c. Cancellation of consecutive source sections proves the product identity and ties the ratios directly to the family under study.

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     H_c(t)=\sum_{n\ge0}\frac{t^{Nn+c}}{(Nn+c)!},\qquad
     A_c(t)=\sum_{n\ge0}\frac{t^{Nn}}{(Nn+c)!},\qquad
     R_c(t)=\frac{A_{c-1}(t)}{A_c(t)}\quad(1\le c<N).
    
R_c'Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 293

    At a nonsingular address 2≤ c<N with pnmid c, differentiation of the section identities gives tR_c'=R_c(1+R_c-1-R_c). Writing q=R_c-1-(c-1) and y=R_c-c, this becomes (tD+c)y=cq+yq-y^2. On degrees divisible by p, the inverse of tD+c divides the degree-d coefficient by the unit d+c. Starting with y_0=0, form y_s+1=(tD+c)^-1(cq+y_sq-y_s^2). Flatness and support are preserved. Indeed, put E_c=tD+c and C_p=pD^p^2-D^p. The commutator is [B_p,E_c]=C_p, and iteration gives B_p^rE_cy=E_cB_p^ry+rC_pB_p^r-…

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     tR_c'=R_c(1+R_{c-1}-R_c).
    
R_{c-1}Use in source · 5 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 293

    At a nonsingular address 2≤ c<N with pnmid c, differentiation of the section identities gives tR_c'=R_c(1+R_c-1-R_c). Writing q=R_c-1-(c-1) and y=R_c-c, this becomes (tD+c)y=cq+yq-y^2. On degrees divisible by p, the inverse of tD+c divides the degree-d coefficient by the unit d+c. Starting with y_0=0, form y_s+1=(tD+c)^-1(cq+y_sq-y_s^2). Flatness and support are preserved. Indeed, put E_c=tD+c and C_p=pD^p^2-D^p. The commutator is [B_p,E_c]=C_p, and iteration gives B_p^rE_cy=E_cB_p^ry+rC_pB_p^r-…

    Exact mathematical passage
    
     tR_c'=R_c(1+R_{c-1}-R_c).
    
r_jUse in source · 2 passages

Search aliases: r_j

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  • Use in source · Read passage · 72_recovered.tex, line 3030

    [Several coprime moduli] Let N_1,…,N_s be pairwise coprime and put N=∏_j=1^sN_j. For compatible addresses r_j, P_N,r = ∏_j=1^sP_N_j,r_j, where r is the unique combined address modulo N.

    Exact mathematical passage
    r_j
R_{j+1}Use in source · 1 passage

Search aliases: R_j+1

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  • Use in source · Read passage · supplement_10.tex, line 178

    Let K_j(t) be the normalized section kernel with constant coefficient one. The constructed unnormalized Euler series is F_N,j(t)=j! K_j(t)^-1. The source/inverse identities and the factorial successor identity prove F_N,j+1=F_N,jR_j+1, F_N,j=F_N,0∏_c=1^jR_c. The product formula is not an assembly hypothesis. Proved flatness of all factors and the original product-closure theorem establish flatness of the full family.

    Exact mathematical passage
    
     F_{N,j+1}=F_{N,j}R_{j+1},\qquad
     F_{N,j}=F_{N,0}\prod_{c=1}^{j}R_c.
    
r_LUse in source · 2 passages

Search aliases: r_L

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  • Use in source · worked_atlas.tex, line 68

    AtlasHeadA compatible root is built one digit at a timeatlas:smooth Take F(x)=x^2-7 over the 3-adic integers and start with x_1=1 modulo three. If F(x_L)=3^Lr_L, the next representative has the form x_L+1=x_L+3^Lt_L. Its correcting digit satisfies r_L+2x_Lt_L≡0pmod3. AtlasImagesmooth_tower For example, F(1)/3=-2 and 2x_1=2, so the next digit is one and x_2=4. Then F(4)/9=1, again giving digit one and x_3=13. At the next stage the residual is divisible by three, so digit zero is correct. A zero d…

    Exact mathematical passage
    F(x_L)=3^Lr_L
R_NUse in source · 2 passages

Search aliases: R_N

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  • Use in source · Read passage · supplement_09.tex, line 54

    Now take the integral cyclic group ring R_N= Z[ Z/N Z], x=[1], x^N=1, and assume Nmid p(p-1). Thus (x^p-1)^p=1. Each genuine consecutive block ∏_i=0^p-1(x-(pb+i)) belongs to (p,x^p-1-1). This follows from the actual polynomial identity ∏_i∈ F_p(X-i)=X^p-X modulo p, not from a hypothesized valuation of the product.

    Exact mathematical passage
    
     R_N=\Z[\Z/N\Z],\qquad x=[1],\qquad x^N=1,
    
R_nUse in source · 8 passages

Search aliases: R_n

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  • Use in source · Read passage · 72_recovered.tex, line 1056

    For even n≥0, define R_n = frac3^n+2-14(n+1)(n+2). Its even exponential generating function is ∑_n evenR_nfract^nn! = fracsinh(2t)sinh t2t^2. Define β_r by fracz^2sinh^2z = ∑_r≥0β_rz^2r.

    Exact mathematical passage
    
    \boxed{
    R_n
    =
    \frac{3^{n+2}-1}{4(n+1)(n+2)}.
    }
    
R_{n-2}Use in source · 1 passage

Search aliases: R_n-2

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  • Use in source · Read passage · 72_recovered.tex, line 1427

    The first three sharp truncations are E_n · ≡ R_n · · pmod 8, E_n · ≡ R_n-frac83binom n2R_n-2 · · pmod 128, E_n · ≡ R_n-frac83binom n2R_n-2 +frac1285binom n4R_n-4 · · pmod 2048.

    Exact mathematical passage
    
    \begin{aligned}
    E_n&\equiv R_n&&\pmod 8,\\
    E_n&\equiv R_n-\frac83\binom n2R_{n-2}&&\pmod {128},\\
    E_n&\equiv R_n-\frac83\binom n2R_{n-2}
    +\frac{128}{5}\binom n4R_{n-4}&&\pmod {2048}.
    \end{aligned}
    
R_{n-2r}Use in source · 4 passages

Search aliases: R_n-2r

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  • Use in source · Read passage · 72_recovered.tex, line 1113

    [Dyadic Euler coordinate pair] For every even n≥0, E_n = ∑_r=0^n/2 β_r2^2r fall n2r R_n-2r. The inverse transform is R_n = ∑_r=0^n/2 frac2^4r+1(2r+2)! fall n2r E_n-2r.

    Exact mathematical passage
    
    \boxed{
    E_n
    =
    \sum_{r=0}^{n/2}
    \beta_r2^{2r}
    \fall n{2r}
    R_{n-2r}.
    }
    
R_{n-4}Use in source · 1 passage

Search aliases: R_n-4

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  • Use in source · Read passage · 72_recovered.tex, line 1427

    The first three sharp truncations are E_n · ≡ R_n · · pmod 8, E_n · ≡ R_n-frac83binom n2R_n-2 · · pmod 128, E_n · ≡ R_n-frac83binom n2R_n-2 +frac1285binom n4R_n-4 · · pmod 2048.

    Exact mathematical passage
    
    \begin{aligned}
    E_n&\equiv R_n&&\pmod 8,\\
    E_n&\equiv R_n-\frac83\binom n2R_{n-2}&&\pmod {128},\\
    E_n&\equiv R_n-\frac83\binom n2R_{n-2}
    +\frac{128}{5}\binom n4R_{n-4}&&\pmod {2048}.
    \end{aligned}
    
R_rUse in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 365

    The current focused obstruction is the admitted ordinary-index family r=(17659+2^18T)2^640-161, T∈ Z_≥0, j_r=161. On its surviving chart define the actual, undepleted function K_r(Z)=2^-178G_r(10202+2^15Z). The complete coefficient theorem gives K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z), a_r,b_r∈ Z_2, u_r∈ Z_2^×, R_r∈ Z_2⟨ Z⟩. Its half derivative is a dyadic isometry. It therefore has a unique critical point c_r∈ Z_2. Set Φ_r^act=K_r(c_r). The earlier independent-undepletion notation for this same sc…

    Exact mathematical passage
    
     K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z),
     \quad a_r,b_r\in\mathbb Z_2,\quad
     u_r\in\mathbb Z_2^\times,\quad
     R_r\in\mathbb Z_2\langle Z\rangle.
     \label{eq:root:quadraticclass}
    
r_{x'}Use in source · 4 passages

Search aliases: r_x'

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  • Use in source · Read passage · 30_lifting.tex, line 133

    [Independence of the integer representative] Under the hypotheses of Theorem~[source reference], let x'=x+q^Ls and write F(x')=q^Lr_x'. Then J_x'=J_x, overliner_x'=overliner_x+J_xoverline s. Consequently the two residual classes coincide in the common quotient S_q^e/im J_x.

    Exact mathematical passage
    F(x')=q^Lr_{x'}
\operatorname{rank}Use in source · 5 passages

Search aliases: rank operatorname

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  • Use in source · Read passage · 71_native.tex, line 31

    Theorem. For this fixed polynomial presentation, the group and distinguished class are independent of the chosen integer representative. The class vanishes if and only if bar x lifts to a solution modulo p^L+1. The obstruction group has p^r-rankJ elements. When lifts exist, their set is a torsor for ker J and contains exactly p^d-rankJ elements.

    Exact mathematical passage
    p^{r-\operatorname{rank}J}
\rankUse in source · 4 passages

Search aliases: rank operatorname

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  • Use in source · Read passage · 30_lifting.tex, line 294

    [Torsor and lift cardinalities] Under the hypotheses of Theorem~[source reference], there is an equivalence T_x≃ L_x,L. Every nonempty fiber is a torsor for ker J_x under addition of correction digits. If q=p is prime and ρ_x=rank_ F_pJ_x, then # L_x,L= p^d-ρ_x, · o_x=0,

    Exact mathematical passage
    \rho_x=\rank_{\F_p}J_x
\operatorname{rank}_{\mathbb F_p}Use in source · 3 passages

Search aliases: rank_ F_p operatorname mathbb

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  • Use in source · Read passage · continuation_02.tex, line 19

    Let p be prime, L≥1, and F∈ Z[X_1,…,X_d]^e, x∈ Z^d, F(x)≡0pmodp^L. Write r_x=F(x)/p^L, b_x=r_xbmod p, A_x=DF(x)bmod p: F_p^d⟶ F_p^e, ρ_x=rank_ F_pA_x. The actual lift set is L_x,L= {y∈( Z/p^L+1 Z)^d: y≡ xpmodp^L, F(y)=0}, where polynomial evaluation in the displayed set is modulo p^L+1. Define O_x= F_p^e/imA_x, o_x=[b_x]. Lean proves # L_x,L= p^d-ρ_x, · o_x=0,

    Exact mathematical passage
    
    r_x=F(x)/p^L,\qquad b_x=r_x\bmod p,
    \qquad A_x=DF(x)\bmod p:\mathbb F_p^d\longrightarrow\mathbb F_p^e,
    \qquad \rho_x=\operatorname{rank}_{\mathbb F_p}A_x.
    
\rank_{\F_p}Use in source · 2 passages

Search aliases: rank_F_p rank_ F_p rank F operatorname mathbb

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  • Use in source · Read passage · 30_lifting.tex, line 294

    [Torsor and lift cardinalities] Under the hypotheses of Theorem~[source reference], there is an equivalence T_x≃ L_x,L. Every nonempty fiber is a torsor for ker J_x under addition of correction digits. If q=p is prime and ρ_x=rank_ F_pJ_x, then # L_x,L= p^d-ρ_x, · o_x=0,

    Exact mathematical passage
    \rho_x=\rank_{\F_p}J_x
\rank_{\FF_p}Use in source · 1 passage

Search aliases: rank_FF_p rank_ F_p rank FF operatorname mathbb

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  • Use in source · Read passage · 72_recovered.tex, line 2588

    [Exact number of linear children] If ρ=rank_ F_poverline M, then a solvable depth L class has exactly p^d-ρ children modulo p^L+1.

    Exact mathematical passage
    \rho=\rank_{\FF_p}\overline M
\rceilUse in source · 1 passage

Search aliases: rceil

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  • Use in source · Read passage · root_scope.tex, line 268

    The largest emergence threshold is max_i t_i=v_2((j-1)!)+lceillog_2 jrceil. For j=2^a, all thresholds equal j-1. There are no roots before that distance and exactly j simple roots at and after it, within the positive-distance range. The equality and pre-wall cases are also resolved: regular labels j≥2 have no roots there; j=1 has one simple odd root at equality.

    Exact mathematical passage
    
     \max_i t_i=v_2((j-1)!)+\lceil\log_2 j\rceil.
    
\ResUse in source · 2 passages

Search aliases: Res operatorname

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  • Use in source · Read passage · 72_recovered.tex, line 2815

    We use the resultant convention Res_y(f,g) = ∏_f(α)=0g(α) when f is monic, with roots counted with multiplicity.

    Exact mathematical passage
    
    \Res_y(f,g)
    =
    \prod_{f(\alpha)=0}g(\alpha)
    
\Res_yUse in source · 2 passages

Search aliases: Res_y Res operatorname

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  • Use in source · Read passage · 72_recovered.tex, line 2815

    We use the resultant convention Res_y(f,g) = ∏_f(α)=0g(α) when f is monic, with roots counted with multiplicity.

    Exact mathematical passage
    
    \Res_y(f,g)
    =
    \prod_{f(\alpha)=0}g(\alpha)
    
\rfloorUse in source · 45 passages

Search aliases: rfloor

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  • Use in source · Read passage · edition_summary.tex, line 16

    The strongest common-step conclusion in width four is particularly compact. For n≥0 and k,H≥1, ∑_i=0^k(-1)^k-ibinom ki E_4(n+iH)^(4,j) ≡0pmod2^ v_2(k!)+kv_2(H)+A_j(k), where A_0(k)=A_1(k)=k, A_2(k)=max{k+4, 2k+lfloor k/2rfloor+1}, A_3(k)=k+2. There is no starting-row threshold. The exponent is exact in addresses zero and one for every order and positive step. It is exact in address two at every power-of-two order. The proof chapters also give mixed-step forms, the cancellation criterion in addre…

    Exact mathematical passage
    
     A_0(k)=A_1(k)=k,\qquad
     A_2(k)=\max\{k+4,\,2k+\lfloor k/2\rfloor+1\},\qquad
     A_3(k)=k+2.
    
\rmUse in source · 1 passage

Search aliases: rm

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  • Use in source · Read passage · supplement_01.tex, line 111

    For N>0 and 0≤ j<N, the independently defined original exponential section is E_N,j(t)=∑_d≡ j (rm mod N)fract^dd!. Using the actual power-series expansion homomorphism z↦ t^N, the code proves E_N,j(t)=fract^jj!K_N,j(t^N), F_N,j(t^N)E_N,j(t)=fract^jj!. Thus the recurrence is connected back to the original normalized EGF definition. The section vanishes at zero when j>0; no inverse of that nonunit section is assumed.

    Exact mathematical passage
    
     E_{N,j}(t)=\sum_{d\equiv j\ ({\rm mod}\ N)}\frac{t^d}{d!}.
    
\boldsymbol sUse in source · 1 passage

Search aliases: s boldsymbol

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  • Use in source · Read passage · 72_recovered.tex, line 2018

    [Nested discrete refinement] For positive integers M,N, (a_r^[M])_s^[N] = a_r+Ms^[MN]. For a mixed radix path (M_1,…,M_ℓ) with digits 0≤ s_i<M_i, the final address is r( s) = s_1+M_1s_2+⋯+ (∏_i=1^ℓ-1M_i)s_ℓ. Only the product radix and accumulated address determine the final branch.

    Exact mathematical passage
    
    \boxed{
    r(\boldsymbol s)
    =
    s_1+M_1s_2+\cdots+
    \left(\prod_{i=1}^{\ell-1}M_i\right)s_\ell.
    }
    
\mathcal SUse in source · 3 passages

Search aliases: S mathcal

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  • Use in source · Read passage · 10_foundations.tex, line 12

    The original arithmetic seed notation S=(A,a_0, O, R,{P_N,j},w,τ_p) records a coefficient ring or module, a seed, allowed operators, relations, residue selectors, a depth weight, and a topology. This notation describes the ingredients of the construction; a category of seed presentations and a notion of equivalence still require separate definitions. We therefore state each theorem for a concrete object. Section~[source reference] discusses the additional structure needed for the original native…

    Exact mathematical passage
    
    \mathcal S=(A,a_0,\mathcal O,\mathcal R,\{P_{N,j}\},w,\tau_p)
    
\mathsf SUse in source · 3 passages

Search aliases: S mathsf

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  • Use in source · Read passage · 72_recovered.tex, line 2063

    [Polyphase shift matrix] Let E be the scalar shift (Ea)_n=a_n+1, and let E_q shift the coarse index. Then P_M(Ea) = S_M(E_q) P_Ma, where S_M(E_q) = 0 · 1 · 0 · ⋯ · 0

    Exact mathematical passage
    
    \boxed{
    \mathcal P_M(Ea)
    =
    \mathsf S_M(E_q)\mathcal P_Ma,
    }
    
s^{[N]}Use in source · 3 passages

Search aliases: s^[N]

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  • Use in source · Read passage · 10_foundations.tex, line 488

    polyphase vocabulary. We explain it through elementary identities with explicit algebraic derivations. These identities have not been separately formalized in Lean for this consolidation. For a sequence a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplication, the phase-r product has a term without a …

    Exact mathematical passage
    
    A(z)=\sum_{r=0}^{M-1}z^rA_r(z^M),\qquad
    \bigl(a_r^{[M]}\bigr)_s^{[N]}=a_{r+Ms}^{[MN]}.
    
S^eUse in source · 3 passages

Search aliases: S^e

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  • Use in source · Read passage · continuation_08.tex, line 16

    Let q>0, L≥1, and m,k,e≥0. Write S= Z/q Z. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0)∈ Z^m+k, F(x)≡0pmodq^L. The actual lower root supplies the unique integral residual r=q^-LF(x)∈ Z^e, bar r=rbmod q∈ S^e. Let the actual reduced Jacobian blocks be J=(D_YF)(x)bmod q:S^m→ S^e, K=(D_TF)(x)bmod q:S^k→ S^e. For a prescribed parameter digit v∈ S^k, choose its canonical integral representative widehat v. Define L_v to be the set of actual residue vectors z∈( Z/q^L+1 Z)^m+k satisfying z≡ xpmodq^L, F(z…

    Exact mathematical passage
    
    r=q^{-L}F(x)\in\mathbb Z^e,\qquad \bar r=r\bmod q\in S^e.
    
S^HUse in source · 5 passages

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  • Use in source · Read passage · strengthening.tex, line 31

    Write S f(n)=f(n+1) and Δ_H=S^H-1. A mixed difference Δ_H_1⋯Δ_H_k acts on the row variable n, rather than on the full Euler subscript. We use v_p(0)=+∞. Orders and starting rows are nonnegative integers; steps are positive integers unless stated otherwise. An empty mixed difference is the original value, and a zero ordinary step makes every nonempty product containing it zero.

    Exact mathematical passage
    \Delta_H=S^H-1
S^kUse in source · 2 passages

Search aliases: S^k

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  • Use in source · Read passage · continuation_08.tex, line 18

    Let q>0, L≥1, and m,k,e≥0. Write S= Z/q Z. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0)∈ Z^m+k, F(x)≡0pmodq^L. The actual lower root supplies the unique integral residual r=q^-LF(x)∈ Z^e, bar r=rbmod q∈ S^e. Let the actual reduced Jacobian blocks be J=(D_YF)(x)bmod q:S^m→ S^e, K=(D_TF)(x)bmod q:S^k→ S^e. For a prescribed parameter digit v∈ S^k, choose its canonical integral representative widehat v. Define L_v to be the set of actual residue vectors z∈( Z/q^L+1 Z)^m+k satisfying z≡ xpmodq^L, F(z…

    Exact mathematical passage
    
    J=(D_YF)(x)\bmod q:S^m\to S^e,
    \qquad K=(D_TF)(x)\bmod q:S^k\to S^e.
    
S^mUse in source · 2 passages

Search aliases: S^m

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  • Use in source · Read passage · continuation_08.tex, line 18

    Let q>0, L≥1, and m,k,e≥0. Write S= Z/q Z. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0)∈ Z^m+k, F(x)≡0pmodq^L. The actual lower root supplies the unique integral residual r=q^-LF(x)∈ Z^e, bar r=rbmod q∈ S^e. Let the actual reduced Jacobian blocks be J=(D_YF)(x)bmod q:S^m→ S^e, K=(D_TF)(x)bmod q:S^k→ S^e. For a prescribed parameter digit v∈ S^k, choose its canonical integral representative widehat v. Define L_v to be the set of actual residue vectors z∈( Z/q^L+1 Z)^m+k satisfying z≡ xpmodq^L, F(z…

    Exact mathematical passage
    
    J=(D_YF)(x)\bmod q:S^m\to S^e,
    \qquad K=(D_TF)(x)\bmod q:S^k\to S^e.
    
S_0Use in source · 3 passages

Search aliases: S_0

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  • Use in source · Read passage · 72_recovered.tex, line 1313

    [The first odd reconstruction] For p=3 and n=2, S_2(3)=1-9+25=17. Since Z_1=1, Z_3=2, and S_0(3)=1, the odd coordinate formula gives E_2 =17-2·9 =-1. The large source value and the depth one correction are both necessary.

    Exact mathematical passage
    S_0(3)=1
S_1Use in source · 1 passage

Search aliases: S_1

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  • Use in source · Read passage · 72_recovered.tex, line 1277

    [Exact change between local source charts] Editorial range correction. This comparison holds for the even source subsequences, equivalently the evenized generating functions H_p. It does not identify their full odd-degree source values. For p=3, the original odd source has A_3,1=S_1(3)=3, whereas every odd dyadic source coefficient is zero; the even kernel L_3K_2 cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity H_p=L_pK_ℓ H_ℓ…

    Exact mathematical passage
    A_{3,1}=S_1(3)=3
s_1Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2018

    [Nested discrete refinement] For positive integers M,N, (a_r^[M])_s^[N] = a_r+Ms^[MN]. For a mixed radix path (M_1,…,M_ℓ) with digits 0≤ s_i<M_i, the final address is r( s) = s_1+M_1s_2+⋯+ (∏_i=1^ℓ-1M_i)s_ℓ. Only the product radix and accumulated address determine the final branch.

    Exact mathematical passage
    
    \boxed{
    r(\boldsymbol s)
    =
    s_1+M_1s_2+\cdots+
    \left(\prod_{i=1}^{\ell-1}M_i\right)s_\ell.
    }
    
S_2Use in source · 2 passages

Search aliases: S_2

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  • Use in source · Read passage · 72_recovered.tex, line 1310

    [The first odd reconstruction] For p=3 and n=2, S_2(3)=1-9+25=17. Since Z_1=1, Z_3=2, and S_0(3)=1, the odd coordinate formula gives E_2 =17-2·9 =-1. The large source value and the depth one correction are both necessary.

    Exact mathematical passage
    
    S_2(3)=1-9+25=17.
    
s_2Use in source · 20 passages

Search aliases: s_2

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  • Use in source · Read passage · 72_recovered.tex, line 2018

    [Nested discrete refinement] For positive integers M,N, (a_r^[M])_s^[N] = a_r+Ms^[MN]. For a mixed radix path (M_1,…,M_ℓ) with digits 0≤ s_i<M_i, the final address is r( s) = s_1+M_1s_2+⋯+ (∏_i=1^ℓ-1M_i)s_ℓ. Only the product radix and accumulated address determine the final branch.

    Exact mathematical passage
    
    \boxed{
    r(\boldsymbol s)
    =
    s_1+M_1s_2+\cdots+
    \left(\prod_{i=1}^{\ell-1}M_i\right)s_\ell.
    }
    
S_{2(r-m)}Use in source · 2 passages

Search aliases: S_2(r-m)

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  • Use in source · Read passage · supplement_02.tex, line 125

    ing3. Exact reconstruction and the triangular matrixTheorem 4.2; manuscript page 11 For odd p, rescaling the already proved inverse law gives sech(pt)cosh(pt)=1. Multiplying the even-source identity by sech^2(pt) therefore proves sech^2(pt)(∑_r≥0S_2r(p)fract^2r(2r)!)=sech t. Actual EGF multiplication now gives the integer coefficient formula E_2r=∑_m=0^rbinom2r2mb_m p^2mS_2(r-m)(p). The formal proof first extracts the sum over all indices and then proves that the odd summands vanish. A checked e…

    Exact mathematical passage
    
     \boxed{E_{2r}=\sum_{m=0}^{r}\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).}
    
S_{2n}Use in source · 3 passages

Search aliases: S_2n

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  • Use in source · Read passage · 20_arithmetic.tex, line 822

    The exponential generating series of the source is the finite sum S_p(t)=∑_a=0^p-1(-1)^a e^(2a+1)t. Multiplication by cosh t telescopes. Since p is odd, S_p(t)=frac1+e^2pt2secht. Taking its even part gives ∑_n≥0S_2n(p)fract^2n(2n)! =cosh^2(pt)secht. Multiplication by the reciprocal sech^2(pt) yields secht. The EGF product rule now gives the reconstruction coefficient by coefficient. Replacing m by n-q produces the displayed lower triangular row. Its diagonal and weights follow from γ_0=1 and the…

    Exact mathematical passage
    
     \sum_{n\ge0}S_{2n}(p)\frac{t^{2n}}{(2n)!}
           =\cosh^2(pt)\operatorname{sech}t.
    
S_{2q}Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 803

    [Classical odd-length reconstruction] For every positive odd integer p and every n∈ N, E_2n=∑_m=0^nbinom2n2mγ_m p^2mS_2(n-m)(p). Equivalently, E_2n=∑_q=0^n V_nqS_2q(p) for the lower triangular matrix V_nq=binom2n2qγ_n-qp^2(n-q) (q≤ n), V_nq=0 (q>n). It has V_nn=1 and p^2(n-q)mid V_nq when q≤ n.

    Exact mathematical passage
    E_{2n}=\sum_{q=0}^n V_{nq}S_{2q}(p)
S_{2r}Use in source · 2 passages

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  • Use in source · Read passage · supplement_02.tex, line 64

    Even projection is identified coefficientwise. For any formal series f, let EvenPart(f)=(f(t)+f(-t))/2. Its nth coefficient is proved to be the coefficient of f when n is even and zero otherwise. Rescaling the inverse law proves sech(-t)=sech(t). The exponential multiplication identity then gives cosh^2(pt)=frac2+e^2pt+e^-2pt4, and hence ∑_r≥0S_2r(p)fract^2r(2r)! =cosh^2(pt) sech t. Both the full and even EGF formulas are derived from the original integer sum. Neither is supplied as a hypothesis…

    Exact mathematical passage
    
     \boxed{\sum_{r\ge0}S_{2r}(p)\frac{t^{2r}}{(2r)!}
           =\cosh^2(pt)\,\sech t.}
    
S_4Use in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 181

    The displayed manuscript row is kernel-checked. The scalar anchors are derived from the same general coefficient recurrence: (E_0,E_2,E_4,E_6,E_8)=(1,-1,5,-61,1385), (b_0,b_1,b_2,b_3,b_4)=(1,-2,16,-272,7936). The literal source sums at p=5 give (S_0,S_2,S_4,S_6,S_8)=(1,49,4705,428689,37665985). Consequently, T_2(5,8)=2 731 001 385, E_8-T_2(5,8)=-2 731 000 000. The source proves both 5^6mid E_8-T_2(5,8), 5^7nmid E_8-T_2(5,8). Thus the guaranteed power is attained in this example. No stronger unif…

    Exact mathematical passage
    
     (S_0,S_2,S_4,S_6,S_8)=(1,49,4705,428689,37665985).
    
S_6Use in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 181

    The displayed manuscript row is kernel-checked. The scalar anchors are derived from the same general coefficient recurrence: (E_0,E_2,E_4,E_6,E_8)=(1,-1,5,-61,1385), (b_0,b_1,b_2,b_3,b_4)=(1,-2,16,-272,7936). The literal source sums at p=5 give (S_0,S_2,S_4,S_6,S_8)=(1,49,4705,428689,37665985). Consequently, T_2(5,8)=2 731 001 385, E_8-T_2(5,8)=-2 731 000 000. The source proves both 5^6mid E_8-T_2(5,8), 5^7nmid E_8-T_2(5,8). Thus the guaranteed power is attained in this example. No stronger unif…

    Exact mathematical passage
    
     (S_0,S_2,S_4,S_6,S_8)=(1,49,4705,428689,37665985).
    
S_8Use in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 181

    The displayed manuscript row is kernel-checked. The scalar anchors are derived from the same general coefficient recurrence: (E_0,E_2,E_4,E_6,E_8)=(1,-1,5,-61,1385), (b_0,b_1,b_2,b_3,b_4)=(1,-2,16,-272,7936). The literal source sums at p=5 give (S_0,S_2,S_4,S_6,S_8)=(1,49,4705,428689,37665985). Consequently, T_2(5,8)=2 731 001 385, E_8-T_2(5,8)=-2 731 000 000. The source proves both 5^6mid E_8-T_2(5,8), 5^7nmid E_8-T_2(5,8). Thus the guaranteed power is attained in this example. No stronger unif…

    Exact mathematical passage
    
     (S_0,S_2,S_4,S_6,S_8)=(1,49,4705,428689,37665985).
    
s_iUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2017

    [Nested discrete refinement] For positive integers M,N, (a_r^[M])_s^[N] = a_r+Ms^[MN]. For a mixed radix path (M_1,…,M_ℓ) with digits 0≤ s_i<M_i, the final address is r( s) = s_1+M_1s_2+⋯+ (∏_i=1^ℓ-1M_i)s_ℓ. Only the product radix and accumulated address determine the final branch.

    Exact mathematical passage
    0\leq s_i<M_i
\mathsf S_MUse in source · 3 passages

Search aliases: S_M mathsf

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  • Use in source · Read passage · 72_recovered.tex, line 2063

    [Polyphase shift matrix] Let E be the scalar shift (Ea)_n=a_n+1, and let E_q shift the coarse index. Then P_M(Ea) = S_M(E_q) P_Ma, where S_M(E_q) = 0 · 1 · 0 · ⋯ · 0

    Exact mathematical passage
    
    \boxed{
    \mathcal P_M(Ea)
    =
    \mathsf S_M(E_q)\mathcal P_Ma,
    }
    
s_nUse in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 44

    First construct the integral solution. For a ring R, a triangular row system T, right-hand side b, and proved unit pivots d_n=T_nn, define s_n=d_n^-1(b_n-∑_q<nT_nqs_q). The new ring-level solver uses well-founded recursion. Its row identities and unique-solution property are proved. The source also proves compatibility of this solver with ring homomorphisms into fields.

    Exact mathematical passage
    
     s_n=d_n^{-1}\left(b_n-\sum_{q<n}T_{nq}s_q\right).
    
S_{n-2m}Use in source · 1 passage

Search aliases: S_n-2m

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  • Use in source · Read passage · supplement_02.tex, line 131

    Substituting the independently proved signed-zigzag identity recovers the manuscript's notation. For every even degree n, E_n=∑_m=0^n/2(-1)^m Z_2m+1binom n2mp^2mS_n-2m(p). The Lean version uses n=2r to avoid ambiguous natural-number division. Every even natural degree has that form. Primality is not used in this equality.

    Exact mathematical passage
    
     \boxed{E_n=\sum_{m=0}^{n/2}(-1)^m Z_{2m+1}\binom n{2m}p^{2m}S_{n-2m}(p).}
    
S_{n-2r}Use in source · 3 passages

Search aliases: S_n-2r

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  • Use in source · Read passage · 72_recovered.tex, line 965

    [Odd modulus Euler coordinate pair] For every positive odd q and every even n≥0, E_n = ∑_r=0^n/2 (-1)^rZ_2r+1 binom n2r q^2rS_n-2r(q). The inverse transform is S_n(q) = E_n + ∑_r=1^n/2 2^2r-1 binom n2r q^2rE_n-2r. Both transforms are lower triangular with diagonal one.

    Exact mathematical passage
    
    \boxed{
    E_n
    =
    \sum_{r=0}^{n/2}
    (-1)^rZ_{2r+1}
    \binom n{2r}
    q^{2r}S_{n-2r}(q).
    }
    
S_{n-2r}^\timesUse in source · 1 passage

Search aliases: S_n-2r^× times

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  • Use in source · Read passage · 72_recovered.tex, line 1825

    [Depleted odd prime reconstruction] For every even n, (1-σ_pp^n)E_n = ∑_r=0^n/2 (-1)^rZ_2r+1 binom n2rp^2rS_n-2r^×(p).

    Exact mathematical passage
    
    \boxed{
    \left(1-\sigma_pp^n\right)E_n
    =
    \sum_{r=0}^{n/2}
    (-1)^rZ_{2r+1}
    \binom n{2r}p^{2r}S_{n-2r}^\times(p).
    }
    
S_{n-r}Use in source · 6 passages

Search aliases: S_n-r

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  • Use in source · Read passage · 10_foundations.tex, line 215

    [Filtered finite reconstruction] For every n,L∈ N and sequences c,S in that ring, · ∑_r=0^nbinom nr c_rp^[w(r)]S_n-r · ≡ ∑_substack0≤ r≤ n w(r)<L binom nr c_rp^[w(r)]S_n-rpmodp^L. Here congruence means divisibility of the difference by p^L in the coefficient ring. In Z_p, it means membership in p^L Z_p.

    Exact mathematical passage
    
    \begin{split}
    &\sum_{r=0}^{n}\binom nr c_rp^{[w(r)]}S_{n-r}\\
    &\hspace{8mm}\equiv
    \sum_{\substack{0\leq r\leq n\\w(r)<L}}
    \binom nr c_rp^{[w(r)]}S_{n-r}\pmod{p^L}.
    \end{split}
    
S_n^\timesUse in source · 1 passage

Search aliases: S_n^× times

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  • Use in source · Read passage · 72_recovered.tex, line 1819

    For an odd prime p, the source S_n(p) contains one nonunit base. Put σ_p=(-1)^(p-1)/2 and S_n^×(p)=S_n(p)-σ_pp^n.

    Exact mathematical passage
    
    S_n^\times(p)=S_n(p)-\sigma_pp^n.
    
S_pUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 15

    Unless a different coefficient ring is specified, a polynomial system is a finite list of integer polynomials. All divisibility conditions on vectors are coordinatewise. For a positive integer q, write S_q= Z/q Z and use an overline for reduction modulo q. For a prime p, write F_p=S_p, normalize |p|_p=p^-1, and put B_L(x)=x+p^L Z_p^d. An integer center is included in this notation by its natural image in Z_p^d. All dimensions below are finite and may be zero. A statement over S_q allows composit…

    Exact mathematical passage
    \F_p=S_p
\mathcal S_pUse in source · 2 passages

Search aliases: S_p mathcal

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  • Use in source · Read passage · 20_arithmetic.tex, line 814

    The exponential generating series of the source is the finite sum S_p(t)=∑_a=0^p-1(-1)^a e^(2a+1)t. Multiplication by cosh t telescopes. Since p is odd, S_p(t)=frac1+e^2pt2secht. Taking its even part gives ∑_n≥0S_2n(p)fract^2n(2n)! =cosh^2(pt)secht. Multiplication by the reciprocal sech^2(pt) yields secht. The EGF product rule now gives the reconstruction coefficient by coefficient. Replacing m by n-q produces the displayed lower triangular row. Its diagonal and weights follow from γ_0=1 and the…

    Exact mathematical passage
    
     \mathcal S_p(t)=\sum_{a=0}^{p-1}(-1)^a e^{(2a+1)t}.
    
s_pUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1039

    [Odd row weight] For an odd prime p, the rth nonzero term in the reconstruction has valuation W_p(n,r) = 2r+v_p(Z_2r+1) +v_pbinom n2r +v_p(S_n-2r(p)). The binomial contribution is v_pbinom n2r = fracs_p(2r)+s_p(n-2r)-s_p(n)p-1, where s_p is the base p digit sum.

    Exact mathematical passage
    
    \boxed{
    \vp\binom n{2r}
    =
    \frac{s_p(2r)+s_p(n-2r)-s_p(n)}{p-1},
    }
    
S_p^eUse in source · 1 passage

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  • Use in source · Read passage · continuation_09.tex, line 123

    For every proposed next parameter digit, the previously checked obstruction is [bar r+K_av]∈ S_p^e/imJ_a. The fixed solved Jacobian is surjective, so this class vanishes for every digit at every stage. The actual lift fiber is a torsor for ker J_a. Rank-nullity gives its cardinality p^m-e.

    Exact mathematical passage
    [\bar r+K_av]\in S_p^e/\operatorname{im}J_a
S_qUse in source · 4 passages

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  • Use in source · Read passage · 30_lifting.tex, line 13

    Unless a different coefficient ring is specified, a polynomial system is a finite list of integer polynomials. All divisibility conditions on vectors are coordinatewise. For a positive integer q, write S_q= Z/q Z and use an overline for reduction modulo q. For a prime p, write F_p=S_p, normalize |p|_p=p^-1, and put B_L(x)=x+p^L Z_p^d. An integer center is included in this notation by its natural image in Z_p^d. All dimensions below are finite and may be zero. A statement over S_q allows composit…

    Exact mathematical passage
    S_q=\Z/q\Z
s_qUse in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 44

    First construct the integral solution. For a ring R, a triangular row system T, right-hand side b, and proved unit pivots d_n=T_nn, define s_n=d_n^-1(b_n-∑_q<nT_nqs_q). The new ring-level solver uses well-founded recursion. Its row identities and unique-solution property are proved. The source also proves compatibility of this solver with ring homomorphisms into fields.

    Exact mathematical passage
    
     s_n=d_n^{-1}\left(b_n-\sum_{q<n}T_{nq}s_q\right).
    
S_q^dUse in source · 7 passages

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  • Use in source · Read passage · 30_lifting.tex, line 28

    For F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, define its evaluated Jacobian by the formal partial derivatives: DF(x)_ij=(∂_jF_i)(x), J_x=DF(x)bmod q:S_q^d⟶ S_q^e. The obstruction therefore uses the Jacobian of the given polynomials. The following algebraic Taylor identity relates this linear map to polynomial evaluation.

    Exact mathematical passage
    
     DF(x)_{ij}=(\partial_jF_i)(x),\qquad
     J_x=DF(x)\bmod q:S_q^d\longrightarrow S_q^e.
    
S_q^eUse in source · 10 passages

Search aliases: S_q^e

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  • Use in source · Read passage · 30_lifting.tex, line 28

    For F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, define its evaluated Jacobian by the formal partial derivatives: DF(x)_ij=(∂_jF_i)(x), J_x=DF(x)bmod q:S_q^d⟶ S_q^e. The obstruction therefore uses the Jacobian of the given polynomials. The following algebraic Taylor identity relates this linear map to polynomial evaluation.

    Exact mathematical passage
    
     DF(x)_{ij}=(\partial_jF_i)(x),\qquad
     J_x=DF(x)\bmod q:S_q^d\longrightarrow S_q^e.
    
S_q^fUse in source · 1 passage

Search aliases: S_q^f

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  • Use in source · Read passage · 30_lifting.tex, line 216

    [Simultaneous polynomial lifting] Let F∈ Z[X_1,…,X_d]^e and G∈ Z[X_1,…,X_d]^f have a common lower root x modulo q^L, where q≥1 and L≥1. Use the integer-domain reduced Jacobians J: Z^d→ S_q^e and K: Z^d→ S_q^f. Suppose u,v∈ Z^d are separately valid corrections for F and G, respectively. Then [u]=[v] in Z^d/(ker J+ker K) if and only if a single y≡ xpmodq^L solves both systems modulo q^L+1. Each separate label is independent of the chosen valid correction. The common quotient is finite.

    Exact mathematical passage
    K:\Z^d\to S_q^f
S_q^kUse in source · 3 passages

Search aliases: S_q^k

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  • Use in source · Read passage · 30_lifting.tex, line 539

    Suppose now that the split into solved variables and parameter variables is fixed in advance. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0), and assume F(x)=q^Lr_x with q≥1 and L≥1. Write the reduced Jacobian as DF(x)bmod q=(J_x K_x), J_x:S_q^m→ S_q^e, K_x:S_q^k→ S_q^e. For a prescribed parameter digit v∈ S_q^k, let L_x,L,v be the set of next residue roots over x whose parameter coordinates are t_0+q^Lwidetilde v modulo q^L+1.

    Exact mathematical passage
    
     DF(x)\bmod q=(J_x\ K_x),\qquad
     J_x:S_q^m\to S_q^e,\quad K_x:S_q^k\to S_q^e.
    
S_q^mUse in source · 4 passages

Search aliases: S_q^m

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  • Use in source · Read passage · 30_lifting.tex, line 539

    Suppose now that the split into solved variables and parameter variables is fixed in advance. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0), and assume F(x)=q^Lr_x with q≥1 and L≥1. Write the reduced Jacobian as DF(x)bmod q=(J_x K_x), J_x:S_q^m→ S_q^e, K_x:S_q^k→ S_q^e. For a prescribed parameter digit v∈ S_q^k, let L_x,L,v be the set of next residue roots over x whose parameter coordinates are t_0+q^Lwidetilde v modulo q^L+1.

    Exact mathematical passage
    
     DF(x)\bmod q=(J_x\ K_x),\qquad
     J_x:S_q^m\to S_q^e,\quad K_x:S_q^k\to S_q^e.
    
s_\ellUse in source · 1 passage

Search aliases: s_ℓ ell

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  • Use in source · Read passage · 72_recovered.tex, line 2018

    [Nested discrete refinement] For positive integers M,N, (a_r^[M])_s^[N] = a_r+Ms^[MN]. For a mixed radix path (M_1,…,M_ℓ) with digits 0≤ s_i<M_i, the final address is r( s) = s_1+M_1s_2+⋯+ (∏_i=1^ℓ-1M_i)s_ℓ. Only the product radix and accumulated address determine the final branch.

    Exact mathematical passage
    
    \boxed{
    r(\boldsymbol s)
    =
    s_1+M_1s_2+\cdots+
    \left(\prod_{i=1}^{\ell-1}M_i\right)s_\ell.
    }
    
\secUse in source · 7 passages

Search aliases: sec

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  • Use in source · Read passage · 20_arithmetic.tex, line 782

    The same inverse coefficients can be written in the signed zigzag notation of the original manuscript. Define Z_r by sec t+tan t=∑_r≥0Z_rfract^rr!. Then γ_m=(-1)^m Z_2m+1 (m∈ N). Indeed, the odd coefficients come from tan t, whose derivative is sec^2t. Changing the sign of the squared variable in the even series for sech^2t gives sec^2t, so its degree-2m Hurwitz coefficient is (-1)^mγ_m. Differentiation identifies this coefficient with Z_2m+1, proving the formula. The Lean bridge is leanKomorebi…

    Exact mathematical passage
    
     \sec t+\tan t=\sum_{r\ge0}Z_r\frac{t^r}{r!}.
    
\sec^2Use in source · 3 passages

Search aliases: sec^2 sec

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  • Use in source · Read passage · 20_arithmetic.tex, line 790

    The same inverse coefficients can be written in the signed zigzag notation of the original manuscript. Define Z_r by sec t+tan t=∑_r≥0Z_rfract^rr!. Then γ_m=(-1)^m Z_2m+1 (m∈ N). Indeed, the odd coefficients come from tan t, whose derivative is sec^2t. Changing the sign of the squared variable in the even series for sech^2t gives sec^2t, so its degree-2m Hurwitz coefficient is (-1)^mγ_m. Differentiation identifies this coefficient with Z_2m+1, proving the formula. The Lean bridge is leanKomorebi…

    Exact mathematical passage
    \sec^2t
\sech^2Use in source · 3 passages

Search aliases: sech^2 sech operatorname

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  • Use in source · Read passage · supplement_02.tex, line 90

    The inverse coefficients are integral by construction. Define the integer binomial convolution B_n=∑_r=0^nbinom nr E_rE_n-r, b_m=B_2m. The product formula for actual EGFs gives ∑_n≥0B_nfract^nn!=sech^2 t. Odd coefficients vanish by parity. In particular, every b_m is an integer; this is the integrality required by the later congruence.

    Exact mathematical passage
    
     \sum_{n\ge0}B_n\frac{t^n}{n!}=\sech^2 t.
    
\sinUse in source · 6 passages

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  • Use in source · Read passage · supplement_02.tex, line 97

    Connecting to the paper's Z_n convention. The paper defines Z_n through sec t+tan t. The Lean construction uses Mathlib's actual formal sine and cosine series and defines sec t=(cos t)^-1, tan t=(sin t)sec t, Z_n=n![t^n](sec t+tan t). It proves the sine and cosine derivative identities, sin^2t+cos^2t=1, and (tan t)'=sec^2t. Sign substitution in the squared variable carries the hyperbolic cosine kernel to the actual cosine kernel. The same homomorphism carries their inverses and their squares. Ex…

    Exact mathematical passage
    
     \sec t=(\cos t)^{-1},\quad \tan t=(\sin t)\sec t,
     \quad Z_n=n![t^n](\sec t+\tan t).
    
\sin^2Use in source · 1 passage

Search aliases: sin^2 sin

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  • Use in source · Read passage · supplement_02.tex, line 101

    Connecting to the paper's Z_n convention. The paper defines Z_n through sec t+tan t. The Lean construction uses Mathlib's actual formal sine and cosine series and defines sec t=(cos t)^-1, tan t=(sin t)sec t, Z_n=n![t^n](sec t+tan t). It proves the sine and cosine derivative identities, sin^2t+cos^2t=1, and (tan t)'=sec^2t. Sign substitution in the squared variable carries the hyperbolic cosine kernel to the actual cosine kernel. The same homomorphism carries their inverses and their squares. Ex…

    Exact mathematical passage
    \sin^2t+\cos^2t=1
\sinh_aUse in source · 1 passage

Search aliases: sinh_a sinh

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  • Use in source · Read passage · supplement_03.tex, line 73

    . targetDyadic.kernel_coefficient_integral, Dyadic.beta_integral targetDyadic.source_valuation, Dyadic.sourcePadic_isUnit pageheading2. Connecting the definitions to actual formal seriesGenerating-function foundations for Theorem 5.3 Work in Q[[t]]. The formal exponential and rescaling homomorphism are the same checked objects used in the second block. Define sinh_a(t)=frace^at-e^-at2, cosh_a(t)=frace^at+e^-at2. Let B(t)=K(t^2)^-1=∑_m≥0β_mt^2m and let F(t)=∑_r≥0R_2rfract^2r(2r)!. These series ar…

    Exact mathematical passage
    
     \sinh_a(t)=\frac{e^{at}-e^{-at}}2,\qquad
     \cosh_a(t)=\frac{e^{at}+e^{-at}}2.
    
\operatorname{Sol}Use in source · 2 passages

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  • Use in source · Read passage · 71_native.tex, line 210

    N=∏_p∈ Sp^L_p, Sol_F(R)={x∈ R^d:F(x)=0}.

    Exact mathematical passage
    N=\prod_{p\in S}p^{L_p},\qquad\operatorname{Sol}_F(R)=\{x\in R^d:F(x)=0\}.
\operatorname{Sol}_FUse in source · 2 passages

Search aliases: Sol_F operatorname

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  • Use in source · Read passage · 71_native.tex, line 210

    N=∏_p∈ Sp^L_p, Sol_F(R)={x∈ R^d:F(x)=0}.

    Exact mathematical passage
    N=\prod_{p\in S}p^{L_p},\qquad\operatorname{Sol}_F(R)=\{x\in R^d:F(x)=0\}.
\sqcupUse in source · 3 passages

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  • Use in source · Read passage · 71_native.tex, line 249

    for maps from A and B to a common set if and only if R is rectangular. In that case a canonical finite choice of common set is the connected-component set C_R of the bipartite graph with vertex set Asqcup B and edges given by R.

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    A\sqcup B
\squareUse in source · 3 passages

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  • Use in source · Read passage · 71_native.tex, line 62

    m x'=x+p^Ls. Since L≥1, its Jacobian is the same modulo p. Dividing (1) by p^L shows that its residual digit is e'=e+Js, so it gives exactly the same quotient class. If t_0 solves (2), all its solutions are t_0+ker J, and no others. The finite cardinalities follow from rank-nullity. This also covers no variables, no equations, zero Jacobian and singular solutions. square

    Exact mathematical passage
    \square
\starhUse in source · 4 passages

Search aliases: starh mathbin⋆ mathbin star

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  • Use in source · Read passage · 72_recovered.tex, line 34

    This article begins with the Hurwitz product (amathbin⋆ b)_n = ∑_r=0^n binom nr a_rb_n-r. Over a field of characteristic zero it is multiplication of exponential generating functions. Over a finite field it realizes the additive group algebra through the exponential modes ϵ_u=(u^n)_n≥0. This finite algebra gives an exact proof of the standard window period. Hurwitz series rings are treated in [source reference]. The finite difference and umbral background may be found in [source reference].

    Exact mathematical passage
    
    (a\starh b)_n
    =
    \sum_{r=0}^{n}
    \binom nr a_rb_{n-r}.
    
\sTwoUse in source · 3 passages

Search aliases: sTwo s_2

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  • Use in source · Read passage · 72_recovered.tex, line 1091

    [Dyadic inverse coefficients] For r≥1, β_r = - frac2^2r(2r-1)B_2r(2r)!. Consequently, v_2(β_r)=s_2(r)-1. Every R_n is a dyadic unit.

    Exact mathematical passage
    
    \boxed{
    \vTwo(\beta_r)=\sTwo(r)-1.
    }
    
\mathcal TUse in source · 10 passages

Search aliases: T mathcal

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  • Use in source · Read passage · 30_lifting.tex, line 277

    At positive polynomial depth, restrict it to the correction fiber T_x={t∈ S_q^d:overliner_x+J_xt=0} and the lift set L_x,L= {y∈( Z/q^L+1 Z)^d:y≡ xpmodq^L, F(y)=0}. The polynomial predicate in the second display is evaluation in the residue ring. Evaluation on canonical integer representatives gives the same predicate, because integer polynomial evaluation respects congruence.

    Exact mathematical passage
    
     \mathcal T_x=\{t\in S_q^d:\overline{r_x}+J_xt=0\}
    
t'Use in source · 2 passages

Search aliases: t'

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  • Use in source · Read passage · continuation_09.tex, line 87

    At depth N≥1, suppose F(a)=q^Nr and a desired next integer parameter target satisfies t'-free(a)=q^Nw, with w∈ Z^k. Surjectivity supplies a solved digit u∈ S_q^m satisfying J_a u=-bar r-K_abar w. Choose an integer representative widetilde u, and set a'=a+q^N(widetilde u,w). The parameter quotient is the supplied integer w. Therefore free(a')=t' exactly. The checked polynomial Taylor identity gives F(a')≡ F(a)+q^NDF(a)(widetilde u,w) ≡0pmodq^N+1. Its quadratic remainder is divisible by q^2N, and …

    Exact mathematical passage
    t'-\operatorname{free}(a)=q^Nw
t^{-c}Use in source · 2 passages

Search aliases: t^-c

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  • Use in source · Read passage · supplement_10.tex, line 113

    Application to the actual singular ratios. The preceding v0.9.0 source independently constructs H_N,c(t)=∑_n≥0fract^Nn+c(Nn+c)!, A_c(t)=t^-cH_N,c(t), R_c=fracA_c-1A_c. The notation t^-cH_N,c means removing its proved initial zero coefficients. For each c=ap, 1≤ a<m, that source proves R_ap(log(1+T))∈ W_p. The new transfer consequently proves flatness of the same rational ratio after scalar extension to Q_p.

    Exact mathematical passage
    
     H_{N,c}(t)=\sum_{n\ge0}\frac{t^{Nn+c}}{(Nn+c)!},\quad
     A_c(t)=t^{-c}H_{N,c}(t),\quad R_c=\frac{A_{c-1}}{A_c}.
    
T^0Use in source · 2 passages

Search aliases: T^0

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  • Use in source · Read passage · 20_arithmetic.tex, line 285

    ^d=(p^lfloor d/prfloora_d) (T^p/p)^lfloor d/prfloorT^dbmod p. The operator preserves these monomials and hence finite weighted polynomials. Each fixed output coefficient depends on only finitely many input coefficients, which extends the result to the formal series. Substitution T=e^t-1 intertwines D with Θ and maps W_p into H_p, because d![t^d]F(e^t-1)=[T^0]Θ^dF. It follows that the singular ratios are flat. This transfer is proved in leanKomorebiPaperIX.WeightedDworkTransfer.singular_ratio_fla…

    Exact mathematical passage
    
     d![t^d]F(e^t-1)=[T^0]\Theta^dF.
    
t^{2N}Use in source · 1 passage

Search aliases: t^2N

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  • Use in source · Read passage · root_scope.tex, line 51

    m≥1, differentiation of zcoth z gives β_m=-frac2^2m(2m-1)B_2m(2m)!, κ_m=-2^4m(2m-1)B_2m. The classical dyadic Bernoulli denominator law v_2(B_2m)=-1 and Legendre's formula yield v_2(β_m)=s_2(m)-1, v_2(κ_m)=4m-1. The source identity ∑_N≥0A(N)fract^2N(2N)! =fracsinh(2t)sinh t2t^2 implies, after multiplication by 4t^2/sinh^2(2t), E_2N=∑_m=0^N κ_mbinom2N2mA(N-m). For a retained depth M≥0, put T_M(N)= E_2N -∑_m=0^min(M,N)κ_mbinom2N2mA(N-m). An empty tail, N≤ M, is exactly zero.

    Exact mathematical passage
    
     \sum_{N\ge0}A(N)\frac{t^{2N}}{(2N)!}
     =\frac{\sinh(2t)\sinh t}{2t^2}
    
t^{2n}Use in source · 7 passages

Search aliases: t^2n

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  • Use in source · Read passage · 20_arithmetic.tex, line 749

    The classical signed Euler numbers have generating function secht=∑_n≥0E_2nt^2n/(2n)!. They are the N=2, j=0 member of [source reference]. The coordinate theorem reconstructs them from finite alternating sums of odd powers.

    Exact mathematical passage
    \operatorname{sech}t=\sum_{n\ge0}E_{2n}t^{2n}/(2n)!
t^{2r}Use in source · 7 passages

Search aliases: t^2r

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  • Use in source · Read passage · 72_recovered.tex, line 948

    [Zigzag inverse kernel] sech^2t = ∑_r≥0 (-1)^rZ_2r+1 fract^2r(2r)!.

    Exact mathematical passage
    
    \boxed{
    \operatorname{sech}^2t
    =
    \sum_{r\geq0}
    (-1)^rZ_{2r+1}
    \frac{t^{2r}}{(2r)!}.
    }
    
t^3Use in source · 1 passage

Search aliases: t^3

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  • Use in source · Read passage · strengthening.tex, line 716

    The address-three source is G_4,3(t)=3(sinh t-sin t)/t^3. Addition identities give sinh t-sin t =2asinh(at)cosh(bt)+2bcosh(at)sinh(bt), cosh t-cos t=2sinh(at)sinh(bt). Consequently D_3(t):=G_4,3(t)F_2(t) =frac32(fraccoth(at)at +fraccoth(bt)bt). This expression is a power series: its two polar terms cancel because a^-2+b^-2=0. Its constant is one, only degrees divisible by four survive, and the Bernoulli expansion gives D_3(t)=∑_i≥0d_ifract^4i(4i)!, d_i=3(-frac14)^i frac2^4i+2B_4i+2(4i+2)(4i+1). …

    Exact mathematical passage
    G_{4,3}(t)=3(\sinh t-\sin t)/t^3
t^4Use in source · 1 passage

Search aliases: t^4

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  • Use in source · Read passage · strengthening.tex, line 918

    For the first contribution, independence gives L(γ_k(P+Q))=∑_a=0^kμ_aμ_k-a. The parity generating series of these convolution moments is (1+t+t^2+t^4+t^8+⋯)^2 =1+t^2+t^4+t^8+⋯ in F_2[[t]]. Its coefficient is one at every k=2^r≥4. Thus the normalized moment in [source reference] is odd and v_2(Δ_1^kC_2(0))=2k+v_2(k!)+k/2. Inserting U^4n=(1+4H_2)^n introduces only corrections with strictly higher valuation, by the filtered bound at orders k+s for s>0. The equality consequently holds at every start…

    Exact mathematical passage
    
     (1+t+t^2+t^4+t^8+\cdots)^2
            =1+t^2+t^4+t^8+\cdots\quad\text{in }\mathbb F_2[[t]].
    
t^{4i}Use in source · 3 passages

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  • Use in source · Read passage · strengthening.tex, line 452

    G_4,1(t)F_0(t) =frac12(fractanh(at)at +fractanh(bt)bt). For example, its numerator follows from sinh t+sin t =2bsinh(at)cosh(bt)+2acosh(at)sinh(bt), while cosh t+cos t=2cosh(at)cosh(bt). Only degrees divisible by four survive. Its actual Hurwitz packets are D_1(t)=∑_i≥0d_ifract^4i(4i)!, d_i=(-frac14)^i frac2^4i+2(2^4i+2-1)B_4i+2(4i+2)(4i+1). The constant is one. Since v_2(4i+2)=1, the elementary Bernoulli bound shows v_2(d_i)≥2i. Hurwitz inversion therefore gives D_1(t)^-1=∑_i≥0h_ifract^4i(4i)!,…

    Exact mathematical passage
    
     D_1(t)=\sum_{i\geq0}d_i\frac{t^{4i}}{(4i)!},\qquad
     d_i=\left(-\frac14\right)^i
     \frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.
     \label{eq:address-one-packets}
    
t^{4n}Use in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 81

    Pack g_j into degrees divisible by four and form its actual EGF. The formal coefficient identity is G_j(t)=∑_n≥0g_j(n)fract^4n(4n)! =j!∑_n≥0fract^4n(4n+j)!. Its inverse is therefore the original normalized packet EGF. The source Kummer property gives flatness of G_j. The proved reciprocal closure gives flatness of G_j^-1.

    Exact mathematical passage
    
    G_j(t)=\sum_{n\ge0}g_j(n)\frac{t^{4n}}{(4n)!}
           =j!\sum_{n\ge0}\frac{t^{4n}}{(4n+j)!}.
    
t^8Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 918

    For the first contribution, independence gives L(γ_k(P+Q))=∑_a=0^kμ_aμ_k-a. The parity generating series of these convolution moments is (1+t+t^2+t^4+t^8+⋯)^2 =1+t^2+t^4+t^8+⋯ in F_2[[t]]. Its coefficient is one at every k=2^r≥4. Thus the normalized moment in [source reference] is odd and v_2(Δ_1^kC_2(0))=2k+v_2(k!)+k/2. Inserting U^4n=(1+4H_2)^n introduces only corrections with strictly higher valuation, by the filtered bound at orders k+s for s>0. The equality consequently holds at every start…

    Exact mathematical passage
    
     (1+t+t^2+t^4+t^8+\cdots)^2
            =1+t^2+t^4+t^8+\cdots\quad\text{in }\mathbb F_2[[t]].
    
T^cUse in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 252

    To pass from the section to the singular ratio, factor out its initial zero: for c=ap, the section has the form H_c(log(1+T))=fracT^cc! V(T), V(0)=1, V∈ W_p. The weight of c! compensates for the coefficient shift by c. The weighted ring is closed under inversion of series with constant coefficient one and under L_T=log(1+T)(1+T)d/dT. With ℓ(T)=log(1+T)/T, interpreted by coefficient shifting, differentiation of the displayed factorization gives R_c(log(1+T))=c(1+T)ℓ(T)+(L_TV)V^-1. Since pℓ(T)∈ W_…

    Exact mathematical passage
    
     H_c(\log(1+T))=\frac{T^c}{c!}\,V(T),\qquad V(0)=1,
     \qquad V\in\mathcal W_p.
    
T^dUse in source · 7 passages

Search aliases: T^d

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  • Use in source · Read passage · 20_arithmetic.tex, line 231

    At the singular addresses c=ap, 1≤ a<m, use the weighted ring W_p={∑_d≥0a_dT^d: p^lfloor d/prfloora_d∈ Z_p}. The key estimate is R_ap(log(1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!)…

    Exact mathematical passage
    
     \mathcal W_p=\left\{\sum_{d\ge0}a_dT^d:
              p^{\lfloor d/p\rfloor}a_d\in\Z_p\right\}.
    
T^{d\bmod p}Use in source · 1 passage

Search aliases: T^dbmod p bmod

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  • Use in source · Read passage · 20_arithmetic.tex, line 276

    mate, use Θ=(1+T)d/dT. Both Θ and (Θ^p^2-Θ^p)/p preserve W_p. For the first operator, its coefficient formula is (d+1)a_d+1+da_d; whenever the floor weight rises, the factor d+1 provides the missing p. For the second operator, the proof uses its full Leibniz correction and the scalar congruence p^2mid b^p^2-b^p. Weighted monomials factor as a_dT^d=(p^lfloor d/prfloora_d) (T^p/p)^lfloor d/prfloorT^dbmod p. The operator preserves these monomials and hence finite weighted polynomials. Each fixed ou…

    Exact mathematical passage
    
     a_dT^d=(p^{\lfloor d/p\rfloor}a_d)
              (T^p/p)^{\lfloor d/p\rfloor}T^{d\bmod p}.
    
t^jUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 582

    For integers N≥1 and 0≤ j<N, define E_N,j(t) = ∑_n≥0 fract^Nn+j(Nn+j)!. The forced factor t^j may be removed without changing the residue address.

    Exact mathematical passage
    t^j
t^NUse in source · 4 passages

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  • Use in source · Read passage · supplement_01.tex, line 103

    The formalization does not stop at a recurrence. In the variable z=t^N, construct K_N,j(z)=∑_n≥0fracj!(Nn+j)!z^n, F_N,j(z)=∑_n≥0fracC_N,j(n)(Nn)!z^n. It proves the factorial coefficient identity, then F_N,jK_N,j=1 and therefore F_N,j=K_N,j^-1.

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    z=t^N
t^{Nn+c}Use in source · 3 passages

Search aliases: t^Nn+c

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  • Use in source · Read passage · 20_arithmetic.tex, line 216

    Set H_c(t)=∑_n≥0fract^Nn+c(Nn+c)!, A_c(t)=∑_n≥0fract^Nn(Nn+c)!, R_c(t)=fracA_c-1(t)A_c(t) (1≤ c<N). The denominator of this last quotient has nonzero constant coefficient. The section identities give R_cH_c=tH_c-1, R_c(0)=c, j! P_N,j= P_N,0∏_c=1^jR_c. Cancellation of consecutive source sections proves the product identity and ties the ratios directly to the family under study.

    Exact mathematical passage
    
     H_c(t)=\sum_{n\ge0}\frac{t^{Nn+c}}{(Nn+c)!},\qquad
     A_c(t)=\sum_{n\ge0}\frac{t^{Nn}}{(Nn+c)!},\qquad
     R_c(t)=\frac{A_{c-1}(t)}{A_c(t)}\quad(1\le c<N).
    
t^{Nn+j}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 576

    For integers N≥1 and 0≤ j<N, define E_N,j(t) = ∑_n≥0 fract^Nn+j(Nn+j)!. The forced factor t^j may be removed without changing the residue address.

    Exact mathematical passage
    
    \Ecal_{N,j}(t)
    =
    \sum_{n\geq0}
    \frac{t^{Nn+j}}{(Nn+j)!}.
    
t^{Nr}Use in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 150

    pageheading4. From these sections to the original Euler packetsActual reciprocal, actual rational sequence, every order and step multiple The factorial-ratio identity is proved coefficient by coefficient: ∑_d≥0g_j(d)fract^dd! =j!∑_r≥0fract^Nr(Nr+j)!. The right side is precisely the normalized source kernel from the first block, not a replacement family. The earlier rational identity for its reciprocal is transported to Q_p[[t]]. A separate encoding theorem identifies the reciprocal's coefficient…

    Exact mathematical passage
    
     \sum_{d\ge0}g_j(d)\frac{t^d}{d!}
     =j!\sum_{r\ge0}\frac{t^{Nr}}{(Nr+j)!}.
    
T^pUse in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 276

    mate, use Θ=(1+T)d/dT. Both Θ and (Θ^p^2-Θ^p)/p preserve W_p. For the first operator, its coefficient formula is (d+1)a_d+1+da_d; whenever the floor weight rises, the factor d+1 provides the missing p. For the second operator, the proof uses its full Leibniz correction and the scalar congruence p^2mid b^p^2-b^p. Weighted monomials factor as a_dT^d=(p^lfloor d/prfloora_d) (T^p/p)^lfloor d/prfloorT^dbmod p. The operator preserves these monomials and hence finite weighted polynomials. Each fixed ou…

    Exact mathematical passage
    
     a_dT^d=(p^{\lfloor d/p\rfloor}a_d)
              (T^p/p)^{\lfloor d/p\rfloor}T^{d\bmod p}.
    
T^rUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1798

    [Weighted convergence] If widehat F(T)=∑_r≥0c_rT^r and v_p(c_r)+rα⟶+∞, then the series converges whenever v_p(T)≥α.

    Exact mathematical passage
    
    \widehat F(T)=\sum_{r\geq0}c_rT^r
    
t^rUse in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 782

    The same inverse coefficients can be written in the signed zigzag notation of the original manuscript. Define Z_r by sec t+tan t=∑_r≥0Z_rfract^rr!. Then γ_m=(-1)^m Z_2m+1 (m∈ N). Indeed, the odd coefficients come from tan t, whose derivative is sec^2t. Changing the sign of the squared variable in the even series for sech^2t gives sec^2t, so its degree-2m Hurwitz coefficient is (-1)^mγ_m. Differentiation identifies this coefficient with Z_2m+1, proving the formula. The Lean bridge is leanKomorebi…

    Exact mathematical passage
    
     \sec t+\tan t=\sum_{r\ge0}Z_r\frac{t^r}{r!}.
    
T_0Use in source · 1 passage

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  • Use in source · Read passage · supplement_03.tex, line 178

    [3pt] 1 · -91019/45 · 4 [3pt] 2 · 398197/45 · 6 [3pt] 3 · -208601/15 · 8 [3pt] 4 · 1385 · 10 The checked nonzero error valuations are E_8-T_0(8)=frac5494445, v_2(E_8-T_0(8))=5, E_8-T_3(8)=frac22937615, v_2(E_8-T_3(8))=15. They exceed the uniform guarantees; these two examples do not establish a universally stronger exponent.

    Exact mathematical passage
    3pt]
    1 & $-91019/45$ & 4\\[3pt]
    2 & $398197/45$ & 6\\[3pt]
    3 & $-208601/15$ & 8\\[3pt]
    4 & $1385$ & 10\\\bottomrule
    \end{tabular}
    \end{center}
    The checked nonzero error valuations are
    \[
     E_8-T_0(8)=\frac{54944}{45},\qquad v_2(E_8-T_0(8))=5,
    
T_2Use in source · 2 passages

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  • Use in source · Read passage · supplement_02.tex, line 185

    The displayed manuscript row is kernel-checked. The scalar anchors are derived from the same general coefficient recurrence: (E_0,E_2,E_4,E_6,E_8)=(1,-1,5,-61,1385), (b_0,b_1,b_2,b_3,b_4)=(1,-2,16,-272,7936). The literal source sums at p=5 give (S_0,S_2,S_4,S_6,S_8)=(1,49,4705,428689,37665985). Consequently, T_2(5,8)=2 731 001 385, E_8-T_2(5,8)=-2 731 000 000. The source proves both 5^6mid E_8-T_2(5,8), 5^7nmid E_8-T_2(5,8). Thus the guaranteed power is attained in this example. No stronger unif…

    Exact mathematical passage
    
     T_2(5,8)=2\,731\,001\,385,\qquad
     E_8-T_2(5,8)=-2\,731\,000\,000.
    
T_{2,M}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1396

    [Optimal dyadic truncation] Let T_2,M(n) = ∑_r=0^min(M,n/2) κ_2,rbinom n2rR_n-2r. Then E_n≡ T_2,M(n)pmod 2^4M+3. More precisely, v_2(E_n-T_2,M(n)) ≥ min_M+1≤ r≤ n/2 ( 4r-1+v_2binom n2r ). The uniform exponent 4M+3 is best possible.

    Exact mathematical passage
    
    T_{2,M}(n)
    =
    \sum_{r=0}^{\min(M,n/2)}
    \kappa_{2,r}\binom n{2r}R_{n-2r}.
    
T_3Use in source · 1 passage

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  • Use in source · Read passage · supplement_03.tex, line 189

    bottomrule The checked nonzero error valuations are E_8-T_0(8)=frac5494445, v_2(E_8-T_0(8))=5, E_8-T_3(8)=frac22937615, v_2(E_8-T_3(8))=15. They exceed the uniform guarantees; these two examples do not establish a universally stronger exponent.

    Exact mathematical passage
    
     E_8-T_3(8)=\frac{229376}{15},\qquad v_2(E_8-T_3(8))=15.
    
t_4Use in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 277

    For example, j=5 has (t_0,…,t_4)=(6,3,4,3,6). The exact root counts at distances 1,2,3,4,5,≥6 are 0,0,2,3,3,5. The older sufficient stages 3,5,7 are not the sharp emergence stages, which are 3,4,6. The first irregular label, j=17, has V≡1+Xpmod2; the regular-wall theorem does not apply. Complete coefficient certificates close distances 1 through 19, and the general separation theorem gives all seventeen simple roots for δ≥20. Those two kinds of evidence together resolve that family.

    Exact mathematical passage
    (t_0,\ldots,t_4)=(6,3,4,3,6)
T_HUse in source · 4 passages

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  • Use in source · Read passage · strengthening.tex, line 175

    [All-row prime-power transfer] Let p be prime, a≥1, N=p^a, and 0≤ j<N. The coefficients C_N,j(n) belong to Z_p for all n≥0. Set A_N,j(n)=(-1)^n C_N,j(n) and T_H=S^H-(-1)^H. For k≥1 and positive H_1,…,H_k, v_p(Δ_H_1⋯Δ_H_kA_N,j(n)) · ≥ k+∑_ν=1^k v_p(H_ν),

    Exact mathematical passage
    T_H=S^H-(-1)^H
T_{H_1}Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 176

    [All-row prime-power transfer] Let p be prime, a≥1, N=p^a, and 0≤ j<N. The coefficients C_N,j(n) belong to Z_p for all n≥0. Set A_N,j(n)=(-1)^n C_N,j(n) and T_H=S^H-(-1)^H. For k≥1 and positive H_1,…,H_k, v_p(Δ_H_1⋯Δ_H_kA_N,j(n)) · ≥ k+∑_ν=1^k v_p(H_ν), v_p(T_H_1⋯ T_H_kE_Nn^(N,j)) · ≥ v_p(j!)+k+∑_ν=1^k v_p(H_ν). At p=2, the ordinary differences of C_N,j and of the original coefficients satisfy the same bounds. These dyadic statements also hold for N=1.

    Exact mathematical passage
    
     v_p(\Delta_{H_1}\cdots\Delta_{H_k}A_{N,j}(n))
      &\geq k+\sum_{\nu=1}^k v_p(H_\nu),
     \label{eq:prime-power-signed-bound}\\
     v_p(T_{H_1}\cdots T_{H_k}E_{Nn}^{(N,j)})
      &\geq v_p(j!)+k+\sum_{\nu=1}^k v_p(H_\nu).
     \label{eq:prime-power-original-twist}
    
T_{H_k}Use in source · 2 passages

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  • Use in source · Read passage · strengthening.tex, line 176

    [All-row prime-power transfer] Let p be prime, a≥1, N=p^a, and 0≤ j<N. The coefficients C_N,j(n) belong to Z_p for all n≥0. Set A_N,j(n)=(-1)^n C_N,j(n) and T_H=S^H-(-1)^H. For k≥1 and positive H_1,…,H_k, v_p(Δ_H_1⋯Δ_H_kA_N,j(n)) · ≥ k+∑_ν=1^k v_p(H_ν), v_p(T_H_1⋯ T_H_kE_Nn^(N,j)) · ≥ v_p(j!)+k+∑_ν=1^k v_p(H_ν). At p=2, the ordinary differences of C_N,j and of the original coefficients satisfy the same bounds. These dyadic statements also hold for N=1.

    Exact mathematical passage
    
     v_p(\Delta_{H_1}\cdots\Delta_{H_k}A_{N,j}(n))
      &\geq k+\sum_{\nu=1}^k v_p(H_\nu),
     \label{eq:prime-power-signed-bound}\\
     v_p(T_{H_1}\cdots T_{H_k}E_{Nn}^{(N,j)})
      &\geq v_p(j!)+k+\sum_{\nu=1}^k v_p(H_\nu).
     \label{eq:prime-power-original-twist}
    
t_iUse in source · 8 passages

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  • Use in source · Read passage · continuation_01.tex, line 51

    For every P∈ Z[X_1,…,X_d], integer a, and x,t∈ Z^d, Lean proves a^2mid P(x+at)-P(x)-a∑_i=1^d(∂_iP)(x)t_i. The proof uses polynomial induction: constants, addition, and multiplication by a variable. The product rule for the actual partial derivatives controls the multiplication step. It does not assume a Taylor identity or a liftability condition.

    Exact mathematical passage
    
    \boxed{
     a^2\mid P(x+at)-P(x)-a\sum_{i=1}^{d}(\partial_iP)(x)t_i.
    }
    
t_LUse in source · 2 passages

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  • Use in source · worked_atlas.tex, line 68

    AtlasHeadA compatible root is built one digit at a timeatlas:smooth Take F(x)=x^2-7 over the 3-adic integers and start with x_1=1 modulo three. If F(x_L)=3^Lr_L, the next representative has the form x_L+1=x_L+3^Lt_L. Its correcting digit satisfies r_L+2x_Lt_L≡0pmod3. AtlasImagesmooth_tower For example, F(1)/3=-2 and 2x_1=2, so the next digit is one and x_2=4. Then F(4)/9=1, again giving digit one and x_3=13. At the next stage the residual is divisible by three, so digit zero is correct. A zero d…

    Exact mathematical passage
    x_{L+1}=x_L+3^Lt_L
\mathcal T_MUse in source · 2 passages

Search aliases: T_M mathcal

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  • Use in source · Read passage · root_scope.tex, line 62

    tion v_2(β_m)=s_2(m)-1, v_2(κ_m)=4m-1. The source identity ∑_N≥0A(N)fract^2N(2N)! =fracsinh(2t)sinh t2t^2 implies, after multiplication by 4t^2/sinh^2(2t), E_2N=∑_m=0^N κ_mbinom2N2mA(N-m). For a retained depth M≥0, put T_M(N)= E_2N -∑_m=0^min(M,N)κ_mbinom2N2mA(N-m). An empty tail, N≤ M, is exactly zero.

    Exact mathematical passage
    
     \mathcal T_M(N)=\mathcal E_{2N}
       -\sum_{m=0}^{\min(M,N)}\kappa_m\binom{2N}{2m}A(N-m).
    
T_{m,p}Use in source · 6 passages

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  • Use in source · Read passage · 72_recovered.tex, line 823

    [Fourier invariance of the standard window operator] Let p be odd, let mmid p-1, and let h=(p-1)/m. Define (T_m,pf)_n = ∑_q=0^nbinommnmqf_q in F_p. Let V_h be the vector space of sequences satisfying f_n+h=f_n (n≥1). Then T_m,p(V_h)=V_h. The restriction of T_m,p to V_h is an automorphism.

    Exact mathematical passage
    
    (T_{m,p}f)_n
    =
    \sum_{q=0}^{n}\binom{mn}{mq}f_q
    \quad\text{in }\FF_p.
    
t_nUse in source · 6 passages

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  • Use in source · Read passage · 30_lifting.tex, line 591

    [Every compatible integer parameter path] In the preceding notation, assume J_x is surjective. For every supplied integer path t_n∈ Z^k satisfying t_0=free(x), t_n+1-t_n∈ q^L+n Z^k, there is a compatible polynomial tower z_n∈ Z^m+k with z_0=x, free(z_n)=t_n, F(z_n)∈ q^L+n Z^e, z_n+1-z_n∈ q^L+n Z^m+k. The free-coordinate equality is exact in integers. Both reduced Jacobian blocks remain fixed, and z_b-z_a∈ q^L+a Z^m+k whenever a≤ b.

    Exact mathematical passage
    t_n\in\Z^k
t_{n+1}Use in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 592

    [Every compatible integer parameter path] In the preceding notation, assume J_x is surjective. For every supplied integer path t_n∈ Z^k satisfying t_0=free(x), t_n+1-t_n∈ q^L+n Z^k, there is a compatible polynomial tower z_n∈ Z^m+k with z_0=x, free(z_n)=t_n, F(z_n)∈ q^L+n Z^e, z_n+1-z_n∈ q^L+n Z^m+k. The free-coordinate equality is exact in integers. Both reduced Jacobian blocks remain fixed, and z_b-z_a∈ q^L+a Z^m+k whenever a≤ b.

    Exact mathematical passage
    
     t_0=\operatorname{free}(x),\qquad
     t_{n+1}-t_n\in q^{L+n}\Z^k,
    
T_{nn}Use in source · 6 passages

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  • Use in source · Read passage · 10_foundations.tex, line 180

    [Forward substitution and denominator clearing] Let T_nq be coefficients over a field with T_nn≠0 for every n. For any sequence b, define a_n=fracb_n-∑_q<nT_nqa_qT_nn. This constructs the unique solution of ∑_q=0^nT_nqa_q=b_n for all n. If T_nq and b_n are integers and the solution is taken over Q, then (∏_r=0^nT_rr)a_n∈ Z.

    Exact mathematical passage
    T_{nn}\ne0
T_{nq}Use in source · 7 passages

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  • Use in source · Read passage · 10_foundations.tex, line 180

    [Forward substitution and denominator clearing] Let T_nq be coefficients over a field with T_nn≠0 for every n. For any sequence b, define a_n=fracb_n-∑_q<nT_nqa_qT_nn. This constructs the unique solution of ∑_q=0^nT_nqa_q=b_n for all n. If T_nq and b_n are integers and the solution is taken over Q, then (∏_r=0^nT_rr)a_n∈ Z.

    Exact mathematical passage
    T_{nq}
\mathcal T_{r-1}Use in source · 2 passages

Search aliases: T_r-1 mathcal

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  • Use in source · Read passage · root_scope.tex, line 117

    Set r=M+1≥1 and define P_q(x)=∏_i=0^q-1(x-i)(2x-(2i+1)), P_0=1, H_r(x)=∑_q≥08^qfracβ_r+qβ_r P_q(x)mathscr A(x-q). The coefficient valuations tend to infinity. At any nonnegative integer x, terms with q>x vanish. Cancelling the factorials in [source reference] therefore gives the exact identity T_r-1(N)=κ_rbinom2N2rH_r(N-r) (N≥ r). This is a finite rational identity at every original Euler row.

    Exact mathematical passage
    
     \mathcal T_{r-1}(N)=\kappa_r\binom{2N}{2r}H_r(N-r)
     \qquad(N\ge r).\label{eq:root:factor}
    
T_{rr}Use in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 188

    em[Forward substitution and denominator clearing] Let T_nq be coefficients over a field with T_nn≠0 for every n. For any sequence b, define a_n=fracb_n-∑_q<nT_nqa_qT_nn. This constructs the unique solution of ∑_q=0^nT_nqa_q=b_n for all n. If T_nq and b_n are integers and the solution is taken over Q, then (∏_r=0^nT_rr)a_n∈ Z.

    Exact mathematical passage
    
    \left(\prod_{r=0}^{n}T_{rr}\right)a_n\in\Z.
    
\mathcal T_xUse in source · 6 passages

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  • Use in source · Read passage · 30_lifting.tex, line 277

    At positive polynomial depth, restrict it to the correction fiber T_x={t∈ S_q^d:overliner_x+J_xt=0} and the lift set L_x,L= {y∈( Z/q^L+1 Z)^d:y≡ xpmodq^L, F(y)=0}. The polynomial predicate in the second display is evaluation in the residue ring. Evaluation on canonical integer representatives gives the same predicate, because integer polynomial evaluation respects congruence.

    Exact mathematical passage
    
     \mathcal T_x=\{t\in S_q^d:\overline{r_x}+J_xt=0\}
    
\tanUse in source · 7 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 782

    The same inverse coefficients can be written in the signed zigzag notation of the original manuscript. Define Z_r by sec t+tan t=∑_r≥0Z_rfract^rr!. Then γ_m=(-1)^m Z_2m+1 (m∈ N). Indeed, the odd coefficients come from tan t, whose derivative is sec^2t. Changing the sign of the squared variable in the even series for sech^2t gives sec^2t, so its degree-2m Hurwitz coefficient is (-1)^mγ_m. Differentiation identifies this coefficient with Z_2m+1, proving the formula. The Lean bridge is leanKomorebi…

    Exact mathematical passage
    
     \sec t+\tan t=\sum_{r\ge0}Z_r\frac{t^r}{r!}.
    
\tanhUse in source · 3 passages

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\operatorname{tr}Use in source · 2 passages

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  • Use in source · Read passage · supplement_06.tex, line 121

    An independent Lucas sequence. Define L_0=2, L_1=1, L_n+2=L_n+L_n+1. Direct finite matrix arithmetic proves C^2=C+I. Multiplication by C^n gives C^n+2=C^n+C^n+1. Taking traces shows that tr(C^n) satisfies the same recurrence and initial values as L_n. Induction yields tr(C^N)=L_N. The implementation separately proves det C=-1, det(C^N)=(-1)^N, L_N=2F_N+1-F_N. No Binet roots or diagonalizability argument is needed.

    Exact mathematical passage
    \operatorname{tr}(C^n)
\mathcal UUse in source · 4 passages

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  • Use in source · Read passage · root_scope.tex, line 382

    [Exact quadratic compression] There is a unique preparation K_r(Z)= U_r(Z) W_r(Z), W_r(Z)=Z^2+2β_rZ+α_r, U_r≡ u_rpmod256. With δ_r=β_r^2-α_r and τ_r= U_r'(c_r)/ U_r(c_r), δ_r=-fracΦ_r^act U_r(c_r) (1-fracτ_r^2Φ_r^act4 U_r(c_r)), v_2(δ_r)=v_2(Φ_r^act). The equality includes the zero case. The same successive monic polynomial division used in Theorem~[source reference], started at precision eight, preserves the quadratic and unit reductions of [source reference]. Write W=Z^2+2β_rZ+α_r and w=W(c_r)…

    Exact mathematical passage
    
     K_r(Z)=\mathcal U_r(Z)\mathcal W_r(Z),\qquad
     \mathcal W_r(Z)=Z^2+2\beta_rZ+\alpha_r,\qquad
     \mathcal U_r\equiv u_r\pmod{256}.
    
u^{(p-1)p^a}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1589

    [Basic odd and dyadic difference gains] For odd p, Δ_(p-1)p^a^ ru^n = u^n(u^(p-1)p^a-1)^r, v_p(Δ_(p-1)p^a^ ru^n)≥(a+1)r. For odd u∈ Z_2^×, Δ_2^a+1^ ru^n = u^n(u^2^a+1-1)^r, v_2(Δ_2^a+1^ ru^n)≥(a+3)r.

    Exact mathematical passage
    
    \boxed{
    \Delta_{(p-1)p^a}^{\,r}u^n
    =
    u^n\left(u^{(p-1)p^a}-1\right)^r,
    \qquad
    \vp\left(\Delta_{(p-1)p^a}^{\,r}u^n\right)\geq(a+1)r.
    }
    
U^{-1}Use in source · 3 passages

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  • Use in source · Read passage · supplement_09.tex, line 131

    integral; mcodeweighted_section. 3. Weighted closure and the normalized denominator. Let w(d)=lfloor d/prfloor. Since w(i)+w(d-i)≤ w(d), the coefficient formula for a product proves closure of W_p under multiplication. Additive closure is immediate. For U∈ W_p with U(0)=1, strong induction in the actual reciprocal coefficient recurrence proves U^-1∈ W_p. This inverse property is a theorem, not an additional field in a structure encoding weighted series.

    Exact mathematical passage
    
     \boxed{U^{-1}\in\Wp.}
    
U^2Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 326

    , and b=(1-i)/2. The exact identity F_0(t)=frac2cosh t+cos t =sech(at)sech(bt) gives, for U=a E+b F, L(U^4n)=C_0(n). Although a and b have negative valuation, their denominators disappear from the following exact polynomial identities: U^2 · = E F+i(P-Q),nonumber H_0:=fracU^4-12 · =P+Q-fracP^22-fracQ^22 +3PQ+i E F(P-Q).

    Exact mathematical passage
    
     U^2&=\mathcal E\mathcal F+i(P-Q),\nonumber\\
     H_0:=\frac{U^4-1}{2}
       &=P+Q-\frac{P^2}{2}-\frac{Q^2}{2}
                     +3PQ+i\mathcal E\mathcal F(P-Q).
     \label{eq:four-euler-H}
    
u^2Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1612

    One step difference multiplies a unit mode by u^H-1. Fermat reduction and lifting the exponent give the odd estimate. For odd u, v_2(u^2^a+1-1) = v_2(u^2-1)+a ≥ a+3.

    Exact mathematical passage
    
    \vTwo\left(u^{2^{a+1}}-1\right)
    =
    \vTwo(u^2-1)+a
    \geq a+3.
    
u^{2^{a+1}}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1599

    [Basic odd and dyadic difference gains] For odd p, Δ_(p-1)p^a^ ru^n = u^n(u^(p-1)p^a-1)^r, v_p(Δ_(p-1)p^a^ ru^n)≥(a+1)r. For odd u∈ Z_2^×, Δ_2^a+1^ ru^n = u^n(u^2^a+1-1)^r, v_2(Δ_2^a+1^ ru^n)≥(a+3)r.

    Exact mathematical passage
    
    \boxed{
    \Delta_{2^{a+1}}^{\,r}u^n
    =
    u^n\left(u^{2^{a+1}}-1\right)^r,
    \qquad
    \vTwo\left(\Delta_{2^{a+1}}^{\,r}u^n\right)\geq(a+3)r.
    }
    
U^4Use in source · 5 passages

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  • Use in source · Read passage · supplement_11.tex, line 88

    For a packed coefficient sequence e, the exact extraction formula is (U^8-U^4)^kpack_4(e)(4n-4k) =Δ_1^k e(n), k≤ n. Here U is the coefficient shift. The nonnegative extraction index is why the first estimate covers k≤ n. The existing saturation lemma extends it to all k with exponent min(k,n), and the step-multiple lemma extends step one to every natural u.

    Exact mathematical passage
    
    \bigl(U^8-U^4\bigr)^k\operatorname{pack}_4(e)(4n-4k)
     =\Delta_1^k e(n),\qquad k\le n.
    
U^{4n}Use in source · 4 passages

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  • Use in source · Read passage · strengthening.tex, line 323

    Put K= Q_2(i), R= Z_2[i], a=(1+i)/2, and b=(1-i)/2. The exact identity F_0(t)=frac2cosh t+cos t =sech(at)sech(bt) gives, for U=a E+b F, L(U^4n)=C_0(n). Although a and b have negative valuation, their denominators disappear from the following exact polynomial identities: U^2 · = E F+i(P-Q),nonumber

    Exact mathematical passage
    \mathcal L(U^{4n})=C_0(n)
U^8Use in source · 1 passage

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  • Use in source · Read passage · supplement_11.tex, line 88

    For a packed coefficient sequence e, the exact extraction formula is (U^8-U^4)^kpack_4(e)(4n-4k) =Δ_1^k e(n), k≤ n. Here U is the coefficient shift. The nonnegative extraction index is why the first estimate covers k≤ n. The existing saturation lemma extends it to all k with exponent min(k,n), and the step-multiple lemma extends step one to every natural u.

    Exact mathematical passage
    
    \bigl(U^8-U^4\bigr)^k\operatorname{pack}_4(e)(4n-4k)
     =\Delta_1^k e(n),\qquad k\le n.
    
U^aUse in source · 3 passages

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  • Use in source · Read passage · supplement_07.tex, line 33

    pageheading1. Coefficient products and reciprocal differentiationA characteristic-p argument without factorial division For sequences f,g: N→ K, where K is a field, define (f⋆ g)(d)=∑_i=0^dbinom di f(i)g(d-i), (U^a f)(d)=f(d+a). These are Hurwitz coefficients. The notation does not divide by d! in F_p. In particular, the implementation is not ordinary power-series multiplication with undefined inverse factorials in positive characteristic.

    Exact mathematical passage
    
     (f\star g)(d)=\sum_{i=0}^{d}\binom di f(i)g(d-i),
     \qquad (U^a f)(d)=f(d+a).
    
U^{a-i}Use in source · 1 passage

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  • Use in source · Read passage · supplement_07.tex, line 40

    The identity sequence is 1(d)=δ_d0. The new source proves the actual higher product rule U^a(f⋆ g)=∑_i=0^abinom ai (U^i f)⋆(U^a-ig). The argument starts with Pascal's identity for the first shift, then inducts on a. All sums are finite at each coefficient.

    Exact mathematical passage
    
     U^a(f\star g)=\sum_{i=0}^{a}\binom ai
           (U^i f)\star(U^{a-i}g).
    
u^HUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1549

    [One unit mode] Let u∈ Z_p^×, let H=Λ_p,L, and put q_u,L=u^H-1∈ p^L Z_p. For every integer address j and every x∈ Z_p, u^Hx+j = u^j∑_r≥0q_u,L^ rbinom xr. The rth Mahler coefficient has valuation at least Lr.

    Exact mathematical passage
    
    q_{u,L}=u^H-1\in p^L\Zp.
    
u^hUse in source · 3 passages

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  • Use in source · Read passage · 10_foundations.tex, line 334

    [Differences of unit modes] Over every commutative ring, Δ_h^s(u^n)=u^n(u^h-1)^s. For any prime p, units u_i∈ Z_p^×, coefficients a_i∈ Z_p, and a finite index set I, the sequence f(n)=∑_i∈ Ia_i u_i^n satisfies Δ_p-1^s f(n)∈ p^s Z_p (n,s∈ N).

    Exact mathematical passage
    
    \Delta_h^s(u^n)=u^n(u^h-1)^s.
    
u^{H_a}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2338

    2^a+1, · p=2. Then Δ_H_a^ ru^n = u^n(u^H_a-1)^r. For odd p, v_p(Δ_H_a^ ru^n) ≥(a+1)r. For p=2 and odd u, v_2(Δ_H_a^ ru^n) ≥(a+3)r. Equivalently, every additional prime power digit increases the certified Mahler slope by one.

    Exact mathematical passage
    
    \boxed{
    \Delta_{H_a}^{\,r}u^n
    =
    u^n\left(u^{H_a}-1\right)^r.
    }
    
u^{Hx+j}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1553

    [One unit mode] Let u∈ Z_p^×, let H=Λ_p,L, and put q_u,L=u^H-1∈ p^L Z_p. For every integer address j and every x∈ Z_p, u^Hx+j = u^j∑_r≥0q_u,L^ rbinom xr. The rth Mahler coefficient has valuation at least Lr.

    Exact mathematical passage
    
    \boxed{
    u^{Hx+j}
    =
    u^j\sum_{r\geq0}q_{u,L}^{\,r}\binom xr.
    }
    
U^iUse in source · 1 passage

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  • Use in source · Read passage · supplement_07.tex, line 40

    The identity sequence is 1(d)=δ_d0. The new source proves the actual higher product rule U^a(f⋆ g)=∑_i=0^abinom ai (U^i f)⋆(U^a-ig). The argument starts with Pascal's identity for the first shift, then inducts on a. All sums are finite at each coefficient.

    Exact mathematical passage
    
     U^a(f\star g)=\sum_{i=0}^{a}\binom ai
           (U^i f)\star(U^{a-i}g).
    
u^jUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1553

    [One unit mode] Let u∈ Z_p^×, let H=Λ_p,L, and put q_u,L=u^H-1∈ p^L Z_p. For every integer address j and every x∈ Z_p, u^Hx+j = u^j∑_r≥0q_u,L^ rbinom xr. The rth Mahler coefficient has valuation at least Lr.

    Exact mathematical passage
    
    \boxed{
    u^{Hx+j}
    =
    u^j\sum_{r\geq0}q_{u,L}^{\,r}\binom xr.
    }
    
u^{j+(p-1)s}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2470

    [Local phase lattice] For a unit mode u^n, the odd prime coordinate gives u^n = u^j+(p-1)s (u^(p-1)p^a)^q. The dyadic coordinate gives u^n = u^j+2s (u^2^a+1)^q. Thus finite character phase, local digit address, and Mahler depth are separate coordinates.

    Exact mathematical passage
    
    \boxed{
    u^n
    =
    u^{j+(p-1)s}
    \left(u^{(p-1)p^a}\right)^q.
    }
    
u^{j+2s}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2479

    [Local phase lattice] For a unit mode u^n, the odd prime coordinate gives u^n = u^j+(p-1)s (u^(p-1)p^a)^q. The dyadic coordinate gives u^n = u^j+2s (u^2^a+1)^q. Thus finite character phase, local digit address, and Mahler depth are separate coordinates.

    Exact mathematical passage
    
    \boxed{
    u^n
    =
    u^{j+2s}
    \left(u^{2^{a+1}}\right)^q.
    }
    
u^mUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 441

    The coefficients c_u in the proof of [source reference] are constant on the cosets of H_m. For u≠0, the value u^m depends only on the coset, and every element of G_m arises. Grouping the modes by cosets gives the displayed expansion.

    Exact mathematical passage
    u^m
u^nUse in source · 19 passages

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  • Use in source · Read passage · 10_foundations.tex, line 334

    [Differences of unit modes] Over every commutative ring, Δ_h^s(u^n)=u^n(u^h-1)^s. For any prime p, units u_i∈ Z_p^×, coefficients a_i∈ Z_p, and a finite index set I, the sequence f(n)=∑_i∈ Ia_i u_i^n satisfies Δ_p-1^s f(n)∈ p^s Z_p (n,s∈ N).

    Exact mathematical passage
    
    \Delta_h^s(u^n)=u^n(u^h-1)^s.
    
u^{n+p-1}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 343

    For positive n, the zero mode contributes nothing. Every nonzero mode satisfies u^p-1=1. Hence b_n+p-1^(m,p) = ∑_u≠0c_uu^n+p-1 = ∑_u≠0c_uu^n = b_n^(m,p).

    Exact mathematical passage
    
    b_{n+p-1}^{(m,p)}
    =
    \sum_{u\neq0}c_uu^{n+p-1}
    =
    \sum_{u\neq0}c_uu^n
    =
    b_n^{(m,p)}.
    
U^{p+L}Use in source · 1 passage

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  • Use in source · Read passage · supplement_07.tex, line 162

    0, · Nnmid d, N=mp. Let L=p(p-1) and δ=U^p+L-U^p=U^p^2-U^p. The source proves the operator dictionary (δ^sX)(d)=Δ_L^s X(d+ps). At the valid nonnegative coefficient degree d=mpn-ps, with s≤ mn, this becomes (δ^sX)(mpn-ps)=Δ_h^s e(n), h=(p-1)/m. The packed sequence and difference operations are definitions. Their coefficient correspondence is proved before any divisibility assumption is introduced.

    Exact mathematical passage
    \delta=U^{p+L}-U^p=U^{p^2}-U^p
u^{p-1}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 342

    For positive n, the zero mode contributes nothing. Every nonzero mode satisfies u^p-1=1. Hence b_n+p-1^(m,p) = ∑_u≠0c_uu^n+p-1 = ∑_u≠0c_uu^n = b_n^(m,p).

    Exact mathematical passage
    u^{p-1}=1
U^{p^2}Use in source · 3 passages

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  • Use in source · Read passage · supplement_07.tex, line 162

    0, · Nnmid d, N=mp. Let L=p(p-1) and δ=U^p+L-U^p=U^p^2-U^p. The source proves the operator dictionary (δ^sX)(d)=Δ_L^s X(d+ps). At the valid nonnegative coefficient degree d=mpn-ps, with s≤ mn, this becomes (δ^sX)(mpn-ps)=Δ_h^s e(n), h=(p-1)/m. The packed sequence and difference operations are definitions. Their coefficient correspondence is proved before any divisibility assumption is introduced.

    Exact mathematical passage
    \delta=U^{p+L}-U^p=U^{p^2}-U^p
U^{q-a}Use in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 33

    pageheading1. The characteristic-zero product ruleThe normalized defect is not an ordinary derivation Let A be a commutative algebra over a characteristic-zero field K, and let U be an actual K-derivation. The source defines B=fracU^p^2-U^pp, M_q(f,g)=∑_a=1^q-1frac1pbinom qa (U^a f)(U^q-ag), q=p,p^2. The coefficients of each M_q are proved integral: the middle entries of a prime-power binomial row are divisible by p. In Lean they are implemented as natural quotients, with exact division proved b…

    Exact mathematical passage
    
     B=\frac{U^{p^2}-U^p}{p},\qquad
     M_q(f,g)=\sum_{a=1}^{q-1}\frac{1}{p}\binom qa
              (U^a f)(U^{q-a}g),\quad q=p,p^2.
    
u^rUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 164

    The nth coefficient is ∑_r=0^n binom nr u^rv^n-r = (u+v)^n by the binomial theorem.

    Exact mathematical passage
    
    \sum_{r=0}^{n}
    \binom nr u^rv^{n-r}
    =
    (u+v)^n
    
U_0Use in source · 3 passages

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  • Use in source · Read passage · 40_regularity.tex, line 127

    To obtain the root equivalence on a product neighborhood, we choose the neighborhoods as follows. The source of the open partial homeomorphism defined by Φ is open around a; its inverse is g. Choose open sets V and U_0 containing y_0 and t_0 such that V× U_0 lies in that source. Restrict U_0 further by the preimage of an open analytic inverse neighborhood under t↦(0,t), the open locus where this same ψ is analytic, and the interior of ψ^-1(V). All of these sets contain t_0. For the last set, use…

    Exact mathematical passage
    U_0
U_1Use in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 38

    [Polynomial Taylor remainder] Let R be a commutative ring, P∈ R[X_1,…,X_d], a∈ R^d, and c∈ R. There is a polynomial Q∈ R[U_1,…,U_d], with Q(0)=0, such that for every u∈ R^d, P(a+cu)=P(a)+c∑_j=1^d(∂_jP)(a)u_j+c^2Q(u). In particular, when R= Z and a,u,c are integral, subtracting the first two terms on the right from P(a+cu) gives a multiple of c^2.

    Exact mathematical passage
    Q\in R[U_1,\ldots,U_d]
U_2Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 736

    For example, U_2(n)≡2pmod32 and U_3(n)≡6pmod8 for every row. At precision 2^10 the least packet periods are 512,512,32,128. A packet displacement T has original degree displacement 4T, as proved by leanKomorebiPaperIX.DyadicFourLeastPeriods.original_degree_exact_displacement. Minimality here concerns packet shifts. Shifts of the full degree-indexed sequence, including its zero coefficients off the packet support, require a separate analysis. The new strengthening chapters determine all repeated-…

    Exact mathematical passage
    U_2(n)\equiv2\pmod{32}
U_3Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 736

    For example, U_2(n)≡2pmod32 and U_3(n)≡6pmod8 for every row. At precision 2^10 the least packet periods are 512,512,32,128. A packet displacement T has original degree displacement 4T, as proved by leanKomorebiPaperIX.DyadicFourLeastPeriods.original_degree_exact_displacement. Minimality here concerns packet shifts. Shifts of the full degree-indexed sequence, including its zero coefficients off the packet support, require a separate analysis. The new strengthening chapters determine all repeated-…

    Exact mathematical passage
    U_3(n)\equiv6\pmod8
U_{4,j}Use in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 467

    For the remainder of the four-packet discussion abbreviate C_j=C_4,j and U_j=U_4,j. Removing the original row cap requires an unshifted coefficient operator.

    Exact mathematical passage
    U_j=U_{4,j}
U_{6,3}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 369

    C_6,3(1)=-frac184, C_6,3(2)=frac7035460, C_6,3(2)-C_6,3(1)=frac64455. The last difference has 3-adic valuation zero. Here h=1, so at n=1 the order-one bound without factorial loss would require valuation at least one. The corrected normalized bound is sharp, since v_3(C_6,3(2)-C_6,3(1))=0=1-v_3(3!). Multiplication by 3! gives U_6,3(2)-U_6,3(1)=frac384455, v_3(U_6,3(2)-U_6,3(1))=1, consistent with the unnormalized theorem. Thus the factorial correction is already necessary at a single step in the…

    Exact mathematical passage
    
     U_{6,3}(2)-U_{6,3}(1)=\frac{384}{455},\qquad
     v_3\bigl(U_{6,3}(2)-U_{6,3}(1)\bigr)=1,
    
U_aUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2741

    [Packet decomposition of a lift obstruction] Suppose F_p^d = ⊕_a∈ AU_a, F_p^r = ⊕_a∈ AV_a, and suppose a linear correction map D: F_p^d→ F_p^r satisfies D(U_a)⊆ V_a for every a. Write D_a=D|_U_a. Then coker D ≅ ⊕_a∈ Acoker D_a. A residual lifts if and only if every packet obstruction vanishes.

    Exact mathematical passage
    
    \FF_p^d
    =
    \bigoplus_{a\in A}U_a,
    \qquad
    \FF_p^r
    =
    \bigoplus_{a\in A}V_a,
    
U_dUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 38

    [Polynomial Taylor remainder] Let R be a commutative ring, P∈ R[X_1,…,X_d], a∈ R^d, and c∈ R. There is a polynomial Q∈ R[U_1,…,U_d], with Q(0)=0, such that for every u∈ R^d, P(a+cu)=P(a)+c∑_j=1^d(∂_jP)(a)u_j+c^2Q(u). In particular, when R= Z and a,u,c are integral, subtracting the first two terms on the right from P(a+cu) gives a multiple of c^2.

    Exact mathematical passage
    Q\in R[U_1,\ldots,U_d]
U_GUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 515

    [Composition of coefficient transports] If U_H and U_G are the triangular transforms associated with unit kernels H and G, then U_HU_G=U_HG. In particular, the inverse of U_H is U_H^-1.

    Exact mathematical passage
    U_G
U_HUse in source · 9 passages

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  • Use in source · Read passage · 10_foundations.tex, line 90

    [Constructed unit-triangular coordinates] Let h: N→ K satisfy h_0=1, write H= E(h) and H^-1= E(k), and define U_Ha= D(H E(a)). Then U_H is an equivalence, with inverse U_H^-1a= D(H^-1 E(a)). Its coefficient formulas are (U_Ha)_n=∑_r=0^nbinom nr h_r a_n-r, (U_H^-1a)_n=∑_r=0^nbinom nr k_r a_n-r. The resulting triangular matrix has diagonal entries 1. The transported selectors widehat P_N,j=U_HP_N,jU_H^-1 are idempotent and pairwise orthogonal, and their sum is the identity.

    Exact mathematical passage
    
    U_Ha=\mathcal D(H\mathcal E(a)).
    
U_H^{-1}Use in source · 5 passages

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  • Use in source · Read passage · 10_foundations.tex, line 94

    [Constructed unit-triangular coordinates] Let h: N→ K satisfy h_0=1, write H= E(h) and H^-1= E(k), and define U_Ha= D(H E(a)). Then U_H is an equivalence, with inverse U_H^-1a= D(H^-1 E(a)). Its coefficient formulas are (U_Ha)_n=∑_r=0^nbinom nr h_r a_n-r, (U_H^-1a)_n=∑_r=0^nbinom nr k_r a_n-r. The resulting triangular matrix has diagonal entries 1. The transported selectors widehat P_N,j=U_HP_N,jU_H^-1 are idempotent and pairwise orthogonal, and their sum is the identity.

    Exact mathematical passage
    U_H^{-1}a=\mathcal D(H^{-1}\mathcal E(a))
U_{H^{-1}}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 521

    [Composition of coefficient transports] If U_H and U_G are the triangular transforms associated with unit kernels H and G, then U_HU_G=U_HG. In particular, the inverse of U_H is U_H^-1.

    Exact mathematical passage
    U_{H^{-1}}
U_{HG}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 516

    [Composition of coefficient transports] If U_H and U_G are the triangular transforms associated with unit kernels H and G, then U_HU_G=U_HG. In particular, the inverse of U_H is U_H^-1.

    Exact mathematical passage
    
    \boxed{
    U_HU_G=U_{HG}.
    }
    
U_iUse in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 52

    Apply polynomial induction using constants, addition, and multiplication by a variable. Constants have zero remainder, and remainders add. For the multiplication step, suppose the assertion is known for P and write D_P(U)=∑_j(∂_jP)(a)U_j. Multiplication of its expansion by a_i+cU_i gives constant term P(a)a_i, linear term c(a_iD_P(U)+P(a)U_i), and remainder c^2(Q(U)(a_i+cU_i)+D_P(U)U_i). The product rule identifies the linear term with the derivative of PX_i. The displayed remainder vanishes at …

    Exact mathematical passage
    a_i+cU_i
u_iUse in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 337

    [Differences of unit modes] Over every commutative ring, Δ_h^s(u^n)=u^n(u^h-1)^s. For any prime p, units u_i∈ Z_p^×, coefficients a_i∈ Z_p, and a finite index set I, the sequence f(n)=∑_i∈ Ia_i u_i^n satisfies Δ_p-1^s f(n)∈ p^s Z_p (n,s∈ N).

    Exact mathematical passage
    u_i\in\Zp^\times
u_i^hUse in source · 1 passage

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  • Use in source · Read passage · supplement_04.tex, line 148

    Finite integral linear combinations. For any finite index set S, coefficients a_i∈ Z_p and units u_i∈ Z_p^×, linearity of the actual iterated operator gives Δ_h^s (∑_i∈ Sa_i u_i^n) =∑_i∈ Sa_i u_i^n(u_i^h-1)^s. At h=p-1, the result is divisible by p^s. The empty sum is permitted. This does not prove that the general congruential Euler packets of Theorem 7.5 admit such a representation.

    Exact mathematical passage
    
     \Delta_h^s\!\left(\sum_{i\in S}a_i u_i^n\right)
     =\sum_{i\in S}a_i u_i^n(u_i^h-1)^s.
    
u_i^nUse in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 338

    [Differences of unit modes] Over every commutative ring, Δ_h^s(u^n)=u^n(u^h-1)^s. For any prime p, units u_i∈ Z_p^×, coefficients a_i∈ Z_p, and a finite index set I, the sequence f(n)=∑_i∈ Ia_i u_i^n satisfies Δ_p-1^s f(n)∈ p^s Z_p (n,s∈ N).

    Exact mathematical passage
    f(n)=\sum_{i\in I}a_i u_i^n
u_i^{p-1}Use in source · 1 passage

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  • Use in source · Read passage · 10_foundations.tex, line 347

    Induction gives the geometric identity. Reduction of a unit to F_p^× gives u_i^p-1=1 modulo p. Taking s powers supplies p^s in each summand. The difference operator is linear over finite sums.

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    u_i^{p-1}=1
U_jUse in source · 11 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 467

    For the remainder of the four-packet discussion abbreviate C_j=C_4,j and U_j=U_4,j. Removing the original row cap requires an unshifted coefficient operator.

    Exact mathematical passage
    U_j=U_{4,j}
u_jUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 40

    [Polynomial Taylor remainder] Let R be a commutative ring, P∈ R[X_1,…,X_d], a∈ R^d, and c∈ R. There is a polynomial Q∈ R[U_1,…,U_d], with Q(0)=0, such that for every u∈ R^d, P(a+cu)=P(a)+c∑_j=1^d(∂_jP)(a)u_j+c^2Q(u). In particular, when R= Z and a,u,c are integral, subtracting the first two terms on the right from P(a+cu) gives a multiple of c^2.

    Exact mathematical passage
    \label{eq:polynomial-taylor}
     P(a+cu)=P(a)+c\sum_{j=1}^d(\partial_jP)(a)u_j+c^2Q(u).
    
U_kUse in source · 2 passages

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  • Use in source · Read passage · root_scope.tex, line 176

    For preparation, start from the monic polynomial g=overline G_r and its zero-one lift W_1, with U_1=1. Suppose G_r-W_kU_k is coefficientwise divisible by 2^k. Its divided reduction is a finite polynomial because G_r is restricted. Divide it by g in F_2[X]: overline(G_r-W_kU_k)/2^k=gq+s, deg s<d_r. With zero-one coefficient lifts, set W_k+1=W_k+2^ks and U_k+1=U_k+2^kq. The error is now divisible by 2^k+1, the polynomial remains monic of degree d_r, and the unit reduces to one. The sequences conve…

    Exact mathematical passage
    G_r-W_kU_k
u_kUse in source · 12 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2865

    [Residue subsequences of a second order recurrence] Suppose x_n+2=Px_n+1-Qx_n over a commutative ring. Define V_0=2, V_1=P, and V_n+2=PV_n+1-QV_n. For every positive integer N and every address j, the subsequence u_k=x_Nk+j satisfies u_k+2=V_Nu_k+1-Q^Nu_k.

    Exact mathematical passage
    
    u_k=x_{Nk+j}
    
U_{k+1}Use in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 183

    or preparation, start from the monic polynomial g=overline G_r and its zero-one lift W_1, with U_1=1. Suppose G_r-W_kU_k is coefficientwise divisible by 2^k. Its divided reduction is a finite polynomial because G_r is restricted. Divide it by g in F_2[X]: overline(G_r-W_kU_k)/2^k=gq+s, deg s<d_r. With zero-one coefficient lifts, set W_k+1=W_k+2^ks and U_k+1=U_k+2^kq. The error is now divisible by 2^k+1, the polynomial remains monic of degree d_r, and the unit reduces to one. The sequences conver…

    Exact mathematical passage
    U_{k+1}=U_k+2^kq
u_{k+1}Use in source · 5 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2869

    [Residue subsequences of a second order recurrence] Suppose x_n+2=Px_n+1-Qx_n over a commutative ring. Define V_0=2, V_1=P, and V_n+2=PV_n+1-QV_n. For every positive integer N and every address j, the subsequence u_k=x_Nk+j satisfies u_k+2=V_Nu_k+1-Q^Nu_k.

    Exact mathematical passage
    
    \boxed{
    u_{k+2}=V_Nu_{k+1}-Q^Nu_k.}
    
u_{k+1}^{(j)}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2792

    [Companion jump recurrence] Every residue state packet satisfies u_k+1^(j)=C^Nu_k^(j). More explicitly, u_k^(j)=C^j(C^N)^kv_0.

    Exact mathematical passage
    
    \boxed{
    u_{k+1}^{(j)}=C^Nu_k^{(j)}.}
    
u_{k+2}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2869

    [Residue subsequences of a second order recurrence] Suppose x_n+2=Px_n+1-Qx_n over a commutative ring. Define V_0=2, V_1=P, and V_n+2=PV_n+1-QV_n. For every positive integer N and every address j, the subsequence u_k=x_Nk+j satisfies u_k+2=V_Nu_k+1-Q^Nu_k.

    Exact mathematical passage
    
    \boxed{
    u_{k+2}=V_Nu_{k+1}-Q^Nu_k.}
    
u_{k+a}Use in source · 4 passages

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  • Use in source · Read passage · supplement_06.tex, line 47

    The jump formula. Associativity and the laws for powers of one matrix give u_k=C^j(C^N)^kv_0=(C^N)^k(C^jv_0). Only powers of the same matrix are interchanged. The argument does not assume that arbitrary matrices commute. In particular, u_k+1=C^Nu_k, u_k+a=(C^N)^a u_k. The step matrix is C^N, not C. A checked example with the Fibonacci matrix and N=2 rejects replacing the jump with a single original step.

    Exact mathematical passage
    
     u_{k+1}=C^Nu_k,\qquad u_{k+a}=(C^N)^a u_k.
    
u_{k+d}Use in source · 2 passages

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  • Use in source · Read passage · supplement_06.tex, line 82

    The complete vector recurrence. Apply the matrix identity to the actual state u_k. The already proved advance identity identifies each D^a u_k with u_k+a. Consequently, ∑_a=0^dχ_D[a]u_k+a=0, u_k+d=-∑_a=0^d-1c_a u_k+a. This is the desired recurrence, with its monic leading coefficient and all earlier coefficients supplied by χ_D. It is not an assumption on the packet sequence.

    Exact mathematical passage
    
     \sum_{a=0}^{d}\chi_D[a]u_{k+a}=0,
     \qquad
     \boxed{u_{k+d}=-\sum_{a=0}^{d-1}c_a u_{k+a}.}
    
u_k^{(j)}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2786

    Let R be a commutative ring. Let v_n+1=Cv_n, C∈ M_d(R). For a positive integer N and an address j∈{0,…,N-1}, define the residue state packet u_k^(j)=v_Nk+j.

    Exact mathematical passage
    
    u_k^{(j)}=v_{Nk+j}.
    
u_{Mk+s}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 54

    Refinement of an arithmetic address. Taking another subsequence with step M and offset s yields u_Mk+s=v_(NM)k+(Ns+j). The formal proof establishes this identity for the actual packet function by checking the index arithmetic. It is not inferred from a diagram or a naming convention.

    Exact mathematical passage
    
     u_{Mk+s}=v_{(NM)k+(Ns+j)}.
    
U_{mp,j}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 571

    [Step-sensitive divisor-branch bound] Under the hypotheses of Theorem~[source reference], for all u,k,n∈ N, Δ_u(p-1)/m^kU_mp,j(n) ∈ p^min(k(1+ν_p(u)),mn) Z_p. For C_mp,j the corresponding valuation lower bound is min(k(1+ν_p(u)),mn)-v_p(j!).

    Exact mathematical passage
    
     \Delta_{u(p-1)/m}^{k}U_{mp,j}(n)
           \in p^{\min(k(1+\nu_p(u)),mn)}\Z_p.
    
U_rUse in source · 3 passages

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  • Use in source · Read passage · root_scope.tex, line 156

    [Exact classical tail valuation] For every r≥1 and N≥ r, v_2( T_r-1(N)) =4r-1+v_2binom Nr+μ_r+v_2G_r(N-r). There are unique factors G_r(X)=W_r(X)U_r(X), where W_r∈ Z_2[X] is monic of degree d_r=max J_r and U_r∈ Z_2⟨ X⟩^× satisfies U_r≡1pmod2. Consequently v_2G_r(x)=v_2W_r(x) for every x∈ Z_2, and deg W_r≤lfloorfracs_2(r)-13rfloor. Equation~[source reference], the exact weight in [source reference], and the digit-sum identity v_2binom2N2r=v_2binom Nr prove [source reference]. The binomial coeffic…

    Exact mathematical passage
    
     G_r(X)=W_r(X)U_r(X),
    
\mathcal U_rUse in source · 4 passages

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  • Use in source · Read passage · root_scope.tex, line 382

    [Exact quadratic compression] There is a unique preparation K_r(Z)= U_r(Z) W_r(Z), W_r(Z)=Z^2+2β_rZ+α_r, U_r≡ u_rpmod256. With δ_r=β_r^2-α_r and τ_r= U_r'(c_r)/ U_r(c_r), δ_r=-fracΦ_r^act U_r(c_r) (1-fracτ_r^2Φ_r^act4 U_r(c_r)), v_2(δ_r)=v_2(Φ_r^act). The equality includes the zero case. The same successive monic polynomial division used in Theorem~[source reference], started at precision eight, preserves the quadratic and unit reductions of [source reference]. Write W=Z^2+2β_rZ+α_r and w=W(c_r)…

    Exact mathematical passage
    
     K_r(Z)=\mathcal U_r(Z)\mathcal W_r(Z),\qquad
     \mathcal W_r(Z)=Z^2+2\beta_rZ+\alpha_r,\qquad
     \mathcal U_r\equiv u_r\pmod{256}.
    
u_rUse in source · 2 passages

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  • Use in source · Read passage · root_scope.tex, line 365

    The current focused obstruction is the admitted ordinary-index family r=(17659+2^18T)2^640-161, T∈ Z_≥0, j_r=161. On its surviving chart define the actual, undepleted function K_r(Z)=2^-178G_r(10202+2^15Z). The complete coefficient theorem gives K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z), a_r,b_r∈ Z_2, u_r∈ Z_2^×, R_r∈ Z_2⟨ Z⟩. Its half derivative is a dyadic isometry. It therefore has a unique critical point c_r∈ Z_2. Set Φ_r^act=K_r(c_r). The earlier independent-undepletion notation for this same sc…

    Exact mathematical passage
    
     K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z),
     \quad a_r,b_r\in\mathbb Z_2,\quad
     u_r\in\mathbb Z_2^\times,\quad
     R_r\in\mathbb Z_2\langle Z\rangle.
     \label{eq:root:quadraticclass}
    
\mathcal U_r'Use in source · 1 passage

Search aliases: U_r' mathcal

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  • Use in source · Read passage · root_scope.tex, line 388

    [Exact quadratic compression] There is a unique preparation K_r(Z)= U_r(Z) W_r(Z), W_r(Z)=Z^2+2β_rZ+α_r, U_r≡ u_rpmod256. With δ_r=β_r^2-α_r and τ_r= U_r'(c_r)/ U_r(c_r), δ_r=-fracΦ_r^act U_r(c_r) (1-fracτ_r^2Φ_r^act4 U_r(c_r)), v_2(δ_r)=v_2(Φ_r^act). The equality includes the zero case. The same successive monic polynomial division used in Theorem~[source reference], started at precision eight, preserves the quadratic and unit reductions of [source reference]. Write W=Z^2+2β_rZ+α_r and w=W(c_r)…

    Exact mathematical passage
    \tau_r=\mathcal U_r'(c_r)/\mathcal U_r(c_r)
U_{r,q}Use in source · 3 passages

Search aliases: U_r,q

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  • Use in source · Read passage · supplement_02.tex, line 138

    The row action is connected to the matrix entries. Reversing the finite sum and using binomial symmetry gives the actual source-index row action E_2r=∑_q=0^rU_r,q(p)S_2q(p), where U_r,q(p)= displaystylebinom2r2qb_r-qp^2(r-q), · q≤ r,

    Exact mathematical passage
    
     E_{2r}=\sum_{q=0}^{r}U_{r,q}(p)S_{2q}(p),
    
U_{r,r}Use in source · 1 passage

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  • Use in source · Read passage · supplement_02.tex, line 149

    0, · q>r. The row-action equality, the diagonal identity U_r,r=1, and vanishing above the diagonal are all checked. The arithmetic weight is also explicit: p^2(r-q)mid U_r,q(p) in Z, 0≤ q≤ r. No inverse-coordinate identity, unit diagonal or weight bound is assumed for these rows.

    Exact mathematical passage
    U_{r,r}=1
\underline qUse in source · 1 passage

Search aliases: underline q underline

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  • Use in source · Read passage · root_scope.tex, line 140

    ad J_r={q:λ_r(q)=μ_r}, G_r=2^-μ_rH_r. Since λ_r(0)=0 and λ_r(q)≥3q+1-s_2(r), every minimizing address satisfies 0≤ q≤lfloorfracs_2(r)-13rfloor. In particular μ_r≤0, and the minimum is found by a finite carry calculation. Reduction of the whole series gives overline G_r(X)=∑_q∈ J_rX^underline q in F_2[X]. Its highest degree is max J_r, with leading coefficient one. Thus the normalization is the actual Gauss normalization.

    Exact mathematical passage
    
     \overline G_r(X)=\sum_{q\in J_r}X^{\underline q}
           \quad\text{in }\mathbb F_2[X].
     \label{eq:root:reduction}
    
V^{-1}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 261

    the form H_c(log(1+T))=fracT^cc! V(T), V(0)=1, V∈ W_p. The weight of c! compensates for the coefficient shift by c. The weighted ring is closed under inversion of series with constant coefficient one and under L_T=log(1+T)(1+T)d/dT. With ℓ(T)=log(1+T)/T, interpreted by coefficient shifting, differentiation of the displayed factorization gives R_c(log(1+T))=c(1+T)ℓ(T)+(L_TV)V^-1. Since pℓ(T)∈ W_p and pmid c, both terms are weighted. This is the singular estimate leanKomorebiPaperIX.SingularRatioA…

    Exact mathematical passage
    
     R_c(\log(1+T))=c(1+T)\ell(T)+(L_TV)V^{-1}.
    
v^nUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 429

    [Exponential form and minimal period] Let G_m=( F_p^×)^m. This cyclic group has order h. There are uniquely determined coefficients d_v∈ F_p, indexed by v∈ G_m, such that e_n^(m,p) = ∑_v∈ G_md_vv^n (n≥1). The minimal positive period is the least common multiple of the orders of those v for which d_v≠0. In particular, it divides h.

    Exact mathematical passage
    
    \boxed{
    e_n^{(m,p)}
    =
    \sum_{v\in G_m}d_vv^n
    \qquad(n\geq1).
    }
    
v^{n-r}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 164

    The nth coefficient is ∑_r=0^n binom nr u^rv^n-r = (u+v)^n by the binomial theorem.

    Exact mathematical passage
    
    \sum_{r=0}^{n}
    \binom nr u^rv^{n-r}
    =
    (u+v)^n
    
V^qUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 887

    For k=2^r≥4, put q=k/2, which is even. Normalize the exact expansion [source reference] by 2^q before reducing modulo two. Its uth auxiliary term has excess u+lfloor(k-u)/2rfloor-q=lfloor u/2rfloor. Only auxiliary indices zero and one can therefore survive: fracγ_k(H_2)2^q ≡ V^qγ_k(X) +ϵ V^q-1Gγ_k-1(X) pmod2 A_2. Here V≡τpmod2 and τ^2≡1pmod2. Since q is even and ϵτ≡ϵpmod2, this reduces to γ_k(X)+ϵ Gγ_k-1(X). Moreover, G≡γ_2(X)+τ Qpmod2. After evaluation, every term with a single τ vanishes, so t…

    Exact mathematical passage
    
     \frac{\gamma_k(H_2)}{2^q}
     \equiv V^q\gamma_k(X)
           +\epsilon V^{q-1}G\gamma_{k-1}(X)
                                    \pmod{2\mathcal A_2}.
     \label{eq:two-power-leading-normalized}
    
V^{q-1}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 887

    For k=2^r≥4, put q=k/2, which is even. Normalize the exact expansion [source reference] by 2^q before reducing modulo two. Its uth auxiliary term has excess u+lfloor(k-u)/2rfloor-q=lfloor u/2rfloor. Only auxiliary indices zero and one can therefore survive: fracγ_k(H_2)2^q ≡ V^qγ_k(X) +ϵ V^q-1Gγ_k-1(X) pmod2 A_2. Here V≡τpmod2 and τ^2≡1pmod2. Since q is even and ϵτ≡ϵpmod2, this reduces to γ_k(X)+ϵ Gγ_k-1(X). Moreover, G≡γ_2(X)+τ Qpmod2. After evaluation, every term with a single τ vanishes, so t…

    Exact mathematical passage
    
     \frac{\gamma_k(H_2)}{2^q}
     \equiv V^q\gamma_k(X)
           +\epsilon V^{q-1}G\gamma_{k-1}(X)
                                    \pmod{2\mathcal A_2}.
     \label{eq:two-power-leading-normalized}
    
v^rUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 443

    Distinct elements of G_m define linearly independent functions on one full period, by the Vandermonde determinant. Therefore a positive integer r is a period exactly when v^r=1 for every active mode. The least such r is the stated least common multiple.

    Exact mathematical passage
    v^r=1
v_{(NM)k+(Ns+j)}Use in source · 1 passage

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  • Use in source · Read passage · supplement_06.tex, line 54

    Refinement of an arithmetic address. Taking another subsequence with step M and offset s yields u_Mk+s=v_(NM)k+(Ns+j). The formal proof establishes this identity for the actual packet function by checking the index arithmetic. It is not inferred from a diagram or a naming convention.

    Exact mathematical passage
    
     u_{Mk+s}=v_{(NM)k+(Ns+j)}.
    
V_0Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2860

    [Residue subsequences of a second order recurrence] Suppose x_n+2=Px_n+1-Qx_n over a commutative ring. Define V_0=2, V_1=P, and V_n+2=PV_n+1-QV_n. For every positive integer N and every address j, the subsequence u_k=x_Nk+j satisfies u_k+2=V_Nu_k+1-Q^Nu_k.

    Exact mathematical passage
    V_0=2
v_0Use in source · 19 passages

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V_1Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2860

    [Residue subsequences of a second order recurrence] Suppose x_n+2=Px_n+1-Qx_n over a commutative ring. Define V_0=2, V_1=P, and V_n+2=PV_n+1-QV_n. For every positive integer N and every address j, the subsequence u_k=x_Nk+j satisfies u_k+2=V_Nu_k+1-Q^Nu_k.

    Exact mathematical passage
    V_1=P
V_2Use in source · 6 passages

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  • Use in source · Read passage · strengthening.tex, line 868

    [Exact address-two power orders] For every starting row n≥0 and all positive mixed steps, v_2(Δ_H_1⋯Δ_H_kC_2(n)) =V_2(k)+∑_ν=1^k v_2(H_ν) whenever k is a positive power of two, where V_2(1)=4, V_2(2)=6, V_2(k)=2k+v_2(k!)+frack2 (k≥4). For the original coefficients, the right side gains one bit.

    Exact mathematical passage
    
     v_2(\Delta_{H_1}\cdots\Delta_{H_k}C_2(n))
           =V_2(k)+\sum_{\nu=1}^k v_2(H_\nu)
     \label{eq:two-power-orders-exact}
    
v_3Use in source · 11 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 365

    ith-normalization-loss Take p=3, m=2, N=6, and j=3. The recurrence gives C_6,3(1)=-frac184, C_6,3(2)=frac7035460, C_6,3(2)-C_6,3(1)=frac64455. The last difference has 3-adic valuation zero. Here h=1, so at n=1 the order-one bound without factorial loss would require valuation at least one. The corrected normalized bound is sharp, since v_3(C_6,3(2)-C_6,3(1))=0=1-v_3(3!). Multiplication by 3! gives U_6,3(2)-U_6,3(1)=frac384455, v_3(U_6,3(2)-U_6,3(1))=1, consistent with the unnormalized theorem. T…

    Exact mathematical passage
    
     v_3\bigl(C_{6,3}(2)-C_{6,3}(1)\bigr)=0=1-v_3(3!).
    
V_aUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2741

    [Packet decomposition of a lift obstruction] Suppose F_p^d = ⊕_a∈ AU_a, F_p^r = ⊕_a∈ AV_a, and suppose a linear correction map D: F_p^d→ F_p^r satisfies D(U_a)⊆ V_a for every a. Write D_a=D|_U_a. Then coker D ≅ ⊕_a∈ Acoker D_a. A residual lifts if and only if every packet obstruction vanishes.

    Exact mathematical passage
    
    \FF_p^d
    =
    \bigoplus_{a\in A}U_a,
    \qquad
    \FF_p^r
    =
    \bigoplus_{a\in A}V_a,
    
V_hUse in source · 6 passages

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  • Use in source · Read passage · 72_recovered.tex, line 829

    [Fourier invariance of the standard window operator] Let p be odd, let mmid p-1, and let h=(p-1)/m. Define (T_m,pf)_n = ∑_q=0^nbinommnmqf_q in F_p. Let V_h be the vector space of sequences satisfying f_n+h=f_n (n≥1). Then T_m,p(V_h)=V_h. The restriction of T_m,p to V_h is an automorphism.

    Exact mathematical passage
    V_h
v_jUse in source · 4 passages

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  • Use in source · Read passage · 40_regularity.tex, line 29

    [Derivative of polynomial evaluation] For P∈ K[X_1,…,X_n], polynomial evaluation is analytic at every point. At a∈ K^n its strict Fr echet derivative is v⟼∑_j=1^n (∂_jP)(a)v_j. For a finite system, its strict derivative is the continuous linear map represented by the matrix of evaluated formal partial derivatives. For the derivative statement, use polynomial induction on constants, addition, and multiplication by a variable. Constants have zero derivative. The sum rule agrees with the additivity…

    Exact mathematical passage
    
     v\longmapsto\sum_{j=1}^n (\partial_jP)(a)v_j.
    
V_NUse in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2869

    [Residue subsequences of a second order recurrence] Suppose x_n+2=Px_n+1-Qx_n over a commutative ring. Define V_0=2, V_1=P, and V_n+2=PV_n+1-QV_n. For every positive integer N and every address j, the subsequence u_k=x_Nk+j satisfies u_k+2=V_Nu_k+1-Q^Nu_k.

    Exact mathematical passage
    
    \boxed{
    u_{k+2}=V_Nu_{k+1}-Q^Nu_k.}
    
V_nUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2861

    [Residue subsequences of a second order recurrence] Suppose x_n+2=Px_n+1-Qx_n over a commutative ring. Define V_0=2, V_1=P, and V_n+2=PV_n+1-QV_n. For every positive integer N and every address j, the subsequence u_k=x_Nk+j satisfies u_k+2=V_Nu_k+1-Q^Nu_k.

    Exact mathematical passage
    
    V_{n+2}=PV_{n+1}-QV_n.
    
v_nUse in source · 8 passages

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V_{n+1}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2861

    [Residue subsequences of a second order recurrence] Suppose x_n+2=Px_n+1-Qx_n over a commutative ring. Define V_0=2, V_1=P, and V_n+2=PV_n+1-QV_n. For every positive integer N and every address j, the subsequence u_k=x_Nk+j satisfies u_k+2=V_Nu_k+1-Q^Nu_k.

    Exact mathematical passage
    
    V_{n+2}=PV_{n+1}-QV_n.
    
v_{n+1}Use in source · 4 passages

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V_{n+2}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2861

    [Residue subsequences of a second order recurrence] Suppose x_n+2=Px_n+1-Qx_n over a commutative ring. Define V_0=2, V_1=P, and V_n+2=PV_n+1-QV_n. For every positive integer N and every address j, the subsequence u_k=x_Nk+j satisfies u_k+2=V_Nu_k+1-Q^Nu_k.

    Exact mathematical passage
    
    V_{n+2}=PV_{n+1}-QV_n.
    
v_{Nk+j}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2786

    Let R be a commutative ring. Let v_n+1=Cv_n, C∈ M_d(R). For a positive integer N and an address j∈{0,…,N-1}, define the residue state packet u_k^(j)=v_Nk+j.

    Exact mathematical passage
    
    u_k^{(j)}=v_{Nk+j}.
    
V_{nn}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 809

    egintheorem[Classical odd-length reconstruction] For every positive odd integer p and every n∈ N, E_2n=∑_m=0^nbinom2n2mγ_m p^2mS_2(n-m)(p). Equivalently, E_2n=∑_q=0^n V_nqS_2q(p) for the lower triangular matrix V_nq=binom2n2qγ_n-qp^2(n-q) (q≤ n), V_nq=0 (q>n). It has V_nn=1 and p^2(n-q)mid V_nq when q≤ n.

    Exact mathematical passage
    V_{nn}=1
V_{nq}Use in source · 3 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 803

    [Classical odd-length reconstruction] For every positive odd integer p and every n∈ N, E_2n=∑_m=0^nbinom2n2mγ_m p^2mS_2(n-m)(p). Equivalently, E_2n=∑_q=0^n V_nqS_2q(p) for the lower triangular matrix V_nq=binom2n2qγ_n-qp^2(n-q) (q≤ n), V_nq=0 (q>n). It has V_nn=1 and p^2(n-q)mid V_nq when q≤ n.

    Exact mathematical passage
    E_{2n}=\sum_{q=0}^n V_{nq}S_{2q}(p)
\vdotsUse in source · 1 passage

Search aliases: vdots

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  • Use in source · Read passage · 72_recovered.tex, line 2071

    [Polyphase shift matrix] Let E be the scalar shift (Ea)_n=a_n+1, and let E_q shift the coarse index. Then P_M(Ea) = S_M(E_q) P_Ma, where S_M(E_q) = 0 · 1 · 0 · ⋯ · 0 0 · 0 · 1 · ⋯ · 0 vdots · · · ddots · vdots 0 · 0 · 0 · ⋯ · 1 E_q · 0 · 0 · ⋯ · 0 . It also satisfies S_M(E_q)^M=E_qI_M.

    Exact mathematical passage
    
    \mathsf S_M(E_q)
    =
    \begin{pmatrix}
    0&1&0&\cdots&0\\
    0&0&1&\cdots&0\\
    \vdots&&&\ddots&\vdots\\
    0&0&0&\cdots&1\\
    E_q&0&0&\cdots&0
    \end{pmatrix}.
    
\vpUse in source · 20 passages

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  • Use in source · Read passage · 72_recovered.tex, line 553

    w(r)<L binom nr c_rp^w(r)S_n-r pmod p^L. When the omitted set is nonempty, the tail has valuation at least min_substack0≤ r≤ n w(r)≥ L ( w(r)+v_pbinom nr+v_p(S_n-r) ).

    Exact mathematical passage
    
    \boxed{
    \min_{\substack{0\leq r\leq n\\w(r)\geq L}}
    \left(
    w(r)+\vp\binom nr+\vp(S_{n-r})
    \right).
    }
    
\vTwoUse in source · 17 passages

Search aliases: vTwo v_2

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  • Use in source · Read passage · 72_recovered.tex, line 90

    The dyadic kernel is substantially sharper than the separate coefficient bound alone suggests. If fracz^2sinh^2z = ∑_r≥0β_rz^2r and κ_2,r=β_r2^2r(2r)!, then v_2(κ_2,r)=4r-1 (r≥1). Consequently, truncation after depth M is uniformly valid modulo 2^4M+3, and this exponent is optimal. The rowwise estimate records the additional binary carries in the binomial coefficient. Classical congruence theory for Euler families is extensive [source reference]. The formulas here are stated as exact coordinate …

    Exact mathematical passage
    
    \vTwo(\kappa_{2,r})=4r-1
    \qquad(r\geq1).
    
\mathcal WUse in source · 9 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 231

    At the singular addresses c=ap, 1≤ a<m, use the weighted ring W_p={∑_d≥0a_dT^d: p^lfloor d/prfloora_d∈ Z_p}. The key estimate is R_ap(log(1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!)…

    Exact mathematical passage
    
     \mathcal W_p=\left\{\sum_{d\ge0}a_dT^d:
              p^{\lfloor d/p\rfloor}a_d\in\Z_p\right\}.
    
wUse in source · 71 passages

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  • Use in source · Read passage · 10_foundations.tex, line 12

    The original arithmetic seed notation S=(A,a_0, O, R,{P_N,j},w,τ_p) records a coefficient ring or module, a seed, allowed operators, relations, residue selectors, a depth weight, and a topology. This notation describes the ingredients of the construction; a category of seed presentations and a notion of equivalence still require separate definitions. We therefore state each theorem for a concrete object. Section~[source reference] discusses the additional structure needed for the original native…

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    \mathcal S=(A,a_0,\mathcal O,\mathcal R,\{P_{N,j}\},w,\tau_p)
    
W'Use in source · 2 passages

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  • Use in source · Read passage · root_scope.tex, line 403

    athrmact). The equality includes the zero case. The same successive monic polynomial division used in Theorem~[source reference], started at precision eight, preserves the quadratic and unit reductions of [source reference]. Write W=Z^2+2β_rZ+α_r and w=W(c_r). Stationarity gives W'(c_r)=-τ_rw, while w=Φ_r^act/ U_r(c_r). The quadratic identity β_r^2-α_r=W'(c_r)^2/4-W(c_r) proves the formula. Since τ_r∈256 Z_2, its bracket is a unit.

    Exact mathematical passage
    W'(c_r)=-\tau_rw
W^kUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 376

    For every starting row, Δ_1^k C_0(n) = L(U^4nW^k) =∑_s=0^nbinom ns L(W^k+s). Each summand lies in 2^k k! Z_2, proving the factorial bound for address zero from row zero onward.

    Exact mathematical passage
    
     \Delta_1^k C_0(n)
       =\mathcal L(U^{4n}W^k)
       =\sum_{s=0}^n\binom ns\mathcal L(W^{k+s}).
     \label{eq:base-row-finite-expansion}
    
W^{k+s}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 376

    For every starting row, Δ_1^k C_0(n) = L(U^4nW^k) =∑_s=0^nbinom ns L(W^k+s). Each summand lies in 2^k k! Z_2, proving the factorial bound for address zero from row zero onward.

    Exact mathematical passage
    
     \Delta_1^k C_0(n)
       =\mathcal L(U^{4n}W^k)
       =\sum_{s=0}^n\binom ns\mathcal L(W^{k+s}).
     \label{eq:base-row-finite-expansion}
    
W^mUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 366

    he first scalar counts partitions into pairs and is an integer. For the last term of H_0, multiply the third identity by (i E F)^m and reduce the Euler powers using the integral quadratic relations. Finally, γ_k(X+Y)=∑_r=0^kγ_r(X)γ_k-r(Y) proves γ_k(H_0)∈ A_0. Applying the functional and putting W=U^4-1=2H_0 gives L(W^m)∈2^m m!R (m≥0). These moments are rational. The identity R∩ Q_2= Z_2, obtained from the linear independence of 1,i over Q_2, descends every such divisibility statement to Z_2. No…

    Exact mathematical passage
    
     \mathcal L(W^m)\in2^m m!R\qquad(m\geq0).
     \label{eq:euler-W-factorial}
    
w^pUse in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 116

    Uniqueness requires a lifting argument. The source proves the actual kernel statement bar x=0 Longleftrightarrow pmid x in Z/p^ℓ Z. For x≡ ypmod p, the checked ideal-power lifting theorem gives x^p^ℓ-1-y^p^ℓ-1∈(p)^ℓ=(0). If w^p-1=1, then w^p=w and hence w^p^k=w for every k. Therefore two roots with the same residue are equal. The decomposition is unique whenever its second factor reduces to one. Multiplicativity of both factors is also proved.

    Exact mathematical passage
    w^p=w
w^{p-1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 116

    Uniqueness requires a lifting argument. The source proves the actual kernel statement bar x=0 Longleftrightarrow pmid x in Z/p^ℓ Z. For x≡ ypmod p, the checked ideal-power lifting theorem gives x^p^ℓ-1-y^p^ℓ-1∈(p)^ℓ=(0). If w^p-1=1, then w^p=w and hence w^p^k=w for every k. Therefore two roots with the same residue are equal. The decomposition is unique whenever its second factor reduces to one. Multiplicativity of both factors is also proved.

    Exact mathematical passage
    w^{p-1}=1
w^{p^k}Use in source · 1 passage

Search aliases: w^p^k

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  • Use in source · Read passage · supplement_05.tex, line 116

    Uniqueness requires a lifting argument. The source proves the actual kernel statement bar x=0 Longleftrightarrow pmid x in Z/p^ℓ Z. For x≡ ypmod p, the checked ideal-power lifting theorem gives x^p^ℓ-1-y^p^ℓ-1∈(p)^ℓ=(0). If w^p-1=1, then w^p=w and hence w^p^k=w for every k. Therefore two roots with the same residue are equal. The decomposition is unique whenever its second factor reduces to one. Multiplicativity of both factors is also proved.

    Exact mathematical passage
    w^{p^k}=w
w^qUse in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 486

    The recovered Annals text develops a broader polyphase vocabulary. We explain it through elementary identities with explicit algebraic derivations. These identities have not been separately formalized in Lean for this consolidation. For a sequence a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplicati…

    Exact mathematical passage
    A_r(w)=\sum_q a_{Mq+r}w^q
w^TUse in source · 3 passages

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W_1Use in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 175

    For preparation, start from the monic polynomial g=overline G_r and its zero-one lift W_1, with U_1=1. Suppose G_r-W_kU_k is coefficientwise divisible by 2^k. Its divided reduction is a finite polynomial because G_r is restricted. Divide it by g in F_2[X]: overline(G_r-W_kU_k)/2^k=gq+s, deg s<d_r. With zero-one coefficient lifts, set W_k+1=W_k+2^ks and U_k+1=U_k+2^kq. The error is now divisible by 2^k+1, the polynomial remains monic of degree d_r, and the unit reduces to one. The sequences conve…

    Exact mathematical passage
    W_1
W_2Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1381

    egative index are omitted. An empty tail is exactly zero, and the minimum over an empty set of term valuations is +∞. These conventions include M≥ n/2. For every r≥1, v_2(κ_2,r)=4r-1. The rth reconstruction term has exact valuation v_2( κ_2,rbinom n2rR_n-2r ) = 4r-1+v_2binom n2r. Equivalently, W_2(n,r) = 4r-1+s_2(r)+s_2(n-2r)-s_2(n).

    Exact mathematical passage
    
    \boxed{
    W_2(n,r)
    =
    4r-1+\sTwo(r)+\sTwo(n-2r)-\sTwo(n).
    }
    
W_kUse in source · 2 passages

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  • Use in source · Read passage · root_scope.tex, line 176

    For preparation, start from the monic polynomial g=overline G_r and its zero-one lift W_1, with U_1=1. Suppose G_r-W_kU_k is coefficientwise divisible by 2^k. Its divided reduction is a finite polynomial because G_r is restricted. Divide it by g in F_2[X]: overline(G_r-W_kU_k)/2^k=gq+s, deg s<d_r. With zero-one coefficient lifts, set W_k+1=W_k+2^ks and U_k+1=U_k+2^kq. The error is now divisible by 2^k+1, the polynomial remains monic of degree d_r, and the unit reduces to one. The sequences conve…

    Exact mathematical passage
    G_r-W_kU_k
W_{k+1}Use in source · 1 passage

Search aliases: W_k+1

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  • Use in source · Read passage · root_scope.tex, line 183

    For preparation, start from the monic polynomial g=overline G_r and its zero-one lift W_1, with U_1=1. Suppose G_r-W_kU_k is coefficientwise divisible by 2^k. Its divided reduction is a finite polynomial because G_r is restricted. Divide it by g in F_2[X]: overline(G_r-W_kU_k)/2^k=gq+s, deg s<d_r. With zero-one coefficient lifts, set W_k+1=W_k+2^ks and U_k+1=U_k+2^kq. The error is now divisible by 2^k+1, the polynomial remains monic of degree d_r, and the unit reduces to one. The sequences conve…

    Exact mathematical passage
    W_{k+1}=W_k+2^ks
W_pUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1029

    [Odd row weight] For an odd prime p, the rth nonzero term in the reconstruction has valuation W_p(n,r) = 2r+v_p(Z_2r+1) +v_pbinom n2r +v_p(S_n-2r(p)). The binomial contribution is v_pbinom n2r = fracs_p(2r)+s_p(n-2r)-s_p(n)p-1, where s_p is the base p digit sum.

    Exact mathematical passage
    
    \boxed{
    W_p(n,r)
    =
    2r+\vp(Z_{2r+1})
    +\vp\binom n{2r}
    +\vp\bigl(S_{n-2r}(p)\bigr).
    }
    
\mathcal W_pUse in source · 7 passages

Search aliases: W_p mathcal

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  • Use in source · Read passage · 20_arithmetic.tex, line 231

    At the singular addresses c=ap, 1≤ a<m, use the weighted ring W_p={∑_d≥0a_dT^d: p^lfloor d/prfloora_d∈ Z_p}. The key estimate is R_ap(log(1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!)…

    Exact mathematical passage
    
     \mathcal W_p=\left\{\sum_{d\ge0}a_dT^d:
              p^{\lfloor d/p\rfloor}a_d\in\Z_p\right\}.
    
W_rUse in source · 6 passages

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  • Use in source · Read passage · root_scope.tex, line 156

    [Exact classical tail valuation] For every r≥1 and N≥ r, v_2( T_r-1(N)) =4r-1+v_2binom Nr+μ_r+v_2G_r(N-r). There are unique factors G_r(X)=W_r(X)U_r(X), where W_r∈ Z_2[X] is monic of degree d_r=max J_r and U_r∈ Z_2⟨ X⟩^× satisfies U_r≡1pmod2. Consequently v_2G_r(x)=v_2W_r(x) for every x∈ Z_2, and deg W_r≤lfloorfracs_2(r)-13rfloor. Equation~[source reference], the exact weight in [source reference], and the digit-sum identity v_2binom2N2r=v_2binom Nr prove [source reference]. The binomial coeffic…

    Exact mathematical passage
    
     G_r(X)=W_r(X)U_r(X),
    
\mathcal W_rUse in source · 2 passages

Search aliases: W_r mathcal

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  • Use in source · Read passage · root_scope.tex, line 382

    [Exact quadratic compression] There is a unique preparation K_r(Z)= U_r(Z) W_r(Z), W_r(Z)=Z^2+2β_rZ+α_r, U_r≡ u_rpmod256. With δ_r=β_r^2-α_r and τ_r= U_r'(c_r)/ U_r(c_r), δ_r=-fracΦ_r^act U_r(c_r) (1-fracτ_r^2Φ_r^act4 U_r(c_r)), v_2(δ_r)=v_2(Φ_r^act). The equality includes the zero case. The same successive monic polynomial division used in Theorem~[source reference], started at precision eight, preserves the quadratic and unit reductions of [source reference]. Write W=Z^2+2β_rZ+α_r and w=W(c_r)…

    Exact mathematical passage
    
     K_r(Z)=\mathcal U_r(Z)\mathcal W_r(Z),\qquad
     \mathcal W_r(Z)=Z^2+2\beta_rZ+\alpha_r,\qquad
     \mathcal U_r\equiv u_r\pmod{256}.
    
\widehat CUse in source · 3 passages

Search aliases: widehat C widehat

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  • Use in source · Read passage · supplement_07.tex, line 119

    pageheading3. Actual packet congruences and sharp boundariesConsequences for the standard window; not the full Theorem 7.5 The fifth block constructed integral lifts widehat C_mp,j(n)∈ Z_p of the original rational coefficients when 0≤ j<p, and proved that their residues equal e_n^(m,p).

    Exact mathematical passage
    \widehat C_{mp,j}(n)\in\Z_p
\widehat C_{mp,j}Use in source · 3 passages

Search aliases: widehat C_mp,j widehat

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  • Use in source · Read passage · supplement_07.tex, line 119

    pageheading3. Actual packet congruences and sharp boundariesConsequences for the standard window; not the full Theorem 7.5 The fifth block constructed integral lifts widehat C_mp,j(n)∈ Z_p of the original rational coefficients when 0≤ j<p, and proved that their residues equal e_n^(m,p).

    Exact mathematical passage
    \widehat C_{mp,j}(n)\in\Z_p
\widehat FUse in source · 3 passages

Search aliases: widehat F widehat

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  • Use in source · Read passage · 72_recovered.tex, line 1777

    [Amice pullback] For a formal exponential generating series F(z)=∑_n≥0a_nfracz^nn!, define widehat F(T)=F(log(1+T)).

    Exact mathematical passage
    
    \widehat F(T)=F(\log(1+T)).
    
\widehat PUse in source · 2 passages

Search aliases: widehat P widehat

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  • Use in source · Read passage · 10_foundations.tex, line 101

    cal E(h) and H^-1= E(k), and define U_Ha= D(H E(a)). Then U_H is an equivalence, with inverse U_H^-1a= D(H^-1 E(a)). Its coefficient formulas are (U_Ha)_n=∑_r=0^nbinom nr h_r a_n-r, (U_H^-1a)_n=∑_r=0^nbinom nr k_r a_n-r. The resulting triangular matrix has diagonal entries 1. The transported selectors widehat P_N,j=U_HP_N,jU_H^-1 are idempotent and pairwise orthogonal, and their sum is the identity.

    Exact mathematical passage
    
    \widehat P_{N,j}=U_HP_{N,j}U_H^{-1}
    
\widehat P_{N,j}Use in source · 2 passages

Search aliases: widehat P_N,j widehat

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  • Use in source · Read passage · 10_foundations.tex, line 101

    cal E(h) and H^-1= E(k), and define U_Ha= D(H E(a)). Then U_H is an equivalence, with inverse U_H^-1a= D(H^-1 E(a)). Its coefficient formulas are (U_Ha)_n=∑_r=0^nbinom nr h_r a_n-r, (U_H^-1a)_n=∑_r=0^nbinom nr k_r a_n-r. The resulting triangular matrix has diagonal entries 1. The transported selectors widehat P_N,j=U_HP_N,jU_H^-1 are idempotent and pairwise orthogonal, and their sum is the identity.

    Exact mathematical passage
    
    \widehat P_{N,j}=U_HP_{N,j}U_H^{-1}
    
\widehat uUse in source · 1 passage

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  • Use in source · Read passage · continuation_08.tex, line 46

    inite solved-block cokernel and its parameter-dependent class by C_x=S^e/imJ, o_x(v)=[bar r+Kv]∈ C_x. Then L_v≠∅iff o_x(v)=0. More precisely, the checked digit-to-lift equivalence is L_v≅{u∈ S^m:Ju=-bar r-Kv}. It sends a solved digit u to the actual next residue represented by x+q^L(widehat u,widehat v). The obstruction depends on the solved block J, not on the range of the full map (J K).

    Exact mathematical passage
    x+q^L(\widehat u,\widehat v)
\widehat vUse in source · 3 passages

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  • Use in source · Read passage · continuation_08.tex, line 22

    x=(y_0,t_0)∈ Z^m+k, F(x)≡0pmodq^L. The actual lower root supplies the unique integral residual r=q^-LF(x)∈ Z^e, bar r=rbmod q∈ S^e. Let the actual reduced Jacobian blocks be J=(D_YF)(x)bmod q:S^m→ S^e, K=(D_TF)(x)bmod q:S^k→ S^e. For a prescribed parameter digit v∈ S^k, choose its canonical integral representative widehat v. Define L_v to be the set of actual residue vectors z∈( Z/q^L+1 Z)^m+k satisfying z≡ xpmodq^L, F(z)≡0pmodq^L+1, z_T≡ t_0+q^Lwidehat vpmodq^L+1. The formal predicate uses cano…

    Exact mathematical passage
    \widehat v
\widehat{\mathbb Z}Use in source · 1 passage

Search aliases: widehat Z widehat mathbb

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  • Use in source · Read passage · 71_native.tex, line 219

    The reason is exact: the CRT ring isomorphism preserves polynomial evaluation, and zero in a finite product is equivalent to zero in every component. For F empty, passing to all compatible finite congruences identifies the profinite completion widehat Z with ∏_p Z_p. The same finite reconstruction is compatible with coefficient projections and integer recurrence matrices because these operations commute with reduction and CRT.

    Exact mathematical passage
    \widehat{\mathbb Z}
\widehat{F'}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1783

    [Derivative conjugacy] widehatF'(T) = (1+T)fracddTwidehat F(T).

    Exact mathematical passage
    
    \boxed{
    \widehat{F'}(T)
    =
    (1+T)\frac{d}{dT}\widehat F(T).
    }
    
\widehat\ZUse in source · 1 passage

Search aliases: widehatZ widehat Z widehat mathbb

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  • Use in source · Read passage · 50_open_questions.tex, line 149

    Finite CRT reconstruction leads to the familiar completed object widehat Z=∏_p Z_p, but does not identify which local tuples come from a single global integer. A complete form of Native 13.3 must therefore specify finite-level compatibility, transition maps and completion, and any additional global descent condition. If recurrence, spectral, or Frobenius information is to supply that condition, it must be expressed through defined maps or constraints on the local data, with a precise account of …

    Exact mathematical passage
    \widehat\Z=\prod_p\Z_p
\widetilde C_{6,3}Use in source · 1 passage

Search aliases: widetilde C_6,3 widetilde

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  • Use in source · Read passage · supplement_01.tex, line 180

    For the unnormalized convention, Δ_1widetilde C_6,3(1)=3! frac64455=frac384455, v_3 (frac384455)=1. The factorial's valuation is also proved: v_3(3!)=1, v_3(Δ_1C_6,3(1)) =1-v_3(3!)=0. This proves sharpness at the displayed normalization boundary. It does not establish the corrected global inequality for all parameters, and does not assert that every nonunit-factorial parameter tuple must violate every no-loss specialization. All valuation calculations here concern nonzero rationals.

    Exact mathematical passage
    
     \Delta_1\widetilde C_{6,3}(1)=3!\,\frac{64}{455}=\frac{384}{455},
     \qquad v_3\!\left(\frac{384}{455}\right)=1.
    
\widetilde C_{N,j}Use in source · 1 passage

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  • Use in source · Read passage · supplement_01.tex, line 168

    Principal targets. targetFiltered.padic_filtered_reconstruction targetStandardWindow.lucas_row_reduction targetStandardWindow.standard_diagonal_unit targetStandardWindow.factorial_valuation_zero pageheading4. Sharp normalization boundarySource item 13.1 The construction fixes the two coefficient conventions: C_N,j(n), widetilde C_N,j(n)=j! C_N,j(n). The witness uses p=3, m=2, N=mp=6, j=3, n=1 and difference step h=(p-1)/m=1. The parameter conditions are checked explicitly, including odd primalit…

    Exact mathematical passage
    
     C_{N,j}(n),\qquad \widetilde C_{N,j}(n)=j!\,C_{N,j}(n).
    
\widetilde eUse in source · 6 passages

Search aliases: widetilde e widetilde

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  • Use in source · Read passage · 72_recovered.tex, line 690

    Fix a prime p, an integer m≥1, and an address 0≤ j<p. Put N=mp. For arbitrary m, let widetilde e_n^(m,p)∈ F_p be the unique sequence satisfying ∑_q=0^n binommnmq widetilde e_q^(m,p) = δ_n0. When mmid p-1, this sequence agrees with the compressed Hurwitz inverse in [source reference].

    Exact mathematical passage
    \widetilde e_n^{(m,p)}\in\FF_p
\widetilde e_n^{(m,p)}Use in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 690

    Fix a prime p, an integer m≥1, and an address 0≤ j<p. Put N=mp. For arbitrary m, let widetilde e_n^(m,p)∈ F_p be the unique sequence satisfying ∑_q=0^n binommnmq widetilde e_q^(m,p) = δ_n0. When mmid p-1, this sequence agrees with the compressed Hurwitz inverse in [source reference].

    Exact mathematical passage
    \widetilde e_n^{(m,p)}\in\FF_p
\widetilde e_q^{(m,p)}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 691

    Fix a prime p, an integer m≥1, and an address 0≤ j<p. Put N=mp. For arbitrary m, let widetilde e_n^(m,p)∈ F_p be the unique sequence satisfying ∑_q=0^n binommnmq widetilde e_q^(m,p) = δ_n0. When mmid p-1, this sequence agrees with the compressed Hurwitz inverse in [source reference].

    Exact mathematical passage
    
    \boxed{
    \sum_{q=0}^{n}
    \binom{mn}{mq}
    \widetilde e_q^{(m,p)}
    =
    \delta_{n0}.
    }
    
\widetilde e_r^{(m,p)}Use in source · 1 passage

Search aliases: widetilde e_r^(m,p) widetilde

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  • Use in source · Read passage · 72_recovered.tex, line 805

    [Rational residual generating function] Under the hypotheses of [source reference], for every standard address j, ∑_n≥0 (C_mp,j(n)bmod p)z^n = 1+ frac ∑_r=1^hwidetilde e_r^(m,p)z^r 1-z^h.

    Exact mathematical passage
    
    \sum_{n\geq0}
    \left(C_{mp,j}(n)\bmod p\right)z^n
    =
    1+
    \frac{
    \sum_{r=1}^{h}\widetilde e_r^{(m,p)}z^r
    }{1-z^h}.
    
\widetilde tUse in source · 5 passages

Search aliases: widetilde t widetilde

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  • Use in source · Read passage · 30_lifting.tex, line 265

    Integer representatives and residue classes at the next level must also be distinguished. For q≥1 and L≥0, the map S_q^d⟶ {y∈( Z/q^L+1 Z)^d:y≡ xpmodq^L}, t⟼ x+q^Lwidetilde t is a bijection, where widetilde t is any integer representative. Independence of the representative and injectivity both follow by cancelling q^L in integer divisibility; surjectivity follows by expressing the lower congruence as a multiple of q^L. This is the equivalence leanKomorebi.ResidueLiftDigits.digitEquivResidue.

    Exact mathematical passage
    
     S_q^d\longrightarrow
     \{y\in(\Z/q^{L+1}\Z)^d:y\equiv x\pmod{q^L}\},
     \qquad t\longmapsto x+q^L\widetilde t
    
\widetilde UUse in source · 2 passages

Search aliases: widetilde U widetilde

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  • Use in source · Read passage · 20_arithmetic.tex, line 1091

    [Exact image of the original coefficients] The continuous extension of the original sequence is widetilde U_j(x)=j!+2^d_jΦ_j(x). It is injective, its image is exactly j!+2^d_j Z_2, and |widetilde U_j(x)-widetilde U_j(y)|_2 =2^-d_j|x-y|_2. Every point of that affine image has exactly one preimage in Z_2.

    Exact mathematical passage
    
     \widetilde U_j(x)=j!+2^{d_j}\Phi_j(x).
    
\widetilde uUse in source · 6 passages

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  • Use in source · Read passage · 30_lifting.tex, line 567

    A candidate for the next lift has the form x+q^L(widetilde u,widetilde v), with solved digit u∈ S_q^m. Its Taylor correction equation is overliner_x+J_xu+K_xv=0. Solvability is exactly vanishing of the displayed cokernel class. Restrict the residue-digit equivalence to the prescribed parameter coordinates. This identifies the lift fiber with the solved-digit fiber. Translation by ker J_x is free and transitive on any nonempty such fiber. The prime-field count follows from rank-nullity as before.

    Exact mathematical passage
    x+q^L(\widetilde u,\widetilde v)
\widetilde U_jUse in source · 2 passages

Search aliases: widetilde U_j widetilde

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  • Use in source · Read passage · 20_arithmetic.tex, line 1091

    [Exact image of the original coefficients] The continuous extension of the original sequence is widetilde U_j(x)=j!+2^d_jΦ_j(x). It is injective, its image is exactly j!+2^d_j Z_2, and |widetilde U_j(x)-widetilde U_j(y)|_2 =2^-d_j|x-y|_2. Every point of that affine image has exactly one preimage in Z_2.

    Exact mathematical passage
    
     \widetilde U_j(x)=j!+2^{d_j}\Phi_j(x).
    
\widetilde vUse in source · 2 passages

Search aliases: widetilde v widetilde

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  • Use in source · Read passage · 30_lifting.tex, line 545

    ixed in advance. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0), and assume F(x)=q^Lr_x with q≥1 and L≥1. Write the reduced Jacobian as DF(x)bmod q=(J_x K_x), J_x:S_q^m→ S_q^e, K_x:S_q^k→ S_q^e. For a prescribed parameter digit v∈ S_q^k, let L_x,L,v be the set of next residue roots over x whose parameter coordinates are t_0+q^Lwidetilde v modulo q^L+1.

    Exact mathematical passage
    t_0+q^L\widetilde v
\WpUse in source · 22 passages

Search aliases: Wp W_p mathcal

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  • Use in source · Read passage · supplement_09.tex, line 111

    Weighted membership. The class used in the formalization is W_p={F∈ Q_p[[T]]: p^lfloor d/prfloorcoefTdF∈ Z_p for every d}. The integral-coordinate estimate from page 2 now proves H_r(log(1+T))∈ W_p for every residue r, including the singular ones. Rational coefficients are mapped into Q_p by the actual coefficient ring homomorphism.

    Exact mathematical passage
    
     \Wp=\left\{F\in\Q_p[[T]]:
     p^{\lfloor d/p\rfloor}\coef{T}{d}F\in\Z_p\ \text{for every }d\right\}.
    
\mathrm{wt}Use in source · 1 passage

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  • Use in source · Read passage · 50_open_questions.tex, line 95

    The original proposed form is X≃ X_2^wt×_C∏_p>2'X_p. It leaves X, the distinguished subsets defining the restricted product, the maps to C, the ambient category, and the universal meaning of ``smallest'' unspecified. In particular, one must decide whether X is a global object or a completed object.

    Exact mathematical passage
    
    X\simeq X_2^{\mathrm{wt}}\times_C\prod_{p>2}'X_p.
    
x'Use in source · 11 passages

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  • Use in source · Read passage · 30_lifting.tex, line 133

    [Independence of the integer representative] Under the hypotheses of Theorem~[source reference], let x'=x+q^Ls and write F(x')=q^Lr_x'. Then J_x'=J_x, overliner_x'=overliner_x+J_xoverline s. Consequently the two residual classes coincide in the common quotient S_q^e/im J_x.

    Exact mathematical passage
    x'=x+q^Ls
x^{-1}Use in source · 1 passage

Search aliases: x^-1

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  • Use in source · Read passage · 72_recovered.tex, line 235

    [Finite inverse formula] If x has augmentation one and n=x-1, then x^-1 = ∑_r=0^p-1(-n)^r.

    Exact mathematical passage
    
    \boxed{
    x^{-1}
    =
    \sum_{r=0}^{p-1}(-n)^r.
    }
    
X^2Use in source · 17 passages

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  • Use in source · Read passage · 30_lifting.tex, line 126

    The affine statement is formalized by leanKomorebi.NativeAffine.finite_affine_obstruction. Positive depth is essential in the nonlinear theorem: for F(X)=X^2-1, q=2, L=0, and x=0, the Jacobian is zero and the residual class is nonzero, yet y=1 is a root modulo two in the same residue modulo one.

    Exact mathematical passage
    F(X)=X^2-1
x^2Use in source · 5 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2721

    [A singular polynomial obstruction] Let p=3 and consider the class x_1=0 modulo three. For F_0(x)=x^2, the residual and the reduced derivative both vanish. Every correction y∈ F_3 gives F_0(3y)≡0pmod9, so the class has three children. For F_1(x)=x^2+3, the reduced derivative is still zero, but the reduced residual is one. Its class in the cokernel is nonzero, so no lift modulo nine exists.

    Exact mathematical passage
    
    F_0(x)=x^2,
    
X^3Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2367

    [Matrix power amplification] Editorial hypothesis correction. Require L≥1. At L=0 the standalone claim is false: for p=3, d=1 and X=2=1+B, one has X^3-1=7∉3 Z_3. With L≥1, every higher binomial term in the retained proof has the required valuation, including p=2. If X=I+p^LB with B∈ M_d( Z_p), then for every a≥0 there is B_a∈ M_d( Z_p) such that X^p^a=I+p^L+aB_a.

    Exact mathematical passage
    X^3-1=7\notin3\Z_3
x^dUse in source · 4 passages

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  • Use in source · Read passage · strengthening.tex, line 478

    Here is the polynomial estimate that supplies one additional bit. For i≥1, put P_i(x)=4^ibinom4x4i. Its constant is zero, and its degree-d coefficient is [x^d]P_i(x)=frac4^i4^d s(4i,d)(4i)!, where s(m,d) is a signed Stirling number of the first kind. In each product contributing to s(4i,d), at most d-1 of the 2i-1 even integers in 1,…,4i-1 can be omitted. Thus v_2(s(4i,d))≥max(0,2i-d). Writing s_2(i) for the number of ones in the binary expansion of i, and using v_2((4i)!)=4i-s_2(i), gives v_2([…

    Exact mathematical passage
    
     [x^d]P_i(x)=\frac{4^i4^d s(4i,d)}{(4i)!},
     \label{eq:scaled-binomial-coefficients}
    
X^mUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 580

    Expand X^m=∑_r=0^mT(m,r)P_r(X). Since XP_r(X)=P_r+1(X)+r(r+1)P_r(X), these integer coefficients satisfy T(m+1,r)=T(m,r-1)+r(r+1)T(m,r), T(0,0)=1. Every r(r+1) is even. Induction, including the boundary indices, proves 2^m-rmid T(m,r) for 0≤ r≤ m. For m>0 the r=0 term vanishes. Each remaining summand of L(P^m/m!) has valuation at least m-r+2v_2(r!)-v_2(m!) =r+s_2(m)-2s_2(r)≥ s_2(m)-1≥0. Here r≥2s_2(r)-1 follows from r≥2^s_2(r)-1 and 2^u≥2u for u≥1. This proves integrality and the stronger bound.

    Exact mathematical passage
    X^m=\sum_{r=0}^mT(m,r)P_r(X)
x^NUse in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 54

    Now take the integral cyclic group ring R_N= Z[ Z/N Z], x=[1], x^N=1, and assume Nmid p(p-1). Thus (x^p-1)^p=1. Each genuine consecutive block ∏_i=0^p-1(x-(pb+i)) belongs to (p,x^p-1-1). This follows from the actual polynomial identity ∏_i∈ F_p(X-i)=X^p-X modulo p, not from a hypothesized valuation of the product.

    Exact mathematical passage
    
     R_N=\Z[\Z/N\Z],\qquad x=[1],\qquad x^N=1,
    
X^pUse in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 62

    Now take the integral cyclic group ring R_N= Z[ Z/N Z], x=[1], x^N=1, and assume Nmid p(p-1). Thus (x^p-1)^p=1. Each genuine consecutive block ∏_i=0^p-1(x-(pb+i)) belongs to (p,x^p-1-1). This follows from the actual polynomial identity ∏_i∈ F_p(X-i)=X^p-X modulo p, not from a hypothesized valuation of the product.

    Exact mathematical passage
    
     \prod_{i\in\F_p}(X-i)=X^p-X
    
x^{p-1}Use in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 237

    At the singular addresses c=ap, 1≤ a<m, use the weighted ring W_p={∑_d≥0a_dT^d: p^lfloor d/prfloora_d∈ Z_p}. The key estimate is R_ap(log(1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!)…

    Exact mathematical passage
    I=(p,x^{p-1}-1)
X^{p^a}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2373

    r Editorial hypothesis correction. Require L≥1. At L=0 the standalone claim is false: for p=3, d=1 and X=2=1+B, one has X^3-1=7∉3 Z_3. With L≥1, every higher binomial term in the retained proof has the required valuation, including p=2. If X=I+p^LB with B∈ M_d( Z_p), then for every a≥0 there is B_a∈ M_d( Z_p) such that X^p^a=I+p^L+aB_a.

    Exact mathematical passage
    
    \boxed{X^{p^a}=I+p^{L+a}B_a.}
    
x^{p^{\ell-1}}Use in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 113

    Uniqueness requires a lifting argument. The source proves the actual kernel statement bar x=0 Longleftrightarrow pmid x in Z/p^ℓ Z. For x≡ ypmod p, the checked ideal-power lifting theorem gives x^p^ℓ-1-y^p^ℓ-1∈(p)^ℓ=(0). If w^p-1=1, then w^p=w and hence w^p^k=w for every k. Therefore two roots with the same residue are equal. The decomposition is unique whenever its second factor reduces to one. Multiplicativity of both factors is also proved.

    Exact mathematical passage
    
     x^{p^{\ell-1}}-y^{p^{\ell-1}}\in(p)^\ell=(0).
    
X^rUse in source · 1 passage

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  • Use in source · Read passage · 10_foundations.tex, line 415

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

    Exact mathematical passage
    \chi_{C^N}(X)=X^d+\sum_{r<d}q_rX^r
X^{\underline q}Use in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 140

    ad J_r={q:λ_r(q)=μ_r}, G_r=2^-μ_rH_r. Since λ_r(0)=0 and λ_r(q)≥3q+1-s_2(r), every minimizing address satisfies 0≤ q≤lfloorfracs_2(r)-13rfloor. In particular μ_r≤0, and the minimum is found by a finite carry calculation. Reduction of the whole series gives overline G_r(X)=∑_q∈ J_rX^underline q in F_2[X]. Its highest degree is max J_r, with leading coefficient one. Thus the normalization is the actual Gauss normalization.

    Exact mathematical passage
    
     \overline G_r(X)=\sum_{q\in J_r}X^{\underline q}
           \quad\text{in }\mathbb F_2[X].
     \label{eq:root:reduction}
    
x^\ellUse in source · 1 passage

Search aliases: x^ℓ ell

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  • Use in source · Read passage · strengthening.tex, line 1245

    [Taylor coefficient bounds and radii] For j=0,1,2,3, the dyadic interpolation has an ordinary power series widetilde C_j(x)=1+∑_ℓ≥1d_j,ℓx^ℓ, v_2(d_j,ℓ)≥ℓ+β_j. Its Taylor series converges for |x|_2<2. At addresses zero and one, v_2(d_j,ℓ)=ℓ (ℓ≥1), R_0=R_1=2. At address two, v_2(d_2,ℓ)≥ max(2ℓ+lfloorfracℓ2rfloor, ℓ+3), R_2=2^5/2. The radius at address three is at least two; its exact radius is not asserted here. The stated exact radii at addresses zero, one, and two hold at every center in Z_2.

    Exact mathematical passage
    
     \widetilde C_j(x)=1+\sum_{\ell\geq1}d_{j,\ell}x^\ell,
     \qquad v_2(d_{j,\ell})\geq\ell+\beta_j.
    
x_0Use in source · 6 passages

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  • Use in source · Read passage · 10_foundations.tex, line 411

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

    Exact mathematical passage
    x_n=C^n x_0
x_0^{m-\ell}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 1290

    For j=0,1, the term m=ℓ in [source reference] has exact valuation ℓ, since s(ℓ,ℓ)=1 and the Newton valuation is exact. Every later term has valuation at least ℓ+1. Thus v_2(d_j,ℓ)=ℓ, and the coefficient radius is exactly two. At a center x_0∈ Z_2, the coefficient of y^ℓ in the translated series is ∑_m≥ℓd_j,mbinom mℓ x_0^m-ℓ. Its initial term again has exact valuation ℓ, and all later terms have at least one more bit. The exact radius is therefore two at every integral center.

    Exact mathematical passage
    
     \sum_{m\geq\ell}d_{j,m}\binom m\ell x_0^{m-\ell}.
     \label{eq:taylor-translation-coefficient}
    
x_1Use in source · 5 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2610

    0 · 0 . The class x_1=(0,0) solves Mx_1≡ bpmod3 for both b_1=(-3,0) and b_2=(0,-3). For b_1, the reduced residual is (1,0), which belongs to the image of overline M. There are three correction classes, given by y_1+y_2=2 pmod3. For b_2, the residual is (0,1), whose cokernel class is nonzero. No lift modulo nine exists because the second row of M remains zero.

    Exact mathematical passage
    x_1=(0,0)
x_2Use in source · 1 passage

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  • Use in source · worked_atlas.tex, line 73

    AtlasHeadA compatible root is built one digit at a timeatlas:smooth Take F(x)=x^2-7 over the 3-adic integers and start with x_1=1 modulo three. If F(x_L)=3^Lr_L, the next representative has the form x_L+1=x_L+3^Lt_L. Its correcting digit satisfies r_L+2x_Lt_L≡0pmod3. AtlasImagesmooth_tower For example, F(1)/3=-2 and 2x_1=2, so the next digit is one and x_2=4. Then F(4)/9=1, again giving digit one and x_3=13. At the next stage the residual is divisible by three, so digit zero is correct. A zero d…

    Exact mathematical passage
    x_2=4
X_2^{\mathrm{wt}}Use in source · 1 passage

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  • Use in source · Read passage · 50_open_questions.tex, line 95

    The original proposed form is X≃ X_2^wt×_C∏_p>2'X_p. It leaves X, the distinguished subsets defining the restricted product, the maps to C, the ambient category, and the universal meaning of ``smallest'' unspecified. In particular, one must decide whether X is a global object or a completed object.

    Exact mathematical passage
    
    X\simeq X_2^{\mathrm{wt}}\times_C\prod_{p>2}'X_p.
    
x_3Use in source · 1 passage

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  • Use in source · worked_atlas.tex, line 73

    t at a timeatlas:smooth Take F(x)=x^2-7 over the 3-adic integers and start with x_1=1 modulo three. If F(x_L)=3^Lr_L, the next representative has the form x_L+1=x_L+3^Lt_L. Its correcting digit satisfies r_L+2x_Lt_L≡0pmod3. AtlasImagesmooth_tower For example, F(1)/3=-2 and 2x_1=2, so the next digit is one and x_2=4. Then F(4)/9=1, again giving digit one and x_3=13. At the next stage the residual is divisible by three, so digit zero is correct. A zero digit is not a stalled construction.

    Exact mathematical passage
    x_3=13
x_AUse in source · 1 passage

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  • Use in source · worked_atlas.tex, line 100

    AtlasHeadA derivative weight can cost one digit of root precisionatlas:weighted Compare the linear equations F(x)=3x=0 and G_A(x)=3x+3^A=0, where A≥2. Their coefficient difference has 3-adic valuation A, but their roots differ by x_A-0=-3^A-1, v_3(x_A)=A-1. AtlasImageweighted_motion The derivative has valuation one. Solving the correction equation divides by three, so one digit of coefficient precision is lost in root displacement. The calculation is exact for every displayed A and also follows …

    Exact mathematical passage
    
     x_A-0=-3^{A-1},\qquad v_3(x_A)=A-1.
    
X_{e+k}Use in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 451

    [Continuous charts with a selected Jacobian minor] Let p be prime, L≥1, F∈ Z[X_1,…,X_e+k]^e, and x∈ Z^e+k satisfy F(x)∈ p^L Z^e. Suppose the full reduced Jacobian is surjective. There is a permutation σ of the coordinates such that, writing σ x=(y_0,t_0) with e solved coordinates and k parameter coordinates, the parameter ball T=t_0+p^L Z_p^k is homeomorphic to the entire root locus R={z∈ B_L(x):F(z)=0}. The inverse homeomorphism is the actual free-coordinate projection z↦free(σ z). If h:T→ R is…

    Exact mathematical passage
    F\in\Z[X_1,\ldots,X_{e+k}]^e
X_{e+q}Use in source · 1 passage

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  • Use in source · Read passage · continuation_05.tex, line 13

    Let p be prime, L≥1, e,q≥0, and F∈ Z[X_1,…,X_e+q]^e, x∈ Z^e+q, F(x)≡0pmodp^L. Assume the actual reduced Jacobian A_x=DF(x)bmod p: F_p^e+q⟶ F_p^e is surjective. All vector congruences are coordinatewise.

    Exact mathematical passage
    
    F\in\mathbb Z[X_1,\ldots,X_{e+q}]^e,\qquad
    x\in\mathbb Z^{e+q},\qquad F(x)\equiv0\pmod{p^L}.
    
X_iUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 58

    s have zero remainder, and remainders add. For the multiplication step, suppose the assertion is known for P and write D_P(U)=∑_j(∂_jP)(a)U_j. Multiplication of its expansion by a_i+cU_i gives constant term P(a)a_i, linear term c(a_iD_P(U)+P(a)U_i), and remainder c^2(Q(U)(a_i+cU_i)+D_P(U)U_i). The product rule identifies the linear term with the derivative of PX_i. The displayed remainder vanishes at U=0. This constructs the polynomial remainder before evaluation at u, even over rings where divi…

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    PX_i
X_jUse in source · 3 passages

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  • Use in source · Read passage · 40_regularity.tex, line 39

    a)v_j. For a finite system, its strict derivative is the continuous linear map represented by the matrix of evaluated formal partial derivatives. For the derivative statement, use polynomial induction on constants, addition, and multiplication by a variable. Constants have zero derivative. The sum rule agrees with the additivity of formal partial differentiation. For PX_j, the product rule gives d(PX_j)_a(v)=a_j dP_a(v)+P(a)v_j, which agrees with the evaluated formal product rule for every coord…

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    PX_j
x_jUse in source · 1 passage

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  • Use in source · Read passage · continuation_11.tex, line 53

    Proof chain. The scalar polynomial derivative bridge is proved by induction on actual multivariate polynomials: constants, sums, and multiplication by a variable. For a scalar polynomial P, its strict derivative is v⟼∑_j frac∂ P∂ x_j(a)v_j. Assembling the coordinates gives the strict derivative of the evaluated system as its actual Jacobian matrix. Polynomial analyticity is supplied by the pinned Mathlib theorem for multivariate polynomial evaluation, followed by finite coordinate assembly.

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    v\longmapsto\sum_j \frac{\partial P}{\partial x_j}(a)v_j.
    
x_{L+1}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 2552

    [One level linear lifting] A class represented by x_L lifts to a solution modulo p^L+1 if and only if Ob^lin_L(x_L)=0. More precisely, a correction y∈ F_p^d gives a lift x_L+1=x_L+p^Ly pmod p^L+1 if and only if overline My=-ρ_L(x_L). When one correction exists, the set of all correction classes is a coset of keroverline M.

    Exact mathematical passage
    
    x_{L+1}=x_L+p^Ly
    \pmod {p^{L+1}}
    
X_nUse in source · 1 passage

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  • Use in source · Read passage · 40_regularity.tex, line 27

    [Derivative of polynomial evaluation] For P∈ K[X_1,…,X_n], polynomial evaluation is analytic at every point. At a∈ K^n its strict Fr echet derivative is v⟼∑_j=1^n (∂_jP)(a)v_j. For a finite system, its strict derivative is the continuous linear map represented by the matrix of evaluated formal partial derivatives. For the derivative statement, use polynomial induction on constants, addition, and multiplication by a variable. Constants have zero derivative. The sum rule agrees with the additivity…

    Exact mathematical passage
    P\in\mathbb K[X_1,\ldots,X_n]
x_nUse in source · 3 passages

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  • Use in source · Read passage · 10_foundations.tex, line 411

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

    Exact mathematical passage
    x_n=C^n x_0
x_{n+1}Use in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 436

    begingroupsmallraggedright Formal correspondence. leanCompanion.companion_jump, leanCompanion.packet_charpoly_recurrence, and leanCompanion.packet_observable_recurrence. The source-sequence versions derive the packet recurrence from the relation x_n+1=Cx_n. For the Fibonacci matrix C=(1 · 1

    Exact mathematical passage
    x_{n+1}=Cx_n
x_{n+2}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2857

    [Residue subsequences of a second order recurrence] Suppose x_n+2=Px_n+1-Qx_n over a commutative ring. Define V_0=2, V_1=P, and V_n+2=PV_n+1-QV_n. For every positive integer N and every address j, the subsequence u_k=x_Nk+j satisfies u_k+2=V_Nu_k+1-Q^Nu_k.

    Exact mathematical passage
    
    x_{n+2}=Px_{n+1}-Qx_n
    
x_{Nk+j}Use in source · 2 passages

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  • Use in source · Read passage · 10_foundations.tex, line 411

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

    Exact mathematical passage
    b_k=x_{Nk+j}
X_pUse in source · 2 passages

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  • Use in source · Read passage · 50_open_questions.tex, line 95

    The original proposed form is X≃ X_2^wt×_C∏_p>2'X_p. It leaves X, the distinguished subsets defining the restricted product, the maps to C, the ambient category, and the universal meaning of ``smallest'' unspecified. In particular, one must decide whether X is a global object or a completed object.

    Exact mathematical passage
    
    X\simeq X_2^{\mathrm{wt}}\times_C\prod_{p>2}'X_p.
    
\xrightarrowUse in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 106

    The formal theorem leanKomorebi.PolynomialObstruction.finite_polynomial_obstruction presents the image using the integer-domain map Z^dxrightarrowDF(x) Z^e→ S_q^e. Reduction onto S_q^d identifies that image with im J_x. The class-preserving comparison is explicit in leanKomorebi.PolynomialLiftCounts.obstructionEquiv; the change of presentation does not change the distinguished residual class.

    Exact mathematical passage
    \Z^d\xrightarrow{DF(x)}\Z^e\to S_q^e
Y^2Use in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 904

    The error exponent cannot uniformly be increased. For F(Y,T)=Y+Y^2+T, H=0, a=0, p=2, s=0, M=1, and v=1, there is an even root y of y+y^2+2=0. Its equation forces y≡2pmod4 and y+2=-y^2, so 4mid y+2, 8nmid y+2. Thus the error depth 2M-3s=2 is attained even for parameter-only motion. The same quadratic with a constant coefficient perturbation gives the coefficient-only boundary. These examples refute a uniform extra error digit; they do not assert optimality for every individual choice of p,s,M.

    Exact mathematical passage
    F(Y,T)=Y+Y^2+T
y^2Use in source · 6 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 297

    At a nonsingular address 2≤ c<N with pnmid c, differentiation of the section identities gives tR_c'=R_c(1+R_c-1-R_c). Writing q=R_c-1-(c-1) and y=R_c-c, this becomes (tD+c)y=cq+yq-y^2. On degrees divisible by p, the inverse of tD+c divides the degree-d coefficient by the unit d+c. Starting with y_0=0, form y_s+1=(tD+c)^-1(cq+y_sq-y_s^2). Flatness and support are preserved. Indeed, put E_c=tD+c and C_p=pD^p^2-D^p. The commutator is [B_p,E_c]=C_p, and iteration gives B_p^rE_cy=E_cB_p^ry+rC_pB_p^r-…

    Exact mathematical passage
    
     (tD+c)y=cq+yq-y^2.
    
y^kUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1641

    [Polynomial exponential Mahler transfer] Let f(n)=P(n)u^n with P∈ Z_p[n] of degree d. Let H=Λ_p,L, fix an address j, and write Q_j(x)=P(Hx+j), y=u^H, q=y-1. Then f(Hx+j)=∑_r≥0c_j,rbinom xr, where c_j,r = u^j ∑_k=0^min(d,r) binom rk Δ^kQ_j(0)y^kq^r-k. Consequently, v_p(c_j,r)≥ Lmax(r-d,0). For R≥ d, truncation at depth R is valid modulo p^L(R+1-d) uniformly on Z_p.

    Exact mathematical passage
    
    \boxed{
    c_{j,r}
    =
    u^j
    \sum_{k=0}^{\min(d,r)}
    \binom rk
    \Delta^kQ_j(0)y^kq^{r-k}.
    }
    
y^mUse in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 846

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    y^m=\chi
y^NUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2828

    [Characteristic polynomial of the jump] Let χ_C(y)=det(yI-C). Then χ_C^N(x) = Res_y(χ_C(y),x-y^N). The identity holds over the coefficient ring after interpreting the resultant universally.

    Exact mathematical passage
    
    \boxed{
    \chi_{C^N}(x)
    =
    \Res_y\bigl(\chi_C(y),x-y^N\bigr).
    }
    
y^pUse in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 223

    Choose a generator g of the cyclic additive group. Then F_p[ F_p^+] ≅ F_p[y]/(y^p-1) = F_p[y]/(y-1)^p. The augmentation ideal is generated by y-1, so its pth power is zero. Every element has the form a+n with a∈ F_p and n∈ J. It is invertible exactly when a≠0, which is the augmentation condition.

    Exact mathematical passage
    
    \FF_p[\FF_p^+]
    \cong
    \FF_p[y]/(y^p-1)
    =
    \FF_p[y]/(y-1)^p.
    
y^{p^{\ell-1}}Use in source · 1 passage

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  • Use in source · Read passage · supplement_05.tex, line 113

    Uniqueness requires a lifting argument. The source proves the actual kernel statement bar x=0 Longleftrightarrow pmid x in Z/p^ℓ Z. For x≡ ypmod p, the checked ideal-power lifting theorem gives x^p^ℓ-1-y^p^ℓ-1∈(p)^ℓ=(0). If w^p-1=1, then w^p=w and hence w^p^k=w for every k. Therefore two roots with the same residue are equal. The decomposition is unique whenever its second factor reduces to one. Multiplicativity of both factors is also proved.

    Exact mathematical passage
    
     x^{p^{\ell-1}}-y^{p^{\ell-1}}\in(p)^\ell=(0).
    
y^\ellUse in source · 1 passage

Search aliases: y^ℓ ell

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  • Use in source · Read passage · strengthening.tex, line 1288

    For j=0,1, the term m=ℓ in [source reference] has exact valuation ℓ, since s(ℓ,ℓ)=1 and the Newton valuation is exact. Every later term has valuation at least ℓ+1. Thus v_2(d_j,ℓ)=ℓ, and the coefficient radius is exactly two. At a center x_0∈ Z_2, the coefficient of y^ℓ in the translated series is ∑_m≥ℓd_j,mbinom mℓ x_0^m-ℓ. Its initial term again has exact valuation ℓ, and all later terms have at least one more bit. The exact radius is therefore two at every integral center.

    Exact mathematical passage
    y^\ell
y_1Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2621

    0 · 0 . The class x_1=(0,0) solves Mx_1≡ bpmod3 for both b_1=(-3,0) and b_2=(0,-3). For b_1, the reduced residual is (1,0), which belongs to the image of overline M. There are three correction classes, given by y_1+y_2=2 pmod3. For b_2, the residual is (0,1), whose cokernel class is nonzero. No lift modulo nine exists because the second row of M remains zero.

    Exact mathematical passage
    
    y_1+y_2=2
    \pmod3.
    
Y_2Use in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 673

    nomialPrescribedPadicTower.exists_prescribed_padic_root_with_tower and leanKomorebi.PolynomialPrescribedPadicTower.exists_unique_prescribed_padic_root. The separate supplied-path theorem leanKomorebi.PolynomialPrescribedPadicTower.exists_root_with_prescribed_path accepts any compatible integer parameter path with limit τ. For m>e, endpoint uniqueness is false in general: the system F(Y_1,Y_2,T)=Y_1 has two distinct roots (0,0,0) and (0,2,0) in the same dyadic ball with identical parameter value.

    Exact mathematical passage
    F(Y_1,Y_2,T)=Y_1
y_2Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2621

    0 · 0 . The class x_1=(0,0) solves Mx_1≡ bpmod3 for both b_1=(-3,0) and b_2=(0,-3). For b_1, the reduced residual is (1,0), which belongs to the image of overline M. There are three correction classes, given by y_1+y_2=2 pmod3. For b_2, the residual is (0,1), whose cokernel class is nonzero. No lift modulo nine exists because the second row of M remains zero.

    Exact mathematical passage
    
    y_1+y_2=2
    \pmod3.
    
Y_mUse in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 534

    Suppose now that the split into solved variables and parameter variables is fixed in advance. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0), and assume F(x)=q^Lr_x with q≥1 and L≥1. Write the reduced Jacobian as DF(x)bmod q=(J_x K_x), J_x:S_q^m→ S_q^e, K_x:S_q^k→ S_q^e. For a prescribed parameter digit v∈ S_q^k, let L_x,L,v be the set of next residue roots over x whose parameter coordinates are t_0+q^Lwidetilde v modulo q^L+1.

    Exact mathematical passage
    
     F\in\Z[Y_1,\ldots,Y_m,T_1,\ldots,T_k]^e,\qquad x=(y_0,t_0),
    
y_nUse in source · 1 passage

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  • Use in source · visual_body.tex, line 370

    [ht] centering includegraphics[width=linewidth]../figures/residue_permutations.pdf captionExact residue permutations modulo 256. The vertical coordinate is y_n=(E_4n^(4,j)-j!)/2^B_j modulo P_j(8). Each panel contains one full least period; every vertical residue occurs once.

    Exact mathematical passage
    y_n=(E_{4n}^{(4,j)}-j!)/2^{B_j}
y_rUse in source · 2 passages

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  • Use in source · Read passage · supplement_10.tex, line 153

    Now define explicit iterations y_0=0, y_r+1=Λ_c^-1(cq+y_rq-y_r^2). If q is flat and p-supported, every iterate has those properties. The crucial link to the actual ratio is an exact finite-prefix theorem: [t^d]y_r=[t^d]y (d≤ r). Indeed, (cq+zq-z^2)-(cq+yq-y^2)=(z-y)(q-z-y). The second factor has zero constant term. Multiplication therefore raises the established agreement prefix by one; coefficientwise division by d+c preserves that prefix.

    Exact mathematical passage
    
     y_0=0,\qquad y_{r+1}=\Lambda_c^{-1}(cq+y_rq-y_r^2).
    
y_{r+1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_10.tex, line 153

    Now define explicit iterations y_0=0, y_r+1=Λ_c^-1(cq+y_rq-y_r^2). If q is flat and p-supported, every iterate has those properties. The crucial link to the actual ratio is an exact finite-prefix theorem: [t^d]y_r=[t^d]y (d≤ r). Indeed, (cq+zq-z^2)-(cq+yq-y^2)=(z-y)(q-z-y). The second factor has zero constant term. Multiplication therefore raises the established agreement prefix by one; coefficientwise division by d+c preserves that prefix.

    Exact mathematical passage
    
     y_0=0,\qquad y_{r+1}=\Lambda_c^{-1}(cq+y_rq-y_r^2).
    
y_r^2Use in source · 1 passage

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  • Use in source · Read passage · supplement_10.tex, line 153

    Now define explicit iterations y_0=0, y_r+1=Λ_c^-1(cq+y_rq-y_r^2). If q is flat and p-supported, every iterate has those properties. The crucial link to the actual ratio is an exact finite-prefix theorem: [t^d]y_r=[t^d]y (d≤ r). Indeed, (cq+zq-z^2)-(cq+yq-y^2)=(z-y)(q-z-y). The second factor has zero constant term. Multiplication therefore raises the established agreement prefix by one; coefficientwise division by d+c preserves that prefix.

    Exact mathematical passage
    
     y_0=0,\qquad y_{r+1}=\Lambda_c^{-1}(cq+y_rq-y_r^2).
    
y_sUse in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 302

    At a nonsingular address 2≤ c<N with pnmid c, differentiation of the section identities gives tR_c'=R_c(1+R_c-1-R_c). Writing q=R_c-1-(c-1) and y=R_c-c, this becomes (tD+c)y=cq+yq-y^2. On degrees divisible by p, the inverse of tD+c divides the degree-d coefficient by the unit d+c. Starting with y_0=0, form y_s+1=(tD+c)^-1(cq+y_sq-y_s^2). Flatness and support are preserved. Indeed, put E_c=tD+c and C_p=pD^p^2-D^p. The commutator is [B_p,E_c]=C_p, and iteration gives B_p^rE_cy=E_cB_p^ry+rC_pB_p^r-…

    Exact mathematical passage
    
     y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).
    
y_{s+1}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 302

    At a nonsingular address 2≤ c<N with pnmid c, differentiation of the section identities gives tR_c'=R_c(1+R_c-1-R_c). Writing q=R_c-1-(c-1) and y=R_c-c, this becomes (tD+c)y=cq+yq-y^2. On degrees divisible by p, the inverse of tD+c divides the degree-d coefficient by the unit d+c. Starting with y_0=0, form y_s+1=(tD+c)^-1(cq+y_sq-y_s^2). Flatness and support are preserved. Indeed, put E_c=tD+c and C_p=pD^p^2-D^p. The commutator is [B_p,E_c]=C_p, and iteration gives B_p^rE_cy=E_cB_p^ry+rC_pB_p^r-…

    Exact mathematical passage
    
     y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).
    
y_s^2Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 302

    At a nonsingular address 2≤ c<N with pnmid c, differentiation of the section identities gives tR_c'=R_c(1+R_c-1-R_c). Writing q=R_c-1-(c-1) and y=R_c-c, this becomes (tD+c)y=cq+yq-y^2. On degrees divisible by p, the inverse of tD+c divides the degree-d coefficient by the unit d+c. Starting with y_0=0, form y_s+1=(tD+c)^-1(cq+y_sq-y_s^2). Flatness and support are preserved. Indeed, put E_c=tD+c and C_p=pD^p^2-D^p. The commutator is [B_p,E_c]=C_p, and iteration gives B_p^rE_cy=E_cB_p^ry+rC_pB_p^r-…

    Exact mathematical passage
    
     y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).
    
ZUse in source · 35 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 781

    The same inverse coefficients can be written in the signed zigzag notation of the original manuscript. Define Z_r by sec t+tan t=∑_r≥0Z_rfract^rr!. Then γ_m=(-1)^m Z_2m+1 (m∈ N). Indeed, the odd coefficients come from tan t, whose derivative is sec^2t. Changing the sign of the squared variable in the even series for sech^2t gives sec^2t, so its degree-2m Hurwitz coefficient is (-1)^mγ_m. Differentiation identifies this coefficient with Z_2m+1, proving the formula. The Lean bridge is leanKomorebi…

    Exact mathematical passage
    Z_r
Z^2Use in source · 4 passages

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  • Use in source · Read passage · root_scope.tex, line 365

    The current focused obstruction is the admitted ordinary-index family r=(17659+2^18T)2^640-161, T∈ Z_≥0, j_r=161. On its surviving chart define the actual, undepleted function K_r(Z)=2^-178G_r(10202+2^15Z). The complete coefficient theorem gives K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z), a_r,b_r∈ Z_2, u_r∈ Z_2^×, R_r∈ Z_2⟨ Z⟩. Its half derivative is a dyadic isometry. It therefore has a unique critical point c_r∈ Z_2. Set Φ_r^act=K_r(c_r). The earlier independent-undepletion notation for this same sc…

    Exact mathematical passage
    
     K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z),
     \quad a_r,b_r\in\mathbb Z_2,\quad
     u_r\in\mathbb Z_2^\times,\quad
     R_r\in\mathbb Z_2\langle Z\rangle.
     \label{eq:root:quadraticclass}
    
z^2Use in source · 7 passages

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  • Use in source · Read passage · 10_foundations.tex, line 518

    kth Newton coefficient of g(q)=f(Mq+s) is therefore c_k(g)=∑_r=k^min(D,Mk+s)c_rΘ_M,s(r,k). Indeed, insert the finite stencil for Δ_1^k g(0) and interchange its two finite sums. Coefficient extraction of the binomial expansion of ((1+z)^M-1)^k gives the kernel. In the binary zero-address case, Θ_2,0(k+j,k)=2^k-jbinom kj (0≤ j≤ k), since (2z+z^2)^k=z^k(2+z)^k. Exact valuations for the general kernel remain to be determined. The recovered infinite analytic extensions and stronger historical identit…

    Exact mathematical passage
    (2z+z^2)^k=z^k(2+z)^k
z^{2n}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 85

    [The Bernoulli boundary address] At N=2, the address j=1 gives a second classical family. Define the Bernoulli numbers by B(z)=fracze^z-1=∑_r≥0B_rfracz^rr!, B_0=1, B_1=-frac12. All quotients here are regular formal series. The identity 2B(z)-B(2z)=fraczsinh z =∑_n≥0C_2,1(n)fracz^2n(2n)! follows by putting the two fractions on a common denominator and using [source reference]. Comparing even coefficients yields C_2,1(n)=(2-2^2n)B_2n (n∈ N). This classical derivation restores the boundary example …

    Exact mathematical passage
    
     2B(z)-B(2z)=\frac{z}{\sinh z}
                =\sum_{n\ge0}C_{2,1}(n)\frac{z^{2n}}{(2n)!}
    
z^{2r}Use in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 80

    The dyadic kernel is substantially sharper than the separate coefficient bound alone suggests. If fracz^2sinh^2z = ∑_r≥0β_rz^2r and κ_2,r=β_r2^2r(2r)!, then v_2(κ_2,r)=4r-1 (r≥1). Consequently, truncation after depth M is uniformly valid modulo 2^4M+3, and this exponent is optimal. The rowwise estimate records the additional binary carries in the binomial coefficient. Classical congruence theory for Euler families is extensive [source reference]. The formulas here are stated as exact coordinate …

    Exact mathematical passage
    
    \frac{z^2}{\sinh^2z}
    =
    \sum_{r\geq0}\beta_rz^{2r}
    
Z^3Use in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 365

    The current focused obstruction is the admitted ordinary-index family r=(17659+2^18T)2^640-161, T∈ Z_≥0, j_r=161. On its surviving chart define the actual, undepleted function K_r(Z)=2^-178G_r(10202+2^15Z). The complete coefficient theorem gives K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z), a_r,b_r∈ Z_2, u_r∈ Z_2^×, R_r∈ Z_2⟨ Z⟩. Its half derivative is a dyadic isometry. It therefore has a unique critical point c_r∈ Z_2. Set Φ_r^act=K_r(c_r). The earlier independent-undepletion notation for this same sc…

    Exact mathematical passage
    
     K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z),
     \quad a_r,b_r\in\mathbb Z_2,\quad
     u_r\in\mathbb Z_2^\times,\quad
     R_r\in\mathbb Z_2\langle Z\rangle.
     \label{eq:root:quadraticclass}
    
\Z^dUse in source · 19 passages

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  • Use in source · Read passage · 30_lifting.tex, line 69

    [Finite polynomial obstruction] Let q≥1, L≥1, x∈ Z^d, and suppose F(x)∈ q^L Z^e. Define the exact integer residual r_x by F(x)=q^Lr_x, and put O_x=S_q^e/im J_x, o_x=[overliner_x]. Then O_x is finite and o_x=0 Longleftrightarrow ∃ y∈ Z^d: y≡ xpmodq^L, F(y)≡0pmodq^L+1. No surjectivity or nonsingularity assumption is required.

    Exact mathematical passage
    x\in\Z^d
z^dUse in source · 1 passage

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  • Use in source · visual_body.tex, line 455

    At every center x_0∈ Z_2, the interpolations admit ordinary Taylor series C_j(x_0+z)=∑_d≥0a_j,d(x_0)z^d. For positive degrees, v_2(a_j,d(x_0)) · =d · · (j=0,1),

    Exact mathematical passage
    C_j(x_0+z)=\sum_{d\ge0}a_{j,d}(x_0)z^d
\Z^eUse in source · 8 passages

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  • Use in source · Read passage · 30_lifting.tex, line 70

    [Finite polynomial obstruction] Let q≥1, L≥1, x∈ Z^d, and suppose F(x)∈ q^L Z^e. Define the exact integer residual r_x by F(x)=q^Lr_x, and put O_x=S_q^e/im J_x, o_x=[overliner_x]. Then O_x is finite and o_x=0 Longleftrightarrow ∃ y∈ Z^d: y≡ xpmodq^L, F(y)≡0pmodq^L+1. No surjectivity or nonsingularity assumption is required.

    Exact mathematical passage
    F(x)\in q^L\Z^e
\Z^{e+k}Use in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 451

    [Continuous charts with a selected Jacobian minor] Let p be prime, L≥1, F∈ Z[X_1,…,X_e+k]^e, and x∈ Z^e+k satisfy F(x)∈ p^L Z^e. Suppose the full reduced Jacobian is surjective. There is a permutation σ of the coordinates such that, writing σ x=(y_0,t_0) with e solved coordinates and k parameter coordinates, the parameter ball T=t_0+p^L Z_p^k is homeomorphic to the entire root locus R={z∈ B_L(x):F(z)=0}. The inverse homeomorphism is the actual free-coordinate projection z↦free(σ z). If h:T→ R is…

    Exact mathematical passage
    x\in\Z^{e+k}
\mathbb Z^{e+q}Use in source · 1 passage

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  • Use in source · Read passage · continuation_05.tex, line 13

    Let p be prime, L≥1, e,q≥0, and F∈ Z[X_1,…,X_e+q]^e, x∈ Z^e+q, F(x)≡0pmodp^L. Assume the actual reduced Jacobian A_x=DF(x)bmod p: F_p^e+q⟶ F_p^e is surjective. All vector congruences are coordinatewise.

    Exact mathematical passage
    
    F\in\mathbb Z[X_1,\ldots,X_{e+q}]^e,\qquad
    x\in\mathbb Z^{e+q},\qquad F(x)\equiv0\pmod{p^L}.
    
\Z^fUse in source · 1 passage

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  • Use in source · Read passage · 30_lifting.tex, line 242

    For affine integer systems F(X)=AX-b and G(X)=BX-c, where A: Z^d→ Z^e and B: Z^d→ Z^f are linear, Corollary~[source reference] holds for q≥1 and every L≥0, including depth zero. Here x is a common root modulo q^L, and J and K are the reductions of A and B with domain Z^d. Given separately valid corrections u,v, a simultaneous next lift exists if and only if their classes in Z^d/(ker J+ker K) agree. The quotient is finite, and each label is independent of the chosen valid correction. Indeed, Prop…

    Exact mathematical passage
    B:\Z^d\to\Z^f
z^hUse in source · 2 passages

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  • Use in source · Read passage · 72_recovered.tex, line 407

    [Rational ordinary generating function] In F_p[[z]], ∑_n≥0e_n^(m,p)z^n = 1+ frac ∑_r=1^he_r^(m,p)z^r 1-z^h.

    Exact mathematical passage
    
    \boxed{
    \sum_{n\geq0}e_n^{(m,p)}z^n
    =
    1+
    \frac{
    \sum_{r=1}^{h}e_r^{(m,p)}z^r
    }{1-z^h}.
    }
    
\Z^kUse in source · 3 passages

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  • Use in source · Read passage · 30_lifting.tex, line 591

    [Every compatible integer parameter path] In the preceding notation, assume J_x is surjective. For every supplied integer path t_n∈ Z^k satisfying t_0=free(x), t_n+1-t_n∈ q^L+n Z^k, there is a compatible polynomial tower z_n∈ Z^m+k with z_0=x, free(z_n)=t_n, F(z_n)∈ q^L+n Z^e, z_n+1-z_n∈ q^L+n Z^m+k. The free-coordinate equality is exact in integers. Both reduced Jacobian blocks remain fixed, and z_b-z_a∈ q^L+a Z^m+k whenever a≤ b.

    Exact mathematical passage
    t_n\in\Z^k
\mathbb Z^kUse in source · 3 passages

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  • Use in source · Read passage · continuation_09.tex, line 30

    Integer prescribed-path theorem. For every supplied path t: N→ Z^k satisfying t(0)=t_0, t(n+1)-t(n)∈ q^L+n Z^k, there is a single tower z: N→ Z^m+k with z(0)=x, free(z(n))=t(n), F(z(n))∈ q^L+n Z^e, z(n+1)-z(n)∈ q^L+n Z^m+k. The parameter equality is exact in integers. For a≤ b, z(b)-z(a)∈ q^L+a Z^m+k, DF(z(n))bmod q=DF(x)bmod q. Consequently J_z(n)=J_x at every level. No later surjectivity or per-step solvability is assumed. This theorem includes positive composite bases and q=1.

    Exact mathematical passage
    t:\mathbb N\to\mathbb Z^k
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  • Use in source · Read passage · 10_foundations.tex, line 518

    kth Newton coefficient of g(q)=f(Mq+s) is therefore c_k(g)=∑_r=k^min(D,Mk+s)c_rΘ_M,s(r,k). Indeed, insert the finite stencil for Δ_1^k g(0) and interchange its two finite sums. Coefficient extraction of the binomial expansion of ((1+z)^M-1)^k gives the kernel. In the binary zero-address case, Θ_2,0(k+j,k)=2^k-jbinom kj (0≤ j≤ k), since (2z+z^2)^k=z^k(2+z)^k. Exact valuations for the general kernel remain to be determined. The recovered infinite analytic extensions and stronger historical identit…

    Exact mathematical passage
    (2z+z^2)^k=z^k(2+z)^k
z^{k+j}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2199

    The lowest degree term of Φ_M(z) is Mz, and its highest degree term is z^M. Multiplication by (1+z)^s gives the two edge coefficients. When M=2 and s=0, Φ_2(z)^k = (2z+z^2)^k = z^k(2+z)^k. Taking the coefficient of z^k+j gives the binary formula.

    Exact mathematical passage
    z^{k+j}
z^MUse in source · 3 passages

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  • Use in source · Read passage · 10_foundations.tex, line 488

    polyphase vocabulary. We explain it through elementary identities with explicit algebraic derivations. These identities have not been separately formalized in Lean for this consolidation. For a sequence a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplication, the phase-r product has a term without a …

    Exact mathematical passage
    
    A(z)=\sum_{r=0}^{M-1}z^rA_r(z^M),\qquad
    \bigl(a_r^{[M]}\bigr)_s^{[N]}=a_{r+Ms}^{[MN]}.
    
\Z^{m+k}Use in source · 3 passages

Search aliases: Z^m+k Z mathbb

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  • Use in source · Read passage · 30_lifting.tex, line 596

    [Every compatible integer parameter path] In the preceding notation, assume J_x is surjective. For every supplied integer path t_n∈ Z^k satisfying t_0=free(x), t_n+1-t_n∈ q^L+n Z^k, there is a compatible polynomial tower z_n∈ Z^m+k with z_0=x, free(z_n)=t_n, F(z_n)∈ q^L+n Z^e, z_n+1-z_n∈ q^L+n Z^m+k. The free-coordinate equality is exact in integers. Both reduced Jacobian blocks remain fixed, and z_b-z_a∈ q^L+a Z^m+k whenever a≤ b.

    Exact mathematical passage
    z_n\in\Z^{m+k}
\mathbb Z^{m+k}Use in source · 5 passages

Search aliases: Z^m+k mathbb

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  • Use in source · Read passage · continuation_08.tex, line 12

    Let q>0, L≥1, and m,k,e≥0. Write S= Z/q Z. Let F∈ Z[Y_1,…,Y_m,T_1,…,T_k]^e, x=(y_0,t_0)∈ Z^m+k, F(x)≡0pmodq^L. The actual lower root supplies the unique integral residual r=q^-LF(x)∈ Z^e, bar r=rbmod q∈ S^e. Let the actual reduced Jacobian blocks be J=(D_YF)(x)bmod q:S^m→ S^e, K=(D_TF)(x)bmod q:S^k→ S^e. For a prescribed parameter digit v∈ S^k, choose its canonical integral representative widehat v. Define L_v to be the set of actual residue vectors z∈( Z/q^L+1 Z)^m+k satisfying z≡ xpmodq^L, F(z…

    Exact mathematical passage
    
    F\in\mathbb Z[Y_1,\ldots,Y_m,T_1,\ldots,T_k]^e,
    \qquad x=(y_0,t_0)\in\mathbb Z^{m+k},
    \qquad F(x)\equiv0\pmod{q^L}.
    
z^nUse in source · 6 passages

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  • Use in source · Read passage · 10_foundations.tex, line 486

    The recovered Annals text develops a broader polyphase vocabulary. We explain it through elementary identities with explicit algebraic derivations. These identities have not been separately formalized in Lean for this consolidation. For a sequence a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplicati…

    Exact mathematical passage
    A(z)=\sum_n a_nz^n
z^pUse in source · 1 passage

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  • Use in source · Read passage · supplement_09.tex, line 31

    sumes primality, the modulus divisibility and the specified residue bounds. It does not assume weighted section membership, reciprocal closure, the ratio identity, a Mahler pairing or the desired higher-difference bound. Pages 2--5 describe the constructed objects and proved bridges. 1. Integral cyclic coefficients. Work first in an arbitrary commutative ring R. If z^p=1, set I=(p,z-1), P=(p). Expanding ((z-1)+1)^p=1 and using divisibility of every interior binomial coefficient in prime row p gi…

    Exact mathematical passage
    z^p=1
z^{p^a}Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 2313

    In characteristic p, (1+z)^p^a-1≡ z^p^a. Substitute this congruence in the definition of Θ_p^a,s.

    Exact mathematical passage
    
    (1+z)^{p^a}-1\equiv z^{p^a}.
    
Z^rUse in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 539

    [Bernoulli divided-power moments] Let Z be the ordinary Bernoulli variable, L(Z^r)=B_r, with exponential transform t/(e^t-1). For P=Z(Z+1) and μ_m= L(P^m/m!), μ_m∈ Z_2, v_2(μ_m)≥ s_2(m)-1 (m≥1). Moreover, μ_m is odd exactly when m=0 or m is a positive power of two.

    Exact mathematical passage
    \mathcal L(Z^r)=B_r
z^rUse in source · 9 passages

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  • Use in source · Read passage · 10_foundations.tex, line 488

    polyphase vocabulary. We explain it through elementary identities with explicit algebraic derivations. These identities have not been separately formalized in Lean for this consolidation. For a sequence a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplication, the phase-r product has a term without a …

    Exact mathematical passage
    
    A(z)=\sum_{r=0}^{M-1}z^rA_r(z^M),\qquad
    \bigl(a_r^{[M]}\bigr)_s^{[N]}=a_{r+Ms}^{[MN]}.
    
z_0Use in source · 4 passages

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  • Use in source · Read passage · 30_lifting.tex, line 338

    [Compatible polynomial tower] Let q≥1, L≥1, F∈ Z[X_1,…,X_d]^e, and x∈ Z^d satisfy F(x)∈ q^L Z^e. Suppose J_x:S_q^d→ S_q^e is surjective. There is a single sequence z_n∈ Z^d such that z_0 · =x, · F(z_n) · ∈ q^L+n Z^e, z_n+1-z_n · ∈ q^L+n Z^d, · J_z_n · =J_x. For a≤ b, it also satisfies z_b-z_a∈ q^L+a Z^d.

    Exact mathematical passage
    
     z_0&=x,& F(z_n)&\in q^{L+n}\Z^e,\\
     z_{n+1}-z_n&\in q^{L+n}\Z^d,& J_{z_n}&=J_x.
    
Z_1Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1313

    [The first odd reconstruction] For p=3 and n=2, S_2(3)=1-9+25=17. Since Z_1=1, Z_3=2, and S_0(3)=1, the odd coordinate formula gives E_2 =17-2·9 =-1. The large source value and the depth one correction are both necessary.

    Exact mathematical passage
    Z_1=1
\mathbb Z_2^\timesUse in source · 3 passages

Search aliases: Z_2^× mathbb times

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  • Use in source · Read passage · 72_recovered.tex, line 1598

    [Basic odd and dyadic difference gains] For odd p, Δ_(p-1)p^a^ ru^n = u^n(u^(p-1)p^a-1)^r, v_p(Δ_(p-1)p^a^ ru^n)≥(a+1)r. For odd u∈ Z_2^×, Δ_2^a+1^ ru^n = u^n(u^2^a+1-1)^r, v_2(Δ_2^a+1^ ru^n)≥(a+3)r.

    Exact mathematical passage
    u\in\mathbb Z_2^\times
\mathbb Z_2^{\times}Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 307

    dd moments vanish. Frobenius's congruence is v_2(∑_r=0^m(-1)^m-rbinom mr Euler_2n+2r) =m+v_2(m!) (n,m≥0). For m=0, the assertion is that every even Euler number is odd. With P=( E^2-1)/2 and γ_m(P)=P^m/m!, the case n=0 says L(γ_m(P))∈ Z_2^× (m≥0). Introduce an independent Euler variable F and Q=( F^2-1)/2.

    Exact mathematical passage
    
     \mathcal L(\gamma_m(P))\in\mathbb Z_2^{\times}
                     \qquad(m\geq0).
     \label{eq:euler-pd-odd-moments}
    
Z_{2m+1}Use in source · 4 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 786

    The same inverse coefficients can be written in the signed zigzag notation of the original manuscript. Define Z_r by sec t+tan t=∑_r≥0Z_rfract^rr!. Then γ_m=(-1)^m Z_2m+1 (m∈ N). Indeed, the odd coefficients come from tan t, whose derivative is sec^2t. Changing the sign of the squared variable in the even series for sech^2t gives sec^2t, so its degree-2m Hurwitz coefficient is (-1)^mγ_m. Differentiation identifies this coefficient with Z_2m+1, proving the formula. The Lean bridge is leanKomorebi…

    Exact mathematical passage
    \label{eq:arith-signed-zigzag}
     \gamma_m=(-1)^m Z_{2m+1}\qquad(m\in\N).
    
Z_{2r+1}Use in source · 7 passages

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  • Use in source · Read passage · 72_recovered.tex, line 948

    [Zigzag inverse kernel] sech^2t = ∑_r≥0 (-1)^rZ_2r+1 fract^2r(2r)!.

    Exact mathematical passage
    
    \boxed{
    \operatorname{sech}^2t
    =
    \sum_{r\geq0}
    (-1)^rZ_{2r+1}
    \frac{t^{2r}}{(2r)!}.
    }
    
Z_3Use in source · 1 passage

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  • Use in source · Read passage · 72_recovered.tex, line 1313

    [The first odd reconstruction] For p=3 and n=2, S_2(3)=1-9+25=17. Since Z_1=1, Z_3=2, and S_0(3)=1, the odd coordinate formula gives E_2 =17-2·9 =-1. The large source value and the depth one correction are both necessary.

    Exact mathematical passage
    Z_3=2
\Z_3Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1675

    :matrix_carmichael_mahler Editorial hypothesis correction. The original displayed statement leaves the vectors untyped. Its integrality conclusion requires v_0,w∈ Z_p^d, L≥1, and H,j,n∈ N with H>0; the value at Hx+j for x∈ Z_p means the matrix-binomial interpolation. With p=3, C=4, L=1, w=1 and v_0=1/3, one has Δ_1a_0=1, which is not in 3 Z_3. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses. A…

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    3\Z_3
z_aUse in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 342

    z_n+1-z_n · ∈ q^L+n Z^d, · J_z_n · =J_x. For a≤ b, it also satisfies z_b-z_a∈ q^L+a Z^d.

    Exact mathematical passage
    z_b-z_a\in q^{L+a}\Z^d
z_bUse in source · 2 passages

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  • Use in source · Read passage · 30_lifting.tex, line 342

    z_n+1-z_n · ∈ q^L+n Z^d, · J_z_n · =J_x. For a≤ b, it also satisfies z_b-z_a∈ q^L+a Z^d.

    Exact mathematical passage
    z_b-z_a\in q^{L+a}\Z^d
z_mUse in source · 1 passage

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  • Use in source · Read passage · continuation_03.tex, line 74

    Compatibility implies, for m≥ n, p^L+nmid z_m-z_n. After casting to Z_p, this bounds each coordinate distance by p^-(L+n), so the sequence is Cauchy. Completeness supplies its limit. Closed norm bounds show that the limit retains the congruence at every stage.

    Exact mathematical passage
    
    p^{L+n}\mid z_m-z_n.
    
Z_nUse in source · 4 passages

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  • Use in source · Read passage · 72_recovered.tex, line 910

    Let q be a positive odd integer. Define S_n(q) = ∑_a=0^q-1 (-1)^a(2a+1)^n. Let the signed Euler numbers and Euler zigzag numbers be defined by secht = ∑_n≥0E_nfract^nn! and sec t+tan t = ∑_n≥0Z_nfract^nn!.

    Exact mathematical passage
    
    \sec t+\tan t
    =
    \sum_{n\geq0}Z_n\frac{t^n}{n!}.
    
z_nUse in source · 25 passages

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  • Use in source · Read passage · 30_lifting.tex, line 337

    [Compatible polynomial tower] Let q≥1, L≥1, F∈ Z[X_1,…,X_d]^e, and x∈ Z^d satisfy F(x)∈ q^L Z^e. Suppose J_x:S_q^d→ S_q^e is surjective. There is a single sequence z_n∈ Z^d such that z_0 · =x, · F(z_n) · ∈ q^L+n Z^e,

    Exact mathematical passage
    z_n\in\Z^d
z_{n+1}Use in source · 8 passages

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  • Use in source · Read passage · 30_lifting.tex, line 338

    [Compatible polynomial tower] Let q≥1, L≥1, F∈ Z[X_1,…,X_d]^e, and x∈ Z^d satisfy F(x)∈ q^L Z^e. Suppose J_x:S_q^d→ S_q^e is surjective. There is a single sequence z_n∈ Z^d such that z_0 · =x, · F(z_n) · ∈ q^L+n Z^e, z_n+1-z_n · ∈ q^L+n Z^d, · J_z_n · =J_x. For a≤ b, it also satisfies z_b-z_a∈ q^L+a Z^d.

    Exact mathematical passage
    
     z_0&=x,& F(z_n)&\in q^{L+n}\Z^e,\\
     z_{n+1}-z_n&\in q^{L+n}\Z^d,& J_{z_n}&=J_x.
    
\Z_p^dUse in source · 2 passages

Search aliases: Z_p^d Z mathbb

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  • Use in source · Read passage · 72_recovered.tex, line 1675

    [Matrix Carmichael and Mahler transfer] Editorial hypothesis correction. The original displayed statement leaves the vectors untyped. Its integrality conclusion requires v_0,w∈ Z_p^d, L≥1, and H,j,n∈ N with H>0; the value at Hx+j for x∈ Z_p means the matrix-binomial interpolation. With p=3, C=4, L=1, w=1 and v_0=1/3, one has Δ_1a_0=1, which is not in 3 Z_3. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected…

    Exact mathematical passage
    v_0,w\in\Z_p^d
\mathbb Z_p^dUse in source · 8 passages

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  • Use in source · Read passage · continuation_03.tex, line 22

    Let p be prime, L≥1, and F∈ Z[X_1,…,X_d]^e, x∈ Z^d, F(x)≡0pmodp^L. Assume the actual reduced Jacobian A_x=DF(x)bmod p: F_p^d⟶ F_p^e is surjective. Then Lean proves that there is a point a∈ Z_p^d, F(a)=0, a≡ xpmodp^L. Here the coefficients of the original integer polynomials are cast into Z_p, and their multivariable evaluation is exactly zero. This is an existence theorem for a fixed polynomial presentation with the stated Jacobian hypothesis.

    Exact mathematical passage
    
    \boxed{a\in\mathbb Z_p^d,\qquad F(a)=0,\qquad
     a\equiv x\pmod{p^L}.}
    
\mathbb Z_p^eUse in source · 2 passages

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  • Use in source · Read passage · continuation_05.tex, line 31

    ((σ z)_i=z_σ(i). Write σ z=(y,t), using e solved coordinates and q free coordinates. Define (y_0,t_0)=σ x, B=x+p^L Z_p^e+q, T=t_0+p^L Z_p^q, R={z∈ B:F(z)=0}. There is a homeomorphism h:Txrightarrow ∼ R, h^-1(z)=free(σ z). Equivalently, a continuous function ϕ:T→ y_0+p^L Z_p^e gives the entire root locus through σ h(t)=(ϕ(t),t). Every parameter has exactly one corresponding root in B. For every N≥0, h(t)≡ h(u)pmodp^Niff t≡ upmodp^N.

    Exact mathematical passage
    \phi:T\to y_0+p^L\mathbb Z_p^e
\mathbb Z_p^{e+k}Use in source · 2 passages

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  • Use in source · Read passage · continuation_09.tex, line 75

    Square solved-block uniqueness. If m=e, every prescribed τ has exactly one root in the original ball: ∃! b∈ x+p^L Z_p^e+k: F(b)=0, free(b)=τ. Thus all valid integer parameter approximations and all lifting choices with the same target have the same endpoint. This does not assert uniqueness of the integer representatives in the tower. All prime statements include p=2.

    Exact mathematical passage
    
    \exists!\,b\in x+p^L\mathbb Z_p^{e+k}:\quad
    F(b)=0,\qquad\operatorname{free}(b)=\tau.
    
\mathbb Z_p^{e+q}Use in source · 1 passage

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  • Use in source · Read passage · continuation_05.tex, line 22

    There is a coordinate permutation σ, with (σ z)_i=z_σ(i). Write σ z=(y,t), using e solved coordinates and q free coordinates. Define (y_0,t_0)=σ x, B=x+p^L Z_p^e+q, T=t_0+p^L Z_p^q, R={z∈ B:F(z)=0}. There is a homeomorphism h:Txrightarrow ∼ R, h^-1(z)=free(σ z). Equivalently, a continuous function ϕ:T→ y_0+p^L Z_p^e gives the entire root locus through σ h(t)=(ϕ(t),t). Every parameter has exactly one corresponding root in B. For every N≥0, h(t)≡ h(u)pmodp^Niff t≡ upmodp^N.

    Exact mathematical passage
    
    (y_0,t_0)=\sigma x,\qquad
    B=x+p^L\mathbb Z_p^{e+q},\qquad
    T=t_0+p^L\mathbb Z_p^q,\qquad
    \mathcal R=\{z\in B:F(z)=0\}.
    
\mathbb Z_p^kUse in source · 2 passages

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  • Use in source · Read passage · continuation_09.tex, line 45

    Arbitrary p-adic target theorem. Now let q=p be prime. For every τ∈ t_0+p^L Z_p^k, there are a compatible integer tower z as above and b∈ Z_p^m+k such that z(n)⟶ b, F(b)=0, free(b)=τ, b-z(n)∈ p^L+n Z_p^m+k (n≥0). A compatible integer parameter path converging to τ is constructed in the proof, with its initial representative fixed exactly. A separate theorem accepts every supplied compatible integer path with limit τ and retains its exact coordinates in the constructed tower.

    Exact mathematical passage
    
    \tau\in t_0+p^L\mathbb Z_p^k,
    
\mathbb Z_p^{m+k}Use in source · 2 passages

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  • Use in source · Read passage · continuation_09.tex, line 48

    Arbitrary p-adic target theorem. Now let q=p be prime. For every τ∈ t_0+p^L Z_p^k, there are a compatible integer tower z as above and b∈ Z_p^m+k such that z(n)⟶ b, F(b)=0, free(b)=τ, b-z(n)∈ p^L+n Z_p^m+k (n≥0). A compatible integer parameter path converging to τ is constructed in the proof, with its initial representative fixed exactly. A separate theorem accepts every supplied compatible integer path with limit τ and retains its exact coordinates in the constructed tower.

    Exact mathematical passage
    b\in\mathbb Z_p^{m+k}
\mathbb Z_p^qUse in source · 1 passage

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  • Use in source · Read passage · continuation_05.tex, line 22

    There is a coordinate permutation σ, with (σ z)_i=z_σ(i). Write σ z=(y,t), using e solved coordinates and q free coordinates. Define (y_0,t_0)=σ x, B=x+p^L Z_p^e+q, T=t_0+p^L Z_p^q, R={z∈ B:F(z)=0}. There is a homeomorphism h:Txrightarrow ∼ R, h^-1(z)=free(σ z). Equivalently, a continuous function ϕ:T→ y_0+p^L Z_p^e gives the entire root locus through σ h(t)=(ϕ(t),t). Every parameter has exactly one corresponding root in B. For every N≥0, h(t)≡ h(u)pmodp^Niff t≡ upmodp^N.

    Exact mathematical passage
    
    (y_0,t_0)=\sigma x,\qquad
    B=x+p^L\mathbb Z_p^{e+q},\qquad
    T=t_0+p^L\mathbb Z_p^q,\qquad
    \mathcal R=\{z\in B:F(z)=0\}.
    
\Z_p^\timesUse in source · 2 passages

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  • Use in source · Read passage · supplement_04.tex, line 125

    Why a p-adic unit supplies one factor of p. Now let u∈ Z_p^×. The actual ring homomorphism π: Z_p⟶ F_p sends u to a nonzero unit. Fermat's finite-field theorem therefore gives π(u)^p-1=1, π(u^p-1-1)=0. Using the proved kernel identity kerπ=p Z_p, the source derives pmid u^p-1-1 inside Z_p. The conclusion is not supplied as an abstract certificate. It follows from the unit hypothesis and the standard residue-field construction.

    Exact mathematical passage
    u\in\Z_p^\times
\mathbb Z_p^\timesUse in source · 1 passage

Search aliases: Z_p^× mathbb times

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  • Use in source · Read passage · 72_recovered.tex, line 2329

    [Scalar polyphase precision amplification] Let u∈ Z_p^×, let a,r≥0, and put H_a= (p-1)p^a, · p odd,

    Exact mathematical passage
    u\in\mathbb Z_p^\times
Z_rUse in source · 2 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 781

    The same inverse coefficients can be written in the signed zigzag notation of the original manuscript. Define Z_r by sec t+tan t=∑_r≥0Z_rfract^rr!. Then γ_m=(-1)^m Z_2m+1 (m∈ N). Indeed, the odd coefficients come from tan t, whose derivative is sec^2t. Changing the sign of the squared variable in the even series for sech^2t gives sec^2t, so its degree-2m Hurwitz coefficient is (-1)^mγ_m. Differentiation identifies this coefficient with Z_2m+1, proving the formula. The Lean bridge is leanKomorebi…

    Exact mathematical passage
    Z_r
z_TUse in source · 1 passage

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  • Use in source · Read passage · continuation_08.tex, line 24

    LF(x)∈ Z^e, bar r=rbmod q∈ S^e. Let the actual reduced Jacobian blocks be J=(D_YF)(x)bmod q:S^m→ S^e, K=(D_TF)(x)bmod q:S^k→ S^e. For a prescribed parameter digit v∈ S^k, choose its canonical integral representative widehat v. Define L_v to be the set of actual residue vectors z∈( Z/q^L+1 Z)^m+k satisfying z≡ xpmodq^L, F(z)≡0pmodq^L+1, z_T≡ t_0+q^Lwidehat vpmodq^L+1. The formal predicate uses canonical integer representatives for evaluation and the lower congruence. The previously checked polyno…

    Exact mathematical passage
    
    z\equiv x\pmod{q^L},\qquad F(z)\equiv0\pmod{q^{L+1}},\qquad
    z_T\equiv t_0+q^L\widehat v\pmod{q^{L+1}}.
    
z_{\sigma(i)}Use in source · 1 passage

Search aliases: z_σ(i) sigma

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  • Use in source · Read passage · continuation_05.tex, line 21

    There is a coordinate permutation σ, with (σ z)_i=z_σ(i). Write σ z=(y,t), using e solved coordinates and q free coordinates. Define (y_0,t_0)=σ x, B=x+p^L Z_p^e+q, T=t_0+p^L Z_p^q, R={z∈ B:F(z)=0}. There is a homeomorphism h:Txrightarrow ∼ R, h^-1(z)=free(σ z). Equivalently, a continuous function ϕ:T→ y_0+p^L Z_p^e gives the entire root locus through σ h(t)=(ϕ(t),t). Every parameter has exactly one corresponding root in B. For every N≥0, h(t)≡ h(u)pmodp^Niff t≡ upmodp^N.

    Exact mathematical passage
    (\sigma z)_i=z_{\sigma(i)}
\mathbb Z_{\ge0}Use in source · 1 passage

Search aliases: Z_≥0 mathbb ge

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  • Use in source · Read passage · root_scope.tex, line 355

    The current focused obstruction is the admitted ordinary-index family r=(17659+2^18T)2^640-161, T∈ Z_≥0, j_r=161. On its surviving chart define the actual, undepleted function K_r(Z)=2^-178G_r(10202+2^15Z). The complete coefficient theorem gives K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z), a_r,b_r∈ Z_2, u_r∈ Z_2^×, R_r∈ Z_2⟨ Z⟩. Its half derivative is a dyadic isometry. It therefore has a unique critical point c_r∈ Z_2. Set Φ_r^act=K_r(c_r). The earlier independent-undepletion notation for this same sc…

    Exact mathematical passage
    
     r=(17659+2^{18}T)2^{640}-161,
     \qquad T\in\mathbb Z_{\ge0},\qquad j_r=161.
     \label{eq:root:focusedfamily}
    
\mathbb Z_{\geq0}Use in source · 1 passage

Search aliases: Z_≥0 mathbb geq

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  • Use in source · Read passage · strengthening.tex, line 970

    [A first-difference criterion] Let C: Z_≥0→ Z_2 have Newton coefficients a_k=Δ_1^kC(0). Suppose for an integer b≥0 that v_2(a_1)=b, v_2(a_k)≥ b+1+lfloorlog_2krfloor (k≥2). Then for every n≥0 and H>0, v_2(C(n+H)-C(n))=b+v_2(H). The Newton interpolation extends continuously to Z_2 and satisfies the same identity for any two distinct dyadic arguments.

    Exact mathematical passage
    C:\mathbb Z_{\geq0}\to\mathbb Z_2
\Zp^eUse in source · 2 passages

Search aliases: Zp^e Z_p^e Zp mathbb

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  • Use in source · Read passage · 30_lifting.tex, line 468

    R={z∈ B_L(x):F(z)=0}. The inverse homeomorphism is the actual free-coordinate projection z↦free(σ z). If h:T→ R is the forward map, then for every N≥0, h(t)≡ h(u)pmodp^N Longleftrightarrow t≡ upmodp^N. Equivalently, σ h(t)=(ϕ(t),t) for a continuous ϕ:T→ y_0+p^L Z_p^e whose graph is the entire reordered root locus.

    Exact mathematical passage
    \phi:T\to y_0+p^L\Zp^e
\Zp^{e+k}Use in source · 2 passages

Search aliases: Zp^e+k Z_p^e+k Zp mathbb

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  • Use in source · Read passage · 30_lifting.tex, line 507

    Now suppose a chosen set of e columns of DF(x)bmod p is independent, and put those coordinates first. Apply the square result to the augmented polynomial map Ψ(Y,T)=(F(Y,T),T). Its reduced Jacobian has upper blocks (D_YF,D_TF) and lower blocks (0,I_k), so it is invertible. For t∈ T, the target (0,t) belongs to Ψ(x)+p^L Z_p^e+k because F(x)≡0pmodp^L. The inverse of Ψ at that target gives the unique root with parameters t. The inverse congruence property gives one direction of [source reference]; …

    Exact mathematical passage
    \Psi(x)+p^L\Zp^{e+k}
\Zp^kUse in source · 3 passages

Search aliases: Zp^k Z_p^k Zp mathbb

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  • Use in source · Read passage · 30_lifting.tex, line 456

    arts with a selected Jacobian minor] Let p be prime, L≥1, F∈ Z[X_1,…,X_e+k]^e, and x∈ Z^e+k satisfy F(x)∈ p^L Z^e. Suppose the full reduced Jacobian is surjective. There is a permutation σ of the coordinates such that, writing σ x=(y_0,t_0) with e solved coordinates and k parameter coordinates, the parameter ball T=t_0+p^L Z_p^k is homeomorphic to the entire root locus R={z∈ B_L(x):F(z)=0}. The inverse homeomorphism is the actual free-coordinate projection z↦free(σ z). If h:T→ R is the forward m…

    Exact mathematical passage
    T=t_0+p^L\Zp^k
\Zp^{m+k}Use in source · 2 passages

Search aliases: Zp^m+k Z_p^m+k Zp mathbb

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  • Use in source · Read passage · 30_lifting.tex, line 629

    [Arbitrary p-adic parameter endpoints] Let q=p be prime in Theorem~[source reference]. For every τ∈ t_0+p^L Z_p^k, there are an integer tower z_n and b∈ Z_p^m+k such that F(b)=0, free(b)=τ, z_n⟶ b, b-z_n∈ p^L+n Z_p^m+k. The initial value is z_0=x. At every stage and for every proposed parameter digit v∈ F_p^k, the corresponding next residue fiber has exactly p^m-e elements. If m=e, then for each τ there is exactly one such root in B_L(x), independently of all choices of integer approximants and …

    Exact mathematical passage
    b\in\Zp^{m+k}
\Zp^\timesUse in source · 2 passages

Search aliases: Zp^× Z_p^× Zp times mathbb times

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  • Use in source · Read passage · 10_foundations.tex, line 337

    [Differences of unit modes] Over every commutative ring, Δ_h^s(u^n)=u^n(u^h-1)^s. For any prime p, units u_i∈ Z_p^×, coefficients a_i∈ Z_p, and a finite index set I, the sequence f(n)=∑_i∈ Ia_i u_i^n satisfies Δ_p-1^s f(n)∈ p^s Z_p (n,s∈ N).

    Exact mathematical passage
    u_i\in\Zp^\times
\ZZUse in source · 9 passages

Search aliases: ZZ Z mathbb

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  • Use in source · Read passage · 72_recovered.tex, line 641

    [A denominator bound] Let D_N,j(n) = ∏_r=1^n binomNr+jNr, D_N,j(0)=1. Then D_N,j(n)C_N,j(n) ∈ Z. For every prime p, v_p(C_N,j(n)) ≥ - ∑_r=1^n v_pbinomNr+jNr.

    Exact mathematical passage
    
    \boxed{
    D_{N,j}(n)C_{N,j}(n)
    \in
    \ZZ.
    }
    
\ZZ^dUse in source · 1 passage

Search aliases: ZZ^d Z^d ZZ mathbb

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  • Use in source · Read passage · 72_recovered.tex, line 2939

    [Base change of an integral recurrence] Let C∈ M_d( Z), v_0∈ Z^d. The integral orbit v_n=C^nv_0 determines, by base change, compatible orbits in every ring R equipped with a homomorphism Z→ R. In particular, the reductions modulo every positive integer and the orbits in every Z_p^d share the same companion matrix, characteristic polynomial, and integer recurrence relations.

    Exact mathematical passage
    
    C\in M_d(\ZZ),
    \qquad
    v_0\in\ZZ^d.
    
|Use in source · 20 passages

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  • Use in source · Read passage · 20_arithmetic.tex, line 717

    [All positive periods and their least element] For r∈ N and 0≤ j<4, a positive integer T is a packet period of U_j modulo 2^r if and only if 2^max(0,r-d_j)mid T. Here a packet period means U_j(n+T)≡ U_j(n)pmod2^r for every n≥0. Consequently the least positive packet periods are c|cccc j · 0 · 1 · 2 · 3 least period · 2^max(0,r-1) · 2^max(0,r-1) · 2^max(0,r-5) · 2^max(0,r-3).

    Exact mathematical passage
    
     \begin{array}{c|cccc}
     j&0&1&2&3\\\hline
     \text{least period}&2^{\max(0,r-1)}&2^{\max(0,r-1)}&
                     2^{\max(0,r-5)}&2^{\max(0,r-3)}.
     \end{array}
    
|_2Use in source · 9 passages

Search aliases: |_2

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  • Use in source · Read passage · 20_arithmetic.tex, line 1051

    [Isometric packet bijections] For 0≤ j<4, define on natural numbers f_j(n)=fracU_j(n)-j!2^d_j, (d_0,d_1,d_2,d_3)=(1,1,5,3). There is a unique continuous map Φ_j: Z_2→ Z_2 whose value at every n∈ N equals the embedded rational f_j(n). It is an isometric bijection and Φ_j(0)=0: |Φ_j(x)-Φ_j(y)|_2=|x-y|_2 (x,y∈ Z_2). In particular, every element of Z_2 has exactly one preimage.

    Exact mathematical passage
    
     |\Phi_j(x)-\Phi_j(y)|_2=|x-y|_2\qquad(x,y\in\Z_2).
    
|_{n=0}Use in source · 2 passages

Search aliases: |_n=0

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  • Use in source · Read passage · supplement_04.tex, line 184

    A sharp unit-mode gain. For p=5 and u=2, Δ_4^2(u^n)|_n=0=(2^4-1)^2=225. The source proves 5^2mid225 and 5^3nmid225. This example attains the guaranteed gain; no additional uniform power of p is claimed.

    Exact mathematical passage
    
     \Delta_4^2(u^n)\big|_{n=0}=(2^4-1)^2=225.
    
|_pUse in source · 3 passages

Search aliases: |_p

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  • Use in source · Read passage · 30_lifting.tex, line 15

    Unless a different coefficient ring is specified, a polynomial system is a finite list of integer polynomials. All divisibility conditions on vectors are coordinatewise. For a positive integer q, write S_q= Z/q Z and use an overline for reduction modulo q. For a prime p, write F_p=S_p, normalize |p|_p=p^-1, and put B_L(x)=x+p^L Z_p^d. An integer center is included in this notation by its natural image in Z_p^d. All dimensions below are finite and may be zero. A statement over S_q allows composit…

    Exact mathematical passage
    |p|_p=p^{-1}
|_{U_a}Use in source · 1 passage

Search aliases: |_U_a

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  • Use in source · Read passage · 72_recovered.tex, line 2754

    [Packet decomposition of a lift obstruction] Suppose F_p^d = ⊕_a∈ AU_a, F_p^r = ⊕_a∈ AV_a, and suppose a linear correction map D: F_p^d→ F_p^r satisfies D(U_a)⊆ V_a for every a. Write D_a=D|_U_a. Then coker D ≅ ⊕_a∈ Acoker D_a. A residual lifts if and only if every packet obstruction vanishes.

    Exact mathematical passage
    D_a=D|_{U_a}
\cdotUse in source · 11 passages

Search aliases: · cdot

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  • Use in source · Read passage · 20_arithmetic.tex, line 456

    For an odd prime, 2mid p-1, so the option m=2 introduces no further case. The sole remaining branch is p=2,m=2,N=4. The all-row theorem proved next gives 2^kmidΔ_1^kC_4,j(n) at every row and address. After multiplying by j!, Lemma~[source reference] gives packet period 2^r-1 and degree displacement 4·2^r-1=2^r+1=qp^r. It also gives integrality. Hence the complete statement follows. The formal degree conversion is leanKomorebiPaperIX.NishibuchiCompletion.original_degree_shift; the all-parameter d…

    Exact mathematical passage
    4\cdot2^{r-1}=2^{r+1}=qp^r
\times_CUse in source · 1 passage

Search aliases: ×_C times

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  • Use in source · Read passage · 50_open_questions.tex, line 95

    The original proposed form is X≃ X_2^wt×_C∏_p>2'X_p. It leaves X, the distinguished subsets defining the restricted product, the maps to C, the ambient category, and the universal meaning of ``smallest'' unspecified. In particular, one must decide whether X is a global object or a completed object.

    Exact mathematical passage
    
    X\simeq X_2^{\mathrm{wt}}\times_C\prod_{p>2}'X_p.
    
\alphaUse in source · 17 passages

Search aliases: α alpha

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  • Use in source · Read passage · 72_recovered.tex, line 1802

    [Weighted convergence] If widehat F(T)=∑_r≥0c_rT^r and v_p(c_r)+rα⟶+∞, then the series converges whenever v_p(T)≥α.

    Exact mathematical passage
    
    \vp(c_r)+r\alpha\longrightarrow+\infty,
    
\alpha_1Use in source · 1 passage

Search aliases: α_1 alpha

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  • Use in source · Read passage · 72_recovered.tex, line 2839

    After passage to an algebraic closure, let α_1,…,α_d be the eigenvalues of C, counted with algebraic multiplicity. The resultant is ∏_i=1^d(x-α_i^N), which is the characteristic polynomial of C^N. Both sides are polynomial expressions with integer coefficients in the entries of C, so the identity descends to every commutative coefficient ring.

    Exact mathematical passage
    \alpha_1,\ldots,\alpha_d
\alpha_dUse in source · 4 passages

Search aliases: α_d alpha

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  • Use in source · Read passage · 72_recovered.tex, line 2839

    After passage to an algebraic closure, let α_1,…,α_d be the eigenvalues of C, counted with algebraic multiplicity. The resultant is ∏_i=1^d(x-α_i^N), which is the characteristic polynomial of C^N. Both sides are polynomial expressions with integer coefficients in the entries of C, so the identity descends to every commutative coefficient ring.

    Exact mathematical passage
    \alpha_1,\ldots,\alpha_d
\alpha_iUse in source · 3 passages

Search aliases: α_i alpha

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  • Use in source · Read passage · root_scope.tex, line 238

    [Regular-wall root law] For every regular label j and every integer δ≥1, v_2G_r(x)= δ, · x∈ B_i, δ<t_i, d_i+v_2(x-α_i), · x∈ B_i, δ≥ t_i. In the second case there is exactly one root α_i in B_i, it is simple, and v_2(α_i-i)=δ-d_i. Thus the number of dyadic roots is exactly #{i:0≤ i<j, t_i≤δ}. If δ<t_i, the unit perturbation has smaller valuation than the skeleton term throughout B_i, so the valuation is exactly δ. Otherwise, after the ball substitution and division by 2^t_i, the coefficientwise …

    Exact mathematical passage
    
     v_2G_r(x)=
     \begin{cases}
      \delta,&x\in B_i,\ \delta<t_i,\\
      d_i+v_2(x-\alpha_i),&x\in B_i,\ \delta\ge t_i.
     \end{cases}
    
\alpha_i^NUse in source · 1 passage

Search aliases: α_i^N alpha

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  • Use in source · Read passage · 72_recovered.tex, line 2840

    After passage to an algebraic closure, let α_1,…,α_d be the eigenvalues of C, counted with algebraic multiplicity. The resultant is ∏_i=1^d(x-α_i^N), which is the characteristic polynomial of C^N. Both sides are polynomial expressions with integer coefficients in the entries of C, so the identity descends to every commutative coefficient ring.

    Exact mathematical passage
    
    \prod_{i=1}^{d}(x-\alpha_i^N),
    
\alpha_rUse in source · 5 passages

Search aliases: α_r alpha

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  • Use in source · Read passage · root_scope.tex, line 382

    [Exact quadratic compression] There is a unique preparation K_r(Z)= U_r(Z) W_r(Z), W_r(Z)=Z^2+2β_rZ+α_r, U_r≡ u_rpmod256. With δ_r=β_r^2-α_r and τ_r= U_r'(c_r)/ U_r(c_r), δ_r=-fracΦ_r^act U_r(c_r) (1-fracτ_r^2Φ_r^act4 U_r(c_r)), v_2(δ_r)=v_2(Φ_r^act). The equality includes the zero case. The same successive monic polynomial division used in Theorem~[source reference], started at precision eight, preserves the quadratic and unit reductions of [source reference]. Write W=Z^2+2β_rZ+α_r and w=W(c_r)…

    Exact mathematical passage
    
     K_r(Z)=\mathcal U_r(Z)\mathcal W_r(Z),\qquad
     \mathcal W_r(Z)=Z^2+2\beta_rZ+\alpha_r,\qquad
     \mathcal U_r\equiv u_r\pmod{256}.
    
\beta_1Use in source · 3 passages

Search aliases: β_1 beta

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  • Use in source · Read passage · supplement_03.tex, line 158

    Principal targets. targetDyadicReconstruction.dyadic_reconstruction targetDyadicReconstruction.dyadic_supercongruence targetDyadicReconstruction.normalized_error_integral pageheading4. Exact examples and adversarial checksExample 5.5 and controls against incorrect normalizations The actual reciprocal recurrence gives (β_0,β_1,β_2,β_3,β_4) =(1,-frac13,frac115,-frac2189,frac1675). These values are obtained from the general formal-series definition. They are not substituted as independent input dat…

    Exact mathematical passage
    
     (\beta_0,\beta_1,\beta_2,\beta_3,\beta_4)
     =\left(1,-\frac13,\frac1{15},-\frac2{189},\frac1{675}\right).
    
\beta_2Use in source · 2 passages

Search aliases: β_2 beta

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  • Use in source · Read passage · supplement_03.tex, line 158

    Principal targets. targetDyadicReconstruction.dyadic_reconstruction targetDyadicReconstruction.dyadic_supercongruence targetDyadicReconstruction.normalized_error_integral pageheading4. Exact examples and adversarial checksExample 5.5 and controls against incorrect normalizations The actual reciprocal recurrence gives (β_0,β_1,β_2,β_3,β_4) =(1,-frac13,frac115,-frac2189,frac1675). These values are obtained from the general formal-series definition. They are not substituted as independent input dat…

    Exact mathematical passage
    
     (\beta_0,\beta_1,\beta_2,\beta_3,\beta_4)
     =\left(1,-\frac13,\frac1{15},-\frac2{189},\frac1{675}\right).
    
\beta_3Use in source · 2 passages

Search aliases: β_3 beta

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  • Use in source · Read passage · supplement_03.tex, line 158

    Principal targets. targetDyadicReconstruction.dyadic_reconstruction targetDyadicReconstruction.dyadic_supercongruence targetDyadicReconstruction.normalized_error_integral pageheading4. Exact examples and adversarial checksExample 5.5 and controls against incorrect normalizations The actual reciprocal recurrence gives (β_0,β_1,β_2,β_3,β_4) =(1,-frac13,frac115,-frac2189,frac1675). These values are obtained from the general formal-series definition. They are not substituted as independent input dat…

    Exact mathematical passage
    
     (\beta_0,\beta_1,\beta_2,\beta_3,\beta_4)
     =\left(1,-\frac13,\frac1{15},-\frac2{189},\frac1{675}\right).
    
\beta_4Use in source · 1 passage

Search aliases: β_4 beta

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  • Use in source · Read passage · supplement_03.tex, line 158

    Principal targets. targetDyadicReconstruction.dyadic_reconstruction targetDyadicReconstruction.dyadic_supercongruence targetDyadicReconstruction.normalized_error_integral pageheading4. Exact examples and adversarial checksExample 5.5 and controls against incorrect normalizations The actual reciprocal recurrence gives (β_0,β_1,β_2,β_3,β_4) =(1,-frac13,frac115,-frac2189,frac1675). These values are obtained from the general formal-series definition. They are not substituted as independent input dat…

    Exact mathematical passage
    
     (\beta_0,\beta_1,\beta_2,\beta_3,\beta_4)
     =\left(1,-\frac13,\frac1{15},-\frac2{189},\frac1{675}\right).
    
\beta_jUse in source · 4 passages

Search aliases: β_j beta

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  • Use in source · Read passage · strengthening.tex, line 260

    [Four-address factorial bound] For j∈{0,1,2,3}, n≥0, k≥1, and positive mixed steps, v_2(Δ_H_1⋯Δ_H_kC_j(n)) ≥ k+v_2(k!)+∑_ν=1^k v_2(H_ν)+β_j, (β_0,β_1,β_2,β_3)=(0,0,3,1). Address two also satisfies v_2(Δ_H_1⋯Δ_H_kC_2(n)) ≥ 2k+v_2(k!)+lfloorfrack2rfloor +∑_ν=1^k v_2(H_ν). For the original coefficients, the offsets in [source reference] are (0,0,4,2). No positive-order offset is asserted for the empty difference.

    Exact mathematical passage
    
     v_2(\Delta_{H_1}\cdots\Delta_{H_k}C_j(n))
     \geq k+v_2(k!)+\sum_{\nu=1}^k v_2(H_\nu)+\beta_j,
     \quad (\beta_0,\beta_1,\beta_2,\beta_3)=(0,0,3,1).
     \label{eq:factorial-four-lower}
    
\beta_rUse in source · 13 passages

Search aliases: β_r beta

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  • Use in source · Read passage · 72_recovered.tex, line 80

    The dyadic kernel is substantially sharper than the separate coefficient bound alone suggests. If fracz^2sinh^2z = ∑_r≥0β_rz^2r and κ_2,r=β_r2^2r(2r)!, then v_2(κ_2,r)=4r-1 (r≥1). Consequently, truncation after depth M is uniformly valid modulo 2^4M+3, and this exponent is optimal. The rowwise estimate records the additional binary carries in the binomial coefficient. Classical congruence theory for Euler families is extensive [source reference]. The formulas here are stated as exact coordinate …

    Exact mathematical passage
    
    \frac{z^2}{\sinh^2z}
    =
    \sum_{r\geq0}\beta_rz^{2r}
    
\beta_{r+q}Use in source · 1 passage

Search aliases: β_r+q beta

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  • Use in source · Read passage · root_scope.tex, line 110

    Set r=M+1≥1 and define P_q(x)=∏_i=0^q-1(x-i)(2x-(2i+1)), P_0=1, H_r(x)=∑_q≥08^qfracβ_r+qβ_r P_q(x)mathscr A(x-q). The coefficient valuations tend to infinity. At any nonnegative integer x, terms with q>x vanish. Cancelling the factorials in [source reference] therefore gives the exact identity T_r-1(N)=κ_rbinom2N2rH_r(N-r) (N≥ r). This is a finite rational identity at every original Euler row.

    Exact mathematical passage
    
     H_r(x)=\sum_{q\ge0}8^q\frac{\beta_{r+q}}{\beta_r}
                        P_q(x)\mathscr A(x-q).
    
\beta_r^2Use in source · 2 passages

Search aliases: β_r^2 beta

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  • Use in source · Read passage · root_scope.tex, line 387

    [Exact quadratic compression] There is a unique preparation K_r(Z)= U_r(Z) W_r(Z), W_r(Z)=Z^2+2β_rZ+α_r, U_r≡ u_rpmod256. With δ_r=β_r^2-α_r and τ_r= U_r'(c_r)/ U_r(c_r), δ_r=-fracΦ_r^act U_r(c_r) (1-fracτ_r^2Φ_r^act4 U_r(c_r)), v_2(δ_r)=v_2(Φ_r^act). The equality includes the zero case. The same successive monic polynomial division used in Theorem~[source reference], started at precision eight, preserves the quadratic and unit reductions of [source reference]. Write W=Z^2+2β_rZ+α_r and w=W(c_r)…

    Exact mathematical passage
    \delta_r=\beta_r^2-\alpha_r
\GammaUse in source · 2 passages

Search aliases: Γ Gamma

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  • Use in source · Read passage · strengthening.tex, line 334

    Let Γ_R(P,Q) be the R-subalgebra generated by all γ_m(P) and γ_r(Q). Adjoin E, F with E^2=1+2P, F^2=1+2Q, and call the resulting algebra A_0. Its elements are finite expressions; no completion is being presumed. It is generated as a Γ_R(P,Q)-module by 1, E, F, E F. The functional maps this module to R. For an even basis term this is [source reference] and independence. For a basis term with an odd Euler factor the moment is zero, since P and Q are even polynomials in their respective variables.

    Exact mathematical passage
    \Gamma_R(P,Q)
\gamma_2Use in source · 7 passages

Search aliases: γ_2 gamma

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  • Use in source · Read passage · strengthening.tex, line 683

    There is a still stronger filtration. Put X=P+Q, τ=i B D, ϵ=1+τ. Then the exact identities are H_2=ϵ X+2G, G=-γ_2(P)-γ_2(Q)+3PQ-τ Q, ϵ^2=2V, V=τ-2X-8PQ. Every divided power of G is integral component by component. The finite addition identity gives γ_k(H_2)=∑_r=0^k2^rγ_r(G) ϵ^k-rγ_k-r(X) ∈2^lfloor k/2rfloor A_2, because the rth term belongs to 2^r+lfloor(k-r)/2rfloor A_2. Applying the functional, rational descent, and the base-index expansion proves the unit-step version of [source reference]. F…

    Exact mathematical passage
    
     H_2=\epsilon X+2G,\quad
     G=-\gamma_2(P)-\gamma_2(Q)+3PQ-\tau Q,\quad
     \epsilon^2=2V,\quad V=\tau-2X-8PQ.
     \label{eq:address-two-filtration-algebra}
    
\gamma_{2k}Use in source · 1 passage

Search aliases: γ_2k gamma

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  • Use in source · Read passage · strengthening.tex, line 803

    wers of G are integral by the same component formulas used above. Thus modulo 2 A_0, γ_k(H_0) equals the divided power of γ_2(X)+ϵ X. The identity γ_r(γ_2(X)) =frac(2r)!2^r r!γ_2r(X) has an odd integer scalar. Every term involving ϵ^j for j≥2 vanishes modulo two. The surviving zero and first powers give γ_k(H_0)≡ γ_2k(X)+ϵγ_2k-1(X) pmod2 A_0. The multiplication coefficient in the second term is 2k-1, also odd. By the odd moments [source reference] and independence, L(γ_m(P+Q))≡ m+1pmod2. The τ-w…

    Exact mathematical passage
    
     \gamma_k(H_0)\equiv
           \gamma_{2k}(X)+\epsilon\gamma_{2k-1}(X)
                                      \pmod{2\mathcal A_0}.
     \label{eq:zero-parity-leading-pd}
    
\gamma_{2k-1}Use in source · 1 passage

Search aliases: γ_2k-1 gamma

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  • Use in source · Read passage · strengthening.tex, line 803

    wers of G are integral by the same component formulas used above. Thus modulo 2 A_0, γ_k(H_0) equals the divided power of γ_2(X)+ϵ X. The identity γ_r(γ_2(X)) =frac(2r)!2^r r!γ_2r(X) has an odd integer scalar. Every term involving ϵ^j for j≥2 vanishes modulo two. The surviving zero and first powers give γ_k(H_0)≡ γ_2k(X)+ϵγ_2k-1(X) pmod2 A_0. The multiplication coefficient in the second term is 2k-1, also odd. By the odd moments [source reference] and independence, L(γ_m(P+Q))≡ m+1pmod2. The τ-w…

    Exact mathematical passage
    
     \gamma_k(H_0)\equiv
           \gamma_{2k}(X)+\epsilon\gamma_{2k-1}(X)
                                      \pmod{2\mathcal A_0}.
     \label{eq:zero-parity-leading-pd}
    
\gamma_{2m}Use in source · 1 passage

Search aliases: γ_2m gamma

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  • Use in source · Read passage · strengthening.tex, line 349

    Every displayed component of H_0 has all its divided powers in A_0. Besides the generators, the needed identities are γ_m(P^2/2) · =frac(2m)!2^m m!γ_2m(P),nonumber γ_m(3PQ) · =3^m m!γ_m(P)γ_m(Q),nonumber γ_m(P-Q) · =∑_r=0^m(-1)^r γ_m-r(P)γ_r(Q). The first scalar counts partitions into pairs and is an integer. For the last term of H_0, multiply the third identity by (i E F)^m and reduce the Euler powers using the integral quadratic relations. Finally, γ_k(X+Y)=∑_r=0^kγ_r(X)γ_k-r(Y) proves γ_k(H_0…

    Exact mathematical passage
    
     \gamma_m(P^2/2)
         &=\frac{(2m)!}{2^m m!}\gamma_{2m}(P),\nonumber\\
     \gamma_m(3PQ)&=3^m m!\gamma_m(P)\gamma_m(Q),\nonumber\\
     \gamma_m(P-Q)&=\sum_{r=0}^m(-1)^r
                       \gamma_{m-r}(P)\gamma_r(Q).
     \label{eq:components-divided-powers}
    
\gamma_{2r}Use in source · 1 passage

Search aliases: γ_2r gamma

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  • Use in source · Read passage · strengthening.tex, line 797

    :four-euler-H gives H_0=γ_2(X)+ϵ X+2G, G=-γ_2(P)-γ_2(Q)+PQ-τ Q, ϵ^2=2(τ-P-Q-2PQ). The divided powers of G are integral by the same component formulas used above. Thus modulo 2 A_0, γ_k(H_0) equals the divided power of γ_2(X)+ϵ X. The identity γ_r(γ_2(X)) =frac(2r)!2^r r!γ_2r(X) has an odd integer scalar. Every term involving ϵ^j for j≥2 vanishes modulo two. The surviving zero and first powers give γ_k(H_0)≡ γ_2k(X)+ϵγ_2k-1(X) pmod2 A_0. The multiplication coefficient in the second term is 2k-1, …

    Exact mathematical passage
    
     \gamma_r(\gamma_2(X))
       =\frac{(2r)!}{2^r r!}\gamma_{2r}(X)
    
\gamma_kUse in source · 9 passages

Search aliases: γ_k gamma

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  • Use in source · Read passage · strengthening.tex, line 361

    γ_m(P-Q) · =∑_r=0^m(-1)^r γ_m-r(P)γ_r(Q). The first scalar counts partitions into pairs and is an integer. For the last term of H_0, multiply the third identity by (i E F)^m and reduce the Euler powers using the integral quadratic relations. Finally, γ_k(X+Y)=∑_r=0^kγ_r(X)γ_k-r(Y) proves γ_k(H_0)∈ A_0. Applying the functional and putting W=U^4-1=2H_0 gives L(W^m)∈2^m m!R (m≥0). These moments are rational. The identity R∩ Q_2= Z_2, obtained from the linear independence of 1,i over Q_2, descends e…

    Exact mathematical passage
    
     \gamma_k(X+Y)=\sum_{r=0}^k\gamma_r(X)\gamma_{k-r}(Y)
    
\gamma_{k-1}Use in source · 4 passages

Search aliases: γ_k-1 gamma

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  • Use in source · Read passage · strengthening.tex, line 887

    For k=2^r≥4, put q=k/2, which is even. Normalize the exact expansion [source reference] by 2^q before reducing modulo two. Its uth auxiliary term has excess u+lfloor(k-u)/2rfloor-q=lfloor u/2rfloor. Only auxiliary indices zero and one can therefore survive: fracγ_k(H_2)2^q ≡ V^qγ_k(X) +ϵ V^q-1Gγ_k-1(X) pmod2 A_2. Here V≡τpmod2 and τ^2≡1pmod2. Since q is even and ϵτ≡ϵpmod2, this reduces to γ_k(X)+ϵ Gγ_k-1(X). Moreover, G≡γ_2(X)+τ Qpmod2. After evaluation, every term with a single τ vanishes, so t…

    Exact mathematical passage
    
     \frac{\gamma_k(H_2)}{2^q}
     \equiv V^q\gamma_k(X)
           +\epsilon V^{q-1}G\gamma_{k-1}(X)
                                    \pmod{2\mathcal A_2}.
     \label{eq:two-power-leading-normalized}
    
\gamma_{k-r}Use in source · 2 passages

Search aliases: γ_k-r gamma

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  • Use in source · Read passage · strengthening.tex, line 361

    γ_m(P-Q) · =∑_r=0^m(-1)^r γ_m-r(P)γ_r(Q). The first scalar counts partitions into pairs and is an integer. For the last term of H_0, multiply the third identity by (i E F)^m and reduce the Euler powers using the integral quadratic relations. Finally, γ_k(X+Y)=∑_r=0^kγ_r(X)γ_k-r(Y) proves γ_k(H_0)∈ A_0. Applying the functional and putting W=U^4-1=2H_0 gives L(W^m)∈2^m m!R (m≥0). These moments are rational. The identity R∩ Q_2= Z_2, obtained from the linear independence of 1,i over Q_2, descends e…

    Exact mathematical passage
    
     \gamma_k(X+Y)=\sum_{r=0}^k\gamma_r(X)\gamma_{k-r}(Y)
    
\gamma_{m-r}Use in source · 1 passage

Search aliases: γ_m-r gamma

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  • Use in source · Read passage · strengthening.tex, line 349

    Every displayed component of H_0 has all its divided powers in A_0. Besides the generators, the needed identities are γ_m(P^2/2) · =frac(2m)!2^m m!γ_2m(P),nonumber γ_m(3PQ) · =3^m m!γ_m(P)γ_m(Q),nonumber γ_m(P-Q) · =∑_r=0^m(-1)^r γ_m-r(P)γ_r(Q). The first scalar counts partitions into pairs and is an integer. For the last term of H_0, multiply the third identity by (i E F)^m and reduce the Euler powers using the integral quadratic relations. Finally, γ_k(X+Y)=∑_r=0^kγ_r(X)γ_k-r(Y) proves γ_k(H_0…

    Exact mathematical passage
    
     \gamma_m(P^2/2)
         &=\frac{(2m)!}{2^m m!}\gamma_{2m}(P),\nonumber\\
     \gamma_m(3PQ)&=3^m m!\gamma_m(P)\gamma_m(Q),\nonumber\\
     \gamma_m(P-Q)&=\sum_{r=0}^m(-1)^r
                       \gamma_{m-r}(P)\gamma_r(Q).
     \label{eq:components-divided-powers}
    
\gamma_{n-q}Use in source · 1 passage

Search aliases: γ_n-q gamma

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  • Use in source · Read passage · 20_arithmetic.tex, line 805

    [Classical odd-length reconstruction] For every positive odd integer p and every n∈ N, E_2n=∑_m=0^nbinom2n2mγ_m p^2mS_2(n-m)(p). Equivalently, E_2n=∑_q=0^n V_nqS_2q(p) for the lower triangular matrix V_nq=binom2n2qγ_n-qp^2(n-q) (q≤ n), V_nq=0 (q>n). It has V_nn=1 and p^2(n-q)mid V_nq when q≤ n.

    Exact mathematical passage
    
     V_{nq}=\binom{2n}{2q}\gamma_{n-q}p^{2(n-q)}\quad(q\le n),
     \qquad V_{nq}=0\quad(q>n).
    
\Gamma_RUse in source · 2 passages

Search aliases: Γ_R Gamma

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  • Use in source · Read passage · strengthening.tex, line 334

    Let Γ_R(P,Q) be the R-subalgebra generated by all γ_m(P) and γ_r(Q). Adjoin E, F with E^2=1+2P, F^2=1+2Q, and call the resulting algebra A_0. Its elements are finite expressions; no completion is being presumed. It is generated as a Γ_R(P,Q)-module by 1, E, F, E F. The functional maps this module to R. For an even basis term this is [source reference] and independence. For a basis term with an odd Euler factor the moment is zero, since P and Q are even polynomials in their respective variables.

    Exact mathematical passage
    \Gamma_R(P,Q)
\gamma_rUse in source · 5 passages

Search aliases: γ_r gamma

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  • Use in source · Read passage · strengthening.tex, line 335

    Let Γ_R(P,Q) be the R-subalgebra generated by all γ_m(P) and γ_r(Q). Adjoin E, F with E^2=1+2P, F^2=1+2Q, and call the resulting algebra A_0. Its elements are finite expressions; no completion is being presumed. It is generated as a Γ_R(P,Q)-module by 1, E, F, E F. The functional maps this module to R. For an even basis term this is [source reference] and independence. For a basis term with an odd Euler factor the moment is zero, since P and Q are even polynomials in their respective variables.

    Exact mathematical passage
    \gamma_r(Q)
\Delta^kUse in source · 2 passages

Search aliases: Δ^k Delta

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  • Use in source · Read passage · 72_recovered.tex, line 1641

    [Polynomial exponential Mahler transfer] Let f(n)=P(n)u^n with P∈ Z_p[n] of degree d. Let H=Λ_p,L, fix an address j, and write Q_j(x)=P(Hx+j), y=u^H, q=y-1. Then f(Hx+j)=∑_r≥0c_j,rbinom xr, where c_j,r = u^j ∑_k=0^min(d,r) binom rk Δ^kQ_j(0)y^kq^r-k. Consequently, v_p(c_j,r)≥ Lmax(r-d,0). For R≥ d, truncation at depth R is valid modulo p^L(R+1-d) uniformly on Z_p.

    Exact mathematical passage
    
    \boxed{
    c_{j,r}
    =
    u^j
    \sum_{k=0}^{\min(d,r)}
    \binom rk
    \Delta^kQ_j(0)y^kq^{r-k}.
    }
    
\Delta^mUse in source · 1 passage

Search aliases: Δ^m Delta

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  • Use in source · Read passage · strengthening.tex, line 390

    The conversion from unit steps to arbitrary mixed steps is used several times, so it is useful to make it explicit. Write Δ=Δ_1. For H>0, Δ_H=∑_m=1^Hbinom HmΔ^m. A mixed term with local orders m_1,…,m_k≥1 and M=∑_ν m_ν therefore has valuation at least M+v_2(M!)+β +∑_ν(v_2(H_ν)-v_2(m_ν)) if the unit-order estimate has offset β for every positive order. The inequalities m-1≥ v_2(m), M≥ k, v_2(M!)≥ v_2(k!) make this at least k+v_2(k!)+β+∑_ν v_2(H_ν). Every expansion is finite. This proves the desir…

    Exact mathematical passage
    
     \Delta_H=\sum_{m=1}^H\binom Hm\Delta^m.
     \label{eq:finite-step-expansion}
    
\Delta^rUse in source · 1 passage

Search aliases: Δ^r Delta

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  • Use in source · Read passage · 72_recovered.tex, line 1461

    [Finite Mahler reconstruction] Given f(0),…,f(N), put c_r=Δ^rf(0). Then f(n)=∑_r=0^nc_rbinom nr (0≤ n≤ N). For every s with 0≤ s≤ N, Δ^sf(n) = ∑_r=s^Nc_rbinom nr-s (0≤ n≤ N-s).

    Exact mathematical passage
    c_r=\Delta^rf(0)
\Delta^sUse in source · 6 passages

Search aliases: Δ^s Delta

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  • Use in source · Read passage · 72_recovered.tex, line 1469

    [Finite Mahler reconstruction] Given f(0),…,f(N), put c_r=Δ^rf(0). Then f(n)=∑_r=0^nc_rbinom nr (0≤ n≤ N). For every s with 0≤ s≤ N, Δ^sf(n) = ∑_r=s^Nc_rbinom nr-s (0≤ n≤ N-s).

    Exact mathematical passage
    
    \boxed{
    \Delta^sf(n)
    =
    \sum_{r=s}^{N}c_r\binom n{r-s}
    \qquad(0\leq n\leq N-s).
    }
    
\delta^sUse in source · 5 passages

Search aliases: δ^s delta

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  • Use in source · Read passage · supplement_07.tex, line 163

    0, · Nnmid d, N=mp. Let L=p(p-1) and δ=U^p+L-U^p=U^p^2-U^p. The source proves the operator dictionary (δ^sX)(d)=Δ_L^s X(d+ps). At the valid nonnegative coefficient degree d=mpn-ps, with s≤ mn, this becomes (δ^sX)(mpn-ps)=Δ_h^s e(n), h=(p-1)/m. The packed sequence and difference operations are definitions. Their coefficient correspondence is proved before any divisibility assumption is introduced.

    Exact mathematical passage
    
     (\delta^sX)(d)=\Delta_L^s X(d+ps).
    
\delta^{s-1}Use in source · 1 passage

Search aliases: δ^s-1 delta

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  • Use in source · Read passage · supplement_08.tex, line 113

    pageheading3. A nonsingular Euler-operator argumentThe standard interval without an analytic power-series-in-the-index premise On Hurwitz coefficients, the actual Euler operator plus c is (Λ_c a)(d)=(d+c)a(d). Direct induction proves its commutator with every defect iterate: δ^sΛ_c a =Λ_cδ^s a +sp(pU^p^2-U^p)δ^s-1a, s≥1. Suppose a is integral and supported at degrees divisible by p, and pnmid c. Every supported pivot d+c is proved to be a unit of Z_p. The defect preserves this support. Induction…

    Exact mathematical passage
    
     \delta^s\Lambda_c a
     =\Lambda_c\delta^s a
      +sp\bigl(pU^{p^2}-U^p\bigr)\delta^{s-1}a,
     \qquad s\ge1.
    
\Delta_{(p-1)p^a}^{\,r}Use in source · 1 passage

Search aliases: Δ_(p-1)p^a^ r Delta

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  • Use in source · Read passage · 72_recovered.tex, line 1589

    [Basic odd and dyadic difference gains] For odd p, Δ_(p-1)p^a^ ru^n = u^n(u^(p-1)p^a-1)^r, v_p(Δ_(p-1)p^a^ ru^n)≥(a+1)r. For odd u∈ Z_2^×, Δ_2^a+1^ ru^n = u^n(u^2^a+1-1)^r, v_2(Δ_2^a+1^ ru^n)≥(a+3)r.

    Exact mathematical passage
    
    \boxed{
    \Delta_{(p-1)p^a}^{\,r}u^n
    =
    u^n\left(u^{(p-1)p^a}-1\right)^r,
    \qquad
    \vp\left(\Delta_{(p-1)p^a}^{\,r}u^n\right)\geq(a+1)r.
    }
    
\delta_0Use in source · 2 passages

Search aliases: δ_0 delta

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  • Use in source · Read passage · 72_recovered.tex, line 842

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    f=d+c\delta_0
\Delta_{12\cdot2^a}^{\,r}Use in source · 1 passage

Search aliases: Δ_12·2^a^ r Delta cdot

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  • Use in source · Read passage · 72_recovered.tex, line 2433

    18 · 11 . Therefore Δ_12·2^a^ rF_n ≡0 pmod 2^(a+3)r. Every dyadic refinement doubles the step and adds one guaranteed binary digit at each Mahler depth.

    Exact mathematical passage
    
    \boxed{
    \Delta_{12\cdot2^a}^{\,r}F_n
    \equiv0
    \pmod {2^{(a+3)r}}.
    }
    
\Delta_1^2Use in source · 3 passages

Search aliases: Δ_1^2 Delta

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  • Use in source · Read passage · supplement_08.tex, line 189

    pageheading5. Examples, verification and the remaining gapThe coverage count is unchanged; the unresolved theorem now has a stronger proved part The original rational recurrence gives C_6,2(1)=-frac128, C_6,2(2)=frac425364, C_6,2(3)=-frac20397592660. Consequently, the new source proves Δ_1^2 C_6,2(1)=-frac9499591235, v_3 (Δ_1^2 C_6,2(1))=2. This attains the bound min(2,2·1)=2. A claim of a third power is rejected. The separate cubic example proves that the normalized differential defect is not a…

    Exact mathematical passage
    
     \Delta_1^2 C_{6,2}(1)=-\frac{949959}{1235},\qquad
     v_3\!\left(\Delta_1^2 C_{6,2}(1)\right)=2.
    
\Delta_1^iUse in source · 3 passages

Search aliases: Δ_1^i Delta

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  • Use in source · Read passage · 20_arithmetic.tex, line 498

    g_2(n) · =frac1(4n+1)(2n+1), · g_3(n) · =frac3(4n+1)(2n+1)(4n+3). Every denominator is odd. An affine sequence with even slope satisfies 2^kmidΔ_1^k f(n) for every k,n. The exact product identity Δ_1^s(fg)(n)=∑_i=0^sbinom si Δ_1^if(n)Δ_1^s-ig(n+i) proves closure of this property under products. Applying the formula with the factors interchanged to fg=1 isolates f(n+s)Δ_1^sg(n). Multiplication by f(n+s)^-1 and induction then prove closure under pointwise unit inverses. Thus all four g_j have the …

    Exact mathematical passage
    
     \Delta_1^s(fg)(n)=\sum_{i=0}^s\binom si
              \Delta_1^if(n)\Delta_1^{s-i}g(n+i)
    
\Delta_1^kUse in source · 27 passages

Search aliases: Δ_1^k Delta

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  • Use in source · Read passage · 10_foundations.tex, line 512

    affine Mahler kernel by Θ_M,s(r,k)=[z^r](1+z)^s((1+z)^M-1)^k =∑_j=0^k(-1)^k-jbinom kjbinomMj+sr. It vanishes unless k≤ r≤ Mk+s. For a finite binomial expansion f(n)=∑_r=0^Dc_rbinom nr, the kth Newton coefficient of g(q)=f(Mq+s) is therefore c_k(g)=∑_r=k^min(D,Mk+s)c_rΘ_M,s(r,k). Indeed, insert the finite stencil for Δ_1^k g(0) and interchange its two finite sums. Coefficient extraction of the binomial expansion of ((1+z)^M-1)^k gives the kernel. In the binary zero-address case, Θ_2,0(k+j,k)=2^k-…

    Exact mathematical passage
    \Delta_1^k g(0)
\Delta_1^{k-r}Use in source · 1 passage

Search aliases: Δ_1^k-r Delta

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  • Use in source · Read passage · strengthening.tex, line 159

    ence identity is Δ_1^kfrac11+sn =frac(-s)^k k!∏_h=0^k(1+s(n+h)). It follows by induction, subtracting adjacent rational functions; each successive subtraction adds the factor -(k+1)s in the numerator. Its denominator is a unit, so it belongs to p^k Z_p. The discrete Leibniz formula Δ_1^k(fg)(n) =∑_r=0^kbinom kr (Δ_1^r f)(n)(Δ_1^k-rg)(n+r) then proves Δ_1^k g_N,j(n)∈ p^k Z_p for every k,n. The coefficient at degree Nn of L_1^kG_N,j is precisely p^-kΔ_1^k g_N,j(n), and the other residue classes va…

    Exact mathematical passage
    
     \Delta_1^k(fg)(n)
     =\sum_{r=0}^k\binom kr
           (\Delta_1^r f)(n)(\Delta_1^{k-r}g)(n+r)
     \label{eq:discrete-leibniz-final}
    
\Delta_1^mUse in source · 4 passages

Search aliases: Δ_1^m Delta

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  • Use in source · Read passage · strengthening.tex, line 203

    To retain the valuation of a step, let B be any sequence satisfying Δ_1^m B∈ p^m Z_p at every order and row. Use the finite operator identity Δ_H=∑_i=1^Hbinom HiΔ_1^i. The relation ibinom Hi=HbinomH-1i-1 implies v_p(p^ibinom Hi) ≥ i+v_p(H)-v_p(i)≥1+v_p(H). Multiplying these identities for the k steps, every term of total order M has its scalar multiplied by p^M, and is divisible by p^k+∑_ν v_p(H_ν). This proves the mixed-step bound. Finally, Δ_H A_N,j(n)=(-1)^n+HT_H C_N,j(n). Iteration contribut…

    Exact mathematical passage
    \Delta_1^m B\in p^m\mathbb Z_p
\Delta_1^rUse in source · 5 passages

Search aliases: Δ_1^r Delta

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  • Use in source · Read passage · 10_foundations.tex, line 249

    For f: N→ R, define the step-h forward difference and its iterates by Δ_h f(n)=f(n+h)-f(n), Δ_h^0 f=f, and put c_r(f)=Δ_1^r f(0). Thus the coefficients are determined by the differences of f.

    Exact mathematical passage
    c_r(f)=\Delta_1^r f(0)
\Delta_1^{s-i}Use in source · 2 passages

Search aliases: Δ_1^s-i Delta

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  • Use in source · Read passage · 20_arithmetic.tex, line 498

    g_2(n) · =frac1(4n+1)(2n+1), · g_3(n) · =frac3(4n+1)(2n+1)(4n+3). Every denominator is odd. An affine sequence with even slope satisfies 2^kmidΔ_1^k f(n) for every k,n. The exact product identity Δ_1^s(fg)(n)=∑_i=0^sbinom si Δ_1^if(n)Δ_1^s-ig(n+i) proves closure of this property under products. Applying the formula with the factors interchanged to fg=1 isolates f(n+s)Δ_1^sg(n). Multiplication by f(n+s)^-1 and induction then prove closure under pointwise unit inverses. Thus all four g_j have the …

    Exact mathematical passage
    
     \Delta_1^s(fg)(n)=\sum_{i=0}^s\binom si
              \Delta_1^if(n)\Delta_1^{s-i}g(n+i)
    
\Delta_2Use in source · 4 passages

Search aliases: Δ_2 Delta

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  • Use in source · Read passage · supplement_04.tex, line 190

    What fails when hypotheses or conventions are dropped. The nonunit u=3 at p=3 gives Δ_2(u^n)|_n=0=8, with no factor of three. A constant unit has every positive-order difference zero, hence infinite extended valuation. The finite-sample counterexample on page 2 distinguishes a true interpolant identity from an unsupported claim about unknown continuation.

    Exact mathematical passage
    \Delta_2(u^n)|_{n=0}=8
\Delta_{24}^{\,r}Use in source · 1 passage

Search aliases: Δ_24^ r Delta

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  • Use in source · Read passage · 72_recovered.tex, line 1759

    1932 · 1194 . For every scalar coordinate of the state sequence C^nv_0, Δ_24^ ra_n ≡0 pmod 24^r. One integral jump therefore gives simultaneous congruence ladders at the primes two and three.

    Exact mathematical passage
    
    \boxed{
    \Delta_{24}^{\,r}a_n
    \equiv0
    \pmod {24^r}.
    }
    
\Delta_{2^{a+1}}^{\,r}Use in source · 1 passage

Search aliases: Δ_2^a+1^ r Delta

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  • Use in source · Read passage · 72_recovered.tex, line 1599

    [Basic odd and dyadic difference gains] For odd p, Δ_(p-1)p^a^ ru^n = u^n(u^(p-1)p^a-1)^r, v_p(Δ_(p-1)p^a^ ru^n)≥(a+1)r. For odd u∈ Z_2^×, Δ_2^a+1^ ru^n = u^n(u^2^a+1-1)^r, v_2(Δ_2^a+1^ ru^n)≥(a+3)r.

    Exact mathematical passage
    
    \boxed{
    \Delta_{2^{a+1}}^{\,r}u^n
    =
    u^n\left(u^{2^{a+1}}-1\right)^r,
    \qquad
    \vTwo\left(\Delta_{2^{a+1}}^{\,r}u^n\right)\geq(a+3)r.
    }
    
\Delta_4Use in source · 1 passage

Search aliases: Δ_4 Delta

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  • Use in source · visual_body.tex, line 154

    For this stencil, the general dyadic bound is 3+v_2(1)+v_2(2)+v_2(4)=6. At rank four the factorial theorem improves it to seven for addresses zero and one. Exact rational evaluation gives v_2(Δ_1Δ_2Δ_4 C_0(0))=7. The order is odd, so the mixed-step equality criterion also proves this valuation at every starting row.

    Exact mathematical passage
    v_2(\Delta_1\Delta_2\Delta_4 C_0(0))=7
\Delta_4^2Use in source · 1 passage

Search aliases: Δ_4^2 Delta

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  • Use in source · Read passage · supplement_04.tex, line 184

    A sharp unit-mode gain. For p=5 and u=2, Δ_4^2(u^n)|_n=0=(2^4-1)^2=225. The source proves 5^2mid225 and 5^3nmid225. This example attains the guaranteed gain; no additional uniform power of p is claimed.

    Exact mathematical passage
    
     \Delta_4^2(u^n)\big|_{n=0}=(2^4-1)^2=225.
    
\Delta_4^sUse in source · 1 passage

Search aliases: Δ_4^s Delta

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  • Use in source · Read passage · 20_arithmetic.tex, line 541

    Packing g_j into degrees divisible by four gives K_4,j. Its source property is precisely charge-one flatness. Inversion gives charge -1, and the charge shift restores charge one for the reciprocal P_4,j. If dec(F)(d)=d![t^d]F, the exact coefficient identity is dec(A_1^sF)(d) =2^-sΔ_4^sdec(F)(d). At d=4n, packing turns the right side into 2^-sΔ_1^sC_j(n). This extraction requires no negative degree and therefore holds at every n, including zero. It proves the unit-step bound leanKomorebiPaperIX.D…

    Exact mathematical passage
    
     \operatorname{dec}(A_1^sF)(d)
           =2^{-s}\Delta_4^s\operatorname{dec}(F)(d).
    
\delta_{d0}Use in source · 3 passages

Search aliases: δ_d0 delta

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  • Use in source · Read passage · supplement_07.tex, line 39

    The identity sequence is 1(d)=δ_d0. The new source proves the actual higher product rule U^a(f⋆ g)=∑_i=0^abinom ai (U^i f)⋆(U^a-ig). The argument starts with Pascal's identity for the first shift, then inducts on a. All sums are finite at each coefficient.

    Exact mathematical passage
    \mathbf1(d)=\delta_{d0}
\Delta_HUse in source · 5 passages

Search aliases: Δ_H Delta

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  • Use in source · Read passage · strengthening.tex, line 31

    Write S f(n)=f(n+1) and Δ_H=S^H-1. A mixed difference Δ_H_1⋯Δ_H_k acts on the row variable n, rather than on the full Euler subscript. We use v_p(0)=+∞. Orders and starting rows are nonnegative integers; steps are positive integers unless stated otherwise. An empty mixed difference is the original value, and a zero ordinary step makes every nonempty product containing it zero.

    Exact mathematical passage
    \Delta_H=S^H-1
\Delta_H^kUse in source · 5 passages

Search aliases: Δ_H^k Delta

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  • Use in source · Read passage · strengthening.tex, line 232

    [Threshold-free dyadic congruence] For N=2^a, a≥0, all permitted j, and n≥0, v_2(Δ_H_1⋯Δ_H_kE_Nn^(N,j)) ≥ v_2(j!)+k+∑_ν=1^k v_2(H_ν). In particular, Δ_H^k gains k(1+v_2(H)) bits before normalization. For r>v_2(j!), the row displacement H=2^max(0,r-v_2(j!)-1) is a sufficient period modulo 2^r from row zero. If r≤ v_2(j!), all original coefficients vanish modulo 2^r.

    Exact mathematical passage
    \Delta_H^k
\Delta_h^{k-mn}Use in source · 1 passage

Search aliases: Δ_h^k-mn Delta

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  • Use in source · Read passage · 20_arithmetic.tex, line 339

    Here is the coefficient extraction explicitly. If a_d=d![t^d]F, then d![t^d]B_p^kF=p^-kΔ_p(p-1)^ka_d+pk. For F=j! P_N,j and k≤ mn, take d=mpn-pk. Support in multiples of mp turns this into p^-kΔ_h^kU_N,j(n), so flatness gives divisibility by p^k. For k>mn, expand Δ_h^k-mn applied to Δ_h^mnU_N,j(n); every shifted row is at least n, so each summand is divisible by p^mn. Finally Δ_uh=((I+Δ_h)^u-I), whose kth power for u>0 is an integral combination of powers of Δ_h of order at least k. The case u=0…

    Exact mathematical passage
    \Delta_h^{k-mn}
\Delta_h^{mn}Use in source · 1 passage

Search aliases: Δ_h^mn Delta

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  • Use in source · Read passage · 20_arithmetic.tex, line 340

    Here is the coefficient extraction explicitly. If a_d=d![t^d]F, then d![t^d]B_p^kF=p^-kΔ_p(p-1)^ka_d+pk. For F=j! P_N,j and k≤ mn, take d=mpn-pk. Support in multiples of mp turns this into p^-kΔ_h^kU_N,j(n), so flatness gives divisibility by p^k. For k>mn, expand Δ_h^k-mn applied to Δ_h^mnU_N,j(n); every shifted row is at least n, so each summand is divisible by p^mn. Finally Δ_uh=((I+Δ_h)^u-I), whose kth power for u>0 is an integral combination of powers of Δ_h of order at least k. The case u=0…

    Exact mathematical passage
    \Delta_h^{mn}U_{N,j}(n)
\Delta_H^rUse in source · 2 passages

Search aliases: Δ_H^r Delta

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  • Use in source · Read passage · 72_recovered.tex, line 1691

    =1, which is not in 3 Z_3. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses. Assume C^H=I+p^LB with B∈ M_d( Z_p). For every address j and every x∈ Z_p, a_Hx+j = ∑_r≥0 p^Lrbinom xr w^TC^jB^rv_0. The series converges on Z_p, and Δ_H^ra_n ≡0 pmod p^Lr.

    Exact mathematical passage
    
    \boxed{
    \Delta_H^ra_n
    \equiv0
    \pmod {p^{Lr}}.
    }
    
\Delta_h^sUse in source · 8 passages

Search aliases: Δ_h^s Delta

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  • Use in source · Read passage · 10_foundations.tex, line 255

    [Finite Newton--Mahler identities] For every n,s,h∈ N, f(n) · =∑_r=0^nc_r(f)binom nr, c_r(Δ_1^s f) · =c_s+r(f), Δ_1^s f(n) · =∑_j=0^nc_s+j(f)binom nj, Δ_h^s f(n) · =∑_j=0^s(-1)^s-jbinom sj f(n+jh). The finite interpolant I_Nf(n)=∑_r=0^Nc_r(f)binom nr agrees with f for n≤ N. It has the same order-s unit-step difference at n whenever n+s≤ N. For arbitrary coefficients c_0,…,c_N, the corresponding binomial interpolant has exactly those coefficients and zero coefficients above N.

    Exact mathematical passage
    
    \begin{aligned}
    f(n)&=\sum_{r=0}^{n}c_r(f)\binom nr,\\
    c_r(\Delta_1^s f)&=c_{s+r}(f),\\
    \Delta_1^s f(n)&=\sum_{j=0}^{n}c_{s+j}(f)\binom nj,\\
    \Delta_h^s f(n)&=\sum_{j=0}^{s}(-1)^{s-j}\binom sj f(n+jh).
    \end{aligned}
    
\Delta_{H_1}Use in source · 11 passages

Search aliases: Δ_H_1 Delta

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  • Use in source · Read passage · strengthening.tex, line 32

    Write S f(n)=f(n+1) and Δ_H=S^H-1. A mixed difference Δ_H_1⋯Δ_H_k acts on the row variable n, rather than on the full Euler subscript. We use v_p(0)=+∞. Orders and starting rows are nonnegative integers; steps are positive integers unless stated otherwise. An empty mixed difference is the original value, and a zero ordinary step makes every nonempty product containing it zero.

    Exact mathematical passage
    \Delta_{H_1}\cdots\Delta_{H_k}
\Delta_{H_a}^{\,r}Use in source · 4 passages

Search aliases: Δ_H_a^ r Delta

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  • Use in source · Read passage · 72_recovered.tex, line 2338

    2^a+1, · p=2. Then Δ_H_a^ ru^n = u^n(u^H_a-1)^r. For odd p, v_p(Δ_H_a^ ru^n) ≥(a+1)r. For p=2 and odd u, v_2(Δ_H_a^ ru^n) ≥(a+3)r. Equivalently, every additional prime power digit increases the certified Mahler slope by one.

    Exact mathematical passage
    
    \boxed{
    \Delta_{H_a}^{\,r}u^n
    =
    u^n\left(u^{H_a}-1\right)^r.
    }
    
\Delta_{H_k}Use in source · 11 passages

Search aliases: Δ_H_k Delta

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  • Use in source · Read passage · strengthening.tex, line 32

    Write S f(n)=f(n+1) and Δ_H=S^H-1. A mixed difference Δ_H_1⋯Δ_H_k acts on the row variable n, rather than on the full Euler subscript. We use v_p(0)=+∞. Orders and starting rows are nonnegative integers; steps are positive integers unless stated otherwise. An empty mixed difference is the original value, and a zero ordinary step makes every nonempty product containing it zero.

    Exact mathematical passage
    \Delta_{H_1}\cdots\Delta_{H_k}
\Delta_L^sUse in source · 1 passage

Search aliases: Δ_L^s Delta

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  • Use in source · Read passage · supplement_07.tex, line 163

    0, · Nnmid d, N=mp. Let L=p(p-1) and δ=U^p+L-U^p=U^p^2-U^p. The source proves the operator dictionary (δ^sX)(d)=Δ_L^s X(d+ps). At the valid nonnegative coefficient degree d=mpn-ps, with s≤ mn, this becomes (δ^sX)(mpn-ps)=Δ_h^s e(n), h=(p-1)/m. The packed sequence and difference operations are definitions. Their coefficient correspondence is proved before any divisibility assumption is introduced.

    Exact mathematical passage
    
     (\delta^sX)(d)=\Delta_L^s X(d+ps).
    
\Delta_{p(p-1)}^kUse in source · 1 passage

Search aliases: Δ_p(p-1)^k Delta

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  • Use in source · Read passage · 20_arithmetic.tex, line 333

    Here is the coefficient extraction explicitly. If a_d=d![t^d]F, then d![t^d]B_p^kF=p^-kΔ_p(p-1)^ka_d+pk. For F=j! P_N,j and k≤ mn, take d=mpn-pk. Support in multiples of mp turns this into p^-kΔ_h^kU_N,j(n), so flatness gives divisibility by p^k. For k>mn, expand Δ_h^k-mn applied to Δ_h^mnU_N,j(n); every shifted row is at least n, so each summand is divisible by p^mn. Finally Δ_uh=((I+Δ_h)^u-I), whose kth power for u>0 is an integral combination of powers of Δ_h of order at least k. The case u=0…

    Exact mathematical passage
    
     d![t^d]B_p^kF=p^{-k}\Delta_{p(p-1)}^ka_{d+pk}.
    
\Delta_{p-1}^sUse in source · 2 passages

Search aliases: Δ_p-1^s Delta

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  • Use in source · Read passage · 10_foundations.tex, line 339

    [Differences of unit modes] Over every commutative ring, Δ_h^s(u^n)=u^n(u^h-1)^s. For any prime p, units u_i∈ Z_p^×, coefficients a_i∈ Z_p, and a finite index set I, the sequence f(n)=∑_i∈ Ia_i u_i^n satisfies Δ_p-1^s f(n)∈ p^s Z_p (n,s∈ N).

    Exact mathematical passage
    
    \Delta_{p-1}^s f(n)\in p^s\Zp
    \qquad(n,s\in\N).
    
\delta_rUse in source · 4 passages

Search aliases: δ_r delta

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  • Use in source · Read passage · root_scope.tex, line 387

    [Exact quadratic compression] There is a unique preparation K_r(Z)= U_r(Z) W_r(Z), W_r(Z)=Z^2+2β_rZ+α_r, U_r≡ u_rpmod256. With δ_r=β_r^2-α_r and τ_r= U_r'(c_r)/ U_r(c_r), δ_r=-fracΦ_r^act U_r(c_r) (1-fracτ_r^2Φ_r^act4 U_r(c_r)), v_2(δ_r)=v_2(Φ_r^act). The equality includes the zero case. The same successive monic polynomial division used in Theorem~[source reference], started at precision eight, preserves the quadratic and unit reductions of [source reference]. Write W=Z^2+2β_rZ+α_r and w=W(c_r)…

    Exact mathematical passage
    \delta_r=\beta_r^2-\alpha_r
\delta_{rs}Use in source · 1 passage

Search aliases: δ_rs delta

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  • Use in source · Read passage · 72_recovered.tex, line 2985

    [Complete residue packet system] For fixed N, the projectors P_N,r satisfy P_N,rP_N,s = δ_rsP_N,r and ∑_rbmod NP_N,r=I.

    Exact mathematical passage
    
    P_{N,r}P_{N,s}
    =
    \delta_{rs}P_{N,r}
    
\Delta_{u(p-1)/m}^{k}Use in source · 1 passage

Search aliases: Δ_u(p-1)/m^k Delta

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  • Use in source · Read passage · 20_arithmetic.tex, line 571

    [Step-sensitive divisor-branch bound] Under the hypotheses of Theorem~[source reference], for all u,k,n∈ N, Δ_u(p-1)/m^kU_mp,j(n) ∈ p^min(k(1+ν_p(u)),mn) Z_p. For C_mp,j the corresponding valuation lower bound is min(k(1+ν_p(u)),mn)-v_p(j!).

    Exact mathematical passage
    
     \Delta_{u(p-1)/m}^{k}U_{mp,j}(n)
           \in p^{\min(k(1+\nu_p(u)),mn)}\Z_p.
    
\Delta_{uh}Use in source · 1 passage

Search aliases: Δ_uh Delta

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  • Use in source · Read passage · 20_arithmetic.tex, line 342

    ]F, then d![t^d]B_p^kF=p^-kΔ_p(p-1)^ka_d+pk. For F=j! P_N,j and k≤ mn, take d=mpn-pk. Support in multiples of mp turns this into p^-kΔ_h^kU_N,j(n), so flatness gives divisibility by p^k. For k>mn, expand Δ_h^k-mn applied to Δ_h^mnU_N,j(n); every shifted row is at least n, so each summand is divisible by p^mn. Finally Δ_uh=((I+Δ_h)^u-I), whose kth power for u>0 is an integral combination of powers of Δ_h of order at least k. The case u=0 gives zero for k>0, and order zero is the original integral…

    Exact mathematical passage
    \Delta_{uh}=((I+\Delta_h)^u-I)
\Delta_{uh}^kUse in source · 1 passage

Search aliases: Δ_uh^k Delta

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  • Use in source · Read passage · 20_arithmetic.tex, line 346

    pplied to Δ_h^mnU_N,j(n); every shifted row is at least n, so each summand is divisible by p^mn. Finally Δ_uh=((I+Δ_h)^u-I), whose kth power for u>0 is an integral combination of powers of Δ_h of order at least k. The case u=0 gives zero for k>0, and order zero is the original integral coefficient. This proves the bound for all u and k. Finally Δ_uh^kU_N,j=j!Δ_uh^kC_N,j gives [source reference], with the zero case retained. The resulting statements are leanKomorebiPaperIX.AllResidueEuler.all_res…

    Exact mathematical passage
    \Delta_{uh}^kU_{N,j}=j!\Delta_{uh}^kC_{N,j}
\epsilonUse in source · 22 passages

Search aliases: ϵ epsilon

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  • Use in source · Read passage · 10_foundations.tex, line 375

    Z/p^L+1 Z)^×. Define ω_L(a)=a^p^L, ⟨ a⟩_L=aω_L(a)^-1. Then a=ω_L(a)⟨ a⟩_L, ω_L(a)^p-1=1, ⟨ a⟩_L≡1pmod p. The two factors are uniquely determined by these properties and are compatible with reduction to lower precision. For a∈( Z/2^L+2 Z)^×, there is instead a unique decomposition a=ϵ(a)⟨ a⟩_4, ϵ(a)∈{1,-1}, ⟨ a⟩_4≡1pmod4.

    Exact mathematical passage
    
    a=\epsilon(a)\langle a\rangle_4,
    \qquad \epsilon(a)\in\{1,-1\},\qquad
    \langle a\rangle_4\equiv1\pmod4.
    
\varepsilonUse in source · 10 passages

Search aliases: ε varepsilon

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  • Use in source · Read passage · 30_lifting.tex, line 767

    [Weighted coefficient deformation] Let F,H∈ R[X_1,…,X_d]^d, F(a)=0, and J=DF(a). Suppose J∈GL_d( Q_p), p^sJ^-1∈ M_d(R), and M>2s. For G=F+p^MH, there is exactly one root b∈ a+p^s+1R^d. It satisfies b-a∈ p^M-sR^d and, in Q_p^d, b-a=-p^MJ^-1H(a)+ε, ε∈ p^2M-3sR^d. Only the original Jacobian DF(a) is subject to an inverse hypothesis.

    Exact mathematical passage
    \label{eq:coefficient-first-order}
     b-a=-p^MJ^{-1}H(a)+\varepsilon,
     \qquad\varepsilon\in p^{2M-3s}R^d.
    
\epsilon^2Use in source · 2 passages

Search aliases: ϵ^2 epsilon

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  • Use in source · Read passage · strengthening.tex, line 683

    There is a still stronger filtration. Put X=P+Q, τ=i B D, ϵ=1+τ. Then the exact identities are H_2=ϵ X+2G, G=-γ_2(P)-γ_2(Q)+3PQ-τ Q, ϵ^2=2V, V=τ-2X-8PQ. Every divided power of G is integral component by component. The finite addition identity gives γ_k(H_2)=∑_r=0^k2^rγ_r(G) ϵ^k-rγ_k-r(X) ∈2^lfloor k/2rfloor A_2, because the rth term belongs to 2^r+lfloor(k-r)/2rfloor A_2. Applying the functional, rational descent, and the base-index expansion proves the unit-step version of [source reference]. F…

    Exact mathematical passage
    
     H_2=\epsilon X+2G,\quad
     G=-\gamma_2(P)-\gamma_2(Q)+3PQ-\tau Q,\quad
     \epsilon^2=2V,\quad V=\tau-2X-8PQ.
     \label{eq:address-two-filtration-algebra}
    
\epsilon^jUse in source · 1 passage

Search aliases: ϵ^j epsilon

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  • Use in source · Read passage · strengthening.tex, line 801

    ϵ^2=2(τ-P-Q-2PQ). The divided powers of G are integral by the same component formulas used above. Thus modulo 2 A_0, γ_k(H_0) equals the divided power of γ_2(X)+ϵ X. The identity γ_r(γ_2(X)) =frac(2r)!2^r r!γ_2r(X) has an odd integer scalar. Every term involving ϵ^j for j≥2 vanishes modulo two. The surviving zero and first powers give γ_k(H_0)≡ γ_2k(X)+ϵγ_2k-1(X) pmod2 A_0. The multiplication coefficient in the second term is 2k-1, also odd. By the odd moments [source reference] and independence…

    Exact mathematical passage
    \epsilon^j
\epsilon^{k-r}Use in source · 1 passage

Search aliases: ϵ^k-r epsilon

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  • Use in source · Read passage · strengthening.tex, line 691

    Put X=P+Q, τ=i B D, ϵ=1+τ. Then the exact identities are H_2=ϵ X+2G, G=-γ_2(P)-γ_2(Q)+3PQ-τ Q, ϵ^2=2V, V=τ-2X-8PQ. Every divided power of G is integral component by component. The finite addition identity gives γ_k(H_2)=∑_r=0^k2^rγ_r(G) ϵ^k-rγ_k-r(X) ∈2^lfloor k/2rfloor A_2, because the rth term belongs to 2^r+lfloor(k-r)/2rfloor A_2. Applying the functional, rational descent, and the base-index expansion proves the unit-step version of [source reference]. For a mixed term of total order M, its …

    Exact mathematical passage
    
     \gamma_k(H_2)=\sum_{r=0}^k2^r\gamma_r(G)
                      \epsilon^{k-r}\gamma_{k-r}(X)
              \in2^{\lfloor k/2\rfloor}\mathcal A_2,
     \label{eq:address-two-filtration-pd}
    
\epsilon_uUse in source · 8 passages

Search aliases: ϵ_u epsilon

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  • Use in source · Read passage · 72_recovered.tex, line 40

    This article begins with the Hurwitz product (amathbin⋆ b)_n = ∑_r=0^n binom nr a_rb_n-r. Over a field of characteristic zero it is multiplication of exponential generating functions. Over a finite field it realizes the additive group algebra through the exponential modes ϵ_u=(u^n)_n≥0. This finite algebra gives an exact proof of the standard window period. Hurwitz series rings are treated in [source reference]. The finite difference and umbral background may be found in [source reference].

    Exact mathematical passage
    \epsilon_u=(u^n)_{n\geq0}
\epsilon_{u+v}Use in source · 1 passage

Search aliases: ϵ_u+v epsilon

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  • Use in source · Read passage · 72_recovered.tex, line 153

    [Addition law for modes] For every u,v∈ R, ϵ_umathbin⋆ϵ_v = ϵ_u+v.

    Exact mathematical passage
    
    \boxed{
    \epsilon_u\starh\epsilon_v
    =
    \epsilon_{u+v}.
    }
    
\epsilon_vUse in source · 1 passage

Search aliases: ϵ_v epsilon

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  • Use in source · Read passage · 72_recovered.tex, line 153

    [Addition law for modes] For every u,v∈ R, ϵ_umathbin⋆ϵ_v = ϵ_u+v.

    Exact mathematical passage
    
    \boxed{
    \epsilon_u\starh\epsilon_v
    =
    \epsilon_{u+v}.
    }
    
\epsilon_\zetaUse in source · 1 passage

Search aliases: ϵ_ζ epsilon zeta

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  • Use in source · Read passage · 72_recovered.tex, line 272

    [Subgroup average] The multiple selector has the mode expansion a^(m) = frac1m ∑_ζ∈ H_m ϵ_ζ.

    Exact mathematical passage
    
    \boxed{
    a^{(m)}
    =
    \frac1m
    \sum_{\zeta\in H_m}
    \epsilon_\zeta.
    }
    
\epsilon_{\eta u}Use in source · 1 passage

Search aliases: ϵ_η u epsilon eta

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  • Use in source · Read passage · 72_recovered.tex, line 323

    Multiplication by an element η∈ H_m gives an automorphism of the additive group and therefore an automorphism ϵ_u⟼ϵ_η u of the mode algebra. The selector a^(m) is fixed by every such automorphism, so its unique inverse is fixed as well.

    Exact mathematical passage
    
    \epsilon_u\longmapsto\epsilon_{\eta u}
    
\varepsilon_\ellUse in source · 2 passages

Search aliases: ε_ℓ varepsilon ell

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  • Use in source · Read passage · supplement_05.tex, line 153

    Why the mod-four branch is separate. Now let ℓ≥2 and a∈( Z/2^ℓ Z)^×. The actual units of Z/4 Z are exhausted by 1 and -1. Define ε_ℓ(a)= 1, · a≡1pmod4, -1, · a≡3pmod4, ⟨ a⟩_2,ℓ=ε_ℓ(a)a. The source uses ℓ=L+2. It proves a=ε_ℓ(a)⟨ a⟩_2,ℓ, ε_ℓ(a)∈{1,-1}, ⟨ a⟩_2,ℓ≡1pmod4. The sign is uniquely determined by the residue modulo four, and cancellation gives uniqueness of the principal factor.

    Exact mathematical passage
    
     \varepsilon_\ell(a)=
     \begin{cases}1,&a\equiv1\pmod4,\\-1,&a\equiv3\pmod4,\end{cases}
     \qquad
     \langle a\rangle_{2,\ell}=\varepsilon_\ell(a)a.
    
\zetaUse in source · 6 passages

Search aliases: ζ zeta

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  • Use in source · Read passage · 72_recovered.tex, line 251

    Fix a divisor m of p-1. Let H_m = {ζ∈ F_p^×:ζ^m=1}. This group has order m.

    Exact mathematical passage
    
    H_m
    =
    \{\zeta\in\FF_p^\times:\zeta^m=1\}.
    
\zeta^mUse in source · 1 passage

Search aliases: ζ^m zeta

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  • Use in source · Read passage · 72_recovered.tex, line 251

    Fix a divisor m of p-1. Let H_m = {ζ∈ F_p^×:ζ^m=1}. This group has order m.

    Exact mathematical passage
    
    H_m
    =
    \{\zeta\in\FF_p^\times:\zeta^m=1\}.
    
\zeta^nUse in source · 1 passage

Search aliases: ζ^n zeta

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  • Use in source · Read passage · 72_recovered.tex, line 285

    The nth coefficient on the right is frac1m ∑_ζ∈ H_mζ^n. The character sum is one when mmid n and zero otherwise.

    Exact mathematical passage
    
    \frac1m
    \sum_{\zeta\in H_m}\zeta^n.
    
\etaUse in source · 8 passages

Search aliases: η eta

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  • Use in source · Read passage · 72_recovered.tex, line 322

    Multiplication by an element η∈ H_m gives an automorphism of the additive group and therefore an automorphism ϵ_u⟼ϵ_η u of the mode algebra. The selector a^(m) is fixed by every such automorphism, so its unique inverse is fixed as well.

    Exact mathematical passage
    \eta\in H_m
\eta^nUse in source · 2 passages

Search aliases: η^n eta

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  • Use in source · Read passage · 72_recovered.tex, line 335

    Write b^(m,p) = ∑_u∈ F_pc_uϵ_u. Invariance gives c_η u=c_u. After the change of variable u↦η u, b_n^(m,p) = η^n b_n^(m,p) for every η∈ H_m. If mnmid n, some η satisfies η^n≠1, so the coefficient vanishes.

    Exact mathematical passage
    
    b_n^{(m,p)}
    =
    \eta^n b_n^{(m,p)}
    
\ThetaUse in source · 37 passages

Search aliases: Θ Theta

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  • Use in source · Read passage · 10_foundations.tex, line 502

    For M≥1 and s,r,k∈ N, define the affine Mahler kernel by Θ_M,s(r,k)=[z^r](1+z)^s((1+z)^M-1)^k =∑_j=0^k(-1)^k-jbinom kjbinomMj+sr. It vanishes unless k≤ r≤ Mk+s. For a finite binomial expansion f(n)=∑_r=0^Dc_rbinom nr, the kth Newton coefficient of g(q)=f(Mq+s) is therefore c_k(g)=∑_r=k^min(D,Mk+s)c_rΘ_M,s(r,k). Indeed, insert the finite stencil for Δ_1^k g(0) and interchange its two finite sums. Coefficient extraction of the binomial expansion of ((1+z)^M-1)^k gives the kernel. In the binary zer…

    Exact mathematical passage
    
    \Theta_{M,s}(r,k)=[z^r](1+z)^s\bigl((1+z)^M-1\bigr)^k
    =\sum_{j=0}^{k}(-1)^{k-j}\binom kj\binom{Mj+s}{r}.
    
\Theta^dUse in source · 1 passage

Search aliases: Θ^d Theta

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  • Use in source · Read passage · 20_arithmetic.tex, line 285

    ^d=(p^lfloor d/prfloora_d) (T^p/p)^lfloor d/prfloorT^dbmod p. The operator preserves these monomials and hence finite weighted polynomials. Each fixed output coefficient depends on only finitely many input coefficients, which extends the result to the formal series. Substitution T=e^t-1 intertwines D with Θ and maps W_p into H_p, because d![t^d]F(e^t-1)=[T^0]Θ^dF. It follows that the singular ratios are flat. This transfer is proved in leanKomorebiPaperIX.WeightedDworkTransfer.singular_ratio_fla…

    Exact mathematical passage
    
     d![t^d]F(e^t-1)=[T^0]\Theta^dF.
    
\Theta^nUse in source · 1 passage

Search aliases: Θ^n Theta

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  • Use in source · Read passage · supplement_10.tex, line 101

    The relevant integral subring is the Hurwitz subring H( Z_p)={f∈ Q_p[[t]]: n![t^n]f∈ Z_p for every n}. It is not coefficientwise integrality of ordinary power-series coefficients. The precise connection is n![t^n] E(F)=[T^0]Θ^nF. Since every iterate of Θ preserves W_p, the right side is integral. Therefore E( W_p)⊆ H( Z_p). Applying the same statement to every B_Θ^sF proves f(log(1+T))∈ W_p Longrightarrow B_D^sf∈ H( Z_p) for every s. This is the proved transfer that was missing in the earlier at…

    Exact mathematical passage
    
     n![t^n]\mathcal E(F)=[T^0]\Theta^nF.
    
\Theta^pUse in source · 2 passages

Search aliases: Θ^p Theta

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  • Use in source · Read passage · 20_arithmetic.tex, line 270

    To deduce flatness from the weighted estimate, use Θ=(1+T)d/dT. Both Θ and (Θ^p^2-Θ^p)/p preserve W_p. For the first operator, its coefficient formula is (d+1)a_d+1+da_d; whenever the floor weight rises, the factor d+1 provides the missing p. For the second operator, the proof uses its full Leibniz correction and the scalar congruence p^2mid b^p^2-b^p. Weighted monomials factor as a_dT^d=(p^lfloor d/prfloora_d) (T^p/p)^lfloor d/prfloorT^dbmod p. The operator preserves these monomials and hence f…

    Exact mathematical passage
    (\Theta^{p^2}-\Theta^p)/p
\Theta^{p^2}Use in source · 2 passages

Search aliases: Θ^p^2 Theta

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  • Use in source · Read passage · 20_arithmetic.tex, line 270

    To deduce flatness from the weighted estimate, use Θ=(1+T)d/dT. Both Θ and (Θ^p^2-Θ^p)/p preserve W_p. For the first operator, its coefficient formula is (d+1)a_d+1+da_d; whenever the floor weight rises, the factor d+1 provides the missing p. For the second operator, the proof uses its full Leibniz correction and the scalar congruence p^2mid b^p^2-b^p. Weighted monomials factor as a_dT^d=(p^lfloor d/prfloora_d) (T^p/p)^lfloor d/prfloorT^dbmod p. The operator preserves these monomials and hence f…

    Exact mathematical passage
    (\Theta^{p^2}-\Theta^p)/p
\Theta_{2,0}Use in source · 3 passages

Search aliases: Θ_2,0 Theta

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  • Use in source · Read passage · 10_foundations.tex, line 515

    or a finite binomial expansion f(n)=∑_r=0^Dc_rbinom nr, the kth Newton coefficient of g(q)=f(Mq+s) is therefore c_k(g)=∑_r=k^min(D,Mk+s)c_rΘ_M,s(r,k). Indeed, insert the finite stencil for Δ_1^k g(0) and interchange its two finite sums. Coefficient extraction of the binomial expansion of ((1+z)^M-1)^k gives the kernel. In the binary zero-address case, Θ_2,0(k+j,k)=2^k-jbinom kj (0≤ j≤ k), since (2z+z^2)^k=z^k(2+z)^k. Exact valuations for the general kernel remain to be determined. The recovered …

    Exact mathematical passage
    
    \Theta_{2,0}(k+j,k)=2^{k-j}\binom kj\qquad(0\leq j\leq k),
    
\Theta_{M,s}Use in source · 9 passages

Search aliases: Θ_M,s Theta

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  • Use in source · Read passage · 10_foundations.tex, line 502

    For M≥1 and s,r,k∈ N, define the affine Mahler kernel by Θ_M,s(r,k)=[z^r](1+z)^s((1+z)^M-1)^k =∑_j=0^k(-1)^k-jbinom kjbinomMj+sr. It vanishes unless k≤ r≤ Mk+s. For a finite binomial expansion f(n)=∑_r=0^Dc_rbinom nr, the kth Newton coefficient of g(q)=f(Mq+s) is therefore c_k(g)=∑_r=k^min(D,Mk+s)c_rΘ_M,s(r,k). Indeed, insert the finite stencil for Δ_1^k g(0) and interchange its two finite sums. Coefficient extraction of the binomial expansion of ((1+z)^M-1)^k gives the kernel. In the binary zer…

    Exact mathematical passage
    
    \Theta_{M,s}(r,k)=[z^r](1+z)^s\bigl((1+z)^M-1\bigr)^k
    =\sum_{j=0}^{k}(-1)^{k-j}\binom kj\binom{Mj+s}{r}.
    
\Theta_{MN,Mt+s}Use in source · 1 passage

Search aliases: Θ_MN,Mt+s Theta

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  • Use in source · Read passage · 72_recovered.tex, line 2242

    [Kernel composition] For positive integers M,N, ∑_k≥0 Θ_M,s(r,k) Θ_N,t(k,ℓ) = Θ_MN,Mt+s(r,ℓ). The sum is finite.

    Exact mathematical passage
    
    \boxed{
    \sum_{k\geq0}
    \Theta_{M,s}(r,k)
    \Theta_{N,t}(k,\ell)
    =
    \Theta_{MN,Mt+s}(r,\ell).
    }
    
\Theta_{N,t}Use in source · 1 passage

Search aliases: Θ_N,t Theta

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  • Use in source · Read passage · 72_recovered.tex, line 2242

    [Kernel composition] For positive integers M,N, ∑_k≥0 Θ_M,s(r,k) Θ_N,t(k,ℓ) = Θ_MN,Mt+s(r,ℓ). The sum is finite.

    Exact mathematical passage
    
    \boxed{
    \sum_{k\geq0}
    \Theta_{M,s}(r,k)
    \Theta_{N,t}(k,\ell)
    =
    \Theta_{MN,Mt+s}(r,\ell).
    }
    
\Theta_{p^a,s}Use in source · 5 passages

Search aliases: Θ_p^a,s Theta

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  • Use in source · Read passage · 72_recovered.tex, line 2286

    but its consequence for transformed Mahler coefficients requires c_r∈ Z_p for every r. Convergence of a Q_p-valued Mahler expansion alone is insufficient. For p=2, M=2, s=0 and f(x)=x/2, one has c_1=1/2, c_2=0, while g(q)=q has d_1=1; thus d_1not≡ c_2pmod2. The coefficient consequence is read with integral original coefficients. Let M=p^a. Then Θ_p^a,s(r,k) ≡ binomsr-p^ak pmod p, where the binomial coefficient is zero outside its natural range. Consequently, d_s,k ≡ ∑_u=0^s binom su c_p^ak+u pmo…

    Exact mathematical passage
    
    \boxed{
    \Theta_{p^a,s}(r,k)
    \equiv
    \binom{s}{r-p^ak}
    \pmod p,
    }
    
\iotaUse in source · 8 passages

Search aliases: ι iota

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  • Use in source · Read passage · 40_regularity.tex, line 195

    [Integral coefficients and centers] Let p be any prime. Suppose F∈ Z_p[Y,T]^e, a∈ Z_p^e+k, and F(a)=0. Let ι: Z_p↪ Q_p be the inclusion. If ι(D_YF(a))∈GL_e( Q_p), the polynomial system obtained by including the original coefficients in Q_p has the analytic local root graph of Theorem~[source reference] at ι(a). The strict derivative at the base is -ι(D_YF(a))^-1ι(D_TF(a)). The open neighborhoods and the output map are in the ambient spaces over Q_p. Coefficient inclusion commutes with evaluation…

    Exact mathematical passage
    \iota:\Zp\hookrightarrow\Qp
\kappaUse in source · 26 passages

Search aliases: κ kappa

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  • Use in source · Read passage · 20_arithmetic.tex, line 951

    [Exact weighted inverse coefficient] Let κ_m=β_m2^2m(2m)!. For every m≥1, κ_m=-2^4m(2m-1)B_2m≠0, v_2(κ_m)=4m-1, where B_2m is the usual even Bernoulli number.

    Exact mathematical passage
    \kappa_m=\beta_m2^{2m}(2m)!
\kappa_0Use in source · 1 passage

Search aliases: κ_0 kappa

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  • Use in source · Read passage · root_scope.tex, line 37

    racte^t-1=∑_n≥0B_nfract^nn!, B_1=-frac12. Throughout this appendix, v_2(2)=1 and v_2(0)=∞. Let s_2(n) be the number of ones in the binary expansion of the nonnegative integer n. Define A(N)=frac3^2N+2-14(2N+1)(2N+2), (frac zsinh z)^2=∑_m≥0β_m z^2m, κ_m=2^2m(2m)!β_m. The initial values are β_0=κ_0=1. For m≥1, differentiation of zcoth z gives β_m=-frac2^2m(2m-1)B_2m(2m)!, κ_m=-2^4m(2m-1)B_2m. The classical dyadic Bernoulli denominator law v_2(B_2m)=-1 and Legendre's formula yield v_2(β_m)=s_2(m)-1…

    Exact mathematical passage
    \beta_0=\kappa_0=1
\kappa_{2,1}Use in source · 1 passage

Search aliases: κ_2,1 kappa

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  • Use in source · Read passage · 72_recovered.tex, line 1332

    [The first nontrivial dyadic row] For n=4, R_4=frac9115, R_2=frac53, R_0=1. The exact kernel coefficients are κ_2,1=-frac83, κ_2,2=frac1285. Therefore E_4 = frac9115 -frac83binom42frac53 +frac1285 =5. The rational source values are dyadic units, while the transfer depth is carried by the kernel.

    Exact mathematical passage
    
    \kappa_{2,1}=-\frac83,
    \qquad
    \kappa_{2,2}=\frac{128}{5}.
    
\kappa_{2,2}Use in source · 1 passage

Search aliases: κ_2,2 kappa

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  • Use in source · Read passage · 72_recovered.tex, line 1332

    [The first nontrivial dyadic row] For n=4, R_4=frac9115, R_2=frac53, R_0=1. The exact kernel coefficients are κ_2,1=-frac83, κ_2,2=frac1285. Therefore E_4 = frac9115 -frac83binom42frac53 +frac1285 =5. The rational source values are dyadic units, while the transfer depth is carried by the kernel.

    Exact mathematical passage
    
    \kappa_{2,1}=-\frac83,
    \qquad
    \kappa_{2,2}=\frac{128}{5}.
    
\kappa_{2,r}Use in source · 8 passages

Search aliases: κ_2,r kappa

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  • Use in source · Read passage · 72_recovered.tex, line 86

    The dyadic kernel is substantially sharper than the separate coefficient bound alone suggests. If fracz^2sinh^2z = ∑_r≥0β_rz^2r and κ_2,r=β_r2^2r(2r)!, then v_2(κ_2,r)=4r-1 (r≥1). Consequently, truncation after depth M is uniformly valid modulo 2^4M+3, and this exponent is optimal. The rowwise estimate records the additional binary carries in the binomial coefficient. Classical congruence theory for Euler families is extensive [source reference]. The formulas here are stated as exact coordinate …

    Exact mathematical passage
    
    \kappa_{2,r}=\beta_r2^{2r}(2r)!,
    
\kappa_mUse in source · 11 passages

Search aliases: κ_m kappa

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  • Use in source · Read passage · 20_arithmetic.tex, line 951

    [Exact weighted inverse coefficient] Let κ_m=β_m2^2m(2m)!. For every m≥1, κ_m=-2^4m(2m-1)B_2m≠0, v_2(κ_m)=4m-1, where B_2m is the usual even Bernoulli number.

    Exact mathematical passage
    \kappa_m=\beta_m2^{2m}(2m)!
\kappa_{M+1}Use in source · 1 passage

Search aliases: κ_M+1 kappa

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  • Use in source · Read passage · 20_arithmetic.tex, line 1006

    At n=M+1 there is exactly one omitted summand. Here binom2n2n=1 and A(0)=1, so the error is κ_M+1. Lemma~[source reference] gives its exact valuation 4M+3 and nonvanishing. The statements are leanKomorebiPaperIX.DyadicSharpTower.dyadic_supercongruence_sharp, leanKomorebiPaperIX.DyadicSharpTower.first_omitted_error, and leanKomorebiPaperIX.DyadicSharpTower.first_omitted_error_valuation.

    Exact mathematical passage
    \kappa_{M+1}
\kappa_{p,r}Use in source · 3 passages

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  • Use in source · Read passage · 72_recovered.tex, line 1166

    0, · p=2 and n odd, and κ_p,r = (-1)^rZ_2r+1p^2r, · p odd, β_r2^2r(2r)!, · p=2.

    Exact mathematical passage
    
    \kappa_{p,r}
    =
    \begin{cases}
    (-1)^rZ_{2r+1}p^{2r},&p\text{ odd},\\
    \beta_r2^{2r}(2r)!,&p=2.
    \end{cases}
    
\kappa_rUse in source · 1 passage

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  • Use in source · Read passage · root_scope.tex, line 117

    Set r=M+1≥1 and define P_q(x)=∏_i=0^q-1(x-i)(2x-(2i+1)), P_0=1, H_r(x)=∑_q≥08^qfracβ_r+qβ_r P_q(x)mathscr A(x-q). The coefficient valuations tend to infinity. At any nonnegative integer x, terms with q>x vanish. Cancelling the factorials in [source reference] therefore gives the exact identity T_r-1(N)=κ_rbinom2N2rH_r(N-r) (N≥ r). This is a finite rational identity at every original Euler row.

    Exact mathematical passage
    
     \mathcal T_{r-1}(N)=\kappa_r\binom{2N}{2r}H_r(N-r)
     \qquad(N\ge r).\label{eq:root:factor}
    
\LambdaUse in source · 12 passages

Search aliases: Λ Lambda

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  • Use in source · Read passage · 72_recovered.tex, line 1538

    For odd p and L≥1, put Λ_p,L=(p-1)p^L-1. For p=2 and L≥3, put Λ_2,L=2^L-2. These are exponents of the unit groups modulo the declared prime powers.

    Exact mathematical passage
    
    \Lambda_{p,L}=(p-1)p^{L-1}.
    
\Lambda_{2,L}Use in source · 1 passage

Search aliases: Λ_2,L Lambda

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  • Use in source · Read passage · 72_recovered.tex, line 1542

    For odd p and L≥1, put Λ_p,L=(p-1)p^L-1. For p=2 and L≥3, put Λ_2,L=2^L-2. These are exponents of the unit groups modulo the declared prime powers.

    Exact mathematical passage
    
    \Lambda_{2,L}=2^{L-2}.
    
\Lambda_cUse in source · 5 passages

Search aliases: Λ_c Lambda

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  • Use in source · Read passage · supplement_08.tex, line 109

    pageheading3. A nonsingular Euler-operator argumentThe standard interval without an analytic power-series-in-the-index premise On Hurwitz coefficients, the actual Euler operator plus c is (Λ_c a)(d)=(d+c)a(d). Direct induction proves its commutator with every defect iterate: δ^sΛ_c a =Λ_cδ^s a +sp(pU^p^2-U^p)δ^s-1a, s≥1. Suppose a is integral and supported at degrees divisible by p, and pnmid c. Every supported pivot d+c is proved to be a unit of Z_p. The defect preserves this support. Induction…

    Exact mathematical passage
    
     (\Lambda_c a)(d)=(d+c)a(d).
    
\Lambda_c^{-1}Use in source · 2 passages

Search aliases: Λ_c^-1 Lambda

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  • Use in source · Read passage · supplement_10.tex, line 147

    When pnmid c, define the actual inverse coefficientwise: [t^d]Λ_c^-1f=frac[t^d]fd+c. On p-supported degrees, d+c is a genuine p-adic unit. The proof establishes Hurwitz integrality of this solution, its support, both inverse identities, and preservation of full flatness using the earlier all-order Euler commutator theorem. The singular pivots c=ap are never declared to be units.

    Exact mathematical passage
    
     [t^d]\Lambda_c^{-1}f=\frac{[t^d]f}{d+c}.
    
\Lambda_{j+1}Use in source · 1 passage

Search aliases: Λ_j+1 Lambda

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  • Use in source · Read passage · supplement_08.tex, line 137

    For j=0, g_0 is the literal N-divisibility indicator. Because Nmid p(p-1), the two shifts in δ g_0 agree and δ g_0=0. It is flat at every order. The actual binomial identity gives Λ_j+1g_j+1=(j+1)g_j. As 1≤ j+1<p, the nonsingular reflection theorem inductively proves flatness of every g_j in the standard window. There is no uniform-convergence assumption in this argument.

    Exact mathematical passage
    
     \Lambda_{j+1}g_{j+1}=(j+1)g_j.
    
\Lambda_{p,L}Use in source · 3 passages

Search aliases: Λ_p,L Lambda

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  • Use in source · Read passage · 72_recovered.tex, line 1538

    For odd p and L≥1, put Λ_p,L=(p-1)p^L-1. For p=2 and L≥3, put Λ_2,L=2^L-2. These are exponents of the unit groups modulo the declared prime powers.

    Exact mathematical passage
    
    \Lambda_{p,L}=(p-1)p^{L-1}.
    
\lambda_{p,r}Use in source · 2 passages

Search aliases: λ_p,r lambda

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  • Use in source · Read passage · 72_recovered.tex, line 1223

    [Unified inverse local transfer] Define λ_p,r = 1, · p odd and r=0, 2^2r-1p^2r, · p odd and r≥1, dfrac2^4r+1(2r+1)(2r+2), · p=2. Then for every prime p and every even n≥0, A_p,n = ∑_r=0^n/2 binom n2r λ_p,r E_n-2r. Equivalently, if L_p(t)=K_p(t)^-1, then H_p(t)=L_p(t)secht, where L_p(t)=cosh^2(pt) for odd p, and L_2(t)=(fracsinh(2t)2t)^2 for p=2.

    Exact mathematical passage
    
    \lambda_{p,r}
    =
    \begin{cases}
    1,&p\text{ odd and }r=0,\\
    2^{2r-1}p^{2r},&p\text{ odd and }r\geq1,\\
    \dfrac{2^{4r+1}}{(2r+1)(2r+2)},&p=2.
    \end{cases}
    
\lambda_rUse in source · 4 passages

Search aliases: λ_r lambda

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  • Use in source · Read passage · root_scope.tex, line 124

    The scalar packet weights have exact valuations λ_r(q)=3q+s_2(r+q)-s_2(r). Put μ_r=min_q≥0λ_r(q), J_r={q:λ_r(q)=μ_r}, G_r=2^-μ_rH_r. Since λ_r(0)=0 and λ_r(q)≥3q+1-s_2(r), every minimizing address satisfies 0≤ q≤lfloorfracs_2(r)-13rfloor. In particular μ_r≤0, and the minimum is found by a finite carry calculation. Reduction of the whole series gives overline G_r(X)=∑_q∈ J_rX^underline q in F_2[X]. Its highest degree is max J_r, with leading coefficient one. Thus the normalization is the actual G…

    Exact mathematical passage
    
     \lambda_r(q)=3q+s_2(r+q)-s_2(r).
    
\muUse in source · 23 passages

Search aliases: μ mu

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  • Use in source · Read passage · 72_recovered.tex, line 843

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    
    d_q=\sum_{\chi\in\mu_h}c_\chi\chi^q.
    
\mu_0Use in source · 1 passage

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  • Use in source · Read passage · strengthening.tex, line 614

    odd: T(m,1)=2^m-1 and v_2(m!)=m-1. For m≥4 a power of two, the r=3 term is even. More explicitly, fracT(m,3)2^m-3=h_m-3(1,3,6), ∑_d≥0h_d(1,3,6)t^d =frac1(1-t)(1-3t)(1-6t). Reduction modulo two gives (1-t)^-2, whose degree m-3 coefficient has parity m-2=0. At m=1,2 the r=3 term is absent, and the r=1 term is again odd. Finally μ_0=1.

    Exact mathematical passage
    \mu_0=1
\mu_aUse in source · 3 passages

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  • Use in source · Read passage · strengthening.tex, line 905

    this reduces to γ_k(X)+ϵ Gγ_k-1(X). Moreover, G≡γ_2(X)+τ Qpmod2. After evaluation, every term with a single τ vanishes, so the second contribution reduces to L((γ_2(X)+Q)γ_k-1(X)). The γ_2(X) product has scalar binomk+12, which is even because 4mid k. The other contribution expands as L(Qγ_k-1(P+Q)) =∑_a=0^k-1(k-a)μ_aμ_k-a. An odd summand would require k-a to be odd and μ_k-a to be odd. By Lemma~[source reference], this requires k-a=1. Then a=k-1 is not a power of two and μ_a is even. Hence the …

    Exact mathematical passage
    
     \mathcal L(Q\gamma_{k-1}(P+Q))
        =\sum_{a=0}^{k-1}(k-a)\mu_a\mu_{k-a}.
     \label{eq:two-power-Q-evaluation}
    
\mu_hUse in source · 2 passages

Search aliases: μ_h mu

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  • Use in source · Read passage · 72_recovered.tex, line 843

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    
    d_q=\sum_{\chi\in\mu_h}c_\chi\chi^q.
    
\mu_{k-a}Use in source · 3 passages

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  • Use in source · Read passage · strengthening.tex, line 905

    this reduces to γ_k(X)+ϵ Gγ_k-1(X). Moreover, G≡γ_2(X)+τ Qpmod2. After evaluation, every term with a single τ vanishes, so the second contribution reduces to L((γ_2(X)+Q)γ_k-1(X)). The γ_2(X) product has scalar binomk+12, which is even because 4mid k. The other contribution expands as L(Qγ_k-1(P+Q)) =∑_a=0^k-1(k-a)μ_aμ_k-a. An odd summand would require k-a to be odd and μ_k-a to be odd. By Lemma~[source reference], this requires k-a=1. Then a=k-1 is not a power of two and μ_a is even. Hence the …

    Exact mathematical passage
    
     \mathcal L(Q\gamma_{k-1}(P+Q))
        =\sum_{a=0}^{k-1}(k-a)\mu_a\mu_{k-a}.
     \label{eq:two-power-Q-evaluation}
    
\mu_mUse in source · 5 passages

Search aliases: μ_m mu

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  • Use in source · Read passage · 72_recovered.tex, line 847

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    
    T_{m,p}(\chi^\bullet)_n
    =
    \frac1m
    \sum_{\zeta\in\mu_m}
    (1+\zeta y)^{mn}.
    
\mu_rUse in source · 4 passages

Search aliases: μ_r mu

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  • Use in source · Read passage · root_scope.tex, line 128

    The scalar packet weights have exact valuations λ_r(q)=3q+s_2(r+q)-s_2(r). Put μ_r=min_q≥0λ_r(q), J_r={q:λ_r(q)=μ_r}, G_r=2^-μ_rH_r. Since λ_r(0)=0 and λ_r(q)≥3q+1-s_2(r), every minimizing address satisfies 0≤ q≤lfloorfracs_2(r)-13rfloor. In particular μ_r≤0, and the minimum is found by a finite carry calculation. Reduction of the whole series gives overline G_r(X)=∑_q∈ J_rX^underline q in F_2[X]. Its highest degree is max J_r, with leading coefficient one. Thus the normalization is the actual G…

    Exact mathematical passage
    
     \mu_r=\min_{q\ge0}\lambda_r(q),\qquad
     J_r=\{q:\lambda_r(q)=\mu_r\},\qquad G_r=2^{-\mu_r}H_r.
    
\nu_2Use in source · 1 passage

Search aliases: ν_2 nu

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  • Use in source · Read passage · 20_arithmetic.tex, line 472

    [All-row dyadic higher differences] For 0≤ j<4 and all u,k,n∈ N, C_j(n) · ∈ Z_2, Δ_u^kC_j(n) · ∈ 2^k(1+ν_2(u)) Z_2, Δ_u^kU_j(n) · ∈ 2^k(1+ν_2(u))+v_2(j!) Z_2. There is no restriction k≤ n and no minimum with the row n.

    Exact mathematical passage
    
     C_j(n)&\in\Z_2,\\
     \Delta_u^kC_j(n)&\in 2^{k(1+\nu_2(u))}\Z_2,\label{eq:arith-allrow-c}\\
     \Delta_u^kU_j(n)&\in
     2^{k(1+\nu_2(u))+v_2(j!)}\Z_2.\label{eq:arith-allrow-u}
    
\piUse in source · 9 passages

Search aliases: π pi

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  • Use in source · Read passage · 40_regularity.tex, line 112

    ather At the base parameter, the strict Fr echet derivative is Dψ(t_0)=-J^-1K. In particular, ψ is differentiable on U. Let g be the analytic local inverse of Φ supplied by Lemma~[source reference]. Since F(a)=0, Φ(a)=(0,t_0) and g(0,t_0)=a. Define ψ(t)=π_Yg(0,t), where π_Y projects to the solved coordinates. This gives the base value and analyticity at t_0. The chain rule and [source reference] give Dψ(t_0)v =π_YJ^-1 · -J^-1K

    Exact mathematical passage
    
     \psi(t)=\pi_Yg(0,t),
    
\pi_YUse in source · 4 passages

Search aliases: π_Y pi

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  • Use in source · Read passage · 40_regularity.tex, line 112

    ather At the base parameter, the strict Fr echet derivative is Dψ(t_0)=-J^-1K. In particular, ψ is differentiable on U. Let g be the analytic local inverse of Φ supplied by Lemma~[source reference]. Since F(a)=0, Φ(a)=(0,t_0) and g(0,t_0)=a. Define ψ(t)=π_Yg(0,t), where π_Y projects to the solved coordinates. This gives the base value and analyticity at t_0. The chain rule and [source reference] give Dψ(t_0)v =π_YJ^-1 · -J^-1K

    Exact mathematical passage
    
     \psi(t)=\pi_Yg(0,t),
    
\rho_pUse in source · 1 passage

Search aliases: ρ_p rho

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  • Use in source · Read passage · continuation_02.tex, line 105

    The earlier obstruction used the integer-domain map J_x=ρ_p∘ DF(x): Z^d→ F_p^e. The new count uses A_x: F_p^d→ F_p^e. Reduction of integer vectors onto finite vectors is surjective, so the two maps have the same image after restricting scalars. Lean constructs a class-preserving integer-linear equivalence F_p^e/imJ_x ≃ F_p^e/imA_x. Consequently the cardinality theorem counts the original obstruction group, preserving its distinguished residual class. It is not merely a formula for a separately i…

    Exact mathematical passage
    
    J_x=\rho_p\circ DF(x):\mathbb Z^d\to\mathbb F_p^e.
    
\rho_xUse in source · 10 passages

Search aliases: ρ_x rho

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  • Use in source · Read passage · 30_lifting.tex, line 294

    [Torsor and lift cardinalities] Under the hypotheses of Theorem~[source reference], there is an equivalence T_x≃ L_x,L. Every nonempty fiber is a torsor for ker J_x under addition of correction digits. If q=p is prime and ρ_x=rank_ F_pJ_x, then # L_x,L= p^d-ρ_x, · o_x=0,

    Exact mathematical passage
    \rho_x=\rank_{\F_p}J_x
\rho_{\ell,k}Use in source · 3 passages

Search aliases: ρ_ℓ,k rho ell

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  • Use in source · Read passage · supplement_05.tex, line 132

    Principal targets. targetFiniteUnits.finite_unit_decomposition, .power_lift_congruence targetFiniteUnits.omega_unique, .decomposition_unique targetFiniteUnits.omega_mul, .principal_mul pageheading3. Precision compatibility and the dyadic signFinite refinements and Proposition 6.2's dyadic branch For 1≤ k≤ℓ, let ρ_ℓ,k: Z/p^ℓ Z⟶ Z/p^k Z be the actual reduction homomorphism. It induces a map on units. The explicit construction satisfies ρ_ℓ,k(ω_ℓ(a)) =ω_k(ρ_ℓ,k(a)), ρ_ℓ,k(⟨ a⟩_ℓ) =⟨ρ_ℓ,k(a)⟩_k. The…

    Exact mathematical passage
    
     \rho_{\ell,k}:\Z/p^\ell\Z\longrightarrow\Z/p^k\Z
    
\sigmaUse in source · 20 passages

Search aliases: σ sigma

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  • Use in source · Read passage · 30_lifting.tex, line 453

    [Continuous charts with a selected Jacobian minor] Let p be prime, L≥1, F∈ Z[X_1,…,X_e+k]^e, and x∈ Z^e+k satisfy F(x)∈ p^L Z^e. Suppose the full reduced Jacobian is surjective. There is a permutation σ of the coordinates such that, writing σ x=(y_0,t_0) with e solved coordinates and k parameter coordinates, the parameter ball T=t_0+p^L Z_p^k is homeomorphic to the entire root locus R={z∈ B_L(x):F(z)=0}. The inverse homeomorphism is the actual free-coordinate projection z↦free(σ z). If h:T→ R is…

    Exact mathematical passage
    \sigma
\sigma^{-1}Use in source · 2 passages

Search aliases: σ^-1 sigma

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  • Use in source · Read passage · 30_lifting.tex, line 516

    Finally, surjectivity of the original rectangular Jacobian guarantees a basis among its columns. Extend the corresponding injection of coordinate indices to a permutation σ. Renaming the variables by σ^-1 transports polynomial evaluation and every formal partial derivative. The fixed-coordinate construction therefore applies to the original system after undoing the permutation.

    Exact mathematical passage
    \sigma^{-1}
\sigma_pUse in source · 7 passages

Search aliases: σ_p sigma

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  • Use in source · Read passage · 72_recovered.tex, line 1815

    For an odd prime p, the source S_n(p) contains one nonunit base. Put σ_p=(-1)^(p-1)/2 and S_n^×(p)=S_n(p)-σ_pp^n.

    Exact mathematical passage
    
    \sigma_p=(-1)^{(p-1)/2}
    
\tauUse in source · 51 passages

Search aliases: τ tau

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  • Use in source · Read passage · 10_foundations.tex, line 12

    The original arithmetic seed notation S=(A,a_0, O, R,{P_N,j},w,τ_p) records a coefficient ring or module, a seed, allowed operators, relations, residue selectors, a depth weight, and a topology. This notation describes the ingredients of the construction; a category of seed presentations and a notion of equivalence still require separate definitions. We therefore state each theorem for a concrete object. Section~[source reference] discusses the additional structure needed for the original native…

    Exact mathematical passage
    
    \mathcal S=(A,a_0,\mathcal O,\mathcal R,\{P_{N,j}\},w,\tau_p)
    
\tau^2Use in source · 1 passage

Search aliases: τ^2 tau

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  • Use in source · Read passage · strengthening.tex, line 894

    ulo two. Its uth auxiliary term has excess u+lfloor(k-u)/2rfloor-q=lfloor u/2rfloor. Only auxiliary indices zero and one can therefore survive: fracγ_k(H_2)2^q ≡ V^qγ_k(X) +ϵ V^q-1Gγ_k-1(X) pmod2 A_2. Here V≡τpmod2 and τ^2≡1pmod2. Since q is even and ϵτ≡ϵpmod2, this reduces to γ_k(X)+ϵ Gγ_k-1(X). Moreover, G≡γ_2(X)+τ Qpmod2. After evaluation, every term with a single τ vanishes, so the second contribution reduces to L((γ_2(X)+Q)γ_k-1(X)). The γ_2(X) product has scalar binomk+12, which is even be…

    Exact mathematical passage
    \tau^2\equiv1\pmod2
\tau_pUse in source · 1 passage

Search aliases: τ_p tau

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  • Use in source · Read passage · 10_foundations.tex, line 12

    The original arithmetic seed notation S=(A,a_0, O, R,{P_N,j},w,τ_p) records a coefficient ring or module, a seed, allowed operators, relations, residue selectors, a depth weight, and a topology. This notation describes the ingredients of the construction; a category of seed presentations and a notion of equivalence still require separate definitions. We therefore state each theorem for a concrete object. Section~[source reference] discusses the additional structure needed for the original native…

    Exact mathematical passage
    
    \mathcal S=(A,a_0,\mathcal O,\mathcal R,\{P_{N,j}\},w,\tau_p)
    
\tau_rUse in source · 3 passages

Search aliases: τ_r tau

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  • Use in source · Read passage · root_scope.tex, line 388

    [Exact quadratic compression] There is a unique preparation K_r(Z)= U_r(Z) W_r(Z), W_r(Z)=Z^2+2β_rZ+α_r, U_r≡ u_rpmod256. With δ_r=β_r^2-α_r and τ_r= U_r'(c_r)/ U_r(c_r), δ_r=-fracΦ_r^act U_r(c_r) (1-fracτ_r^2Φ_r^act4 U_r(c_r)), v_2(δ_r)=v_2(Φ_r^act). The equality includes the zero case. The same successive monic polynomial division used in Theorem~[source reference], started at precision eight, preserves the quadratic and unit reductions of [source reference]. Write W=Z^2+2β_rZ+α_r and w=W(c_r)…

    Exact mathematical passage
    \tau_r=\mathcal U_r'(c_r)/\mathcal U_r(c_r)
\tau_r^2Use in source · 1 passage

Search aliases: τ_r^2 tau

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  • Use in source · Read passage · root_scope.tex, line 389

    [Exact quadratic compression] There is a unique preparation K_r(Z)= U_r(Z) W_r(Z), W_r(Z)=Z^2+2β_rZ+α_r, U_r≡ u_rpmod256. With δ_r=β_r^2-α_r and τ_r= U_r'(c_r)/ U_r(c_r), δ_r=-fracΦ_r^act U_r(c_r) (1-fracτ_r^2Φ_r^act4 U_r(c_r)), v_2(δ_r)=v_2(Φ_r^act). The equality includes the zero case. The same successive monic polynomial division used in Theorem~[source reference], started at precision eight, preserves the quadratic and unit reductions of [source reference]. Write W=Z^2+2β_rZ+α_r and w=W(c_r)…

    Exact mathematical passage
    
     \delta_r=-\frac{\Phi_r^{\mathrm{act}}}{\mathcal U_r(c_r)}
      \left(1-\frac{\tau_r^2\Phi_r^{\mathrm{act}}}{4\mathcal U_r(c_r)}\right),
     \qquad v_2(\delta_r)=v_2(\Phi_r^{\mathrm{act}}).
     \label{eq:root:discriminant}
    
\PhiUse in source · 35 passages

Search aliases: Φ Phi

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  • Use in source · Read passage · 20_arithmetic.tex, line 1048

    [Isometric packet bijections] For 0≤ j<4, define on natural numbers f_j(n)=fracU_j(n)-j!2^d_j, (d_0,d_1,d_2,d_3)=(1,1,5,3). There is a unique continuous map Φ_j: Z_2→ Z_2 whose value at every n∈ N equals the embedded rational f_j(n). It is an isometric bijection and Φ_j(0)=0: |Φ_j(x)-Φ_j(y)|_2=|x-y|_2 (x,y∈ Z_2). In particular, every element of Z_2 has exactly one preimage.

    Exact mathematical passage
    \Phi_j:\Z_2\to\Z_2
\phiUse in source · 6 passages

Search aliases: ϕ phi

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  • Use in source · Read passage · 30_lifting.tex, line 467

    rphic to the entire root locus R={z∈ B_L(x):F(z)=0}. The inverse homeomorphism is the actual free-coordinate projection z↦free(σ z). If h:T→ R is the forward map, then for every N≥0, h(t)≡ h(u)pmodp^N Longleftrightarrow t≡ upmodp^N. Equivalently, σ h(t)=(ϕ(t),t) for a continuous ϕ:T→ y_0+p^L Z_p^e whose graph is the entire reordered root locus.

    Exact mathematical passage
    \sigma h(t)=(\phi(t),t)
\varphiUse in source · 1 passage

Search aliases: φ varphi

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  • Use in source · Read passage · supplement_05.tex, line 89

    The root equation and residue are derived. Euler's unit-group exponent gives ω_ℓ(a)^p-1 =a^p^ℓ-1(p-1)=a^φ(p^ℓ)=1. Reduction to the actual field F_p commutes with powers. Its Frobenius identity gives overlineω_ℓ(a) =bar a^p^ℓ-1=bar a. Consequently, a=ω_ℓ(a)⟨ a⟩_ℓ, ω_ℓ(a)^p-1=1, overline⟨ a⟩_ℓ=1. The manuscript's finite existence conclusion is obtained directly. The proof does not assume a Hensel lift or an already supplied root representative.

    Exact mathematical passage
    
     \omega_\ell(a)^{p-1}
     =a^{p^{\ell-1}(p-1)}=a^{\varphi(p^\ell)}=1.
    
\Phi_2Use in source · 1 passage

Search aliases: Φ_2 Phi

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  • Use in source · Read passage · 72_recovered.tex, line 2192

    The lowest degree term of Φ_M(z) is Mz, and its highest degree term is z^M. Multiplication by (1+z)^s gives the two edge coefficients. When M=2 and s=0, Φ_2(z)^k = (2z+z^2)^k = z^k(2+z)^k. Taking the coefficient of z^k+j gives the binary formula.

    Exact mathematical passage
    
    \Phi_2(z)^k
    =
    (2z+z^2)^k
    =
    z^k(2+z)^k.
    
\Phi_jUse in source · 6 passages

Search aliases: Φ_j Phi

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  • Use in source · Read passage · 20_arithmetic.tex, line 1048

    [Isometric packet bijections] For 0≤ j<4, define on natural numbers f_j(n)=fracU_j(n)-j!2^d_j, (d_0,d_1,d_2,d_3)=(1,1,5,3). There is a unique continuous map Φ_j: Z_2→ Z_2 whose value at every n∈ N equals the embedded rational f_j(n). It is an isometric bijection and Φ_j(0)=0: |Φ_j(x)-Φ_j(y)|_2=|x-y|_2 (x,y∈ Z_2). In particular, every element of Z_2 has exactly one preimage.

    Exact mathematical passage
    \Phi_j:\Z_2\to\Z_2
\Phi_MUse in source · 4 passages

Search aliases: Φ_M Phi

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  • Use in source · Read passage · 72_recovered.tex, line 2137

    Let Φ_M(z)=(1+z)^M-1. For M≥1, s≥0, and r,k≥0, define Θ_M,s(r,k) = [z^r](1+z)^sΦ_M(z)^k. Equivalently, Θ_M,s(r,k) = ∑_j=0^k (-1)^k-jbinom kjbinomMj+sr. It vanishes unless k≤ r≤ Mk+s.

    Exact mathematical passage
    
    \Phi_M(z)=(1+z)^M-1.
    
\Phi_rUse in source · 1 passage

Search aliases: Φ_r Phi

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  • Use in source · Read passage · root_scope.tex, line 376

    K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z), a_r,b_r∈ Z_2, u_r∈ Z_2^×, R_r∈ Z_2⟨ Z⟩. Its half derivative is a dyadic isometry. It therefore has a unique critical point c_r∈ Z_2. Set Φ_r^act=K_r(c_r). The earlier independent-undepletion notation for this same scalar is Φ_r(2^2r-25). Excluding a multiple root on the surviving chart is exactly the assertion Φ_r^act≠0.

    Exact mathematical passage
    \Phi_r(2^{2r-25})
\Phi_{r,m}^{\mathrm{act}}Use in source · 2 passages

Search aliases: Φ_r,m^act Phi mathrm

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  • Use in source · Read passage · root_scope.tex, line 431

    The latest continuation retains the exact Bernoulli undepletion factor e_r,q=frac1-2^2r-11-2^2r-14^q, q≥0. It is a dyadic unit, including e_r,0=1. Multiplication by this factor preserves the all-q error of the shifted Jacobi moment model. Consequently the actual model critical values satisfy, for every m≥25, v_2(Φ_r,m^act-Φ_r^act) ≥ L_m=8m-4s_2(m)-178, L_m⟶∞. The estimate includes every packet and controls motion of the critical point. There is no ceiling at the undepletion exponent 2r-25. For e…

    Exact mathematical passage
    
     v_2\bigl(\Phi_{r,m}^{\mathrm{act}}-\Phi_r^{\mathrm{act}}\bigr)
     \ge L_m=8m-4s_2(m)-178,
     \qquad L_m\longrightarrow\infty.
     \label{eq:root:actualapprox}
    
\Phi_r^{\mathrm{act}}Use in source · 6 passages

Search aliases: Φ_r^act Phi mathrm

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  • Use in source · Read passage · root_scope.tex, line 374

    ion K_r(Z)=2^-178G_r(10202+2^15Z). The complete coefficient theorem gives K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z), a_r,b_r∈ Z_2, u_r∈ Z_2^×, R_r∈ Z_2⟨ Z⟩. Its half derivative is a dyadic isometry. It therefore has a unique critical point c_r∈ Z_2. Set Φ_r^act=K_r(c_r). The earlier independent-undepletion notation for this same scalar is Φ_r(2^2r-25). Excluding a multiple root on the surviving chart is exactly the assertion Φ_r^act≠0.

    Exact mathematical passage
    \Phi_r^{\mathrm{act}}=K_r(c_r)
\Phi_{x,L}Use in source · 2 passages

Search aliases: Φ_x,L Phi

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  • Use in source · Read passage · continuation_02.tex, line 51

    For every positive integer q, including composite moduli and modulus one, and every L≥0, the checked map Φ_x,L:( Z/q Z)^d⟶ {ybmod q^L+1:y≡ xpmodq^L}, t⟼ x+q^Lwidetilde t is a bijection. Here widetilde t is the canonical integer representative of the digit vector. The entire residue fiber therefore has q^d elements. Injectivity is proved by cancelling q^L in the corresponding integer divisibility statement; surjectivity comes from the lower-residue congruence.

    Exact mathematical passage
    
    \Phi_{x,L}:(\mathbb Z/q\mathbb Z)^d\longrightarrow
    \{y\bmod q^{L+1}:y\equiv x\pmod{q^L}\},
    \qquad t\longmapsto x+q^L\widetilde t
    
\chi^\bulletUse in source · 1 passage

Search aliases: χ^bullet chi bullet

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  • Use in source · Read passage · 72_recovered.tex, line 847

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    
    T_{m,p}(\chi^\bullet)_n
    =
    \frac1m
    \sum_{\zeta\in\mu_m}
    (1+\zeta y)^{mn}.
    
\chi^qUse in source · 1 passage

Search aliases: χ^q chi

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  • Use in source · Read passage · 72_recovered.tex, line 843

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    
    d_q=\sum_{\chi\in\mu_h}c_\chi\chi^q.
    
\chi_CUse in source · 3 passages

Search aliases: χ_C chi

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  • Use in source · Read passage · 72_recovered.tex, line 2824

    [Characteristic polynomial of the jump] Let χ_C(y)=det(yI-C). Then χ_C^N(x) = Res_y(χ_C(y),x-y^N). The identity holds over the coefficient ring after interpreting the resultant universally.

    Exact mathematical passage
    
    \chi_C(y)=\det(yI-C).
    
\chi_{C^N}Use in source · 4 passages

Search aliases: χ_C^N chi

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  • Use in source · Read passage · 10_foundations.tex, line 415

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

    Exact mathematical passage
    \chi_{C^N}(X)=X^d+\sum_{r<d}q_rX^r
\PsiUse in source · 4 passages

Search aliases: Ψ Psi

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  • Use in source · Read passage · 30_lifting.tex, line 502

    Now suppose a chosen set of e columns of DF(x)bmod p is independent, and put those coordinates first. Apply the square result to the augmented polynomial map Ψ(Y,T)=(F(Y,T),T). Its reduced Jacobian has upper blocks (D_YF,D_TF) and lower blocks (0,I_k), so it is invertible. For t∈ T, the target (0,t) belongs to Ψ(x)+p^L Z_p^e+k because F(x)≡0pmodp^L. The inverse of Ψ at that target gives the unique root with parameters t. The inverse congruence property gives one direction of [source reference]; …

    Exact mathematical passage
    
     \Psi(Y,T)=(F(Y,T),T).
    
\psi^{-1}Use in source · 1 passage

Search aliases: ψ^-1 psi

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  • Use in source · Read passage · 40_regularity.tex, line 131

    rhoods as follows. The source of the open partial homeomorphism defined by Φ is open around a; its inverse is g. Choose open sets V and U_0 containing y_0 and t_0 such that V× U_0 lies in that source. Restrict U_0 further by the preimage of an open analytic inverse neighborhood under t↦(0,t), the open locus where this same ψ is analytic, and the interior of ψ^-1(V). All of these sets contain t_0. For the last set, use the continuity of ψ at t_0 and ψ(t_0)=y_0∈ V. Their intersection is an open ne…

    Exact mathematical passage
    \psi^{-1}(V)
\OmegaUse in source · 1 passage

Search aliases: Ω Omega

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  • Use in source · Read passage · 71_native.tex, line 134

    coker(Hom_B(Ω_P/ Z⊗_P B, F_p)⟶Hom_B(I/I^2, F_p)).

    Exact mathematical passage
    \operatorname{coker}\left(\operatorname{Hom}_B(\Omega_{P/\mathbb Z}\otimes_P B,\mathbb F_p)\longrightarrow\operatorname{Hom}_B(I/I^2,\mathbb F_p)\right).
\omegaUse in source · 9 passages

Search aliases: ω omega

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  • Use in source · Read passage · 10_foundations.tex, line 364

    [Finite unit coordinates] Let p be prime, L≥0, and a∈( Z/p^L+1 Z)^×. Define ω_L(a)=a^p^L, ⟨ a⟩_L=aω_L(a)^-1. Then a=ω_L(a)⟨ a⟩_L, ω_L(a)^p-1=1, ⟨ a⟩_L≡1pmod p. The two factors are uniquely determined by these properties and are compatible with reduction to lower precision. For a∈( Z/2^L+2 Z)^×, there is instead a unique decomposition a=ϵ(a)⟨ a⟩_4, ϵ(a)∈{1,-1}, ⟨ a⟩_4≡1pmod4.

    Exact mathematical passage
    
    \omega_L(a)=a^{p^L},\qquad \langle a\rangle_L=a\omega_L(a)^{-1}.
    
\omega_2Use in source · 1 passage

Search aliases: ω_2 omega

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  • Use in source · Read passage · supplement_05.tex, line 119

    For p=3,ℓ=2,a=2, the checked values are ω_2(2)=8, ⟨2⟩_2=7, 8^2=1,quad8·7=2 in Z/9 Z. The unlifted representative 2 is not a square root of one modulo nine.

    Exact mathematical passage
    
     \omega_2(2)=8,\qquad \langle2\rangle_2=7,\qquad
     8^2=1,\quad8\cdot7=2\quad\text{in }\Z/9\Z.
    
\omega_kUse in source · 1 passage

Search aliases: ω_k omega

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  • Use in source · Read passage · supplement_05.tex, line 136

    a_unique, .decomposition_unique targetFiniteUnits.omega_mul, .principal_mul pageheading3. Precision compatibility and the dyadic signFinite refinements and Proposition 6.2's dyadic branch For 1≤ k≤ℓ, let ρ_ℓ,k: Z/p^ℓ Z⟶ Z/p^k Z be the actual reduction homomorphism. It induces a map on units. The explicit construction satisfies ρ_ℓ,k(ω_ℓ(a)) =ω_k(ρ_ℓ,k(a)), ρ_ℓ,k(⟨ a⟩_ℓ) =⟨ρ_ℓ,k(a)⟩_k. The proof uses the exponent factorization p^ℓ-1=p^k-1p^ℓ-k and the already proved Frobenius invariance of a lift…

    Exact mathematical passage
    
     \boxed{\rho_{\ell,k}(\omega_\ell(a))
     =\omega_k(\rho_{\ell,k}(a)),}
    
\omega_LUse in source · 3 passages

Search aliases: ω_L omega

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  • Use in source · Read passage · 10_foundations.tex, line 364

    [Finite unit coordinates] Let p be prime, L≥0, and a∈( Z/p^L+1 Z)^×. Define ω_L(a)=a^p^L, ⟨ a⟩_L=aω_L(a)^-1. Then a=ω_L(a)⟨ a⟩_L, ω_L(a)^p-1=1, ⟨ a⟩_L≡1pmod p. The two factors are uniquely determined by these properties and are compatible with reduction to lower precision. For a∈( Z/2^L+2 Z)^×, there is instead a unique decomposition a=ϵ(a)⟨ a⟩_4, ϵ(a)∈{1,-1}, ⟨ a⟩_4≡1pmod4.

    Exact mathematical passage
    
    \omega_L(a)=a^{p^L},\qquad \langle a\rangle_L=a\omega_L(a)^{-1}.
    
\Omega_{P/\mathbb Z}Use in source · 1 passage

Search aliases: Ω_P/ Z Omega mathbb

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  • Use in source · Read passage · 71_native.tex, line 134

    coker(Hom_B(Ω_P/ Z⊗_P B, F_p)⟶Hom_B(I/I^2, F_p)).

    Exact mathematical passage
    \operatorname{coker}\left(\operatorname{Hom}_B(\Omega_{P/\mathbb Z}\otimes_P B,\mathbb F_p)\longrightarrow\operatorname{Hom}_B(I/I^2,\mathbb F_p)\right).
\omega_\ellUse in source · 4 passages

Search aliases: ω_ℓ omega ell

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  • Use in source · Read passage · supplement_05.tex, line 81

    ffPadic_coe, .standard_window_integral targetWindowIndependence.residue_independence, .reduced_unique targetWindowIndependence.two_window_addresses_congruent pageheading2. Explicit finite prime-power decompositionProposition 6.2; odd-prime branch Let p be prime, ℓ≥1, and a∈( Z/p^ℓ Z)^×. Define actual units in the same ring by ω_ℓ(a)=a^p^ℓ-1, ⟨ a⟩_ℓ=a ω_ℓ(a)^-1. The Lean source writes ℓ=L+1, so zero precision is excluded without truncated natural-number subtraction.

    Exact mathematical passage
    
     \boxed{\omega_\ell(a)=a^{p^{\ell-1}},\qquad
     \langle a\rangle_\ell=a\,\omega_\ell(a)^{-1}.}
    
\ellUse in source · 59 passages

Search aliases: ℓ ell

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  • Use in source · Read passage · 20_arithmetic.tex, line 259

    To pass from the section to the singular ratio, factor out its initial zero: for c=ap, the section has the form H_c(log(1+T))=fracT^cc! V(T), V(0)=1, V∈ W_p. The weight of c! compensates for the coefficient shift by c. The weighted ring is closed under inversion of series with constant coefficient one and under L_T=log(1+T)(1+T)d/dT. With ℓ(T)=log(1+T)/T, interpreted by coefficient shifting, differentiation of the displayed factorization gives R_c(log(1+T))=c(1+T)ℓ(T)+(L_TV)V^-1. Since pℓ(T)∈ W_…

    Exact mathematical passage
    \ell(T)=\log(1+T)/T
\ell^mUse in source · 1 passage

Search aliases: ℓ^m ell

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  • Use in source · Read passage · 72_recovered.tex, line 1905

    Put y=x+1 and let ℓ=log 9. Since v_2(ℓ)=3, frac9^y-18y = fracℓ8 ∑_m≥0 fracℓ^m(m+1)!y^m. The coefficient of y^m has valuation 3m-v_2((m+1)!), which is nonnegative and tends to positive infinity. The quotient therefore belongs to Z_2⟨ y⟩. The factor 2x+1=1+2x is a unit in Z_2⟨ x⟩, with inverse ∑_m≥0(-2x)^m. Hence mathscr R is integral and analytic. The integer values follow from the definition of R_2n.

    Exact mathematical passage
    
    \frac{9^y-1}{8y}
    =
    \frac{\ell}{8}
    \sum_{m\geq0}
    \frac{\ell^m}{(m+1)!}y^m.
    
\twoheadrightarrowUse in source · 1 passage

Search aliases: ↠ twoheadrightarrow

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  • Use in source · Read passage · 71_native.tex, line 129

    The group is finite. Choose a finite presentation P= Z[X_1,…,X_d]↠ B with finitely generated kernel I. The presentation description in https://stacks.math.columbia.edu/tag/0GPUStacks Project, Remark 91.2.4, tag 0GPU identifies the group as

    Exact mathematical passage
    P=\mathbb Z[X_1,\ldots,X_d]\twoheadrightarrow B
\hookrightarrowUse in source · 5 passages

Search aliases: ↪ hookrightarrow

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  • Use in source · Read passage · 30_lifting.tex, line 649

    Set the first integer parameter approximation equal to the prescribed t_0. For n≥1, use canonical representatives of τbmod p^L+n. The initial congruence and truncation precision give τ-t_n∈ p^L+n Z_p^k. Subtracting consecutive relations gives compatibility in Z_p. Faithfulness of integer divisibility under Z↪ Z_p gives compatibility in Z, and the same norm bounds give t_n→τ. Theorem~[source reference] and completion now produce b with exact polynomial vanishing and all retained precisions. Each …

    Exact mathematical passage
    \Z\hookrightarrow\Zp
\partialUse in source · 8 passages

Search aliases: ∂ partial

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  • Use in source · Read passage · 30_lifting.tex, line 28

    For F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, define its evaluated Jacobian by the formal partial derivatives: DF(x)_ij=(∂_jF_i)(x), J_x=DF(x)bmod q:S_q^d⟶ S_q^e. The obstruction therefore uses the Jacobian of the given polynomials. The following algebraic Taylor identity relates this linear map to polynomial evaluation.

    Exact mathematical passage
    
     DF(x)_{ij}=(\partial_jF_i)(x),\qquad
     J_x=DF(x)\bmod q:S_q^d\longrightarrow S_q^e.
    
\partial_iUse in source · 1 passage

Search aliases: ∂_i partial

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  • Use in source · Read passage · continuation_01.tex, line 51

    For every P∈ Z[X_1,…,X_d], integer a, and x,t∈ Z^d, Lean proves a^2mid P(x+at)-P(x)-a∑_i=1^d(∂_iP)(x)t_i. The proof uses polynomial induction: constants, addition, and multiplication by a variable. The product rule for the actual partial derivatives controls the multiplication step. It does not assume a Taylor identity or a liftability condition.

    Exact mathematical passage
    
    \boxed{
     a^2\mid P(x+at)-P(x)-a\sum_{i=1}^{d}(\partial_iP)(x)t_i.
    }
    
\partial_jUse in source · 5 passages

Search aliases: ∂_j partial

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  • Use in source · Read passage · 30_lifting.tex, line 28

    For F=(F_1,…,F_e)∈ Z[X_1,…,X_d]^e, define its evaluated Jacobian by the formal partial derivatives: DF(x)_ij=(∂_jF_i)(x), J_x=DF(x)bmod q:S_q^d⟶ S_q^e. The obstruction therefore uses the Jacobian of the given polynomials. The following algebraic Taylor identity relates this linear map to polynomial evaluation.

    Exact mathematical passage
    
     DF(x)_{ij}=(\partial_jF_i)(x),\qquad
     J_x=DF(x)\bmod q:S_q^d\longrightarrow S_q^e.
    
\existsUse in source · 9 passages

Search aliases: ∃ exists

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  • Use in source · Read passage · 30_lifting.tex, line 76

    [Finite polynomial obstruction] Let q≥1, L≥1, x∈ Z^d, and suppose F(x)∈ q^L Z^e. Define the exact integer residual r_x by F(x)=q^Lr_x, and put O_x=S_q^e/im J_x, o_x=[overliner_x]. Then O_x is finite and o_x=0 Longleftrightarrow ∃ y∈ Z^d: y≡ xpmodq^L, F(y)≡0pmodq^L+1. No surjectivity or nonsingularity assumption is required.

    Exact mathematical passage
    \label{eq:finite-obstruction}
     o_x=0\quad\Longleftrightarrow\quad
     \exists y\in\Z^d:\quad
     y\equiv x\pmod{q^L},\qquad F(y)\equiv0\pmod{q^{L+1}}.
    
\varnothingUse in source · 1 passage

Search aliases: ∅ varnothing

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  • Use in source · Read passage · continuation_08.tex, line 41

    Define the finite solved-block cokernel and its parameter-dependent class by C_x=S^e/imJ, o_x(v)=[bar r+Kv]∈ C_x. Then L_v≠∅iff o_x(v)=0. More precisely, the checked digit-to-lift equivalence is L_v≅{u∈ S^m:Ju=-bar r-Kv}. It sends a solved digit u to the actual next residue represented by x+q^L(widehat u,widehat v). The obstruction depends on the solved block J, not on the range of the full map (J K).

    Exact mathematical passage
    
    \boxed{\mathcal L_v\ne\varnothing\iff o_x(v)=0.}
    
\notinUse in source · 3 passages

Search aliases: ∉ notin

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  • Use in source · Read passage · 72_recovered.tex, line 2367

    [Matrix power amplification] Editorial hypothesis correction. Require L≥1. At L=0 the standalone claim is false: for p=3, d=1 and X=2=1+B, one has X^3-1=7∉3 Z_3. With L≥1, every higher binomial term in the retained proof has the required valuation, including p=2. If X=I+p^LB with B∈ M_d( Z_p), then for every a≥0 there is B_a∈ M_d( Z_p) such that X^p^a=I+p^L+aB_a.

    Exact mathematical passage
    X^3-1=7\notin3\Z_3
\prodUse in source · 35 passages

Search aliases: ∏ prod

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  • Use in source · Read passage · 10_foundations.tex, line 188

    em[Forward substitution and denominator clearing] Let T_nq be coefficients over a field with T_nn≠0 for every n. For any sequence b, define a_n=fracb_n-∑_q<nT_nqa_qT_nn. This constructs the unique solution of ∑_q=0^nT_nqa_q=b_n for all n. If T_nq and b_n are integers and the solution is taken over Q, then (∏_r=0^nT_rr)a_n∈ Z.

    Exact mathematical passage
    
    \left(\prod_{r=0}^{n}T_{rr}\right)a_n\in\Z.
    
\prod_{a=0}^{r-1}Use in source · 2 passages

Search aliases: ∏_a=0^r-1 prod

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  • Use in source · Read passage · strengthening.tex, line 552

    Define the central-factorial basis polynomials P_r(X)=∏_a=0^r-1(X-a(a+1)). At X=Z(Z+1), they have the exact factorization P_r(Z(Z+1)) =∏_a=0^r-1(Z-a)(Z+a+1) =(2r)!binomZ+r2r. The Bernoulli transform gives L((1+T)^Z)=log(1+T)/T, whence L(binom Zj)=(-1)^j/(j+1). Finite Vandermonde expansion therefore gives L(binomZ+r2r) · =(-1)^r∑_q=0^r(-1)^qbinom rqfrac1r+q+1nonumber

    Exact mathematical passage
    
     P_r(X)=\prod_{a=0}^{r-1}(X-a(a+1)).
    
\prod_{c=1}^{j}Use in source · 2 passages

Search aliases: ∏_c=1^j prod

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  • Use in source · Read passage · 20_arithmetic.tex, line 223

    Set H_c(t)=∑_n≥0fract^Nn+c(Nn+c)!, A_c(t)=∑_n≥0fract^Nn(Nn+c)!, R_c(t)=fracA_c-1(t)A_c(t) (1≤ c<N). The denominator of this last quotient has nonzero constant coefficient. The section identities give R_cH_c=tH_c-1, R_c(0)=c, j! P_N,j= P_N,0∏_c=1^jR_c. Cancellation of consecutive source sections proves the product identity and ties the ratios directly to the family under study.

    Exact mathematical passage
    
     R_cH_c=tH_{c-1},\qquad R_c(0)=c,\qquad
     j!\mathcal P_{N,j}=\mathcal P_{N,0}\prod_{c=1}^{j}R_c.
    
\prod_{f(\alpha)=0}Use in source · 1 passage

Search aliases: ∏_f(α)=0 prod alpha

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  • Use in source · Read passage · 72_recovered.tex, line 2815

    We use the resultant convention Res_y(f,g) = ∏_f(α)=0g(α) when f is monic, with roots counted with multiplicity.

    Exact mathematical passage
    
    \Res_y(f,g)
    =
    \prod_{f(\alpha)=0}g(\alpha)
    
\prod_{h=0}^kUse in source · 1 passage

Search aliases: ∏_h=0^k prod

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  • Use in source · Read passage · strengthening.tex, line 150

    For s∈ p Z_p, an exact finite-difference identity is Δ_1^kfrac11+sn =frac(-s)^k k!∏_h=0^k(1+s(n+h)). It follows by induction, subtracting adjacent rational functions; each successive subtraction adds the factor -(k+1)s in the numerator. Its denominator is a unit, so it belongs to p^k Z_p. The discrete Leibniz formula Δ_1^k(fg)(n) =∑_r=0^kbinom kr (Δ_1^r f)(n)(Δ_1^k-rg)(n+r) then proves Δ_1^k g_N,j(n)∈ p^k Z_p for every k,n. The coefficient at degree Nn of L_1^kG_N,j is precisely p^-kΔ_1^k g_N,j(…

    Exact mathematical passage
    
     \Delta_1^k\frac1{1+sn}
      =\frac{(-s)^k k!}{\prod_{h=0}^k(1+s(n+h))}.
     \label{eq:affine-reciprocal-exact}
    
\prod_{h=1}^{2^a-1}Use in source · 1 passage

Search aliases: ∏_h=1^2^a-1 prod

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  • Use in source · Read passage · strengthening.tex, line 1139

    at least two, since i≤ j<N/2. Hence g(1)≡1pmod4 and g(2)≡1pmod8. The first two Newton coefficients from the reciprocal recurrence are a_1=-g(1)-1, a_2=binom2NNg(1)^2-g(2)+2g(1)+1. The first has valuation exactly one. Also binom2NN≡6pmod8 for every N=2^a≥2. To see this, write B_a=binom2^a+12^a =2∏_h=1^2^a-1(1+frac2^ah). The even-h factors give B_a-1; the odd-h factors are one modulo four for a≥2. Since B_a-1 has valuation one, B_a≡ B_a-1pmod8, and B_1=6. Therefore a_2≡6-1+2+1=0pmod8. The global M…

    Exact mathematical passage
    
     B_a=\binom{2^{a+1}}{2^a}
           =2\prod_{h=1}^{2^a-1}\left(1+\frac{2^a}{h}\right).
    
\prod_iUse in source · 1 passage

Search aliases: ∏_i prod

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  • Use in source · Read passage · strengthening.tex, line 101

    e with unspecified remainders. On the tensor of the two factors use the mutually commuting operators L_λ⊗1, 1⊗ L_μ, c_iD^i⊗ D^N-i (1≤ i<N). Applying multiplication after the kth power of their sum gives exactly L_λ+μ^k(AB). Expanding it yields terms indexed by a,b,e_1,…,e_N-1≥0, with a+b+∑_i e_i=k: frack!a!b!∏_i e_i!∏_i c_i^e_i (D^α L_λ^a A)(D^β L_μ^b B), α=∑_i i e_i, β=∑_i(N-i)e_i. Every scalar is integral. If A and B have the stated flatness types, both factors in every term belong to H_p, as …

    Exact mathematical passage
    
     \frac{k!}{a!b!\prod_i e_i!}\prod_i c_i^{e_i}
     (D^\alpha L_\lambda^a A)(D^\beta L_\mu^b B),
     \quad
     \alpha=\sum_i i e_i,\quad
     \beta=\sum_i(N-i)e_i.
     \label{eq:translated-multinomial}
    
\prod_{i=0}^{d-1}Use in source · 2 passages

Search aliases: ∏_i=0^d-1 prod

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  • Use in source · Read passage · 20_arithmetic.tex, line 240

    _p={∑_d≥0a_dT^d: p^lfloor d/prfloora_d∈ Z_p}. The key estimate is R_ap(log(1+T))∈ W_p. Its arithmetic input comes from the cyclic group ring Z[ Z/N Z]. For its canonical element x and I=(p,x^p-1-1), the relation (x^p-1)^p=1 implies I^p⊆(p)I. Induction yields I^q⊆(p^v_p(q!)). Grouping the falling factorial (x)_d=∏_i=0^d-1(x-i) into consecutive blocks of length p then gives an integral residue-coordinate factor p^v_p(lfloor d/prfloor!). The identity v_p(d!)=lfloor d/prfloor+v_p(lfloor d/prfloor!) …

    Exact mathematical passage
    (x)_d=\prod_{i=0}^{d-1}(x-i)
\prod_{i=0}^{p-1}Use in source · 1 passage

Search aliases: ∏_i=0^p-1 prod

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  • Use in source · Read passage · supplement_09.tex, line 58

    Now take the integral cyclic group ring R_N= Z[ Z/N Z], x=[1], x^N=1, and assume Nmid p(p-1). Thus (x^p-1)^p=1. Each genuine consecutive block ∏_i=0^p-1(x-(pb+i)) belongs to (p,x^p-1-1). This follows from the actual polynomial identity ∏_i∈ F_p(X-i)=X^p-X modulo p, not from a hypothesized valuation of the product.

    Exact mathematical passage
    
     \prod_{i=0}^{p-1}\bigl(x-(pb+i)\bigr)
    
\prod_{i=0}^{q-1}Use in source · 1 passage

Search aliases: ∏_i=0^q-1 prod

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  • Use in source · Read passage · root_scope.tex, line 107

    Set r=M+1≥1 and define P_q(x)=∏_i=0^q-1(x-i)(2x-(2i+1)), P_0=1, H_r(x)=∑_q≥08^qfracβ_r+qβ_r P_q(x)mathscr A(x-q). The coefficient valuations tend to infinity. At any nonnegative integer x, terms with q>x vanish. Cancelling the factorials in [source reference] therefore gives the exact identity T_r-1(N)=κ_rbinom2N2rH_r(N-r) (N≥ r). This is a finite rational identity at every original Euler row.

    Exact mathematical passage
    
     P_q(x)=\prod_{i=0}^{q-1}(x-i)(2x-(2i+1)),\qquad P_0=1,
    
\prod_{i=1}^{d}Use in source · 1 passage

Search aliases: ∏_i=1^d prod

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  • Use in source · Read passage · 72_recovered.tex, line 2840

    After passage to an algebraic closure, let α_1,…,α_d be the eigenvalues of C, counted with algebraic multiplicity. The resultant is ∏_i=1^d(x-α_i^N), which is the characteristic polynomial of C^N. Both sides are polynomial expressions with integer coefficients in the entries of C, so the identity descends to every commutative coefficient ring.

    Exact mathematical passage
    
    \prod_{i=1}^{d}(x-\alpha_i^N),
    
\prod_{i=1}^jUse in source · 1 passage

Search aliases: ∏_i=1^j prod

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  • Use in source · Read passage · strengthening.tex, line 140

    The packed factorial coefficient of G_N,j is g_N,j(n)=fracj!(Nn)!(Nn+j)! =∏_i=1^j(1+frac Ni n)^-1. The empty product is one. If N=p^a and 1≤ i<N, then v_p(i)<a, so N/i∈ p Z_p. Every affine denominator in [source reference] is therefore a unit on Z_p.

    Exact mathematical passage
    
     g_{N,j}(n)=\frac{j!(Nn)!}{(Nn+j)!}
       =\prod_{i=1}^j\left(1+\frac Ni n\right)^{-1}.
     \label{eq:actual-prime-power-kernel}
    
\prod_{i=1}^{\ell-1}Use in source · 1 passage

Search aliases: ∏_i=1^ℓ-1 prod ell

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  • Use in source · Read passage · 72_recovered.tex, line 2018

    [Nested discrete refinement] For positive integers M,N, (a_r^[M])_s^[N] = a_r+Ms^[MN]. For a mixed radix path (M_1,…,M_ℓ) with digits 0≤ s_i<M_i, the final address is r( s) = s_1+M_1s_2+⋯+ (∏_i=1^ℓ-1M_i)s_ℓ. Only the product radix and accumulated address determine the final branch.

    Exact mathematical passage
    
    \boxed{
    r(\boldsymbol s)
    =
    s_1+M_1s_2+\cdots+
    \left(\prod_{i=1}^{\ell-1}M_i\right)s_\ell.
    }
    
\prod_{i\in\F_p}Use in source · 1 passage

Search aliases: ∏_i∈F_p ∏_i∈ F_p prod in F prod in mathbb

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  • Use in source · Read passage · supplement_09.tex, line 62

    Now take the integral cyclic group ring R_N= Z[ Z/N Z], x=[1], x^N=1, and assume Nmid p(p-1). Thus (x^p-1)^p=1. Each genuine consecutive block ∏_i=0^p-1(x-(pb+i)) belongs to (p,x^p-1-1). This follows from the actual polynomial identity ∏_i∈ F_p(X-i)=X^p-X modulo p, not from a hypothesized valuation of the product.

    Exact mathematical passage
    
     \prod_{i\in\F_p}(X-i)=X^p-X
    
\prod_{j=1}^{s}Use in source · 4 passages

Search aliases: ∏_j=1^s prod

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  • Use in source · Read passage · 72_recovered.tex, line 3027

    [Several coprime moduli] Let N_1,…,N_s be pairwise coprime and put N=∏_j=1^sN_j. For compatible addresses r_j, P_N,r = ∏_j=1^sP_N_j,r_j, where r is the unique combined address modulo N.

    Exact mathematical passage
    
    N=\prod_{j=1}^{s}N_j.
    
\prod_pUse in source · 3 passages

Search aliases: ∏_p prod

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  • Use in source · Read passage · 50_open_questions.tex, line 128

    Why the global/completed distinction is necessary. Consider constant sequences with integer initial value. Their global parameter space is Z, while their local parameter spaces are Z_p. Under the canonical local maps, the image of the global objects is the diagonal D={(n)_p:n∈ Z}⊆∏_p Z_p. This diagonal is dense: a basic open set specifies finitely many congruences, and the Chinese remainder theorem provides an integer satisfying them. It is proper: the tuple with 2-adic coordinate zero and all o…

    Exact mathematical passage
    
    D=\{(n)_p:n\in\Z\}\subseteq\prod_p\Z_p.
    
\prod_{p>2}Use in source · 2 passages

Search aliases: ∏_p>2 prod

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  • Use in source · Read passage · 50_open_questions.tex, line 138

    Consequently this canonical global diagonal cannot be an equality locus {(a,b):f(a)=g(b)} ⊆ Z_2×∏_p>2 Z_p for continuous maps to a Hausdorff space C. Such an equality locus is closed, as the inverse image of the closed diagonal in C× C, whereas D is proper and dense. This topological argument, retained from the native progress note, concerns the stated topological interpretation. It has not been formalized in Lean for this manuscript.

    Exact mathematical passage
    
    \{(a,b):f(a)=g(b)\}
    \subseteq \Z_2\times\prod_{p>2}\Z_p
    
\prod_{p>2}'Use in source · 1 passage

Search aliases: ∏_p>2' prod

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  • Use in source · Read passage · 50_open_questions.tex, line 95

    The original proposed form is X≃ X_2^wt×_C∏_p>2'X_p. It leaves X, the distinguished subsets defining the restricted product, the maps to C, the ambient category, and the universal meaning of ``smallest'' unspecified. In particular, one must decide whether X is a global object or a completed object.

    Exact mathematical passage
    
    X\simeq X_2^{\mathrm{wt}}\times_C\prod_{p>2}'X_p.
    
\prod_{p\in S}Use in source · 3 passages

Search aliases: ∏_p∈ S prod in

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  • Use in source · Read passage · 71_native.tex, line 210

    N=∏_p∈ Sp^L_p, Sol_F(R)={x∈ R^d:F(x)=0}.

    Exact mathematical passage
    N=\prod_{p\in S}p^{L_p},\qquad\operatorname{Sol}_F(R)=\{x\in R^d:F(x)=0\}.
\prod_rUse in source · 1 passage

Search aliases: ∏_r prod

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  • Use in source · Read passage · 72_recovered.tex, line 704

    [Lucas congruence] If a=a_0+a_1p+⋯ and b=b_0+b_1p+⋯ are base p expansions, then binom ab ≡ ∏_rbinoma_rb_r pmod p. This is the classical theorem of Lucas [source reference].

    Exact mathematical passage
    
    \binom ab
    \equiv
    \prod_r\binom{a_r}{b_r}
    \pmod p.
    
\prod_{r<n}Use in source · 1 passage

Search aliases: ∏_r<n prod

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  • Use in source · Read passage · 10_foundations.tex, line 197

    The recursion uses only smaller indices, and substitution verifies each row. If a second solution agrees through row n-1, subtracting row n and cancelling T_nn proves equality at n. For integrality, use induction and promote every earlier clearing product to ∏_r<nT_rr. Multiplication by the additional factor T_nn cancels the only new division.

    Exact mathematical passage
    \prod_{r<n}T_{rr}
\prod_{r=0}^{n}Use in source · 1 passage

Search aliases: ∏_r=0^n prod

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  • Use in source · Read passage · 10_foundations.tex, line 188

    em[Forward substitution and denominator clearing] Let T_nq be coefficients over a field with T_nn≠0 for every n. For any sequence b, define a_n=fracb_n-∑_q<nT_nqa_qT_nn. This constructs the unique solution of ∑_q=0^nT_nqa_q=b_n for all n. If T_nq and b_n are integers and the solution is taken over Q, then (∏_r=0^nT_rr)a_n∈ Z.

    Exact mathematical passage
    
    \left(\prod_{r=0}^{n}T_{rr}\right)a_n\in\Z.
    
\prod_{r=1}^{n}Use in source · 3 passages

Search aliases: ∏_r=1^n prod

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  • Use in source · Read passage · 20_arithmetic.tex, line 59

    [A denominator bound for every address] For N≥1, 0≤ j<N, and n∈ N, set D_N,j(n)=∏_r=1^nbinomNr+jNr, D_N,j(0)=1. Then D_N,j(n)C_N,j(n)∈ Z.

    Exact mathematical passage
    
     D_{N,j}(n)=\prod_{r=1}^{n}\binom{Nr+j}{Nr},\qquad D_{N,j}(0)=1.
    
\sum_{a=0}^kUse in source · 1 passage

Search aliases: ∑_a=0^k sum

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  • Use in source · Read passage · strengthening.tex, line 916

    For the first contribution, independence gives L(γ_k(P+Q))=∑_a=0^kμ_aμ_k-a. The parity generating series of these convolution moments is (1+t+t^2+t^4+t^8+⋯)^2 =1+t^2+t^4+t^8+⋯ in F_2[[t]]. Its coefficient is one at every k=2^r≥4. Thus the normalized moment in [source reference] is odd and v_2(Δ_1^kC_2(0))=2k+v_2(k!)+k/2. Inserting U^4n=(1+4H_2)^n introduces only corrections with strictly higher valuation, by the filtered bound at orders k+s for s>0. The equality consequently holds at every start…

    Exact mathematical passage
    \mathcal L(\gamma_k(P+Q))=\sum_{a=0}^k\mu_a\mu_{k-a}
\sum_{a=0}^{k-1}Use in source · 1 passage

Search aliases: ∑_a=0^k-1 sum

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  • Use in source · Read passage · strengthening.tex, line 905

    this reduces to γ_k(X)+ϵ Gγ_k-1(X). Moreover, G≡γ_2(X)+τ Qpmod2. After evaluation, every term with a single τ vanishes, so the second contribution reduces to L((γ_2(X)+Q)γ_k-1(X)). The γ_2(X) product has scalar binomk+12, which is even because 4mid k. The other contribution expands as L(Qγ_k-1(P+Q)) =∑_a=0^k-1(k-a)μ_aμ_k-a. An odd summand would require k-a to be odd and μ_k-a to be odd. By Lemma~[source reference], this requires k-a=1. Then a=k-1 is not a power of two and μ_a is even. Hence the …

    Exact mathematical passage
    
     \mathcal L(Q\gamma_{k-1}(P+Q))
        =\sum_{a=0}^{k-1}(k-a)\mu_a\mu_{k-a}.
     \label{eq:two-power-Q-evaluation}
    
\sum_{a=0}^{q-1}Use in source · 2 passages

Search aliases: ∑_a=0^q-1 sum

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  • Use in source · Read passage · 72_recovered.tex, line 897

    Let q be a positive odd integer. Define S_n(q) = ∑_a=0^q-1 (-1)^a(2a+1)^n. Let the signed Euler numbers and Euler zigzag numbers be defined by secht = ∑_n≥0E_nfract^nn! and sec t+tan t = ∑_n≥0Z_nfract^nn!.

    Exact mathematical passage
    
    S_n(q)
    =
    \sum_{a=0}^{q-1}
    (-1)^a(2a+1)^n.
    
\sum_{a=0}^{u-1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 162

    l closure therefore gives flatness of the actual packed Euler coefficient sequence. The earlier coefficient-extraction theorem now has a proved antecedent for every j<p, and yields p^min(k,mn)midΔ_h^k C_mp,j(n) in Z_p, h=(p-1)/m. A new finite telescoping argument proves the extension to arbitrary natural multiples of h. Its first-step identity is f(n+uh)-f(n)=∑_a=0^u-1Δ_h f(n+ah). Commutation of differences and induction on their order transfer any fixed ideal bound valid on a forward tail. The …

    Exact mathematical passage
    
     f(n+uh)-f(n)=\sum_{a=0}^{u-1}\Delta_h f(n+ah).
    
\sum_{a=1}^{q-1}Use in source · 1 passage

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  • Use in source · Read passage · supplement_08.tex, line 33

    pageheading1. The characteristic-zero product ruleThe normalized defect is not an ordinary derivation Let A be a commutative algebra over a characteristic-zero field K, and let U be an actual K-derivation. The source defines B=fracU^p^2-U^pp, M_q(f,g)=∑_a=1^q-1frac1pbinom qa (U^a f)(U^q-ag), q=p,p^2. The coefficients of each M_q are proved integral: the middle entries of a prime-power binomial row are divisible by p. In Lean they are implemented as natural quotients, with exact division proved b…

    Exact mathematical passage
    
     B=\frac{U^{p^2}-U^p}{p},\qquad
     M_q(f,g)=\sum_{a=1}^{q-1}\frac{1}{p}\binom qa
              (U^a f)(U^{q-a}g),\quad q=p,p^2.
    
\sum_{A\subseteq\{1,\ldots,k\}}Use in source · 1 passage

Search aliases: ∑_A⊆{1,…,k} sum subseteq ldots

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  • Use in source · visual_body.tex, line 133

    Write Sf(n)=f(n+1) and Δ_H=S^H-1. Mixed differences commute. Their exact finite expression is Δ_H_1⋯Δ_H_kf(n) =∑_A⊆{1,…,k} (-1)^k-|A| f (n+∑_i∈ AH_i). Repeated subset offsets are combined with their signed multiplicities.

    Exact mathematical passage
    
     \Delta_{H_1}\cdots\Delta_{H_k}f(n)
     =\sum_{A\subseteq\{1,\ldots,k\}}
     (-1)^{k-|A|}
     f\!\left(n+\sum_{i\in A}H_i\right).
    
\sum_{d\equiv j\ ({\rm mod}\ N)}Use in source · 1 passage

Search aliases: ∑_d≡ j (rm mod N) sum equiv rm

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  • Use in source · Read passage · supplement_01.tex, line 111

    For N>0 and 0≤ j<N, the independently defined original exponential section is E_N,j(t)=∑_d≡ j (rm mod N)fract^dd!. Using the actual power-series expansion homomorphism z↦ t^N, the code proves E_N,j(t)=fract^jj!K_N,j(t^N), F_N,j(t^N)E_N,j(t)=fract^jj!. Thus the recurrence is connected back to the original normalized EGF definition. The section vanishes at zero when j>0; no inverse of that nonunit section is assumed.

    Exact mathematical passage
    
     E_{N,j}(t)=\sum_{d\equiv j\ ({\rm mod}\ N)}\frac{t^d}{d!}.
    
\sum_{h=0}^kUse in source · 1 passage

Search aliases: ∑_h=0^k sum

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  • Use in source · Read passage · strengthening.tex, line 1066

    3 · 4 · 161644544/4849845 · 15 They are finite rational evaluations of [source reference], followed by a_k=∑_h=0^k(-1)^k-hbinom kh C_j(h). For addresses zero and one, k≥3 is covered by the generic bound; for address two, k≥7 is covered; for address three, k≥5 is covered. In each range the elementary inequality is k-lfloorlog_2krfloor≥ b_j+1. Lemma~[source reference] proves the normalized identity. Multiplication by j! proves the original identity.

    Exact mathematical passage
    a_k=\sum_{h=0}^k(-1)^{k-h}\binom kh C_j(h)
\sum_iUse in source · 6 passages

Search aliases: ∑_i sum

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  • Use in source · Read passage · strengthening.tex, line 100

    formula, rather than a product rule with unspecified remainders. On the tensor of the two factors use the mutually commuting operators L_λ⊗1, 1⊗ L_μ, c_iD^i⊗ D^N-i (1≤ i<N). Applying multiplication after the kth power of their sum gives exactly L_λ+μ^k(AB). Expanding it yields terms indexed by a,b,e_1,…,e_N-1≥0, with a+b+∑_i e_i=k: frack!a!b!∏_i e_i!∏_i c_i^e_i (D^α L_λ^a A)(D^β L_μ^b B), α=∑_i i e_i, β=∑_i(N-i)e_i. Every scalar is integral. If A and B have the stated flatness types, both factor…

    Exact mathematical passage
    a+b+\sum_i e_i=k
\sum_{i<d}Use in source · 1 passage

Search aliases: ∑_i<d sum

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  • Use in source · Read passage · supplement_07.tex, line 54

    Suppose g(0)=1. The source constructs f by the full triangular recurrence f(d)=δ_d0-∑_i<dbinom di f(i)g(d-i) and proves f⋆ g=1. It also proves cancellation: if a⋆ g=0, then a=0, by induction on the first nonzero coefficient.

    Exact mathematical passage
    
     f(d)=\delta_{d0}-\sum_{i<d}\binom di f(i)g(d-i)
    
\sum_{i=0}^{a}Use in source · 1 passage

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  • Use in source · Read passage · supplement_07.tex, line 40

    The identity sequence is 1(d)=δ_d0. The new source proves the actual higher product rule U^a(f⋆ g)=∑_i=0^abinom ai (U^i f)⋆(U^a-ig). The argument starts with Pascal's identity for the first shift, then inducts on a. All sums are finite at each coefficient.

    Exact mathematical passage
    
     U^a(f\star g)=\sum_{i=0}^{a}\binom ai
           (U^i f)\star(U^{a-i}g).
    
\sum_{i=0}^dUse in source · 1 passage

Search aliases: ∑_i=0^d sum

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  • Use in source · Read passage · strengthening.tex, line 57

    Products preserve H_p because the coefficient at factorial degree d is (AB)_d=∑_i=0^dbinom di a_i b_d-i. The derivative D acts by the coefficient shift (DA)_d=a_d+1. Consequently every nonnegative power of D preserves the lattice. If A∈ H_p has unit constant term, its reciprocal belongs to the same lattice: with B=A^-1, b_0=a_0^-1, b_d=-a_0^-1∑_i=1^dbinom di a_i b_d-i. If A is supported at degrees divisible by N, this recurrence also proves that B has the same support.

    Exact mathematical passage
    
     (AB)_d=\sum_{i=0}^d\binom di a_i b_{d-i}.
    
\sum_{i=0}^{d}Use in source · 1 passage

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  • Use in source · Read passage · supplement_07.tex, line 33

    pageheading1. Coefficient products and reciprocal differentiationA characteristic-p argument without factorial division For sequences f,g: N→ K, where K is a field, define (f⋆ g)(d)=∑_i=0^dbinom di f(i)g(d-i), (U^a f)(d)=f(d+a). These are Hurwitz coefficients. The notation does not divide by d! in F_p. In particular, the implementation is not ordinary power-series multiplication with undefined inverse factorials in positive characteristic.

    Exact mathematical passage
    
     (f\star g)(d)=\sum_{i=0}^{d}\binom di f(i)g(d-i),
     \qquad (U^a f)(d)=f(d+a).
    
\sum_{i=0}^kUse in source · 3 passages

Search aliases: ∑_i=0^k sum

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  • Use in source · Read passage · edition_summary.tex, line 11

    The strongest common-step conclusion in width four is particularly compact. For n≥0 and k,H≥1, ∑_i=0^k(-1)^k-ibinom ki E_4(n+iH)^(4,j) ≡0pmod2^ v_2(k!)+kv_2(H)+A_j(k), where A_0(k)=A_1(k)=k, A_2(k)=max{k+4, 2k+lfloor k/2rfloor+1}, A_3(k)=k+2. There is no starting-row threshold. The exponent is exact in addresses zero and one for every order and positive step. It is exact in address two at every power-of-two order. The proof chapters also give mixed-step forms, the cancellation criterion in addre…

    Exact mathematical passage
    
     \sum_{i=0}^k(-1)^{k-i}\binom ki E_{4(n+iH)}^{(4,j)}
     \equiv0\pmod{2^{\,v_2(k!)+kv_2(H)+A_j(k)}},
    
\sum_{i=0}^{n}Use in source · 1 passage

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  • Use in source · Read passage · 20_arithmetic.tex, line 628

    Finite Newton reconstruction at an arbitrary row gives Δ_1^kf(n)=∑_i=0^nbinom ni a_k+i. The monotonicity of lfloorlog_2krfloor therefore preserves the higher-coefficient bound after translation. In addition, Δ_1f(n)-a_1∈2^c+1 Z_2, so the first difference is nonzero of exact valuation c.

    Exact mathematical passage
    
     \Delta_1^kf(n)=\sum_{i=0}^{n}\binom ni a_{k+i}.
    
\sum_{i=0}^sUse in source · 2 passages

Search aliases: ∑_i=0^s sum

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  • Use in source · Read passage · 20_arithmetic.tex, line 498

    g_2(n) · =frac1(4n+1)(2n+1), · g_3(n) · =frac3(4n+1)(2n+1)(4n+3). Every denominator is odd. An affine sequence with even slope satisfies 2^kmidΔ_1^k f(n) for every k,n. The exact product identity Δ_1^s(fg)(n)=∑_i=0^sbinom si Δ_1^if(n)Δ_1^s-ig(n+i) proves closure of this property under products. Applying the formula with the factors interchanged to fg=1 isolates f(n+s)Δ_1^sg(n). Multiplication by f(n+s)^-1 and induction then prove closure under pointwise unit inverses. Thus all four g_j have the …

    Exact mathematical passage
    
     \Delta_1^s(fg)(n)=\sum_{i=0}^s\binom si
              \Delta_1^if(n)\Delta_1^{s-i}g(n+i)
    
\sum_{i=1}^dUse in source · 1 passage

Search aliases: ∑_i=1^d sum

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  • Use in source · Read passage · strengthening.tex, line 64

    Products preserve H_p because the coefficient at factorial degree d is (AB)_d=∑_i=0^dbinom di a_i b_d-i. The derivative D acts by the coefficient shift (DA)_d=a_d+1. Consequently every nonnegative power of D preserves the lattice. If A∈ H_p has unit constant term, its reciprocal belongs to the same lattice: with B=A^-1, b_0=a_0^-1, b_d=-a_0^-1∑_i=1^dbinom di a_i b_d-i. If A is supported at degrees divisible by N, this recurrence also proves that B has the same support.

    Exact mathematical passage
    
     b_0=a_0^{-1},\qquad
     b_d=-a_0^{-1}\sum_{i=1}^d\binom di a_i b_{d-i}.
     \label{eq:hurwitz-reciprocal-recurrence}
    
\sum_{i=1}^{d}Use in source · 1 passage

Search aliases: ∑_i=1^d sum

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  • Use in source · Read passage · continuation_01.tex, line 51

    For every P∈ Z[X_1,…,X_d], integer a, and x,t∈ Z^d, Lean proves a^2mid P(x+at)-P(x)-a∑_i=1^d(∂_iP)(x)t_i. The proof uses polynomial induction: constants, addition, and multiplication by a variable. The product rule for the actual partial derivatives controls the multiplication step. It does not assume a Taylor identity or a liftability condition.

    Exact mathematical passage
    
    \boxed{
     a^2\mid P(x+at)-P(x)-a\sum_{i=1}^{d}(\partial_iP)(x)t_i.
    }
    
\sum_{i=1}^HUse in source · 1 passage

Search aliases: ∑_i=1^H sum

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  • Use in source · Read passage · strengthening.tex, line 205

    To retain the valuation of a step, let B be any sequence satisfying Δ_1^m B∈ p^m Z_p at every order and row. Use the finite operator identity Δ_H=∑_i=1^Hbinom HiΔ_1^i. The relation ibinom Hi=HbinomH-1i-1 implies v_p(p^ibinom Hi) ≥ i+v_p(H)-v_p(i)≥1+v_p(H). Multiplying these identities for the k steps, every term of total order M has its scalar multiplied by p^M, and is divisible by p^k+∑_ν v_p(H_ν). This proves the mixed-step bound. Finally, Δ_H A_N,j(n)=(-1)^n+HT_H C_N,j(n). Iteration contribut…

    Exact mathematical passage
    
     \Delta_H=\sum_{i=1}^H\binom Hi\Delta_1^i.
    
\sum_{i=1}^kUse in source · 2 passages

Search aliases: ∑_i=1^k sum

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  • Use in source · Read passage · strengthening.tex, line 989

    At a natural argument the exact Newton expansion is finite. Vandermonde gives binomn+Hk-binom nk =∑_i=1^kbinom Hibinom nk-i. For 1≤ i≤ H, v_2(binom Hi)≥ v_2(H)-v_2(i) ≥ v_2(H)-lfloorlog_2krfloor; terms with i>H vanish. Thus each Newton summand with k≥2 has valuation at least b+v_2(H)+1. The k=1 summand is Ha_1, of exact valuation b+v_2(H), proving [source reference].

    Exact mathematical passage
    
     \binom{n+H}{k}-\binom nk
       =\sum_{i=1}^k\binom Hi\binom n{k-i}.
    
\sum_{i=1}^{N-1}Use in source · 1 passage

Search aliases: ∑_i=1^N-1 sum

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  • Use in source · Read passage · strengthening.tex, line 75

    [Translated product formula] For N=p^a, a≥1, and λ,μ∈ Z_p, L_λ+μ(AB) =(L_λ A)B+A(L_μ B) +∑_i=1^N-1c_i(D^iA)(D^N-iB), c_i=frac1pbinom Ni∈ Z. Flatness types add under multiplication.

    Exact mathematical passage
    
     L_{\lambda+\mu}(AB)
      =(L_\lambda A)B+A(L_\mu B)
           +\sum_{i=1}^{N-1}c_i(D^iA)(D^{N-i}B),
     \qquad c_i=\frac1p\binom Ni\in\mathbb Z.
     \label{eq:translated-product}
    
\sum_{i\in A}Use in source · 1 passage

Search aliases: ∑_i∈ A sum in

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  • Use in source · visual_body.tex, line 133

    Write Sf(n)=f(n+1) and Δ_H=S^H-1. Mixed differences commute. Their exact finite expression is Δ_H_1⋯Δ_H_kf(n) =∑_A⊆{1,…,k} (-1)^k-|A| f (n+∑_i∈ AH_i). Repeated subset offsets are combined with their signed multiplicities.

    Exact mathematical passage
    
     \Delta_{H_1}\cdots\Delta_{H_k}f(n)
     =\sum_{A\subseteq\{1,\ldots,k\}}
     (-1)^{k-|A|}
     f\!\left(n+\sum_{i\in A}H_i\right).
    
\sum_{i\in I}Use in source · 1 passage

Search aliases: ∑_i∈ I sum in

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  • Use in source · Read passage · 10_foundations.tex, line 338

    [Differences of unit modes] Over every commutative ring, Δ_h^s(u^n)=u^n(u^h-1)^s. For any prime p, units u_i∈ Z_p^×, coefficients a_i∈ Z_p, and a finite index set I, the sequence f(n)=∑_i∈ Ia_i u_i^n satisfies Δ_p-1^s f(n)∈ p^s Z_p (n,s∈ N).

    Exact mathematical passage
    f(n)=\sum_{i\in I}a_i u_i^n
\sum_{i\in S}Use in source · 1 passage

Search aliases: ∑_i∈ S sum in

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  • Use in source · Read passage · supplement_04.tex, line 148

    Finite integral linear combinations. For any finite index set S, coefficients a_i∈ Z_p and units u_i∈ Z_p^×, linearity of the actual iterated operator gives Δ_h^s (∑_i∈ Sa_i u_i^n) =∑_i∈ Sa_i u_i^n(u_i^h-1)^s. At h=p-1, the result is divisible by p^s. The empty sum is permitted. This does not prove that the general congruential Euler packets of Theorem 7.5 admit such a representation.

    Exact mathematical passage
    
     \Delta_h^s\!\left(\sum_{i\in S}a_i u_i^n\right)
     =\sum_{i\in S}a_i u_i^n(u_i^h-1)^s.
    
\sum_{i\geq0}Use in source · 3 passages

Search aliases: ∑_i≥0 sum geq

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  • Use in source · Read passage · strengthening.tex, line 452

    G_4,1(t)F_0(t) =frac12(fractanh(at)at +fractanh(bt)bt). For example, its numerator follows from sinh t+sin t =2bsinh(at)cosh(bt)+2acosh(at)sinh(bt), while cosh t+cos t=2cosh(at)cosh(bt). Only degrees divisible by four survive. Its actual Hurwitz packets are D_1(t)=∑_i≥0d_ifract^4i(4i)!, d_i=(-frac14)^i frac2^4i+2(2^4i+2-1)B_4i+2(4i+2)(4i+1). The constant is one. Since v_2(4i+2)=1, the elementary Bernoulli bound shows v_2(d_i)≥2i. Hurwitz inversion therefore gives D_1(t)^-1=∑_i≥0h_ifract^4i(4i)!,…

    Exact mathematical passage
    
     D_1(t)=\sum_{i\geq0}d_i\frac{t^{4i}}{(4i)!},\qquad
     d_i=\left(-\frac14\right)^i
     \frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.
     \label{eq:address-one-packets}
    
\sum_{i\geq1}Use in source · 2 passages

Search aliases: ∑_i≥1 sum geq

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  • Use in source · Read passage · strengthening.tex, line 468

    the elementary Bernoulli bound shows v_2(d_i)≥2i. Hurwitz inversion therefore gives D_1(t)^-1=∑_i≥0h_ifract^4i(4i)!, h_0=1, v_2(h_i)≥2i. This follows by induction from the inverse recurrence: for packet index i, every product d_rh_i-r has valuation at least 2i, and its binomial multiplier is integral. Because F_1=F_0D_1^-1, C_1(n)-C_0(n) =∑_i≥1h_ibinom4n4iC_0(n-i). At a natural n this is a finite sum, with i≤ n.

    Exact mathematical passage
    
     C_1(n)-C_0(n)
         =\sum_{i\geq1}h_i\binom{4n}{4i}C_0(n-i).
     \label{eq:address-one-exact-transfer}
    
\sum_jUse in source · 2 passages

Search aliases: ∑_j sum

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  • Use in source · Read passage · 30_lifting.tex, line 51

    Apply polynomial induction using constants, addition, and multiplication by a variable. Constants have zero remainder, and remainders add. For the multiplication step, suppose the assertion is known for P and write D_P(U)=∑_j(∂_jP)(a)U_j. Multiplication of its expansion by a_i+cU_i gives constant term P(a)a_i, linear term c(a_iD_P(U)+P(a)U_i), and remainder c^2(Q(U)(a_i+cU_i)+D_P(U)U_i). The product rule identifies the linear term with the derivative of PX_i. The displayed remainder vanishes at …

    Exact mathematical passage
    D_P(U)=\sum_j(\partial_jP)(a)U_j
\sum_{j=0}^{k}Use in source · 2 passages

Search aliases: ∑_j=0^k sum

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  • Use in source · Read passage · 10_foundations.tex, line 502

    For M≥1 and s,r,k∈ N, define the affine Mahler kernel by Θ_M,s(r,k)=[z^r](1+z)^s((1+z)^M-1)^k =∑_j=0^k(-1)^k-jbinom kjbinomMj+sr. It vanishes unless k≤ r≤ Mk+s. For a finite binomial expansion f(n)=∑_r=0^Dc_rbinom nr, the kth Newton coefficient of g(q)=f(Mq+s) is therefore c_k(g)=∑_r=k^min(D,Mk+s)c_rΘ_M,s(r,k). Indeed, insert the finite stencil for Δ_1^k g(0) and interchange its two finite sums. Coefficient extraction of the binomial expansion of ((1+z)^M-1)^k gives the kernel. In the binary zer…

    Exact mathematical passage
    
    \Theta_{M,s}(r,k)=[z^r](1+z)^s\bigl((1+z)^M-1\bigr)^k
    =\sum_{j=0}^{k}(-1)^{k-j}\binom kj\binom{Mj+s}{r}.
    
\sum_{j=0}^{n}Use in source · 2 passages

Search aliases: ∑_j=0^n sum

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  • Use in source · Read passage · 10_foundations.tex, line 255

    [Finite Newton--Mahler identities] For every n,s,h∈ N, f(n) · =∑_r=0^nc_r(f)binom nr, c_r(Δ_1^s f) · =c_s+r(f), Δ_1^s f(n) · =∑_j=0^nc_s+j(f)binom nj, Δ_h^s f(n) · =∑_j=0^s(-1)^s-jbinom sj f(n+jh). The finite interpolant I_Nf(n)=∑_r=0^Nc_r(f)binom nr agrees with f for n≤ N. It has the same order-s unit-step difference at n whenever n+s≤ N. For arbitrary coefficients c_0,…,c_N, the corresponding binomial interpolant has exactly those coefficients and zero coefficients above N.

    Exact mathematical passage
    
    \begin{aligned}
    f(n)&=\sum_{r=0}^{n}c_r(f)\binom nr,\\
    c_r(\Delta_1^s f)&=c_{s+r}(f),\\
    \Delta_1^s f(n)&=\sum_{j=0}^{n}c_{s+j}(f)\binom nj,\\
    \Delta_h^s f(n)&=\sum_{j=0}^{s}(-1)^{s-j}\binom sj f(n+jh).
    \end{aligned}
    
\sum_{j=0}^{N-1}Use in source · 2 passages

Search aliases: ∑_j=0^N-1 sum

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  • Use in source · Read passage · 10_foundations.tex, line 45

    [Residue reconstruction] The selectors satisfy P_N,j^2=P_N,j, P_N,iP_N,j=0 (i≠ j), ∑_j=0^N-1P_N,j=I. If Nmid M and M>0, then P_N,ja=∑_substack0≤ r<M

    Exact mathematical passage
    
    P_{N,j}^2=P_{N,j},\qquad P_{N,i}P_{N,j}=0\quad(i\ne j),
    \qquad \sum_{j=0}^{N-1}P_{N,j}=I.
    
\sum_{j=0}^{s}Use in source · 2 passages

Search aliases: ∑_j=0^s sum

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  • Use in source · Read passage · 10_foundations.tex, line 255

    [Finite Newton--Mahler identities] For every n,s,h∈ N, f(n) · =∑_r=0^nc_r(f)binom nr, c_r(Δ_1^s f) · =c_s+r(f), Δ_1^s f(n) · =∑_j=0^nc_s+j(f)binom nj, Δ_h^s f(n) · =∑_j=0^s(-1)^s-jbinom sj f(n+jh). The finite interpolant I_Nf(n)=∑_r=0^Nc_r(f)binom nr agrees with f for n≤ N. It has the same order-s unit-step difference at n whenever n+s≤ N. For arbitrary coefficients c_0,…,c_N, the corresponding binomial interpolant has exactly those coefficients and zero coefficients above N.

    Exact mathematical passage
    
    \begin{aligned}
    f(n)&=\sum_{r=0}^{n}c_r(f)\binom nr,\\
    c_r(\Delta_1^s f)&=c_{s+r}(f),\\
    \Delta_1^s f(n)&=\sum_{j=0}^{n}c_{s+j}(f)\binom nj,\\
    \Delta_h^s f(n)&=\sum_{j=0}^{s}(-1)^{s-j}\binom sj f(n+jh).
    \end{aligned}
    
\sum_{j=1}^dUse in source · 1 passage

Search aliases: ∑_j=1^d sum

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  • Use in source · Read passage · 30_lifting.tex, line 40

    [Polynomial Taylor remainder] Let R be a commutative ring, P∈ R[X_1,…,X_d], a∈ R^d, and c∈ R. There is a polynomial Q∈ R[U_1,…,U_d], with Q(0)=0, such that for every u∈ R^d, P(a+cu)=P(a)+c∑_j=1^d(∂_jP)(a)u_j+c^2Q(u). In particular, when R= Z and a,u,c are integral, subtracting the first two terms on the right from P(a+cu) gives a multiple of c^2.

    Exact mathematical passage
    \label{eq:polynomial-taylor}
     P(a+cu)=P(a)+c\sum_{j=1}^d(\partial_jP)(a)u_j+c^2Q(u).
    
\sum_{j=1}^nUse in source · 1 passage

Search aliases: ∑_j=1^n sum

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  • Use in source · Read passage · 40_regularity.tex, line 29

    [Derivative of polynomial evaluation] For P∈ K[X_1,…,X_n], polynomial evaluation is analytic at every point. At a∈ K^n its strict Fr echet derivative is v⟼∑_j=1^n (∂_jP)(a)v_j. For a finite system, its strict derivative is the continuous linear map represented by the matrix of evaluated formal partial derivatives. For the derivative statement, use polynomial induction on constants, addition, and multiplication by a variable. Constants have zero derivative. The sum rule agrees with the additivity…

    Exact mathematical passage
    
     v\longmapsto\sum_{j=1}^n (\partial_jP)(a)v_j.
    
\sum_{k=0}^{\min(d,r)}Use in source · 1 passage

Search aliases: ∑_k=0^min(d,r) sum min

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  • Use in source · Read passage · 72_recovered.tex, line 1641

    [Polynomial exponential Mahler transfer] Let f(n)=P(n)u^n with P∈ Z_p[n] of degree d. Let H=Λ_p,L, fix an address j, and write Q_j(x)=P(Hx+j), y=u^H, q=y-1. Then f(Hx+j)=∑_r≥0c_j,rbinom xr, where c_j,r = u^j ∑_k=0^min(d,r) binom rk Δ^kQ_j(0)y^kq^r-k. Consequently, v_p(c_j,r)≥ Lmax(r-d,0). For R≥ d, truncation at depth R is valid modulo p^L(R+1-d) uniformly on Z_p.

    Exact mathematical passage
    
    \boxed{
    c_{j,r}
    =
    u^j
    \sum_{k=0}^{\min(d,r)}
    \binom rk
    \Delta^kQ_j(0)y^kq^{r-k}.
    }
    
\sum_{k=0}^{n}Use in source · 1 passage

Search aliases: ∑_k=0^n sum

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  • Use in source · worked_atlas.tex, line 56

    AtlasHeadNewton coordinates turn precision into a finite computationatlas:newton Let b_k=Δ_1^kC_4,0(0). The first coefficients obtained from the difference triangle are inputgenerated/worked_newton_values.tex The exact Newton identity recovers each nonnegative integer row: C_4,0(n)=∑_k=0^nb_kbinom nk. AtlasImagenewton_cutoff For this address the proved law is v_2(b_k)=2k-s_2(k) for k≥1. Every k≥5 therefore has v_2(b_k)≥8. Since binom nk is integral, those terms vanish modulo 256. Thus, for every…

    Exact mathematical passage
    
     C_{4,0}(n)=\sum_{k=0}^{n}b_k\binom nk.
    
\sum_{k=1}^{p^{r-1}}Use in source · 2 passages

Search aliases: ∑_k=1^p^r-1 sum

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  • Use in source · Read passage · 20_arithmetic.tex, line 424

    The finite Newton expansion gives f(n+p^r-1h)-f(n) =∑_k=1^p^r-1binomp^r-1kΔ_h^k f(n). Terms with k≥ r have the required factor from the difference. For 1≤ k<r, the identity kbinomp^r-1k=p^r-1binomp^r-1-1k-1 and v_p(k)<k imply p^rmidbinomp^r-1k p^k in Z. Mapping that integer divisibility to R proves the claim term by term. The argument includes r=1. The declaration is leanKomorebiPaperIX.PeriodFromDifferences.prime_power_period.

    Exact mathematical passage
    
     f(n+p^{r-1}h)-f(n)
     =\sum_{k=1}^{p^{r-1}}\binom{p^{r-1}}k\Delta_h^k f(n).
    
\sum_{k=2}^{T}Use in source · 1 passage

Search aliases: ∑_k=2^T sum

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  • Use in source · Read passage · 20_arithmetic.tex, line 637

    Now expand the displacement: f(n+T)-f(n)=TΔ_1f(n) +∑_k=2^Tbinom TkΔ_1^kf(n). For 1≤ k≤ T, the binomial denominator identity implies v_2binom Tk≥ v_2(T)-v_2(k) ≥ v_2(T)-lfloorlog_2krfloor. Every term of the remaining sum consequently has valuation at least v_2(T)+c+1. Its sum cannot cancel the leading term, which has exact valuation v_2(T)+c. This proves nonvanishing and the claimed valuation using only finite sums. The general theorem is leanKomorebiPaperIX.NewtonDominance.exact_displacement.

    Exact mathematical passage
    
     f(n+T)-f(n)=T\Delta_1f(n)
              +\sum_{k=2}^{T}\binom Tk\Delta_1^kf(n).
    
\sum_{k\geq0}Use in source · 3 passages

Search aliases: ∑_k≥0 sum geq

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  • Use in source · Read passage · 72_recovered.tex, line 2212

    [Mahler polyphase transfer] Let f(x)=∑_r≥0c_rbinom xr be a convergent Mahler expansion. For g_s(q)=f(Mq+s), one has g_s(q)=∑_k≥0d_s,kbinom qk, where d_s,k = ∑_r=k^Mk+s c_rΘ_M,s(r,k). Every refined coefficient uses a finite window of the original Mahler coefficients.

    Exact mathematical passage
    
    \boxed{
    g_s(q)=\sum_{k\geq0}d_{s,k}\binom qk,}
    
\sum_{m=0}^{\min(M,N)}Use in source · 1 passage

Search aliases: ∑_m=0^min(M,N) sum min

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  • Use in source · Read passage · root_scope.tex, line 62

    tion v_2(β_m)=s_2(m)-1, v_2(κ_m)=4m-1. The source identity ∑_N≥0A(N)fract^2N(2N)! =fracsinh(2t)sinh t2t^2 implies, after multiplication by 4t^2/sinh^2(2t), E_2N=∑_m=0^N κ_mbinom2N2mA(N-m). For a retained depth M≥0, put T_M(N)= E_2N -∑_m=0^min(M,N)κ_mbinom2N2mA(N-m). An empty tail, N≤ M, is exactly zero.

    Exact mathematical passage
    
     \mathcal T_M(N)=\mathcal E_{2N}
       -\sum_{m=0}^{\min(M,N)}\kappa_m\binom{2N}{2m}A(N-m).
    
\sum_{m=0}^{\min(M,r)}Use in source · 2 passages

Search aliases: ∑_m=0^min(M,r) sum min

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  • Use in source · Read passage · supplement_02.tex, line 163

    al targets. targetOddEuler.odd_reconstruction_series targetOddEuler.odd_euler_reconstruction targetOddEuler.odd_euler_reconstruction_zigzag targetOddEuler.row_action, OddEuler.row_diagonal, OddEuler.row_weight pageheading4. Uniform truncation and a sharp exact exampleCorollary 4.3 and Example 4.4 For row 2r and depth M, the actual truncation is T_M(p,2r)=∑_m=0^min(M,r)binom2r2mb_m p^2mS_2(r-m)(p). Every omitted term has m≥ M+1. Its other factors are integers, so the first-block omitted-tail theo…

    Exact mathematical passage
    
     T_M(p,2r)=\sum_{m=0}^{\min(M,r)}\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).
    
\sum_{m=0}^NUse in source · 1 passage

Search aliases: ∑_m=0^N sum

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  • Use in source · Read passage · root_scope.tex, line 56

    The classical dyadic Bernoulli denominator law v_2(B_2m)=-1 and Legendre's formula yield v_2(β_m)=s_2(m)-1, v_2(κ_m)=4m-1. The source identity ∑_N≥0A(N)fract^2N(2N)! =fracsinh(2t)sinh t2t^2 implies, after multiplication by 4t^2/sinh^2(2t), E_2N=∑_m=0^N κ_mbinom2N2mA(N-m). For a retained depth M≥0, put T_M(N)= E_2N -∑_m=0^min(M,N)κ_mbinom2N2mA(N-m). An empty tail, N≤ M, is exactly zero.

    Exact mathematical passage
    
     \mathcal E_{2N}=\sum_{m=0}^N
        \kappa_m\binom{2N}{2m}A(N-m).
     \label{eq:root:packet}
    
\sum_{m=0}^{n}Use in source · 2 passages

Search aliases: ∑_m=0^n sum

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  • Use in source · Read passage · 20_arithmetic.tex, line 800

    [Classical odd-length reconstruction] For every positive odd integer p and every n∈ N, E_2n=∑_m=0^nbinom2n2mγ_m p^2mS_2(n-m)(p). Equivalently, E_2n=∑_q=0^n V_nqS_2q(p) for the lower triangular matrix V_nq=binom2n2qγ_n-qp^2(n-q) (q≤ n), V_nq=0 (q>n). It has V_nn=1 and p^2(n-q)mid V_nq when q≤ n.

    Exact mathematical passage
    
     E_{2n}=\sum_{m=0}^{n}\binom{2n}{2m}\gamma_m p^{2m}S_{2(n-m)}(p).
    
\sum_{m=0}^{n/2}Use in source · 1 passage

Search aliases: ∑_m=0^n/2 sum

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  • Use in source · Read passage · supplement_02.tex, line 131

    Substituting the independently proved signed-zigzag identity recovers the manuscript's notation. For every even degree n, E_n=∑_m=0^n/2(-1)^m Z_2m+1binom n2mp^2mS_n-2m(p). The Lean version uses n=2r to avoid ambiguous natural-number division. Every even natural degree has that form. Primality is not used in this equality.

    Exact mathematical passage
    
     \boxed{E_n=\sum_{m=0}^{n/2}(-1)^m Z_{2m+1}\binom n{2m}p^{2m}S_{n-2m}(p).}
    
\sum_{m=0}^{r}Use in source · 2 passages

Search aliases: ∑_m=0^r sum

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  • Use in source · Read passage · supplement_02.tex, line 125

    ing3. Exact reconstruction and the triangular matrixTheorem 4.2; manuscript page 11 For odd p, rescaling the already proved inverse law gives sech(pt)cosh(pt)=1. Multiplying the even-source identity by sech^2(pt) therefore proves sech^2(pt)(∑_r≥0S_2r(p)fract^2r(2r)!)=sech t. Actual EGF multiplication now gives the integer coefficient formula E_2r=∑_m=0^rbinom2r2mb_m p^2mS_2(r-m)(p). The formal proof first extracts the sum over all indices and then proves that the odd summands vanish. A checked e…

    Exact mathematical passage
    
     \boxed{E_{2r}=\sum_{m=0}^{r}\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).}
    
\sum_{m=1}^HUse in source · 1 passage

Search aliases: ∑_m=1^H sum

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  • Use in source · Read passage · strengthening.tex, line 390

    The conversion from unit steps to arbitrary mixed steps is used several times, so it is useful to make it explicit. Write Δ=Δ_1. For H>0, Δ_H=∑_m=1^Hbinom HmΔ^m. A mixed term with local orders m_1,…,m_k≥1 and M=∑_ν m_ν therefore has valuation at least M+v_2(M!)+β +∑_ν(v_2(H_ν)-v_2(m_ν)) if the unit-order estimate has offset β for every positive order. The inequalities m-1≥ v_2(m), M≥ k, v_2(M!)≥ v_2(k!) make this at least k+v_2(k!)+β+∑_ν v_2(H_ν). Every expansion is finite. This proves the desir…

    Exact mathematical passage
    
     \Delta_H=\sum_{m=1}^H\binom Hm\Delta^m.
     \label{eq:finite-step-expansion}
    
\sum_{m=k}^{n+k}Use in source · 1 passage

Search aliases: ∑_m=k^n+k sum

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  • Use in source · Read passage · strengthening.tex, line 667

    eciprocal are C_2(0)=1, C_2(1)=-frac115, C_2(2)=frac1345, a_1=-frac1615, a_2=frac6445, where a_m=Δ_1^m C_2(0). Thus v_2(a_1)=4 and v_2(a_2)=6. For m≥3, [source reference] bounds a_m by 2m+v_2(m!). The finite shifted Newton formula Δ_1^kC_2(n)=∑_m=k^n+ka_mbinom nm-k therefore gives v_2(Δ_1^kC_2(n))≥ k+v_2(k!)+3 for every k≥1. For k≥3 this is 2k≥ k+3; for k=1,2 it uses the exact initial coefficients and the higher tail bound. The finite step expansion proves the offset three in [source reference].

    Exact mathematical passage
    
     \Delta_1^kC_2(n)=\sum_{m=k}^{n+k}a_m\binom n{m-k}
     \label{eq:shifted-newton-finite}
    
\sum_{m\geq0}Use in source · 5 passages

Search aliases: ∑_m≥0 sum geq

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  • Use in source · Read passage · 72_recovered.tex, line 1188

    [Unified local Euler transfer] For every prime p and every even n≥0, E_n = ∑_r=0^n/2 binom n2r κ_p,r A_p,n-2r. If H_p(t) = ∑_m≥0A_p,2mfract^2m(2m)! and K_p(t) = ∑_r≥0κ_p,rfract^2r(2r)!, then K_p(t)H_p(t)=secht. For odd p, H_p(t)=cosh^2(pt)secht, K_p(t)=sech^2(pt). For p=2, H_2(t)=fracsinh(2t)sinh t2t^2, K_2(t)=frac4t^2sinh^2(2t).

    Exact mathematical passage
    
    H_p(t)
    =
    \sum_{m\geq0}A_{p,2m}\frac{t^{2m}}{(2m)!}
    
\sum_{m\ge1}Use in source · 1 passage

Search aliases: ∑_m≥1 sum ge

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  • Use in source · Read passage · 20_arithmetic.tex, line 966

    Let Q(z)=zcoth z, the regular formal series with constant coefficient one. The elementary differential identity (frac zsinh z)^2=Q(z)-zQ'(z) and the Bernoulli expansion Q(z)=1+∑_m≥1frac2^2mB_2m(2m)!z^2m give β_m=-2^2m(2m-1)B_2m/(2m)!. Multiplication by 2^2m(2m)! proves the identity for κ_m. The Bernoulli denominator theorem gives v_2(B_2m)=-1 for m>0: its denominator contains each prime q with (q-1)mid 2m exactly once, including 2[source reference]. Since 2m-1 is odd, the valuation is exactly 4m…

    Exact mathematical passage
    
     Q(z)=1+\sum_{m\ge1}\frac{2^{2m}B_{2m}}{(2m)!}z^{2m}
    
\sum_{m\geq\ell}Use in source · 2 passages

Search aliases: ∑_m≥ℓ sum geq ell

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  • Use in source · Read passage · strengthening.tex, line 1272

    Put a_m=Δ_1^mC_j(0) and expand the falling factorial in the Newton series. The ordinary coefficients are d_j,ℓ=∑_m≥ℓfraca_mm!s(m,ℓ). Since s(m,ℓ) is integral and v_2(a_m/m!)≥ m+β_j, the coefficient sum converges and has valuation at least ℓ+β_j. The sum exchange is justified on Z_2 by these uniform geometric coefficient bounds. The ordinary power series agrees there with the Newton interpolation and converges whenever |x|_2<2.

    Exact mathematical passage
    
     d_{j,\ell}=\sum_{m\geq\ell}\frac{a_m}{m!}s(m,\ell).
     \label{eq:newton-taylor-stirling}
    
\sum_nUse in source · 1 passage

Search aliases: ∑_n sum

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  • Use in source · Read passage · 10_foundations.tex, line 486

    The recovered Annals text develops a broader polyphase vocabulary. We explain it through elementary identities with explicit algebraic derivations. These identities have not been separately formalized in Lean for this consolidation. For a sequence a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplicati…

    Exact mathematical passage
    A(z)=\sum_n a_nz^n
\sum_{n\text{ even}}Use in source · 1 passage

Search aliases: ∑_n even sum text

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  • Use in source · Read passage · 72_recovered.tex, line 1064

    For even n≥0, define R_n = frac3^n+2-14(n+1)(n+2). Its even exponential generating function is ∑_n evenR_nfract^nn! = fracsinh(2t)sinh t2t^2. Define β_r by fracz^2sinh^2z = ∑_r≥0β_rz^2r.

    Exact mathematical passage
    
    \boxed{
    \sum_{n\text{ even}}R_n\frac{t^n}{n!}
    =
    \frac{\sinh(2t)\sinh t}{2t^2}.
    }
    
\sum_{N\ge0}Use in source · 1 passage

Search aliases: ∑_N≥0 sum ge

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  • Use in source · Read passage · root_scope.tex, line 51

    m≥1, differentiation of zcoth z gives β_m=-frac2^2m(2m-1)B_2m(2m)!, κ_m=-2^4m(2m-1)B_2m. The classical dyadic Bernoulli denominator law v_2(B_2m)=-1 and Legendre's formula yield v_2(β_m)=s_2(m)-1, v_2(κ_m)=4m-1. The source identity ∑_N≥0A(N)fract^2N(2N)! =fracsinh(2t)sinh t2t^2 implies, after multiplication by 4t^2/sinh^2(2t), E_2N=∑_m=0^N κ_mbinom2N2mA(N-m). For a retained depth M≥0, put T_M(N)= E_2N -∑_m=0^min(M,N)κ_mbinom2N2mA(N-m). An empty tail, N≤ M, is exactly zero.

    Exact mathematical passage
    
     \sum_{N\ge0}A(N)\frac{t^{2N}}{(2N)!}
     =\frac{\sinh(2t)\sinh t}{2t^2}
    
\sum_{n\geq0}Use in source · 15 passages

Search aliases: ∑_n≥0 sum geq

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  • Use in source · Read passage · 10_foundations.tex, line 78

    Over a characteristic-zero field K, put E(a)=∑_n≥0fraca_nn!t^n, D(A)_n=n![t^n]A. These inverse maps between sequences and K[[t]] provide formal exponential coordinates. No analytic convergence is needed for the coefficient identities below.

    Exact mathematical passage
    
    \mathcal E(a)=\sum_{n\geq0}\frac{a_n}{n!}t^n,
    \qquad \mathcal D(A)_n=n![t^n]A.
    
\sum_qUse in source · 1 passage

Search aliases: ∑_q sum

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  • Use in source · Read passage · 10_foundations.tex, line 486

    The recovered Annals text develops a broader polyphase vocabulary. We explain it through elementary identities with explicit algebraic derivations. These identities have not been separately formalized in Lean for this consolidation. For a sequence a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplicati…

    Exact mathematical passage
    A_r(w)=\sum_q a_{Mq+r}w^q
\sum_{q<n}Use in source · 3 passages

Search aliases: ∑_q<n sum

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  • Use in source · Read passage · 10_foundations.tex, line 182

    [Forward substitution and denominator clearing] Let T_nq be coefficients over a field with T_nn≠0 for every n. For any sequence b, define a_n=fracb_n-∑_q<nT_nqa_qT_nn. This constructs the unique solution of ∑_q=0^nT_nqa_q=b_n for all n. If T_nq and b_n are integers and the solution is taken over Q, then (∏_r=0^nT_rr)a_n∈ Z.

    Exact mathematical passage
    
    a_n=\frac{b_n-\sum_{q<n}T_{nq}a_q}{T_{nn}}.
    
\sum_{q=0}^nUse in source · 1 passage

Search aliases: ∑_q=0^n sum

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  • Use in source · Read passage · 20_arithmetic.tex, line 803

    [Classical odd-length reconstruction] For every positive odd integer p and every n∈ N, E_2n=∑_m=0^nbinom2n2mγ_m p^2mS_2(n-m)(p). Equivalently, E_2n=∑_q=0^n V_nqS_2q(p) for the lower triangular matrix V_nq=binom2n2qγ_n-qp^2(n-q) (q≤ n), V_nq=0 (q>n). It has V_nn=1 and p^2(n-q)mid V_nq when q≤ n.

    Exact mathematical passage
    E_{2n}=\sum_{q=0}^n V_{nq}S_{2q}(p)
\sum_{q=0}^{n-1}Use in source · 2 passages

Search aliases: ∑_q=0^n-1 sum

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  • Use in source · Read passage · 20_arithmetic.tex, line 39

    eleq:arith-kernel K_N,j(t)=j!∑_n≥0fract^Nn(Nn+j)!, P_N,j(t)=∑_n≥0C_N,j(n)fract^Nn(Nn)!. Their product is one. Clearing factorials in its coefficient of t^Nn gives the triangular system ∑_q=0^nbinomNn+jNqC_N,j(q)=δ_n0. In particular, for n>0, C_N,j(n)=-binomNn+jNn^-1 ∑_q=0^n-1binomNn+jNqC_N,j(q). The diagonal binomial coefficient is positive, so this recurrence determines the sequence uniquely. The Lean construction begins with this recurrence and derives the reciprocal identity from it. The resu…

    Exact mathematical passage
    
     C_{N,j}(n)=-\binom{Nn+j}{Nn}^{-1}
           \sum_{q=0}^{n-1}\binom{Nn+j}{Nq}C_{N,j}(q).
    
\sum_{q=0}^rUse in source · 1 passage

Search aliases: ∑_q=0^r sum

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  • Use in source · Read passage · strengthening.tex, line 566

    X)=∏_a=0^r-1(X-a(a+1)). At X=Z(Z+1), they have the exact factorization P_r(Z(Z+1)) =∏_a=0^r-1(Z-a)(Z+a+1) =(2r)!binomZ+r2r. The Bernoulli transform gives L((1+T)^Z)=log(1+T)/T, whence L(binom Zj)=(-1)^j/(j+1). Finite Vandermonde expansion therefore gives L(binomZ+r2r) · =(-1)^r∑_q=0^r(-1)^qbinom rqfrac1r+q+1nonumber · =(-1)^rfrac(r!)^2(2r+1)!. The last equality is the rational identity obtained by expanding ∫_0^1t^r(1-t)^r dt. This use of an ordinary finite integral makes no assertion about a po…

    Exact mathematical passage
    
     \mathcal L\left(\binom{Z+r}{2r}\right)
     &=(-1)^r\sum_{q=0}^r(-1)^q\binom rq\frac1{r+q+1}\nonumber\\
     &=(-1)^r\frac{(r!)^2}{(2r+1)!}.
     \label{eq:central-factorial-evaluation}
    
\sum_{q=0}^{r}Use in source · 1 passage

Search aliases: ∑_q=0^r sum

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  • Use in source · Read passage · supplement_02.tex, line 138

    The row action is connected to the matrix entries. Reversing the finite sum and using binomial symmetry gives the actual source-index row action E_2r=∑_q=0^rU_r,q(p)S_2q(p), where U_r,q(p)= displaystylebinom2r2qb_r-qp^2(r-q), · q≤ r,

    Exact mathematical passage
    
     E_{2r}=\sum_{q=0}^{r}U_{r,q}(p)S_{2q}(p),
    
\sum_{q=1}^nUse in source · 1 passage

Search aliases: ∑_q=1^n sum

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  • Use in source · Read passage · strengthening.tex, line 1031

    The actual reciprocal recurrence is C_N,j(n)=-∑_q=1^nbinomNnNq g_N,j(q)C_N,j(n-q), C_N,j(0)=1. For N=4, the four kernels are 1, frac14q+1, frac1(4q+1)(2q+1), frac3(4q+1)(2q+1)(4q+3). The threshold-free theorem gives v_2(a_k)≥ k in every address. The following exact values cover all orders where that bound does not already imply [source reference]: c r l r j · k · a_k · v_2(a_k)

    Exact mathematical passage
    
     C_{N,j}(n)=-\sum_{q=1}^n\binom{Nn}{Nq}
               g_{N,j}(q)C_{N,j}(n-q),\qquad C_{N,j}(0)=1.
     \label{eq:exact-reciprocal-row-recurrence}
    
\sum_{q\in J_r}Use in source · 1 passage

Search aliases: ∑_q∈ J_r sum in

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  • Use in source · Read passage · root_scope.tex, line 140

    ad J_r={q:λ_r(q)=μ_r}, G_r=2^-μ_rH_r. Since λ_r(0)=0 and λ_r(q)≥3q+1-s_2(r), every minimizing address satisfies 0≤ q≤lfloorfracs_2(r)-13rfloor. In particular μ_r≤0, and the minimum is found by a finite carry calculation. Reduction of the whole series gives overline G_r(X)=∑_q∈ J_rX^underline q in F_2[X]. Its highest degree is max J_r, with leading coefficient one. Thus the normalization is the actual Gauss normalization.

    Exact mathematical passage
    
     \overline G_r(X)=\sum_{q\in J_r}X^{\underline q}
           \quad\text{in }\mathbb F_2[X].
     \label{eq:root:reduction}
    
\sum_{q\ge0}Use in source · 1 passage

Search aliases: ∑_q≥0 sum ge

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  • Use in source · Read passage · root_scope.tex, line 110

    Set r=M+1≥1 and define P_q(x)=∏_i=0^q-1(x-i)(2x-(2i+1)), P_0=1, H_r(x)=∑_q≥08^qfracβ_r+qβ_r P_q(x)mathscr A(x-q). The coefficient valuations tend to infinity. At any nonnegative integer x, terms with q>x vanish. Cancelling the factorials in [source reference] therefore gives the exact identity T_r-1(N)=κ_rbinom2N2rH_r(N-r) (N≥ r). This is a finite rational identity at every original Euler row.

    Exact mathematical passage
    
     H_r(x)=\sum_{q\ge0}8^q\frac{\beta_{r+q}}{\beta_r}
                        P_q(x)\mathscr A(x-q).
    
\sum_{q\geq0}Use in source · 1 passage

Search aliases: ∑_q≥0 sum geq

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  • Use in source · Read passage · 72_recovered.tex, line 1993

    [Exact discrete polyphase reconstruction] The analysis map P_M is a bijection between sequences and ordered families of M sequences. If A(z)=∑_n≥0a_nz^n and A_r^[M](w)=∑_q≥0a_Mq+rw^q, then A(z)=∑_r=0^M-1z^rA_r^[M](z^M).

    Exact mathematical passage
    
    A_r^{[M]}(w)=\sum_{q\geq0}a_{Mq+r}w^q,
    
\sum_rUse in source · 1 passage

Search aliases: ∑_r sum

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  • Use in source · Read passage · strengthening.tex, line 498

    g s_2(i) for the number of ones in the binary expansion of i, and using v_2((4i)!)=4i-s_2(i), gives v_2([x^d]P_i) ≥2d-2i+s_2(i)+max(0,2i-d) ≥ d+s_2(i)≥ d+1 (d≥1). For k≥1, Δ_1^k x^d vanishes when d<k and is an integer polynomial with values divisible by k!. To see the latter, expand x^d=∑_r{d r} r!binom xr and apply Δ_1^kbinom xr=binomxr-k. It follows that Δ_1^kP_i(x)∈2^k+1k! Z_2 (k≥0, x∈ Z_2). For k=0 the bound follows directly from the coefficient estimate.

    Exact mathematical passage
    x^d=\sum_r\left\{\begin{smallmatrix}d\\r\end{smallmatrix}\right\}
    r!\binom xr
\sum_{r<d}Use in source · 1 passage

Search aliases: ∑_r<d sum

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  • Use in source · Read passage · 10_foundations.tex, line 415

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

    Exact mathematical passage
    \chi_{C^N}(X)=X^d+\sum_{r<d}q_rX^r
\sum_{r<n}Use in source · 1 passage

Search aliases: ∑_r<n sum

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  • Use in source · Read passage · 10_foundations.tex, line 157

    Pascal's identity proves the first product rule, and induction proves the higher rule. The intermediate binomial coefficients binom pr vanish in characteristic p. For the inverse, solve row n using the coefficient of b_n, which is g_0=1: b_n=1_n-∑_r<nbinom nr b_r g_n-r. Subtract the first and pth product rules for b⋆ g=1. The assumed identity for g leaves (D^p b-Db)⋆ g=0. A second triangular induction shows that convolution with a sequence having constant coefficient 1 has trivial kernel.

    Exact mathematical passage
    
    b_n=\mathbf1_n-\sum_{r<n}\binom nr b_r g_{n-r}.
    
\sum_{r=0}^{D}Use in source · 1 passage

Search aliases: ∑_r=0^D sum

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  • Use in source · Read passage · 10_foundations.tex, line 507

    For M≥1 and s,r,k∈ N, define the affine Mahler kernel by Θ_M,s(r,k)=[z^r](1+z)^s((1+z)^M-1)^k =∑_j=0^k(-1)^k-jbinom kjbinomMj+sr. It vanishes unless k≤ r≤ Mk+s. For a finite binomial expansion f(n)=∑_r=0^Dc_rbinom nr, the kth Newton coefficient of g(q)=f(Mq+s) is therefore c_k(g)=∑_r=k^min(D,Mk+s)c_rΘ_M,s(r,k). Indeed, insert the finite stencil for Δ_1^k g(0) and interchange its two finite sums. Coefficient extraction of the binomial expansion of ((1+z)^M-1)^k gives the kernel. In the binary zer…

    Exact mathematical passage
    f(n)=\sum_{r=0}^{D}c_r\binom nr
\sum_{r=0}^{d}Use in source · 2 passages

Search aliases: ∑_r=0^d sum

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  • Use in source · Read passage · 10_foundations.tex, line 140

    [Hurwitz product rules] For d≥0, D^d(f⋆ g)=∑_r=0^dbinom dr(D^r f)⋆(D^d-rg). If R has prime characteristic p, then D^p(f⋆ g)=f⋆ D^p g+D^p f⋆ g. Over a field, if g_0=1, a sequence b is constructed recursively with b⋆ g=1. If also D^p g=Dg in characteristic p, then D^p b=Db.

    Exact mathematical passage
    
    D^d(f\star g)=\sum_{r=0}^{d}\binom dr(D^r f)\star(D^{d-r}g).
    
\sum_{r=0}^{d-1}Use in source · 2 passages

Search aliases: ∑_r=0^d-1 sum

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  • Use in source · Read passage · 10_foundations.tex, line 416

    [Companion jumps and packet recurrences] Let C be a d× d matrix over a commutative ring and x_n=C^n x_0. For N,j,k∈ N, put b_k=x_Nk+j. Then b_k=C^j(C^N)^kx_0, b_k+1=C^Nb_k. If χ_C^N(X)=X^d+∑_r<dq_rX^r, then b_k+d=-∑_r=0^d-1q_r b_k+r. Every coordinate and every linear scalar observable satisfies the same recurrence. No invertibility or diagonalization hypothesis is required.

    Exact mathematical passage
    
    b_{k+d}=-\sum_{r=0}^{d-1}q_r b_{k+r}.
    
\sum_{r=0}^kUse in source · 3 passages

Search aliases: ∑_r=0^k sum

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  • Use in source · Read passage · strengthening.tex, line 159

    ence identity is Δ_1^kfrac11+sn =frac(-s)^k k!∏_h=0^k(1+s(n+h)). It follows by induction, subtracting adjacent rational functions; each successive subtraction adds the factor -(k+1)s in the numerator. Its denominator is a unit, so it belongs to p^k Z_p. The discrete Leibniz formula Δ_1^k(fg)(n) =∑_r=0^kbinom kr (Δ_1^r f)(n)(Δ_1^k-rg)(n+r) then proves Δ_1^k g_N,j(n)∈ p^k Z_p for every k,n. The coefficient at degree Nn of L_1^kG_N,j is precisely p^-kΔ_1^k g_N,j(n), and the other residue classes va…

    Exact mathematical passage
    
     \Delta_1^k(fg)(n)
     =\sum_{r=0}^k\binom kr
           (\Delta_1^r f)(n)(\Delta_1^{k-r}g)(n+r)
     \label{eq:discrete-leibniz-final}
    
\sum_{r=0}^{M-1}Use in source · 2 passages

Search aliases: ∑_r=0^M-1 sum

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  • Use in source · Read passage · 10_foundations.tex, line 488

    polyphase vocabulary. We explain it through elementary identities with explicit algebraic derivations. These identities have not been separately formalized in Lean for this consolidation. For a sequence a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplication, the phase-r product has a term without a …

    Exact mathematical passage
    
    A(z)=\sum_{r=0}^{M-1}z^rA_r(z^M),\qquad
    \bigl(a_r^{[M]}\bigr)_s^{[N]}=a_{r+Ms}^{[MN]}.
    
\sum_{r=0}^{\min(M,n/2)}Use in source · 2 passages

Search aliases: ∑_r=0^min(M,n/2) sum min

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  • Use in source · Read passage · 72_recovered.tex, line 1010

    [Odd modulus truncation] For every M≥0, every positive odd q, and every even n, E_n ≡ ∑_r=0^min(M,n/2) (-1)^rZ_2r+1 binom n2rq^2rS_n-2r(q) pmod q^2M+2. If pmid q and a=v_p(q), the same truncation is valid modulo p^2a(M+1).

    Exact mathematical passage
    
    \boxed{
    E_n
    \equiv
    \sum_{r=0}^{\min(M,n/2)}
    (-1)^rZ_{2r+1}
    \binom n{2r}q^{2r}S_{n-2r}(q)
    \pmod {q^{2M+2}}.
    }
    
\sum_{r=0}^{N}Use in source · 3 passages

Search aliases: ∑_r=0^N sum

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  • Use in source · Read passage · 10_foundations.tex, line 264

    Δ_h^s f(n) · =∑_j=0^s(-1)^s-jbinom sj f(n+jh). The finite interpolant I_Nf(n)=∑_r=0^Nc_r(f)binom nr agrees with f for n≤ N. It has the same order-s unit-step difference at n whenever n+s≤ N. For arbitrary coefficients c_0,…,c_N, the corresponding binomial interpolant has exactly those coefficients and zero coefficients above N.

    Exact mathematical passage
    I_Nf(n)=\sum_{r=0}^{N}c_r(f)\binom nr
\sum_{r=0}^{n-1}Use in source · 1 passage

Search aliases: ∑_r=0^n-1 sum

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  • Use in source · worked_atlas.tex, line 16

    tlasFigDir/#1.pdf AtlasHeadCompute the reciprocal before estimating itatlas:reciprocal For width N and address j, the normalized source and its reciprocal are G_N,j(t)=j!∑_n≥0fract^Nn(Nn+j)!, G_N,j(t)^-1=∑_n≥0C_N,j(n)fract^Nn(Nn)!. Multiplying the two series and clearing the factorials gives an exact triangular recurrence: C_N,j(0)=1, C_N,j(n)=-fracdisplaystyle∑_r=0^n-1 binomNn+jNrC_N,j(r)binomNn+jNn. AtlasImagereciprocal_rows The table is for N=4. The first two address-zero rows illustrate the …

    Exact mathematical passage
    
     C_{N,j}(0)=1,\qquad
     C_{N,j}(n)=-\frac{\displaystyle\sum_{r=0}^{n-1}
     \binom{Nn+j}{Nr}C_{N,j}(r)}{\binom{Nn+j}{Nn}}.
    
\sum_{r=0}^{n/2}Use in source · 6 passages

Search aliases: ∑_r=0^n/2 sum

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  • Use in source · Read passage · 72_recovered.tex, line 965

    [Odd modulus Euler coordinate pair] For every positive odd q and every even n≥0, E_n = ∑_r=0^n/2 (-1)^rZ_2r+1 binom n2r q^2rS_n-2r(q). The inverse transform is S_n(q) = E_n + ∑_r=1^n/2 2^2r-1 binom n2r q^2rE_n-2r. Both transforms are lower triangular with diagonal one.

    Exact mathematical passage
    
    \boxed{
    E_n
    =
    \sum_{r=0}^{n/2}
    (-1)^rZ_{2r+1}
    \binom n{2r}
    q^{2r}S_{n-2r}(q).
    }
    
\sum_{r=0}^{p-1}Use in source · 1 passage

Search aliases: ∑_r=0^p-1 sum

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  • Use in source · Read passage · 72_recovered.tex, line 235

    [Finite inverse formula] If x has augmentation one and n=x-1, then x^-1 = ∑_r=0^p-1(-n)^r.

    Exact mathematical passage
    
    \boxed{
    x^{-1}
    =
    \sum_{r=0}^{p-1}(-n)^r.
    }
    
\sum_{r=0}^{x}Use in source · 1 passage

Search aliases: ∑_r=0^x sum

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  • Use in source · Read passage · 72_recovered.tex, line 1719

    /3 in the one-prime example above loses the advertised digit; invertibility of C alone does not restore it. The original proof is retained under these hypotheses. Let Q=∏_p∈ Sp^L_p for a finite set of primes. Let C∈ M_d( Z) be invertible modulo Q, and choose H such that C^H=I+QB with B∈ M_d( Z). Then for every nonnegative integer x, a_Hx+j = ∑_r=0^xQ^rbinom xr w^TC^jB^rv_0. For every r, Δ_H^ra_n≡0pmod Q^r.

    Exact mathematical passage
    
    \boxed{
    a_{Hx+j}
    =
    \sum_{r=0}^{x}Q^r\binom xr
    w^TC^jB^rv_0.
    }
    
\sum_{r=1}^{h}Use in source · 2 passages

Search aliases: ∑_r=1^h sum

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  • Use in source · Read passage · 72_recovered.tex, line 407

    [Rational ordinary generating function] In F_p[[z]], ∑_n≥0e_n^(m,p)z^n = 1+ frac ∑_r=1^he_r^(m,p)z^r 1-z^h.

    Exact mathematical passage
    
    \boxed{
    \sum_{n\geq0}e_n^{(m,p)}z^n
    =
    1+
    \frac{
    \sum_{r=1}^{h}e_r^{(m,p)}z^r
    }{1-z^h}.
    }
    
\sum_{r=1}^{n}Use in source · 1 passage

Search aliases: ∑_r=1^n sum

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  • Use in source · Read passage · 72_recovered.tex, line 649

    [A denominator bound] Let D_N,j(n) = ∏_r=1^n binomNr+jNr, D_N,j(0)=1. Then D_N,j(n)C_N,j(n) ∈ Z. For every prime p, v_p(C_N,j(n)) ≥ - ∑_r=1^n v_pbinomNr+jNr.

    Exact mathematical passage
    
    \vp(C_{N,j}(n))
    \geq
    -
    \sum_{r=1}^{n}
    \vp\binom{Nr+j}{Nr}.
    
\sum_{r=1}^{n/2}Use in source · 1 passage

Search aliases: ∑_r=1^n/2 sum

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  • Use in source · Read passage · 72_recovered.tex, line 976

    [Odd modulus Euler coordinate pair] For every positive odd q and every even n≥0, E_n = ∑_r=0^n/2 (-1)^rZ_2r+1 binom n2r q^2rS_n-2r(q). The inverse transform is S_n(q) = E_n + ∑_r=1^n/2 2^2r-1 binom n2r q^2rE_n-2r. Both transforms are lower triangular with diagonal one.

    Exact mathematical passage
    
    \boxed{
    S_n(q)
    =
    E_n
    +
    \sum_{r=1}^{n/2}
    2^{2r-1}
    \binom n{2r}
    q^{2r}E_{n-2r}.
    }
    
\sum_{r=k}^{\min(D,Mk+s)}Use in source · 1 passage

Search aliases: ∑_r=k^min(D,Mk+s) sum min

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  • Use in source · Read passage · 10_foundations.tex, line 509

    For M≥1 and s,r,k∈ N, define the affine Mahler kernel by Θ_M,s(r,k)=[z^r](1+z)^s((1+z)^M-1)^k =∑_j=0^k(-1)^k-jbinom kjbinomMj+sr. It vanishes unless k≤ r≤ Mk+s. For a finite binomial expansion f(n)=∑_r=0^Dc_rbinom nr, the kth Newton coefficient of g(q)=f(Mq+s) is therefore c_k(g)=∑_r=k^min(D,Mk+s)c_rΘ_M,s(r,k). Indeed, insert the finite stencil for Δ_1^k g(0) and interchange its two finite sums. Coefficient extraction of the binomial expansion of ((1+z)^M-1)^k gives the kernel. In the binary zer…

    Exact mathematical passage
    
    c_k(g)=\sum_{r=k}^{\min(D,Mk+s)}c_r\Theta_{M,s}(r,k).
    
\sum_{r=k}^{Mk+s}Use in source · 1 passage

Search aliases: ∑_r=k^Mk+s sum

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  • Use in source · Read passage · 72_recovered.tex, line 2217

    [Mahler polyphase transfer] Let f(x)=∑_r≥0c_rbinom xr be a convergent Mahler expansion. For g_s(q)=f(Mq+s), one has g_s(q)=∑_k≥0d_s,kbinom qk, where d_s,k = ∑_r=k^Mk+s c_rΘ_M,s(r,k). Every refined coefficient uses a finite window of the original Mahler coefficients.

    Exact mathematical passage
    
    \boxed{
    d_{s,k}
    =
    \sum_{r=k}^{Mk+s}
    c_r\Theta_{M,s}(r,k).
    }
    
\sum_{r=s}^{N}Use in source · 2 passages

Search aliases: ∑_r=s^N sum

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  • Use in source · Read passage · 72_recovered.tex, line 1469

    [Finite Mahler reconstruction] Given f(0),…,f(N), put c_r=Δ^rf(0). Then f(n)=∑_r=0^nc_rbinom nr (0≤ n≤ N). For every s with 0≤ s≤ N, Δ^sf(n) = ∑_r=s^Nc_rbinom nr-s (0≤ n≤ N-s).

    Exact mathematical passage
    
    \boxed{
    \Delta^sf(n)
    =
    \sum_{r=s}^{N}c_r\binom n{r-s}
    \qquad(0\leq n\leq N-s).
    }
    
\sum_{r\bmod N}Use in source · 1 passage

Search aliases: ∑_rbmod N sum bmod

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  • Use in source · Read passage · 72_recovered.tex, line 2991

    [Complete residue packet system] For fixed N, the projectors P_N,r satisfy P_N,rP_N,s = δ_rsP_N,r and ∑_rbmod NP_N,r=I.

    Exact mathematical passage
    
    \boxed{
    \sum_{r\bmod N}P_{N,r}=I.
    }
    
\sum_{r\ge0}Use in source · 5 passages

Search aliases: ∑_r≥0 sum ge

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  • Use in source · Read passage · 20_arithmetic.tex, line 80

    [The Bernoulli boundary address] At N=2, the address j=1 gives a second classical family. Define the Bernoulli numbers by B(z)=fracze^z-1=∑_r≥0B_rfracz^rr!, B_0=1, B_1=-frac12. All quotients here are regular formal series. The identity 2B(z)-B(2z)=fraczsinh z =∑_n≥0C_2,1(n)fracz^2n(2n)! follows by putting the two fractions on a common denominator and using [source reference]. Comparing even coefficients yields C_2,1(n)=(2-2^2n)B_2n (n∈ N). This classical derivation restores the boundary example …

    Exact mathematical passage
    
     B(z)=\frac{z}{e^z-1}=\sum_{r\ge0}B_r\frac{z^r}{r!},
     \qquad B_0=1,\quad B_1=-\frac12.
    
\sum_{r\geq0}Use in source · 17 passages

Search aliases: ∑_r≥0 sum geq

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  • Use in source · Read passage · 72_recovered.tex, line 80

    The dyadic kernel is substantially sharper than the separate coefficient bound alone suggests. If fracz^2sinh^2z = ∑_r≥0β_rz^2r and κ_2,r=β_r2^2r(2r)!, then v_2(κ_2,r)=4r-1 (r≥1). Consequently, truncation after depth M is uniformly valid modulo 2^4M+3, and this exponent is optimal. The rowwise estimate records the additional binary carries in the binomial coefficient. Classical congruence theory for Euler families is extensive [source reference]. The formulas here are stated as exact coordinate …

    Exact mathematical passage
    
    \frac{z^2}{\sinh^2z}
    =
    \sum_{r\geq0}\beta_rz^{2r}
    
\sum_{r\geq1}Use in source · 1 passage

Search aliases: ∑_r≥1 sum geq

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  • Use in source · Read passage · 72_recovered.tex, line 1000

    The even source identity gives secht = sech^2(qt) ∑_n≥0S_2n(q)fract^2n(2n)!. Insert [source reference] and compare coefficients. For the inverse, multiply secht by cosh^2(qt) = 1+ ∑_r≥12^2r-1q^2rfract^2r(2r)!.

    Exact mathematical passage
    
    \cosh^2(qt)
    =
    1+
    \sum_{r\geq1}2^{2r-1}q^{2r}\frac{t^{2r}}{(2r)!}.
    
\sum_{s=0}^nUse in source · 1 passage

Search aliases: ∑_s=0^n sum

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  • Use in source · Read passage · strengthening.tex, line 376

    For every starting row, Δ_1^k C_0(n) = L(U^4nW^k) =∑_s=0^nbinom ns L(W^k+s). Each summand lies in 2^k k! Z_2, proving the factorial bound for address zero from row zero onward.

    Exact mathematical passage
    
     \Delta_1^k C_0(n)
       =\mathcal L(U^{4n}W^k)
       =\sum_{s=0}^n\binom ns\mathcal L(W^{k+s}).
     \label{eq:base-row-finite-expansion}
    
\sum_{\substack{0\leq a<p\\2a+1\neq p}}Use in source · 1 passage

Search aliases: ∑_substack0≤ a<p 2a+1≠ p sum substack leq neq

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  • Use in source · Read passage · 72_recovered.tex, line 1841

    Fix an even address j∈{0,2,…,p-3}. For x∈ Z_p, define mathscr E_p,j^×(x) = ∑_r≥0 · (-1)^rZ_2r+1p^2r binom(p-1)x+j2r · × ∑_substack0≤ a<p 2a+1≠ p (-1)^a (2a+1)^j-2r ((2a+1)^p-1)^x.

    Exact mathematical passage
    
    \begin{aligned}
    \mathscr E_{p,j}^\times(x)
    =
    \sum_{r\geq0}
    &(-1)^rZ_{2r+1}p^{2r}
    \binom{(p-1)x+j}{2r}\\
    &\times
    \sum_{\substack{0\leq a<p\\2a+1\neq p}}
    (-1)^a
    (2a+1)^{j-2r}
    \left((2a+1)^{p-1}\right)^x.
    \end{aligned}
    
\sum_{\substack{0\le r<M\\r\bmod N=j}}Use in source · 1 passage

Search aliases: ∑_substack0≤ r<M rbmod N=j sum substack le r bmod

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  • Use in source · Read passage · supplement_01.tex, line 47

    0, · otherwise. The source proves idempotence for every selector, orthogonality for distinct standard residue labels, and, when N>0, ∑_j=0^N-1P_N,ja=a. If Nmid M and M>0, it also proves P_N,ja=∑_substack0≤ r<M rbmod N=jP_M,ra. The formal power-series versions are proved coefficientwise for actual mcodePowerSeries objects. There is no finite truncation of the sequence domain in these theorems.

    Exact mathematical passage
    
     P_{N,j}a=\sum_{\substack{0\le r<M\\r\bmod N=j}}P_{M,r}a.
    
\sum_{\substack{0\leq r<M\\r\equiv j\pmod N}}Use in source · 1 passage

Search aliases: ∑_substack0≤ r<M r≡ jpmod N sum substack leq r equiv pmod

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  • Use in source · Read passage · 10_foundations.tex, line 50

    [Residue reconstruction] The selectors satisfy P_N,j^2=P_N,j, P_N,iP_N,j=0 (i≠ j), ∑_j=0^N-1P_N,j=I. If Nmid M and M>0, then P_N,ja=∑_substack0≤ r<M r≡ jpmod NP_M,ra. The same identities hold coefficientwise for formal power series.

    Exact mathematical passage
    
    P_{N,j}a=\sum_{\substack{0\leq r<M\\r\equiv j\pmod N}}P_{M,r}a.
    
\sum_{\substack{0\le r\le n\\w(r)<L}}Use in source · 1 passage

Search aliases: ∑_substack0≤ r≤ n w(r)<L sum substack le le w

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  • Use in source · Read passage · supplement_01.tex, line 135

    0, · w=∞. The infinity convention expresses a vanishing coefficient, not an infinite exponent operation. For any finite index set, every term with weight at least L is divisible by p^L. A finite-sum proof yields p^L bigm| ∑_r=0^nbinom nr c_rp^[w(r)]S_n-r -∑_substack0≤ r≤ n w(r)<L binom nr c_rp^[w(r)]S_n-r. The result is specialized to actual p-adic integers R= Z_p for prime p. Divisibility there is precisely the asserted membership in p^L Z_p.

    Exact mathematical passage
    
     \boxed{p^L\ \bigm|\ 
     \sum_{r=0}^{n}\binom nr c_rp^{[w(r)]}S_{n-r}
     -\sum_{\substack{0\le r\le n\\w(r)<L}}
         \binom nr c_rp^{[w(r)]}S_{n-r}.}
    
\sum_{\substack{0\leq r\leq n\\w(r)<L}}Use in source · 2 passages

Search aliases: ∑_substack0≤ r≤ n w(r)<L sum substack leq leq w

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  • Use in source · Read passage · 10_foundations.tex, line 215

    [Filtered finite reconstruction] For every n,L∈ N and sequences c,S in that ring, · ∑_r=0^nbinom nr c_rp^[w(r)]S_n-r · ≡ ∑_substack0≤ r≤ n w(r)<L binom nr c_rp^[w(r)]S_n-rpmodp^L. Here congruence means divisibility of the difference by p^L in the coefficient ring. In Z_p, it means membership in p^L Z_p.

    Exact mathematical passage
    
    \begin{split}
    &\sum_{r=0}^{n}\binom nr c_rp^{[w(r)]}S_{n-r}\\
    &\hspace{8mm}\equiv
    \sum_{\substack{0\leq r\leq n\\w(r)<L}}
    \binom nr c_rp^{[w(r)]}S_{n-r}\pmod{p^L}.
    \end{split}
    
\sum_uUse in source · 1 passage

Search aliases: ∑_u sum

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  • Use in source · Read passage · 72_recovered.tex, line 210

    [Augmentation criterion] Let x=∑_u∈ F_pc_u[u] ∈ F_p[ F_p^+]. Then x is a unit if and only if aug(x) = ∑_uc_u is nonzero. If J is the augmentation ideal, then J^p=0.

    Exact mathematical passage
    
    \operatorname{aug}(x)
    =
    \sum_uc_u
    
\sum_{u+v=r}Use in source · 2 passages

Search aliases: ∑_u+v=r sum

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  • Use in source · Read passage · 10_foundations.tex, line 494

    a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplication, the phase-r product has a term without a carry and a term with one carry: C_r(w)=∑_u+v=rA_u(w)B_v(w) +w∑_u+v=r+MA_u(w)B_v(w), 0≤ u,v<M. Adding the two phase addresses gives the formula. It describes ordinary Cauchy multiplication; Hurwitz convo…

    Exact mathematical passage
    
    C_r(w)=\sum_{u+v=r}A_u(w)B_v(w)
    +w\sum_{u+v=r+M}A_u(w)B_v(w),\qquad 0\leq u,v<M.
    
\sum_{u+v=r+M}Use in source · 2 passages

Search aliases: ∑_u+v=r+M sum

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  • Use in source · Read passage · 10_foundations.tex, line 494

    a and M≥1, let a_r^[M](q)=a_Mq+r for 0≤ r<M. If A(z)=∑_n a_nz^n and A_r(w)=∑_q a_Mq+rw^q are ordinary generating functions, then unique quotient and remainder give A(z)=∑_r=0^M-1z^rA_r(z^M), (a_r^[M])_s^[N]=a_r+Ms^[MN]. For ordinary Cauchy multiplication, the phase-r product has a term without a carry and a term with one carry: C_r(w)=∑_u+v=rA_u(w)B_v(w) +w∑_u+v=r+MA_u(w)B_v(w), 0≤ u,v<M. Adding the two phase addresses gives the formula. It describes ordinary Cauchy multiplication; Hurwitz convo…

    Exact mathematical passage
    
    C_r(w)=\sum_{u+v=r}A_u(w)B_v(w)
    +w\sum_{u+v=r+M}A_u(w)B_v(w),\qquad 0\leq u,v<M.
    
\sum_{u=0}^{s}Use in source · 1 passage

Search aliases: ∑_u=0^s sum

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  • Use in source · Read passage · 72_recovered.tex, line 2295

    insufficient. For p=2, M=2, s=0 and f(x)=x/2, one has c_1=1/2, c_2=0, while g(q)=q has d_1=1; thus d_1not≡ c_2pmod2. The coefficient consequence is read with integral original coefficients. Let M=p^a. Then Θ_p^a,s(r,k) ≡ binomsr-p^ak pmod p, where the binomial coefficient is zero outside its natural range. Consequently, d_s,k ≡ ∑_u=0^s binom su c_p^ak+u pmod p. In particular, d_0,k≡ c_p^akpmod p.

    Exact mathematical passage
    
    \boxed{
    d_{s,k}
    \equiv
    \sum_{u=0}^{s}
    \binom su
    c_{p^ak+u}
    \pmod p.
    }
    
\sum_{u\in\FF_p}Use in source · 3 passages

Search aliases: ∑_u∈FF_p ∑_u∈ F_p sum in FF sum in mathbb

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  • Use in source · Read passage · 72_recovered.tex, line 192

    The product law follows from [source reference]. To prove injectivity, suppose ∑_u∈ F_pc_uϵ_u=0. Looking at the coordinates n=0,1,…,p-1 gives a Vandermonde system with matrix (u^n)_0≤ n≤ p-1, u∈ F_p. Its determinant is the product of all differences between distinct field elements, so it is nonzero. Hence every c_u vanishes.

    Exact mathematical passage
    
    \sum_{u\in\FF_p}c_u\epsilon_u=0.
    
\sum_{u\neq0}Use in source · 1 passage

Search aliases: ∑_u≠0 sum neq

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  • Use in source · Read passage · 72_recovered.tex, line 343

    For positive n, the zero mode contributes nothing. Every nonzero mode satisfies u^p-1=1. Hence b_n+p-1^(m,p) = ∑_u≠0c_uu^n+p-1 = ∑_u≠0c_uu^n = b_n^(m,p).

    Exact mathematical passage
    
    b_{n+p-1}^{(m,p)}
    =
    \sum_{u\neq0}c_uu^{n+p-1}
    =
    \sum_{u\neq0}c_uu^n
    =
    b_n^{(m,p)}.
    
\sum_{v\in G_m}Use in source · 1 passage

Search aliases: ∑_v∈ G_m sum in

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  • Use in source · Read passage · 72_recovered.tex, line 429

    [Exponential form and minimal period] Let G_m=( F_p^×)^m. This cyclic group has order h. There are uniquely determined coefficients d_v∈ F_p, indexed by v∈ G_m, such that e_n^(m,p) = ∑_v∈ G_md_vv^n (n≥1). The minimal positive period is the least common multiple of the orders of those v for which d_v≠0. In particular, it divides h.

    Exact mathematical passage
    
    \boxed{
    e_n^{(m,p)}
    =
    \sum_{v\in G_m}d_vv^n
    \qquad(n\geq1).
    }
    
\sum_{\zeta\in H_m}Use in source · 2 passages

Search aliases: ∑_ζ∈ H_m sum zeta in

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  • Use in source · Read passage · 72_recovered.tex, line 272

    [Subgroup average] The multiple selector has the mode expansion a^(m) = frac1m ∑_ζ∈ H_m ϵ_ζ.

    Exact mathematical passage
    
    \boxed{
    a^{(m)}
    =
    \frac1m
    \sum_{\zeta\in H_m}
    \epsilon_\zeta.
    }
    
\sum_{\zeta\in\mu_m}Use in source · 1 passage

Search aliases: ∑_ζ∈μ_m sum zeta in mu

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  • Use in source · Read passage · 72_recovered.tex, line 847

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    
    T_{m,p}(\chi^\bullet)_n
    =
    \frac1m
    \sum_{\zeta\in\mu_m}
    (1+\zeta y)^{mn}.
    
\sum_\nuUse in source · 9 passages

Search aliases: ∑_ν sum nu

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  • Use in source · Read passage · strengthening.tex, line 215

    ng Δ_1^m B∈ p^m Z_p at every order and row. Use the finite operator identity Δ_H=∑_i=1^Hbinom HiΔ_1^i. The relation ibinom Hi=HbinomH-1i-1 implies v_p(p^ibinom Hi) ≥ i+v_p(H)-v_p(i)≥1+v_p(H). Multiplying these identities for the k steps, every term of total order M has its scalar multiplied by p^M, and is divisible by p^k+∑_ν v_p(H_ν). This proves the mixed-step bound. Finally, Δ_H A_N,j(n)=(-1)^n+HT_H C_N,j(n). Iteration contributes only a global sign. Multiplying by j! proves [source reference…

    Exact mathematical passage
    p^{k+\sum_\nu v_p(H_\nu)}
\sum_{\nu=1}^kUse in source · 7 passages

Search aliases: ∑_ν=1^k sum nu

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  • Use in source · Read passage · strengthening.tex, line 176

    [All-row prime-power transfer] Let p be prime, a≥1, N=p^a, and 0≤ j<N. The coefficients C_N,j(n) belong to Z_p for all n≥0. Set A_N,j(n)=(-1)^n C_N,j(n) and T_H=S^H-(-1)^H. For k≥1 and positive H_1,…,H_k, v_p(Δ_H_1⋯Δ_H_kA_N,j(n)) · ≥ k+∑_ν=1^k v_p(H_ν), v_p(T_H_1⋯ T_H_kE_Nn^(N,j)) · ≥ v_p(j!)+k+∑_ν=1^k v_p(H_ν). At p=2, the ordinary differences of C_N,j and of the original coefficients satisfy the same bounds. These dyadic statements also hold for N=1.

    Exact mathematical passage
    
     v_p(\Delta_{H_1}\cdots\Delta_{H_k}A_{N,j}(n))
      &\geq k+\sum_{\nu=1}^k v_p(H_\nu),
     \label{eq:prime-power-signed-bound}\\
     v_p(T_{H_1}\cdots T_{H_k}E_{Nn}^{(N,j)})
      &\geq v_p(j!)+k+\sum_{\nu=1}^k v_p(H_\nu).
     \label{eq:prime-power-original-twist}
    
\sum_{\chi\in\mu_h}Use in source · 1 passage

Search aliases: ∑_χ∈μ_h sum chi in mu

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  • Use in source · Read passage · 72_recovered.tex, line 843

    Choose d_0=f_h and d_n=f_n for n≥1. Then d is fully h periodic and f=d+cδ_0 for c=f_0-f_h. Since T_m,pδ_0 is the constant sequence one, it is enough to treat d. Expand d_q=∑_χ∈μ_hc_χχ^q. The mth power map on F_p^× has image μ_h. Choose y with y^m=χ. The root filter gives T_m,p(χ^bullet)_n = frac1m ∑_ζ∈μ_m (1+ζ y)^mn. If 1+ζ y=0, the corresponding term vanishes for positive n. Otherwise ((1+ζ y)^m)^h = (1+ζ y)^p-1 =1. Every summand is therefore h periodic for positive n. Thus T_m,p preserves V_h.…

    Exact mathematical passage
    
    d_q=\sum_{\chi\in\mu_h}c_\chi\chi^q.
    
\sum_{\ell\geq1}Use in source · 1 passage

Search aliases: ∑_ℓ≥1 sum ell geq

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  • Use in source · Read passage · strengthening.tex, line 1245

    [Taylor coefficient bounds and radii] For j=0,1,2,3, the dyadic interpolation has an ordinary power series widetilde C_j(x)=1+∑_ℓ≥1d_j,ℓx^ℓ, v_2(d_j,ℓ)≥ℓ+β_j. Its Taylor series converges for |x|_2<2. At addresses zero and one, v_2(d_j,ℓ)=ℓ (ℓ≥1), R_0=R_1=2. At address two, v_2(d_2,ℓ)≥ max(2ℓ+lfloorfracℓ2rfloor, ℓ+3), R_2=2^5/2. The radius at address three is at least two; its exact radius is not asserted here. The stated exact radii at addresses zero, one, and two hold at every center in Z_2.

    Exact mathematical passage
    
     \widetilde C_j(x)=1+\sum_{\ell\geq1}d_{j,\ell}x^\ell,
     \qquad v_2(d_{j,\ell})\geq\ell+\beta_j.
    
\capUse in source · 2 passages

Search aliases: ∩ cap

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  • Use in source · Read passage · continuation_10.tex, line 42

    The full-point theorem constructs b=(y,τ) and also proves b-a∈ p^M-sR^e+k∩ p^s+1R^e+k. Parameter equality b_T=τ is exact. Uniqueness concerns every root in the stated solved-coordinate ball with these parameters, not only roots obtained by the construction. No global uniqueness is claimed.

    Exact mathematical passage
    
    b-a\in p^{M-s}R^{e+k}\cap p^{s+1}R^{e+k}.
    
\cupUse in source · 4 passages

Search aliases: ∪ cup

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  • Use in source · Read passage · 10_foundations.tex, line 209

    To isolate the filtration argument, let p be any element of a commutative ring and let w: N→ N∪{∞}. Set p^[r]=p^r for finite r and p^[∞]=0. The latter notation means the coefficient vanishes.

    Exact mathematical passage
    w:\N\to\N\cup\{\infty\}
\intUse in source · 1 passage

Search aliases: ∫ int

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  • Use in source · Read passage · strengthening.tex, line 573

    · =(-1)^rfrac(r!)^2(2r+1)!. The last equality is the rational identity obtained by expanding ∫_0^1t^r(1-t)^r dt. This use of an ordinary finite integral makes no assertion about a positive or bounded p-adic measure. In particular, L(P_r(P))=(-1)^rfrac(r!)^22r+1.

    Exact mathematical passage
    \int_0^1t^r(1-t)^r\,dt
\int_0^1Use in source · 1 passage

Search aliases: ∫_0^1 int

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  • Use in source · Read passage · strengthening.tex, line 573

    · =(-1)^rfrac(r!)^2(2r+1)!. The last equality is the rational identity obtained by expanding ∫_0^1t^r(1-t)^r dt. This use of an ordinary finite integral makes no assertion about a positive or bounded p-adic measure. In particular, L(P_r(P))=(-1)^rfrac(r!)^22r+1.

    Exact mathematical passage
    \int_0^1t^r(1-t)^r\,dt
\simUse in source · 1 passage

Search aliases: ∼ sim

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  • Use in source · Read passage · continuation_05.tex, line 27

    There is a coordinate permutation σ, with (σ z)_i=z_σ(i). Write σ z=(y,t), using e solved coordinates and q free coordinates. Define (y_0,t_0)=σ x, B=x+p^L Z_p^e+q, T=t_0+p^L Z_p^q, R={z∈ B:F(z)=0}. There is a homeomorphism h:Txrightarrow ∼ R, h^-1(z)=free(σ z). Equivalently, a continuous function ϕ:T→ y_0+p^L Z_p^e gives the entire root locus through σ h(t)=(ϕ(t),t). Every parameter has exactly one corresponding root in B. For every N≥0, h(t)≡ h(u)pmodp^Niff t≡ upmodp^N.

    Exact mathematical passage
    
    h:T\xrightarrow{\ \sim\ }\mathcal R,\qquad
    h^{-1}(z)=\operatorname{free}(\sigma z).
    
\simeqUse in source · 7 passages

Search aliases: ≃ simeq

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  • Use in source · Read passage · 10_foundations.tex, line 452

    [Dyadic--odd packet factorization] Let a≥0 and let M be a positive odd integer. The Chinese remainder map is a ring isomorphism Z/(2^aM) Z≃ Z/2^a Z× Z/M Z. For 0≤ r<2^aM and any coefficient sequence f, P_2^aM,rf =P_2^a,rbmod 2^a P_M,rbmod Mf. The same identity holds for formal-series coefficient selectors.

    Exact mathematical passage
    
    \Z/(2^aM)\Z\simeq\Z/2^a\Z\times\Z/M\Z.
    
\neqUse in source · 8 passages

Search aliases: ≠ neq

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  • Use in source · Read passage · 72_recovered.tex, line 230

    Choose a generator g of the cyclic additive group. Then F_p[ F_p^+] ≅ F_p[y]/(y^p-1) = F_p[y]/(y-1)^p. The augmentation ideal is generated by y-1, so its pth power is zero. Every element has the form a+n with a∈ F_p and n∈ J. It is invertible exactly when a≠0, which is the augmentation condition.

    Exact mathematical passage
    a\neq0
\leqUse in source · 71 passages

Search aliases: ≤ leq

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  • Use in source · Read passage · 10_foundations.tex, line 29

    Let R be a commutative ring, a: N→ R, N≥1, and 0≤ j<N. Define (P_N,ja)(n)= a(n), · n≡ jpmod N,

    Exact mathematical passage
    0\leq j<N
\bigoplusUse in source · 2 passages

Search aliases: ⊕ bigoplus

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  • Use in source · Read passage · 72_recovered.tex, line 2741

    [Packet decomposition of a lift obstruction] Suppose F_p^d = ⊕_a∈ AU_a, F_p^r = ⊕_a∈ AV_a, and suppose a linear correction map D: F_p^d→ F_p^r satisfies D(U_a)⊆ V_a for every a. Write D_a=D|_U_a. Then coker D ≅ ⊕_a∈ Acoker D_a. A residual lifts if and only if every packet obstruction vanishes.

    Exact mathematical passage
    
    \FF_p^d
    =
    \bigoplus_{a\in A}U_a,
    \qquad
    \FF_p^r
    =
    \bigoplus_{a\in A}V_a,
    
\bigoplus_{a\in A}Use in source · 2 passages

Search aliases: ⊕_a∈ A bigoplus in

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  • Use in source · Read passage · 72_recovered.tex, line 2741

    [Packet decomposition of a lift obstruction] Suppose F_p^d = ⊕_a∈ AU_a, F_p^r = ⊕_a∈ AV_a, and suppose a linear correction map D: F_p^d→ F_p^r satisfies D(U_a)⊆ V_a for every a. Write D_a=D|_U_a. Then coker D ≅ ⊕_a∈ Acoker D_a. A residual lifts if and only if every packet obstruction vanishes.

    Exact mathematical passage
    
    \FF_p^d
    =
    \bigoplus_{a\in A}U_a,
    \qquad
    \FF_p^r
    =
    \bigoplus_{a\in A}V_a,
    
\otimesUse in source · 2 passages

Search aliases: ⊗ otimes

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  • Use in source · Read passage · 71_native.tex, line 134

    coker(Hom_B(Ω_P/ Z⊗_P B, F_p)⟶Hom_B(I/I^2, F_p)).

    Exact mathematical passage
    \operatorname{coker}\left(\operatorname{Hom}_B(\Omega_{P/\mathbb Z}\otimes_P B,\mathbb F_p)\longrightarrow\operatorname{Hom}_B(I/I^2,\mathbb F_p)\right).
\otimes_PUse in source · 1 passage

Search aliases: ⊗_P otimes

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  • Use in source · Read passage · 71_native.tex, line 134

    coker(Hom_B(Ω_P/ Z⊗_P B, F_p)⟶Hom_B(I/I^2, F_p)).

    Exact mathematical passage
    \operatorname{coker}\left(\operatorname{Hom}_B(\Omega_{P/\mathbb Z}\otimes_P B,\mathbb F_p)\longrightarrow\operatorname{Hom}_B(I/I^2,\mathbb F_p)\right).
\cdotsUse in source · 17 passages

Search aliases: ⋯ cdots

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  • Use in source · Read passage · 72_recovered.tex, line 703

    [Lucas congruence] If a=a_0+a_1p+⋯ and b=b_0+b_1p+⋯ are base p expansions, then binom ab ≡ ∏_rbinoma_rb_r pmod p. This is the classical theorem of Lucas [source reference].

    Exact mathematical passage
    a=a_0+a_1p+\cdots
\rangle^\timesUse in source · 3 passages

Search aliases: ⟩^× rangle times

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  • Use in source · Read passage · root_scope.tex, line 103

    to zero dyadically. Extend A by mathscr A(x)=frac9^x+1-18(x+1)(2x+1), mathscr A(-1)=-fraclog98. The apparent singularity at x=-1 is removable. Since v_2(log9)=3, its convergent expansion is mathscr A(x)=fraclog9/81+2x ∑_h≥0frac(log9)^h(h+1)!(x+1)^h. It follows coefficientwise that mathscr A∈ Z_2⟨ x⟩^×, mathscr A≡1pmod2, and mathscr A(N)=A(N).

    Exact mathematical passage
    \mathscr A\in\mathbb Z_2\langle x\rangle^\times
\rangle_2Use in source · 1 passage

Search aliases: ⟩_2 rangle

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  • Use in source · Read passage · supplement_05.tex, line 119

    For p=3,ℓ=2,a=2, the checked values are ω_2(2)=8, ⟨2⟩_2=7, 8^2=1,quad8·7=2 in Z/9 Z. The unlifted representative 2 is not a square root of one modulo nine.

    Exact mathematical passage
    
     \omega_2(2)=8,\qquad \langle2\rangle_2=7,\qquad
     8^2=1,\quad8\cdot7=2\quad\text{in }\Z/9\Z.
    
\rangle_{2,\ell}Use in source · 2 passages

Search aliases: ⟩_2,ℓ rangle ell

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  • Use in source · Read passage · supplement_05.tex, line 153

    Why the mod-four branch is separate. Now let ℓ≥2 and a∈( Z/2^ℓ Z)^×. The actual units of Z/4 Z are exhausted by 1 and -1. Define ε_ℓ(a)= 1, · a≡1pmod4, -1, · a≡3pmod4, ⟨ a⟩_2,ℓ=ε_ℓ(a)a. The source uses ℓ=L+2. It proves a=ε_ℓ(a)⟨ a⟩_2,ℓ, ε_ℓ(a)∈{1,-1}, ⟨ a⟩_2,ℓ≡1pmod4. The sign is uniquely determined by the residue modulo four, and cancellation gives uniqueness of the principal factor.

    Exact mathematical passage
    
     \varepsilon_\ell(a)=
     \begin{cases}1,&a\equiv1\pmod4,\\-1,&a\equiv3\pmod4,\end{cases}
     \qquad
     \langle a\rangle_{2,\ell}=\varepsilon_\ell(a)a.
    
\rangle_4Use in source · 1 passage

Search aliases: ⟩_4 rangle

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  • Use in source · Read passage · 10_foundations.tex, line 375

    Z/p^L+1 Z)^×. Define ω_L(a)=a^p^L, ⟨ a⟩_L=aω_L(a)^-1. Then a=ω_L(a)⟨ a⟩_L, ω_L(a)^p-1=1, ⟨ a⟩_L≡1pmod p. The two factors are uniquely determined by these properties and are compatible with reduction to lower precision. For a∈( Z/2^L+2 Z)^×, there is instead a unique decomposition a=ϵ(a)⟨ a⟩_4, ϵ(a)∈{1,-1}, ⟨ a⟩_4≡1pmod4.

    Exact mathematical passage
    
    a=\epsilon(a)\langle a\rangle_4,
    \qquad \epsilon(a)\in\{1,-1\},\qquad
    \langle a\rangle_4\equiv1\pmod4.
    
\rangle_kUse in source · 1 passage

Search aliases: ⟩_k rangle

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  • Use in source · Read passage · supplement_05.tex, line 140

    pal_mul pageheading3. Precision compatibility and the dyadic signFinite refinements and Proposition 6.2's dyadic branch For 1≤ k≤ℓ, let ρ_ℓ,k: Z/p^ℓ Z⟶ Z/p^k Z be the actual reduction homomorphism. It induces a map on units. The explicit construction satisfies ρ_ℓ,k(ω_ℓ(a)) =ω_k(ρ_ℓ,k(a)), ρ_ℓ,k(⟨ a⟩_ℓ) =⟨ρ_ℓ,k(a)⟩_k. The proof uses the exponent factorization p^ℓ-1=p^k-1p^ℓ-k and the already proved Frobenius invariance of a lifted root. Principal factors also satisfy ⟨ a⟩_ℓ^ p^ℓ-1=1. At precisio…

    Exact mathematical passage
    
     \boxed{\rho_{\ell,k}(\langle a\rangle_\ell)
     =\langle\rho_{\ell,k}(a)\rangle_k.}
    
\rangle_LUse in source · 2 passages

Search aliases: ⟩_L rangle

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  • Use in source · Read passage · 10_foundations.tex, line 364

    [Finite unit coordinates] Let p be prime, L≥0, and a∈( Z/p^L+1 Z)^×. Define ω_L(a)=a^p^L, ⟨ a⟩_L=aω_L(a)^-1. Then a=ω_L(a)⟨ a⟩_L, ω_L(a)^p-1=1, ⟨ a⟩_L≡1pmod p. The two factors are uniquely determined by these properties and are compatible with reduction to lower precision. For a∈( Z/2^L+2 Z)^×, there is instead a unique decomposition a=ϵ(a)⟨ a⟩_4, ϵ(a)∈{1,-1}, ⟨ a⟩_4≡1pmod4.

    Exact mathematical passage
    
    \omega_L(a)=a^{p^L},\qquad \langle a\rangle_L=a\omega_L(a)^{-1}.
    
\rangle_\ellUse in source · 3 passages

Search aliases: ⟩_ℓ rangle ell

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  • Use in source · Read passage · supplement_05.tex, line 81

    ffPadic_coe, .standard_window_integral targetWindowIndependence.residue_independence, .reduced_unique targetWindowIndependence.two_window_addresses_congruent pageheading2. Explicit finite prime-power decompositionProposition 6.2; odd-prime branch Let p be prime, ℓ≥1, and a∈( Z/p^ℓ Z)^×. Define actual units in the same ring by ω_ℓ(a)=a^p^ℓ-1, ⟨ a⟩_ℓ=a ω_ℓ(a)^-1. The Lean source writes ℓ=L+1, so zero precision is excluded without truncated natural-number subtraction.

    Exact mathematical passage
    
     \boxed{\omega_\ell(a)=a^{p^{\ell-1}},\qquad
     \langle a\rangle_\ell=a\,\omega_\ell(a)^{-1}.}
    
\rangle_\ell^{\,p^{\ell-1}}Use in source · 1 passage

Search aliases: ⟩_ℓ^ p^ℓ-1 rangle ell ell

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  • Use in source · Read passage · supplement_05.tex, line 146

    hism. It induces a map on units. The explicit construction satisfies ρ_ℓ,k(ω_ℓ(a)) =ω_k(ρ_ℓ,k(a)), ρ_ℓ,k(⟨ a⟩_ℓ) =⟨ρ_ℓ,k(a)⟩_k. The proof uses the exponent factorization p^ℓ-1=p^k-1p^ℓ-k and the already proved Frobenius invariance of a lifted root. Principal factors also satisfy ⟨ a⟩_ℓ^ p^ℓ-1=1. At precision one the principal factor is exactly one. These are coherent finite certificates; no infinite inverse limit or topological unit-group equivalence is constructed in this release.

    Exact mathematical passage
    
     \langle a\rangle_\ell^{\,p^{\ell-1}}=1.
    
\longmapstoUse in source · 8 passages

Search aliases: ⟼ longmapsto

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  • Use in source · Read passage · 30_lifting.tex, line 265

    Integer representatives and residue classes at the next level must also be distinguished. For q≥1 and L≥0, the map S_q^d⟶ {y∈( Z/q^L+1 Z)^d:y≡ xpmodq^L}, t⟼ x+q^Lwidetilde t is a bijection, where widetilde t is any integer representative. Independence of the representative and injectivity both follow by cancelling q^L in integer divisibility; surjectivity follows by expressing the lower congruence as a multiple of q^L. This is the equivalence leanKomorebi.ResidueLiftDigits.digitEquivResidue.

    Exact mathematical passage
    
     S_q^d\longrightarrow
     \{y\in(\Z/q^{L+1}\Z)^d:y\equiv x\pmod{q^L}\},
     \qquad t\longmapsto x+q^L\widetilde t
    

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