Notation by work
Special Functions
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974 symbol spellings in 2254 mathematical passages across 37 selected source files. Subscripts and superscripts have their own entries alongside their base symbols.
The same symbol can have several meanings here. A passage classified as a definition may define another part of its expression. Read the linked context before assigning a meaning.
All 974 entries are available below.
+Introduction passage · 437 passages
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Introduction passage · chapters_sequences.tex, line 70
At degree four, E_4=5 contributes 5t^4/4! to secant. At degree five, E_5=16 contributes 16t^5/5! to tangent. The first tangent entry corresponds to degree one. Parity packets can identify familiar functions, provided the factorial normalization and sequence offset are kept. OEIS A000111, A000364, and A000182. The tangent entry has OEIS initial index one; τ_m denotes its degree-2m+1 value. Exact series checked through degree eighteen.
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2m+1
=Introduction passage · 911 passages
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Introduction passage · chapters_sequences.tex, line 68
ler_updown ∑_n≥0E_nfract^nn!=sec t+tan t, E_2m=s_m, E_2m+1=τ_m The coefficient table separates even and odd positions of the Euler up/down family. Even positions are positive secant numbers. Odd positions are tangent numbers. The factorial row makes clear that these are exponential-generating-function numbers, not ordinary Taylor coefficients. At degree four, E_4=5 contributes 5t^4/4! to secant. At degree five, E_5=16 contributes 16t^5/5! to tangent. The first tangent entry corresponds to degree…
Exact mathematical passage
E_4=5
AIntroduction passage · 101 passages
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Introduction passage · Read passage · coupled_visual.tex, line 188
There exist nonzero complex constants A,B with V(z)=A U(Bz) if and only if D=0. On either surface, put s=sqrtk/p>0. The scalar family is defined by T_x,y'(z)=1+T_x,y(xz)T_x,y(yz), T_x,y(0)=0. It gives the explicit reduction U(z)=sqrts T_q,sr(z/sqrts), V(z)=s^-1/2T_q,sr(sqrts z). Both effective scales lie in (0,1). Equality U=V is more restrictive: it occurs exactly when {q,r}={a,b}.
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A,B
aIntroduction passage · 255 passages
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Introduction passage · Read passage · notation.tex, line 3
FrontChapterNotation and glossary Notation is defined locally when each family is introduced. The same letter can serve different roles in separate chapters. In particular, a is an arithmetic weight parameter in Chapter~[source reference], a contraction in Chapter~[source reference], and a spectral shift in Appendix~[source reference].
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a
a^\rhoIntroduction passage · 5 passages
Search aliases: a^ρ rho
Introduction passage · Read passage · notation.tex, line 28
item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ a^ρ · b^ρ=1. A positive right eigenvector gives the proportions of the two type vectors. item[Arithmetic depth A_a,β,Q.] The entire sum ∑_n≥2, (n,Q)=1μ(n)^2n^-2a(1-n^-β)^2s, with a>1/2, β>0, Q≥1 integer and the real logarithm in the complex power. The atom n=1 is excluded at every depth. item[Möbius function and modulus radical.] μ(1)=1. For a product of k distinct primes, μ(n)=…
Exact mathematical passage
\operatorname{spr}\begin{pmatrix}q^\rho&r^\rho\\a^\rho&b^\rho\end{pmatrix}=1
BIntroduction passage · 92 passages
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Introduction passage · Read passage · coupled_visual.tex, line 188
There exist nonzero complex constants A,B with V(z)=A U(Bz) if and only if D=0. On either surface, put s=sqrtk/p>0. The scalar family is defined by T_x,y'(z)=1+T_x,y(xz)T_x,y(yz), T_x,y(0)=0. It gives the explicit reduction U(z)=sqrts T_q,sr(z/sqrts), V(z)=s^-1/2T_q,sr(sqrts z). Both effective scales lie in (0,1). Equality U=V is more restrictive: it occurs exactly when {q,r}={a,b}.
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A,B
bIntroduction passage · 150 passages
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Introduction passage · Read passage · notation.tex, line 24
item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ
Exact mathematical passage
V'=1+U(az)V(bz)
b^\rhoIntroduction passage · 7 passages
Search aliases: b^ρ rho
Introduction passage · Read passage · notation.tex, line 28
item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ a^ρ · b^ρ=1. A positive right eigenvector gives the proportions of the two type vectors. item[Arithmetic depth A_a,β,Q.] The entire sum ∑_n≥2, (n,Q)=1μ(n)^2n^-2a(1-n^-β)^2s, with a>1/2, β>0, Q≥1 integer and the real logarithm in the complex power. The atom n=1 is excluded at every depth. item[Möbius function and modulus radical.] μ(1)=1. For a product of k distinct primes, μ(n)=…
Exact mathematical passage
\operatorname{spr}\begin{pmatrix}q^\rho&r^\rho\\a^\rho&b^\rho\end{pmatrix}=1
DIntroduction passage · 72 passages
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Introduction passage · Read passage · sieve_full.tex, line 28
Write D(s)=ζ(s)/ζ(2s). For arbitrary finite complex strengths and Re s>1, absolute convergence permits expansion of the finite prime product. The local squarefree Euler factors give
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D(s)=\zeta(s)/\zeta(2s)
dIntroduction passage · 107 passages
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Introduction passage · Read passage · resolution_full.tex, line 78
Write f=f_3/4 for the normalized diagonal function, and f(z)=∑_n≥0d_nz^2n+1. The inherited tap continuity theorem gives
Exact mathematical passage
f(z)=\sum_{n\ge0}d_nz^{2n+1}
d_nIntroduction passage · 20 passages
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Introduction passage · Read passage · resolution_full.tex, line 78
Write f=f_3/4 for the normalized diagonal function, and f(z)=∑_n≥0d_nz^2n+1. The inherited tap continuity theorem gives
Exact mathematical passage
f(z)=\sum_{n\ge0}d_nz^{2n+1}
EIntroduction passage · 23 passages
Search aliases: E
Introduction passage · chapters_sequences.tex, line 68
ler_updown ∑_n≥0E_nfract^nn!=sec t+tan t, E_2m=s_m, E_2m+1=τ_m The coefficient table separates even and odd positions of the Euler up/down family. Even positions are positive secant numbers. Odd positions are tangent numbers. The factorial row makes clear that these are exponential-generating-function numbers, not ordinary Taylor coefficients. At degree four, E_4=5 contributes 5t^4/4! to secant. At degree five, E_5=16 contributes 16t^5/5! to tangent. The first tangent entry corresponds to degree…
Exact mathematical passage
E_4=5
E_4Introduction passage · 1 passage
Search aliases: E_4
Introduction passage · chapters_sequences.tex, line 68
ler_updown ∑_n≥0E_nfract^nn!=sec t+tan t, E_2m=s_m, E_2m+1=τ_m The coefficient table separates even and odd positions of the Euler up/down family. Even positions are positive secant numbers. Odd positions are tangent numbers. The factorial row makes clear that these are exponential-generating-function numbers, not ordinary Taylor coefficients. At degree four, E_4=5 contributes 5t^4/4! to secant. At degree five, E_5=16 contributes 16t^5/5! to tangent. The first tangent entry corresponds to degree…
Exact mathematical passage
E_4=5
E_5Introduction passage · 1 passage
Search aliases: E_5
Introduction passage · chapters_sequences.tex, line 68
E_2m=s_m, E_2m+1=τ_m The coefficient table separates even and odd positions of the Euler up/down family. Even positions are positive secant numbers. Odd positions are tangent numbers. The factorial row makes clear that these are exponential-generating-function numbers, not ordinary Taylor coefficients. At degree four, E_4=5 contributes 5t^4/4! to secant. At degree five, E_5=16 contributes 16t^5/5! to tangent. The first tangent entry corresponds to degree one. Parity packets can identify familiar…
Exact mathematical passage
E_5=16
FIntroduction passage · 97 passages
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Introduction passage · Read passage · notation.tex, line 14
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
Exact mathematical passage
F_\Theta'=1+\sum_jh_jF_\Theta(q_jz)F_\Theta(r_jz)
fIntroduction passage · 60 passages
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Introduction passage · Read passage · notation.tex, line 16
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
Exact mathematical passage
M_f(R)=\max_{|z|=R}|f(z)|
f_{3/4}Introduction passage · 6 passages
Search aliases: f_3/4
Introduction passage · Read passage · resolution_full.tex, line 77
Write f=f_3/4 for the normalized diagonal function, and f(z)=∑_n≥0d_nz^2n+1. The inherited tap continuity theorem gives
Exact mathematical passage
f=f_{3/4}
F_\ThetaIntroduction passage · 5 passages
Search aliases: F_Θ Theta
Introduction passage · Read passage · notation.tex, line 14
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
Exact mathematical passage
F_\Theta'=1+\sum_jh_jF_\Theta(q_jz)F_\Theta(r_jz)
F_\Theta'Introduction passage · 2 passages
Search aliases: F_Θ' Theta
Introduction passage · Read passage · notation.tex, line 14
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
Exact mathematical passage
F_\Theta'=1+\sum_jh_jF_\Theta(q_jz)F_\Theta(r_jz)
hIntroduction passage · 84 passages
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Introduction passage · Read passage · notation.tex, line 13
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
Exact mathematical passage
h_j\ge0
h_jIntroduction passage · 24 passages
Search aliases: h_j
Introduction passage · Read passage · notation.tex, line 13
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
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h_j\ge0
jIntroduction passage · 211 passages
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Introduction passage · Read passage · notation.tex, line 13
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
Exact mathematical passage
h_j\ge0
kIntroduction passage · 122 passages
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Introduction passage · Read passage · coupled_inverse_full.tex, line 55
If these conditions hold, put s=sqrtk/p>0, and let T_x,y denote the normalized scalar solution T'=1+T(xz)T(yz), T(0)=0. Then
Exact mathematical passage
s=\sqrt{k/p}>0
\limsupIntroduction passage · 6 passages
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Introduction passage · Read passage · notation.tex, line 18
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
Exact mathematical passage
\rho(f)=\limsup_{R\to\infty}\frac{\log\log M_f(R)}{\log R}.
\limsup_{R\to\infty}Introduction passage · 6 passages
Search aliases: limsup_R→∞ limsup to infty
Introduction passage · Read passage · notation.tex, line 18
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
Exact mathematical passage
\rho(f)=\limsup_{R\to\infty}\frac{\log\log M_f(R)}{\log R}.
\logIntroduction passage · 126 passages
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Introduction passage · Read passage · notation.tex, line 18
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
Exact mathematical passage
\rho(f)=\limsup_{R\to\infty}\frac{\log\log M_f(R)}{\log R}.
MIntroduction passage · 72 passages
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Introduction passage · Read passage · notation.tex, line 16
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
Exact mathematical passage
M_f(R)=\max_{|z|=R}|f(z)|
mIntroduction passage · 122 passages
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Introduction passage · chapters_sequences.tex, line 70
dinary Taylor coefficients. At degree four, E_4=5 contributes 5t^4/4! to secant. At degree five, E_5=16 contributes 16t^5/5! to tangent. The first tangent entry corresponds to degree one. Parity packets can identify familiar functions, provided the factorial normalization and sequence offset are kept. OEIS A000111, A000364, and A000182. The tangent entry has OEIS initial index one; τ_m denotes its degree-2m+1 value. Exact series checked through degree eighteen.
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\tau_m
M_fIntroduction passage · 5 passages
Search aliases: M_f
Introduction passage · Read passage · notation.tex, line 16
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
Exact mathematical passage
M_f(R)=\max_{|z|=R}|f(z)|
\maxIntroduction passage · 43 passages
Search aliases: max
Introduction passage · Read passage · notation.tex, line 16
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
Exact mathematical passage
M_f(R)=\max_{|z|=R}|f(z)|
\max_{|z|=R}Introduction passage · 4 passages
Search aliases: max_|z|=R max
Introduction passage · Read passage · notation.tex, line 16
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
Exact mathematical passage
M_f(R)=\max_{|z|=R}|f(z)|
nIntroduction passage · 274 passages
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Introduction passage · Read passage · resolution_full.tex, line 78
Write f=f_3/4 for the normalized diagonal function, and f(z)=∑_n≥0d_nz^2n+1. The inherited tap continuity theorem gives
Exact mathematical passage
f(z)=\sum_{n\ge0}d_nz^{2n+1}
PIntroduction passage · 43 passages
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Introduction passage · Read passage · notation.tex, line 26
item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ
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\operatorname{spr}P
pIntroduction passage · 225 passages
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Introduction passage · Read passage · coupled_inverse_full.tex, line 55
If these conditions hold, put s=sqrtk/p>0, and let T_x,y denote the normalized scalar solution T'=1+T(xz)T(yz), T(0)=0. Then
Exact mathematical passage
s=\sqrt{k/p}>0
qIntroduction passage · 273 passages
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Introduction passage · Read passage · notation.tex, line 13
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
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0<q_j,r_j<1
q^\rhoIntroduction passage · 19 passages
Search aliases: q^ρ rho
Introduction passage · Read passage · notation.tex, line 28
item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ a^ρ · b^ρ=1. A positive right eigenvector gives the proportions of the two type vectors. item[Arithmetic depth A_a,β,Q.] The entire sum ∑_n≥2, (n,Q)=1μ(n)^2n^-2a(1-n^-β)^2s, with a>1/2, β>0, Q≥1 integer and the real logarithm in the complex power. The atom n=1 is excluded at every depth. item[Möbius function and modulus radical.] μ(1)=1. For a product of k distinct primes, μ(n)=…
Exact mathematical passage
\operatorname{spr}\begin{pmatrix}q^\rho&r^\rho\\a^\rho&b^\rho\end{pmatrix}=1
q_jIntroduction passage · 22 passages
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Introduction passage · Read passage · notation.tex, line 13
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
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0<q_j,r_j<1
RIntroduction passage · 164 passages
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Introduction passage · Read passage · notation.tex, line 16
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
Exact mathematical passage
M_f(R)=\max_{|z|=R}|f(z)|
rIntroduction passage · 268 passages
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Introduction passage · Read passage · notation.tex, line 13
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
Exact mathematical passage
0<q_j,r_j<1
R^\rhoIntroduction passage · 21 passages
Search aliases: R^ρ rho
Introduction passage · Read passage · notation.tex, line 22
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
Exact mathematical passage
\log M_f(R)/R^\rho
r^\rhoIntroduction passage · 15 passages
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Introduction passage · Read passage · notation.tex, line 28
item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ a^ρ · b^ρ=1. A positive right eigenvector gives the proportions of the two type vectors. item[Arithmetic depth A_a,β,Q.] The entire sum ∑_n≥2, (n,Q)=1μ(n)^2n^-2a(1-n^-β)^2s, with a>1/2, β>0, Q≥1 integer and the real logarithm in the complex power. The atom n=1 is excluded at every depth. item[Möbius function and modulus radical.] μ(1)=1. For a product of k distinct primes, μ(n)=…
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\operatorname{spr}\begin{pmatrix}q^\rho&r^\rho\\a^\rho&b^\rho\end{pmatrix}=1
r_jIntroduction passage · 26 passages
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Introduction passage · Read passage · notation.tex, line 13
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
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0<q_j,r_j<1
\ReIntroduction passage · 24 passages
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Introduction passage · Read passage · sieve_full.tex, line 29
Write D(s)=ζ(s)/ζ(2s). For arbitrary finite complex strengths and Re s>1, absolute convergence permits expansion of the finite prime product. The local squarefree Euler factors give
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\Re s>1
sIntroduction passage · 154 passages
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Introduction passage · Read passage · sieve_full.tex, line 28
Write D(s)=ζ(s)/ζ(2s). For arbitrary finite complex strengths and Re s>1, absolute convergence permits expansion of the finite prime product. The local squarefree Euler factors give
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D(s)=\zeta(s)/\zeta(2s)
s^{-1/2}Introduction passage · 2 passages
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Introduction passage · Read passage · coupled_visual.tex, line 193
There exist nonzero complex constants A,B with V(z)=A U(Bz) if and only if D=0. On either surface, put s=sqrtk/p>0. The scalar family is defined by T_x,y'(z)=1+T_x,y(xz)T_x,y(yz), T_x,y(0)=0. It gives the explicit reduction U(z)=sqrts T_q,sr(z/sqrts), V(z)=s^-1/2T_q,sr(sqrts z). Both effective scales lie in (0,1). Equality U=V is more restrictive: it occurs exactly when {q,r}={a,b}.
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U(z)=\sqrt{s}\,T_{q,sr}(z/\sqrt{s}),\qquad V(z)=s^{-1/2}T_{q,sr}(\sqrt{s}\,z).
\operatorname{spr}Introduction passage · 5 passages
Search aliases: spr operatorname
Introduction passage · Read passage · notation.tex, line 26
item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ
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\operatorname{spr}P
TIntroduction passage · 83 passages
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Introduction passage · Read passage · coupled_inverse_full.tex, line 55
If these conditions hold, put s=sqrtk/p>0, and let T_x,y denote the normalized scalar solution T'=1+T(xz)T(yz), T(0)=0. Then
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T_{x,y}
tIntroduction passage · 130 passages
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Introduction passage · chapters_sequences.tex, line 68
≥0E_nfract^nn!=sec t+tan t, E_2m=s_m, E_2m+1=τ_m The coefficient table separates even and odd positions of the Euler up/down family. Even positions are positive secant numbers. Odd positions are tangent numbers. The factorial row makes clear that these are exponential-generating-function numbers, not ordinary Taylor coefficients. At degree four, E_4=5 contributes 5t^4/4! to secant. At degree five, E_5=16 contributes 16t^5/5! to tangent. The first tangent entry corresponds to degree one. Parity p…
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5t^4/4!
T'Introduction passage · 4 passages
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Introduction passage · Read passage · coupled_inverse_full.tex, line 56
If these conditions hold, put s=sqrtk/p>0, and let T_x,y denote the normalized scalar solution T'=1+T(xz)T(yz), T(0)=0. Then
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T'=1+T(xz)T(yz)
t^4Introduction passage · 1 passage
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Introduction passage · chapters_sequences.tex, line 68
≥0E_nfract^nn!=sec t+tan t, E_2m=s_m, E_2m+1=τ_m The coefficient table separates even and odd positions of the Euler up/down family. Even positions are positive secant numbers. Odd positions are tangent numbers. The factorial row makes clear that these are exponential-generating-function numbers, not ordinary Taylor coefficients. At degree four, E_4=5 contributes 5t^4/4! to secant. At degree five, E_5=16 contributes 16t^5/5! to tangent. The first tangent entry corresponds to degree one. Parity p…
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5t^4/4!
t^5Introduction passage · 1 passage
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Introduction passage · chapters_sequences.tex, line 68
d E_2m+1=τ_m The coefficient table separates even and odd positions of the Euler up/down family. Even positions are positive secant numbers. Odd positions are tangent numbers. The factorial row makes clear that these are exponential-generating-function numbers, not ordinary Taylor coefficients. At degree four, E_4=5 contributes 5t^4/4! to secant. At degree five, E_5=16 contributes 16t^5/5! to tangent. The first tangent entry corresponds to degree one. Parity packets can identify familiar functio…
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16t^5/5!
T_{q,sr}Introduction passage · 3 passages
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Introduction passage · Read passage · coupled_visual.tex, line 193
There exist nonzero complex constants A,B with V(z)=A U(Bz) if and only if D=0. On either surface, put s=sqrtk/p>0. The scalar family is defined by T_x,y'(z)=1+T_x,y(xz)T_x,y(yz), T_x,y(0)=0. It gives the explicit reduction U(z)=sqrts T_q,sr(z/sqrts), V(z)=s^-1/2T_q,sr(sqrts z). Both effective scales lie in (0,1). Equality U=V is more restrictive: it occurs exactly when {q,r}={a,b}.
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U(z)=\sqrt{s}\,T_{q,sr}(z/\sqrt{s}),\qquad V(z)=s^{-1/2}T_{q,sr}(\sqrt{s}\,z).
T_{x,y}Introduction passage · 3 passages
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Introduction passage · Read passage · coupled_inverse_full.tex, line 55
If these conditions hold, put s=sqrtk/p>0, and let T_x,y denote the normalized scalar solution T'=1+T(xz)T(yz), T(0)=0. Then
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T_{x,y}
T_{x,y}'Introduction passage · 1 passage
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Introduction passage · Read passage · coupled_visual.tex, line 191
There exist nonzero complex constants A,B with V(z)=A U(Bz) if and only if D=0. On either surface, put s=sqrtk/p>0. The scalar family is defined by T_x,y'(z)=1+T_x,y(xz)T_x,y(yz), T_x,y(0)=0. It gives the explicit reduction U(z)=sqrts T_q,sr(z/sqrts), V(z)=s^-1/2T_q,sr(sqrts z). Both effective scales lie in (0,1). Equality U=V is more restrictive: it occurs exactly when {q,r}={a,b}.
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T_{x,y}'(z)=1+T_{x,y}(xz)T_{x,y}(yz)
UIntroduction passage · 62 passages
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Introduction passage · Read passage · notation.tex, line 23
item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ
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(U,V)
U'Introduction passage · 6 passages
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Introduction passage · Read passage · notation.tex, line 24
item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ
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U'=1+U(qz)V(rz)
VIntroduction passage · 54 passages
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Introduction passage · Read passage · notation.tex, line 23
item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ
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(U,V)
V'Introduction passage · 4 passages
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Introduction passage · Read passage · notation.tex, line 24
item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ
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V'=1+U(az)V(bz)
xIntroduction passage · 179 passages
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Introduction passage · Read passage · coupled_inverse_full.tex, line 55
If these conditions hold, put s=sqrtk/p>0, and let T_x,y denote the normalized scalar solution T'=1+T(xz)T(yz), T(0)=0. Then
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T_{x,y}
yIntroduction passage · 33 passages
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Introduction passage · Read passage · coupled_inverse_full.tex, line 55
If these conditions hold, put s=sqrtk/p>0, and let T_x,y denote the normalized scalar solution T'=1+T(xz)T(yz), T(0)=0. Then
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T_{x,y}
zIntroduction passage · 131 passages
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Introduction passage · Read passage · notation.tex, line 14
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
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F_\Theta'=1+\sum_jh_jF_\Theta(q_jz)F_\Theta(r_jz)
z^{2n+1}Introduction passage · 12 passages
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Introduction passage · Read passage · resolution_full.tex, line 78
Write f=f_3/4 for the normalized diagonal function, and f(z)=∑_n≥0d_nz^2n+1. The inherited tap continuity theorem gives
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f(z)=\sum_{n\ge0}d_nz^{2n+1}
|Introduction passage · 97 passages
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Introduction passage · Read passage · notation.tex, line 16
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
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M_f(R)=\max_{|z|=R}|f(z)|
\zetaIntroduction passage · 13 passages
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Introduction passage · Read passage · sieve_full.tex, line 28
Write D(s)=ζ(s)/ζ(2s). For arbitrary finite complex strengths and Re s>1, absolute convergence permits expansion of the finite prime product. The local squarefree Euler factors give
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D(s)=\zeta(s)/\zeta(2s)
\ThetaIntroduction passage · 5 passages
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Introduction passage · Read passage · notation.tex, line 14
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
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F_\Theta'=1+\sum_jh_jF_\Theta(q_jz)F_\Theta(r_jz)
\rhoIntroduction passage · 105 passages
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Introduction passage · Read passage · notation.tex, line 18
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
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\rho(f)=\limsup_{R\to\infty}\frac{\log\log M_f(R)}{\log R}.
\tauIntroduction passage · 9 passages
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Introduction passage · chapters_sequences.tex, line 70
dinary Taylor coefficients. At degree four, E_4=5 contributes 5t^4/4! to secant. At degree five, E_5=16 contributes 16t^5/5! to tangent. The first tangent entry corresponds to degree one. Parity packets can identify familiar functions, provided the factorial normalization and sequence offset are kept. OEIS A000111, A000364, and A000182. The tangent entry has OEIS initial index one; τ_m denotes its degree-2m+1 value. Exact series checked through degree eighteen.
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\tau_m
\tau_mIntroduction passage · 3 passages
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Introduction passage · chapters_sequences.tex, line 70
dinary Taylor coefficients. At degree four, E_4=5 contributes 5t^4/4! to secant. At degree five, E_5=16 contributes 16t^5/5! to tangent. The first tangent entry corresponds to degree one. Parity packets can identify familiar functions, provided the factorial normalization and sequence offset are kept. OEIS A000111, A000364, and A000182. The tangent entry has OEIS initial index one; τ_m denotes its degree-2m+1 value. Exact series checked through degree eighteen.
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\tau_m
\toIntroduction passage · 27 passages
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Introduction passage · Read passage · notation.tex, line 18
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
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\rho(f)=\limsup_{R\to\infty}\frac{\log\log M_f(R)}{\log R}.
\inIntroduction passage · 49 passages
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Introduction passage · Read passage · notation.tex, line 25
item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exponent satisfies sprq^ρ · r^ρ
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q,r,a,b\in(0,1)
\sumIntroduction passage · 157 passages
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Introduction passage · Read passage · notation.tex, line 13
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
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\sum_jh_j=1
\sum_jIntroduction passage · 31 passages
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Introduction passage · Read passage · notation.tex, line 13
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
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\sum_jh_j=1
\sum_{n\ge0}Introduction passage · 16 passages
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Introduction passage · Read passage · resolution_full.tex, line 78
Write f=f_3/4 for the normalized diagonal function, and f(z)=∑_n≥0d_nz^2n+1. The inherited tap continuity theorem gives
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f(z)=\sum_{n\ge0}d_nz^{2n+1}
\inftyIntroduction passage · 44 passages
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Introduction passage · Read passage · notation.tex, line 18
item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coupled pair (U,V).] The normalized entire solution of U'=1+U(qz)V(rz), V'=1+U(az)V(bz) and U(0)=V(0)=0, with q,r,a,b∈(0,1). item[Perron balance.] sprP denotes the spectral radius of a matrix. The characteristic exp…
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\rho(f)=\limsup_{R\to\infty}\frac{\log\log M_f(R)}{\log R}.
\geIntroduction passage · 212 passages
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Introduction passage · Read passage · notation.tex, line 13
item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglog M_f(R)log R. At positive finite order ρ, the lower and upper types used here are the liminf and limsup of log M_f(R)/R^ρ. Their equality is not assumed. item[Coup…
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h_j\ge0
*Use in source · 7 passages
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Use in source · Read passage · main.tex, line 252
ion. Right: initial squarefree frequencies and weights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If p_* is the least prime not dividing Q, then τ=-2log(1-p_*^-β), M=Z_Q(2a)-1, |A(s)|≤ M e^τmax(-Re s,0). This proves local uniform convergence and entire continuation. The firs…
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p_*
\mathbf1Use in source · 7 passages
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Use in source · Read passage · main.tex, line 360
Prime sieves and signed measures. pageheadingFAMILY 3An infinite prime shelf changes the packets For a prime p and λ≥-1, let x_n=(1-n^-2)^2, ν_p,λ=∑_n≥1μ(n)^2n^-4 (1+λ1_pmid n)δ_x_n. Define monic orthogonal polynomials P_m^(k;p,λ) for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the or…
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x_n=(1-n^{-2})^2,\qquad \nu_{p,\lambda}=\sum_{n\ge1}\mu(n)^2n^{-4} (1+\lambda\mathbf1_{p\mid n})\delta_{x_n}.
\mathbf1_{p\mid n}Use in source · 6 passages
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Use in source · Read passage · main.tex, line 360
Prime sieves and signed measures. pageheadingFAMILY 3An infinite prime shelf changes the packets For a prime p and λ≥-1, let x_n=(1-n^-2)^2, ν_p,λ=∑_n≥1μ(n)^2n^-4 (1+λ1_pmid n)δ_x_n. Define monic orthogonal polynomials P_m^(k;p,λ) for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the or…
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x_n=(1-n^{-2})^2,\qquad \nu_{p,\lambda}=\sum_{n\ge1}\mu(n)^2n^{-4} (1+\lambda\mathbf1_{p\mid n})\delta_{x_n}.
\mathbf1_{q\mid n}Use in source · 1 passage
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Use in source · Read passage · sieve_full.tex, line 187
Fix p∈ F, with all other strengths at least -1, and write κ(n)=∏_q∈ F∖{p}(1+λ_q1_qmid n). Separate the positive measure into its p-free and p-divisible parts:
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\kappa(n)=\prod_{q\in F\setminus\{p\}}(1+\lambda_q\mathbf1_{q\mid n})
\AUse in source · 6 passages
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Use in source · Read passage · main.tex, line 230
MonographPartIIArithmetic depth and signed measuresComplex parameters, positive kernels and the boundaries of positivity. Arithmetic at complex depth. pageheadingFAMILY 2Arithmetic energy at complex depth For real a>1/2, β>0, and integer Q≥1, define A_a,β,Q(s)=∑_substackn≥2 (n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At d…
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\boxed{\A(s)=\sum_{\substack{n\ge2\\(n,Q)=1}} \mu(n)^2n^{-2a}(1-n^{-\beta})^{2s},\qquad s\in\mathbb C.}
a'Use in source · 1 passage
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Use in source · Read passage · arithmetic_full.tex, line 48
a=a', β=β', rad(Q)=rad(Q').
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a=a',\qquad \beta=\beta',\qquad \operatorname{rad}(Q)=\operatorname{rad}(Q').
A^{(j)}Use in source · 2 passages
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Use in source · Read passage · main.tex, line 282
claimNo polynomial-coefficient linear ODE Suppose ∑_j=0^rP_j(s)A^(j)(s)=0, where the P_j are polynomials and are not all zero. At s=-t, divide by the exponential of the first support atom. The remaining tail is exponentially smaller, so the polynomial ∑_jP_j(-t)(-λ_p_*)^j tends to zero exponentially and must vanish identically. Remove that atom and repeat. Every support frequency therefore satisfies ∑_jP_j(s)(-λ_n)^j=0 for all s. For fixed s, this polynomial in λ has infinitely many distinct roo…
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\sum_{j=0}^{r}P_j(s)A^{(j)}(s)=0
A^{(k)}Use in source · 1 passage
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Use in source · Read passage · arithmetic_full.tex, line 402
The monomial moment matrices also have infinite rank: m_k=(-1)^kA^(k)(0)=∑_jw_jλ_j^k, and for every nonzero polynomial P, ∑_jw_j|P(λ_j)|^2>0, because a nonzero polynomial cannot vanish at infinitely many distinct frequencies. All finite derivative Hankel matrices are strictly positive definite.
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m_k=(-1)^kA^{(k)}(0)=\sum_jw_j\lambda_j^k
a^2Use in source · 8 passages
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Use in source · Read passage · sieve_full.tex, line 115
L_a(x)=∏_n=1^a-1fracx-x_nx_a-x_n, C_a=L_a(1)=∏_n=1^a-1frac1-x_nx_a-x_n>0, r_a=frac1-x_a+11-x_a =fraca^4(2(a+1)^2-1)(a+1)^4(2a^2-1)∈(0,1), f_a,m(x)=L_a(x)(frac1-x1-x_a)^m, m=0,1,2,…. tag2
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r_a=\frac{1-x_{a+1}}{1-x_a} =\frac{a^4\bigl(2(a+1)^2-1\bigr)}{(a+1)^4(2a^2-1)}\in(0,1),
a^{2i+1}Use in source · 4 passages
Search aliases: a^2i+1
Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
a^3Use in source · 1 passage
Search aliases: a^3
Use in source · Read passage · sieve_full.tex, line 133
∑_n>a|μ(n)^2n^-4κ_F(n)| ≤fracC_F3a^3. tag3
Exact mathematical passage
\sum_{n>a}|\mu(n)^2n^{-4}\kappa_F(n)| \le\frac{C_F}{3a^3}. \tag{3}
a^4Use in source · 2 passages
Search aliases: a^4
Use in source · Read passage · sieve_full.tex, line 115
L_a(x)=∏_n=1^a-1fracx-x_nx_a-x_n, C_a=L_a(1)=∏_n=1^a-1frac1-x_nx_a-x_n>0, r_a=frac1-x_a+11-x_a =fraca^4(2(a+1)^2-1)(a+1)^4(2a^2-1)∈(0,1), f_a,m(x)=L_a(x)(frac1-x1-x_a)^m, m=0,1,2,…. tag2
Exact mathematical passage
r_a=\frac{1-x_{a+1}}{1-x_a} =\frac{a^4\bigl(2(a+1)^2-1\bigr)}{(a+1)^4(2a^2-1)}\in(0,1),
a^kUse in source · 5 passages
Search aliases: a^k
Use in source · Read passage · spectral_continuation.tex, line 89
ageheadingORBIT SPECTRA / SHIFTSA Hurwitz-type deformation and its determinant For real a≥0, define initially Z_N(s;a)=∑_j(w_j+a)^-s, Re s>1/N. The rightward shift preserves avoidance of the negative branch cut. Normalize the product with E_N(a) and use log E_N(a+t) in the Mellin argument. For a>0 there are genuine simple poles at s=1/N-k, k≥0: Res_s=1/N-k Z_N(s;a) =fracsin(π/N)Nπfrac(1-1/N)_kk!a^k. Indeed the large-radius derivative expands as frac1N(a+t)^1/N-1 =frac1N∑_k≥0frac(-1)^k(1-1/N)_kk!…
Exact mathematical passage
\boxed{\operatorname{Res}_{s=1/N-k}\mathcal Z_N(s;a) =\frac{\sin(\pi/N)}{N\pi}\frac{(1-1/N)_k}{k!}a^k.}
a^sUse in source · 3 passages
Search aliases: a^s
Use in source · Read passage · coupled_growth_full.tex, line 113
P(s)=q^s · r^s a^s · b^s.
Exact mathematical passage
P(s)=\begin{pmatrix}q^s&r^s\\a^s&b^s\end{pmatrix}.
A_0Use in source · 6 passages
Search aliases: A_0
Use in source · chapters_sequences.tex, line 89
17n+5)A_n-n^3A_n-1 The prefix annotation begins the Apéry sequence defined by A_n=∑_k=0^nbinom nk^2binomn+kk^2. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at n=2. Using A_1=5 and A_2=73 gives 27A_3=535·73-8·5=39015. Therefore A_3=1445. The recurrence applies for n≥1, with A_0=1. This is a three-term recurrence with polynomial coefficients. Its weights change with n, so it is not a constant-tap filter. OEIS A005259…
Exact mathematical passage
A_0=1
A_1Use in source · 8 passages
Search aliases: A_1
Use in source · chapters_sequences.tex, line 89
lessonSEQUENCES / 13Apéry uses taps that depend on the indexS13_apery_recurrence (n+1)^3A_n+1=(2n+1)(17n^2+17n+5)A_n-n^3A_n-1 The prefix annotation begins the Apéry sequence defined by A_n=∑_k=0^nbinom nk^2binomn+kk^2. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at n=2. Using A_1=5 and A_2=73 gives 27A_3=535·73-8·5=39015. Therefore A_3=1445. The recurrence applies for n≥1, with A_0=1. This is a three-term recurrenc…
Exact mathematical passage
A_1=5
A_{12}Use in source · 1 passage
Search aliases: A_12
Use in source · Read passage · sequence_full.tex, line 416
tightlist item 18 catalog rows through A_12, independently compared with dense formal Picard iteration through degree 25. item Bivariate polynomials through A_8, with 540 nonzero monomials checked for integer positivity, symmetry, parity and the degree bound. item Tangent coefficients independently obtained by division of sine and cosine, plus hyperbolic tangent, scaled hyperbolic sine and linear anchors. item Root-filter multipliers reduced exactly in cyclotomic fields for bases one through sev…
Exact mathematical passage
A_{12}
A_2Use in source · 3 passages
Search aliases: A_2
Use in source · chapters_sequences.tex, line 89
onSEQUENCES / 13Apéry uses taps that depend on the indexS13_apery_recurrence (n+1)^3A_n+1=(2n+1)(17n^2+17n+5)A_n-n^3A_n-1 The prefix annotation begins the Apéry sequence defined by A_n=∑_k=0^nbinom nk^2binomn+kk^2. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at n=2. Using A_1=5 and A_2=73 gives 27A_3=535·73-8·5=39015. Therefore A_3=1445. The recurrence applies for n≥1, with A_0=1. This is a three-term recurrence wi…
Exact mathematical passage
A_2=73
a_{2n}Use in source · 1 passage
Search aliases: a_2n
Use in source · Read passage · sequence_full.tex, line 30
The two coefficient systems are related by A_n=(2n+1)!c_n. In particular, A_n is the derivative of order 2n+1 at zero. The coefficient c_n, the derivative A_n, and the full EGF sequence a_m with a_2n+1=A_n, a_2n=0 are different objects. None of the code silently substitutes one for another.
Exact mathematical passage
a_{2n+1}=A_n, a_{2n}=0
a_{2n+1}Use in source · 1 passage
Search aliases: a_2n+1
Use in source · Read passage · sequence_full.tex, line 30
The two coefficient systems are related by A_n=(2n+1)!c_n. In particular, A_n is the derivative of order 2n+1 at zero. The coefficient c_n, the derivative A_n, and the full EGF sequence a_m with a_2n+1=A_n, a_2n=0 are different objects. None of the code silently substitutes one for another.
Exact mathematical passage
a_{2n+1}=A_n, a_{2n}=0
A_3Use in source · 5 passages
Search aliases: A_3
Use in source · chapters_sequences.tex, line 89
13Apéry uses taps that depend on the indexS13_apery_recurrence (n+1)^3A_n+1=(2n+1)(17n^2+17n+5)A_n-n^3A_n-1 The prefix annotation begins the Apéry sequence defined by A_n=∑_k=0^nbinom nk^2binomn+kk^2. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at n=2. Using A_1=5 and A_2=73 gives 27A_3=535·73-8·5=39015. Therefore A_3=1445. The recurrence applies for n≥1, with A_0=1. This is a three-term recurrence with polynomial …
Exact mathematical passage
27A_3=535\cdot73-8\cdot5=39015
A_4Use in source · 1 passage
Search aliases: A_4
Use in source · Read passage · sequence_full.tex, line 454
reports no failed assertion in this run. Independent audit identified two API input-validation defects: a fractional residue was accepted by the section operator, and an empty phase matrix accepted base zero. Both are now rejected and have passing regression checks. These defects affected invalid inputs, not the displayed identities. A draft prose example initially transcribed A_3(1,2) and A_4(1,2) incorrectly; the exact catalog comparison exposed the mismatch and corrected them to 35200 and 380…
Exact mathematical passage
A_4(1,2)
A_8Use in source · 1 passage
Search aliases: A_8
Use in source · Read passage · sequence_full.tex, line 419
tightlist item 18 catalog rows through A_12, independently compared with dense formal Picard iteration through degree 25. item Bivariate polynomials through A_8, with 540 nonzero monomials checked for integer positivity, symmetry, parity and the degree bound. item Tangent coefficients independently obtained by division of sine and cosine, plus hyperbolic tangent, scaled hyperbolic sine and linear anchors. item Root-filter multipliers reduced exactly in cyclotomic fields for bases one through sev…
Exact mathematical passage
A_8
A_{a+j\beta/2,\beta,Q}Use in source · 1 passage
Search aliases: A_a+jβ/2,β,Q beta beta
Use in source · Read passage · main.tex, line 271
pageheadingFAMILY 2 / CALCULUSClosed shifts, infinite differential complexity The Euler product for Z_Q shows that only rad(Q)=∏_pmid Qp matters. Expansion of the depth factor gives A_a,β,Q(s+1)=A_a,β,Q(s)-2A_a+β/2,β,Q(s) +A_a+β,β,Q(s). The fractional transfer and Newton formulas are A_a,β,Q(s+t)=∑_j≥0(-1)^jbinom2tjA_a+jβ/2,β,Q(s), A_a,β,Q(t)=∑_j≥0(-1)^jbinom2tj[Z_Q(2a+jβ)-1]. They converge absolutely and locally uniformly: binomial coefficients have at most polynomial growth on compact paramete…
Exact mathematical passage
\A(s+t)=\sum_{j\ge0}(-1)^j\binom{2t}{j}A_{a+j\beta/2,\beta,Q}(s),
A_{a+\beta,\beta,Q}Use in source · 1 passage
Search aliases: A_a+β,β,Q beta beta
Use in source · Read passage · main.tex, line 266
pageheadingFAMILY 2 / CALCULUSClosed shifts, infinite differential complexity The Euler product for Z_Q shows that only rad(Q)=∏_pmid Qp matters. Expansion of the depth factor gives A_a,β,Q(s+1)=A_a,β,Q(s)-2A_a+β/2,β,Q(s) +A_a+β,β,Q(s). The fractional transfer and Newton formulas are A_a,β,Q(s+t)=∑_j≥0(-1)^jbinom2tjA_a+jβ/2,β,Q(s), A_a,β,Q(t)=∑_j≥0(-1)^jbinom2tj[Z_Q(2a+jβ)-1]. They converge absolutely and locally uniformly: binomial coefficients have at most polynomial growth on compact paramete…
Exact mathematical passage
\boxed{\A(s+1)=\A(s)-2A_{a+\beta/2,\beta,Q}(s) +A_{a+\beta,\beta,Q}(s).}
A_{a+\beta/2,\beta,Q}Use in source · 1 passage
Search aliases: A_a+β/2,β,Q beta beta
Use in source · Read passage · main.tex, line 266
pageheadingFAMILY 2 / CALCULUSClosed shifts, infinite differential complexity The Euler product for Z_Q shows that only rad(Q)=∏_pmid Qp matters. Expansion of the depth factor gives A_a,β,Q(s+1)=A_a,β,Q(s)-2A_a+β/2,β,Q(s) +A_a+β,β,Q(s). The fractional transfer and Newton formulas are A_a,β,Q(s+t)=∑_j≥0(-1)^jbinom2tjA_a+jβ/2,β,Q(s), A_a,β,Q(t)=∑_j≥0(-1)^jbinom2tj[Z_Q(2a+jβ)-1]. They converge absolutely and locally uniformly: binomial coefficients have at most polynomial growth on compact paramete…
Exact mathematical passage
\boxed{\A(s+1)=\A(s)-2A_{a+\beta/2,\beta,Q}(s) +A_{a+\beta,\beta,Q}(s).}
A_{a,\beta,Q}Use in source · 5 passages
Search aliases: A_a,β,Q beta
Use in source · Read passage · notation.tex, line 30
item[Arithmetic depth A_a,β,Q.] The entire sum ∑_n≥2, (n,Q)=1μ(n)^2n^-2a(1-n^-β)^2s, with a>1/2, β>0, Q≥1 integer and the real logarithm in the complex power. The atom n=1 is excluded at every depth. item[Möbius function and modulus radical.] μ(1)=1. For a product of k distinct primes, μ(n)=(-1)^k. If a prime square divides n, then μ(n)=0. The radical rad(Q)=∏_pmid Qp records the excluded prime set, with rad(1)=1. item[Squarefree Dirichlet sum.] For v>1, Z_Q(v)=ζ(v)ζ(2v)^-1∏_pmid Q(1+p^-v)^-1. T…
Exact mathematical passage
A_{a,\beta,Q}
A_{a,\beta,Q}^{(k)}Use in source · 2 passages
Search aliases: A_a,β,Q^(k) beta
Use in source · Read passage · arithmetic_full.tex, line 318
displaystyle sup_|s|≤ R|A_a,β,Q^(k)(s)-A_a,β,Qp^(k)(s)| ≤ λ_p^k e^Rλ_p fracZ_Q(2a)p^2a+1.
Exact mathematical passage
\boxed{\displaystyle \sup_{|s|\le R}|A_{a,\beta,Q}^{(k)}(s)-A_{a,\beta,Qp}^{(k)}(s)| \le \lambda_p^k e^{R\lambda_p} \frac{Z_Q(2a)}{p^{2a}+1}.}
A_{a,\beta,Qp}^{(k)}Use in source · 2 passages
Search aliases: A_a,β,Qp^(k) beta
Use in source · Read passage · arithmetic_full.tex, line 318
displaystyle sup_|s|≤ R|A_a,β,Q^(k)(s)-A_a,β,Qp^(k)(s)| ≤ λ_p^k e^Rλ_p fracZ_Q(2a)p^2a+1.
Exact mathematical passage
\boxed{\displaystyle \sup_{|s|\le R}|A_{a,\beta,Q}^{(k)}(s)-A_{a,\beta,Qp}^{(k)}(s)| \le \lambda_p^k e^{R\lambda_p} \frac{Z_Q(2a)}{p^{2a}+1}.}
a_{bn+j}Use in source · 5 passages
Search aliases: a_bn+j
Use in source · Read passage · sequence_full.tex, line 286
The full-position section and the decimated lane are distinct arrays. A section keeps the original positions, inserting zeros outside the residue. A lane x_j[n]=a_bn+j changes the index to n.
Exact mathematical passage
x_j[n]=a_{bn+j}
a_{bn+j-t}Use in source · 1 passage
Search aliases: a_bn+j-t
Use in source · Read passage · sequence_full.tex, line 369
Proof. A term h_t a_bn+j-t is in phase l exactly when t+l≡ jpmod b. Write d=(t+l-j)/b. That term is h_t x_l[n-d], whose OGF is h_tw^dX_l(w) by zero extension. Summing over taps and phases gives the matrix. square
Exact mathematical passage
h_t a_{bn+j-t}
a_{bn+l}Use in source · 1 passage
Search aliases: a_bn+l
Use in source · Read passage · sequence_full.tex, line 356
For an OGF sequence a_m, define lanes x_l[n]=a_bn+l and X_l(w)=∑_nx_l[n]w^n. Apply finite ordinary taps and obtain y_j[n]=b_bn+j. Then the exact matrix identity is
Exact mathematical passage
x_l[n]=a_{bn+l}
A_iUse in source · 6 passages
Search aliases: A_i
Use in source · Read passage · sequence_full.tex, line 35
A_0=1, A_n=∑_i=0^n-1binom2n2i+1 q^2i+1r^2(n-1-i)+1A_i A_n-1-i, n≥1.
Exact mathematical passage
\boxed{ A_0=1,\qquad A_n=\sum_{i=0}^{n-1}\binom{2n}{2i+1} q^{2i+1}r^{2(n-1-i)+1}A_i A_{n-1-i},\quad n\ge1.}
A_jUse in source · 2 passages
Search aliases: A_j
Use in source · Read passage · sequence_full.tex, line 43
Proof. Multiply the ordinary Taylor recurrence by (2n+1)!. The factor attached to A_iA_j, where i+j=n-1, is (2n)!/[(2i+1)!(2j+1)!]=binom2n2i+1. This also proves the recurrence directly in the EGF convention. square
Exact mathematical passage
A_iA_j
a_jUse in source · 5 passages
Search aliases: a_j
Use in source · Read passage · main.tex, line 504
1/N and the finite-order zero-counting theorem give absolute local uniform convergence in the stated half-plane. Since the order is below one and d_N(0)=1, the Hadamard factorization has genus zero: d_N(w)=∏_j(1-w/w_j). At positive integers, a log derivative gives -wfracd_N'(w)d_N(w)=∑_k≥1 Z_N(k)w^k. If a_n=(-1)^n/(Nn)! and b_k= Z_N(k), then b_k=-ka_k-∑_j=1^k-1a_jb_k-j. Thus every positive integer value is rational without numerical zero finding.
Exact mathematical passage
b_k=-ka_k-\sum_{j=1}^{k-1}a_jb_{k-j}.
a_kUse in source · 2 passages
Search aliases: a_k
Use in source · Read passage · main.tex, line 504
1/N and the finite-order zero-counting theorem give absolute local uniform convergence in the stated half-plane. Since the order is below one and d_N(0)=1, the Hadamard factorization has genus zero: d_N(w)=∏_j(1-w/w_j). At positive integers, a log derivative gives -wfracd_N'(w)d_N(w)=∑_k≥1 Z_N(k)w^k. If a_n=(-1)^n/(Nn)! and b_k= Z_N(k), then b_k=-ka_k-∑_j=1^k-1a_jb_k-j. Thus every positive integer value is rational without numerical zero finding.
Exact mathematical passage
b_k=-ka_k-\sum_{j=1}^{k-1}a_jb_{k-j}.
A_mUse in source · 7 passages
Search aliases: A_m
Use in source · Read passage · main.tex, line 393
pageheadingFAMILY 3 / BOUNDARYFinite positivity can conceal a signed defect Let ν_nonp and ν_p be the restrictions of the undeformed measure ν_p,0 to indices not divisible by p and divisible by p, respectively. Then ν_p,λ=ν_nonp+(1+λ)ν_p. Let A_m,B_m be the order-m Hankel matrices of these two unscaled positive measures. Both are positive definite. Thus τ_m(λ)=det(A_m+(1+λ)B_m) =det A_m∏_i=1^m[1+(1+λ)η_i], where η_i>0 are the eigenvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and…
Exact mathematical passage
A_m,B_m
a_mUse in source · 9 passages
Search aliases: a_m
Use in source · Read passage · sequence_full.tex, line 30
The two coefficient systems are related by A_n=(2n+1)!c_n. In particular, A_n is the derivative of order 2n+1 at zero. The coefficient c_n, the derivative A_n, and the full EGF sequence a_m with a_2n+1=A_n, a_2n=0 are different objects. None of the code silently substitutes one for another.
Exact mathematical passage
a_m
a_{m+k}Use in source · 1 passage
Search aliases: a_m+k
Use in source · Read passage · sequence_polyphase.tex, line 101
claimThree operations that must be kept distinct For an EGF E_a(z)=∑_ma_mz^m/m!, differentiation transports future coefficients: E_a^(k)=∑_ma_m+kz^m/m!. Causal ordinary taps instead act by repeated zero-constant integration: E_b=∑_ℓ h_ℓ I^ℓ E_a. Multiplication of two EGFs gives Hurwitz convolution, b_m=∑_k=0^mbinom mk a_kh_m-k. It is not ordinary sequence FIR convolution. Intrinsic nonlinear feedback taps alter the defining family and are a fourth operation with their own coefficient recurrence.
Exact mathematical passage
E_a^{(k)}=\sum_ma_{m+k}z^m/m!
a_{m+t}Use in source · 1 passage
Search aliases: a_m+t
Use in source · Read passage · sequence_full.tex, line 328
Proof. The EGF of the delayed sequence is ∑_m≥ ta_m-tz^m/m!=I^tG(z). Summing taps gives the formula. Conversely, G^(t) has coefficient sequence a_m+t, so derivatives sample future coefficients and do not implement these past delays. square
Exact mathematical passage
a_{m+t}
a_{m-t}Use in source · 3 passages
Search aliases: a_m-t
Use in source · Read passage · sequence_full.tex, line 313
b_m=∑_t=0^L-1h_ta_m-t.
Exact mathematical passage
b_m=\sum_{t=0}^{L-1}h_ta_{m-t}.
a_{m-\ell}Use in source · 1 passage
Search aliases: a_m-ℓ ell
Use in source · Read passage · sequence_polyphase.tex, line 75
For causal FIR taps h_ℓ, b_m=∑_ℓ h_ℓ a_m-ℓ with zero extension to negative indices, split A(z) and B(z) into b phases. Then B_i(t)=∑_jH_ij(t)A_j(t), where H_ij(t)=∑_substackℓ≥0
Exact mathematical passage
b_m=\sum_\ell h_\ell a_{m-\ell}
A_m^{-1/2}Use in source · 2 passages
Search aliases: A_m^-1/2
Use in source · Read passage · main.tex, line 400
re ν_p,0 to indices not divisible by p and divisible by p, respectively. Then ν_p,λ=ν_nonp+(1+λ)ν_p. Let A_m,B_m be the order-m Hankel matrices of these two unscaled positive measures. Both are positive definite. Thus τ_m(λ)=det(A_m+(1+λ)B_m) =det A_m∏_i=1^m[1+(1+λ)η_i], where η_i>0 are the eigenvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and below -1; possible coefficient poles occur there, although numerator cancellation is possible. fig05_sieve_boundaryThe line is a proved b…
Exact mathematical passage
A_m^{-1/2}B_mA_m^{-1/2}
A_nUse in source · 27 passages
Search aliases: A_n
Use in source · chapters_sequences.tex, line 88
lessonSEQUENCES / 13Apéry uses taps that depend on the indexS13_apery_recurrence (n+1)^3A_n+1=(2n+1)(17n^2+17n+5)A_n-n^3A_n-1 The prefix annotation begins the Apéry sequence defined by A_n=∑_k=0^nbinom nk^2binomn+kk^2. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at n=2. Using A_1=5 and A_2=73 gives 27A_3=535·73-8·5=39015. Therefore A_3=1445. The recurrence applies for n≥1, with A_0=1. This is a three-term recurrenc…
Exact mathematical passage
A_n=\sum_{k=0}^{n}\binom nk^2\binom{n+k}{k}^2
a_nUse in source · 4 passages
Search aliases: a_n
Use in source · chapters_sequences.tex, line 11
lessonSEQUENCES / 02Longer recurrence memoryS02_tribonacci_tetranacci T_n=T_n-1+T_n-2+T_n-3, Q_n=∑_j=1^4Q_n-j The seed table fixes the OEIS convention. Tribonacci starts 0,0,1; tetranacci starts 0,0,0,1. The curves use log_10(1+a_n) so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence. Tribonacci gives T_5=2+1+1=4. Tetranacci gives Q_7=4+2+1+1=8. The recurrence first applies at n=3 for T and n=4 for Q. These finite-memory recurre…
Exact mathematical passage
\log_{10}(1+a_n)
a_{N,k}Use in source · 1 passage
Search aliases: a_N,k
Use in source · Read passage · spectral_full.tex, line 218
G_N(t)=∑_k≥0a_N,kt^k, g_N(x)=∑_m≥1N a_N,m-1x^Nm-1. tagS12
Exact mathematical passage
G_N(t)=\sum_{k\ge0}a_{N,k}t^k, \qquad g_N(x)=\sum_{m\ge1}N a_{N,m-1}x^{Nm-1}. \tag{S12}
a_{N,m-1}Use in source · 4 passages
Search aliases: a_N,m-1
Use in source · Read passage · spectral_full.tex, line 218
G_N(t)=∑_k≥0a_N,kt^k, g_N(x)=∑_m≥1N a_N,m-1x^Nm-1. tagS12
Exact mathematical passage
G_N(t)=\sum_{k\ge0}a_{N,k}t^k, \qquad g_N(x)=\sum_{m\ge1}N a_{N,m-1}x^{Nm-1}. \tag{S12}
A_{n-1-i}Use in source · 2 passages
Search aliases: A_n-1-i
Use in source · Read passage · sequence_full.tex, line 35
A_0=1, A_n=∑_i=0^n-1binom2n2i+1 q^2i+1r^2(n-1-i)+1A_i A_n-1-i, n≥1.
Exact mathematical passage
\boxed{ A_0=1,\qquad A_n=\sum_{i=0}^{n-1}\binom{2n}{2i+1} q^{2i+1}r^{2(n-1-i)+1}A_i A_{n-1-i},\quad n\ge1.}
A_pUse in source · 2 passages
Search aliases: A_p
Use in source · Read passage · resolution_full.tex, line 225
For a prime p, let D_p=A_1-A_p. This positive difference consists exactly of the squarefree atoms divisible by p. Its largest frequency is
Exact mathematical passage
D_p=A_1-A_p
A_QUse in source · 3 passages
Search aliases: A_Q
Use in source · Read passage · arithmetic_inverse.tex, line 85
und for every derivative. fig11_prime_instabilityThe proved compact-window envelope shrinks as the excluded prime increases. At a=β=2, Q=1, R=5, excluding p=10007 changes the entire window by at most approximately 1.0749×10^-16. The depth-zero difference is known exactly. The radical differs for every such prime, while all bounded-window derivatives converge to those of A_Q. Without a bound on excluded primes, there is no uniformly stable inverse from these windows to the radical parameter.
Exact mathematical passage
A_Q
A_{\rm tail}Use in source · 2 passages
Search aliases: A_rm tail rm
Use in source · Read passage · sieve_full.tex, line 239
A_rm tail=D(4)fracp^4p^4+1 ∏_q≠ p(1+fracλ_qq^4+1) -∑_1≤ n<pμ(n)^2n^-4κ(n)>0,
Exact mathematical passage
A_{\rm tail}=D(4)\frac{p^4}{p^4+1} \prod_{q\ne p}\left(1+\frac{\lambda_q}{q^4+1}\right) -\sum_{1\le n<p}\mu(n)^2n^{-4}\kappa(n)>0,
A_{\chi,\psi}Use in source · 5 passages
Search aliases: A_χ,ψ chi psi
Use in source · Read passage · arithmetic_full.tex, line 415
A_χ,ψ(s)=∑_n≥2μ(n)^2 η(n)n^-2a(1-n^-β)^2s.
Exact mathematical passage
A_{\chi,\psi}(s)=\sum_{n\ge2}\mu(n)^2 \eta(n)n^{-2a}(1-n^{-\beta})^{2s}.
\argUse in source · 7 passages
Search aliases: arg
Use in source · Read passage · main.tex, line 314
Let α=frac2a-1β, δ_Q=frac1ζ(2)∏_pmid Qfracpp+1. As |s|→∞ in any closed sector |arg s|≤π/2-ϵ with 0<ϵ<π/2, A_a,β,Q(s)=fracδ_QβB(α,2s+1) +O_a,β,Q,ϵ(|s|^-(2a-1/2)/β). In particular, A_a,β,Q(s)∼fracδ_QΓ(α)β(2s)^-α. The principal power is used in the right half-plane. The beta expression is an asymptotic continuous-density model, not an identity with the arithmetic sum.
Exact mathematical passage
|\arg s|\le\pi/2-\epsilon
\operatorname{Arg}Use in source · 4 passages
Search aliases: Arg operatorname
Use in source · Read passage · main.tex, line 486
bits pageheadingFAMILY 4One spectral address for each rotation orbit For integer N≥2, define the classical Mittag--Leffler seed~[source reference] d_N(w)=∑_n≥0frac(-1)^nw^n(Nn)!=E_N,1(-w), D_N(z)=d_N(z^N). Let w_j be the nonzero zeros of d_N, counted with multiplicity. Each corresponds to one rotation orbit of N zeros of D_N. Using Argw∈(-π,π], define Z_N(s)=∑_jw_j^-s, Re s>1/N. fig06_orbit_zerosFor N=4, the four real/imaginary zeros in each orbit share one positive coordinate w=z^4. The plot sh…
Exact mathematical passage
\operatorname{Arg}w\in(-\pi,\pi]
\asympUse in source · 1 passage
Search aliases: asymp
Use in source · Read passage · arithmetic_full.tex, line 119
For completeness, the first estimate follows initially from j=δ_Qn_j+O_Q(sqrtn_j), which gives n_jasymp_Qj; substituting this back improves the error to O_Q(sqrt j). The second uses -log(1-u)=u(1+O(u)) as udownarrow0.
Exact mathematical passage
n_j\asymp_Qj
\asymp_QUse in source · 1 passage
Search aliases: asymp_Q asymp
Use in source · Read passage · arithmetic_full.tex, line 119
For completeness, the first estimate follows initially from j=δ_Qn_j+O_Q(sqrtn_j), which gives n_jasymp_Qj; substituting this back improves the error to O_Q(sqrt j). The second uses -log(1-u)=u(1+O(u)) as udownarrow0.
Exact mathematical passage
n_j\asymp_Qj
B^{\mathsf T}Use in source · 1 passage
Search aliases: B^ T mathsf
Use in source · Read passage · coupled_structure_full.tex, line 133
It has rank one, but is not symmetric: B-B^ T=ℓ_1wedgeℓ_2≠0. At four distinct nonzero nodes, polynomial interpolation lets their vector values vary independently, so terms supported at (az,bz) cannot cancel this coefficient. A scalar autonomous channel at the same node pair would require B=κℓ⊗ℓ, a symmetric rank-one tensor, after projecting the output onto that channel. For any output row with a nonzero coefficient on this term, the alternating part remains nonzero. The other node pair gives the…
Exact mathematical passage
B-B^{\mathsf T}=\ell_1\wedge\ell_2\ne0
b^{-1}Use in source · 1 passage
Search aliases: b^-1
Use in source · Read passage · sequence_full.tex, line 276
This holds as a formal series and wherever the original series converges. Indeed the coefficient multiplier is the geometric sum b^-1∑_k=0^b-1ω^k(m-j), equal to one or zero according to the residue. It proves all the projector identities
Exact mathematical passage
b^{-1}\sum_{k=0}^{b-1}\omega^{k(m-j)}
B^2Use in source · 3 passages
Search aliases: B^2
Use in source · Read passage · main.tex, line 148
Suppose such an equality held. Normalization gives AB=1. Write p=qr, u=q^2+r^2, and let ρ be the common order. The first two nontrivial coefficients and the growth law force B^2s^2=p, s^2=u/2, 2s^ρ=q^ρ+r^ρ=1. Consequently 2(fracq^2+r^22)^ρ/2=q^ρ+r^ρ. For ρ≠2, strict convexity or concavity of x^ρ/2 forces q=r. For the exceptional case ρ=2, we have u=1, s^2=1/2, B^2=2p. The two z^7 coefficients become fracp^3(3-p^2)315 and frac11p^31260. Their equality forces p=1/2 and again (q-r)^2=u-2p=0. This c…
Exact mathematical passage
B^2s^2=p,\qquad s^2=u/2,\qquad 2s^\rho=q^\rho+r^\rho=1.
b^2Use in source · 8 passages
Search aliases: b^2
Use in source · REPORT.md, line 39
D=k^2u_2-p^2v_2 =fracpkar15(q^2-b^2)(qb-ar).
Exact mathematical passage
D=k^2u_2-p^2v_2 =\frac{pkar}{15}(q^2-b^2)(qb-ar).
b^{2(n-1-i)+1}Use in source · 2 passages
Search aliases: b^2(n-1-i)+1
Use in source · Read passage · coupled_growth_full.tex, line 36
u_n · =frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1u_i v_n-1-i, v_n · =frac12n+1∑_i=0^n-1 a^2i+1b^2(n-1-i)+1u_i v_n-1-i.
Exact mathematical passage
\begin{aligned} u_n&=\frac1{2n+1}\sum_{i=0}^{n-1} q^{2i+1}r^{2(n-1-i)+1}u_i v_{n-1-i},\\ v_n&=\frac1{2n+1}\sum_{i=0}^{n-1} a^{2i+1}b^{2(n-1-i)+1}u_i v_{n-1-i}. \end{aligned}
b^{2(n-i)+1}Use in source · 1 passage
Search aliases: b^2(n-i)+1
Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
b^{2j+1}Use in source · 1 passage
Search aliases: b^2j+1
Use in source · Read passage · coupled_lab_full.tex, line 25
A_n=∑_i+j=n-1binom2n2i+1q^2i+1r^2j+1A_iB_j, B_n=∑_i+j=n-1binom2n2i+1a^2i+1b^2j+1A_iB_j,
Exact mathematical passage
A_n=\sum_{i+j=n-1}\binom{2n}{2i+1}q^{2i+1}r^{2j+1}A_iB_j, \qquad B_n=\sum_{i+j=n-1}\binom{2n}{2i+1}a^{2i+1}b^{2j+1}A_iB_j,
b^4Use in source · 1 passage
Search aliases: b^4
Use in source · Read passage · coupled_inverse_full.tex, line 20
u_1=frac p3, v_1=frac k3, u_2=frac p15(p q^2+k r^2), v_2=frac k15(p a^2+k b^2), u_3=frac p7(q^4u_2+p^2u_1v_1+r^4v_2), v_3=frac k7(a^4u_2+k^2u_1v_1+b^4v_2).
Exact mathematical passage
u_3=\frac p7(q^4u_2+p^2u_1v_1+r^4v_2),\qquad v_3=\frac k7(a^4u_2+k^2u_1v_1+b^4v_2).
b^sUse in source · 5 passages
Search aliases: b^s
Use in source · Read passage · coupled_growth_full.tex, line 113
P(s)=q^s · r^s a^s · b^s.
Exact mathematical passage
P(s)=\begin{pmatrix}q^s&r^s\\a^s&b^s\end{pmatrix}.
B_+Use in source · 11 passages
Search aliases: B_+
Use in source · Read passage · dilation_full.tex, line 111
Q=max_jmax(q_j,r_j), m=min_jmin(q_j,r_j), Δ=log(1/m), B_+=max_jlogfrac1q_jr_j+log2.
Exact mathematical passage
Q=\max_j\max(q_j,r_j),\quad m=\min_j\min(q_j,r_j),\quad \Delta=\log(1/m),\quad B_+=\max_j\log\frac1{q_jr_j}+\log2.
B_0Use in source · 1 passage
Search aliases: B_0
Use in source · Read passage · coupled_lab_full.tex, line 31
with A_0=B_0=1. Coefficients with even derivative index vanish. Swapping the components sends the parameter tuple (q,r,a,b) to (b,a,r,q). Identical rows (a,b)=(q,r), or exchanged rows (a,b)=(r,q), yield the scalar reduction U=V=T_q,r.
Exact mathematical passage
A_0=B_0=1
B_1Use in source · 1 passage
Search aliases: B_1
Use in source · chapters_sequences.tex, line 74
lessonSEQUENCES / 11Bernoulli data need both coordinatesS11_bernoulli_pair fracte^t-1=∑_n≥0B_nfract^nn!, B_n=fracp_nq_n, q_n>0 The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as 0/1, and this guide uses B_1=-1/2. At n=10, p_10=5 and q_10=66, so B_10=5/66. Its Taylor coefficient is (5/66)/10!. At n=12, B_12=-691/2730. An OEIS integer numerator sequen…
Exact mathematical passage
B_1=-1/2
B_{10}Use in source · 1 passage
Search aliases: B_10
Use in source · chapters_sequences.tex, line 75
inatesS11_bernoulli_pair fracte^t-1=∑_n≥0B_nfract^nn!, B_n=fracp_nq_n, q_n>0 The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as 0/1, and this guide uses B_1=-1/2. At n=10, p_10=5 and q_10=66, so B_10=5/66. Its Taylor coefficient is (5/66)/10!. At n=12, B_12=-691/2730. An OEIS integer numerator sequence does not by itself specify the rational family …
Exact mathematical passage
B_{10}=5/66
B_{12}Use in source · 1 passage
Search aliases: B_12
Use in source · chapters_sequences.tex, line 75
t^nn!, B_n=fracp_nq_n, q_n>0 The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as 0/1, and this guide uses B_1=-1/2. At n=10, p_10=5 and q_10=66, so B_10=5/66. Its Taylor coefficient is (5/66)/10!. At n=12, B_12=-691/2730. An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient. OEIS A02…
Exact mathematical passage
B_{12}=-691/2730
b_2Use in source · 2 passages
Search aliases: b_2
Use in source · Read passage · sequence_full.tex, line 341
This is a binomial convolution. For example, with a_m=1 and taps h=(1,1), ordinary taps give b_2=2, while the Hurwitz product gives b_2=3. This explicit negative control catches a common normalization error.
Exact mathematical passage
b_2=2
B_4Use in source · 1 passage
Search aliases: B_4
Use in source · chapters_sequences.tex, line 33
lessonSEQUENCES / 05Normalize central binomial coefficientsS05_binomial_catalan B_n=binom2nn, C_n=fracB_nn+1 Each table column shows one central binomial coefficient and its Catalan normalization. The curves extend the comparison through index fourteen on a base-ten logarithmic axis. Division by n+1 is exact here and gives an integer at every index. For n=4, B_4=binom84=70 and C_4=70/5=14. For n=5, 252/6=42. The divisor changes with the index. Normalization is an operator on coefficients. It nee…
Exact mathematical passage
B_4=\binom84=70
B_5Use in source · 1 passage
Search aliases: B_5
Use in source · chapters_sequences.tex, line 47
row sumsS07_bell_stirling B_n=∑_k=0^nS(n,k), ∑_n≥0B_nfract^nn!=exp(e^t-1) Each matrix row contains Stirling numbers of the second kind, counting partitions into k nonempty blocks. Summing across the row gives the Bell number at its right. The generating function is exponential, so its coefficient is B_n/n!. The fifth row contains 0,1,15,25,10,1. Its sum is 52=B_5. Equivalently, apply the Stirling transform to the constant input sequence 1,1,1,…. A triangular transform uses all relevant input pos…
Exact mathematical passage
52=B_5
b_{bn+j}Use in source · 1 passage
Search aliases: b_bn+j
Use in source · Read passage · sequence_full.tex, line 358
For an OGF sequence a_m, define lanes x_l[n]=a_bn+l and X_l(w)=∑_nx_l[n]w^n. Apply finite ordinary taps and obtain y_j[n]=b_bn+j. Then the exact matrix identity is
Exact mathematical passage
y_j[n]=b_{bn+j}
B_iUse in source · 1 passage
Search aliases: B_i
Use in source · Read passage · sequence_polyphase.tex, line 77
For causal FIR taps h_ℓ, b_m=∑_ℓ h_ℓ a_m-ℓ with zero extension to negative indices, split A(z) and B(z) into b phases. Then B_i(t)=∑_jH_ij(t)A_j(t), where H_ij(t)=∑_substackℓ≥0
Exact mathematical passage
B_i(t)=\sum_jH_{ij}(t)A_j(t)
b_iUse in source · 3 passages
Search aliases: b_i
Use in source · Read passage · resolution_full.tex, line 399
ℓ_i(x)=∏_j≠ ifracx-y_jy_i-y_j, b_i=∏_j≠ ifracmax(y_j,1-y_j)|y_i-y_j|, K_d=∑_i=0^dn_i^4b_i^2.
Exact mathematical passage
\ell_i(x)=\prod_{j\ne i}\frac{x-y_j}{y_i-y_j}, \quad b_i=\prod_{j\ne i}\frac{\max(y_j,1-y_j)}{|y_i-y_j|}, \quad K_d=\sum_{i=0}^{d}n_i^4b_i^2.
b_i^2Use in source · 2 passages
Search aliases: b_i^2
Use in source · Read passage · resolution_full.tex, line 399
ℓ_i(x)=∏_j≠ ifracx-y_jy_i-y_j, b_i=∏_j≠ ifracmax(y_j,1-y_j)|y_i-y_j|, K_d=∑_i=0^dn_i^4b_i^2.
Exact mathematical passage
\ell_i(x)=\prod_{j\ne i}\frac{x-y_j}{y_i-y_j}, \quad b_i=\prod_{j\ne i}\frac{\max(y_j,1-y_j)}{|y_i-y_j|}, \quad K_d=\sum_{i=0}^{d}n_i^4b_i^2.
B_jUse in source · 2 passages
Search aliases: B_j
Use in source · Read passage · coupled_lab_full.tex, line 25
A_n=∑_i+j=n-1binom2n2i+1q^2i+1r^2j+1A_iB_j, B_n=∑_i+j=n-1binom2n2i+1a^2i+1b^2j+1A_iB_j,
Exact mathematical passage
A_n=\sum_{i+j=n-1}\binom{2n}{2i+1}q^{2i+1}r^{2j+1}A_iB_j, \qquad B_n=\sum_{i+j=n-1}\binom{2n}{2i+1}a^{2i+1}b^{2j+1}A_iB_j,
b_jUse in source · 4 passages
Search aliases: b_j
Use in source · Read passage · dilation_full.tex, line 143
Let a_j=log(1/q_j), b_j=log(1/r_j), and G(t)=log M(e^t)+t+B_+. For t≥1+Δ, all child moduli are at least one, so the positive integral and weight normalization give
Exact mathematical passage
b_j=\log(1/r_j)
b_kUse in source · 2 passages
Search aliases: b_k
Use in source · Read passage · main.tex, line 503
he bound |d_N(w)|≤ e^|w|^1/N and the finite-order zero-counting theorem give absolute local uniform convergence in the stated half-plane. Since the order is below one and d_N(0)=1, the Hadamard factorization has genus zero: d_N(w)=∏_j(1-w/w_j). At positive integers, a log derivative gives -wfracd_N'(w)d_N(w)=∑_k≥1 Z_N(k)w^k. If a_n=(-1)^n/(Nn)! and b_k= Z_N(k), then b_k=-ka_k-∑_j=1^k-1a_jb_k-j. Thus every positive integer value is rational without numerical zero finding.
Exact mathematical passage
b_k=\mathcal Z_N(k)
b_{k-j}Use in source · 1 passage
Search aliases: b_k-j
Use in source · Read passage · main.tex, line 504
1/N and the finite-order zero-counting theorem give absolute local uniform convergence in the stated half-plane. Since the order is below one and d_N(0)=1, the Hadamard factorization has genus zero: d_N(w)=∏_j(1-w/w_j). At positive integers, a log derivative gives -wfracd_N'(w)d_N(w)=∑_k≥1 Z_N(k)w^k. If a_n=(-1)^n/(Nn)! and b_k= Z_N(k), then b_k=-ka_k-∑_j=1^k-1a_jb_k-j. Thus every positive integer value is rational without numerical zero finding.
Exact mathematical passage
b_k=-ka_k-\sum_{j=1}^{k-1}a_jb_{k-j}.
B_mUse in source · 9 passages
Search aliases: B_m
Use in source · Read passage · main.tex, line 393
pageheadingFAMILY 3 / BOUNDARYFinite positivity can conceal a signed defect Let ν_nonp and ν_p be the restrictions of the undeformed measure ν_p,0 to indices not divisible by p and divisible by p, respectively. Then ν_p,λ=ν_nonp+(1+λ)ν_p. Let A_m,B_m be the order-m Hankel matrices of these two unscaled positive measures. Both are positive definite. Thus τ_m(λ)=det(A_m+(1+λ)B_m) =det A_m∏_i=1^m[1+(1+λ)η_i], where η_i>0 are the eigenvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and…
Exact mathematical passage
A_m,B_m
b_mUse in source · 4 passages
Search aliases: b_m
Use in source · Read passage · sequence_full.tex, line 313
b_m=∑_t=0^L-1h_ta_m-t.
Exact mathematical passage
b_m=\sum_{t=0}^{L-1}h_ta_{m-t}.
b_{m-j}Use in source · 1 passage
Search aliases: b_m-j
Use in source · Read passage · spectral_full.tex, line 295
b_m:= Z_N(m) =-m c_m-∑_j=1^m-1c_jb_m-j. tagS17
Exact mathematical passage
b_m:=\mathcal Z_N(m) =-m c_m-\sum_{j=1}^{m-1}c_jb_{m-j}. \tag{S17}
B_nUse in source · 11 passages
Search aliases: B_n
Use in source · chapters_sequences.tex, line 46
lessonSEQUENCES / 07Bell numbers are Stirling row sumsS07_bell_stirling B_n=∑_k=0^nS(n,k), ∑_n≥0B_nfract^nn!=exp(e^t-1) Each matrix row contains Stirling numbers of the second kind, counting partitions into k nonempty blocks. Summing across the row gives the Bell number at its right. The generating function is exponential, so its coefficient is B_n/n!. The fifth row contains 0,1,15,25,10,1. Its sum is 52=B_5. Equivalently, apply the Stirling transform to the constant input sequence 1,1,1,…. A tr…
Exact mathematical passage
B_n/n!
B_{N,0}Use in source · 2 passages
Search aliases: B_N,0
Use in source · Read passage · spectral_full.tex, line 275
B_N,0(s)=∫_0^1x^-Nsg_N(x) dx +∫_1^∞ x^-Ns(g_N(x)-1) dx+frac1Ns-1. tagS15
Exact mathematical passage
B_{N,0}(s)=\int_0^1x^{-Ns}g_N(x)\,dx +\int_1^\infty x^{-Ns}(g_N(x)-1)\,dx+\frac1{Ns-1}. \tag{S15}
B_{N,a}Use in source · 2 passages
Search aliases: B_N,a
Use in source · Read passage · spectral_full.tex, line 464
B_N,a(0)=∫_0^1G_N,a(t)dt +∫_1^∞[G_N,a(t)-α t^α-1]dt-1.
Exact mathematical passage
B_{N,a}(0)=\int_0^1G_{N,a}(t)dt +\int_1^\infty[G_{N,a}(t)-\alpha t^{\alpha-1}]dt-1.
B_{N,K}Use in source · 2 passages
Search aliases: B_N,K
Use in source · Read passage · spectral_full.tex, line 226
B_N,K(s)= · ∫_0^1x^-Ns[g_N(x)-∑_m=1^KN a_N,m-1x^Nm-1]dx · +∑_m=1^Kfraca_N,m-1m-s +∫_1^∞ x^-Ns[g_N(x)-1]dx +frac1Ns-1. tagS13
Exact mathematical passage
\begin{aligned} B_{N,K}(s)={}& \int_0^1x^{-Ns}\left[g_N(x)-\sum_{m=1}^{K}N a_{N,m-1}x^{Nm-1}\right]dx\\ &+\sum_{m=1}^{K}\frac{a_{N,m-1}}{m-s} +\int_1^\infty x^{-Ns}\bigl[g_N(x)-1\bigr]dx +\frac1{Ns-1}. \end{aligned} \tag{S13}
B_{n-1-i}Use in source · 1 passage
Search aliases: B_n-1-i
Use in source · Read passage · coupled_growth_full.tex, line 94
A_n · =∑_i=0^n-1binom2n2i+1 q^2i+1r^2(n-1-i)+1A_iB_n-1-i, B_n · =∑_i=0^n-1binom2n2i+1 a^2i+1b^2(n-1-i)+1A_iB_n-1-i.
Exact mathematical passage
\begin{aligned} A_n&=\sum_{i=0}^{n-1}\binom{2n}{2i+1} q^{2i+1}r^{2(n-1-i)+1}A_iB_{n-1-i},\\ B_n&=\sum_{i=0}^{n-1}\binom{2n}{2i+1} a^{2i+1}b^{2(n-1-i)+1}A_iB_{n-1-i}. \end{aligned}
B_{\rm EGF}Use in source · 1 passage
Search aliases: B_rm EGF rm
Use in source · Read passage · sequence_full.tex, line 321
B_rm EGF(z)=∑_t=0^L-1h_t I^tG(z), IG(z)=∫_0^zG(w) dw.
Exact mathematical passage
\boxed{B_{\rm EGF}(z)=\sum_{t=0}^{L-1}h_t I^tG(z),\qquad IG(z)=\int_0^zG(w)\,dw.}
CUse in source · 65 passages
Search aliases: C
Use in source · chapters_sequences.tex, line 33
S / 05Normalize central binomial coefficientsS05_binomial_catalan B_n=binom2nn, C_n=fracB_nn+1 Each table column shows one central binomial coefficient and its Catalan normalization. The curves extend the comparison through index fourteen on a base-ten logarithmic axis. Division by n+1 is exact here and gives an integer at every index. For n=4, B_4=binom84=70 and C_4=70/5=14. For n=5, 252/6=42. The divisor changes with the index. Normalization is an operator on coefficients. It need not be a fix…
Exact mathematical passage
C_4=70/5=14
\mathbb CUse in source · 5 passages
Search aliases: C mathbb
Use in source · Read passage · main.tex, line 230
MonographPartIIArithmetic depth and signed measuresComplex parameters, positive kernels and the boundaries of positivity. Arithmetic at complex depth. pageheadingFAMILY 2Arithmetic energy at complex depth For real a>1/2, β>0, and integer Q≥1, define A_a,β,Q(s)=∑_substackn≥2 (n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At d…
Exact mathematical passage
\boxed{\A(s)=\sum_{\substack{n\ge2\\(n,Q)=1}} \mu(n)^2n^{-2a}(1-n^{-\beta})^{2s},\qquad s\in\mathbb C.}
cUse in source · 99 passages
Search aliases: c
Use in source · REPORT.md, line 28
|∑_n≥ Nc_nz^2n+1| ≤ frac|z|ρ^N1-ρ, |z|^2<2.
Exact mathematical passage
\left|\sum_{n\ge N}c_nz^{2n+1}\right| \le \frac{|z|\rho^N}{1-\rho},\qquad |z|^2<2.
C^2Use in source · 1 passage
Search aliases: C^2
Use in source · Read passage · resolution_visual.tex, line 172
claimThe certificate uses rational arithmetic throughout Take S={1,3,5,7} and define P(x)=∏_n∈ Sfracx-x_nx_2-x_n (frac1-x1-x_2)^3. Its degree is seven. All remaining positive nodes begin at n=11. With C=756482048/20777538825 and r=3856/102487, a complete tail bound gives Q_10^-12(P)≤-frac10^-1216 +fracC^2r^63·7^3 <-5.8845708899×10^-14. The inequality is certified as a comparison of exact rational numbers. The first four zeros remove the dominant positive contributions. The factor (1-x)^3 suppres…
Exact mathematical passage
Q_{10^{-12}}(P)\le-\frac{10^{-12}}{16} +\frac{C^2r^6}{3\cdot7^3} <-5.8845708899\times10^{-14}.
\mathbb C^2Use in source · 1 passage
Search aliases: C^2 mathbb
Use in source · Read passage · coupled_structure_full.tex, line 19
A nonzero constant row c=(c_1,c_2)∈ C^2 is an autonomous scalar projection if some scalar functional H satisfies
Exact mathematical passage
c=(c_1,c_2)\in\mathbb C^2
c_+Use in source · 12 passages
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Use in source · Read passage · dilation_full.tex, line 28
Theorem A.1. There are explicit constants 0<c_-≤ c_+<∞ such that
Exact mathematical passage
0<c_-\le c_+<\infty
c_-Use in source · 16 passages
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Use in source · Read passage · dilation_full.tex, line 28
Theorem A.1. There are explicit constants 0<c_-≤ c_+<∞ such that
Exact mathematical passage
0<c_-\le c_+<\infty
C_0Use in source · 9 passages
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Use in source · Read passage · resolution_full.tex, line 154
log F_0(R)≤ C_0sqrt R (R≥ e), C_0=e^-1/2(1+22log2)≈9.85566103.
Exact mathematical passage
\log F_0(R)\le C_0\sqrt R\qquad(R\ge e), \qquad C_0=e^{-1/2}(1+22\log2)\approx9.85566103.
c_0Use in source · 1 passage
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Use in source · Read passage · main.tex, line 89
of the defining feedback. The function is symmetric in q,r, but unequal scales retain left/right address information. fig01_asymmetric_functionsComputed Taylor curves. Each entire curve uses terms through z^321; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent. For T_q,r(z)=∑_n≥0c_nz^2n+1, c_0=1, coefficient matching gives c_n=frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1c_ic_n-1-i, n≥1. In particular, T_q,r(z)=z+fracqr3z…
Exact mathematical passage
c_0=1
c_1Use in source · 14 passages
Search aliases: c_1
Use in source · Read passage · main.tex, line 103
The parameters can be recovered, up to swapping, from c_1,c_2. Put p=3c_1 and u=15c_2/p^2. Then q,r are the roots of x^2-sqrtu+2p x+p=0.
Exact mathematical passage
c_1,c_2
c_2Use in source · 11 passages
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Use in source · Read passage · main.tex, line 103
The parameters can be recovered, up to swapping, from c_1,c_2. Put p=3c_1 and u=15c_2/p^2. Then q,r are the roots of x^2-sqrtu+2p x+p=0.
Exact mathematical passage
c_1,c_2
C_4Use in source · 1 passage
Search aliases: C_4
Use in source · chapters_sequences.tex, line 33
S / 05Normalize central binomial coefficientsS05_binomial_catalan B_n=binom2nn, C_n=fracB_nn+1 Each table column shows one central binomial coefficient and its Catalan normalization. The curves extend the comparison through index fourteen on a base-ten logarithmic axis. Division by n+1 is exact here and gives an integer at every index. For n=4, B_4=binom84=70 and C_4=70/5=14. For n=5, 252/6=42. The divisor changes with the index. Normalization is an operator on coefficients. It need not be a fix…
Exact mathematical passage
C_4=70/5=14
C_5Use in source · 2 passages
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Use in source · chapters_sequences.tex, line 39
lessonSEQUENCES / 06Catalan convolution rebuilds the next termS06_catalan_convolution C_n+1=∑_k=0^nC_kC_n-k, C(z)=1+zC(z)^2 The five bars are the five products contributing to C_5. The input indices add to four, and their sum supplies the next coefficient. Here C(z)=∑_n≥0C_nz^n is an ordinary generating function. The contributions are 1·14, 1·5, 2·2, 5·1, and 14·1. Their sum is 14+5+4+5+14=42=C_5. A nonlinear generating-function equation becomes an exact convolution identity among coefficients. …
Exact mathematical passage
C_5
C_aUse in source · 4 passages
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Use in source · Read passage · multi_prime.tex, line 68
pageheadingPRIME SIEVES / CONSTRUCTIVE WITNESSEvery negative prime atom has a finite detector For an integer a≥2, let L_a(x)=∏_n=1^a-1fracx-x_nx_a-x_n, C_a=L_a(1), r_a=frac1-x_a+11-x_a<1, f_a,m(x)=L_a(x)(frac1-x1-x_a)^m. This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects …
Exact mathematical passage
L_a(x)=\prod_{n=1}^{a-1}\frac{x-x_n}{x_a-x_n},\quad C_a=L_a(1),\quad r_a=\frac{1-x_{a+1}}{1-x_a}<1,
C_FUse in source · 9 passages
Search aliases: C_F
Use in source · Read passage · multi_prime.tex, line 80
x_a<1, f_a,m(x)=L_a(x)(frac1-x1-x_a)^m. This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects the signed defect. Set C_F=∏_q∈ Fmax(1,|1+λ_q|). The summable tail gives ∫ f_p,m^2ddν ≤frac1+λ_pp^4+ fracC_FC_p^23p^3r_p^2m. Hence the quadratic form is negative whenever pC_FC_p^2r…
Exact mathematical passage
C_F=\prod_{q\in F}\max(1,|1+\lambda_q|)
c_iUse in source · 1 passage
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Use in source · Read passage · main.tex, line 90
on is symmetric in q,r, but unequal scales retain left/right address information. fig01_asymmetric_functionsComputed Taylor curves. Each entire curve uses terms through z^321; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent. For T_q,r(z)=∑_n≥0c_nz^2n+1, c_0=1, coefficient matching gives c_n=frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1c_ic_n-1-i, n≥1. In particular, T_q,r(z)=z+fracqr3z^3+ fracq^2r^2(q^2+r^2)15z^5+O(z^7).…
Exact mathematical passage
c_n=\frac{1}{2n+1}\sum_{i=0}^{n-1} q^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\qquad n\ge1.
c_jUse in source · 4 passages
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Use in source · Read passage · arithmetic_full.tex, line 364
For c=(c_j)∈ℓ^2, define
Exact mathematical passage
c=(c_j)\in\ell^2
c_kUse in source · 1 passage
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Use in source · Read passage · arithmetic_full.tex, line 383
M<∞, so the same spectral peeling argument, now for complex coefficients, forces the first coefficient and then every coefficient to vanish whenever Uc=0. Positivity was needed to recover unknown frequencies in AI-1; injectivity here only needs absolute summability and the known ordered support. Specifically, after dividing by the largest remaining exponential, the tail is bounded by e^-Δ_jt∑_k>j|c_k|sqrtw_k, which tends to zero.
Exact mathematical passage
e^{-\Delta_jt}\sum_{k>j}|c_k|\sqrt{w_k}
c_mUse in source · 1 passage
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Use in source · Read passage · spectral_full.tex, line 295
b_m:= Z_N(m) =-m c_m-∑_j=1^m-1c_jb_m-j. tagS17
Exact mathematical passage
b_m:=\mathcal Z_N(m) =-m c_m-\sum_{j=1}^{m-1}c_jb_{m-j}. \tag{S17}
c_{mj}Use in source · 1 passage
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Use in source · Read passage · sieve_full.tex, line 320
If at least two strengths in a positive multiplicative family are nonzero, its entire depth-zero monic polynomial sequence cannot equal any one-prime sequence on the same variable and node system. A direct proof avoids importing a determinacy theorem. Write P_m=x^m+∑_j<mc_mjx^j. The equations ∫ P_m dν=0 recursively determine every normalized moment from the monic sequence. Equal sequences therefore give equal normalized Stieltjes series for |z|>1. Analytic continuation on the connected complemen…
Exact mathematical passage
P_m=x^m+\sum_{j<m}c_{mj}x^j
C_nUse in source · 1 passage
Search aliases: C_n
Use in source · chapters_sequences.tex, line 39
lessonSEQUENCES / 06Catalan convolution rebuilds the next termS06_catalan_convolution C_n+1=∑_k=0^nC_kC_n-k, C(z)=1+zC(z)^2 The five bars are the five products contributing to C_5. The input indices add to four, and their sum supplies the next coefficient. Here C(z)=∑_n≥0C_nz^n is an ordinary generating function. The contributions are 1·14, 1·5, 2·2, 5·1, and 14·1. Their sum is 14+5+4+5+14=42=C_5. A nonlinear generating-function equation becomes an exact convolution identity among coefficients. …
Exact mathematical passage
C(z)=\sum_{n\ge0}C_nz^n
c_NUse in source · 5 passages
Search aliases: c_N
Use in source · Read passage · spectral_continuation.tex, line 44
sector around negative w, choose the root x=(-w)^1/N near the positive real ray. The first exponential dominates the other N-1 terms by a uniform real-part gap at large radius, so this sector is eventually zero free. The finitely many remaining zeros avoid the negative real ray because d_N(-t)>0 for t≥0. Thus all zero arguments have a uniform gap from the branch cut. In particular |t+w_j|≥ c_N(t+|w_j|) for t≥0.
Exact mathematical passage
|t+w_j|\ge c_N(t+|w_j|)
c_nUse in source · 21 passages
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Use in source · REPORT.md, line 28
|∑_n≥ Nc_nz^2n+1| ≤ frac|z|ρ^N1-ρ, |z|^2<2.
Exact mathematical passage
\left|\sum_{n\ge N}c_nz^{2n+1}\right| \le \frac{|z|\rho^N}{1-\rho},\qquad |z|^2<2.
c_{n-1}Use in source · 1 passage
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Use in source · Read passage · sequence_full.tex, line 129
c_n≥fracq r^2n-12n+1c_n-1 Longrightarrow c_n≥fracq^n r^n^23·5⋯(2n+1).
Exact mathematical passage
c_n\ge\frac{q r^{2n-1}}{2n+1}c_{n-1} \quad\Longrightarrow\quad c_n\ge\frac{q^n r^{n^2}}{3\cdot5\cdots(2n+1)}.
c_{n-1-i}Use in source · 1 passage
Search aliases: c_n-1-i
Use in source · Read passage · main.tex, line 90
on is symmetric in q,r, but unequal scales retain left/right address information. fig01_asymmetric_functionsComputed Taylor curves. Each entire curve uses terms through z^321; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent. For T_q,r(z)=∑_n≥0c_nz^2n+1, c_0=1, coefficient matching gives c_n=frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1c_ic_n-1-i, n≥1. In particular, T_q,r(z)=z+fracqr3z^3+ fracq^2r^2(q^2+r^2)15z^5+O(z^7).…
Exact mathematical passage
c_n=\frac{1}{2n+1}\sum_{i=0}^{n-1} q^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\qquad n\ge1.
c_n^{1/(2n+1)}Use in source · 1 passage
Search aliases: c_n^1/(2n+1)
Use in source · Read passage · sequence_full.tex, line 137
c_n^1/(2n+1)≥ q^n/(2n+1)r^n^2/(2n+1)(2n+1)^-n/(2n+1)⟶∞.
Exact mathematical passage
c_n^{1/(2n+1)}\ge q^{n/(2n+1)}r^{n^2/(2n+1)}(2n+1)^{-n/(2n+1)}\longrightarrow\infty.
C_p^2Use in source · 9 passages
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Use in source · Read passage · multi_prime.tex, line 81
This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects the signed defect. Set C_F=∏_q∈ Fmax(1,|1+λ_q|). The summable tail gives ∫ f_p,m^2ddν ≤frac1+λ_pp^4+ fracC_FC_p^23p^3r_p^2m. Hence the quadratic form is negative whenever pC_FC_p^2r_p^2m/3<-(1+λ_p). For rational strengths…
Exact mathematical passage
\int f_{p,m}^2\dd\nu \le\frac{1+\lambda_p}{p^4}+ \frac{C_FC_p^2}{3p^3}r_p^{2m}.
C_QUse in source · 4 passages
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Use in source · Read passage · main.tex, line 335
(n,Q)=1μ(n)^2 =δ_Q x+O_Q(sqrtx). Integrate against g_s(x)=x^-2a(1-x^-β)^2s on [1,∞). The density term becomes the beta integral after y=x^-β. Partial integration bounds the error by C_Q∫_1^∞sqrtx|g_s'(x)| d x. The two derivative terms are beta integrals with decay exponent (2a-1/2)/β; |s|/Re s stays bounded in the stated sector. This gives the error term and then the leading asymptotic.
Exact mathematical passage
C_Q\int_1^\infty\sqrt{x}|g_s'(x)|\dd x
C_Q^2Use in source · 1 passage
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Use in source · Read passage · arithmetic_full.tex, line 206
(fracsqrtC_Q^2+4δ_Q(j-1)-C_Q2δ_Q)^2 ≤ n_j≤ (fracsqrtC_Q^2+4δ_Q(j+1)+C_Q2δ_Q)^2.
Exact mathematical passage
\left(\frac{\sqrt{C_Q^2+4\delta_Q(j-1)}-C_Q}{2\delta_Q}\right)^2 \le n_j\le \left(\frac{\sqrt{C_Q^2+4\delta_Q(j+1)}+C_Q}{2\delta_Q}\right)^2.
C_SUse in source · 3 passages
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Use in source · Read passage · resolution_full.tex, line 455
P_S,m(x)=∏_n∈ Sfracx-x_nx_2-x_n (frac1-x1-x_2)^m, C_S=∏_n∈ Sfrac1-x_n|x_2-x_n|, r_S=frac1-x_n_+1-x_2<1.
Exact mathematical passage
C_S=\prod_{n\in S}\frac{1-x_n}{|x_2-x_n|}, \qquad r_S=\frac{1-x_{n_+}}{1-x_2}<1.
C_S^2Use in source · 2 passages
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Use in source · Read passage · resolution_full.tex, line 467
Q_δ(P_S,m)≤-fracδ16 +fracC_S^23K^3r_S^2m. tagRA-10
Exact mathematical passage
\boxed{Q_\delta(P_{S,m})\le-\frac\delta{16} +\frac{C_S^2}{3K^3}r_S^{2m}.} \tag{RA-10}
c_\varepsilonUse in source · 8 passages
Search aliases: c_ε varepsilon
Use in source · Read passage · resolution_full.tex, line 144
log F_ε(R)≥ c_ε(R+3/ε)^ρ ≥ c_ε R^ρ (R≥2), c_ε=fraclog2(4/ε+8/3)^ρ.
Exact mathematical passage
\log F_\varepsilon(R)\ge c_\varepsilon(R+3/\varepsilon)^\rho \ge c_\varepsilon R^\rho\qquad(R\ge2),
C_\kappaUse in source · 5 passages
Search aliases: C_κ kappa
Use in source · Read passage · multi_prime.tex, line 106
deg f<m frac∫ f^2ddν_p-free ∫ f^2ddν_p-divisible, δ_mdownarrow0. The detector yields, for C_κ=∏_q≠ pmax(1,1+λ_q), δ_p+m≤fracpC_κ C_p^23r_p^2m. The denominator retains the atom p; all lower nodes in the numerator are killed. This extends the prime-two bound to every prime and to simultaneous positive or deletion strengths. It does not assert simple roots or global multivariate real stability.
Exact mathematical passage
C_\kappa=\prod_{q\ne p}\max(1,1+\lambda_q)
\cosUse in source · 7 passages
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Use in source · Read passage · main.tex, line 527
claimTwo exact Hurwitz-zeta reductions For N=2, d_2(w)=cossqrt w, whose zeros are w_j=π^2(j+1/2)^2, j≥0. Therefore Z_2(s)=π^-2sζ(2s,1/2). For N=4, D_4(z)=cos(z/sqrt2)cosh(z/sqrt2), and the rotation quotient has one zero w_j=(sqrt2π)^4(j+1/2)^4 per orbit. Hence Z_4(s)=(sqrt2π)^-4sζ(4s,1/2). These anchors are classical reductions. The general spectral-zeta mechanism also has substantial antecedents, including hypergeometric zero zeta functions.
Exact mathematical passage
d_2(w)=\cos\sqrt w
\cos^2Use in source · 1 passage
Search aliases: cos^2 cos
Use in source · Read passage · spectral_full.tex, line 152
To check the elementary estimate, write w=re^iθ: |t+w|^2=(t+r)^2-4trsin^2(θ/2)≥(t+r)^2cos^2(θ/2).
Exact mathematical passage
|t+w|^2=(t+r)^2-4tr\sin^2(\theta/2)\ge(t+r)^2\cos^2(\theta/2)
\coshUse in source · 5 passages
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Use in source · Read passage · main.tex, line 533
claimTwo exact Hurwitz-zeta reductions For N=2, d_2(w)=cossqrt w, whose zeros are w_j=π^2(j+1/2)^2, j≥0. Therefore Z_2(s)=π^-2sζ(2s,1/2). For N=4, D_4(z)=cos(z/sqrt2)cosh(z/sqrt2), and the rotation quotient has one zero w_j=(sqrt2π)^4(j+1/2)^4 per orbit. Hence Z_4(s)=(sqrt2π)^-4sζ(4s,1/2). These anchors are classical reductions. The general spectral-zeta mechanism also has substantial antecedents, including hypergeometric zero zeta functions.
Exact mathematical passage
D_4(z)=\cos(z/\sqrt2)\cosh(z/\sqrt2),
d'Use in source · 1 passage
Search aliases: d'
Use in source · Read passage · coupled_structure_full.tex, line 196
w'=2+tfrac12(w(sz)^2-d(sz)^2), d'=0.
Exact mathematical passage
w'=2+\tfrac12\bigl(w(sz)^2-d(sz)^2\bigr),\qquad d'=0.
d^{-2}Use in source · 1 passage
Search aliases: d^-2
Use in source · Read passage · arithmetic_full.tex, line 201
2=∑_d^2mid nμ(d). For (d,Q)=1, the number of m≤ x/d^2 coprime to Q differs from (φ(Q)/Q)x/d^2 by at most 2^ω(Q). Summing over at most sqrt x values of d gives the first part of C_Q. The omitted main-term tail is at most (φ(Q)/Q)x/lfloorsqrt xrfloor≤2(φ(Q)/Q)sqrt x. Removing n=1 adds one. The main-term density is (φ(Q)/Q)∑_(d,Q)=1μ(d)d^-2=δ_Q by its absolutely convergent Euler product.
Exact mathematical passage
(\varphi(Q)/Q)\sum_{(d,Q)=1}\mu(d)d^{-2}=\delta_Q
d^2Use in source · 4 passages
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Use in source · Read passage · arithmetic_full.tex, line 194
To verify the constants, expand μ(n)^2=∑_d^2mid nμ(d). For (d,Q)=1, the number of m≤ x/d^2 coprime to Q differs from (φ(Q)/Q)x/d^2 by at most 2^ω(Q). Summing over at most sqrt x values of d gives the first part of C_Q. The omitted main-term tail is at most (φ(Q)/Q)x/lfloorsqrt xrfloor≤2(φ(Q)/Q)sqrt x. Removing n=1 adds one. The main-term density is (φ(Q)/Q)∑_(d,Q)=1μ(d)d^-2=δ_Q by its absolutely convergent Euler product.
Exact mathematical passage
\mu(n)^2=\sum_{d^2\mid n}\mu(d)
D^{n(n+1)}Use in source · 4 passages
Search aliases: D^n(n+1)
Use in source · Read passage · sequence_full.tex, line 105
B_n=H^nD^n(n+1)A_n∈ Z hboxfor all n.
Exact mathematical passage
\boxed{B_n=H^nD^{n(n+1)}A_n\in\mathbb Z\quad\hbox{for all }n.}
D_0Use in source · 2 passages
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Use in source · chapters_sequences.tex, line 60
lessonSEQUENCES / 09Derangements show signed cancellationS09_derangements D_n=n!∑_k=0^nfrac(-1)^kk!, D_n=(n-1)(D_n-1+D_n-2) (n≥2) Positive and negative bars are the inclusion-exclusion contributions at n=6. Their large magnitudes cancel to a smaller integer. The dashed line records the final sum rather than an individual contribution. Here D_0=1 and D_1=0. The sum is 720-720+360-120+30-6+1=265. The recurrence independently gives D_6=5(44+9)=265. Signed transforms must retain cancellation. Magnit…
Exact mathematical passage
D_0=1
D_0^{n(n+1)}Use in source · 1 passage
Search aliases: D_0^n(n+1)
Use in source · Read passage · coupled_visual.tex, line 162
These derivative arrays are polynomials with nonnegative integer coefficients in the four scale variables. Their total degree is at most n(n+1). If every parameter denominator divides D_0, then D_0^n(n+1)A_n, D_0^n(n+1)B_n hboxare integers. This uniform clearing factor makes exact rational families available to integer sequence experiments. It need not be the smallest possible factor.
Exact mathematical passage
D_0^{n(n+1)}A_n,\qquad D_0^{n(n+1)}B_n \quad\hbox{are integers}.
D_1Use in source · 1 passage
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Use in source · chapters_sequences.tex, line 60
lessonSEQUENCES / 09Derangements show signed cancellationS09_derangements D_n=n!∑_k=0^nfrac(-1)^kk!, D_n=(n-1)(D_n-1+D_n-2) (n≥2) Positive and negative bars are the inclusion-exclusion contributions at n=6. Their large magnitudes cancel to a smaller integer. The dashed line records the final sum rather than an individual contribution. Here D_0=1 and D_1=0. The sum is 720-720+360-120+30-6+1=265. The recurrence independently gives D_6=5(44+9)=265. Signed transforms must retain cancellation. Magnit…
Exact mathematical passage
D_1=0
d_2Use in source · 1 passage
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Use in source · Read passage · main.tex, line 527
claimTwo exact Hurwitz-zeta reductions For N=2, d_2(w)=cossqrt w, whose zeros are w_j=π^2(j+1/2)^2, j≥0. Therefore Z_2(s)=π^-2sζ(2s,1/2). For N=4, D_4(z)=cos(z/sqrt2)cosh(z/sqrt2), and the rotation quotient has one zero w_j=(sqrt2π)^4(j+1/2)^4 per orbit. Hence Z_4(s)=(sqrt2π)^-4sζ(4s,1/2). These anchors are classical reductions. The general spectral-zeta mechanism also has substantial antecedents, including hypergeometric zero zeta functions.
Exact mathematical passage
d_2(w)=\cos\sqrt w
D_4Use in source · 1 passage
Search aliases: D_4
Use in source · Read passage · main.tex, line 533
claimTwo exact Hurwitz-zeta reductions For N=2, d_2(w)=cossqrt w, whose zeros are w_j=π^2(j+1/2)^2, j≥0. Therefore Z_2(s)=π^-2sζ(2s,1/2). For N=4, D_4(z)=cos(z/sqrt2)cosh(z/sqrt2), and the rotation quotient has one zero w_j=(sqrt2π)^4(j+1/2)^4 per orbit. Hence Z_4(s)=(sqrt2π)^-4sζ(4s,1/2). These anchors are classical reductions. The general spectral-zeta mechanism also has substantial antecedents, including hypergeometric zero zeta functions.
Exact mathematical passage
D_4(z)=\cos(z/\sqrt2)\cosh(z/\sqrt2),
D_6Use in source · 1 passage
Search aliases: D_6
Use in source · chapters_sequences.tex, line 61
D_n=n!∑_k=0^nfrac(-1)^kk!, D_n=(n-1)(D_n-1+D_n-2) (n≥2) Positive and negative bars are the inclusion-exclusion contributions at n=6. Their large magnitudes cancel to a smaller integer. The dashed line records the final sum rather than an individual contribution. Here D_0=1 and D_1=0. The sum is 720-720+360-120+30-6+1=265. The recurrence independently gives D_6=5(44+9)=265. Signed transforms must retain cancellation. Magnitudes alone do not determine the result. OEIS A000166. Factorial sum, varia…
Exact mathematical passage
D_6=5(44+9)=265
D_NUse in source · 4 passages
Search aliases: D_N
Use in source · Read passage · main.tex, line 480
MonographPartIIIOrbit spectra and exact coordinatesSpectral continuation and the coefficient language of polyphase transforms. Spectral functions of zero orbits. pageheadingFAMILY 4One spectral address for each rotation orbit For integer N≥2, define the classical Mittag--Leffler seed~[source reference] d_N(w)=∑_n≥0frac(-1)^nw^n(Nn)!=E_N,1(-w), D_N(z)=d_N(z^N). Let w_j be the nonzero zeros of d_N, counted with multiplicity. Each corresponds to one rotation orbit of N zeros of D_N. Using Argw∈(-π,…
Exact mathematical passage
d_N(w)=\sum_{n\ge0}\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w), \qquad D_N(z)=d_N(z^N).
d_NUse in source · 21 passages
Search aliases: d_N
Use in source · Read passage · main.tex, line 480
MonographPartIIIOrbit spectra and exact coordinatesSpectral continuation and the coefficient language of polyphase transforms. Spectral functions of zero orbits. pageheadingFAMILY 4One spectral address for each rotation orbit For integer N≥2, define the classical Mittag--Leffler seed~[source reference] d_N(w)=∑_n≥0frac(-1)^nw^n(Nn)!=E_N,1(-w), D_N(z)=d_N(z^N). Let w_j be the nonzero zeros of d_N, counted with multiplicity. Each corresponds to one rotation orbit of N zeros of D_N. Using Argw∈(-π,…
Exact mathematical passage
d_N(w)=\sum_{n\ge0}\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w), \qquad D_N(z)=d_N(z^N).
d_N'Use in source · 1 passage
Search aliases: d_N'
Use in source · Read passage · main.tex, line 500
ts. Multiplicity is retained by the definition; simplicity is not assumed in general. The bound |d_N(w)|≤ e^|w|^1/N and the finite-order zero-counting theorem give absolute local uniform convergence in the stated half-plane. Since the order is below one and d_N(0)=1, the Hadamard factorization has genus zero: d_N(w)=∏_j(1-w/w_j). At positive integers, a log derivative gives -wfracd_N'(w)d_N(w)=∑_k≥1 Z_N(k)w^k. If a_n=(-1)^n/(Nn)! and b_k= Z_N(k), then b_k=-ka_k-∑_j=1^k-1a_jb_k-j. Thus every posi…
Exact mathematical passage
-w\frac{d_N'(w)}{d_N(w)}=\sum_{k\ge1}\mathcal Z_N(k)w^k.
D_pUse in source · 11 passages
Search aliases: D_p
Use in source · Read passage · resolution_full.tex, line 225
For a prime p, let D_p=A_1-A_p. This positive difference consists exactly of the squarefree atoms divisible by p. Its largest frequency is
Exact mathematical passage
D_p=A_1-A_p
D_p^{(k)}Use in source · 2 passages
Search aliases: D_p^(k)
Use in source · Read passage · resolution_full.tex, line 242
sup_|s|≤ R|D_p^(k)(s)| ≤ λ_p^k e^Rλ_pfracD(4)p^4+1. tagRA-4
Exact mathematical passage
\sup_{|s|\le R}|D_p^{(k)}(s)| \le \lambda_p^k e^{R\lambda_p}\frac{D(4)}{p^4+1}. \tag{RA-4}
\ddUse in source · 13 passages
Search aliases: dd dd d
Use in source · Read passage · main.tex, line 335
(n,Q)=1μ(n)^2 =δ_Q x+O_Q(sqrtx). Integrate against g_s(x)=x^-2a(1-x^-β)^2s on [1,∞). The density term becomes the beta integral after y=x^-β. Partial integration bounds the error by C_Q∫_1^∞sqrtx|g_s'(x)| d x. The two derivative terms are beta integrals with decay exponent (2a-1/2)/β; |s|/Re s stays bounded in the stated sector. This gives the error term and then the leading asymptotic.
Exact mathematical passage
C_Q\int_1^\infty\sqrt{x}|g_s'(x)|\dd x
\degUse in source · 3 passages
Search aliases: deg
Use in source · Read passage · main.tex, line 407
B_mA_m^-1/2. Every determinant root is real and below -1; possible coefficient poles occur there, although numerator cancellation is possible. fig05_sieve_boundaryThe line is a proved bound for p=2; dots are ordinary high-precision pencil roots. A defect λ=-1.001 eventually violates positivity even though it can pass low-order checks. The nearest root is -1-δ_m, with δ_m=min_substackf∈ R[x]∖{0} deg f<m frac∫ f^2 dν_nonp∫ f^2 dν_p>0, δ_mdownarrow0. To prove the limit, approximate a continuous bum…
Exact mathematical passage
\delta_m=\min_{\substack{f\in\mathbb R[x]\setminus\{0\}\\\deg f<m}} \frac{\int f^2\dd\nu_{\mathrm{non}p}}{\int f^2\dd\nu_p}>0, \qquad\delta_m\downarrow0.
\detUse in source · 4 passages
Search aliases: det
Use in source · Read passage · main.tex, line 396
ARYFinite positivity can conceal a signed defect Let ν_nonp and ν_p be the restrictions of the undeformed measure ν_p,0 to indices not divisible by p and divisible by p, respectively. Then ν_p,λ=ν_nonp+(1+λ)ν_p. Let A_m,B_m be the order-m Hankel matrices of these two unscaled positive measures. Both are positive definite. Thus τ_m(λ)=det(A_m+(1+λ)B_m) =det A_m∏_i=1^m[1+(1+λ)η_i], where η_i>0 are the eigenvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and below -1; possible coeffic…
Exact mathematical passage
\tau_m(\lambda)=\det(A_m+(1+\lambda)B_m) =\det A_m\prod_{i=1}^{m}[1+(1+\lambda)\eta_i],
\downarrowUse in source · 6 passages
Search aliases: downarrow
Use in source · Read passage · main.tex, line 214
ontinuously on every fixed compact set, while its growth order jumps when the second channel becomes active. For 0<ε≤1 the order is log2/log(4/3); at ε=0 it is 1/2. Coefficients depend continuously on ε and are dominated by those of the entire f_3/4. Dominated convergence of the Taylor series therefore proves uniform convergence on compact sets as εdownarrow0.
Exact mathematical passage
\varepsilon\downarrow0
eUse in source · 76 passages
Search aliases: e
Use in source · chapters_sequences.tex, line 77
stored as 0/1, and this guide uses B_1=-1/2. At n=10, p_10=5 and q_10=66, so B_10=5/66. Its Taylor coefficient is (5/66)/10!. At n=12, B_12=-691/2730. An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient. OEIS A027641 and A027642, indexed from zero. Exact rational recurrence and multiplication by (e^t-1)/t checked through degree eighteen.
Exact mathematical passage
(e^t-1)/t
e^{(\lambda_1-\lambda_j)t}Use in source · 2 passages
Search aliases: e^(λ_1-λ_j)t lambda lambda
Use in source · Read passage · arithmetic_full.tex, line 299
η κ_j(t), κ_j(t)=fracA(-t)R_j(-t) ∼fracw_1w_je^(λ_1-λ_j)t.
Exact mathematical passage
\eta\,\kappa_j(t),\qquad \kappa_j(t)=\frac{A(-t)}{R_j(-t)} \sim\frac{w_1}{w_j}e^{(\lambda_1-\lambda_j)t}.
e^{(\lambda_2-\lambda_p)T}Use in source · 2 passages
Search aliases: e^(λ_2-λ_p)T lambda lambda
Use in source · Read passage · resolution_full.tex, line 356
fracA_1(-T)D_p(-T) ≥fracp^4+116D(4)e^(λ_2-λ_p)T. tagRA-8
Exact mathematical passage
\boxed{\frac{A_1(-T)}{D_p(-T)} \ge\frac{p^4+1}{16D(4)}e^{(\lambda_2-\lambda_p)T}.} \tag{RA-8}
e^{(\omega^j-1)x}Use in source · 1 passage
Search aliases: e^(ω^j-1)x omega
Use in source · Read passage · spectral_full.tex, line 70
h_N(x)=∑_j=1^N-1e^(ω^j-1)x, c_N=1-cos(2π/N)>0.
Exact mathematical passage
h_N(x)=\sum_{j=1}^{N-1}e^{(\omega^j-1)x}, \qquad c_N=1-\cos(2\pi/N)>0.
e^{-(\lambda_j-\lambda_{j+1})t}Use in source · 1 passage
Search aliases: e^-(λ_j-λ_j+1)t lambda lambda
Use in source · Read passage · arithmetic_full.tex, line 75
R_j(-t)=w_je^λ_jt(1+r_j(t)), 0≤ r_j(t)≤fracM_j+1w_j e^-(λ_j-λ_j+1)t.
Exact mathematical passage
R_j(-t)=w_je^{\lambda_jt}(1+r_j(t)),\qquad 0\le r_j(t)\le\frac{M_{j+1}}{w_j} e^{-(\lambda_j-\lambda_{j+1})t}.
e^{-1/2}Use in source · 2 passages
Search aliases: e^-1/2
Use in source · Read passage · resolution_full.tex, line 154
log F_0(R)≤ C_0sqrt R (R≥ e), C_0=e^-1/2(1+22log2)≈9.85566103.
Exact mathematical passage
\log F_0(R)\le C_0\sqrt R\qquad(R\ge e), \qquad C_0=e^{-1/2}(1+22\log2)\approx9.85566103.
e^{-B(w)}Use in source · 1 passage
Search aliases: e^-B(w)
Use in source · Read passage · sequence_full.tex, line 202
U(z)=e^B(z)∫_0^z e^-B(w) dw, B(z)=∫_0^zT(qw) dw.
Exact mathematical passage
U(z)=e^{B(z)}\int_0^z e^{-B(w)}\,dw, \qquad B(z)=\int_0^zT(qw)\,dw.
e^{-c_N(t+a)^\alpha}Use in source · 1 passage
Search aliases: e^-c_N(t+a)^α alpha
Use in source · Read passage · spectral_full.tex, line 389
G_N,a(t)=α(t+a)^α-1 +O ((t+a)^α-1e^-c_N(t+a)^α). tagT4
Exact mathematical passage
G_{N,a}(t)=\alpha(t+a)^{\alpha-1} +O\!\left((t+a)^{\alpha-1}e^{-c_N(t+a)^\alpha}\right). \tag{T4}
e^{-c_Nx}Use in source · 2 passages
Search aliases: e^-c_Nx
Use in source · Read passage · spectral_full.tex, line 77
F_N(x)=frace^xN(1+h_N(x)), |h_N(x)|≤(N-1)e^-c_Nx, |h_N'(x)|≤2(N-1)e^-c_Nx. tagS2
Exact mathematical passage
F_N(x)=\frac{e^x}{N}(1+h_N(x)),\qquad |h_N(x)|\le(N-1)e^{-c_Nx},\qquad |h_N'(x)|\le2(N-1)e^{-c_Nx}. \tag{S2}
e^{-Ns\log x}Use in source · 1 passage
Search aliases: e^-Nslog x log
Use in source · Read passage · spectral_full.tex, line 199
For x>0, x^-Ns=e^-Nslog x uses the real logarithm. Multiplicities are already incorporated in (S4), so no simple-zero assumption entered the proof.
Exact mathematical passage
x^{-Ns}=e^{-Ns\log x}
e^{-s\operatorname{Log}w}Use in source · 1 passage
Search aliases: e^-sLogw operatorname
Use in source · Read passage · spectral_full.tex, line 178
∫_0^∞fract^-st+w dt =fracπsin(π s)e^-sLogw, 0<Re s<1, |arg w|<π. tagS10
Exact mathematical passage
\int_0^\infty\frac{t^{-s}}{t+w}\,dt =\frac{\pi}{\sin(\pi s)}e^{-s\operatorname{Log}w}, \quad 0<\Re s<1,\quad |\arg w|<\pi. \tag{S10}
e^{-s\operatorname{Log}w_j}Use in source · 1 passage
Search aliases: e^-sLogw_j operatorname
Use in source · Read passage · spectral_full.tex, line 22
Z_N(s)=∑_j e^-sLogw_j
Exact mathematical passage
\mathcal Z_N(s)=\sum_j e^{-s\operatorname{Log}w_j}
e^{-x}Use in source · 1 passage
Search aliases: e^-x
Use in source · Read passage · spectral_full.tex, line 84
Since 1+h_N(x)=Ne^-xF_N(x)>0, the real logarithm is well defined for all x≥0. At infinity,
Exact mathematical passage
1+h_N(x)=Ne^{-x}F_N(x)>0
e^{-\gamma_N|x|}Use in source · 1 passage
Search aliases: e^-γ_N|x| gamma
Use in source · Read passage · spectral_full.tex, line 132
Indeed, the nearest competing root is separated from the positive direction by at least 3π/(2N), while the distinguished root is within π/(2N). Cosine decreases on [0,π]. Thus, once (N-1)e^-γ_N|x|<1, the distinguished term e^x in (S1) is larger in modulus than the sum of all competitors, and d_N(w)≠0. Consequently, sufficiently large zeros avoid the entire closed left half-plane. There are only finitely many remaining zeros in a bounded disk. None lies on the negative ray, since d_N(-t)>0 for t≥…
Exact mathematical passage
(N-1)e^{-\gamma_N|x|}<1
e^{-\Delta_jt}Use in source · 4 passages
Search aliases: e^-Δ_jt Delta
Use in source · Read passage · arithmetic_full.tex, line 232
0≤λ_j-widehatλ_j(t,h) ≤frac1hlog (1+fracM_j+1w_je^-Δ_jt),
Exact mathematical passage
0\le\lambda_j-\widehat\lambda_j(t,h) \le\frac1h\log\!\left(1+\frac{M_{j+1}}{w_j}e^{-\Delta_jt}\right),
e^{-\kappa_N t^{1/N}}Use in source · 1 passage
Search aliases: e^-κ_N t^1/N kappa
Use in source · Read passage · spectral_continuation.tex, line 29
pageheadingORBIT SPECTRA / LARGE RADIUSA finite exponential sum controls the tail Put E_N(t)=d_N(-t)=∑_n≥0t^n/(Nn)! and ω=e^2π i/N. For x≥0, E_N(x^N)=frac1N∑_j=0^N-1e^ω^jx. Root filtering proves this identity term by term: only powers divisible by N remain. The positive real exponential is dominant, so log E_N(t)=t^1/N-log N+O(e^-κ_N t^1/N), κ_N=1-cos(2π/N)>0, with corresponding differentiated estimates on the positive ray. fig14_spectral_tailComputed logarithmic derivatives of the finite expone…
Exact mathematical passage
\log E_N(t)=t^{1/N}-\log N+O(e^{-\kappa_N t^{1/N}}), \quad \kappa_N=1-\cos(2\pi/N)>0,
e^{-\lambda_j/2}Use in source · 2 passages
Search aliases: e^-λ_j/2 lambda
Use in source · Read passage · arithmetic_full.tex, line 149
displaystyle n_j=(1-e^-λ_j/2)^-1/β, a=-fraclog w_j2log n_j.
Exact mathematical passage
\boxed{\displaystyle n_j=(1-e^{-\lambda_j/2})^{-1/\beta},\qquad a=-\frac{\log w_j}{2\log n_j}.}
e^{-\lambda_j\overline t}Use in source · 1 passage
Search aliases: e^-λ_joverline t lambda overline
Use in source · Read passage · arithmetic_full.tex, line 394
Using an inner product linear in the first argument, the coefficient vector of K(·,t) is (sqrtw_je^-λ_joverline t)_j, so ⟨ Uc,K(·,t)⟩=(Uc)(t). This also verifies the sign and conjugation convention directly. Arbitrarily large finite Gram matrices at distinct complex depths are strictly positive, as proved in the base route. Thus the feature space has infinite rank, not a finite-dimensional Prony model.
Exact mathematical passage
(\sqrt{w_j}e^{-\lambda_j\overline t})_j
e^{-\lambda_js}Use in source · 4 passages
Search aliases: e^-λ_js lambda
Use in source · Read passage · arithmetic_full.tex, line 63
λ_j=lim_t→∞fraclog R_j(-t)t, w_j=lim_t→∞e^-λ_jtR_j(-t), R_j+1(s)=R_j(s)-w_je^-λ_js.
Exact mathematical passage
\lambda_j=\lim_{t\to\infty}\frac{\log R_j(-t)}t, \qquad w_j=\lim_{t\to\infty}e^{-\lambda_jt}R_j(-t), \qquad R_{j+1}(s)=R_j(s)-w_je^{-\lambda_js}.
e^{-\lambda_jt}Use in source · 5 passages
Search aliases: e^-λ_jt lambda
Use in source · Read passage · arithmetic_full.tex, line 63
λ_j=lim_t→∞fraclog R_j(-t)t, w_j=lim_t→∞e^-λ_jtR_j(-t), R_j+1(s)=R_j(s)-w_je^-λ_js.
Exact mathematical passage
\lambda_j=\lim_{t\to\infty}\frac{\log R_j(-t)}t, \qquad w_j=\lim_{t\to\infty}e^{-\lambda_jt}R_j(-t), \qquad R_{j+1}(s)=R_j(s)-w_je^{-\lambda_js}.
e^{-\lambda_ns}Use in source · 1 passage
Search aliases: e^-λ_ns lambda
Use in source · Read passage · main.tex, line 248
mid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit working precision. Right: initial squarefree frequencies and weights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If p_* is the least prime not dividin…
Exact mathematical passage
A(s)=\sum_{n\in\mathcal S_Q}w_ne^{-\lambda_ns},\qquad w_n=n^{-2a}.
e^{-\rho}Use in source · 3 passages
Search aliases: e^-ρ rho
Use in source · Read passage · dilation_full.tex, line 125
c_+=e^-ρ(V_N+1+Δ+B_+).
Exact mathematical passage
c_+=e^{-\rho}(V_N+1+\Delta+B_+).
e^{2\pi i/b}Use in source · 2 passages
Search aliases: e^2π i/b pi
Use in source · Read passage · sequence_full.tex, line 266
For an EGF G(z)=∑_m≥0a_m z^m/m!, an integer base b≥1, a residue 0≤ j<b, and ω=e^2π i/b, define
Exact mathematical passage
\omega=e^{2\pi i/b}
e^{2\pi i/N}Use in source · 2 passages
Search aliases: e^2π i/N pi
Use in source · Read passage · spectral_continuation.tex, line 23
pageheadingORBIT SPECTRA / LARGE RADIUSA finite exponential sum controls the tail Put E_N(t)=d_N(-t)=∑_n≥0t^n/(Nn)! and ω=e^2π i/N. For x≥0, E_N(x^N)=frac1N∑_j=0^N-1e^ω^jx. Root filtering proves this identity term by term: only powers divisible by N remain. The positive real exponential is dominant, so log E_N(t)=t^1/N-log N+O(e^-κ_N t^1/N), κ_N=1-cos(2π/N)>0, with corresponding differentiated estimates on the positive ray. fig14_spectral_tailComputed logarithmic derivatives of the finite expone…
Exact mathematical passage
\omega=e^{2\pi i/N}
e^{B(z)}Use in source · 1 passage
Search aliases: e^B(z)
Use in source · Read passage · sequence_full.tex, line 202
U(z)=e^B(z)∫_0^z e^-B(w) dw, B(z)=∫_0^zT(qw) dw.
Exact mathematical passage
U(z)=e^{B(z)}\int_0^z e^{-B(w)}\,dw, \qquad B(z)=\int_0^zT(qw)\,dw.
e^{h\varepsilon}Use in source · 1 passage
Search aliases: e^hε varepsilon
Use in source · Read passage · arithmetic_full.tex, line 253
t≥max{0,frac1Δ_j logfracM_j+1w_j(e^hε-1)}.
Exact mathematical passage
t\ge\max\left\{0,\frac1{\Delta_j} \log\frac{M_{j+1}}{w_j(e^{h\varepsilon}-1)}\right\}.
e^{i\theta}Use in source · 1 passage
Search aliases: e^iθ theta
Use in source · Read passage · spectral_full.tex, line 151
To check the elementary estimate, write w=re^iθ: |t+w|^2=(t+r)^2-4trsin^2(θ/2)≥(t+r)^2cos^2(θ/2).
Exact mathematical passage
w=re^{i\theta}
e^{R\lambda_p}Use in source · 4 passages
Search aliases: e^Rλ_p lambda
Use in source · Read passage · arithmetic_full.tex, line 318
displaystyle sup_|s|≤ R|A_a,β,Q^(k)(s)-A_a,β,Qp^(k)(s)| ≤ λ_p^k e^Rλ_p fracZ_Q(2a)p^2a+1.
Exact mathematical passage
\boxed{\displaystyle \sup_{|s|\le R}|A_{a,\beta,Q}^{(k)}(s)-A_{a,\beta,Qp}^{(k)}(s)| \le \lambda_p^k e^{R\lambda_p} \frac{Z_Q(2a)}{p^{2a}+1}.}
e^tUse in source · 9 passages
Search aliases: e^t
Use in source · chapters_sequences.tex, line 77
stored as 0/1, and this guide uses B_1=-1/2. At n=10, p_10=5 and q_10=66, so B_10=5/66. Its Taylor coefficient is (5/66)/10!. At n=12, B_12=-691/2730. An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient. OEIS A027641 and A027642, indexed from zero. Exact rational recurrence and multiplication by (e^t-1)/t checked through degree eighteen.
Exact mathematical passage
(e^t-1)/t
e^{t-a_j}Use in source · 1 passage
Search aliases: e^t-a_j
Use in source · Read passage · dilation_full.tex, line 147
log M(e^t)≤ t+log2+ max_j{log M(e^t-a_j)+log M(e^t-b_j)}.
Exact mathematical passage
\log M(e^t)\le t+\log2+ \max_j\{\log M(e^{t-a_j})+\log M(e^{t-b_j})\}.
e^{t-b_j}Use in source · 1 passage
Search aliases: e^t-b_j
Use in source · Read passage · dilation_full.tex, line 147
log M(e^t)≤ t+log2+ max_j{log M(e^t-a_j)+log M(e^t-b_j)}.
Exact mathematical passage
\log M(e^t)\le t+\log2+ \max_j\{\log M(e^{t-a_j})+\log M(e^{t-b_j})\}.
e^{t-\log(1/q)}Use in source · 1 passage
Search aliases: e^t-log(1/q) log
Use in source · Read passage · coupled_growth_full.tex, line 319
H_1(e^t)≤ t+log2 +H_1(e^t-log(1/q))+H_2(e^t-log(1/r)).
Exact mathematical passage
H_1(e^t)\le t+\log2 +H_1(e^{t-\log(1/q)})+H_2(e^{t-\log(1/r)}).
e^{t-\log(1/r)}Use in source · 1 passage
Search aliases: e^t-log(1/r) log
Use in source · Read passage · coupled_growth_full.tex, line 319
H_1(e^t)≤ t+log2 +H_1(e^t-log(1/q))+H_2(e^t-log(1/r)).
Exact mathematical passage
H_1(e^t)\le t+\log2 +H_1(e^{t-\log(1/q)})+H_2(e^{t-\log(1/r)}).
e^{t\lambda_{L+1}}Use in source · 1 passage
Search aliases: e^tλ_L+1 lambda
Use in source · Read passage · arithmetic_full.tex, line 269
E_L(t)=e^tλ_L+1fracL^1-2a2a-1.
Exact mathematical passage
E_L(t)=e^{t\lambda_{L+1}}\frac{L^{1-2a}}{2a-1}.
e^xUse in source · 2 passages
Search aliases: e^x
Use in source · Read passage · spectral_full.tex, line 77
F_N(x)=frace^xN(1+h_N(x)), |h_N(x)|≤(N-1)e^-c_Nx, |h_N'(x)|≤2(N-1)e^-c_Nx. tagS2
Exact mathematical passage
F_N(x)=\frac{e^x}{N}(1+h_N(x)),\qquad |h_N(x)|\le(N-1)e^{-c_Nx},\qquad |h_N'(x)|\le2(N-1)e^{-c_Nx}. \tag{S2}
e^{|w|^{1/N}}Use in source · 1 passage
Search aliases: e^|w|^1/N
Use in source · Read passage · main.tex, line 493
one rotation orbit of N zeros of D_N. Using Argw∈(-π,π], define Z_N(s)=∑_jw_j^-s, Re s>1/N. fig06_orbit_zerosFor N=4, the four real/imaginary zeros in each orbit share one positive coordinate w=z^4. The plot shows the first six orbits. Multiplicity is retained by the definition; simplicity is not assumed in general. The bound |d_N(w)|≤ e^|w|^1/N and the finite-order zero-counting theorem give absolute local uniform convergence in the stated half-plane. Since the order is below one and d_N(0)=1, …
Exact mathematical passage
|d_N(w)|\le e^{|w|^{1/N}}
e^{\lambda_jt}Use in source · 1 passage
Search aliases: e^λ_jt lambda
Use in source · Read passage · arithmetic_full.tex, line 75
R_j(-t)=w_je^λ_jt(1+r_j(t)), 0≤ r_j(t)≤fracM_j+1w_j e^-(λ_j-λ_j+1)t.
Exact mathematical passage
R_j(-t)=w_je^{\lambda_jt}(1+r_j(t)),\qquad 0\le r_j(t)\le\frac{M_{j+1}}{w_j} e^{-(\lambda_j-\lambda_{j+1})t}.
e^{\lambda_pT}Use in source · 2 passages
Search aliases: e^λ_pT lambda
Use in source · Read passage · resolution_full.tex, line 289
p^-4e^λ_pT≤ D_p(-T) ≤fracD(4)p^4+1e^λ_pT (T≥0). tagRA-6
Exact mathematical passage
\boxed{p^{-4}e^{\lambda_pT}\le D_p(-T) \le\frac{D(4)}{p^4+1}e^{\lambda_pT}\qquad(T\ge0).} \tag{RA-6}
e^{\rho t}Use in source · 3 passages
Search aliases: e^ρ t rho
Use in source · Read passage · dilation_full.tex, line 158
On the seed, G(t)≤ V_N+1+Δ+B_+≤ c_+e^ρ t. Each channel satisfies q_j^ρ+r_j^ρ≤1. Induction in intervals of length min_jmin(a_j,b_j)>0 therefore propagates the seed comparison, giving the explicit upper bound
Exact mathematical passage
G(t)\le V_N+1+\Delta+B_+\le c_+e^{\rho t}
e^{\tau\max(-\Re s,0)}Use in source · 1 passage
Search aliases: e^τmax(-Re s,0) tau max Re
Use in source · Read passage · main.tex, line 256
the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If p_* is the least prime not dividing Q, then τ=-2log(1-p_*^-β), M=Z_Q(2a)-1, |A(s)|≤ M e^τmax(-Re s,0). This proves local uniform convergence and entire continuation. The first atom dominates A(-t) as t→∞, so the entire function has exact order one and exponential type τ.
Exact mathematical passage
|A(s)|\le M e^{\tau\max(-\Re s,0)}.
e^{\omega^j x}Use in source · 1 passage
Search aliases: e^ω^j x omega
Use in source · Read passage · spectral_full.tex, line 59
F_N(x):=d_N(-x^N) =frac1N∑_j=0^N-1e^ω^j x =∑_n=0^∞fracx^Nn(Nn)!. tagS1
Exact mathematical passage
F_N(x):=d_N(-x^N) =\frac1N\sum_{j=0}^{N-1}e^{\omega^j x} =\sum_{n=0}^{\infty}\frac{x^{Nn}}{(Nn)!}. \tag{S1}
e^{\omega^jx}Use in source · 1 passage
Search aliases: e^ω^jx omega
Use in source · Read passage · spectral_continuation.tex, line 24
pageheadingORBIT SPECTRA / LARGE RADIUSA finite exponential sum controls the tail Put E_N(t)=d_N(-t)=∑_n≥0t^n/(Nn)! and ω=e^2π i/N. For x≥0, E_N(x^N)=frac1N∑_j=0^N-1e^ω^jx. Root filtering proves this identity term by term: only powers divisible by N remain. The positive real exponential is dominant, so log E_N(t)=t^1/N-log N+O(e^-κ_N t^1/N), κ_N=1-cos(2π/N)>0, with corresponding differentiated estimates on the positive ray. fig14_spectral_tailComputed logarithmic derivatives of the finite expone…
Exact mathematical passage
E_N(x^N)=\frac1N\sum_{j=0}^{N-1}e^{\omega^jx}.
E_aUse in source · 2 passages
Search aliases: E_a
Use in source · Read passage · sequence_polyphase.tex, line 100
claimThree operations that must be kept distinct For an EGF E_a(z)=∑_ma_mz^m/m!, differentiation transports future coefficients: E_a^(k)=∑_ma_m+kz^m/m!. Causal ordinary taps instead act by repeated zero-constant integration: E_b=∑_ℓ h_ℓ I^ℓ E_a. Multiplication of two EGFs gives Hurwitz convolution, b_m=∑_k=0^mbinom mk a_kh_m-k. It is not ordinary sequence FIR convolution. Intrinsic nonlinear feedback taps alter the defining family and are a fourth operation with their own coefficient recurrence.
Exact mathematical passage
E_a(z)=\sum_ma_mz^m/m!
E_a^{(k)}Use in source · 1 passage
Search aliases: E_a^(k)
Use in source · Read passage · sequence_polyphase.tex, line 101
claimThree operations that must be kept distinct For an EGF E_a(z)=∑_ma_mz^m/m!, differentiation transports future coefficients: E_a^(k)=∑_ma_m+kz^m/m!. Causal ordinary taps instead act by repeated zero-constant integration: E_b=∑_ℓ h_ℓ I^ℓ E_a. Multiplication of two EGFs gives Hurwitz convolution, b_m=∑_k=0^mbinom mk a_kh_m-k. It is not ordinary sequence FIR convolution. Intrinsic nonlinear feedback taps alter the defining family and are a fourth operation with their own coefficient recurrence.
Exact mathematical passage
E_a^{(k)}=\sum_ma_{m+k}z^m/m!
E_bUse in source · 1 passage
Search aliases: E_b
Use in source · Read passage · sequence_polyphase.tex, line 103
claimThree operations that must be kept distinct For an EGF E_a(z)=∑_ma_mz^m/m!, differentiation transports future coefficients: E_a^(k)=∑_ma_m+kz^m/m!. Causal ordinary taps instead act by repeated zero-constant integration: E_b=∑_ℓ h_ℓ I^ℓ E_a. Multiplication of two EGFs gives Hurwitz convolution, b_m=∑_k=0^mbinom mk a_kh_m-k. It is not ordinary sequence FIR convolution. Intrinsic nonlinear feedback taps alter the defining family and are a fourth operation with their own coefficient recurrence.
Exact mathematical passage
E_b=\sum_\ell h_\ell I^\ell E_a.
E_LUse in source · 2 passages
Search aliases: E_L
Use in source · Read passage · arithmetic_full.tex, line 269
E_L(t)=e^tλ_L+1fracL^1-2a2a-1.
Exact mathematical passage
E_L(t)=e^{t\lambda_{L+1}}\frac{L^{1-2a}}{2a-1}.
E_NUse in source · 10 passages
Search aliases: E_N
Use in source · Read passage · spectral_continuation.tex, line 22
pageheadingORBIT SPECTRA / LARGE RADIUSA finite exponential sum controls the tail Put E_N(t)=d_N(-t)=∑_n≥0t^n/(Nn)! and ω=e^2π i/N. For x≥0, E_N(x^N)=frac1N∑_j=0^N-1e^ω^jx. Root filtering proves this identity term by term: only powers divisible by N remain. The positive real exponential is dominant, so log E_N(t)=t^1/N-log N+O(e^-κ_N t^1/N), κ_N=1-cos(2π/N)>0, with corresponding differentiated estimates on the positive ray. fig14_spectral_tailComputed logarithmic derivatives of the finite expone…
Exact mathematical passage
E_N(t)=d_N(-t)=\sum_{n\ge0}t^n/(Nn)!
E_{N,1}Use in source · 6 passages
Search aliases: E_N,1
Use in source · Read passage · main.tex, line 480
MonographPartIIIOrbit spectra and exact coordinatesSpectral continuation and the coefficient language of polyphase transforms. Spectral functions of zero orbits. pageheadingFAMILY 4One spectral address for each rotation orbit For integer N≥2, define the classical Mittag--Leffler seed~[source reference] d_N(w)=∑_n≥0frac(-1)^nw^n(Nn)!=E_N,1(-w), D_N(z)=d_N(z^N). Let w_j be the nonzero zeros of d_N, counted with multiplicity. Each corresponds to one rotation orbit of N zeros of D_N. Using Argw∈(-π,…
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d_N(w)=\sum_{n\ge0}\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w), \qquad D_N(z)=d_N(z^N).
\expUse in source · 9 passages
Search aliases: exp
Use in source · Read passage · arithmetic_full.tex, line 19
A_a,β,Q(s)=∑_substackn≥2 μ(n)^2=1 (n,Q)=1 n^-2aexp(-λ_n s), λ_n=-2log(1-n^-β).
Exact mathematical passage
A_{a,\beta,Q}(s)=\sum_{\substack{n\ge2\\ \mu(n)^2=1\\ (n,Q)=1}} n^{-2a}\exp(-\lambda_n s),\qquad \lambda_n=-2\log(1-n^{-\beta}).
\mathcal FUse in source · 3 passages
Search aliases: F mathcal
Use in source · Read passage · coupled_structure_full.tex, line 13
F[X](z)= 1+u(qz)v(rz) 1+u(az)v(bz). tagS1
Exact mathematical passage
\mathcal F[X](z)= \begin{pmatrix}1+u(qz)v(rz)\\1+u(az)v(bz)\end{pmatrix}. \tag{S1}
F'Use in source · 4 passages
Search aliases: F'
Use in source · Read passage · dilation_full.tex, line 13
F'(z)=1+∑_j=1^Jh_jF(q_jz)F(r_jz), F(0)=0,
Exact mathematical passage
F'(z)=1+\sum_{j=1}^{J}h_jF(q_jz)F(r_jz),\qquad F(0)=0,
f'Use in source · 1 passage
Search aliases: f'
Use in source · Read passage · resolution_full.tex, line 81
sup_|z|≤ R|F_ε(z)-F_0(z)| ≤ ε L(R), L(R)=∑_n≥1n d_nR^2n+1 =fracRf'(R)-f(R)2. tagRA-1
Exact mathematical passage
\sup_{|z|\le R}|F_\varepsilon(z)-F_0(z)| \le \varepsilon L(R),\qquad L(R)=\sum_{n\ge1}n d_nR^{2n+1} =\frac{Rf'(R)-f(R)}2. \tag{RA-1}
f^2Use in source · 3 passages
Search aliases: f^2
Use in source · Read passage · main.tex, line 407
B_mA_m^-1/2. Every determinant root is real and below -1; possible coefficient poles occur there, although numerator cancellation is possible. fig05_sieve_boundaryThe line is a proved bound for p=2; dots are ordinary high-precision pencil roots. A defect λ=-1.001 eventually violates positivity even though it can pass low-order checks. The nearest root is -1-δ_m, with δ_m=min_substackf∈ R[x]∖{0} deg f<m frac∫ f^2 dν_nonp∫ f^2 dν_p>0, δ_mdownarrow0. To prove the limit, approximate a continuous bum…
Exact mathematical passage
\delta_m=\min_{\substack{f\in\mathbb R[x]\setminus\{0\}\\\deg f<m}} \frac{\int f^2\dd\nu_{\mathrm{non}p}}{\int f^2\dd\nu_p}>0, \qquad\delta_m\downarrow0.
F_0Use in source · 9 passages
Search aliases: F_0
Use in source · chapters_sequences.tex, line 4
startvolumeSequence atlasOEIS families and exact transforms lessonSEQUENCES / 01Two seeds, one Fibonacci recurrenceS01_fibonacci_lucas F_n+2=F_n+1+F_n, L_n+2=L_n+1+L_n, L_n^2-5F_n^2=4(-1)^n The two families share the same recurrence and have different seeds: (F_0,F_1)=(0,1) and (L_0,L_1)=(2,1). The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude. At n=7, F_7=13 and L_7=29. Thus 29^2-5·13^2=841-845=-4. Comparing terms at different indices w…
Exact mathematical passage
(F_0,F_1)=(0,1)
F_1Use in source · 4 passages
Search aliases: F_1
Use in source · chapters_sequences.tex, line 4
startvolumeSequence atlasOEIS families and exact transforms lessonSEQUENCES / 01Two seeds, one Fibonacci recurrenceS01_fibonacci_lucas F_n+2=F_n+1+F_n, L_n+2=L_n+1+L_n, L_n^2-5F_n^2=4(-1)^n The two families share the same recurrence and have different seeds: (F_0,F_1)=(0,1) and (L_0,L_1)=(2,1). The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude. At n=7, F_7=13 and L_7=29. Thus 29^2-5·13^2=841-845=-4. Comparing terms at different indices w…
Exact mathematical passage
(F_0,F_1)=(0,1)
f_{1/2}Use in source · 1 passage
Search aliases: f_1/2
Use in source · Read passage · main.tex, line 171
claimA worked order-one member T_1/3,2/3(z)=z+frac227z^3+ frac42187z^5+frac23211160261z^7+O(z^9). It has order one, and T_1/3,2/3(1)≈1.0759239897408249. The diagonal point q=r=1/2 gives f_1/2(z)=sqrt2sinh(z/sqrt2). Every off-diagonal member is separated from all rescaled diagonal members by the theorem above.
Exact mathematical passage
f_{1/2}(z)=\sqrt2\sinh(z/\sqrt2)
f_{1/4}Use in source · 2 passages
Search aliases: f_1/4
Use in source · Read passage · dilation_full.tex, line 263
sup_|z|≤ R|F_ε(z)-f_1/4(z)| ≤fracε2{R f_3/4'(R)-f_3/4(R)}.
Exact mathematical passage
\sup_{|z|\le R}|F_\varepsilon(z)-f_{1/4}(z)| \le\frac{\varepsilon}{2}\{R f_{3/4}'(R)-f_{3/4}(R)\}.
F_{10^{-12}}Use in source · 1 passage
Search aliases: F_10^-12
Use in source · Read passage · resolution_full.tex, line 119
sup_|z|≤1|F_10^-12(z)-F_0(z)|<2.41890267×10^-13,
Exact mathematical passage
\boxed{\sup_{|z|\le1}|F_{10^{-12}}(z)-F_0(z)|<2.41890267\times10^{-13},}
F_2Use in source · 2 passages
Search aliases: F_2
Use in source · Read passage · coupled_growth_full.tex, line 170
Set F_1=U, F_2=V, and H_i(R)=log F_i(R). Since F_i(R)≥ R, these logarithms are positive for R>1. The type convention used here is
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F_2=V
f_{3/4}'Use in source · 2 passages
Search aliases: f_3/4'
Use in source · Read passage · dilation_full.tex, line 263
sup_|z|≤ R|F_ε(z)-f_1/4(z)| ≤fracε2{R f_3/4'(R)-f_3/4(R)}.
Exact mathematical passage
\sup_{|z|\le R}|F_\varepsilon(z)-f_{1/4}(z)| \le\frac{\varepsilon}{2}\{R f_{3/4}'(R)-f_{3/4}(R)\}.
F_7Use in source · 1 passage
Search aliases: F_7
Use in source · chapters_sequences.tex, line 5
nsforms lessonSEQUENCES / 01Two seeds, one Fibonacci recurrenceS01_fibonacci_lucas F_n+2=F_n+1+F_n, L_n+2=L_n+1+L_n, L_n^2-5F_n^2=4(-1)^n The two families share the same recurrence and have different seeds: (F_0,F_1)=(0,1) and (L_0,L_1)=(2,1). The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude. At n=7, F_7=13 and L_7=29. Thus 29^2-5·13^2=841-845=-4. Comparing terms at different indices would destroy this identity. A recurrence operator an…
Exact mathematical passage
F_7=13
f_{a,m}Use in source · 4 passages
Search aliases: f_a,m
Use in source · Read passage · multi_prime.tex, line 72
pageheadingPRIME SIEVES / CONSTRUCTIVE WITNESSEvery negative prime atom has a finite detector For an integer a≥2, let L_a(x)=∏_n=1^a-1fracx-x_nx_a-x_n, C_a=L_a(1), r_a=frac1-x_a+11-x_a<1, f_a,m(x)=L_a(x)(frac1-x1-x_a)^m. This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects …
Exact mathematical passage
f_{a,m}(x)=L_a(x)\left(\frac{1-x}{1-x_a}\right)^m.
F_gUse in source · 2 passages
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Use in source · Read passage · dilation_full.tex, line 217
sup_|z|≤ R|F_h(z)-F_g(z)| ≤frac d4{R f_Q'(R)-f_Q(R)}.
Exact mathematical passage
\boxed{\sup_{|z|\le R}|F_h(z)-F_g(z)| \le\frac d4\{R f_Q'(R)-f_Q(R)\}.}
F_hUse in source · 2 passages
Search aliases: F_h
Use in source · Read passage · dilation_full.tex, line 217
sup_|z|≤ R|F_h(z)-F_g(z)| ≤frac d4{R f_Q'(R)-f_Q(R)}.
Exact mathematical passage
\boxed{\sup_{|z|\le R}|F_h(z)-F_g(z)| \le\frac d4\{R f_Q'(R)-f_Q(R)\}.}
F_iUse in source · 7 passages
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Use in source · Read passage · coupled_growth_full.tex, line 170
Set F_1=U, F_2=V, and H_i(R)=log F_i(R). Since F_i(R)≥ R, these logarithms are positive for R>1. The type convention used here is
Exact mathematical passage
H_i(R)=\log F_i(R)
F_NUse in source · 11 passages
Search aliases: F_N
Use in source · Read passage · spectral_full.tex, line 59
F_N(x):=d_N(-x^N) =frac1N∑_j=0^N-1e^ω^j x =∑_n=0^∞fracx^Nn(Nn)!. tagS1
Exact mathematical passage
F_N(x):=d_N(-x^N) =\frac1N\sum_{j=0}^{N-1}e^{\omega^j x} =\sum_{n=0}^{\infty}\frac{x^{Nn}}{(Nn)!}. \tag{S1}
f_nUse in source · 1 passage
Search aliases: f_n
Use in source · chapters_sequences.tex, line 109
lessonSEQUENCES / 16Sums and differences form a polynomial ladderS16_polynomial_ladder T_n=∑_j=1^nj, H_n=∑_j=1^nT_j, Δ T_n=n, Δ H_n=T_n The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, Δ f_n=f_n-f_n-1. All plotted values use their displayed common index. At n=4, T_4=10 and H_4=20. The differences are 10-6=4 and 20-10=10. Separately, 4^2-3^2=7 and 4^3-3…
Exact mathematical passage
\Delta f_n=f_n-f_{n-1}
F_N'Use in source · 1 passage
Search aliases: F_N'
Use in source · Read passage · spectral_full.tex, line 87
log F_N(x)=x-log N+O(e^-c_Nx), g_N(x):=fracF_N'(x)F_N(x) =1+O(e^-c_Nx). tagS3
Exact mathematical passage
\log F_N(x)=x-\log N+O(e^{-c_Nx}), \qquad g_N(x):=\frac{F_N'(x)}{F_N(x)} =1+O(e^{-c_Nx}). \tag{S3}
f_{n-1}Use in source · 1 passage
Search aliases: f_n-1
Use in source · chapters_sequences.tex, line 109
lessonSEQUENCES / 16Sums and differences form a polynomial ladderS16_polynomial_ladder T_n=∑_j=1^nj, H_n=∑_j=1^nT_j, Δ T_n=n, Δ H_n=T_n The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, Δ f_n=f_n-f_n-1. All plotted values use their displayed common index. At n=4, T_4=10 and H_4=20. The differences are 10-6=4 and 20-10=10. Separately, 4^2-3^2=7 and 4^3-3…
Exact mathematical passage
\Delta f_n=f_n-f_{n-1}
f_{p,m}Use in source · 3 passages
Search aliases: f_p,m
Use in source · Read passage · sieve_full.tex, line 141
∫ f_p,m(x)^2 dν_F,λ(x) ≤frac1+λ_pp^4 +fracC_F C_p^23p^3r_p^2m. tag4
Exact mathematical passage
\int f_{p,m}(x)^2\,d\nu_{F,\boldsymbol\lambda}(x) \le\frac{1+\lambda_p}{p^4} +\frac{C_F C_p^2}{3p^3}r_p^{2m}. \tag{4}
f_{p,m}^2Use in source · 1 passage
Search aliases: f_p,m^2
Use in source · Read passage · multi_prime.tex, line 81
This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects the signed defect. Set C_F=∏_q∈ Fmax(1,|1+λ_q|). The summable tail gives ∫ f_p,m^2ddν ≤frac1+λ_pp^4+ fracC_FC_p^23p^3r_p^2m. Hence the quadratic form is negative whenever pC_FC_p^2r_p^2m/3<-(1+λ_p). For rational strengths…
Exact mathematical passage
\int f_{p,m}^2\dd\nu \le\frac{1+\lambda_p}{p^4}+ \frac{C_FC_p^2}{3p^3}r_p^{2m}.
f_QUse in source · 8 passages
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Use in source · Read passage · dilation_full.tex, line 193
where f_Q(z)=∑ d_n(Q)z^2n+1 is the established diagonal function. Consequently, on every disk |z|≤ R,
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f_Q(z)=\sum d_n(Q)z^{2n+1}
f_qUse in source · 3 passages
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Use in source · Read passage · main.tex, line 61
claimAn established diagonal antecedent The equation f_q'(z)=1+f_q(qz)^2, f_q(0)=0, is already present in M endez~[source reference]. It remains an important baseline, but its definition is not a discovery of this project. The asymmetric extension keeps two separate contractions and permits a proved distinction from every amplitude/argument rescaling of that baseline.
Exact mathematical passage
f_q'(z)=1+f_q(qz)^2
f_Q'Use in source · 4 passages
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Use in source · Read passage · dilation_full.tex, line 217
sup_|z|≤ R|F_h(z)-F_g(z)| ≤frac d4{R f_Q'(R)-f_Q(R)}.
Exact mathematical passage
\boxed{\sup_{|z|\le R}|F_h(z)-F_g(z)| \le\frac d4\{R f_Q'(R)-f_Q(R)\}.}
f_q'Use in source · 1 passage
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Use in source · Read passage · main.tex, line 61
claimAn established diagonal antecedent The equation f_q'(z)=1+f_q(qz)^2, f_q(0)=0, is already present in M endez~[source reference]. It remains an important baseline, but its definition is not a discovery of this project. The asymmetric extension keeps two separate contractions and permits a proved distinction from every amplitude/argument rescaling of that baseline.
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f_q'(z)=1+f_q(qz)^2
f_sUse in source · 2 passages
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Use in source · Read passage · main.tex, line 142
pageheadingFAMILY 1 / SEPARATIONMore than a rescaled one-scale seed claimNonrescaling theorem If q≠ r, then T_q,r cannot equal A f_s(Bz) for nonzero constants A,B and any real 0<s<1, where f_s'=1+f_s(sz)^2 is the known family.
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A f_s(Bz)
f_s'Use in source · 1 passage
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Use in source · Read passage · main.tex, line 143
pageheadingFAMILY 1 / SEPARATIONMore than a rescaled one-scale seed claimNonrescaling theorem If q≠ r, then T_q,r cannot equal A f_s(Bz) for nonzero constants A,B and any real 0<s<1, where f_s'=1+f_s(sz)^2 is the known family.
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f_s'=1+f_s(sz)^2
F_\varepsilonUse in source · 11 passages
Search aliases: F_ε varepsilon
Use in source · Read passage · dilation_full.tex, line 261
Take the two diagonal channels (q,r)=(1/4,1/4) and (3/4,3/4), with weights 1-ε and ε, respectively. For 0≤ε≤1, write the resulting function as F_ε. The quantitative bound gives
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F_\varepsilon
F_\varepsilon'Use in source · 2 passages
Search aliases: F_ε' varepsilon
Use in source · Read passage · resolution_full.tex, line 55
F_ε'(z)=1+(1-ε)F_ε(z/4)^2 +ε F_ε(3z/4)^2, F_ε(0)=0, 0≤ε≤1.
Exact mathematical passage
F_\varepsilon'(z)=1+(1-\varepsilon)F_\varepsilon(z/4)^2 +\varepsilon F_\varepsilon(3z/4)^2, \qquad F_\varepsilon(0)=0, \qquad 0\le\varepsilon\le1.
GUse in source · 30 passages
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Use in source · Read passage · references.tex, line 38
bibitemDLMF NIST Digital Library of Mathematical Functions. Sections 5.5 and 5.12 give the gamma reflection and beta identities. Section 10.46 defines the Mittag--Leffler function. Sections 16.8 and 16.21 give the generalized hypergeometric and Meijer G differential equations. Section 18.2 develops general orthogonal-polynomial theory. https://dlmf.nist.gov/
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G
gUse in source · 29 passages
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Use in source · chapters_sequences.tex, line 103
esS15_binary_families w(n)=popcount(n), t(n)=w(n)bmod2, g(n)=2^w(n) Each column has a four-bit address, its binary weight, its Thue--Morse parity, and its Gould value. Reading a final zero bit preserves weight; reading a final one adds one. Parity toggles and the Gould value doubles in the latter case. The address 13=1101_2 has weight three, so t(13)=1 and g(13)=8. Address 12=1100_2 has weight two, parity zero, and Gould value four. The same state statistic can feed several exact outputs. Binary…
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g(13)=8
G^{(t)}Use in source · 1 passage
Search aliases: G^(t)
Use in source · Read passage · sequence_full.tex, line 328
Proof. The EGF of the delayed sequence is ∑_m≥ ta_m-tz^m/m!=I^tG(z). Summing taps gives the formula. Conversely, G^(t) has coefficient sequence a_m+t, so derivatives sample future coefficients and do not implement these past delays. square
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G^{(t)}
G_1Use in source · 2 passages
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Use in source · Read passage · coupled_growth_full.tex, line 326
G_1(t)≤ G_1(t-log(1/q))+G_2(t-log(1/r)).
Exact mathematical passage
G_1(t)\le G_1(t-\log(1/q))+G_2(t-\log(1/r)).
G_2Use in source · 2 passages
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Use in source · Read passage · coupled_growth_full.tex, line 326
G_1(t)≤ G_1(t-log(1/q))+G_2(t-log(1/r)).
Exact mathematical passage
G_1(t)\le G_1(t-\log(1/q))+G_2(t-\log(1/r)).
G_aUse in source · 2 passages
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Use in source · Read passage · sequence_full.tex, line 332
For two EGFs G_a,G_h, ordinary function multiplication instead gives the Hurwitz product
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G_a,G_h
G_hUse in source · 2 passages
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Use in source · Read passage · sequence_full.tex, line 332
For two EGFs G_a,G_h, ordinary function multiplication instead gives the Hurwitz product
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G_a,G_h
G_iUse in source · 2 passages
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Use in source · Read passage · coupled_growth_full.tex, line 312
G_i(t)=H_i(e^t)+t+D.
Exact mathematical passage
G_i(t)=H_i(e^t)+t+D.
G_jUse in source · 2 passages
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Use in source · Read passage · sequence_polyphase.tex, line 61
pageheadingEXACT LABORATORY / PROJECTORSPolyphase sections preserve the full sequence For an ordinary generating function G(z)=∑_m≥0a_mz^m and integer b≥1, write G(z)=∑_j=0^b-1z^jG_j(z^b), G_j(t)=∑_n≥0a_bn+jt^n. The root projector is P_b,jG(z)=frac1b∑_k=0^b-1ω^-jkG(ω^kz) =z^jG_j(z^b), ω=e^2π i/b. The same projector works on an EGF, but the section retains its original factorials: ∑_na_bn+jz^bn+j/(bn+j)!. Removing those denominators changes the representation and requires an explicit transport. R…
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G(z)=\sum_{j=0}^{b-1}z^jG_j(z^b),\qquad G_j(t)=\sum_{n\ge0}a_{bn+j}t^n.
G_NUse in source · 4 passages
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Use in source · Read passage · spectral_full.tex, line 158
Let H_N(t)=d_N(-t)=E_N,1(t), and G_N(t)=H_N'(t)/H_N(t). Logarithmic differentiation of (S4), locally uniformly off the zeros, gives
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G_N(t)=H_N'(t)/H_N(t)
g_NUse in source · 6 passages
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Use in source · Read passage · spectral_full.tex, line 87
log F_N(x)=x-log N+O(e^-c_Nx), g_N(x):=fracF_N'(x)F_N(x) =1+O(e^-c_Nx). tagS3
Exact mathematical passage
\log F_N(x)=x-\log N+O(e^{-c_Nx}), \qquad g_N(x):=\frac{F_N'(x)}{F_N(x)} =1+O(e^{-c_Nx}). \tag{S3}
G_{N,a}Use in source · 7 passages
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Use in source · Read passage · spectral_full.tex, line 377
This follows by taking the ratio of the two convergent products in (S4), not by introducing an unverified exponential factor. Its logarithmic derivative is G_N,a(t)=H_N'(a+t)/H_N(a+t). The same dominated beta-integral argument gives
Exact mathematical passage
G_{N,a}(t)=H_N'(a+t)/H_N(a+t)
g_sUse in source · 2 passages
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Use in source · Read passage · main.tex, line 332
(n,Q)=1μ(n)^2 =δ_Q x+O_Q(sqrtx). Integrate against g_s(x)=x^-2a(1-x^-β)^2s on [1,∞). The density term becomes the beta integral after y=x^-β. Partial integration bounds the error by C_Q∫_1^∞sqrtx|g_s'(x)| d x. The two derivative terms are beta integrals with decay exponent (2a-1/2)/β; |s|/Re s stays bounded in the stated sector. This gives the error term and then the leading asymptotic.
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g_s(x)=x^{-2a}(1-x^{-\beta})^{2s}
g_s'Use in source · 2 passages
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Use in source · Read passage · main.tex, line 335
(n,Q)=1μ(n)^2 =δ_Q x+O_Q(sqrtx). Integrate against g_s(x)=x^-2a(1-x^-β)^2s on [1,∞). The density term becomes the beta integral after y=x^-β. Partial integration bounds the error by C_Q∫_1^∞sqrtx|g_s'(x)| d x. The two derivative terms are beta integrals with decay exponent (2a-1/2)/β; |s|/Re s stays bounded in the stated sector. This gives the error term and then the leading asymptotic.
Exact mathematical passage
C_Q\int_1^\infty\sqrt{x}|g_s'(x)|\dd x
HUse in source · 35 passages
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Use in source · chapters_sequences.tex, line 110
_ladder T_n=∑_j=1^nj, H_n=∑_j=1^nT_j, Δ T_n=n, Δ H_n=T_n The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, Δ f_n=f_n-f_n-1. All plotted values use their displayed common index. At n=4, T_4=10 and H_4=20. The differences are 10-6=4 and 20-10=10. Separately, 4^2-3^2=7 and 4^3-3^3=37. Prefix summation raises polynomial degree; differencing lowers it. These…
Exact mathematical passage
H_4=20
\mathcal HUse in source · 5 passages
Search aliases: H mathcal
Use in source · Read passage · arithmetic_full.tex, line 385
The image H_a,β,Q=Uℓ^2, with Uc _ H= c _ℓ^2, is therefore a complete Hilbert space of entire functions. Its reproducing kernel is
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\mathcal H_{a,\beta,Q}=U\ell^2
h^{-1}Use in source · 1 passage
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Use in source · Read passage · arithmetic_full.tex, line 245
Proof. The preceding positive remainder formula gives widehatλ_j=λ_j+h^-1[log(1+r_j(t+h))-log(1+r_j(t))]. Each tail factor decreases, so 0≤ r_j(t+h)≤ r_j(t). The two inequalities now follow from the bound on r_j(t). For the amplitude, multiply the positive tail bound by e^-λ_jt.
Exact mathematical passage
\widehat\lambda_j=\lambda_j+h^{-1}[\log(1+r_j(t+h))-\log(1+r_j(t))]
H^nUse in source · 2 passages
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Use in source · Read passage · sequence_full.tex, line 105
B_n=H^nD^n(n+1)A_n∈ Z hboxfor all n.
Exact mathematical passage
\boxed{B_n=H^nD^{n(n+1)}A_n\in\mathbb Z\quad\hbox{for all }n.}
h_0Use in source · 2 passages
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Use in source · Read passage · sequence_full.tex, line 310
Let h_0,…,h_L-1 be a finite tap vector and let a_m=0 for m<0. Ordinary taps form
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h_0,\ldots,h_{L-1}
H_{0,2}Use in source · 1 passage
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Use in source · Read passage · sequence_polyphase.tex, line 97
-1 · 2 · 1 . The entries arise from the same congruence condition. For instance an input in phase two shifted by one lands in phase zero of the next block, contributing 2t to H_0,2.
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H_{0,2}
H_1Use in source · 3 passages
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Use in source · Read passage · coupled_growth_full.tex, line 242
H_1(R) · ≥ c_-{w_1[q(R-1)+C]^ρ +w_2[r(R-1)+C]^ρ} · ≥ c_-(q^ρ w_1+r^ρ w_2)(R+C)^ρ =c_-w_1(R+C)^ρ.
Exact mathematical passage
\begin{aligned} H_1(R)&\ge c_-\{w_1[q(R-1)+C]^\rho +w_2[r(R-1)+C]^\rho\}\\ &\ge c_-(q^\rho w_1+r^\rho w_2)(R+C)^\rho =c_-w_1(R+C)^\rho. \end{aligned}
h_1Use in source · 2 passages
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Use in source · Read passage · dilation_full.tex, line 181
c_n(h,q,r)∈ Q[h_1,…,h_J,q_1,r_1,…,q_J,r_J].
Exact mathematical passage
c_n(h,q,r)\in\mathbb Q[h_1,\ldots,h_J,q_1,r_1,\ldots,q_J,r_J].
H_2Use in source · 2 passages
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Use in source · Read passage · coupled_growth_full.tex, line 319
H_1(e^t)≤ t+log2 +H_1(e^t-log(1/q))+H_2(e^t-log(1/r)).
Exact mathematical passage
H_1(e^t)\le t+\log2 +H_1(e^{t-\log(1/q)})+H_2(e^{t-\log(1/r)}).
H_4Use in source · 1 passage
Search aliases: H_4
Use in source · chapters_sequences.tex, line 110
_ladder T_n=∑_j=1^nj, H_n=∑_j=1^nT_j, Δ T_n=n, Δ H_n=T_n The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, Δ f_n=f_n-f_n-1. All plotted values use their displayed common index. At n=4, T_4=10 and H_4=20. The differences are 10-6=4 and 20-10=10. Separately, 4^2-3^2=7 and 4^3-3^3=37. Prefix summation raises polynomial degree; differencing lowers it. These…
Exact mathematical passage
H_4=20
\mathcal H_{a,\beta,Q}Use in source · 1 passage
Search aliases: H_a,β,Q mathcal beta
Use in source · Read passage · arithmetic_full.tex, line 385
The image H_a,β,Q=Uℓ^2, with Uc _ H= c _ℓ^2, is therefore a complete Hilbert space of entire functions. Its reproducing kernel is
Exact mathematical passage
\mathcal H_{a,\beta,Q}=U\ell^2
H_iUse in source · 6 passages
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Use in source · Read passage · coupled_growth_full.tex, line 170
Set F_1=U, F_2=V, and H_i(R)=log F_i(R). Since F_i(R)≥ R, these logarithms are positive for R>1. The type convention used here is
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H_i(R)=\log F_i(R)
H_{ij}Use in source · 2 passages
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Use in source · Read passage · sequence_polyphase.tex, line 77
For causal FIR taps h_ℓ, b_m=∑_ℓ h_ℓ a_m-ℓ with zero extension to negative indices, split A(z) and B(z) into b phases. Then B_i(t)=∑_jH_ij(t)A_j(t), where H_ij(t)=∑_substackℓ≥0
Exact mathematical passage
B_i(t)=\sum_jH_{ij}(t)A_j(t)
h_JUse in source · 1 passage
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Use in source · Read passage · dilation_full.tex, line 181
c_n(h,q,r)∈ Q[h_1,…,h_J,q_1,r_1,…,q_J,r_J].
Exact mathematical passage
c_n(h,q,r)\in\mathbb Q[h_1,\ldots,h_J,q_1,r_1,\ldots,q_J,r_J].
H_{jl}Use in source · 1 passage
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Use in source · Read passage · sequence_full.tex, line 360
Y_j(w)=∑_l=0^b-1H_jl(w)X_l(w), H_jl(w)=∑_substack0≤ t<L t+l≡ j (b) h_t w^(t+l-j)/b.
Exact mathematical passage
\boxed{Y_j(w)=\sum_{l=0}^{b-1}H_{jl}(w)X_l(w),\qquad H_{jl}(w)=\sum_{\substack{0\le t<L\\t+l\equiv j\,(b)}} h_t w^{(t+l-j)/b}.}
h_{L-1}Use in source · 1 passage
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Use in source · Read passage · sequence_full.tex, line 310
Let h_0,…,h_L-1 be a finite tap vector and let a_m=0 for m<0. Ordinary taps form
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h_0,\ldots,h_{L-1}
H_mUse in source · 1 passage
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Use in source · Read passage · notation.tex, line 48
item[Hankel order and polynomial degree.] For moments M_j=∫ x^jddν, the order-m matrix is H_m=[M_i+j]_0≤ i,j<m. It tests quadratic forms of real polynomials of degree at most m-1. At depth k, the matrix is [M_k+i+j]_0≤ i,j<m. An order-five test therefore covers degree four, not degree five. item[Signed quadratic form.] For real polynomials, Q_ν(f)=∫ f(x)^2ddν(x). On a positive measure it is a squared norm. On a signed measure it may be negative and is not a norm. item[Orbit spectral zeta Z_N.] L…
Exact mathematical passage
H_m=[M_{i+j}]_{0\le i,j<m}
h_{m-k}Use in source · 1 passage
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Use in source · Read passage · sequence_polyphase.tex, line 107
claimThree operations that must be kept distinct For an EGF E_a(z)=∑_ma_mz^m/m!, differentiation transports future coefficients: E_a^(k)=∑_ma_m+kz^m/m!. Causal ordinary taps instead act by repeated zero-constant integration: E_b=∑_ℓ h_ℓ I^ℓ E_a. Multiplication of two EGFs gives Hurwitz convolution, b_m=∑_k=0^mbinom mk a_kh_m-k. It is not ordinary sequence FIR convolution. Intrinsic nonlinear feedback taps alter the defining family and are a fourth operation with their own coefficient recurrence.
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b_m=\sum_{k=0}^m\binom mk a_kh_{m-k}
H_NUse in source · 7 passages
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Use in source · Read passage · spectral_full.tex, line 158
Let H_N(t)=d_N(-t)=E_N,1(t), and G_N(t)=H_N'(t)/H_N(t). Logarithmic differentiation of (S4), locally uniformly off the zeros, gives
Exact mathematical passage
H_N(t)=d_N(-t)=E_{N,1}(t)
h_NUse in source · 3 passages
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Use in source · Read passage · spectral_full.tex, line 70
h_N(x)=∑_j=1^N-1e^(ω^j-1)x, c_N=1-cos(2π/N)>0.
Exact mathematical passage
h_N(x)=\sum_{j=1}^{N-1}e^{(\omega^j-1)x}, \qquad c_N=1-\cos(2\pi/N)>0.
H_N'Use in source · 2 passages
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Use in source · Read passage · spectral_full.tex, line 158
Let H_N(t)=d_N(-t)=E_N,1(t), and G_N(t)=H_N'(t)/H_N(t). Logarithmic differentiation of (S4), locally uniformly off the zeros, gives
Exact mathematical passage
G_N(t)=H_N'(t)/H_N(t)
h_N'Use in source · 1 passage
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Use in source · Read passage · spectral_full.tex, line 77
F_N(x)=frace^xN(1+h_N(x)), |h_N(x)|≤(N-1)e^-c_Nx, |h_N'(x)|≤2(N-1)e^-c_Nx. tagS2
Exact mathematical passage
F_N(x)=\frac{e^x}{N}(1+h_N(x)),\qquad |h_N(x)|\le(N-1)e^{-c_Nx},\qquad |h_N'(x)|\le2(N-1)e^{-c_Nx}. \tag{S2}
h_tUse in source · 13 passages
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Use in source · Read passage · sequence_full.tex, line 89
For rational channel tuples (h_t,q_t,r_t), use
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(h_t,q_t,r_t)
h_\ellUse in source · 4 passages
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Use in source · Read passage · sequence_polyphase.tex, line 75
For causal FIR taps h_ℓ, b_m=∑_ℓ h_ℓ a_m-ℓ with zero extension to negative indices, split A(z) and B(z) into b phases. Then B_i(t)=∑_jH_ij(t)A_j(t), where H_ij(t)=∑_substackℓ≥0
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h_\ell
\hboxUse in source · 2 passages
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Use in source · Read passage · sequence_full.tex, line 105
B_n=H^nD^n(n+1)A_n∈ Z hboxfor all n.
Exact mathematical passage
\boxed{B_n=H^nD^{n(n+1)}A_n\in\mathbb Z\quad\hbox{for all }n.}
IUse in source · 10 passages
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Use in source · Read passage · arithmetic_full.tex, line 484
At real positive depth, the matrix (A_χ,ψ(s)) on any d distinct characters is positive semidefinite and has identical diagonal A_a,β,Q(s). Its normalized matrix tends to I_d, with operator norm error O_a,β,Q,d(s^-1/(2β)). Indeed, the off-diagonal entry bound controls the maximum row and column sums, and thus the operator norm. This gives quantitative separation of a fixed character bank. It is consistent with a complex or signed off-diagonal entry and makes no complete-monotonicity assertion for…
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I_d
iUse in source · 75 passages
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Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
I^tUse in source · 2 passages
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Use in source · Read passage · sequence_full.tex, line 321
B_rm EGF(z)=∑_t=0^L-1h_t I^tG(z), IG(z)=∫_0^zG(w) dw.
Exact mathematical passage
\boxed{B_{\rm EGF}(z)=\sum_{t=0}^{L-1}h_t I^tG(z),\qquad IG(z)=\int_0^zG(w)\,dw.}
I^\ellUse in source · 1 passage
Search aliases: I^ℓ ell
Use in source · Read passage · sequence_polyphase.tex, line 103
claimThree operations that must be kept distinct For an EGF E_a(z)=∑_ma_mz^m/m!, differentiation transports future coefficients: E_a^(k)=∑_ma_m+kz^m/m!. Causal ordinary taps instead act by repeated zero-constant integration: E_b=∑_ℓ h_ℓ I^ℓ E_a. Multiplication of two EGFs gives Hurwitz convolution, b_m=∑_k=0^mbinom mk a_kh_m-k. It is not ordinary sequence FIR convolution. Intrinsic nonlinear feedback taps alter the defining family and are a fourth operation with their own coefficient recurrence.
Exact mathematical passage
E_b=\sum_\ell h_\ell I^\ell E_a.
I_dUse in source · 1 passage
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Use in source · Read passage · arithmetic_full.tex, line 484
At real positive depth, the matrix (A_χ,ψ(s)) on any d distinct characters is positive semidefinite and has identical diagonal A_a,β,Q(s). Its normalized matrix tends to I_d, with operator norm error O_a,β,Q,d(s^-1/(2β)). Indeed, the off-diagonal entry bound controls the maximum row and column sums, and thus the operator norm. This gives quantitative separation of a fixed character bank. It is consistent with a complex or signed off-diagonal entry and makes no complete-monotonicity assertion for…
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I_d
I_KUse in source · 3 passages
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Use in source · Read passage · spectral_continuation.tex, line 61
Near zero write f(t)=(log E_N)'(t)=∑_k≥1α_kt^k-1, where α_k=(-1)^k+1 Z_N(k). For any integer K≥1, subtract K terms on [0,1]: I_K(s)= · ∫_0^1t^-s[f(t)-∑_k=1^Kα_kt^k-1]dd t +∑_k=1^Kfracα_kk-s · +∫_1^∞ t^-s[f(t)-frac1N t^1/N-1]dd t +frac1N(s-1/N). The first integral is holomorphic for Re s<K+1. The second is entire because its remainder decays exponentially in t^1/N. The factor sinπ s cancels the positive-integer poles. Increasing K covers the plane, leaving only 1/N. At nonpositive integers I_K is…
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\begin{aligned} I_K(s)={}&\int_0^1t^{-s}\left[f(t)-\sum_{k=1}^K\alpha_kt^{k-1}\right]\dd t +\sum_{k=1}^K\frac{\alpha_k}{k-s}\\ &+\int_1^\infty t^{-s}\left[f(t)-\frac1N t^{1/N-1}\right]\dd t +\frac{1}{N(s-1/N)}. \end{aligned}
JUse in source · 8 passages
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Use in source · chapters_sequences.tex, line 18
lessonSEQUENCES / 03Jacobsthal separates two modesS03_jacobsthal J_0=0, J_1=1, J_n+2=J_n+1+2J_n=frac2^n+2-(-1)^n+23 The rising bars are Jacobsthal numbers. The right panel removes their growing mode: 2^n-3J_n=(-1)^n. What remains is an alternating mode of constant magnitude. Both modes belong to the same second-order recurrence. At n=6, J_6=21 and 2^6-3·21=1. At n=7, J_7=43 and 2^7-3·43=-1. A recurrence can combine growth and oscillation. Splitting the modes explains the parity correction. OEIS …
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2^n-3J_n=(-1)^n
j^{-1}Use in source · 1 passage
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Use in source · Read passage · arithmetic_full.tex, line 138
The floor contributes an additional O(j^-1), which is absorbed by the displayed error. These are fixed-parameter asymptotics. They do not give a constant uniform in arbitrary moduli.
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O(j^{-1})
j^{-1/2}Use in source · 3 passages
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Use in source · Read passage · arithmetic_full.tex, line 111
n_j=fracjδ_Q(1+O_Q(j^-1/2)), λ_j=2δ_Q^βj^-β (1+O_Q,β(j^-1/2+j^-β)).
Exact mathematical passage
n_j=\frac{j}{\delta_Q}\left(1+O_Q(j^{-1/2})\right), \quad \lambda_j=2\delta_Q^{\beta}j^{-\beta} \left(1+O_{Q,\beta}(j^{-1/2}+j^{-\beta})\right).
j^{-\beta}Use in source · 3 passages
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Use in source · Read passage · arithmetic_full.tex, line 111
n_j=fracjδ_Q(1+O_Q(j^-1/2)), λ_j=2δ_Q^βj^-β (1+O_Q,β(j^-1/2+j^-β)).
Exact mathematical passage
n_j=\frac{j}{\delta_Q}\left(1+O_Q(j^{-1/2})\right), \quad \lambda_j=2\delta_Q^{\beta}j^{-\beta} \left(1+O_{Q,\beta}(j^{-1/2}+j^{-\beta})\right).
J_6Use in source · 1 passage
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Use in source · chapters_sequences.tex, line 19
lessonSEQUENCES / 03Jacobsthal separates two modesS03_jacobsthal J_0=0, J_1=1, J_n+2=J_n+1+2J_n=frac2^n+2-(-1)^n+23 The rising bars are Jacobsthal numbers. The right panel removes their growing mode: 2^n-3J_n=(-1)^n. What remains is an alternating mode of constant magnitude. Both modes belong to the same second-order recurrence. At n=6, J_6=21 and 2^6-3·21=1. At n=7, J_7=43 and 2^7-3·43=-1. A recurrence can combine growth and oscillation. Splitting the modes explains the parity correction. OEIS …
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J_6=21
J_7Use in source · 1 passage
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Use in source · chapters_sequences.tex, line 19
03Jacobsthal separates two modesS03_jacobsthal J_0=0, J_1=1, J_n+2=J_n+1+2J_n=frac2^n+2-(-1)^n+23 The rising bars are Jacobsthal numbers. The right panel removes their growing mode: 2^n-3J_n=(-1)^n. What remains is an alternating mode of constant magnitude. Both modes belong to the same second-order recurrence. At n=6, J_6=21 and 2^6-3·21=1. At n=7, J_7=43 and 2^7-3·43=-1. A recurrence can combine growth and oscillation. Splitting the modes explains the parity correction. OEIS A001045, indexed f…
Exact mathematical passage
J_7=43
J_nUse in source · 1 passage
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Use in source · chapters_sequences.tex, line 18
lessonSEQUENCES / 03Jacobsthal separates two modesS03_jacobsthal J_0=0, J_1=1, J_n+2=J_n+1+2J_n=frac2^n+2-(-1)^n+23 The rising bars are Jacobsthal numbers. The right panel removes their growing mode: 2^n-3J_n=(-1)^n. What remains is an alternating mode of constant magnitude. Both modes belong to the same second-order recurrence. At n=6, J_6=21 and 2^6-3·21=1. At n=7, J_7=43 and 2^7-3·43=-1. A recurrence can combine growth and oscillation. Splitting the modes explains the parity correction. OEIS …
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2^n-3J_n=(-1)^n
KUse in source · 48 passages
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Use in source · Read passage · main.tex, line 297
claimStrict positive complex kernel For distinct complex points s_i, K_ij=A(s_i+overlines_j) is strictly positive definite. A zero quadratic form would force a finite exponential sum to vanish at every λ_n. Their accumulation at zero invokes the identity theorem; derivatives at zero then give a nonsingular Vandermonde system. On the real axis, (-1)^jA^(j)(s)>0 for every j≥0.
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K_{ij}=A(s_i+\overline{s_j})
k^2Use in source · 9 passages
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Use in source · chapters_sequences.tex, line 88
lessonSEQUENCES / 13Apéry uses taps that depend on the indexS13_apery_recurrence (n+1)^3A_n+1=(2n+1)(17n^2+17n+5)A_n-n^3A_n-1 The prefix annotation begins the Apéry sequence defined by A_n=∑_k=0^nbinom nk^2binomn+kk^2. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at n=2. Using A_1=5 and A_2=73 gives 27A_3=535·73-8·5=39015. Therefore A_3=1445. The recurrence applies for n≥1, with A_0=1. This is a three-term recurrenc…
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A_n=\sum_{k=0}^{n}\binom nk^2\binom{n+k}{k}^2
K^3Use in source · 3 passages
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Use in source · Read passage · resolution_full.tex, line 465
This polynomial has degree |S|+m, equals one at x_2, and kills every positive atom through the cutoff. At each remaining positive node, the product of the absolute root factors is at most C_S, and the last factor is at most r_S^m. Dropping all other negative atoms and bounding the positive tail by the integral ∑_n>Kn^-4≤1/(3K^3) gives
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\sum_{n>K}n^{-4}\le1/(3K^3)
k^3Use in source · 3 passages
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Use in source · Read passage · coupled_inverse_full.tex, line 138
p y^2-frac15v_2ky+k^3=0,
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p y^2-\frac{15v_2}{k}y+k^3=0,
K_{-1}Use in source · 1 passage
Search aliases: K_-1
Use in source · Read passage · main.tex, line 450
claimAn endpoint mass is numerically delicate For the undeformed depth-zero measure, set ν_θ=ν+θδ_0, θ≥-1. Write h_j=∫ P_j(x)^2 dν(x) and K_m(x,y)=∑_j=0^mP_j(x)P_j(y)/h_j. The Uvarov formula~[source reference] is P_m^[θ](x)=P_m(x)- fracθ P_m(0)1+θ K_m-1(0,0)K_m-1(x,0), with K_-1=0 for degree zero. The mass at zero is recovered by 1≤frac1K_m(0,0)≤1+(M_0-1)(7/16)^2m. The lower bound comes from the zero atom, whose mass is one. The test polynomial (1-x)^m gives the upper bound since every other nod…
Exact mathematical passage
K_{-1}=0
K_4Use in source · 4 passages
Search aliases: K_4
Use in source · Read passage · resolution_full.tex, line 435
K_4≈1.50393441996×10^11, 12/K_4≈7.9790713217×10^-11>10^-12.
Exact mathematical passage
K_4\approx1.50393441996\times10^{11},\qquad 12/K_4\approx7.9790713217\times10^{-11}>10^{-12}.
K_dUse in source · 4 passages
Search aliases: K_d
Use in source · Read passage · resolution_full.tex, line 399
ℓ_i(x)=∏_j≠ ifracx-y_jy_i-y_j, b_i=∏_j≠ ifracmax(y_j,1-y_j)|y_i-y_j|, K_d=∑_i=0^dn_i^4b_i^2.
Exact mathematical passage
\ell_i(x)=\prod_{j\ne i}\frac{x-y_j}{y_i-y_j}, \quad b_i=\prod_{j\ne i}\frac{\max(y_j,1-y_j)}{|y_i-y_j|}, \quad K_d=\sum_{i=0}^{d}n_i^4b_i^2.
K_{ij}Use in source · 1 passage
Search aliases: K_ij
Use in source · Read passage · main.tex, line 297
claimStrict positive complex kernel For distinct complex points s_i, K_ij=A(s_i+overlines_j) is strictly positive definite. A zero quadratic form would force a finite exponential sum to vanish at every λ_n. Their accumulation at zero invokes the identity theorem; derivatives at zero then give a nonsingular Vandermonde system. On the real axis, (-1)^jA^(j)(s)>0 for every j≥0.
Exact mathematical passage
K_{ij}=A(s_i+\overline{s_j})
K_mUse in source · 4 passages
Search aliases: K_m
Use in source · Read passage · main.tex, line 444
claimAn endpoint mass is numerically delicate For the undeformed depth-zero measure, set ν_θ=ν+θδ_0, θ≥-1. Write h_j=∫ P_j(x)^2 dν(x) and K_m(x,y)=∑_j=0^mP_j(x)P_j(y)/h_j. The Uvarov formula~[source reference] is P_m^[θ](x)=P_m(x)- fracθ P_m(0)1+θ K_m-1(0,0)K_m-1(x,0), with K_-1=0 for degree zero. The mass at zero is recovered by 1≤frac1K_m(0,0)≤1+(M_0-1)(7/16)^2m. The lower bound comes from the zero atom, whose mass is one. The test polynomial (1-x)^m gives the upper bound since every other nod…
Exact mathematical passage
K_m(x,y)=\sum_{j=0}^{m}P_j(x)P_j(y)/h_j
K_{m-1}Use in source · 1 passage
Search aliases: K_m-1
Use in source · Read passage · main.tex, line 446
claimAn endpoint mass is numerically delicate For the undeformed depth-zero measure, set ν_θ=ν+θδ_0, θ≥-1. Write h_j=∫ P_j(x)^2 dν(x) and K_m(x,y)=∑_j=0^mP_j(x)P_j(y)/h_j. The Uvarov formula~[source reference] is P_m^[θ](x)=P_m(x)- fracθ P_m(0)1+θ K_m-1(0,0)K_m-1(x,0), with K_-1=0 for degree zero. The mass at zero is recovered by 1≤frac1K_m(0,0)≤1+(M_0-1)(7/16)^2m. The lower bound comes from the zero atom, whose mass is one. The test polynomial (1-x)^m gives the upper bound since every other nod…
Exact mathematical passage
P_m^{[\theta]}(x)=P_m(x)- \frac{\theta P_m(0)}{1+\theta K_{m-1}(0,0)}K_{m-1}(x,0),
LUse in source · 36 passages
Search aliases: L
Use in source · chapters_sequences.tex, line 4
startvolumeSequence atlasOEIS families and exact transforms lessonSEQUENCES / 01Two seeds, one Fibonacci recurrenceS01_fibonacci_lucas F_n+2=F_n+1+F_n, L_n+2=L_n+1+L_n, L_n^2-5F_n^2=4(-1)^n The two families share the same recurrence and have different seeds: (F_0,F_1)=(0,1) and (L_0,L_1)=(2,1). The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude. At n=7, F_7=13 and L_7=29. Thus 29^2-5·13^2=841-845=-4. Comparing terms at different indices w…
Exact mathematical passage
(L_0,L_1)=(2,1)
lUse in source · 12 passages
Search aliases: l
Use in source · Read passage · sequence_full.tex, line 105
B_n=H^nD^n(n+1)A_n∈ Z hboxfor all n.
Exact mathematical passage
\boxed{B_n=H^nD^{n(n+1)}A_n\in\mathbb Z\quad\hbox{for all }n.}
L^{1-2a}Use in source · 1 passage
Search aliases: L^1-2a
Use in source · Read passage · arithmetic_full.tex, line 269
E_L(t)=e^tλ_L+1fracL^1-2a2a-1.
Exact mathematical passage
E_L(t)=e^{t\lambda_{L+1}}\frac{L^{1-2a}}{2a-1}.
L_0Use in source · 1 passage
Search aliases: L_0
Use in source · chapters_sequences.tex, line 4
startvolumeSequence atlasOEIS families and exact transforms lessonSEQUENCES / 01Two seeds, one Fibonacci recurrenceS01_fibonacci_lucas F_n+2=F_n+1+F_n, L_n+2=L_n+1+L_n, L_n^2-5F_n^2=4(-1)^n The two families share the same recurrence and have different seeds: (F_0,F_1)=(0,1) and (L_0,L_1)=(2,1). The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude. At n=7, F_7=13 and L_7=29. Thus 29^2-5·13^2=841-845=-4. Comparing terms at different indices w…
Exact mathematical passage
(L_0,L_1)=(2,1)
L_1Use in source · 1 passage
Search aliases: L_1
Use in source · chapters_sequences.tex, line 4
startvolumeSequence atlasOEIS families and exact transforms lessonSEQUENCES / 01Two seeds, one Fibonacci recurrenceS01_fibonacci_lucas F_n+2=F_n+1+F_n, L_n+2=L_n+1+L_n, L_n^2-5F_n^2=4(-1)^n The two families share the same recurrence and have different seeds: (F_0,F_1)=(0,1) and (L_0,L_1)=(2,1). The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude. At n=7, F_7=13 and L_7=29. Thus 29^2-5·13^2=841-845=-4. Comparing terms at different indices w…
Exact mathematical passage
(L_0,L_1)=(2,1)
L_7Use in source · 1 passage
Search aliases: L_7
Use in source · chapters_sequences.tex, line 5
sonSEQUENCES / 01Two seeds, one Fibonacci recurrenceS01_fibonacci_lucas F_n+2=F_n+1+F_n, L_n+2=L_n+1+L_n, L_n^2-5F_n^2=4(-1)^n The two families share the same recurrence and have different seeds: (F_0,F_1)=(0,1) and (L_0,L_1)=(2,1). The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude. At n=7, F_7=13 and L_7=29. Thus 29^2-5·13^2=841-845=-4. Comparing terms at different indices would destroy this identity. A recurrence operator and its initi…
Exact mathematical passage
L_7=29
L_aUse in source · 5 passages
Search aliases: L_a
Use in source · Read passage · multi_prime.tex, line 68
pageheadingPRIME SIEVES / CONSTRUCTIVE WITNESSEvery negative prime atom has a finite detector For an integer a≥2, let L_a(x)=∏_n=1^a-1fracx-x_nx_a-x_n, C_a=L_a(1), r_a=frac1-x_a+11-x_a<1, f_a,m(x)=L_a(x)(frac1-x1-x_a)^m. This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects …
Exact mathematical passage
L_a(x)=\prod_{n=1}^{a-1}\frac{x-x_n}{x_a-x_n},\quad C_a=L_a(1),\quad r_a=\frac{1-x_{a+1}}{1-x_a}<1,
\lfloorUse in source · 2 passages
Search aliases: lfloor
Use in source · Read passage · arithmetic_full.tex, line 133
-fraclog(λ_lfloor cjrfloor/λ_j)log c =β+O_Q,β,c(j^-1/2+j^-β).
Exact mathematical passage
-\frac{\log(\lambda_{\lfloor cj\rfloor}/\lambda_j)}{\log c} =\beta+O_{Q,\beta,c}(j^{-1/2}+j^{-\beta}).
\limUse in source · 9 passages
Search aliases: lim
Use in source · Read passage · arithmetic_full.tex, line 63
λ_j=lim_t→∞fraclog R_j(-t)t, w_j=lim_t→∞e^-λ_jtR_j(-t), R_j+1(s)=R_j(s)-w_je^-λ_js.
Exact mathematical passage
\lambda_j=\lim_{t\to\infty}\frac{\log R_j(-t)}t, \qquad w_j=\lim_{t\to\infty}e^{-\lambda_jt}R_j(-t), \qquad R_{j+1}(s)=R_j(s)-w_je^{-\lambda_js}.
\lim_{j\to\infty}Use in source · 2 passages
Search aliases: lim_j→∞ lim to infty
Use in source · Read passage · arithmetic_full.tex, line 125
displaystyle β=-lim_j→∞fraclogλ_jlog j.
Exact mathematical passage
\boxed{\displaystyle \beta=-\lim_{j\to\infty}\frac{\log\lambda_j}{\log j}.}
\lim_{R\to\infty}Use in source · 4 passages
Search aliases: lim_R→∞ lim to infty
Use in source · Read passage · dilation_full.tex, line 39
lim_R→∞fracloglog M(R)log R=ρ.
Exact mathematical passage
\lim_{R\to\infty}\frac{\log\log M(R)}{\log R}=\rho.
\lim_{T\to\infty}Use in source · 1 passage
Search aliases: lim_T→∞ lim to infty
Use in source · Read passage · spectral_continuation.tex, line 75
At zero the finite parts give I_K(0)=lim_T→∞[log E_N(T)-T^1/N]=-log N. Since (sinπ s/π)'_s=0=1, this equals Z_N'(0). Absolute convergence at a smaller real exponent would give a holomorphic series across 1/N, contradicting the pole and proving the exact abscissa.
Exact mathematical passage
I_K(0)=\lim_{T\to\infty}[\log E_N(T)-T^{1/N}]=-\log N.
\lim_{t\to\infty}Use in source · 2 passages
Search aliases: lim_t→∞ lim to infty
Use in source · Read passage · arithmetic_full.tex, line 63
λ_j=lim_t→∞fraclog R_j(-t)t, w_j=lim_t→∞e^-λ_jtR_j(-t), R_j+1(s)=R_j(s)-w_je^-λ_js.
Exact mathematical passage
\lambda_j=\lim_{t\to\infty}\frac{\log R_j(-t)}t, \qquad w_j=\lim_{t\to\infty}e^{-\lambda_jt}R_j(-t), \qquad R_{j+1}(s)=R_j(s)-w_je^{-\lambda_js}.
\liminfUse in source · 4 passages
Search aliases: liminf
Use in source · Read passage · dilation_full.tex, line 31
c_-≤liminf_R→∞fraclog M(R)R^ρ ≤limsup_R→∞fraclog M(R)R^ρ≤ c_+.
Exact mathematical passage
c_-\le\liminf_{R\to\infty}\frac{\log M(R)}{R^\rho} \le\limsup_{R\to\infty}\frac{\log M(R)}{R^\rho}\le c_+.
\liminf_{R\to\infty}Use in source · 4 passages
Search aliases: liminf_R→∞ liminf to infty
Use in source · Read passage · dilation_full.tex, line 31
c_-≤liminf_R→∞fraclog M(R)R^ρ ≤limsup_R→∞fraclog M(R)R^ρ≤ c_+.
Exact mathematical passage
c_-\le\liminf_{R\to\infty}\frac{\log M(R)}{R^\rho} \le\limsup_{R\to\infty}\frac{\log M(R)}{R^\rho}\le c_+.
\operatorname{Log}Use in source · 2 passages
Search aliases: Log operatorname
Use in source · Read passage · spectral_full.tex, line 22
Z_N(s)=∑_j e^-sLogw_j
Exact mathematical passage
\mathcal Z_N(s)=\sum_j e^{-s\operatorname{Log}w_j}
\log_{10}Use in source · 2 passages
Search aliases: log_10 log
Use in source · chapters_sequences.tex, line 11
lessonSEQUENCES / 02Longer recurrence memoryS02_tribonacci_tetranacci T_n=T_n-1+T_n-2+T_n-3, Q_n=∑_j=1^4Q_n-j The seed table fixes the OEIS convention. Tribonacci starts 0,0,1; tetranacci starts 0,0,0,1. The curves use log_10(1+a_n) so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence. Tribonacci gives T_5=2+1+1=4. Tetranacci gives Q_7=4+2+1+1=8. The recurrence first applies at n=3 for T and n=4 for Q. These finite-memory recurre…
Exact mathematical passage
\log_{10}(1+a_n)
\LongleftrightarrowUse in source · 2 passages
Search aliases: Longleftrightarrow
Use in source · Read passage · coupled_growth_full.tex, line 431
U=V Longleftrightarrow {q,r}={a,b}.
Exact mathematical passage
U=V\quad\Longleftrightarrow\quad\{q,r\}=\{a,b\}.
\LongrightarrowUse in source · 1 passage
Search aliases: Longrightarrow
Use in source · Read passage · sequence_full.tex, line 129
c_n≥fracq r^2n-12n+1c_n-1 Longrightarrow c_n≥fracq^n r^n^23·5⋯(2n+1).
Exact mathematical passage
c_n\ge\frac{q r^{2n-1}}{2n+1}c_{n-1} \quad\Longrightarrow\quad c_n\ge\frac{q^n r^{n^2}}{3\cdot5\cdots(2n+1)}.
M_0Use in source · 7 passages
Search aliases: M_0
Use in source · Read passage · main.tex, line 378
ight: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P_1=x-fracM_1M_0, P_2=x^2-fracM_0M_3-M_1M_2Δx+ fracM_1M_3-M_2^2Δ. The infinitely many positive nodes make every Hankel matrix strictly positive definite throughout λ≥-1. Hence…
Exact mathematical passage
\Delta=M_0M_2-M_1^2
M_1Use in source · 1 passage
Search aliases: M_1
Use in source · Read passage · main.tex, line 379
s. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P_1=x-fracM_1M_0, P_2=x^2-fracM_0M_3-M_1M_2Δx+ fracM_1M_3-M_2^2Δ. The infinitely many positive nodes make every Hankel matrix strictly positive definite throughout λ≥-1. Hence all degrees exist, with…
Exact mathematical passage
P_1=x-\frac{M_1}{M_0},\qquad P_2=x^2-\frac{M_0M_3-M_1M_2}{\Delta}x+ \frac{M_1M_3-M_2^2}{\Delta}.
M_1^2Use in source · 1 passage
Search aliases: M_1^2
Use in source · Read passage · main.tex, line 378
ight: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P_1=x-fracM_1M_0, P_2=x^2-fracM_0M_3-M_1M_2Δx+ fracM_1M_3-M_2^2Δ. The infinitely many positive nodes make every Hankel matrix strictly positive definite throughout λ≥-1. Hence…
Exact mathematical passage
\Delta=M_0M_2-M_1^2
M_2Use in source · 2 passages
Search aliases: M_2
Use in source · Read passage · main.tex, line 378
ight: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P_1=x-fracM_1M_0, P_2=x^2-fracM_0M_3-M_1M_2Δx+ fracM_1M_3-M_2^2Δ. The infinitely many positive nodes make every Hankel matrix strictly positive definite throughout λ≥-1. Hence…
Exact mathematical passage
\Delta=M_0M_2-M_1^2
M_2^2Use in source · 1 passage
Search aliases: M_2^2
Use in source · Read passage · main.tex, line 379
s. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P_1=x-fracM_1M_0, P_2=x^2-fracM_0M_3-M_1M_2Δx+ fracM_1M_3-M_2^2Δ. The infinitely many positive nodes make every Hankel matrix strictly positive definite throughout λ≥-1. Hence all degrees exist, with…
Exact mathematical passage
P_1=x-\frac{M_1}{M_0},\qquad P_2=x^2-\frac{M_0M_3-M_1M_2}{\Delta}x+ \frac{M_1M_3-M_2^2}{\Delta}.
M_3Use in source · 1 passage
Search aliases: M_3
Use in source · Read passage · main.tex, line 379
s. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P_1=x-fracM_1M_0, P_2=x^2-fracM_0M_3-M_1M_2Δx+ fracM_1M_3-M_2^2Δ. The infinitely many positive nodes make every Hankel matrix strictly positive definite throughout λ≥-1. Hence all degrees exist, with…
Exact mathematical passage
P_1=x-\frac{M_1}{M_0},\qquad P_2=x^2-\frac{M_0M_3-M_1M_2}{\Delta}x+ \frac{M_1M_3-M_2^2}{\Delta}.
M_FUse in source · 1 passage
Search aliases: M_F
Use in source · Read passage · dilation_growth.tex, line 8
pageheadingDILATION / GROWTH THEOREMThe exact order has positive lower type The single-channel result extends to finite positive feedback banks. Here a bank has weights h_j>0 with ∑_jh_j=1, contractions 0<q_j,r_j<1, and normalized solution F'(z)=1+∑_jh_jF(q_jz)F(r_jz), F(0)=0. Write M(R)=M_F(R) and let q_j^ρ_j+r_j^ρ_j=1. Then, with ρ=max_jρ_j, 0<liminf_R→∞fraclog M(R)R^ρ ≤limsup_R→∞fraclog M(R)R^ρ<∞. Thus logarithmic growth is trapped between two positive multiples of R^ρ. In particular, lim_R→∞…
Exact mathematical passage
M(R)=M_F(R)
M_{F_i}Use in source · 3 passages
Search aliases: M_F_i
Use in source · Read passage · coupled_growth_full.tex, line 174
underlineτ_i=liminf_R→∞fraclog M_F_i(R)R^ρ, overlineτ_i=limsup_R→∞fraclog M_F_i(R)R^ρ.
Exact mathematical passage
\underline\tau_i=\liminf_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho}, \qquad \overline\tau_i=\limsup_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho}.
M_{i+j}Use in source · 2 passages
Search aliases: M_i+j
Use in source · Read passage · notation.tex, line 48
item[Hankel order and polynomial degree.] For moments M_j=∫ x^jddν, the order-m matrix is H_m=[M_i+j]_0≤ i,j<m. It tests quadratic forms of real polynomials of degree at most m-1. At depth k, the matrix is [M_k+i+j]_0≤ i,j<m. An order-five test therefore covers degree four, not degree five. item[Signed quadratic form.] For real polynomials, Q_ν(f)=∫ f(x)^2ddν(x). On a positive measure it is a squared norm. On a signed measure it may be negative and is not a norm. item[Orbit spectral zeta Z_N.] L…
Exact mathematical passage
H_m=[M_{i+j}]_{0\le i,j<m}
M_jUse in source · 5 passages
Search aliases: M_j
Use in source · Read passage · notation.tex, line 47
item[Hankel order and polynomial degree.] For moments M_j=∫ x^jddν, the order-m matrix is H_m=[M_i+j]_0≤ i,j<m. It tests quadratic forms of real polynomials of degree at most m-1. At depth k, the matrix is [M_k+i+j]_0≤ i,j<m. An order-five test therefore covers degree four, not degree five. item[Signed quadratic form.] For real polynomials, Q_ν(f)=∫ f(x)^2ddν(x). On a positive measure it is a squared norm. On a signed measure it may be negative and is not a norm. item[Orbit spectral zeta Z_N.] L…
Exact mathematical passage
M_j=\int x^j\dd\nu
M_{j+1}Use in source · 7 passages
Search aliases: M_j+1
Use in source · Read passage · arithmetic_full.tex, line 72
Every limit exists and recovers exactly the j-th support frequency and weight. Indeed, if M_j+1=∑_k>jw_k, then for every t≥0,
Exact mathematical passage
M_{j+1}=\sum_{k>j}w_k
M_j^{p,\lambda}Use in source · 1 passage
Search aliases: M_j^p,λ lambda
Use in source · Read passage · main.tex, line 372
for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P…
Exact mathematical passage
M_j^{p,\lambda}=\sum_{r=0}^{2j}(-1)^r\binom{2j}{r} \frac{\zeta(4+2r)}{\zeta(8+4r)} \left(1+\frac{\lambda}{p^{4+2r}+1}\right).
m_kUse in source · 1 passage
Search aliases: m_k
Use in source · Read passage · arithmetic_full.tex, line 402
The monomial moment matrices also have infinite rank: m_k=(-1)^kA^(k)(0)=∑_jw_jλ_j^k, and for every nonzero polynomial P, ∑_jw_j|P(λ_j)|^2>0, because a nonzero polynomial cannot vanish at infinitely many distinct frequencies. All finite derivative Hankel matrices are strictly positive definite.
Exact mathematical passage
m_k=(-1)^kA^{(k)}(0)=\sum_jw_j\lambda_j^k
M_{k+i+j}Use in source · 2 passages
Search aliases: M_k+i+j
Use in source · Read passage · main.tex, line 432
eheadingFAMILY 3 / TRANSPORTClassical operations with arithmetic coordinates The depth tilt x^k is a Christoffel transformation. Suppressing p,λ, P_m^(k+1)(x)=fracP_m+1^(k)(x) -[P_m+1^(k)(0)/P_m^(k)(0)]P_m^(k)(x)x. The numerator vanishes at zero. All roots are positive, so the displayed denominator coefficient is nonzero. For shifted Hankel determinants τ_m^(k)=det[M_k+i+j]_i,j<m, with τ_0^(k)=1, τ_m^(k)τ_m^(k+2)-(τ_m^(k+1))^2 =τ_m+1^(k)τ_m-1^(k+2)>0, m≥1. This is the Desnanot--Jacobi identity a…
Exact mathematical passage
\tau_m^{(k)}=\det[M_{k+i+j}]_{i,j<m}
M_{k+j}Use in source · 1 passage
Search aliases: M_k+j
Use in source · Read passage · sieve_full.tex, line 53
Depth-k moments are M_k+j. With v=π^-2, the even-zeta formula makes every D(4+2r) a rational multiple of v^r+2. Thus the moments belong to Q[v,λ_p:p∈ F], and monic orthogonal-polynomial coefficients belong to Q(v,λ_p:p∈ F). This is an exact field statement, with no decimal recognition step.
Exact mathematical passage
M_{k+j}
M_{T}Use in source · 1 passage
Search aliases: M_T
Use in source · Read passage · main.tex, line 112
pageheadingFAMILY 1 / GROWTHA scale-balance atlas ρ(T_q,r)=ρ where q^ρ+r^ρ=1. Here M_f(R)=max_|z|=R|f(z)| and ρ(f)=limsup_R→∞loglog M_f(R)/log R is the order of an entire function. Moreover log M_T(R)=O(R^ρ): the upper type is finite. smallfig02_order_atlasThe displayed values numerically solve the proved characteristic equation. Colors saturate at order 7; the upper corner contains larger orders. The dashed diagonal is the previously known one-scale family. @lll@ Scales · Order · Interpretation
Exact mathematical passage
\log M_{T}(R)=O(R^\rho)
M_UUse in source · 2 passages
Search aliases: M_U
Use in source · Read passage · coupled_growth_full.tex, line 165
M_U(R)=max_|z|=R|U(z)|=U(R), M_V(R)=V(R).
Exact mathematical passage
M_U(R)=\max_{|z|=R}|U(z)|=U(R),\qquad M_V(R)=V(R).
M_VUse in source · 2 passages
Search aliases: M_V
Use in source · Read passage · coupled_growth_full.tex, line 165
M_U(R)=max_|z|=R|U(z)|=U(R), M_V(R)=V(R).
Exact mathematical passage
M_U(R)=\max_{|z|=R}|U(z)|=U(R),\qquad M_V(R)=V(R).
\max_jUse in source · 7 passages
Search aliases: max_j max
Use in source · Read passage · dilation_full.tex, line 111
Q=max_jmax(q_j,r_j), m=min_jmin(q_j,r_j), Δ=log(1/m), B_+=max_jlogfrac1q_jr_j+log2.
Exact mathematical passage
Q=\max_j\max(q_j,r_j),\quad m=\min_j\min(q_j,r_j),\quad \Delta=\log(1/m),\quad B_+=\max_j\log\frac1{q_jr_j}+\log2.
\max_{j\text{ active}}Use in source · 1 passage
Search aliases: max_j active max text
Use in source · Read passage · main.tex, line 200
y many channels j=1,…,J, with h_j>0, ∑_jh_j=1, and 0<q_j,r_j<1. Define F_Θ'(z)=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz), F_Θ(0)=0. These taps act inside the nonlinear definition. With μ_1=∑_jh_jq_jr_j and μ_2=∑_jh_jq_jr_j(q_j^2+r_j^2), F_Θ(z)=z+fracμ_13z^3+fracμ_1μ_215z^5+O(z^7). If q_j^ρ_j+r_j^ρ_j=1, then ρ(F_Θ)=max_j activeρ_j. The positive weights alter local coefficients. The most expansive active channel determines the asymptotic order. A zero-weight channel is removed before taking the maximum. fig07_we…
Exact mathematical passage
\boxed{\rho(F_\Theta)=\max_{j\text{ active}}\rho_j.}
\max_{j:h_j>0}Use in source · 1 passage
Search aliases: max_j:h_j>0 max
Use in source · Read passage · dilation_full.tex, line 21
ρ=max_j:h_j>0ρ_j, q_j^ρ_j+r_j^ρ_j=1.
Exact mathematical passage
\rho=\max_{j:h_j>0}\rho_j,\qquad q_j^{\rho_j}+r_j^{\rho_j}=1.
\max_{|z|\le R}Use in source · 2 passages
Search aliases: max_|z|≤ R max le
Use in source · Read passage · coupled_lab_full.tex, line 75
max_|z|≤ R|F(z)-F_N(z)| ≤ B(S)(frac RS)^2N+3.
Exact mathematical passage
\max_{|z|\le R}|F(z)-F_N(z)| \le B(S)\left(\frac RS\right)^{2N+3}.
\max_{|z|\le x}Use in source · 1 passage
Search aliases: max_|z|≤ x max le
Use in source · Read passage · coupled_lab_full.tex, line 59
Analytic continuation by the integral equations expands each known disk by the factor 1/Q, so both solutions are entire. Positive coefficients give max_|z|≤ x|F(z)|=F(x) for either component. The integral equations therefore imply
Exact mathematical passage
\max_{|z|\le x}|F(z)|=F(x)
\midUse in source · 20 passages
Search aliases: mid
Use in source · Read passage · main.tex, line 238
(n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At depth zero the corresponding mass-one atom is absent. In particular, A(0)=Z_Q(2a)-1, where Z_Q(v)=∑_substackn≥1 (n,Q)=1μ(n)^2n^-v =fracζ(v)ζ(2v)∏_pmid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit wor…
Exact mathematical passage
Z_Q(v)=\sum_{\substack{n\ge1\\(n,Q)=1}}\mu(n)^2n^{-v} =\frac{\zeta(v)}{\zeta(2v)}\prod_{p\mid Q}(1+p^{-v})^{-1},\qquad v>1.
\minUse in source · 10 passages
Search aliases: min
Use in source · Read passage · main.tex, line 407
B_mA_m^-1/2. Every determinant root is real and below -1; possible coefficient poles occur there, although numerator cancellation is possible. fig05_sieve_boundaryThe line is a proved bound for p=2; dots are ordinary high-precision pencil roots. A defect λ=-1.001 eventually violates positivity even though it can pass low-order checks. The nearest root is -1-δ_m, with δ_m=min_substackf∈ R[x]∖{0} deg f<m frac∫ f^2 dν_nonp∫ f^2 dν_p>0, δ_mdownarrow0. To prove the limit, approximate a continuous bum…
Exact mathematical passage
\delta_m=\min_{\substack{f\in\mathbb R[x]\setminus\{0\}\\\deg f<m}} \frac{\int f^2\dd\nu_{\mathrm{non}p}}{\int f^2\dd\nu_p}>0, \qquad\delta_m\downarrow0.
\min_{0\ne f\in\mathbb R[x],\ \deg f<m}Use in source · 1 passage
Search aliases: min_0≠ f∈ R[x], deg f<m min ne in mathbb deg
Use in source · Read passage · sieve_full.tex, line 212
δ_m=min_0≠ f∈ R[x], deg f<m frac∫ f^2 dν_p-free∫ f^2 dν_p-divisible>0. tag6
Exact mathematical passage
\delta_m=\min_{0\ne f\in\mathbb R[x],\ \deg f<m} \frac{\int f^2\,d\nu_{p\text{-free}}}{\int f^2\,d\nu_{p\text{-divisible}}}>0. \tag{6}
\min_jUse in source · 3 passages
Search aliases: min_j min
Use in source · Read passage · dilation_full.tex, line 111
Q=max_jmax(q_j,r_j), m=min_jmin(q_j,r_j), Δ=log(1/m), B_+=max_jlogfrac1q_jr_j+log2.
Exact mathematical passage
Q=\max_j\max(q_j,r_j),\quad m=\min_j\min(q_j,r_j),\quad \Delta=\log(1/m),\quad B_+=\max_j\log\frac1{q_jr_j}+\log2.
\min_{\substack{f\in\mathbb R[x]\setminus\{0\}\\\deg f<m}}Use in source · 2 passages
Search aliases: min_substackf∈ R[x]∖{0} deg f<m min substack in mathbb setminus deg
Use in source · Read passage · main.tex, line 407
B_mA_m^-1/2. Every determinant root is real and below -1; possible coefficient poles occur there, although numerator cancellation is possible. fig05_sieve_boundaryThe line is a proved bound for p=2; dots are ordinary high-precision pencil roots. A defect λ=-1.001 eventually violates positivity even though it can pass low-order checks. The nearest root is -1-δ_m, with δ_m=min_substackf∈ R[x]∖{0} deg f<m frac∫ f^2 dν_nonp∫ f^2 dν_p>0, δ_mdownarrow0. To prove the limit, approximate a continuous bum…
Exact mathematical passage
\delta_m=\min_{\substack{f\in\mathbb R[x]\setminus\{0\}\\\deg f<m}} \frac{\int f^2\dd\nu_{\mathrm{non}p}}{\int f^2\dd\nu_p}>0, \qquad\delta_m\downarrow0.
NUse in source · 195 passages
Search aliases: N
Use in source · REPORT.md, line 28
|∑_n≥ Nc_nz^2n+1| ≤ frac|z|ρ^N1-ρ, |z|^2<2.
Exact mathematical passage
\left|\sum_{n\ge N}c_nz^{2n+1}\right| \le \frac{|z|\rho^N}{1-\rho},\qquad |z|^2<2.
N^{-1}Use in source · 1 passage
Search aliases: N^-1
Use in source · Read passage · spectral_full.tex, line 67
This equality follows by expanding the exponentials and using N^-1∑_jω^jm=1 when Nmid m, and zero otherwise. For real x≥0, the power series makes F_N(x)>0. Put
Exact mathematical passage
N^{-1}\sum_j\omega^{jm}=1
n^{-2}Use in source · 8 passages
Search aliases: n^-2
Use in source · Read passage · main.tex, line 360
Prime sieves and signed measures. pageheadingFAMILY 3An infinite prime shelf changes the packets For a prime p and λ≥-1, let x_n=(1-n^-2)^2, ν_p,λ=∑_n≥1μ(n)^2n^-4 (1+λ1_pmid n)δ_x_n. Define monic orthogonal polynomials P_m^(k;p,λ) for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the or…
Exact mathematical passage
x_n=(1-n^{-2})^2,\qquad \nu_{p,\lambda}=\sum_{n\ge1}\mu(n)^2n^{-4} (1+\lambda\mathbf1_{p\mid n})\delta_{x_n}.
n^{-2a}Use in source · 7 passages
Search aliases: n^-2a
Use in source · Read passage · main.tex, line 230
MonographPartIIArithmetic depth and signed measuresComplex parameters, positive kernels and the boundaries of positivity. Arithmetic at complex depth. pageheadingFAMILY 2Arithmetic energy at complex depth For real a>1/2, β>0, and integer Q≥1, define A_a,β,Q(s)=∑_substackn≥2 (n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At d…
Exact mathematical passage
\boxed{\A(s)=\sum_{\substack{n\ge2\\(n,Q)=1}} \mu(n)^2n^{-2a}(1-n^{-\beta})^{2s},\qquad s\in\mathbb C.}
n^{-4}Use in source · 13 passages
Search aliases: n^-4
Use in source · Read passage · main.tex, line 360
Prime sieves and signed measures. pageheadingFAMILY 3An infinite prime shelf changes the packets For a prime p and λ≥-1, let x_n=(1-n^-2)^2, ν_p,λ=∑_n≥1μ(n)^2n^-4 (1+λ1_pmid n)δ_x_n. Define monic orthogonal polynomials P_m^(k;p,λ) for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the or…
Exact mathematical passage
x_n=(1-n^{-2})^2,\qquad \nu_{p,\lambda}=\sum_{n\ge1}\mu(n)^2n^{-4} (1+\lambda\mathbf1_{p\mid n})\delta_{x_n}.
n^{-s}Use in source · 1 passage
Search aliases: n^-s
Use in source · Read passage · sieve_full.tex, line 32
Z_F(s;λ) =∑_n≥1μ(n)^2n^-sκ_F(n;λ) =D(s)∏_p∈ F(1+fracλ_pp^s+1).
Exact mathematical passage
Z_F(s;\boldsymbol\lambda) =\sum_{n\ge1}\mu(n)^2n^{-s}\kappa_F(n;\boldsymbol\lambda) =D(s)\prod_{p\in F}\left(1+\frac{\lambda_p}{p^s+1}\right).
n^{-v}Use in source · 1 passage
Search aliases: n^-v
Use in source · Read passage · main.tex, line 238
(n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At depth zero the corresponding mass-one atom is absent. In particular, A(0)=Z_Q(2a)-1, where Z_Q(v)=∑_substackn≥1 (n,Q)=1μ(n)^2n^-v =fracζ(v)ζ(2v)∏_pmid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit wor…
Exact mathematical passage
Z_Q(v)=\sum_{\substack{n\ge1\\(n,Q)=1}}\mu(n)^2n^{-v} =\frac{\zeta(v)}{\zeta(2v)}\prod_{p\mid Q}(1+p^{-v})^{-1},\qquad v>1.
n^{-\beta}Use in source · 6 passages
Search aliases: n^-β beta
Use in source · Read passage · main.tex, line 230
MonographPartIIArithmetic depth and signed measuresComplex parameters, positive kernels and the boundaries of positivity. Arithmetic at complex depth. pageheadingFAMILY 2Arithmetic energy at complex depth For real a>1/2, β>0, and integer Q≥1, define A_a,β,Q(s)=∑_substackn≥2 (n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At d…
Exact mathematical passage
\boxed{\A(s)=\sum_{\substack{n\ge2\\(n,Q)=1}} \mu(n)^2n^{-2a}(1-n^{-\beta})^{2s},\qquad s\in\mathbb C.}
n^2Use in source · 3 passages
Search aliases: n^2
Use in source · Read passage · sequence_full.tex, line 129
c_n≥fracq r^2n-12n+1c_n-1 Longrightarrow c_n≥fracq^n r^n^23·5⋯(2n+1).
Exact mathematical passage
c_n\ge\frac{q r^{2n-1}}{2n+1}c_{n-1} \quad\Longrightarrow\quad c_n\ge\frac{q^n r^{n^2}}{3\cdot5\cdots(2n+1)}.
n^4Use in source · 1 passage
Search aliases: n^4
Use in source · Read passage · resolution_visual.tex, line 148
pageheadingRESOLUTION ATLAS / NEGATIVE ATOMSA degree-seven witness exposes a signed defect Let x_n=(1-n^-2)^2, and change the prime-two sieve strength to λ_2=-1-δ. For a real polynomial f, the associated quadratic form is Q_δ(f)=∑_substackn≥1 μ(n)^2=1 n odd fracf(x_n)^2n^4 -δ∑_substackn≥1 μ(n)^2=1 n even fracf(x_n)^2n^4. Every even squarefree atom has negative mass. At δ=10^-12, the polynomial below certifies that this is detectable at finite degree.
Exact mathematical passage
Q_\delta(f)=\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ odd}}} \frac{f(x_n)^2}{n^4} -\delta\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ even}}} \frac{f(x_n)^2}{n^4}.
n_+Use in source · 4 passages
Search aliases: n_+
Use in source · Read passage · resolution_full.tex, line 449
Let S be exactly the set of all odd squarefree labels n≤ K, where K is an odd squarefree integer, and let n_+ be the next such label. Assume n_+>2, and define
Exact mathematical passage
n_+
n_0Use in source · 1 passage
Search aliases: n_0
Use in source · Read passage · resolution_full.tex, line 396
Choose d+1 distinct odd squarefree labels n_0,…,n_d, including 1, and set y_i=x_n_i. Their Lagrange polynomials are
Exact mathematical passage
n_0,\ldots,n_d
n_1Use in source · 2 passages
Search aliases: n_1
Use in source · Read passage · arithmetic_full.tex, line 26
The logarithm is real. The omitted n=1 atom is not restored in any inverse formula. Let n_1<n_2<⋯ be the support integers, and abbreviate λ_j=λ_n_j, w_j=n_j^-2a. Then λ_1>λ_2>⋯>0, λ_j→0, and ∑_jw_j=M=Z_Q(2a)-1<∞, where
Exact mathematical passage
n_1<n_2<\cdots
n_2Use in source · 2 passages
Search aliases: n_2
Use in source · Read passage · arithmetic_full.tex, line 26
The logarithm is real. The omitted n=1 atom is not restored in any inverse formula. Let n_1<n_2<⋯ be the support integers, and abbreviate λ_j=λ_n_j, w_j=n_j^-2a. Then λ_1>λ_2>⋯>0, λ_j→0, and ∑_jw_j=M=Z_Q(2a)-1<∞, where
Exact mathematical passage
n_1<n_2<\cdots
n_dUse in source · 1 passage
Search aliases: n_d
Use in source · Read passage · resolution_full.tex, line 396
Choose d+1 distinct odd squarefree labels n_0,…,n_d, including 1, and set y_i=x_n_i. Their Lagrange polynomials are
Exact mathematical passage
n_0,\ldots,n_d
n_iUse in source · 3 passages
Search aliases: n_i
Use in source · Read passage · resolution_full.tex, line 397
Choose d+1 distinct odd squarefree labels n_0,…,n_d, including 1, and set y_i=x_n_i. Their Lagrange polynomials are
Exact mathematical passage
y_i=x_{n_i}
n_i^{-4}Use in source · 3 passages
Search aliases: n_i^-4
Use in source · Read passage · resolution_full.tex, line 410
|f(x)|^2≤ (∑_i=0^dn_i^-4f(y_i)^2) (∑_i=0^dn_i^4ℓ_i(x)^2) ≤ K_d∑_i=0^dn_i^-4f(y_i)^2.
Exact mathematical passage
|f(x)|^2\le \left(\sum_{i=0}^{d}n_i^{-4}f(y_i)^2\right) \left(\sum_{i=0}^{d}n_i^4\ell_i(x)^2\right) \le K_d\sum_{i=0}^{d}n_i^{-4}f(y_i)^2.
n_i^4Use in source · 3 passages
Search aliases: n_i^4
Use in source · Read passage · resolution_full.tex, line 399
ℓ_i(x)=∏_j≠ ifracx-y_jy_i-y_j, b_i=∏_j≠ ifracmax(y_j,1-y_j)|y_i-y_j|, K_d=∑_i=0^dn_i^4b_i^2.
Exact mathematical passage
\ell_i(x)=\prod_{j\ne i}\frac{x-y_j}{y_i-y_j}, \quad b_i=\prod_{j\ne i}\frac{\max(y_j,1-y_j)}{|y_i-y_j|}, \quad K_d=\sum_{i=0}^{d}n_i^4b_i^2.
n_jUse in source · 13 passages
Search aliases: n_j
Use in source · Read passage · arithmetic_full.tex, line 27
The logarithm is real. The omitted n=1 atom is not restored in any inverse formula. Let n_1<n_2<⋯ be the support integers, and abbreviate λ_j=λ_n_j, w_j=n_j^-2a. Then λ_1>λ_2>⋯>0, λ_j→0, and ∑_jw_j=M=Z_Q(2a)-1<∞, where
Exact mathematical passage
\lambda_j=\lambda_{n_j}
n_j^{-2a}Use in source · 2 passages
Search aliases: n_j^-2a
Use in source · Read passage · arithmetic_full.tex, line 27
The logarithm is real. The omitted n=1 atom is not restored in any inverse formula. Let n_1<n_2<⋯ be the support integers, and abbreviate λ_j=λ_n_j, w_j=n_j^-2a. Then λ_1>λ_2>⋯>0, λ_j→0, and ∑_jw_j=M=Z_Q(2a)-1<∞, where
Exact mathematical passage
w_j=n_j^{-2a}
n_j^{-\beta}Use in source · 1 passage
Search aliases: n_j^-β beta
Use in source · Read passage · arithmetic_inverse.tex, line 12
Let the support integers be n_1<n_2<⋯, frequencies λ_j=-2log(1-n_j^-β), and weights w_j=n_j^-2a. Starting at R_1=A, peel the ordered atoms: λ_j=lim_t→∞fraclog R_j(-t)t, w_j=lim_t→∞e^-λ_jtR_j(-t), R_j+1(s)=R_j(s)-w_je^-λ_js. Every limit follows from the isolated largest remaining frequency and the positive summable tail. Thus the exact function determines its spectral list.
Exact mathematical passage
\lambda_j=-2\log(1-n_j^{-\beta})
N_QUse in source · 4 passages
Search aliases: N_Q
Use in source · Read passage · arithmetic_full.tex, line 104
N_Q(x):=#{2≤ n≤ x:μ(n)^2=1,(n,Q)=1} =δ_Qx+O_Q(sqrtx).
Exact mathematical passage
N_Q(x):=\#\{2\le n\le x:\mu(n)^2=1,(n,Q)=1\} =\delta_Qx+O_Q(\sqrt{x}).
\nmidUse in source · 3 passages
Search aliases: nmid
Use in source · Read passage · arithmetic_full.tex, line 165
A prime p lies in the recovered integer support exactly when pnmid Q: primes are squarefree, and no other restriction is imposed. Hence
Exact mathematical passage
p\nmid Q
\mathrm{non}Use in source · 3 passages
Search aliases: non mathrm
Use in source · Read passage · main.tex, line 389
pageheadingFAMILY 3 / BOUNDARYFinite positivity can conceal a signed defect Let ν_nonp and ν_p be the restrictions of the undeformed measure ν_p,0 to indices not divisible by p and divisible by p, respectively. Then ν_p,λ=ν_nonp+(1+λ)ν_p. Let A_m,B_m be the order-m Hankel matrices of these two unscaled positive measures. Both are positive definite. Thus τ_m(λ)=det(A_m+(1+λ)B_m) =det A_m∏_i=1^m[1+(1+λ)η_i], where η_i>0 are the eigenvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and…
Exact mathematical passage
\nu_{\mathrm{non}p}
OUse in source · 34 passages
Search aliases: O
Use in source · Read passage · main.tex, line 95
nctionsComputed Taylor curves. Each entire curve uses terms through z^321; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent. For T_q,r(z)=∑_n≥0c_nz^2n+1, c_0=1, coefficient matching gives c_n=frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1c_ic_n-1-i, n≥1. In particular, T_q,r(z)=z+fracqr3z^3+ fracq^2r^2(q^2+r^2)15z^5+O(z^7). The coefficients are positive and bounded by tangent coefficients, giving a convergent germ. If the …
Exact mathematical passage
T_{q,r}(z)=z+\frac{qr}{3}z^3+ \frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).
oUse in source · 1 passage
Search aliases: o
Use in source · Read passage · sequence_full.tex, line 105
B_n=H^nD^n(n+1)A_n∈ Z hboxfor all n.
Exact mathematical passage
\boxed{B_n=H^nD^{n(n+1)}A_n\in\mathbb Z\quad\hbox{for all }n.}
O_{a,\beta,Q,d}Use in source · 1 passage
Search aliases: O_a,β,Q,d beta
Use in source · Read passage · arithmetic_full.tex, line 485
At real positive depth, the matrix (A_χ,ψ(s)) on any d distinct characters is positive semidefinite and has identical diagonal A_a,β,Q(s). Its normalized matrix tends to I_d, with operator norm error O_a,β,Q,d(s^-1/(2β)). Indeed, the off-diagonal entry bound controls the maximum row and column sums, and thus the operator norm. This gives quantitative separation of a fixed character bank. It is consistent with a complex or signed off-diagonal entry and makes no complete-monotonicity assertion for…
Exact mathematical passage
O_{a,\beta,Q,d}(s^{-1/(2\beta)})
O_{a,\beta,Q,\epsilon}Use in source · 2 passages
Search aliases: O_a,β,Q,ϵ beta epsilon
Use in source · Read passage · main.tex, line 315
Let α=frac2a-1β, δ_Q=frac1ζ(2)∏_pmid Qfracpp+1. As |s|→∞ in any closed sector |arg s|≤π/2-ϵ with 0<ϵ<π/2, A_a,β,Q(s)=fracδ_QβB(α,2s+1) +O_a,β,Q,ϵ(|s|^-(2a-1/2)/β). In particular, A_a,β,Q(s)∼fracδ_QΓ(α)β(2s)^-α. The principal power is used in the right half-plane. The beta expression is an asymptotic continuous-density model, not an identity with the arithmetic sum.
Exact mathematical passage
\boxed{\A(s)=\frac{\delta_Q}{\beta}B(\alpha,2s+1) +O_{a,\beta,Q,\epsilon}\left(|s|^{-(2a-1/2)/\beta}\right).}
O_{a,\beta,Q,\varepsilon}Use in source · 1 passage
Search aliases: O_a,β,Q,ε beta varepsilon
Use in source · Read passage · arithmetic_full.tex, line 441
A_χ,ψ(s) =O_a,β,Q,ε(|s|^-(2a-1/2)/β), fracA_χ,ψ(s)A_a,β,Q(s) =O_a,β,Q,ε(|s|^-1/(2β)).
Exact mathematical passage
A_{\chi,\psi}(s) =O_{a,\beta,Q,\varepsilon}\bigl(|s|^{-(2a-1/2)/\beta}\bigr), \qquad \frac{A_{\chi,\psi}(s)}{A_{a,\beta,Q}(s)} =O_{a,\beta,Q,\varepsilon}\bigl(|s|^{-1/(2\beta)}\bigr).
O_{a,\beta,\varepsilon}Use in source · 1 passage
Search aliases: O_a,β,ε beta varepsilon
Use in source · Read passage · arithmetic_full.tex, line 474
which is O_a,β,ε(|s|^-γ) because u≥ |s|sinε. The principal diagonal has the nonzero leading term δ_QΓ(α)β^-1(2s)^-α, α=(2a-1)/β, by the base sector theorem. Its modulus is bounded below by a positive fixed-sector multiple of |s|^-α for sufficiently large |s|. Division gives the second bound, since γ-α=1/(2β).
Exact mathematical passage
O_{a,\beta,\varepsilon}(|s|^{-\gamma})
O_QUse in source · 8 passages
Search aliases: O_Q
Use in source · Read passage · main.tex, line 328
claimProof mechanism Squarefree inclusion--exclusion gives ∑_substackn≤ x (n,Q)=1μ(n)^2 =δ_Q x+O_Q(sqrtx). Integrate against g_s(x)=x^-2a(1-x^-β)^2s on [1,∞). The density term becomes the beta integral after y=x^-β. Partial integration bounds the error by C_Q∫_1^∞sqrtx|g_s'(x)| d x. The two derivative terms are beta integrals with decay exponent (2a-1/2)/β; |s|/Re s stays bounded in the stated sector. This gives the error term and then the leading asymptotic.
Exact mathematical passage
\sum_{\substack{n\le x\\(n,Q)=1}}\mu(n)^2 =\delta_Q x+O_Q(\sqrt{x}).
O_{Q,\beta}Use in source · 2 passages
Search aliases: O_Q,β beta
Use in source · Read passage · arithmetic_full.tex, line 111
n_j=fracjδ_Q(1+O_Q(j^-1/2)), λ_j=2δ_Q^βj^-β (1+O_Q,β(j^-1/2+j^-β)).
Exact mathematical passage
n_j=\frac{j}{\delta_Q}\left(1+O_Q(j^{-1/2})\right), \quad \lambda_j=2\delta_Q^{\beta}j^{-\beta} \left(1+O_{Q,\beta}(j^{-1/2}+j^{-\beta})\right).
O_{Q,\beta,c}Use in source · 1 passage
Search aliases: O_Q,β,c beta
Use in source · Read passage · arithmetic_full.tex, line 133
-fraclog(λ_lfloor cjrfloor/λ_j)log c =β+O_Q,β,c(j^-1/2+j^-β).
Exact mathematical passage
-\frac{\log(\lambda_{\lfloor cj\rfloor}/\lambda_j)}{\log c} =\beta+O_{Q,\beta,c}(j^{-1/2}+j^{-\beta}).
\overline tUse in source · 4 passages
Search aliases: overline t overline
Use in source · Read passage · arithmetic_full.tex, line 389
K(s,t)=A(s+overline t).
Exact mathematical passage
K(s,t)=A(s+\overline t).
\overline{s_j}Use in source · 1 passage
Search aliases: overlines_j overline
Use in source · Read passage · main.tex, line 297
claimStrict positive complex kernel For distinct complex points s_i, K_ij=A(s_i+overlines_j) is strictly positive definite. A zero quadratic form would force a finite exponential sum to vanish at every λ_n. Their accumulation at zero invokes the identity theorem; derivatives at zero then give a nonsingular Vandermonde system. On the real axis, (-1)^jA^(j)(s)>0 for every j≥0.
Exact mathematical passage
K_{ij}=A(s_i+\overline{s_j})
\overline\tauUse in source · 8 passages
Search aliases: overlineτ overline tau
Use in source · Read passage · coupled_growth_full.tex, line 174
underlineτ_i=liminf_R→∞fraclog M_F_i(R)R^ρ, overlineτ_i=limsup_R→∞fraclog M_F_i(R)R^ρ.
Exact mathematical passage
\underline\tau_i=\liminf_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho}, \qquad \overline\tau_i=\limsup_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho}.
\overline{\boldsymbol\tau}Use in source · 1 passage
Search aliases: overlineτ overline boldsymbol tau
Use in source · Read passage · coupled_visual.tex, line 81
For F_1=U and F_2=V, define underlineτ_i=liminf_R→∞fraclog M_F_i(R)R^ρ, overlineτ_i=limsup_R→∞fraclog M_F_i(R)R^ρ. If P(ρ)w=w with w>0, then both type vectors have precisely these Perron proportions: underlineτ=underlineτ w, overlineτ=overlineτ w, qquad0<underlineτ≤overlineτ<∞. For the worked pair, both the lower type and the upper type of V are twice the corresponding type of U. Either both ordinary type limits exist or neither does. Equality of the lower and upper types remains open. The finit…
Exact mathematical passage
\boxed{\underline{\boldsymbol\tau}=\underline\tau\,w, \qquad\overline{\boldsymbol\tau}=\overline\tau\,w, \qquad0<\underline\tau\le\overline\tau<\infty.}
\overline\tau_1Use in source · 2 passages
Search aliases: overlineτ_1 overline tau
Use in source · Read passage · coupled_growth_full.tex, line 363
(underlineτ_1,underlineτ_2)^T · =underlineτ w, (overlineτ_1,overlineτ_2)^T · =overlineτ w.
Exact mathematical passage
\boxed{\begin{aligned} (\underline\tau_1,\underline\tau_2)^T&=\underline\tau\,w,\\ (\overline\tau_1,\overline\tau_2)^T&=\overline\tau\,w. \end{aligned}}
\overline\tau_2Use in source · 2 passages
Search aliases: overlineτ_2 overline tau
Use in source · Read passage · coupled_growth_full.tex, line 363
(underlineτ_1,underlineτ_2)^T · =underlineτ w, (overlineτ_1,overlineτ_2)^T · =overlineτ w.
Exact mathematical passage
\boxed{\begin{aligned} (\underline\tau_1,\underline\tau_2)^T&=\underline\tau\,w,\\ (\overline\tau_1,\overline\tau_2)^T&=\overline\tau\,w. \end{aligned}}
\overline\tau_iUse in source · 3 passages
Search aliases: overlineτ_i overline tau
Use in source · Read passage · coupled_growth_full.tex, line 174
underlineτ_i=liminf_R→∞fraclog M_F_i(R)R^ρ, overlineτ_i=limsup_R→∞fraclog M_F_i(R)R^ρ.
Exact mathematical passage
\underline\tau_i=\liminf_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho}, \qquad \overline\tau_i=\limsup_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho}.
\overline\psiUse in source · 2 passages
Search aliases: overlineψ overline psi
Use in source · Read passage · arithmetic_full.tex, line 413
Let χ,ψ be Dirichlet characters modulo the same fixed integer Q, extended by zero outside the units, and put η=χoverlineψ. The base extension is
Exact mathematical passage
\eta=\chi\overline\psi
\mathcal PUse in source · 2 passages
Search aliases: P mathcal
Use in source · Read passage · sequence_full.tex, line 268
P_b,jG(z)=frac1b∑_k=0^b-1ω^-jkG(ω^kz) =∑_m≡ j (b)a_mfracz^mm!.
Exact mathematical passage
\boxed{ \mathcal P_{b,j}G(z)=\frac1b\sum_{k=0}^{b-1}\omega^{-jk}G(\omega^kz) =\sum_{m\equiv j\,(b)}a_m\frac{z^m}{m!}.}
p^{-2}Use in source · 5 passages
Search aliases: p^-2
Use in source · Read passage · resolution_full.tex, line 229
λ_p=-2log(1-p^-2),
Exact mathematical passage
\lambda_p=-2\log(1-p^{-2}),
p^{-2a}Use in source · 1 passage
Search aliases: p^-2a
Use in source · Read passage · arithmetic_full.tex, line 341
There is no n=1 term in this difference. Bound the k-th derivative of every exponential by λ_p^ke^Rλ_p and sum. Since λ_p→0 and p^-2a→0, the bound tends to zero. This proves the derivative convergence and the claimed instability.
Exact mathematical passage
p^{-2a}\to0
p^{-4}Use in source · 6 passages
Search aliases: p^-4
Use in source · Read passage · main.tex, line 414
deg f<m frac∫ f^2 dν_nonp∫ f^2 dν_p>0, δ_mdownarrow0. To prove the limit, approximate a continuous bump supported at the isolated node x_p by polynomials on [0,1]. The numerator tends to zero while the denominator retains the p^-4 atom. Hence every λ<-1 eventually produces a negative finite-degree quadratic form.
Exact mathematical passage
p^{-4}
p^{-4-2k}Use in source · 1 passage
Search aliases: p^-4-2k
Use in source · Read passage · resolution_full.tex, line 250
This is the inherited arithmetic inverse theorem, now read as a precision requirement. For fixed R, the envelope is of order p^-4-2k. Each additional derivative suppresses the large-prime difference even further in absolute units. This does not rule out methods that rescale derivatives and account for their corresponding noise amplification.
Exact mathematical passage
p^{-4-2k}
p^{-v}Use in source · 3 passages
Search aliases: p^-v
Use in source · Read passage · main.tex, line 238
(n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At depth zero the corresponding mass-one atom is absent. In particular, A(0)=Z_Q(2a)-1, where Z_Q(v)=∑_substackn≥1 (n,Q)=1μ(n)^2n^-v =fracζ(v)ζ(2v)∏_pmid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit wor…
Exact mathematical passage
Z_Q(v)=\sum_{\substack{n\ge1\\(n,Q)=1}}\mu(n)^2n^{-v} =\frac{\zeta(v)}{\zeta(2v)}\prod_{p\mid Q}(1+p^{-v})^{-1},\qquad v>1.
p^2Use in source · 16 passages
Search aliases: p^2
Use in source · REPORT.md, line 39
D=k^2u_2-p^2v_2 =fracpkar15(q^2-b^2)(qb-ar).
Exact mathematical passage
D=k^2u_2-p^2v_2 =\frac{pkar}{15}(q^2-b^2)(qb-ar).
p^{2a}Use in source · 3 passages
Search aliases: p^2a
Use in source · Read passage · arithmetic_full.tex, line 318
displaystyle sup_|s|≤ R|A_a,β,Q^(k)(s)-A_a,β,Qp^(k)(s)| ≤ λ_p^k e^Rλ_p fracZ_Q(2a)p^2a+1.
Exact mathematical passage
\boxed{\displaystyle \sup_{|s|\le R}|A_{a,\beta,Q}^{(k)}(s)-A_{a,\beta,Qp}^{(k)}(s)| \le \lambda_p^k e^{R\lambda_p} \frac{Z_Q(2a)}{p^{2a}+1}.}
p^3Use in source · 5 passages
Search aliases: p^3
Use in source · Read passage · main.tex, line 158
he common order. The first two nontrivial coefficients and the growth law force B^2s^2=p, s^2=u/2, 2s^ρ=q^ρ+r^ρ=1. Consequently 2(fracq^2+r^22)^ρ/2=q^ρ+r^ρ. For ρ≠2, strict convexity or concavity of x^ρ/2 forces q=r. For the exceptional case ρ=2, we have u=1, s^2=1/2, B^2=2p. The two z^7 coefficients become fracp^3(3-p^2)315 and frac11p^31260. Their equality forces p=1/2 and again (q-r)^2=u-2p=0. This completes the exceptional-order case.
Exact mathematical passage
\frac{p^3(3-p^2)}{315}\quad\text{and}\quad\frac{11p^3}{1260}.
p^4Use in source · 24 passages
Search aliases: p^4
Use in source · Read passage · main.tex, line 423
For p=2, test with f(x)=x(1-x)^m-2, m≥2: δ_m≤frac409681(frac16M_017-1) (frac272567)^2(m-2), M_0=105π^-4. This follows from x_2=9/16 and x_n≥64/81 at nonzero odd nodes. Exactly δ_1=p^4; the limit is an infinite-resolution phenomenon.
Exact mathematical passage
\delta_1=p^4
p^{4+2r}Use in source · 3 passages
Search aliases: p^4+2r
Use in source · Read passage · main.tex, line 372
for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P…
Exact mathematical passage
M_j^{p,\lambda}=\sum_{r=0}^{2j}(-1)^r\binom{2j}{r} \frac{\zeta(4+2r)}{\zeta(8+4r)} \left(1+\frac{\lambda}{p^{4+2r}+1}\right).
p^sUse in source · 2 passages
Search aliases: p^s
Use in source · Read passage · sieve_full.tex, line 32
Z_F(s;λ) =∑_n≥1μ(n)^2n^-sκ_F(n;λ) =D(s)∏_p∈ F(1+fracλ_pp^s+1).
Exact mathematical passage
Z_F(s;\boldsymbol\lambda) =\sum_{n\ge1}\mu(n)^2n^{-s}\kappa_F(n;\boldsymbol\lambda) =D(s)\prod_{p\in F}\left(1+\frac{\lambda_p}{p^s+1}\right).
p^\rhoUse in source · 1 passage
Search aliases: p^ρ rho
Use in source · Read passage · coupled_growth_full.tex, line 392
Here p(R-1) tends to infinity continuously and [p(R-1)]^ρ/R^ρ→ p^ρ, which justifies each scaled liminf. The liminf of a sum is at least the sum of the liminfs. All quantities are positive and finite by Theorem F.2.
Exact mathematical passage
[p(R-1)]^\rho/R^\rho\to p^\rho
p_*Use in source · 2 passages
Search aliases: p_*
Use in source · Read passage · main.tex, line 252
ion. Right: initial squarefree frequencies and weights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If p_* is the least prime not dividing Q, then τ=-2log(1-p_*^-β), M=Z_Q(2a)-1, |A(s)|≤ M e^τmax(-Re s,0). This proves local uniform convergence and entire continuation. The firs…
Exact mathematical passage
p_*
p_*^{-\beta}Use in source · 1 passage
Search aliases: p_*^-β beta
Use in source · Read passage · main.tex, line 253
ights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If p_* is the least prime not dividing Q, then τ=-2log(1-p_*^-β), M=Z_Q(2a)-1, |A(s)|≤ M e^τmax(-Re s,0). This proves local uniform convergence and entire continuation. The first atom dominates A(-t) as t→∞, so the entire func…
Exact mathematical passage
\tau=-2\log(1-p_*^{-\beta}),\qquad M=\Z(2a)-1,
P_1Use in source · 3 passages
Search aliases: P_1
Use in source · Read passage · main.tex, line 379
s. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P_1=x-fracM_1M_0, P_2=x^2-fracM_0M_3-M_1M_2Δx+ fracM_1M_3-M_2^2Δ. The infinitely many positive nodes make every Hankel matrix strictly positive definite throughout λ≥-1. Hence all degrees exist, with…
Exact mathematical passage
P_1=x-\frac{M_1}{M_0},\qquad P_2=x^2-\frac{M_0M_3-M_1M_2}{\Delta}x+ \frac{M_1M_3-M_2^2}{\Delta}.
p_{10}Use in source · 1 passage
Search aliases: p_10
Use in source · chapters_sequences.tex, line 75
Bernoulli data need both coordinatesS11_bernoulli_pair fracte^t-1=∑_n≥0B_nfract^nn!, B_n=fracp_nq_n, q_n>0 The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as 0/1, and this guide uses B_1=-1/2. At n=10, p_10=5 and q_10=66, so B_10=5/66. Its Taylor coefficient is (5/66)/10!. At n=12, B_12=-691/2730. An OEIS integer numerator sequence does not by itsel…
Exact mathematical passage
p_{10}=5
P_2Use in source · 3 passages
Search aliases: P_2
Use in source · Read passage · main.tex, line 379
s. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P_1=x-fracM_1M_0, P_2=x^2-fracM_0M_3-M_1M_2Δx+ fracM_1M_3-M_2^2Δ. The infinitely many positive nodes make every Hankel matrix strictly positive definite throughout λ≥-1. Hence all degrees exist, with…
Exact mathematical passage
P_1=x-\frac{M_1}{M_0},\qquad P_2=x^2-\frac{M_0M_3-M_1M_2}{\Delta}x+ \frac{M_1M_3-M_2^2}{\Delta}.
P_{b,j}Use in source · 1 passage
Search aliases: P_b,j
Use in source · Read passage · sequence_polyphase.tex, line 66
pageheadingEXACT LABORATORY / PROJECTORSPolyphase sections preserve the full sequence For an ordinary generating function G(z)=∑_m≥0a_mz^m and integer b≥1, write G(z)=∑_j=0^b-1z^jG_j(z^b), G_j(t)=∑_n≥0a_bn+jt^n. The root projector is P_b,jG(z)=frac1b∑_k=0^b-1ω^-jkG(ω^kz) =z^jG_j(z^b), ω=e^2π i/b. The same projector works on an EGF, but the section retains its original factorials: ∑_na_bn+jz^bn+j/(bn+j)!. Removing those denominators changes the representation and requires an explicit transport. R…
Exact mathematical passage
P_{b,j}G(z)=\frac1b\sum_{k=0}^{b-1}\omega^{-jk}G(\omega^kz) =z^jG_j(z^b),\qquad\omega=e^{2\pi i/b}.
\mathcal P_{b,j}Use in source · 2 passages
Search aliases: P_b,j mathcal
Use in source · Read passage · sequence_full.tex, line 268
P_b,jG(z)=frac1b∑_k=0^b-1ω^-jkG(ω^kz) =∑_m≡ j (b)a_mfracz^mm!.
Exact mathematical passage
\boxed{ \mathcal P_{b,j}G(z)=\frac1b\sum_{k=0}^{b-1}\omega^{-jk}G(\omega^kz) =\sum_{m\equiv j\,(b)}a_m\frac{z^m}{m!}.}
\mathcal P_{b,l}Use in source · 1 passage
Search aliases: P_b,l mathcal
Use in source · Read passage · sequence_full.tex, line 279
∑_j=0^b-1 P_b,j=I, P_b,j P_b,l=δ_jl P_b,j.
Exact mathematical passage
\sum_{j=0}^{b-1}\mathcal P_{b,j}=I,\qquad \mathcal P_{b,j}\mathcal P_{b,l}=\delta_{jl}\mathcal P_{b,j}.
P_jUse in source · 7 passages
Search aliases: P_j
Use in source · Read passage · main.tex, line 282
claimNo polynomial-coefficient linear ODE Suppose ∑_j=0^rP_j(s)A^(j)(s)=0, where the P_j are polynomials and are not all zero. At s=-t, divide by the exponential of the first support atom. The remaining tail is exponentially smaller, so the polynomial ∑_jP_j(-t)(-λ_p_*)^j tends to zero exponentially and must vanish identically. Remove that atom and repeat. Every support frequency therefore satisfies ∑_jP_j(s)(-λ_n)^j=0 for all s. For fixed s, this polynomial in λ has infinitely many distinct roo…
Exact mathematical passage
\sum_{j=0}^{r}P_j(s)A^{(j)}(s)=0
P_mUse in source · 3 passages
Search aliases: P_m
Use in source · Read passage · main.tex, line 446
claimAn endpoint mass is numerically delicate For the undeformed depth-zero measure, set ν_θ=ν+θδ_0, θ≥-1. Write h_j=∫ P_j(x)^2 dν(x) and K_m(x,y)=∑_j=0^mP_j(x)P_j(y)/h_j. The Uvarov formula~[source reference] is P_m^[θ](x)=P_m(x)- fracθ P_m(0)1+θ K_m-1(0,0)K_m-1(x,0), with K_-1=0 for degree zero. The mass at zero is recovered by 1≤frac1K_m(0,0)≤1+(M_0-1)(7/16)^2m. The lower bound comes from the zero atom, whose mass is one. The test polynomial (1-x)^m gives the upper bound since every other nod…
Exact mathematical passage
P_m^{[\theta]}(x)=P_m(x)- \frac{\theta P_m(0)}{1+\theta K_{m-1}(0,0)}K_{m-1}(x,0),
P_{m+1}^{(k)}Use in source · 1 passage
Search aliases: P_m+1^(k)
Use in source · Read passage · main.tex, line 427
pageheadingFAMILY 3 / TRANSPORTClassical operations with arithmetic coordinates The depth tilt x^k is a Christoffel transformation. Suppressing p,λ, P_m^(k+1)(x)=fracP_m+1^(k)(x) -[P_m+1^(k)(0)/P_m^(k)(0)]P_m^(k)(x)x. The numerator vanishes at zero. All roots are positive, so the displayed denominator coefficient is nonzero. For shifted Hankel determinants τ_m^(k)=det[M_k+i+j]_i,j<m, with τ_0^(k)=1, τ_m^(k)τ_m^(k+2)-(τ_m^(k+1))^2 =τ_m+1^(k)τ_m-1^(k+2)>0, m≥1. This is the Desnanot--Jacobi identit…
Exact mathematical passage
P_m^{(k+1)}(x)=\frac{P_{m+1}^{(k)}(x) -[P_{m+1}^{(k)}(0)/P_m^{(k)}(0)]P_m^{(k)}(x)}{x}.
P_m^{(k)}Use in source · 1 passage
Search aliases: P_m^(k)
Use in source · Read passage · main.tex, line 427
pageheadingFAMILY 3 / TRANSPORTClassical operations with arithmetic coordinates The depth tilt x^k is a Christoffel transformation. Suppressing p,λ, P_m^(k+1)(x)=fracP_m+1^(k)(x) -[P_m+1^(k)(0)/P_m^(k)(0)]P_m^(k)(x)x. The numerator vanishes at zero. All roots are positive, so the displayed denominator coefficient is nonzero. For shifted Hankel determinants τ_m^(k)=det[M_k+i+j]_i,j<m, with τ_0^(k)=1, τ_m^(k)τ_m^(k+2)-(τ_m^(k+1))^2 =τ_m+1^(k)τ_m-1^(k+2)>0, m≥1. This is the Desnanot--Jacobi identit…
Exact mathematical passage
P_m^{(k+1)}(x)=\frac{P_{m+1}^{(k)}(x) -[P_{m+1}^{(k)}(0)/P_m^{(k)}(0)]P_m^{(k)}(x)}{x}.
P_m^{(k+1)}Use in source · 1 passage
Search aliases: P_m^(k+1)
Use in source · Read passage · main.tex, line 427
pageheadingFAMILY 3 / TRANSPORTClassical operations with arithmetic coordinates The depth tilt x^k is a Christoffel transformation. Suppressing p,λ, P_m^(k+1)(x)=fracP_m+1^(k)(x) -[P_m+1^(k)(0)/P_m^(k)(0)]P_m^(k)(x)x. The numerator vanishes at zero. All roots are positive, so the displayed denominator coefficient is nonzero. For shifted Hankel determinants τ_m^(k)=det[M_k+i+j]_i,j<m, with τ_0^(k)=1, τ_m^(k)τ_m^(k+2)-(τ_m^(k+1))^2 =τ_m+1^(k)τ_m-1^(k+2)>0, m≥1. This is the Desnanot--Jacobi identit…
Exact mathematical passage
P_m^{(k+1)}(x)=\frac{P_{m+1}^{(k)}(x) -[P_{m+1}^{(k)}(0)/P_m^{(k)}(0)]P_m^{(k)}(x)}{x}.
P_m^{(k;F,\boldsymbol\lambda)}Use in source · 1 passage
Search aliases: P_m^(k;F,λ) boldsymbol lambda
Use in source · Read passage · sieve_full.tex, line 93
The positive monic families P_m^(k;F,λ) consequently exist at all degrees throughout the closed orthant. Their Hankel determinants never vanish there, including at simultaneous deletion endpoints. Each finite-degree coefficient is analytic in a neighborhood of every point of that orthant. This neighborhood is a finite-degree statement; the next theorem explains why no common enlargement of the positivity domain works at all degrees.
Exact mathematical passage
P_m^{(k;F,\boldsymbol\lambda)}
P_m^{(k;p,\lambda)}Use in source · 1 passage
Search aliases: P_m^(k;p,λ) lambda
Use in source · Read passage · main.tex, line 365
Prime sieves and signed measures. pageheadingFAMILY 3An infinite prime shelf changes the packets For a prime p and λ≥-1, let x_n=(1-n^-2)^2, ν_p,λ=∑_n≥1μ(n)^2n^-4 (1+λ1_pmid n)δ_x_n. Define monic orthogonal polynomials P_m^(k;p,λ) for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the or…
Exact mathematical passage
P_m^{(k;p,\lambda)}
P_m^{[\theta]}Use in source · 1 passage
Search aliases: P_m^[θ] theta
Use in source · Read passage · main.tex, line 446
claimAn endpoint mass is numerically delicate For the undeformed depth-zero measure, set ν_θ=ν+θδ_0, θ≥-1. Write h_j=∫ P_j(x)^2 dν(x) and K_m(x,y)=∑_j=0^mP_j(x)P_j(y)/h_j. The Uvarov formula~[source reference] is P_m^[θ](x)=P_m(x)- fracθ P_m(0)1+θ K_m-1(0,0)K_m-1(x,0), with K_-1=0 for degree zero. The mass at zero is recovered by 1≤frac1K_m(0,0)≤1+(M_0-1)(7/16)^2m. The lower bound comes from the zero atom, whose mass is one. The test polynomial (1-x)^m gives the upper bound since every other nod…
Exact mathematical passage
P_m^{[\theta]}(x)=P_m(x)- \frac{\theta P_m(0)}{1+\theta K_{m-1}(0,0)}K_{m-1}(x,0),
P_{S,3}Use in source · 2 passages
Search aliases: P_S,3
Use in source · Read passage · resolution_full.tex, line 497
and P_S,3 has degree seven. The exact rational right side of (RA-10) is strictly negative and is approximately
Exact mathematical passage
P_{S,3}
P_{S,m}Use in source · 2 passages
Search aliases: P_S,m
Use in source · Read passage · resolution_full.tex, line 452
P_S,m(x)=∏_n∈ Sfracx-x_nx_2-x_n (frac1-x1-x_2)^m, C_S=∏_n∈ Sfrac1-x_n|x_2-x_n|, r_S=frac1-x_n_+1-x_2<1.
Exact mathematical passage
P_{S,m}(x)=\prod_{n\in S}\frac{x-x_n}{x_2-x_n} \left(\frac{1-x}{1-x_2}\right)^m,
\pmodUse in source · 3 passages
Search aliases: pmod
Use in source · Read passage · sequence_full.tex, line 51
u,v≥ n, u≡ v≡ npmod2, u+v≤ n(n+1).
Exact mathematical passage
u,v\ge n,\qquad u\equiv v\equiv n\pmod2,\qquad u+v\le n(n+1).
\preceqUse in source · 2 passages
Search aliases: preceq
Use in source · Read passage · dilation_full.tex, line 226
0preceq c_n(h)preceq d_n(Q)(∑_jh_j)^n.
Exact mathematical passage
0\preceq c_n(h)\preceq d_n(Q)\Bigl(\sum_jh_j\Bigr)^n.
\mathrm{prime}Use in source · 1 passage
Search aliases: prime mathrm
Use in source · Read passage · arithmetic_full.tex, line 168
displaystyle rad(Q)=∏_substackp prime p∉{n_j:j≥1}p.
Exact mathematical passage
\boxed{\displaystyle \operatorname{rad}(Q)=\prod_{\substack{p\ \mathrm{prime}\\ p\notin\{n_j:j\ge1\}}}p.}
QUse in source · 156 passages
Search aliases: Q
Use in source · chapters_sequences.tex, line 12
ryS02_tribonacci_tetranacci T_n=T_n-1+T_n-2+T_n-3, Q_n=∑_j=1^4Q_n-j The seed table fixes the OEIS convention. Tribonacci starts 0,0,1; tetranacci starts 0,0,0,1. The curves use log_10(1+a_n) so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence. Tribonacci gives T_5=2+1+1=4. Tetranacci gives Q_7=4+2+1+1=8. The recurrence first applies at n=3 for T and n=4 for Q. These finite-memory recurrences have three and four feedback taps. Th…
Exact mathematical passage
Q_7=4+2+1+1=8
\mathbb QUse in source · 5 passages
Search aliases: Q mathbb
Use in source · Read passage · main.tex, line 377
two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P_1=x-fracM_1M_0, P_2=x^2-fracM_0M_3-M_1M_2Δx+ fracM_1M_3-M_2^2Δ. The infinitely many positive nodes make every Hankel matrix strictly positi…
Exact mathematical passage
\mathbb Q(\pi^{-2},\lambda)
Q'Use in source · 1 passage
Search aliases: Q'
Use in source · Read passage · arithmetic_full.tex, line 48
a=a', β=β', rad(Q)=rad(Q').
Exact mathematical passage
a=a',\qquad \beta=\beta',\qquad \operatorname{rad}(Q)=\operatorname{rad}(Q').
Q^{-K}Use in source · 1 passage
Search aliases: Q^-K
Use in source · Read passage · dilation_growth.tex, line 65
claimExplicit upper constants are available too Set Q=max_jmax(q_j,r_j), m=min_jmin(q_j,r_j), Δ=log(1/m), and B_+=max_jlog(1/(q_jr_j))+log2. Choose an integer K with Q^-K≥3/m and put V_K=(2^K+1-1)log2+(2^K+1-K-2)log(1/Q), c_+=e^-ρ(V_K+1+Δ+B_+). Then log M(R)≤ c_+R^ρ-log R-B_+ for R≥ e. The tangent majorant M(1)<2, followed by the integral recurrence on radii Q^-k, bounds the initial seed by V_K. The logarithmic-radius comparison then supplies the displayed upper constant. The full induction is i…
Exact mathematical passage
Q^{-K}\ge3/m
Q^{-k}Use in source · 5 passages
Search aliases: Q^-k
Use in source · Read passage · dilation_full.tex, line 131
These constants are intentionally conservative, and require no numerical evaluation of an unknown solution. The tangent majorant gives M(1)<2. At R_k=Q^-k, the integral equation gives
Exact mathematical passage
R_k=Q^{-k}
Q^{-N}Use in source · 2 passages
Search aliases: Q^-N
Use in source · Read passage · dilation_full.tex, line 118
Choose an integer N≥0 with Q^-N≥3/m, and put
Exact mathematical passage
Q^{-N}\ge3/m
q^2Use in source · 17 passages
Search aliases: q^2
Use in source · REPORT.md, line 39
D=k^2u_2-p^2v_2 =fracpkar15(q^2-b^2)(qb-ar).
Exact mathematical passage
D=k^2u_2-p^2v_2 =\frac{pkar}{15}(q^2-b^2)(qb-ar).
q^{2i+1}Use in source · 10 passages
Search aliases: q^2i+1
Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
q^3Use in source · 4 passages
Search aliases: q^3
Use in source · Read passage · sequence_full.tex, line 71
A_0= · 1, A_1= · 2qr, A_2= · 8q^2r^2(q^2+r^2), A_3= · q^3r^3[48(q^2+r^2)(q^4+r^4)+80q^2r^2].
Exact mathematical passage
\begin{aligned} A_0={}&1,\\ A_1={}&2qr,\\ A_2={}&8q^2r^2(q^2+r^2),\\ A_3={}&q^3r^3\bigl[48(q^2+r^2)(q^4+r^4)+80q^2r^2\bigr]. \end{aligned}
q^4Use in source · 4 passages
Search aliases: q^4
Use in source · Read passage · sequence_full.tex, line 71
A_0= · 1, A_1= · 2qr, A_2= · 8q^2r^2(q^2+r^2), A_3= · q^3r^3[48(q^2+r^2)(q^4+r^4)+80q^2r^2].
Exact mathematical passage
\begin{aligned} A_0={}&1,\\ A_1={}&2qr,\\ A_2={}&8q^2r^2(q^2+r^2),\\ A_3={}&q^3r^3\bigl[48(q^2+r^2)(q^4+r^4)+80q^2r^2\bigr]. \end{aligned}
Q^kUse in source · 1 passage
Search aliases: Q^k
Use in source · Read passage · coupled_lab_full.tex, line 66
Start at the first member Q^kx<1 of this finite contraction ladder, use x/(1-x^2) there, and repeatedly apply B(x)=x(1+B(Qx)^2) backward. This defines the rational pathdisk_majorant bound for rational parameters and radius. The implementation has explicit recursion and integer size guards, which raise errors rather than returning a weakened result.
Exact mathematical passage
Q^kx<1
q^nUse in source · 3 passages
Search aliases: q^n
Use in source · Read passage · sequence_full.tex, line 129
c_n≥fracq r^2n-12n+1c_n-1 Longrightarrow c_n≥fracq^n r^n^23·5⋯(2n+1).
Exact mathematical passage
c_n\ge\frac{q r^{2n-1}}{2n+1}c_{n-1} \quad\Longrightarrow\quad c_n\ge\frac{q^n r^{n^2}}{3\cdot5\cdots(2n+1)}.
q^{n/(2n+1)}Use in source · 1 passage
Search aliases: q^n/(2n+1)
Use in source · Read passage · sequence_full.tex, line 137
c_n^1/(2n+1)≥ q^n/(2n+1)r^n^2/(2n+1)(2n+1)^-n/(2n+1)⟶∞.
Exact mathematical passage
c_n^{1/(2n+1)}\ge q^{n/(2n+1)}r^{n^2/(2n+1)}(2n+1)^{-n/(2n+1)}\longrightarrow\infty.
q^sUse in source · 5 passages
Search aliases: q^s
Use in source · Read passage · coupled_growth_full.tex, line 113
P(s)=q^s · r^s a^s · b^s.
Exact mathematical passage
P(s)=\begin{pmatrix}q^s&r^s\\a^s&b^s\end{pmatrix}.
q^uUse in source · 1 passage
Search aliases: q^u
Use in source · Read passage · sequence_full.tex, line 49
Polynomial theorem. For every n, A_n∈ Z[q,r], all its nonzero coefficients are positive integers, and A_n(q,r)=A_n(r,q). Every monomial q^u r^v satisfies
Exact mathematical passage
q^u r^v
Q_*Use in source · 4 passages
Search aliases: Q_*
Use in source · Read passage · dilation_full.tex, line 97
For a rigorous propagation, put Q_* =max(q,r)<1 and d=(1-Q_*)B+Q_*L>0. Every child of every R≥ B is at most R-d. Induction on the intervals [B+kd,B+(k+1)d], beginning with the already controlled seed, proves the comparison for all larger radii. Continuity deals with common endpoints. All power comparisons use only monotonicity of x^ρ for positive x; no convexity assumption is made, so the argument also covers ρ<1.
Exact mathematical passage
Q_* =\max(q,r)<1
q_1Use in source · 1 passage
Search aliases: q_1
Use in source · Read passage · dilation_full.tex, line 181
c_n(h,q,r)∈ Q[h_1,…,h_J,q_1,r_1,…,q_J,r_J].
Exact mathematical passage
c_n(h,q,r)\in\mathbb Q[h_1,\ldots,h_J,q_1,r_1,\ldots,q_J,r_J].
q_{10}Use in source · 1 passage
Search aliases: q_10
Use in source · chapters_sequences.tex, line 75
need both coordinatesS11_bernoulli_pair fracte^t-1=∑_n≥0B_nfract^nn!, B_n=fracp_nq_n, q_n>0 The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as 0/1, and this guide uses B_1=-1/2. At n=10, p_10=5 and q_10=66, so B_10=5/66. Its Taylor coefficient is (5/66)/10!. At n=12, B_12=-691/2730. An OEIS integer numerator sequence does not by itself specify the r…
Exact mathematical passage
q_{10}=66
Q_{10^{-12}}Use in source · 2 passages
Search aliases: Q_10^-12
Use in source · Read passage · resolution_full.tex, line 500
Q_10^-12(P_S,3)≤-5.8845708899×10^-14.
Exact mathematical passage
\boxed{Q_{10^{-12}}(P_{S,3})\le-5.8845708899\times10^{-14}.}
Q_7Use in source · 1 passage
Search aliases: Q_7
Use in source · chapters_sequences.tex, line 12
ryS02_tribonacci_tetranacci T_n=T_n-1+T_n-2+T_n-3, Q_n=∑_j=1^4Q_n-j The seed table fixes the OEIS convention. Tribonacci starts 0,0,1; tetranacci starts 0,0,0,1. The curves use log_10(1+a_n) so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence. Tribonacci gives T_5=2+1+1=4. Tetranacci gives Q_7=4+2+1+1=8. The recurrence first applies at n=3 for T and n=4 for Q. These finite-memory recurrences have three and four feedback taps. Th…
Exact mathematical passage
Q_7=4+2+1+1=8
Q_cUse in source · 4 passages
Search aliases: Q_c
Use in source · Read passage · coupled_structure_full.tex, line 61
Q_c[u,v](z)=c_1u(qz)v(rz)+c_2u(az)v(bz).
Exact mathematical passage
Q_c[u,v](z)=c_1u(qz)v(rz)+c_2u(az)v(bz).
q_JUse in source · 1 passage
Search aliases: q_J
Use in source · Read passage · dilation_full.tex, line 181
c_n(h,q,r)∈ Q[h_1,…,h_J,q_1,r_1,…,q_J,r_J].
Exact mathematical passage
c_n(h,q,r)\in\mathbb Q[h_1,\ldots,h_J,q_1,r_1,\ldots,q_J,r_J].
q_j^2Use in source · 1 passage
Search aliases: q_j^2
Use in source · Read passage · main.tex, line 195
pageheadingFAMILY 1 / TAPSFeedback banks and weak-channel sensitivity Choose finitely many channels j=1,…,J, with h_j>0, ∑_jh_j=1, and 0<q_j,r_j<1. Define F_Θ'(z)=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz), F_Θ(0)=0. These taps act inside the nonlinear definition. With μ_1=∑_jh_jq_jr_j and μ_2=∑_jh_jq_jr_j(q_j^2+r_j^2), F_Θ(z)=z+fracμ_13z^3+fracμ_1μ_215z^5+O(z^7). If q_j^ρ_j+r_j^ρ_j=1, then ρ(F_Θ)=max_j activeρ_j. The positive weights alter local coefficients. The most expansive active channel determines the as…
Exact mathematical passage
\mu_2=\sum_jh_jq_jr_j(q_j^2+r_j^2)
q_j^\rhoUse in source · 1 passage
Search aliases: q_j^ρ rho
Use in source · Read passage · dilation_full.tex, line 159
On the seed, G(t)≤ V_N+1+Δ+B_+≤ c_+e^ρ t. Each channel satisfies q_j^ρ+r_j^ρ≤1. Induction in intervals of length min_jmin(a_j,b_j)>0 therefore propagates the seed comparison, giving the explicit upper bound
Exact mathematical passage
q_j^\rho+r_j^\rho\le1
q_j^{\rho_j}Use in source · 3 passages
Search aliases: q_j^ρ_j rho
Use in source · Read passage · main.tex, line 199
vity Choose finitely many channels j=1,…,J, with h_j>0, ∑_jh_j=1, and 0<q_j,r_j<1. Define F_Θ'(z)=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz), F_Θ(0)=0. These taps act inside the nonlinear definition. With μ_1=∑_jh_jq_jr_j and μ_2=∑_jh_jq_jr_j(q_j^2+r_j^2), F_Θ(z)=z+fracμ_13z^3+fracμ_1μ_215z^5+O(z^7). If q_j^ρ_j+r_j^ρ_j=1, then ρ(F_Θ)=max_j activeρ_j. The positive weights alter local coefficients. The most expansive active channel determines the asymptotic order. A zero-weight channel is removed before taking th…
Exact mathematical passage
q_j^{\rho_j}+r_j^{\rho_j}=1
q_tUse in source · 4 passages
Search aliases: q_t
Use in source · Read passage · sequence_full.tex, line 89
For rational channel tuples (h_t,q_t,r_t), use
Exact mathematical passage
(h_t,q_t,r_t)
Q_\deltaUse in source · 5 passages
Search aliases: Q_δ delta
Use in source · Read passage · resolution_full.tex, line 380
Q_δ(f)=∑_substackn≥1 μ(n)^2=1 n oddn^-4f(x_n)^2 -δ∑_substackn≥1 μ(n)^2=1 n evenn^-4f(x_n)^2.
Exact mathematical passage
Q_\delta(f)=\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ odd}}}n^{-4}f(x_n)^2 -\delta\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ even}}}n^{-4}f(x_n)^2.
Q_\nuUse in source · 1 passage
Search aliases: Q_ν nu
Use in source · Read passage · notation.tex, line 53
item[Signed quadratic form.] For real polynomials, Q_ν(f)=∫ f(x)^2ddν(x). On a positive measure it is a squared norm. On a signed measure it may be negative and is not a norm. item[Orbit spectral zeta Z_N.] Let d_N(w)=∑_n≥0(-1)^nw^n/(Nn)! and D_N(z)=d_N(z^N), N≥2. The series Z_N(s)=∑_jw_j^-s initially runs over the zeros w_j of d_N, counted with multiplicity, for Re s>1/N. Each w_j represents one N-fold rotation orbit of zeros of D_N. Powers use Argw∈(-π,π]. item[Rising factorial.] (s)_k=s(s+1)⋯…
Exact mathematical passage
Q_\nu(f)=\int f(x)^2\dd\nu(x)
\mathbb RUse in source · 3 passages
Search aliases: R mathbb
Use in source · Read passage · main.tex, line 407
B_mA_m^-1/2. Every determinant root is real and below -1; possible coefficient poles occur there, although numerator cancellation is possible. fig05_sieve_boundaryThe line is a proved bound for p=2; dots are ordinary high-precision pencil roots. A defect λ=-1.001 eventually violates positivity even though it can pass low-order checks. The nearest root is -1-δ_m, with δ_m=min_substackf∈ R[x]∖{0} deg f<m frac∫ f^2 dν_nonp∫ f^2 dν_p>0, δ_mdownarrow0. To prove the limit, approximate a continuous bum…
Exact mathematical passage
\delta_m=\min_{\substack{f\in\mathbb R[x]\setminus\{0\}\\\deg f<m}} \frac{\int f^2\dd\nu_{\mathrm{non}p}}{\int f^2\dd\nu_p}>0, \qquad\delta_m\downarrow0.
R^2Use in source · 2 passages
Search aliases: R^2
Use in source · Read passage · sequence_full.tex, line 224
log M(R)≥frac R2exp(fracq^3R^28-frac q2).
Exact mathematical passage
\boxed{ \log M(R)\ge\frac R2\exp\left(\frac{q^3R^2}{8}-\frac q2\right).}
r^2Use in source · 12 passages
Search aliases: r^2
Use in source · Read passage · main.tex, line 95
nctionsComputed Taylor curves. Each entire curve uses terms through z^321; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent. For T_q,r(z)=∑_n≥0c_nz^2n+1, c_0=1, coefficient matching gives c_n=frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1c_ic_n-1-i, n≥1. In particular, T_q,r(z)=z+fracqr3z^3+ fracq^2r^2(q^2+r^2)15z^5+O(z^7). The coefficients are positive and bounded by tangent coefficients, giving a convergent germ. If the …
Exact mathematical passage
T_{q,r}(z)=z+\frac{qr}{3}z^3+ \frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).
r^{2(n-1-i)+1}Use in source · 5 passages
Search aliases: r^2(n-1-i)+1
Use in source · Read passage · main.tex, line 90
on is symmetric in q,r, but unequal scales retain left/right address information. fig01_asymmetric_functionsComputed Taylor curves. Each entire curve uses terms through z^321; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent. For T_q,r(z)=∑_n≥0c_nz^2n+1, c_0=1, coefficient matching gives c_n=frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1c_ic_n-1-i, n≥1. In particular, T_q,r(z)=z+fracqr3z^3+ fracq^2r^2(q^2+r^2)15z^5+O(z^7).…
Exact mathematical passage
c_n=\frac{1}{2n+1}\sum_{i=0}^{n-1} q^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\qquad n\ge1.
r^{2(n-i)+1}Use in source · 1 passage
Search aliases: r^2(n-i)+1
Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
r^{2j+1}Use in source · 4 passages
Search aliases: r^2j+1
Use in source · Read passage · main.tex, line 177
claimA combinatorial address formula Sum over full ordered binary trees with n internal vertices. A vertex with i internal vertices in its left subtree and j in its right contributes fracq^2i+1r^2j+12(i+j+1)+1. Multiply these factors over internal vertices, then sum over trees. Root splitting reproduces the coefficient recurrence exactly. Unequal contractions therefore record left and right path lengths separately.
Exact mathematical passage
\frac{q^{2i+1}r^{2j+1}}{2(i+j+1)+1}.
R^{2n+1}Use in source · 8 passages
Search aliases: R^2n+1
Use in source · Read passage · dilation_full.tex, line 196
∑_n≥0|c_n(h,q,r)|R^2n+1 ≤ W^-1/2f_Q(sqrt W R)<∞.
Exact mathematical passage
\sum_{n\ge0}|c_n(h,q,r)|R^{2n+1} \le W^{-1/2}f_Q(\sqrt W R)<\infty.
r^{2n-1}Use in source · 1 passage
Search aliases: r^2n-1
Use in source · Read passage · sequence_full.tex, line 129
c_n≥fracq r^2n-12n+1c_n-1 Longrightarrow c_n≥fracq^n r^n^23·5⋯(2n+1).
Exact mathematical passage
c_n\ge\frac{q r^{2n-1}}{2n+1}c_{n-1} \quad\Longrightarrow\quad c_n\ge\frac{q^n r^{n^2}}{3\cdot5\cdots(2n+1)}.
r^3Use in source · 2 passages
Search aliases: r^3
Use in source · Read passage · sequence_full.tex, line 71
A_0= · 1, A_1= · 2qr, A_2= · 8q^2r^2(q^2+r^2), A_3= · q^3r^3[48(q^2+r^2)(q^4+r^4)+80q^2r^2].
Exact mathematical passage
\begin{aligned} A_0={}&1,\\ A_1={}&2qr,\\ A_2={}&8q^2r^2(q^2+r^2),\\ A_3={}&q^3r^3\bigl[48(q^2+r^2)(q^4+r^4)+80q^2r^2\bigr]. \end{aligned}
r^4Use in source · 3 passages
Search aliases: r^4
Use in source · Read passage · sequence_full.tex, line 71
A_0= · 1, A_1= · 2qr, A_2= · 8q^2r^2(q^2+r^2), A_3= · q^3r^3[48(q^2+r^2)(q^4+r^4)+80q^2r^2].
Exact mathematical passage
\begin{aligned} A_0={}&1,\\ A_1={}&2qr,\\ A_2={}&8q^2r^2(q^2+r^2),\\ A_3={}&q^3r^3\bigl[48(q^2+r^2)(q^4+r^4)+80q^2r^2\bigr]. \end{aligned}
r^6Use in source · 1 passage
Search aliases: r^6
Use in source · Read passage · resolution_visual.tex, line 172
claimThe certificate uses rational arithmetic throughout Take S={1,3,5,7} and define P(x)=∏_n∈ Sfracx-x_nx_2-x_n (frac1-x1-x_2)^3. Its degree is seven. All remaining positive nodes begin at n=11. With C=756482048/20777538825 and r=3856/102487, a complete tail bound gives Q_10^-12(P)≤-frac10^-1216 +fracC^2r^63·7^3 <-5.8845708899×10^-14. The inequality is certified as a comparison of exact rational numbers. The first four zeros remove the dominant positive contributions. The factor (1-x)^3 suppres…
Exact mathematical passage
Q_{10^{-12}}(P)\le-\frac{10^{-12}}{16} +\frac{C^2r^6}{3\cdot7^3} <-5.8845708899\times10^{-14}.
R^mUse in source · 1 passage
Search aliases: R^m
Use in source · Read passage · coupled_lab_full.tex, line 80
For every omitted exponent m≥2N+3, use R^m≤ (R/S)^2N+3S^m, sum positive terms, and bound their sum by F(S)≤ B(S). No extra geometric denominator is needed. For rational x≥0, pathcertified_evaluation returns the exact interval [F_N(x),F_N(x)+B(S)(x/S)^2N+3] for each component.
Exact mathematical passage
R^m\le (R/S)^{2N+3}S^m
r^{n^2}Use in source · 2 passages
Search aliases: r^n^2
Use in source · Read passage · sequence_full.tex, line 129
c_n≥fracq r^2n-12n+1c_n-1 Longrightarrow c_n≥fracq^n r^n^23·5⋯(2n+1).
Exact mathematical passage
c_n\ge\frac{q r^{2n-1}}{2n+1}c_{n-1} \quad\Longrightarrow\quad c_n\ge\frac{q^n r^{n^2}}{3\cdot5\cdots(2n+1)}.
r^{n^2/(2n+1)}Use in source · 1 passage
Search aliases: r^n^2/(2n+1)
Use in source · Read passage · sequence_full.tex, line 137
c_n^1/(2n+1)≥ q^n/(2n+1)r^n^2/(2n+1)(2n+1)^-n/(2n+1)⟶∞.
Exact mathematical passage
c_n^{1/(2n+1)}\ge q^{n/(2n+1)}r^{n^2/(2n+1)}(2n+1)^{-n/(2n+1)}\longrightarrow\infty.
r^sUse in source · 4 passages
Search aliases: r^s
Use in source · Read passage · coupled_growth_full.tex, line 113
P(s)=q^s · r^s a^s · b^s.
Exact mathematical passage
P(s)=\begin{pmatrix}q^s&r^s\\a^s&b^s\end{pmatrix}.
r^vUse in source · 1 passage
Search aliases: r^v
Use in source · Read passage · sequence_full.tex, line 49
Polynomial theorem. For every n, A_n∈ Z[q,r], all its nonzero coefficients are positive integers, and A_n(q,r)=A_n(r,q). Every monomial q^u r^v satisfies
Exact mathematical passage
q^u r^v
R^{\rho_\alpha}Use in source · 1 passage
Search aliases: R^ρ_α rho alpha
Use in source · Read passage · main.tex, line 134
claimWhy this equation appears Nonnegative coefficients give M(R)=T_q,r(R). The integral equation implies M(R)≤ R[1+M(qR)M(rR)], M(R)≥(1-α)R M(α qR)M(α rR),quad0<α<1. Taking logarithms turns product feedback into a sum over contracted scales. Comparison with CR^ρ proves the upper bound; comparison with cR^ρ_α, where (α q)^ρ_α+ (α r)^ρ_α=1, proves the lower bound. Letting αuparrow1 gives the same order. The complete interval-induction proof is given in Appendix~[source reference].
Exact mathematical passage
cR^{\rho_\alpha}
r_0Use in source · 2 passages
Search aliases: r_0
Use in source · Read passage · arithmetic_full.tex, line 276
Indeed all omitted frequencies are at most λ_L+1, and ∑_n>Ln^-2a≤∫_L^∞ x^-2a dx. Ignoring squarefreeness and coprimality makes this a valid upper bound. If r_0,r_1 are the two positive truncated residual sums at t,t+h, the true secant belongs to
Exact mathematical passage
r_0,r_1
R_1Use in source · 2 passages
Search aliases: R_1
Use in source · Read passage · arithmetic_full.tex, line 61
Starting with R_1=A, define successively
Exact mathematical passage
R_1=A
r_1Use in source · 3 passages
Search aliases: r_1
Use in source · Read passage · arithmetic_full.tex, line 276
Indeed all omitted frequencies are at most λ_L+1, and ∑_n>Ln^-2a≤∫_L^∞ x^-2a dx. Ignoring squarefreeness and coprimality makes this a valid upper bound. If r_0,r_1 are the two positive truncated residual sums at t,t+h, the true secant belongs to
Exact mathematical passage
r_0,r_1
R_2Use in source · 5 passages
Search aliases: R_2
Use in source · Read passage · resolution_full.tex, line 165
R_2(ε)=max{ e,(frac2C_0c_ε)^1/(ρ-1/2), (frac2log2c_ε)^1/ρ}. tagRA-3
Exact mathematical passage
R_2(\varepsilon)=\max\left\{ e,\left(\frac{2C_0}{c_\varepsilon}\right)^{1/(\rho-1/2)}, \left(\frac{2\log2}{c_\varepsilon}\right)^{1/\rho}\right\}. \tag{RA-3}
r_aUse in source · 2 passages
Search aliases: r_a
Use in source · Read passage · multi_prime.tex, line 68
pageheadingPRIME SIEVES / CONSTRUCTIVE WITNESSEvery negative prime atom has a finite detector For an integer a≥2, let L_a(x)=∏_n=1^a-1fracx-x_nx_a-x_n, C_a=L_a(1), r_a=frac1-x_a+11-x_a<1, f_a,m(x)=L_a(x)(frac1-x1-x_a)^m. This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects …
Exact mathematical passage
L_a(x)=\prod_{n=1}^{a-1}\frac{x-x_n}{x_a-x_n},\quad C_a=L_a(1),\quad r_a=\frac{1-x_{a+1}}{1-x_a}<1,
r_a^mUse in source · 2 passages
Search aliases: r_a^m
Use in source · Read passage · multi_prime.tex, line 76
pageheadingPRIME SIEVES / CONSTRUCTIVE WITNESSEvery negative prime atom has a finite detector For an integer a≥2, let L_a(x)=∏_n=1^a-1fracx-x_nx_a-x_n, C_a=L_a(1), r_a=frac1-x_a+11-x_a<1, f_a,m(x)=L_a(x)(frac1-x1-x_a)^m. This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects …
Exact mathematical passage
C_ar_a^m
R_jUse in source · 9 passages
Search aliases: R_j
Use in source · Read passage · arithmetic_full.tex, line 63
λ_j=lim_t→∞fraclog R_j(-t)t, w_j=lim_t→∞e^-λ_jtR_j(-t), R_j+1(s)=R_j(s)-w_je^-λ_js.
Exact mathematical passage
\lambda_j=\lim_{t\to\infty}\frac{\log R_j(-t)}t, \qquad w_j=\lim_{t\to\infty}e^{-\lambda_jt}R_j(-t), \qquad R_{j+1}(s)=R_j(s)-w_je^{-\lambda_js}.
r_JUse in source · 1 passage
Search aliases: r_J
Use in source · Read passage · dilation_full.tex, line 181
c_n(h,q,r)∈ Q[h_1,…,h_J,q_1,r_1,…,q_J,r_J].
Exact mathematical passage
c_n(h,q,r)\in\mathbb Q[h_1,\ldots,h_J,q_1,r_1,\ldots,q_J,r_J].
R_{j+1}Use in source · 2 passages
Search aliases: R_j+1
Use in source · Read passage · arithmetic_full.tex, line 63
λ_j=lim_t→∞fraclog R_j(-t)t, w_j=lim_t→∞e^-λ_jtR_j(-t), R_j+1(s)=R_j(s)-w_je^-λ_js.
Exact mathematical passage
\lambda_j=\lim_{t\to\infty}\frac{\log R_j(-t)}t, \qquad w_j=\lim_{t\to\infty}e^{-\lambda_jt}R_j(-t), \qquad R_{j+1}(s)=R_j(s)-w_je^{-\lambda_js}.
r_j^2Use in source · 1 passage
Search aliases: r_j^2
Use in source · Read passage · main.tex, line 195
pageheadingFAMILY 1 / TAPSFeedback banks and weak-channel sensitivity Choose finitely many channels j=1,…,J, with h_j>0, ∑_jh_j=1, and 0<q_j,r_j<1. Define F_Θ'(z)=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz), F_Θ(0)=0. These taps act inside the nonlinear definition. With μ_1=∑_jh_jq_jr_j and μ_2=∑_jh_jq_jr_j(q_j^2+r_j^2), F_Θ(z)=z+fracμ_13z^3+fracμ_1μ_215z^5+O(z^7). If q_j^ρ_j+r_j^ρ_j=1, then ρ(F_Θ)=max_j activeρ_j. The positive weights alter local coefficients. The most expansive active channel determines the as…
Exact mathematical passage
\mu_2=\sum_jh_jq_jr_j(q_j^2+r_j^2)
r_j^\rhoUse in source · 1 passage
Search aliases: r_j^ρ rho
Use in source · Read passage · dilation_full.tex, line 159
On the seed, G(t)≤ V_N+1+Δ+B_+≤ c_+e^ρ t. Each channel satisfies q_j^ρ+r_j^ρ≤1. Induction in intervals of length min_jmin(a_j,b_j)>0 therefore propagates the seed comparison, giving the explicit upper bound
Exact mathematical passage
q_j^\rho+r_j^\rho\le1
r_j^{\rho_j}Use in source · 3 passages
Search aliases: r_j^ρ_j rho
Use in source · Read passage · main.tex, line 199
vity Choose finitely many channels j=1,…,J, with h_j>0, ∑_jh_j=1, and 0<q_j,r_j<1. Define F_Θ'(z)=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz), F_Θ(0)=0. These taps act inside the nonlinear definition. With μ_1=∑_jh_jq_jr_j and μ_2=∑_jh_jq_jr_j(q_j^2+r_j^2), F_Θ(z)=z+fracμ_13z^3+fracμ_1μ_215z^5+O(z^7). If q_j^ρ_j+r_j^ρ_j=1, then ρ(F_Θ)=max_j activeρ_j. The positive weights alter local coefficients. The most expansive active channel determines the asymptotic order. A zero-weight channel is removed before taking th…
Exact mathematical passage
q_j^{\rho_j}+r_j^{\rho_j}=1
R_kUse in source · 5 passages
Search aliases: R_k
Use in source · Read passage · dilation_full.tex, line 131
These constants are intentionally conservative, and require no numerical evaluation of an unknown solution. The tangent majorant gives M(1)<2. At R_k=Q^-k, the integral equation gives
Exact mathematical passage
R_k=Q^{-k}
R_{k-1}Use in source · 1 passage
Search aliases: R_k-1
Use in source · Read passage · dilation_full.tex, line 133
M(R_k)≤ R_k[1+M(R_k-1)^2].
Exact mathematical passage
M(R_k)\le R_k[1+M(R_{k-1})^2].
R_NUse in source · 2 passages
Search aliases: R_N
Use in source · Read passage · dilation_full.tex, line 141
Define U_0=2, U_k=Q^-k(1+U_k-1^2). Then M(R_k)≤ U_k, U_k≥1, and log U_k≤ klog(1/Q)+log2+2log U_k-1. Summing this elementary recurrence gives log U_N≤ V_N. Since e<3, the logarithmic-radius seed 1≤ t≤1+Δ has e^t≤ e/m<3/m≤ R_N, and therefore log M(e^t)≤ V_N.
Exact mathematical passage
e^t\le e/m<3/m\le R_N
r_pUse in source · 1 passage
Search aliases: r_p
Use in source · Read passage · sieve_full.tex, line 159
All quantities in (5) are rational when the strengths are rational. It provides an exact finite certificate without calculating Hankel eigenvalues or relying on floating-point root finding. A suitable finite m exists because r_p<1.
Exact mathematical passage
r_p<1
r_p^{2m}Use in source · 9 passages
Search aliases: r_p^2m
Use in source · Read passage · multi_prime.tex, line 81
This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects the signed defect. Set C_F=∏_q∈ Fmax(1,|1+λ_q|). The summable tail gives ∫ f_p,m^2ddν ≤frac1+λ_pp^4+ fracC_FC_p^23p^3r_p^2m. Hence the quadratic form is negative whenever pC_FC_p^2r_p^2m/3<-(1+λ_p). For rational strengths…
Exact mathematical passage
\int f_{p,m}^2\dd\nu \le\frac{1+\lambda_p}{p^4}+ \frac{C_FC_p^2}{3p^3}r_p^{2m}.
r_SUse in source · 2 passages
Search aliases: r_S
Use in source · Read passage · resolution_full.tex, line 455
P_S,m(x)=∏_n∈ Sfracx-x_nx_2-x_n (frac1-x1-x_2)^m, C_S=∏_n∈ Sfrac1-x_n|x_2-x_n|, r_S=frac1-x_n_+1-x_2<1.
Exact mathematical passage
C_S=\prod_{n\in S}\frac{1-x_n}{|x_2-x_n|}, \qquad r_S=\frac{1-x_{n_+}}{1-x_2}<1.
r_S^{2m}Use in source · 2 passages
Search aliases: r_S^2m
Use in source · Read passage · resolution_full.tex, line 467
Q_δ(P_S,m)≤-fracδ16 +fracC_S^23K^3r_S^2m. tagRA-10
Exact mathematical passage
\boxed{Q_\delta(P_{S,m})\le-\frac\delta{16} +\frac{C_S^2}{3K^3}r_S^{2m}.} \tag{RA-10}
r_S^mUse in source · 1 passage
Search aliases: r_S^m
Use in source · Read passage · resolution_full.tex, line 463
This polynomial has degree |S|+m, equals one at x_2, and kills every positive atom through the cutoff. At each remaining positive node, the product of the absolute root factors is at most C_S, and the last factor is at most r_S^m. Dropping all other negative atoms and bounding the positive tail by the integral ∑_n>Kn^-4≤1/(3K^3) gives
Exact mathematical passage
r_S^m
r_tUse in source · 4 passages
Search aliases: r_t
Use in source · Read passage · sequence_full.tex, line 89
For rational channel tuples (h_t,q_t,r_t), use
Exact mathematical passage
(h_t,q_t,r_t)
\operatorname{rad}Use in source · 7 passages
Search aliases: rad operatorname
Use in source · Read passage · main.tex, line 265
pageheadingFAMILY 2 / CALCULUSClosed shifts, infinite differential complexity The Euler product for Z_Q shows that only rad(Q)=∏_pmid Qp matters. Expansion of the depth factor gives A_a,β,Q(s+1)=A_a,β,Q(s)-2A_a+β/2,β,Q(s) +A_a+β,β,Q(s). The fractional transfer and Newton formulas are A_a,β,Q(s+t)=∑_j≥0(-1)^jbinom2tjA_a+jβ/2,β,Q(s), A_a,β,Q(t)=∑_j≥0(-1)^jbinom2tj[Z_Q(2a+jβ)-1]. They converge absolutely and locally uniformly: binomial coefficients have at most polynomial growth on compact paramete…
Exact mathematical passage
\operatorname{rad}(Q)=\prod_{p\mid Q}p
\operatorname{Res}Use in source · 4 passages
Search aliases: Res operatorname
Use in source · Read passage · spectral_continuation.tex, line 4
pageheadingORBIT SPECTRA / CONTINUATION THEOREMGlobal meromorphic continuation For every integer N≥2, the orbit zeta has a meromorphic continuation to the whole complex plane with exactly one pole: s=1/N, Res_s=1/N Z_N(s)=fracsin(π/N)Nπ. Its exact absolute-convergence abscissa is 1/N. At nonpositive integers, Z_N(-m)=0 (m≥0), Z_N'(0)=-log N. The zeta-regularized product of the orbit coordinates is consequently exp[- Z_N'(0)]=N. fig15_spectral_polesPole and zero locations follow from the proved c…
Exact mathematical passage
\boxed{s=1/N,\qquad \operatorname{Res}_{s=1/N}\mathcal Z_N(s)=\frac{\sin(\pi/N)}{N\pi}.}
\operatorname{Res}_{s=1/N}Use in source · 2 passages
Search aliases: Res_s=1/N operatorname
Use in source · Read passage · spectral_continuation.tex, line 4
pageheadingORBIT SPECTRA / CONTINUATION THEOREMGlobal meromorphic continuation For every integer N≥2, the orbit zeta has a meromorphic continuation to the whole complex plane with exactly one pole: s=1/N, Res_s=1/N Z_N(s)=fracsin(π/N)Nπ. Its exact absolute-convergence abscissa is 1/N. At nonpositive integers, Z_N(-m)=0 (m≥0), Z_N'(0)=-log N. The zeta-regularized product of the orbit coordinates is consequently exp[- Z_N'(0)]=N. fig15_spectral_polesPole and zero locations follow from the proved c…
Exact mathematical passage
\boxed{s=1/N,\qquad \operatorname{Res}_{s=1/N}\mathcal Z_N(s)=\frac{\sin(\pi/N)}{N\pi}.}
\operatorname{Res}_{s=1/N-k}Use in source · 1 passage
Search aliases: Res_s=1/N-k operatorname
Use in source · Read passage · spectral_continuation.tex, line 89
ageheadingORBIT SPECTRA / SHIFTSA Hurwitz-type deformation and its determinant For real a≥0, define initially Z_N(s;a)=∑_j(w_j+a)^-s, Re s>1/N. The rightward shift preserves avoidance of the negative branch cut. Normalize the product with E_N(a) and use log E_N(a+t) in the Mellin argument. For a>0 there are genuine simple poles at s=1/N-k, k≥0: Res_s=1/N-k Z_N(s;a) =fracsin(π/N)Nπfrac(1-1/N)_kk!a^k. Indeed the large-radius derivative expands as frac1N(a+t)^1/N-1 =frac1N∑_k≥0frac(-1)^k(1-1/N)_kk!…
Exact mathematical passage
\boxed{\operatorname{Res}_{s=1/N-k}\mathcal Z_N(s;a) =\frac{\sin(\pi/N)}{N\pi}\frac{(1-1/N)_k}{k!}a^k.}
\operatorname{Res}_{s=\alpha-k}Use in source · 1 passage
Search aliases: Res_s=α-k operatorname alpha
Use in source · Read passage · spectral_full.tex, line 433
Res_s=α-k Z_N(s;a) · =fracαsin(π(α-k))π binomα-1ka^k · =fracsin(π/N)Nπ frac(1-1/N)_kk!a^k. tagT7
Exact mathematical passage
\begin{aligned} \operatorname{Res}_{s=\alpha-k}\mathcal Z_N(s;a) &=\frac{\alpha\sin(\pi(\alpha-k))}{\pi} \binom{\alpha-1}{k}a^k\\ &=\boxed{\frac{\sin(\pi/N)}{N\pi} \frac{(1-1/N)_k}{k!}a^k.} \end{aligned} \tag{T7}
\rfloorUse in source · 2 passages
Search aliases: rfloor
Use in source · Read passage · arithmetic_full.tex, line 133
-fraclog(λ_lfloor cjrfloor/λ_j)log c =β+O_Q,β,c(j^-1/2+j^-β).
Exact mathematical passage
-\frac{\log(\lambda_{\lfloor cj\rfloor}/\lambda_j)}{\log c} =\beta+O_{Q,\beta,c}(j^{-1/2}+j^{-\beta}).
\rmUse in source · 4 passages
Search aliases: rm
Use in source · Read passage · sequence_full.tex, line 321
B_rm EGF(z)=∑_t=0^L-1h_t I^tG(z), IG(z)=∫_0^zG(w) dw.
Exact mathematical passage
\boxed{B_{\rm EGF}(z)=\sum_{t=0}^{L-1}h_t I^tG(z),\qquad IG(z)=\int_0^zG(w)\,dw.}
SUse in source · 34 passages
Search aliases: S
Use in source · Read passage · main.tex, line 461
claimThe resolvent is non-D-finite For λ≥-1, the Stieltjes function S_p,λ(z)=∫(z-x)^-1 dν_p,λ(x) has infinitely many isolated poles at the positive-weight nodes x_n→1. A solution of a finite rational-coefficient linear ODE has only finitely many possible finite singular points. The germ at z=2 therefore is not D-finite. This is a general infinite-atomic-resolvent mechanism, specialized here to arithmetic data.
Exact mathematical passage
S_{p,\lambda}(z)=\int(z-x)^{-1}\dd\nu_{p,\lambda}(x)
\mathcal SUse in source · 2 passages
Search aliases: S mathcal
Use in source · Read passage · main.tex, line 246
(n,Q)=1μ(n)^2n^-v =fracζ(v)ζ(2v)∏_pmid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit working precision. Right: initial squarefree frequencies and weights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If…
Exact mathematical passage
\mathcal S_Q=\{n\ge2:\mu(n)^2=1,\ (n,Q)=1\}
s^{-1}Use in source · 1 passage
Search aliases: s^-1
Use in source · Read passage · coupled_inverse_full.tex, line 96
When qb=ar, one has a=sq and b=sr, giving equality in the displayed order. When q=b, one has a=s^2r, giving equality after interchanging the factors. The pair also has the required initial values and derivatives. Uniqueness of the formal coefficient recursion identifies it with the coupled solution. Finally, V(z)=s^-1U(sz), proving rescaling.
Exact mathematical passage
V(z)=s^{-1}U(sz)
s^{-1/(2\beta)}Use in source · 1 passage
Search aliases: s^-1/(2β) beta
Use in source · Read passage · arithmetic_full.tex, line 485
At real positive depth, the matrix (A_χ,ψ(s)) on any d distinct characters is positive semidefinite and has identical diagonal A_a,β,Q(s). Its normalized matrix tends to I_d, with operator norm error O_a,β,Q,d(s^-1/(2β)). Indeed, the off-diagonal entry bound controls the maximum row and column sums, and thus the operator norm. This gives quantitative separation of a fixed character bank. It is consistent with a complex or signed off-diagonal entry and makes no complete-monotonicity assertion for…
Exact mathematical passage
O_{a,\beta,Q,d}(s^{-1/(2\beta)})
s^2Use in source · 3 passages
Search aliases: s^2
Use in source · Read passage · main.tex, line 148
Suppose such an equality held. Normalization gives AB=1. Write p=qr, u=q^2+r^2, and let ρ be the common order. The first two nontrivial coefficients and the growth law force B^2s^2=p, s^2=u/2, 2s^ρ=q^ρ+r^ρ=1. Consequently 2(fracq^2+r^22)^ρ/2=q^ρ+r^ρ. For ρ≠2, strict convexity or concavity of x^ρ/2 forces q=r. For the exceptional case ρ=2, we have u=1, s^2=1/2, B^2=2p. The two z^7 coefficients become fracp^3(3-p^2)315 and frac11p^31260. Their equality forces p=1/2 and again (q-r)^2=u-2p=0. This c…
Exact mathematical passage
B^2s^2=p,\qquad s^2=u/2,\qquad 2s^\rho=q^\rho+r^\rho=1.
s^3Use in source · 1 passage
Search aliases: s^3
Use in source · Read passage · spectral_full.tex, line 315
Since sin(π s)/π=s+O(s^3), (S14) implies Z_N(0)=0 and Z_N'(0)=-log N.
Exact mathematical passage
\sin(\pi s)/\pi=s+O(s^3)
s^{3/2}Use in source · 1 passage
Search aliases: s^3/2
Use in source · Read passage · coupled_inverse_full.tex, line 87
{qsqrts, s^3/2r} ={a/sqrts, bsqrts}.
Exact mathematical passage
\{q\sqrt{s},\;s^{3/2}r\} =\{a/\sqrt{s},\;b\sqrt{s}\}.
S^mUse in source · 1 passage
Search aliases: S^m
Use in source · Read passage · coupled_lab_full.tex, line 80
For every omitted exponent m≥2N+3, use R^m≤ (R/S)^2N+3S^m, sum positive terms, and bound their sum by F(S)≤ B(S). No extra geometric denominator is needed. For rational x≥0, pathcertified_evaluation returns the exact interval [F_N(x),F_N(x)+B(S)(x/S)^2N+3] for each component.
Exact mathematical passage
R^m\le (R/S)^{2N+3}S^m
s^\rhoUse in source · 1 passage
Search aliases: s^ρ rho
Use in source · Read passage · main.tex, line 148
Suppose such an equality held. Normalization gives AB=1. Write p=qr, u=q^2+r^2, and let ρ be the common order. The first two nontrivial coefficients and the growth law force B^2s^2=p, s^2=u/2, 2s^ρ=q^ρ+r^ρ=1. Consequently 2(fracq^2+r^22)^ρ/2=q^ρ+r^ρ. For ρ≠2, strict convexity or concavity of x^ρ/2 forces q=r. For the exceptional case ρ=2, we have u=1, s^2=1/2, B^2=2p. The two z^7 coefficients become fracp^3(3-p^2)315 and frac11p^31260. Their equality forces p=1/2 and again (q-r)^2=u-2p=0. This c…
Exact mathematical passage
B^2s^2=p,\qquad s^2=u/2,\qquad 2s^\rho=q^\rho+r^\rho=1.
s_iUse in source · 2 passages
Search aliases: s_i
Use in source · Read passage · main.tex, line 297
claimStrict positive complex kernel For distinct complex points s_i, K_ij=A(s_i+overlines_j) is strictly positive definite. A zero quadratic form would force a finite exponential sum to vanish at every λ_n. Their accumulation at zero invokes the identity theorem; derivatives at zero then give a nonsingular Vandermonde system. On the real axis, (-1)^jA^(j)(s)>0 for every j≥0.
Exact mathematical passage
s_i
S_{p,\lambda}Use in source · 1 passage
Search aliases: S_p,λ lambda
Use in source · Read passage · main.tex, line 461
claimThe resolvent is non-D-finite For λ≥-1, the Stieltjes function S_p,λ(z)=∫(z-x)^-1 dν_p,λ(x) has infinitely many isolated poles at the positive-weight nodes x_n→1. A solution of a finite rational-coefficient linear ODE has only finitely many possible finite singular points. The germ at z=2 therefore is not D-finite. This is a general infinite-atomic-resolvent mechanism, specialized here to arithmetic data.
Exact mathematical passage
S_{p,\lambda}(z)=\int(z-x)^{-1}\dd\nu_{p,\lambda}(x)
\mathcal S_QUse in source · 2 passages
Search aliases: S_Q mathcal
Use in source · Read passage · main.tex, line 246
(n,Q)=1μ(n)^2n^-v =fracζ(v)ζ(2v)∏_pmid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit working precision. Right: initial squarefree frequencies and weights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If…
Exact mathematical passage
\mathcal S_Q=\{n\ge2:\mu(n)^2=1,\ (n,Q)=1\}
\sinUse in source · 23 passages
Search aliases: sin
Use in source · Read passage · arithmetic_full.tex, line 475
which is O_a,β,ε(|s|^-γ) because u≥ |s|sinε. The principal diagonal has the nonzero leading term δ_QΓ(α)β^-1(2s)^-α, α=(2a-1)/β, by the base sector theorem. Its modulus is bounded below by a positive fixed-sector multiple of |s|^-α for sufficiently large |s|. Division gives the second bound, since γ-α=1/(2β).
Exact mathematical passage
u\ge |s|\sin\varepsilon
\sin^2Use in source · 1 passage
Search aliases: sin^2 sin
Use in source · Read passage · spectral_full.tex, line 152
To check the elementary estimate, write w=re^iθ: |t+w|^2=(t+r)^2-4trsin^2(θ/2)≥(t+r)^2cos^2(θ/2).
Exact mathematical passage
|t+w|^2=(t+r)^2-4tr\sin^2(\theta/2)\ge(t+r)^2\cos^2(\theta/2)
\sinhUse in source · 2 passages
Search aliases: sinh
Use in source · Read passage · main.tex, line 171
claimA worked order-one member T_1/3,2/3(z)=z+frac227z^3+ frac42187z^5+frac23211160261z^7+O(z^9). It has order one, and T_1/3,2/3(1)≈1.0759239897408249. The diagonal point q=r=1/2 gives f_1/2(z)=sqrt2sinh(z/sqrt2). Every off-diagonal member is separated from all rescaled diagonal members by the theorem above.
Exact mathematical passage
f_{1/2}(z)=\sqrt2\sinh(z/\sqrt2)
\squareUse in source · 13 passages
Search aliases: square
Use in source · Read passage · sequence_full.tex, line 45
Proof. Multiply the ordinary Taylor recurrence by (2n+1)!. The factor attached to A_iA_j, where i+j=n-1, is (2n)!/[(2i+1)!(2j+1)!]=binom2n2i+1. This also proves the recurrence directly in the EGF convention. square
Exact mathematical passage
\square
\supUse in source · 11 passages
Search aliases: sup
Use in source · Read passage · arithmetic_full.tex, line 318
displaystyle sup_|s|≤ R|A_a,β,Q^(k)(s)-A_a,β,Qp^(k)(s)| ≤ λ_p^k e^Rλ_p fracZ_Q(2a)p^2a+1.
Exact mathematical passage
\boxed{\displaystyle \sup_{|s|\le R}|A_{a,\beta,Q}^{(k)}(s)-A_{a,\beta,Qp}^{(k)}(s)| \le \lambda_p^k e^{R\lambda_p} \frac{Z_Q(2a)}{p^{2a}+1}.}
\sup_{|s|\le R}Use in source · 3 passages
Search aliases: sup_|s|≤ R sup le
Use in source · Read passage · arithmetic_full.tex, line 318
displaystyle sup_|s|≤ R|A_a,β,Q^(k)(s)-A_a,β,Qp^(k)(s)| ≤ λ_p^k e^Rλ_p fracZ_Q(2a)p^2a+1.
Exact mathematical passage
\boxed{\displaystyle \sup_{|s|\le R}|A_{a,\beta,Q}^{(k)}(s)-A_{a,\beta,Qp}^{(k)}(s)| \le \lambda_p^k e^{R\lambda_p} \frac{Z_Q(2a)}{p^{2a}+1}.}
\sup_{|s|\le5}Use in source · 2 passages
Search aliases: sup_|s|≤5 sup le
Use in source · Read passage · resolution_full.tex, line 258
sup_|s|≤5|D_p(s)| <frac43(p^4+1)(1-p^-2)^-10. tagRA-5
Exact mathematical passage
\boxed{\sup_{|s|\le5}|D_p(s)| <\frac{4}{3(p^4+1)}(1-p^{-2})^{-10}.} \tag{RA-5}
\sup_{|z|\le R}Use in source · 5 passages
Search aliases: sup_|z|≤ R sup le
Use in source · Read passage · dilation_full.tex, line 217
sup_|z|≤ R|F_h(z)-F_g(z)| ≤frac d4{R f_Q'(R)-f_Q(R)}.
Exact mathematical passage
\boxed{\sup_{|z|\le R}|F_h(z)-F_g(z)| \le\frac d4\{R f_Q'(R)-f_Q(R)\}.}
\sup_{|z|\le1}Use in source · 1 passage
Search aliases: sup_|z|≤1 sup le
Use in source · Read passage · resolution_full.tex, line 119
sup_|z|≤1|F_10^-12(z)-F_0(z)|<2.41890267×10^-13,
Exact mathematical passage
\boxed{\sup_{|z|\le1}|F_{10^{-12}}(z)-F_0(z)|<2.41890267\times10^{-13},}
\mathsf TUse in source · 1 passage
Search aliases: T mathsf
Use in source · Read passage · coupled_structure_full.tex, line 133
It has rank one, but is not symmetric: B-B^ T=ℓ_1wedgeℓ_2≠0. At four distinct nonzero nodes, polynomial interpolation lets their vector values vary independently, so terms supported at (az,bz) cannot cancel this coefficient. A scalar autonomous channel at the same node pair would require B=κℓ⊗ℓ, a symmetric rank-one tensor, after projecting the output onto that channel. For any output row with a nonzero coefficient on this term, the alternating part remains nonzero. The other node pair gives the…
Exact mathematical passage
B-B^{\mathsf T}=\ell_1\wedge\ell_2\ne0
T'_{q,r}Use in source · 1 passage
Search aliases: T'_q,r
Use in source · Read passage · notation.tex, line 10
item[Contracted feedback T_q,r.] The normalized odd entire solution of T'_q,r(z)=1+T_q,r(qz)T_q,r(rz), T_q,r(0)=0, for 0<q,r<1. The diagonal notation is f_q=T_q,q. item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglo…
Exact mathematical passage
T'_{q,r}(z)=1+T_{q,r}(qz)T_{q,r}(rz)
t^{(j+\ell-i)/b}Use in source · 1 passage
Search aliases: t^(j+ℓ-i)/b ell
Use in source · Read passage · sequence_polyphase.tex, line 78
For causal FIR taps h_ℓ, b_m=∑_ℓ h_ℓ a_m-ℓ with zero extension to negative indices, split A(z) and B(z) into b phases. Then B_i(t)=∑_jH_ij(t)A_j(t), where H_ij(t)=∑_substackℓ≥0 j+ℓ≡ ipmod b h_ℓ t^(j+ℓ-i)/b. The exponent is the nonnegative carry to the next phase block. This is an exact coefficient law, with no approximation or loss of terms.
Exact mathematical passage
\boxed{H_{ij}(t)=\sum_{\substack{\ell\ge0\\j+\ell\equiv i\pmod b}} h_\ell t^{(j+\ell-i)/b}.}
t^{-s}Use in source · 5 passages
Search aliases: t^-s
Use in source · Read passage · spectral_continuation.tex, line 49
pageheadingORBIT SPECTRA / MELLIN PROOFThe two ends have different jobs For 1/N<Re s<1, the genus-zero log derivative and the angular gap give absolute interchange in Z_N(s)=fracsinπ sπ ∫_0^∞ t^-sfracddtlog E_N(t)dd t. Each term uses the principal-branch beta integral ∫_0^∞ t^-s(t+w)^-1dd t=π w^-s/sinπ s. The denominator estimate bounds the absolute integral by a constant times |w|^-Re s, whose sum converges in this strip.
Exact mathematical passage
\mathcal Z_N(s)=\frac{\sin\pi s}{\pi} \int_0^\infty t^{-s}\frac{d}{dt}\log E_N(t)\dd t.
t^{-\sigma}Use in source · 1 passage
Search aliases: t^-σ sigma
Use in source · Read passage · spectral_full.tex, line 169
∑_j∫_0^∞fract^-σ|t+w_j| dt ≤fracπcsin(πσ)∑_j|w_j|^-σ<∞. tagS9
Exact mathematical passage
\sum_j\int_0^\infty\frac{t^{-\sigma}}{|t+w_j|}\,dt \le\frac{\pi}{c\sin(\pi\sigma)}\sum_j|w_j|^{-\sigma}<\infty. \tag{S9}
T^{1/N}Use in source · 1 passage
Search aliases: T^1/N
Use in source · Read passage · spectral_continuation.tex, line 75
At zero the finite parts give I_K(0)=lim_T→∞[log E_N(T)-T^1/N]=-log N. Since (sinπ s/π)'_s=0=1, this equals Z_N'(0). Absolute convergence at a smaller real exponent would give a holomorphic series across 1/N, contradicting the pole and proving the exact abscissa.
Exact mathematical passage
I_K(0)=\lim_{T\to\infty}[\log E_N(T)-T^{1/N}]=-\log N.
t^{1/N}Use in source · 2 passages
Search aliases: t^1/N
Use in source · Read passage · spectral_continuation.tex, line 29
pageheadingORBIT SPECTRA / LARGE RADIUSA finite exponential sum controls the tail Put E_N(t)=d_N(-t)=∑_n≥0t^n/(Nn)! and ω=e^2π i/N. For x≥0, E_N(x^N)=frac1N∑_j=0^N-1e^ω^jx. Root filtering proves this identity term by term: only powers divisible by N remain. The positive real exponential is dominant, so log E_N(t)=t^1/N-log N+O(e^-κ_N t^1/N), κ_N=1-cos(2π/N)>0, with corresponding differentiated estimates on the positive ray. fig14_spectral_tailComputed logarithmic derivatives of the finite expone…
Exact mathematical passage
\log E_N(t)=t^{1/N}-\log N+O(e^{-\kappa_N t^{1/N}}), \quad \kappa_N=1-\cos(2\pi/N)>0,
t^{1/N-1}Use in source · 1 passage
Search aliases: t^1/N-1
Use in source · Read passage · spectral_continuation.tex, line 61
Near zero write f(t)=(log E_N)'(t)=∑_k≥1α_kt^k-1, where α_k=(-1)^k+1 Z_N(k). For any integer K≥1, subtract K terms on [0,1]: I_K(s)= · ∫_0^1t^-s[f(t)-∑_k=1^Kα_kt^k-1]dd t +∑_k=1^Kfracα_kk-s · +∫_1^∞ t^-s[f(t)-frac1N t^1/N-1]dd t +frac1N(s-1/N). The first integral is holomorphic for Re s<K+1. The second is entire because its remainder decays exponentially in t^1/N. The factor sinπ s cancels the positive-integer poles. Increasing K covers the plane, leaving only 1/N. At nonpositive integers I_K is…
Exact mathematical passage
\begin{aligned} I_K(s)={}&\int_0^1t^{-s}\left[f(t)-\sum_{k=1}^K\alpha_kt^{k-1}\right]\dd t +\sum_{k=1}^K\frac{\alpha_k}{k-s}\\ &+\int_1^\infty t^{-s}\left[f(t)-\frac1N t^{1/N-1}\right]\dd t +\frac{1}{N(s-1/N)}. \end{aligned}
t^{1/N-1-k}Use in source · 1 passage
Search aliases: t^1/N-1-k
Use in source · Read passage · spectral_continuation.tex, line 94
^-s, Re s>1/N. The rightward shift preserves avoidance of the negative branch cut. Normalize the product with E_N(a) and use log E_N(a+t) in the Mellin argument. For a>0 there are genuine simple poles at s=1/N-k, k≥0: Res_s=1/N-k Z_N(s;a) =fracsin(π/N)Nπfrac(1-1/N)_kk!a^k. Indeed the large-radius derivative expands as frac1N(a+t)^1/N-1 =frac1N∑_k≥0frac(-1)^k(1-1/N)_kk!a^kt^1/N-1-k as an asymptotic series. The alternating sign is canceled by sin(π(1/N-k))=(-1)^ksin(π/N). At a=0 all secondary resi…
Exact mathematical passage
\frac1N(a+t)^{1/N-1} =\frac1N\sum_{k\ge0}\frac{(-1)^k(1-1/N)_k}{k!}a^kt^{1/N-1-k}
t^2Use in source · 1 passage
Search aliases: t^2
Use in source · Read passage · resolution_full.tex, line 98
For N≥1 and 0≤ R≤1, put t=3R/4. If (N+2)t^2≤ N+1, then n t^2n+1 is nonincreasing for n≥ N+1, and hence
Exact mathematical passage
(N+2)t^2\le N+1
t^{2n+1}Use in source · 2 passages
Search aliases: t^2n+1
Use in source · Read passage · resolution_full.tex, line 98
For N≥1 and 0≤ R≤1, put t=3R/4. If (N+2)t^2≤ N+1, then n t^2n+1 is nonincreasing for n≥ N+1, and hence
Exact mathematical passage
n t^{2n+1}
t^{2N+3}Use in source · 1 passage
Search aliases: t^2N+3
Use in source · Read passage · resolution_full.tex, line 101
0≤ L(R)-∑_n=1^Nn d_nR^2n+1 ≤12(N+1)t^2N+3. tagRA-2
Exact mathematical passage
0\le L(R)-\sum_{n=1}^{N}n d_nR^{2n+1} \le12(N+1)t^{2N+3}. \tag{RA-2}
t^kUse in source · 1 passage
Search aliases: t^k
Use in source · Read passage · spectral_full.tex, line 218
G_N(t)=∑_k≥0a_N,kt^k, g_N(x)=∑_m≥1N a_N,m-1x^Nm-1. tagS12
Exact mathematical passage
G_N(t)=\sum_{k\ge0}a_{N,k}t^k, \qquad g_N(x)=\sum_{m\ge1}N a_{N,m-1}x^{Nm-1}. \tag{S12}
t^{k-1}Use in source · 2 passages
Search aliases: t^k-1
Use in source · Read passage · spectral_continuation.tex, line 58
Near zero write f(t)=(log E_N)'(t)=∑_k≥1α_kt^k-1, where α_k=(-1)^k+1 Z_N(k). For any integer K≥1, subtract K terms on [0,1]: I_K(s)= · ∫_0^1t^-s[f(t)-∑_k=1^Kα_kt^k-1]dd t +∑_k=1^Kfracα_kk-s
Exact mathematical passage
f(t)=(\log E_N)'(t)=\sum_{k\ge1}\alpha_kt^{k-1}
t^nUse in source · 2 passages
Search aliases: t^n
Use in source · Read passage · sequence_polyphase.tex, line 61
pageheadingEXACT LABORATORY / PROJECTORSPolyphase sections preserve the full sequence For an ordinary generating function G(z)=∑_m≥0a_mz^m and integer b≥1, write G(z)=∑_j=0^b-1z^jG_j(z^b), G_j(t)=∑_n≥0a_bn+jt^n. The root projector is P_b,jG(z)=frac1b∑_k=0^b-1ω^-jkG(ω^kz) =z^jG_j(z^b), ω=e^2π i/b. The same projector works on an EGF, but the section retains its original factorials: ∑_na_bn+jz^bn+j/(bn+j)!. Removing those denominators changes the representation and requires an explicit transport. R…
Exact mathematical passage
G(z)=\sum_{j=0}^{b-1}z^jG_j(z^b),\qquad G_j(t)=\sum_{n\ge0}a_{bn+j}t^n.
t^\alphaUse in source · 1 passage
Search aliases: t^α alpha
Use in source · Read passage · spectral_full.tex, line 469
Integrating exactly and using log H_N(a+t)-t^α→-log N gives
Exact mathematical passage
\log H_N(a+t)-t^\alpha\to-\log N
t^{\alpha-1}Use in source · 2 passages
Search aliases: t^α-1 alpha
Use in source · Read passage · spectral_full.tex, line 461
Finally, the one-term subtraction is already convergent at s=0, since G_N,a(t)-α t^α-1=O(t^α-2) and α<1. Let
Exact mathematical passage
G_{N,a}(t)-\alpha t^{\alpha-1}=O(t^{\alpha-2})
t^{\alpha-2}Use in source · 1 passage
Search aliases: t^α-2 alpha
Use in source · Read passage · spectral_full.tex, line 461
Finally, the one-term subtraction is already convergent at s=0, since G_N,a(t)-α t^α-1=O(t^α-2) and α<1. Let
Exact mathematical passage
G_{N,a}(t)-\alpha t^{\alpha-1}=O(t^{\alpha-2})
t^{\alpha-k-1}Use in source · 1 passage
Search aliases: t^α-k-1 alpha
Use in source · Read passage · spectral_full.tex, line 398
G_N,a(t)=∑_k=0^Kαbinomα-1ka^k t^α-k-1 +O(t^α-K-2). tagT5
Exact mathematical passage
G_{N,a}(t)=\sum_{k=0}^{K}\alpha\binom{\alpha-1}{k}a^k t^{\alpha-k-1} +O(t^{\alpha-K-2}). \tag{T5}
t^{\alpha-K-2}Use in source · 1 passage
Search aliases: t^α-K-2 alpha
Use in source · Read passage · spectral_full.tex, line 398
G_N,a(t)=∑_k=0^Kαbinomα-1ka^k t^α-k-1 +O(t^α-K-2). tagT5
Exact mathematical passage
G_{N,a}(t)=\sum_{k=0}^{K}\alpha\binom{\alpha-1}{k}a^k t^{\alpha-k-1} +O(t^{\alpha-K-2}). \tag{T5}
T^{\alpha-k-s}Use in source · 1 passage
Search aliases: T^α-k-s alpha
Use in source · Read passage · spectral_full.tex, line 409
fracαbinomα-1ka^k T^α-k-ss-α+k. tagT6
Exact mathematical passage
\frac{\alpha\binom{\alpha-1}{k}a^k T^{\alpha-k-s}}{s-\alpha+k}. \tag{T6}
T_+Use in source · 10 passages
Search aliases: T_+
Use in source · Read passage · resolution_full.tex, line 299
T_-=fraclog(η(p^4+1)/D(4))λ_p, T_+=fraclog(η p^4)λ_p. tagRA-7
Exact mathematical passage
T_-=\frac{\log\bigl(\eta(p^4+1)/D(4)\bigr)}{\lambda_p}, \qquad T_+=\frac{\log(\eta p^4)}{\lambda_p}. \tag{RA-7}
T_-Use in source · 8 passages
Search aliases: T_-
Use in source · Read passage · resolution_full.tex, line 299
T_-=fraclog(η(p^4+1)/D(4))λ_p, T_+=fraclog(η p^4)λ_p. tagRA-7
Exact mathematical passage
T_-=\frac{\log\bigl(\eta(p^4+1)/D(4)\bigr)}{\lambda_p}, \qquad T_+=\frac{\log(\eta p^4)}{\lambda_p}. \tag{RA-7}
T_{1/3,2/3}Use in source · 2 passages
Search aliases: T_1/3,2/3
Use in source · Read passage · main.tex, line 165
claimA worked order-one member T_1/3,2/3(z)=z+frac227z^3+ frac42187z^5+frac23211160261z^7+O(z^9). It has order one, and T_1/3,2/3(1)≈1.0759239897408249. The diagonal point q=r=1/2 gives f_1/2(z)=sqrt2sinh(z/sqrt2). Every off-diagonal member is separated from all rescaled diagonal members by the theorem above.
Exact mathematical passage
T_{1/3,2/3}(z)=z+\frac{2}{27}z^3+ \frac{4}{2187}z^5+\frac{232}{11160261}z^7+O(z^9).
T_4Use in source · 1 passage
Search aliases: T_4
Use in source · chapters_sequences.tex, line 110
16_polynomial_ladder T_n=∑_j=1^nj, H_n=∑_j=1^nT_j, Δ T_n=n, Δ H_n=T_n The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, Δ f_n=f_n-f_n-1. All plotted values use their displayed common index. At n=4, T_4=10 and H_4=20. The differences are 10-6=4 and 20-10=10. Separately, 4^2-3^2=7 and 4^3-3^3=37. Prefix summation raises polynomial degree; differencing low…
Exact mathematical passage
T_4=10
T_5Use in source · 1 passage
Search aliases: T_5
Use in source · chapters_sequences.tex, line 12
CES / 02Longer recurrence memoryS02_tribonacci_tetranacci T_n=T_n-1+T_n-2+T_n-3, Q_n=∑_j=1^4Q_n-j The seed table fixes the OEIS convention. Tribonacci starts 0,0,1; tetranacci starts 0,0,0,1. The curves use log_10(1+a_n) so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence. Tribonacci gives T_5=2+1+1=4. Tetranacci gives Q_7=4+2+1+1=8. The recurrence first applies at n=3 for T and n=4 for Q. These finite-memory recurrences have th…
Exact mathematical passage
T_5=2+1+1=4
T_{q,q}Use in source · 1 passage
Search aliases: T_q,q
Use in source · Read passage · notation.tex, line 11
item[Contracted feedback T_q,r.] The normalized odd entire solution of T'_q,r(z)=1+T_q,r(qz)T_q,r(rz), T_q,r(0)=0, for 0<q,r<1. The diagonal notation is f_q=T_q,q. item[Feedback bank F_Θ.] A finite collection of channels with h_j≥0, ∑_jh_j=1 and 0<q_j,r_j<1, entering F_Θ'=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz). Zero-weight channels are removed before taking an active-channel maximum. item[Maximum modulus and entire order.] M_f(R)=max_|z|=R|f(z)|. For a nonpolynomial entire function, ρ(f)=limsup_R→∞fracloglo…
Exact mathematical passage
f_q=T_{q,q}
T_{q,r}Use in source · 14 passages
Search aliases: T_q,r
Use in source · Read passage · main.tex, line 81
mainmatter MonographPartIEntire functions and their growthContractions, coefficients and nonlinear feedback. Contracted nonlinear feedback. pageheadingFAMILY 1Two scales inside the function For real 0<q,r<1, define the normalized odd solution T_q,r'(z)=1+T_q,r(qz)T_q,r(rz), T_q,r(0)=0. The contractions are part of the defining feedback. The function is symmetric in q,r, but unequal scales retain left/right address information. fig01_asymmetric_functionsComputed Taylor curves. Each entire curve u…
Exact mathematical passage
\boxed{T_{q,r}'(z)=1+T_{q,r}(qz)T_{q,r}(rz),\qquad T_{q,r}(0)=0.}
T_{q,r}'Use in source · 1 passage
Search aliases: T_q,r'
Use in source · Read passage · main.tex, line 81
mainmatter MonographPartIEntire functions and their growthContractions, coefficients and nonlinear feedback. Contracted nonlinear feedback. pageheadingFAMILY 1Two scales inside the function For real 0<q,r<1, define the normalized odd solution T_q,r'(z)=1+T_q,r(qz)T_q,r(rz), T_q,r(0)=0. The contractions are part of the defining feedback. The function is symmetric in q,r, but unequal scales retain left/right address information. fig01_asymmetric_functionsComputed Taylor curves. Each entire curve u…
Exact mathematical passage
\boxed{T_{q,r}'(z)=1+T_{q,r}(qz)T_{q,r}(rz),\qquad T_{q,r}(0)=0.}
\tanUse in source · 1 passage
Search aliases: tan
Use in source · Read passage · resolution_full.tex, line 93
f(4/3)≤tan(4/3)≤12.
Exact mathematical passage
f(4/3)\le\tan(4/3)\le12.
uUse in source · 61 passages
Search aliases: u
Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
u'Use in source · 2 passages
Search aliases: u'
Use in source · Read passage · references.tex, line 25
bibitemDascaliuc2025 Radu Dascaliuc, Tuan N. Pham, Enrique Thomann and Edward C. Waymire. Transformation of Stochastic Recursions and Critical Phenomena in the Analysis of the Aldous-Shields-Athreya Cascade and Related Mean Flow Equations. arXiv:2411.00629v2 (2025). The nonlinear equation u'=-u+u(α t)^2 is an adjacent model, not the equation studied here. https://arxiv.org/abs/2411.00629v2
Exact mathematical passage
u'=-u+u(\alpha t)^2
U_0Use in source · 1 passage
Search aliases: U_0
Use in source · Read passage · dilation_full.tex, line 137
Define U_0=2, U_k=Q^-k(1+U_k-1^2). Then M(R_k)≤ U_k, U_k≥1, and log U_k≤ klog(1/Q)+log2+2log U_k-1. Summing this elementary recurrence gives log U_N≤ V_N. Since e<3, the logarithmic-radius seed 1≤ t≤1+Δ has e^t≤ e/m<3/m≤ R_N, and therefore log M(e^t)≤ V_N.
Exact mathematical passage
U_0=2
u_0Use in source · 2 passages
Search aliases: u_0
Use in source · Read passage · coupled_growth_full.tex, line 34
Then u_0=v_0=1, and, for n≥1,
Exact mathematical passage
u_0=v_0=1
u_1Use in source · 7 passages
Search aliases: u_1
Use in source · Read passage · coupled_inverse_full.tex, line 11
Write U(z)=z+u_1z^3+u_2z^5+u_3z^7+⋯ and V(z)=z+v_1z^3+v_2z^5+v_3z^7+⋯. Put p=qr, k=ab. Coefficient matching gives
Exact mathematical passage
U(z)=z+u_1z^3+u_2z^5+u_3z^7+\cdots
u_2Use in source · 15 passages
Search aliases: u_2
Use in source · REPORT.md, line 39
D=k^2u_2-p^2v_2 =fracpkar15(q^2-b^2)(qb-ar).
Exact mathematical passage
D=k^2u_2-p^2v_2 =\frac{pkar}{15}(q^2-b^2)(qb-ar).
u_3Use in source · 7 passages
Search aliases: u_3
Use in source · Read passage · coupled_inverse_full.tex, line 11
Write U(z)=z+u_1z^3+u_2z^5+u_3z^7+⋯ and V(z)=z+v_1z^3+v_2z^5+v_3z^7+⋯. Put p=qr, k=ab. Coefficient matching gives
Exact mathematical passage
U(z)=z+u_1z^3+u_2z^5+u_3z^7+\cdots
u_iUse in source · 4 passages
Search aliases: u_i
Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
U_kUse in source · 4 passages
Search aliases: U_k
Use in source · Read passage · dilation_full.tex, line 137
Define U_0=2, U_k=Q^-k(1+U_k-1^2). Then M(R_k)≤ U_k, U_k≥1, and log U_k≤ klog(1/Q)+log2+2log U_k-1. Summing this elementary recurrence gives log U_N≤ V_N. Since e<3, the logarithmic-radius seed 1≤ t≤1+Δ has e^t≤ e/m<3/m≤ R_N, and therefore log M(e^t)≤ V_N.
Exact mathematical passage
U_k=Q^{-k}(1+U_{k-1}^2)
U_{k-1}Use in source · 1 passage
Search aliases: U_k-1
Use in source · Read passage · dilation_full.tex, line 138
Define U_0=2, U_k=Q^-k(1+U_k-1^2). Then M(R_k)≤ U_k, U_k≥1, and log U_k≤ klog(1/Q)+log2+2log U_k-1. Summing this elementary recurrence gives log U_N≤ V_N. Since e<3, the logarithmic-radius seed 1≤ t≤1+Δ has e^t≤ e/m<3/m≤ R_N, and therefore log M(e^t)≤ V_N.
Exact mathematical passage
\log U_k\le k\log(1/Q)+\log2+2\log U_{k-1}
U_{k-1}^2Use in source · 1 passage
Search aliases: U_k-1^2
Use in source · Read passage · dilation_full.tex, line 137
Define U_0=2, U_k=Q^-k(1+U_k-1^2). Then M(R_k)≤ U_k, U_k≥1, and log U_k≤ klog(1/Q)+log2+2log U_k-1. Summing this elementary recurrence gives log U_N≤ V_N. Since e<3, the logarithmic-radius seed 1≤ t≤1+Δ has e^t≤ e/m<3/m≤ R_N, and therefore log M(e^t)≤ V_N.
Exact mathematical passage
U_k=Q^{-k}(1+U_{k-1}^2)
U_NUse in source · 1 passage
Search aliases: U_N
Use in source · Read passage · dilation_full.tex, line 139
Define U_0=2, U_k=Q^-k(1+U_k-1^2). Then M(R_k)≤ U_k, U_k≥1, and log U_k≤ klog(1/Q)+log2+2log U_k-1. Summing this elementary recurrence gives log U_N≤ V_N. Since e<3, the logarithmic-radius seed 1≤ t≤1+Δ has e^t≤ e/m<3/m≤ R_N, and therefore log M(e^t)≤ V_N.
Exact mathematical passage
\log U_N\le V_N
u_nUse in source · 9 passages
Search aliases: u_n
Use in source · REPORT.md, line 22
0<u_n≤ 2^-n, 0<v_n≤ 2^-n.
Exact mathematical passage
0<u_n\le 2^{-n},\qquad 0<v_n\le 2^{-n}.
u_{n+1}Use in source · 1 passage
Search aliases: u_n+1
Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
U_{p,j}Use in source · 2 passages
Search aliases: U_p,j
Use in source · Read passage · multi_prime.tex, line 28
The exact coordinate formula is M_j(λ)=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) ∏_p∈ F(1+fracλ_pp^4+2r+1). Coefficients of the monic families lie in Q(π^-2,λ_p:p∈ F). For two primes, the overlap term is visible: M_j=M_j(0,0)+λ_pU_p,j+λ_qU_q,j +λ_pλ_qU_pq,j. Here U_pq,j>0 is the pq-divisible submeasure moment. The family does not reduce to adding separate prime responses without their overlap.
Exact mathematical passage
M_j=M_j(0,0)+\lambda_pU_{p,j}+\lambda_qU_{q,j} +\lambda_p\lambda_qU_{pq,j}.
U_{pq,j}Use in source · 4 passages
Search aliases: U_pq,j
Use in source · Read passage · multi_prime.tex, line 28
The exact coordinate formula is M_j(λ)=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) ∏_p∈ F(1+fracλ_pp^4+2r+1). Coefficients of the monic families lie in Q(π^-2,λ_p:p∈ F). For two primes, the overlap term is visible: M_j=M_j(0,0)+λ_pU_p,j+λ_qU_q,j +λ_pλ_qU_pq,j. Here U_pq,j>0 is the pq-divisible submeasure moment. The family does not reduce to adding separate prime responses without their overlap.
Exact mathematical passage
M_j=M_j(0,0)+\lambda_pU_{p,j}+\lambda_qU_{q,j} +\lambda_p\lambda_qU_{pq,j}.
U_{q,j}Use in source · 2 passages
Search aliases: U_q,j
Use in source · Read passage · multi_prime.tex, line 28
The exact coordinate formula is M_j(λ)=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) ∏_p∈ F(1+fracλ_pp^4+2r+1). Coefficients of the monic families lie in Q(π^-2,λ_p:p∈ F). For two primes, the overlap term is visible: M_j=M_j(0,0)+λ_pU_p,j+λ_qU_q,j +λ_pλ_qU_pq,j. Here U_pq,j>0 is the pq-divisible submeasure moment. The family does not reduce to adding separate prime responses without their overlap.
Exact mathematical passage
M_j=M_j(0,0)+\lambda_pU_{p,j}+\lambda_qU_{q,j} +\lambda_p\lambda_qU_{pq,j}.
\underline tUse in source · 1 passage
Search aliases: underline t underline
Use in source · Read passage · coupled_growth_full.tex, line 385
underline t≥ P(ρ)underline t, underline t=(underlineτ_1,underlineτ_2)^T.
Exact mathematical passage
\underline t\ge P(\rho)\underline t, \qquad \underline t=(\underline\tau_1,\underline\tau_2)^T.
\underline\tauUse in source · 9 passages
Search aliases: underlineτ underline tau
Use in source · Read passage · coupled_growth_full.tex, line 174
underlineτ_i=liminf_R→∞fraclog M_F_i(R)R^ρ, overlineτ_i=limsup_R→∞fraclog M_F_i(R)R^ρ.
Exact mathematical passage
\underline\tau_i=\liminf_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho}, \qquad \overline\tau_i=\limsup_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho}.
\underline{\boldsymbol\tau}Use in source · 1 passage
Search aliases: underlineτ underline boldsymbol tau
Use in source · Read passage · coupled_visual.tex, line 81
For F_1=U and F_2=V, define underlineτ_i=liminf_R→∞fraclog M_F_i(R)R^ρ, overlineτ_i=limsup_R→∞fraclog M_F_i(R)R^ρ. If P(ρ)w=w with w>0, then both type vectors have precisely these Perron proportions: underlineτ=underlineτ w, overlineτ=overlineτ w, qquad0<underlineτ≤overlineτ<∞. For the worked pair, both the lower type and the upper type of V are twice the corresponding type of U. Either both ordinary type limits exist or neither does. Equality of the lower and upper types remains open. The finit…
Exact mathematical passage
\boxed{\underline{\boldsymbol\tau}=\underline\tau\,w, \qquad\overline{\boldsymbol\tau}=\overline\tau\,w, \qquad0<\underline\tau\le\overline\tau<\infty.}
\underline\tau_1Use in source · 3 passages
Search aliases: underlineτ_1 underline tau
Use in source · Read passage · coupled_growth_full.tex, line 363
(underlineτ_1,underlineτ_2)^T · =underlineτ w, (overlineτ_1,overlineτ_2)^T · =overlineτ w.
Exact mathematical passage
\boxed{\begin{aligned} (\underline\tau_1,\underline\tau_2)^T&=\underline\tau\,w,\\ (\overline\tau_1,\overline\tau_2)^T&=\overline\tau\,w. \end{aligned}}
\underline\tau_2Use in source · 3 passages
Search aliases: underlineτ_2 underline tau
Use in source · Read passage · coupled_growth_full.tex, line 363
(underlineτ_1,underlineτ_2)^T · =underlineτ w, (overlineτ_1,overlineτ_2)^T · =overlineτ w.
Exact mathematical passage
\boxed{\begin{aligned} (\underline\tau_1,\underline\tau_2)^T&=\underline\tau\,w,\\ (\overline\tau_1,\overline\tau_2)^T&=\overline\tau\,w. \end{aligned}}
\underline\tau_iUse in source · 3 passages
Search aliases: underlineτ_i underline tau
Use in source · Read passage · coupled_growth_full.tex, line 174
underlineτ_i=liminf_R→∞fraclog M_F_i(R)R^ρ, overlineτ_i=limsup_R→∞fraclog M_F_i(R)R^ρ.
Exact mathematical passage
\underline\tau_i=\liminf_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho}, \qquad \overline\tau_i=\limsup_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho}.
\uparrowUse in source · 2 passages
Search aliases: uparrow
Use in source · Read passage · evidence_notes.tex, line 20
The earlier moving-width lower argument used a fixed α<1; it proved orders below ρ and then recovered the order by αuparrow1, but did not establish positive lower type at ρ. This is a limitation of that argument, not a counterexample. The fixed-width comparison closes precisely that missing implication. The evidence archive retains execution failures, their causes and the subsequent corrected runs.
Exact mathematical passage
\alpha\uparrow1
vUse in source · 59 passages
Search aliases: v
Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
v'Use in source · 1 passage
Search aliases: v'
Use in source · Read passage · coupled_structure_full.tex, line 205
Zero scales lie outside Theorem H.1. On zero-constant inputs, a product vanishes identically if either of its scales is zero. If both products vanish, both equations are already u'=v'=1. If only the first product vanishes, the projection u is autonomous, regardless of whether the second pair agrees with the first. Thus extending Theorem H.1 to zero scales would be false. Scale one and positive scales greater than one do not affect the algebraic operator theorem, although the contraction-domain e…
Exact mathematical passage
u'=v'=1
v^{r+2}Use in source · 1 passage
Search aliases: v^r+2
Use in source · Read passage · sieve_full.tex, line 54
Depth-k moments are M_k+j. With v=π^-2, the even-zeta formula makes every D(4+2r) a rational multiple of v^r+2. Thus the moments belong to Q[v,λ_p:p∈ F], and monic orthogonal-polynomial coefficients belong to Q(v,λ_p:p∈ F). This is an exact field statement, with no decimal recognition step.
Exact mathematical passage
v^{r+2}
v_0Use in source · 2 passages
Search aliases: v_0
Use in source · Read passage · coupled_growth_full.tex, line 34
Then u_0=v_0=1, and, for n≥1,
Exact mathematical passage
u_0=v_0=1
v_1Use in source · 7 passages
Search aliases: v_1
Use in source · Read passage · coupled_inverse_full.tex, line 12
Write U(z)=z+u_1z^3+u_2z^5+u_3z^7+⋯ and V(z)=z+v_1z^3+v_2z^5+v_3z^7+⋯. Put p=qr, k=ab. Coefficient matching gives
Exact mathematical passage
V(z)=z+v_1z^3+v_2z^5+v_3z^7+\cdots
V_2Use in source · 2 passages
Search aliases: V_2
Use in source · Read passage · resolution_full.tex, line 160
The value of C_0 comes directly from the earlier explicit constants: V_2=15log2, Δ=2log2, and B_+=5log2. Thus V_2+1+Δ+B_+=1+22log2.
Exact mathematical passage
V_2=15\log2
v_2Use in source · 14 passages
Search aliases: v_2
Use in source · REPORT.md, line 39
D=k^2u_2-p^2v_2 =fracpkar15(q^2-b^2)(qb-ar).
Exact mathematical passage
D=k^2u_2-p^2v_2 =\frac{pkar}{15}(q^2-b^2)(qb-ar).
v_3Use in source · 5 passages
Search aliases: v_3
Use in source · Read passage · coupled_inverse_full.tex, line 12
Write U(z)=z+u_1z^3+u_2z^5+u_3z^7+⋯ and V(z)=z+v_1z^3+v_2z^5+v_3z^7+⋯. Put p=qr, k=ab. Coefficient matching gives
Exact mathematical passage
V(z)=z+v_1z^3+v_2z^5+v_3z^7+\cdots
v_jUse in source · 3 passages
Search aliases: v_j
Use in source · Read passage · dilation_full.tex, line 238
Put v=h-g. Its positive and negative parts both have sum d/2, since ∑_jv_j=0. At every point of the segment between g and h, all partial derivatives lie in the interval [0,n d_n(Q)]. Hence
Exact mathematical passage
\sum_jv_j=0
V_KUse in source · 3 passages
Search aliases: V_K
Use in source · Read passage · dilation_growth.tex, line 66
claimExplicit upper constants are available too Set Q=max_jmax(q_j,r_j), m=min_jmin(q_j,r_j), Δ=log(1/m), and B_+=max_jlog(1/(q_jr_j))+log2. Choose an integer K with Q^-K≥3/m and put V_K=(2^K+1-1)log2+(2^K+1-K-2)log(1/Q), c_+=e^-ρ(V_K+1+Δ+B_+). Then log M(R)≤ c_+R^ρ-log R-B_+ for R≥ e. The tangent majorant M(1)<2, followed by the integral recurrence on radii Q^-k, bounds the initial seed by V_K. The logarithmic-radius comparison then supplies the displayed upper constant. The full induction is i…
Exact mathematical passage
V_K=(2^{K+1}-1)\log2+(2^{K+1}-K-2)\log(1/Q),
V_NUse in source · 5 passages
Search aliases: V_N
Use in source · Read passage · dilation_full.tex, line 120
V_N=(2^N+1-1)log2+ (2^N+1-N-2)log(1/Q),
Exact mathematical passage
V_N=(2^{N+1}-1)\log2+ (2^{N+1}-N-2)\log(1/Q),
v_nUse in source · 9 passages
Search aliases: v_n
Use in source · REPORT.md, line 22
0<u_n≤ 2^-n, 0<v_n≤ 2^-n.
Exact mathematical passage
0<u_n\le 2^{-n},\qquad 0<v_n\le 2^{-n}.
v_{n+1}Use in source · 1 passage
Search aliases: v_n+1
Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
v_{n-1-i}Use in source · 1 passage
Search aliases: v_n-1-i
Use in source · Read passage · coupled_growth_full.tex, line 36
u_n · =frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1u_i v_n-1-i, v_n · =frac12n+1∑_i=0^n-1 a^2i+1b^2(n-1-i)+1u_i v_n-1-i.
Exact mathematical passage
\begin{aligned} u_n&=\frac1{2n+1}\sum_{i=0}^{n-1} q^{2i+1}r^{2(n-1-i)+1}u_i v_{n-1-i},\\ v_n&=\frac1{2n+1}\sum_{i=0}^{n-1} a^{2i+1}b^{2(n-1-i)+1}u_i v_{n-1-i}. \end{aligned}
v_{n-i}Use in source · 1 passage
Search aliases: v_n-i
Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
v_pUse in source · 1 passage
Search aliases: v_p
Use in source · Read passage · arithmetic_full.tex, line 55
The function itself therefore encodes the weight exponent, the depth exponent, and the finite excluded prime set. It does not encode valuations v_p(Q).
Exact mathematical passage
v_p(Q)
WUse in source · 14 passages
Search aliases: W
Use in source · Read passage · dilation_full.tex, line 186
Each is homogeneous of degree n in the weights. If a parameter set has |q_j|,|r_j|≤ Q<1 and ∑_j|h_j|≤ W, with W>0, coefficient induction gives
Exact mathematical passage
\sum_j|h_j|\le W
wUse in source · 118 passages
Search aliases: w
Use in source · Read passage · main.tex, line 248
mid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit working precision. Right: initial squarefree frequencies and weights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If p_* is the least prime not dividin…
Exact mathematical passage
A(s)=\sum_{n\in\mathcal S_Q}w_ne^{-\lambda_ns},\qquad w_n=n^{-2a}.
w'Use in source · 1 passage
Search aliases: w'
Use in source · Read passage · coupled_structure_full.tex, line 196
w'=2+tfrac12(w(sz)^2-d(sz)^2), d'=0.
Exact mathematical passage
w'=2+\tfrac12\bigl(w(sz)^2-d(sz)^2\bigr),\qquad d'=0.
w^{(t+l-j)/b}Use in source · 1 passage
Search aliases: w^(t+l-j)/b
Use in source · Read passage · sequence_full.tex, line 360
Y_j(w)=∑_l=0^b-1H_jl(w)X_l(w), H_jl(w)=∑_substack0≤ t<L t+l≡ j (b) h_t w^(t+l-j)/b.
Exact mathematical passage
\boxed{Y_j(w)=\sum_{l=0}^{b-1}H_{jl}(w)X_l(w),\qquad H_{jl}(w)=\sum_{\substack{0\le t<L\\t+l\equiv j\,(b)}} h_t w^{(t+l-j)/b}.}
W^{-1/2}Use in source · 2 passages
Search aliases: W^-1/2
Use in source · Read passage · dilation_full.tex, line 196
∑_n≥0|c_n(h,q,r)|R^2n+1 ≤ W^-1/2f_Q(sqrt W R)<∞.
Exact mathematical passage
\sum_{n\ge0}|c_n(h,q,r)|R^{2n+1} \le W^{-1/2}f_Q(\sqrt W R)<\infty.
w^{-s}Use in source · 1 passage
Search aliases: w^-s
Use in source · Read passage · spectral_continuation.tex, line 54
pageheadingORBIT SPECTRA / MELLIN PROOFThe two ends have different jobs For 1/N<Re s<1, the genus-zero log derivative and the angular gap give absolute interchange in Z_N(s)=fracsinπ sπ ∫_0^∞ t^-sfracddtlog E_N(t)dd t. Each term uses the principal-branch beta integral ∫_0^∞ t^-s(t+w)^-1dd t=π w^-s/sinπ s. The denominator estimate bounds the absolute integral by a constant times |w|^-Re s, whose sum converges in this strip.
Exact mathematical passage
\int_0^\infty t^{-s}(t+w)^{-1}\dd t=\pi w^{-s}/\sin\pi s
w^dUse in source · 1 passage
Search aliases: w^d
Use in source · Read passage · sequence_full.tex, line 371
Proof. A term h_t a_bn+j-t is in phase l exactly when t+l≡ jpmod b. Write d=(t+l-j)/b. That term is h_t x_l[n-d], whose OGF is h_tw^dX_l(w) by zero extension. Summing over taps and phases gives the matrix. square
Exact mathematical passage
h_tw^dX_l(w)
w^kUse in source · 1 passage
Search aliases: w^k
Use in source · Read passage · main.tex, line 500
ts. Multiplicity is retained by the definition; simplicity is not assumed in general. The bound |d_N(w)|≤ e^|w|^1/N and the finite-order zero-counting theorem give absolute local uniform convergence in the stated half-plane. Since the order is below one and d_N(0)=1, the Hadamard factorization has genus zero: d_N(w)=∏_j(1-w/w_j). At positive integers, a log derivative gives -wfracd_N'(w)d_N(w)=∑_k≥1 Z_N(k)w^k. If a_n=(-1)^n/(Nn)! and b_k= Z_N(k), then b_k=-ka_k-∑_j=1^k-1a_jb_k-j. Thus every posi…
Exact mathematical passage
-w\frac{d_N'(w)}{d_N(w)}=\sum_{k\ge1}\mathcal Z_N(k)w^k.
W^nUse in source · 2 passages
Search aliases: W^n
Use in source · Read passage · dilation_full.tex, line 189
|c_n(h,q,r)|≤ W^n d_n(Q),
Exact mathematical passage
|c_n(h,q,r)|\le W^n d_n(Q),
w^nUse in source · 5 passages
Search aliases: w^n
Use in source · Read passage · main.tex, line 480
MonographPartIIIOrbit spectra and exact coordinatesSpectral continuation and the coefficient language of polyphase transforms. Spectral functions of zero orbits. pageheadingFAMILY 4One spectral address for each rotation orbit For integer N≥2, define the classical Mittag--Leffler seed~[source reference] d_N(w)=∑_n≥0frac(-1)^nw^n(Nn)!=E_N,1(-w), D_N(z)=d_N(z^N). Let w_j be the nonzero zeros of d_N, counted with multiplicity. Each corresponds to one rotation orbit of N zeros of D_N. Using Argw∈(-π,…
Exact mathematical passage
d_N(w)=\sum_{n\ge0}\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w), \qquad D_N(z)=d_N(z^N).
w_1Use in source · 7 passages
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Use in source · Read passage · arithmetic_full.tex, line 299
η κ_j(t), κ_j(t)=fracA(-t)R_j(-t) ∼fracw_1w_je^(λ_1-λ_j)t.
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\eta\,\kappa_j(t),\qquad \kappa_j(t)=\frac{A(-t)}{R_j(-t)} \sim\frac{w_1}{w_j}e^{(\lambda_1-\lambda_j)t}.
w_2Use in source · 5 passages
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Use in source · Read passage · coupled_growth_full.tex, line 133
Both coordinates of w are positive, max(w_1,w_2)=1, and direct multiplication gives
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\max(w_1,w_2)=1
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Use in source · Read passage · coupled_growth_full.tex, line 184
0<c_-w_i≤underlineτ_i≤overlineτ_i≤ c_+w_i<∞, i=1,2.
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0<c_-w_i\le\underline\tau_i\le\overline\tau_i\le c_+w_i<\infty, \qquad i=1,2.
w_jUse in source · 43 passages
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Use in source · Read passage · main.tex, line 484
act coordinatesSpectral continuation and the coefficient language of polyphase transforms. Spectral functions of zero orbits. pageheadingFAMILY 4One spectral address for each rotation orbit For integer N≥2, define the classical Mittag--Leffler seed~[source reference] d_N(w)=∑_n≥0frac(-1)^nw^n(Nn)!=E_N,1(-w), D_N(z)=d_N(z^N). Let w_j be the nonzero zeros of d_N, counted with multiplicity. Each corresponds to one rotation orbit of N zeros of D_N. Using Argw∈(-π,π], define Z_N(s)=∑_jw_j^-s, Re s>1/…
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w_j
w_j^{-s}Use in source · 2 passages
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Use in source · Read passage · main.tex, line 487
4One spectral address for each rotation orbit For integer N≥2, define the classical Mittag--Leffler seed~[source reference] d_N(w)=∑_n≥0frac(-1)^nw^n(Nn)!=E_N,1(-w), D_N(z)=d_N(z^N). Let w_j be the nonzero zeros of d_N, counted with multiplicity. Each corresponds to one rotation orbit of N zeros of D_N. Using Argw∈(-π,π], define Z_N(s)=∑_jw_j^-s, Re s>1/N. fig06_orbit_zerosFor N=4, the four real/imaginary zeros in each orbit share one positive coordinate w=z^4. The plot shows the first six orbit…
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\boxed{\mathcal Z_N(s)=\sum_jw_j^{-s},\qquad\Re s>1/N.}
w_kUse in source · 3 passages
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Use in source · Read passage · arithmetic_full.tex, line 72
Every limit exists and recovers exactly the j-th support frequency and weight. Indeed, if M_j+1=∑_k>jw_k, then for every t≥0,
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M_{j+1}=\sum_{k>j}w_k
W_NUse in source · 4 passages
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Use in source · Read passage · coupled_growth_full.tex, line 279
W_N=(2^N+1-1)log2 +(2^N+1-N-2)log(1/Q),
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W_N=(2^{N+1}-1)\log2 +(2^{N+1}-N-2)\log(1/Q),
w_nUse in source · 1 passage
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Use in source · Read passage · main.tex, line 248
mid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit working precision. Right: initial squarefree frequencies and weights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If p_* is the least prime not dividin…
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A(s)=\sum_{n\in\mathcal S_Q}w_ne^{-\lambda_ns},\qquad w_n=n^{-2a}.
\wedgeUse in source · 1 passage
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Use in source · Read passage · coupled_structure_full.tex, line 133
It has rank one, but is not symmetric: B-B^ T=ℓ_1wedgeℓ_2≠0. At four distinct nonzero nodes, polynomial interpolation lets their vector values vary independently, so terms supported at (az,bz) cannot cancel this coefficient. A scalar autonomous channel at the same node pair would require B=κℓ⊗ℓ, a symmetric rank-one tensor, after projecting the output onto that channel. For any output row with a nonzero coefficient on this term, the alternating part remains nonzero. The other node pair gives the…
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B-B^{\mathsf T}=\ell_1\wedge\ell_2\ne0
\widehat\lambdaUse in source · 5 passages
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Use in source · Read passage · arithmetic_full.tex, line 225
widehatλ_j(t,h) =fraclog R_j(-t-h)-log R_j(-t)h.
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\widehat\lambda_j(t,h) =\frac{\log R_j(-t-h)-\log R_j(-t)}h.
\widehat\lambda_jUse in source · 5 passages
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Use in source · Read passage · arithmetic_full.tex, line 225
widehatλ_j(t,h) =fraclog R_j(-t-h)-log R_j(-t)h.
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\widehat\lambda_j(t,h) =\frac{\log R_j(-t-h)-\log R_j(-t)}h.
\widetilde wUse in source · 1 passage
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Use in source · Read passage · coupled_growth_full.tex, line 127
widetilde w=(r^ρ,1-q^ρ)^T, w=fracwidetilde wmax(r^ρ,1-q^ρ).
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\widetilde w=(r^\rho,1-q^\rho)^T, \qquad w=\frac{\widetilde w}{\max(r^\rho,1-q^\rho)}.
\widetilde\nuUse in source · 1 passage
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Use in source · Read passage · sieve_full.tex, line 288
widetildeν_F,γ =∑_n≥1μ(n)^2n^-4 (1+∑_p∈ Fγ_p1_pmid n)δ_x_n.
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\widetilde\nu_{F,\boldsymbol\gamma} =\sum_{n\ge1}\mu(n)^2n^{-4} \left(1+\sum_{p\in F}\gamma_p\mathbf1_{p\mid n}\right)\delta_{x_n}.
\widetilde\nu_{F,\boldsymbol\gamma}Use in source · 1 passage
Search aliases: widetildeν_F,γ widetilde nu boldsymbol gamma
Use in source · Read passage · sieve_full.tex, line 288
widetildeν_F,γ =∑_n≥1μ(n)^2n^-4 (1+∑_p∈ Fγ_p1_pmid n)δ_x_n.
Exact mathematical passage
\widetilde\nu_{F,\boldsymbol\gamma} =\sum_{n\ge1}\mu(n)^2n^{-4} \left(1+\sum_{p\in F}\gamma_p\mathbf1_{p\mid n}\right)\delta_{x_n}.
XUse in source · 15 passages
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Use in source · REPORT.md, line 48
(x-X)(pX-ky)=0.
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(x-X)(pX-ky)=0.
x'Use in source · 5 passages
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Use in source · Read passage · coupled_inverse_full.tex, line 121
This leaves at most two positive candidates. If there are two distinct positive roots x,x', their product is xx'=kp. The corresponding seventh-coefficient values differ by
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x,x'
x'^2Use in source · 2 passages
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Use in source · Read passage · coupled_inverse_full.tex, line 124
u_3(x)-u_3(x') =frac p7(x^2-x'^2) (u_2-fracp^2k^2v_2) =fracpD7k^2(x^2-x'^2).
Exact mathematical passage
u_3(x)-u_3(x') =\frac p7(x^2-x'^2) \left(u_2-\frac{p^2}{k^2}v_2\right) =\frac{pD}{7k^2}(x^2-x'^2).
x^{-2a}Use in source · 3 passages
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Use in source · Read passage · main.tex, line 332
(n,Q)=1μ(n)^2 =δ_Q x+O_Q(sqrtx). Integrate against g_s(x)=x^-2a(1-x^-β)^2s on [1,∞). The density term becomes the beta integral after y=x^-β. Partial integration bounds the error by C_Q∫_1^∞sqrtx|g_s'(x)| d x. The two derivative terms are beta integrals with decay exponent (2a-1/2)/β; |s|/Re s stays bounded in the stated sector. This gives the error term and then the leading asymptotic.
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g_s(x)=x^{-2a}(1-x^{-\beta})^{2s}
x^{-Ns}Use in source · 4 passages
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Use in source · Read passage · spectral_full.tex, line 192
Z_N(s)=fracsin(π s)π ∫_0^∞ x^-Nsg_N(x) dx, frac1N<Re s<1. tagS11
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\boxed{\mathcal Z_N(s)=\frac{\sin(\pi s)}{\pi} \int_0^\infty x^{-Ns}g_N(x)\,dx, \qquad \frac1N<\Re s<1.} \tag{S11}
x^{-\beta}Use in source · 4 passages
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Use in source · Read passage · main.tex, line 332
(n,Q)=1μ(n)^2 =δ_Q x+O_Q(sqrtx). Integrate against g_s(x)=x^-2a(1-x^-β)^2s on [1,∞). The density term becomes the beta integral after y=x^-β. Partial integration bounds the error by C_Q∫_1^∞sqrtx|g_s'(x)| d x. The two derivative terms are beta integrals with decay exponent (2a-1/2)/β; |s|/Re s stays bounded in the stated sector. This gives the error term and then the leading asymptotic.
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g_s(x)=x^{-2a}(1-x^{-\beta})^{2s}
X^2Use in source · 1 passage
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Use in source · REPORT.md, line 54
(k^2u-p^2v)(x^2-X^2)=0.
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(k^2u-p^2v)(x^2-X^2)=0.
x^2Use in source · 16 passages
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Use in source · REPORT.md, line 54
(k^2u-p^2v)(x^2-X^2)=0.
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(k^2u-p^2v)(x^2-X^2)=0.
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Use in source · Read passage · notation.tex, line 47
item[Hankel order and polynomial degree.] For moments M_j=∫ x^jddν, the order-m matrix is H_m=[M_i+j]_0≤ i,j<m. It tests quadratic forms of real polynomials of degree at most m-1. At depth k, the matrix is [M_k+i+j]_0≤ i,j<m. An order-five test therefore covers degree four, not degree five. item[Signed quadratic form.] For real polynomials, Q_ν(f)=∫ f(x)^2ddν(x). On a positive measure it is a squared norm. On a signed measure it may be negative and is not a norm. item[Orbit spectral zeta Z_N.] L…
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M_j=\int x^j\dd\nu
x^kUse in source · 6 passages
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Use in source · Read passage · main.tex, line 366
Prime sieves and signed measures. pageheadingFAMILY 3An infinite prime shelf changes the packets For a prime p and λ≥-1, let x_n=(1-n^-2)^2, ν_p,λ=∑_n≥1μ(n)^2n^-4 (1+λ1_pmid n)δ_x_n. Define monic orthogonal polynomials P_m^(k;p,λ) for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the or…
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x^k\dd\nu_{p,\lambda}
x^mUse in source · 1 passage
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Use in source · Read passage · sieve_full.tex, line 320
If at least two strengths in a positive multiplicative family are nonzero, its entire depth-zero monic polynomial sequence cannot equal any one-prime sequence on the same variable and node system. A direct proof avoids importing a determinacy theorem. Write P_m=x^m+∑_j<mc_mjx^j. The equations ∫ P_m dν=0 recursively determine every normalized moment from the monic sequence. Equal sequences therefore give equal normalized Stieltjes series for |z|>1. Analytic continuation on the connected complemen…
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P_m=x^m+\sum_{j<m}c_{mj}x^j
x^NUse in source · 3 passages
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Use in source · Read passage · spectral_continuation.tex, line 24
pageheadingORBIT SPECTRA / LARGE RADIUSA finite exponential sum controls the tail Put E_N(t)=d_N(-t)=∑_n≥0t^n/(Nn)! and ω=e^2π i/N. For x≥0, E_N(x^N)=frac1N∑_j=0^N-1e^ω^jx. Root filtering proves this identity term by term: only powers divisible by N remain. The positive real exponential is dominant, so log E_N(t)=t^1/N-log N+O(e^-κ_N t^1/N), κ_N=1-cos(2π/N)>0, with corresponding differentiated estimates on the positive ray. fig14_spectral_tailComputed logarithmic derivatives of the finite expone…
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E_N(x^N)=\frac1N\sum_{j=0}^{N-1}e^{\omega^jx}.
x^{N(K+1)-1}Use in source · 1 passage
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Use in source · Read passage · spectral_full.tex, line 238
The first integral is holomorphic for Re s<K+1. This includes local uniform domination, since the Taylor remainder is O(x^N(K+1)-1). The last integral is entire in s, because (S3) gives exponential decay against every power of x, uniformly on compact s-sets, including after any number of s-derivatives. The rational terms exhibit the only possible poles: positive integers 1,…,K and 1/N.
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O(x^{N(K+1)-1})
x^{N-1}Use in source · 1 passage
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Use in source · Read passage · spectral_full.tex, line 94
At the origin g_N(x)=O(x^N-1). Its only Taylor powers are x^Nm-1, with m≥1.
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g_N(x)=O(x^{N-1})
x^{Nm-1}Use in source · 3 passages
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Use in source · Read passage · spectral_full.tex, line 95
At the origin g_N(x)=O(x^N-1). Its only Taylor powers are x^Nm-1, with m≥1.
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x^{Nm-1}
x^{Nn}Use in source · 1 passage
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Use in source · Read passage · spectral_full.tex, line 59
F_N(x):=d_N(-x^N) =frac1N∑_j=0^N-1e^ω^j x =∑_n=0^∞fracx^Nn(Nn)!. tagS1
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F_N(x):=d_N(-x^N) =\frac1N\sum_{j=0}^{N-1}e^{\omega^j x} =\sum_{n=0}^{\infty}\frac{x^{Nn}}{(Nn)!}. \tag{S1}
x^\rhoUse in source · 1 passage
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Use in source · Read passage · dilation_full.tex, line 102
For a rigorous propagation, put Q_* =max(q,r)<1 and d=(1-Q_*)B+Q_*L>0. Every child of every R≥ B is at most R-d. Induction on the intervals [B+kd,B+(k+1)d], beginning with the already controlled seed, proves the comparison for all larger radii. Continuity deals with common endpoints. All power comparisons use only monotonicity of x^ρ for positive x; no convexity assumption is made, so the argument also covers ρ<1.
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x^\rho
x^{\rho/2}Use in source · 1 passage
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Use in source · Read passage · main.tex, line 155
Suppose such an equality held. Normalization gives AB=1. Write p=qr, u=q^2+r^2, and let ρ be the common order. The first two nontrivial coefficients and the growth law force B^2s^2=p, s^2=u/2, 2s^ρ=q^ρ+r^ρ=1. Consequently 2(fracq^2+r^22)^ρ/2=q^ρ+r^ρ. For ρ≠2, strict convexity or concavity of x^ρ/2 forces q=r. For the exceptional case ρ=2, we have u=1, s^2=1/2, B^2=2p. The two z^7 coefficients become fracp^3(3-p^2)315 and frac11p^31260. Their equality forces p=1/2 and again (q-r)^2=u-2p=0. This c…
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x^{\rho/2}
X_0Use in source · 1 passage
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Use in source · Read passage · sequence_full.tex, line 376
Y_0 Y_1 =h_0 · h_1w h_1 · h_0 X_0 X_1.
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\begin{pmatrix}Y_0\\Y_1\end{pmatrix} =\begin{pmatrix}h_0&h_1w\\h_1&h_0\end{pmatrix} \begin{pmatrix}X_0\\X_1\end{pmatrix}.
X_1Use in source · 1 passage
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Use in source · Read passage · sequence_full.tex, line 376
Y_0 Y_1 =h_0 · h_1w h_1 · h_0 X_0 X_1.
Exact mathematical passage
\begin{pmatrix}Y_0\\Y_1\end{pmatrix} =\begin{pmatrix}h_0&h_1w\\h_1&h_0\end{pmatrix} \begin{pmatrix}X_0\\X_1\end{pmatrix}.
x_1Use in source · 1 passage
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Use in source · Read passage · sieve_full.tex, line 24
The mass at x_1=0 is exactly 1. A strength λ_p=-1 deletes every squarefree atom divisible by p. Several deletions commute. The depth-k family uses x^k dν, for integers k≥0.
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x_1=0
x_2Use in source · 6 passages
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Use in source · chapters_sequences.tex, line 26
lessonSEQUENCES / 04A Pell pair shares one unitS04_sqrt3_pell x_n+y_nsqrt3=(2+sqrt3)^n, x_n^2-3y_n^2=1 The table keeps both coordinates at the same index. Multiplication by 2+sqrt3 advances the pair using the displayed integer update. Its determinant is one, and it preserves the quadratic norm. The zero-index pair is (1,0). Starting from (x_2,y_2)=(7,4) gives (x_3,y_3)=(2·7+3·4,7+2·4)=(26,15). The norm is 26^2-3·15^2=1. A shared recurrence state can retain an exact invariant while both coordinat…
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(x_2,y_2)=(7,4)
x_3Use in source · 1 passage
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Use in source · chapters_sequences.tex, line 26
lessonSEQUENCES / 04A Pell pair shares one unitS04_sqrt3_pell x_n+y_nsqrt3=(2+sqrt3)^n, x_n^2-3y_n^2=1 The table keeps both coordinates at the same index. Multiplication by 2+sqrt3 advances the pair using the displayed integer update. Its determinant is one, and it preserves the quadratic norm. The zero-index pair is (1,0). Starting from (x_2,y_2)=(7,4) gives (x_3,y_3)=(2·7+3·4,7+2·4)=(26,15). The norm is 26^2-3·15^2=1. A shared recurrence state can retain an exact invariant while both coordinat…
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(x_3,y_3)=(2\cdot7+3\cdot4,7+2\cdot4)=(26,15)
x_aUse in source · 8 passages
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Use in source · Read passage · multi_prime.tex, line 68
pageheadingPRIME SIEVES / CONSTRUCTIVE WITNESSEvery negative prime atom has a finite detector For an integer a≥2, let L_a(x)=∏_n=1^a-1fracx-x_nx_a-x_n, C_a=L_a(1), r_a=frac1-x_a+11-x_a<1, f_a,m(x)=L_a(x)(frac1-x1-x_a)^m. This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects …
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L_a(x)=\prod_{n=1}^{a-1}\frac{x-x_n}{x_a-x_n},\quad C_a=L_a(1),\quad r_a=\frac{1-x_{a+1}}{1-x_a}<1,
x_{a+1}Use in source · 3 passages
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Use in source · Read passage · multi_prime.tex, line 68
pageheadingPRIME SIEVES / CONSTRUCTIVE WITNESSEvery negative prime atom has a finite detector For an integer a≥2, let L_a(x)=∏_n=1^a-1fracx-x_nx_a-x_n, C_a=L_a(1), r_a=frac1-x_a+11-x_a<1, f_a,m(x)=L_a(x)(frac1-x1-x_a)^m. This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects …
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L_a(x)=\prod_{n=1}^{a-1}\frac{x-x_n}{x_a-x_n},\quad C_a=L_a(1),\quad r_a=\frac{1-x_{a+1}}{1-x_a}<1,
X_jUse in source · 2 passages
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Use in source · Read passage · sequence_full.tex, line 289
For an ordinary generating function X(z)=∑_ma_m z^m, define X_j(w)=∑_na_bn+jw^n. Then
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X_j(w)=\sum_na_{bn+j}w^n
x_jUse in source · 1 passage
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Use in source · Read passage · sequence_full.tex, line 286
The full-position section and the decimated lane are distinct arrays. A section keeps the original positions, inserting zeros outside the residue. A lane x_j[n]=a_bn+j changes the index to n.
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x_j[n]=a_{bn+j}
X_lUse in source · 3 passages
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Use in source · Read passage · sequence_full.tex, line 357
For an OGF sequence a_m, define lanes x_l[n]=a_bn+l and X_l(w)=∑_nx_l[n]w^n. Apply finite ordinary taps and obtain y_j[n]=b_bn+j. Then the exact matrix identity is
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X_l(w)=\sum_nx_l[n]w^n
x_lUse in source · 3 passages
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Use in source · Read passage · sequence_full.tex, line 356
For an OGF sequence a_m, define lanes x_l[n]=a_bn+l and X_l(w)=∑_nx_l[n]w^n. Apply finite ordinary taps and obtain y_j[n]=b_bn+j. Then the exact matrix identity is
Exact mathematical passage
x_l[n]=a_{bn+l}
x_nUse in source · 19 passages
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Use in source · Read passage · main.tex, line 360
Prime sieves and signed measures. pageheadingFAMILY 3An infinite prime shelf changes the packets For a prime p and λ≥-1, let x_n=(1-n^-2)^2, ν_p,λ=∑_n≥1μ(n)^2n^-4 (1+λ1_pmid n)δ_x_n. Define monic orthogonal polynomials P_m^(k;p,λ) for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the or…
Exact mathematical passage
x_n=(1-n^{-2})^2,\qquad \nu_{p,\lambda}=\sum_{n\ge1}\mu(n)^2n^{-4} (1+\lambda\mathbf1_{p\mid n})\delta_{x_n}.
x_n^{i+j}Use in source · 1 passage
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Use in source · Read passage · sieve_full.tex, line 191
A_m=[∑_pnmid nμ(n)^2n^-4κ(n)x_n^i+j]_i,j<m, B_m=[∑_pmid nμ(n)^2n^-4κ(n)x_n^i+j]_i,j<m.
Exact mathematical passage
A_m=[\sum_{p\nmid n}\mu(n)^2n^{-4}\kappa(n)x_n^{i+j}]_{i,j<m}, \quad B_m=[\sum_{p\mid n}\mu(n)^2n^{-4}\kappa(n)x_n^{i+j}]_{i,j<m}.
x_n^jUse in source · 1 passage
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Use in source · Read passage · sieve_full.tex, line 67
where U_pq,j=∑_pqmid nμ(n)^2n^-4x_n^j>0. More generally, with all other strengths at least -1, every mixed moment derivative in a set of distinct primes is strictly positive. It is the integral over atoms divisible by those primes, with the remaining nonnegative prime factors. Infinitely many such atoms remain after any finite collection of deletions. Pure second derivatives in any one strength vanish because the moments are separately affine.
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U_{pq,j}=\sum_{pq\mid n}\mu(n)^2n^{-4}x_n^j>0
x_{n_+}Use in source · 1 passage
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Use in source · Read passage · resolution_full.tex, line 455
P_S,m(x)=∏_n∈ Sfracx-x_nx_2-x_n (frac1-x1-x_2)^m, C_S=∏_n∈ Sfrac1-x_n|x_2-x_n|, r_S=frac1-x_n_+1-x_2<1.
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C_S=\prod_{n\in S}\frac{1-x_n}{|x_2-x_n|}, \qquad r_S=\frac{1-x_{n_+}}{1-x_2}<1.
x_{n_i}Use in source · 2 passages
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Use in source · Read passage · resolution_full.tex, line 397
Choose d+1 distinct odd squarefree labels n_0,…,n_d, including 1, and set y_i=x_n_i. Their Lagrange polynomials are
Exact mathematical passage
y_i=x_{n_i}
x_pUse in source · 2 passages
Search aliases: x_p
Use in source · Read passage · main.tex, line 413
deg f<m frac∫ f^2 dν_nonp∫ f^2 dν_p>0, δ_mdownarrow0. To prove the limit, approximate a continuous bump supported at the isolated node x_p by polynomials on [0,1]. The numerator tends to zero while the denominator retains the p^-4 atom. Hence every λ<-1 eventually produces a negative finite-degree quadratic form.
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x_p
x_p^{-k}Use in source · 1 passage
Search aliases: x_p^-k
Use in source · Read passage · sieve_full.tex, line 260
The same argument applies to every nonnegative depth, replacing each integral by its x^k tilt. The denominator's explicit atom contribution becomes p^-4x_p^k, and (7) acquires the factor x_p^-k.
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x_p^{-k}
x_p^kUse in source · 4 passages
Search aliases: x_p^k
Use in source · Read passage · multi_prime.tex, line 90
od_q∈ Fmax(1,|1+λ_q|). The summable tail gives ∫ f_p,m^2ddν ≤frac1+λ_pp^4+ fracC_FC_p^23p^3r_p^2m. Hence the quadratic form is negative whenever pC_FC_p^2r_p^2m/3<-(1+λ_p). For rational strengths this is an exact rational certificate. It needs no numerical eigenvalue calculation. A depth tilt x^k replaces the negative atom contribution by (1+λ_p)x_p^k/p^4.
Exact mathematical passage
(1+\lambda_p)x_p^k/p^4
YUse in source · 2 passages
Search aliases: Y
Use in source · Read passage · sequence_full.tex, line 360
Y_j(w)=∑_l=0^b-1H_jl(w)X_l(w), H_jl(w)=∑_substack0≤ t<L t+l≡ j (b) h_t w^(t+l-j)/b.
Exact mathematical passage
\boxed{Y_j(w)=\sum_{l=0}^{b-1}H_{jl}(w)X_l(w),\qquad H_{jl}(w)=\sum_{\substack{0\le t<L\\t+l\equiv j\,(b)}} h_t w^{(t+l-j)/b}.}
y'Use in source · 2 passages
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Use in source · Read passage · coupled_inverse_full.tex, line 142
with root product yy'=k^3/p. The two candidate seventh coefficients have difference
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yy'=k^3/p
y'^2Use in source · 1 passage
Search aliases: y'^2
Use in source · Read passage · coupled_inverse_full.tex, line 145
v_3(y)-v_3(y') =frac k7(y^2-y'^2) (u_2-fracp^2k^2v_2) =fracD7k(y^2-y'^2),
Exact mathematical passage
v_3(y)-v_3(y') =\frac k7(y^2-y'^2) \left(u_2-\frac{p^2}{k^2}v_2\right) =\frac{D}{7k}(y^2-y'^2),
y^2Use in source · 3 passages
Search aliases: y^2
Use in source · Read passage · coupled_inverse_full.tex, line 138
p y^2-frac15v_2ky+k^3=0,
Exact mathematical passage
p y^2-\frac{15v_2}{k}y+k^3=0,
Y_0Use in source · 1 passage
Search aliases: Y_0
Use in source · Read passage · sequence_full.tex, line 376
Y_0 Y_1 =h_0 · h_1w h_1 · h_0 X_0 X_1.
Exact mathematical passage
\begin{pmatrix}Y_0\\Y_1\end{pmatrix} =\begin{pmatrix}h_0&h_1w\\h_1&h_0\end{pmatrix} \begin{pmatrix}X_0\\X_1\end{pmatrix}.
Y_1Use in source · 1 passage
Search aliases: Y_1
Use in source · Read passage · sequence_full.tex, line 376
Y_0 Y_1 =h_0 · h_1w h_1 · h_0 X_0 X_1.
Exact mathematical passage
\begin{pmatrix}Y_0\\Y_1\end{pmatrix} =\begin{pmatrix}h_0&h_1w\\h_1&h_0\end{pmatrix} \begin{pmatrix}X_0\\X_1\end{pmatrix}.
y_2Use in source · 1 passage
Search aliases: y_2
Use in source · chapters_sequences.tex, line 26
lessonSEQUENCES / 04A Pell pair shares one unitS04_sqrt3_pell x_n+y_nsqrt3=(2+sqrt3)^n, x_n^2-3y_n^2=1 The table keeps both coordinates at the same index. Multiplication by 2+sqrt3 advances the pair using the displayed integer update. Its determinant is one, and it preserves the quadratic norm. The zero-index pair is (1,0). Starting from (x_2,y_2)=(7,4) gives (x_3,y_3)=(2·7+3·4,7+2·4)=(26,15). The norm is 26^2-3·15^2=1. A shared recurrence state can retain an exact invariant while both coordinat…
Exact mathematical passage
(x_2,y_2)=(7,4)
y_3Use in source · 1 passage
Search aliases: y_3
Use in source · chapters_sequences.tex, line 26
lessonSEQUENCES / 04A Pell pair shares one unitS04_sqrt3_pell x_n+y_nsqrt3=(2+sqrt3)^n, x_n^2-3y_n^2=1 The table keeps both coordinates at the same index. Multiplication by 2+sqrt3 advances the pair using the displayed integer update. Its determinant is one, and it preserves the quadratic norm. The zero-index pair is (1,0). Starting from (x_2,y_2)=(7,4) gives (x_3,y_3)=(2·7+3·4,7+2·4)=(26,15). The norm is 26^2-3·15^2=1. A shared recurrence state can retain an exact invariant while both coordinat…
Exact mathematical passage
(x_3,y_3)=(2\cdot7+3\cdot4,7+2\cdot4)=(26,15)
y_iUse in source · 7 passages
Search aliases: y_i
Use in source · Read passage · resolution_full.tex, line 397
Choose d+1 distinct odd squarefree labels n_0,…,n_d, including 1, and set y_i=x_n_i. Their Lagrange polynomials are
Exact mathematical passage
y_i=x_{n_i}
Y_jUse in source · 1 passage
Search aliases: Y_j
Use in source · Read passage · sequence_full.tex, line 360
Y_j(w)=∑_l=0^b-1H_jl(w)X_l(w), H_jl(w)=∑_substack0≤ t<L t+l≡ j (b) h_t w^(t+l-j)/b.
Exact mathematical passage
\boxed{Y_j(w)=\sum_{l=0}^{b-1}H_{jl}(w)X_l(w),\qquad H_{jl}(w)=\sum_{\substack{0\le t<L\\t+l\equiv j\,(b)}} h_t w^{(t+l-j)/b}.}
y_jUse in source · 3 passages
Search aliases: y_j
Use in source · Read passage · sequence_full.tex, line 358
For an OGF sequence a_m, define lanes x_l[n]=a_bn+l and X_l(w)=∑_nx_l[n]w^n. Apply finite ordinary taps and obtain y_j[n]=b_bn+j. Then the exact matrix identity is
Exact mathematical passage
y_j[n]=b_{bn+j}
ZUse in source · 18 passages
Search aliases: Z
Use in source · Read passage · main.tex, line 237
(n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At depth zero the corresponding mass-one atom is absent. In particular, A(0)=Z_Q(2a)-1, where Z_Q(v)=∑_substackn≥1
Exact mathematical passage
A(0)=Z_Q(2a)-1
\ZUse in source · 2 passages
Search aliases: Z Z_Q
Use in source · Read passage · main.tex, line 253
ights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If p_* is the least prime not dividing Q, then τ=-2log(1-p_*^-β), M=Z_Q(2a)-1, |A(s)|≤ M e^τmax(-Re s,0). This proves local uniform convergence and entire continuation. The first atom dominates A(-t) as t→∞, so the entire func…
Exact mathematical passage
\tau=-2\log(1-p_*^{-\beta}),\qquad M=\Z(2a)-1,
\mathbb ZUse in source · 4 passages
Search aliases: Z mathbb
Use in source · Read passage · sequence_full.tex, line 47
Polynomial theorem. For every n, A_n∈ Z[q,r], all its nonzero coefficients are positive integers, and A_n(q,r)=A_n(r,q). Every monomial q^u r^v satisfies
Exact mathematical passage
A_n\in\mathbb Z[q,r]
\mathcal ZUse in source · 38 passages
Search aliases: Z mathcal
Use in source · Read passage · main.tex, line 487
4One spectral address for each rotation orbit For integer N≥2, define the classical Mittag--Leffler seed~[source reference] d_N(w)=∑_n≥0frac(-1)^nw^n(Nn)!=E_N,1(-w), D_N(z)=d_N(z^N). Let w_j be the nonzero zeros of d_N, counted with multiplicity. Each corresponds to one rotation orbit of N zeros of D_N. Using Argw∈(-π,π], define Z_N(s)=∑_jw_j^-s, Re s>1/N. fig06_orbit_zerosFor N=4, the four real/imaginary zeros in each orbit share one positive coordinate w=z^4. The plot shows the first six orbit…
Exact mathematical passage
\boxed{\mathcal Z_N(s)=\sum_jw_j^{-s},\qquad\Re s>1/N.}
z^2Use in source · 2 passages
Search aliases: z^2
Use in source · Read passage · coupled_structure_full.tex, line 105
Taking successively (w,g)=(z,z),(z^2,z),(z,z^2) yields qr=ab, q^2r=a^2b, and qr^2=ab^2. Nonzero scales then give q=a and r=b. Conversely these equalities make the two components of (S1) identical on every input, so (1,-1) F[X]=0. The only possible autonomous rows therefore span a one-dimensional space, which cannot be the two rows of an invertible matrix. square
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(w,g)=(z,z),(z^2,z),(z,z^2)
z^{201}Use in source · 1 passage
Search aliases: z^201
Use in source · Read passage · resolution_full.tex, line 134
How to read the figure. The left curves divide the difference by the tap weight, so the relevant small quantity is the plotted height multiplied by ε. The right panel records the proved order, which is determined by arbitrarily large radii. The curve computation uses powers through z^201, with a normalized analytic omitted-tail bound below 5.26×10^-23 on the unit interval. Ordinary floating rounding is separate from that analytic tail bound.
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z^{201}
z^3Use in source · 11 passages
Search aliases: z^3
Use in source · Read passage · main.tex, line 95
nctionsComputed Taylor curves. Each entire curve uses terms through z^321; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent. For T_q,r(z)=∑_n≥0c_nz^2n+1, c_0=1, coefficient matching gives c_n=frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1c_ic_n-1-i, n≥1. In particular, T_q,r(z)=z+fracqr3z^3+ fracq^2r^2(q^2+r^2)15z^5+O(z^7). The coefficients are positive and bounded by tangent coefficients, giving a convergent germ. If the …
Exact mathematical passage
T_{q,r}(z)=z+\frac{qr}{3}z^3+ \frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).
z^{321}Use in source · 1 passage
Search aliases: z^321
Use in source · Read passage · main.tex, line 87
1Two scales inside the function For real 0<q,r<1, define the normalized odd solution T_q,r'(z)=1+T_q,r(qz)T_q,r(rz), T_q,r(0)=0. The contractions are part of the defining feedback. The function is symmetric in q,r, but unequal scales retain left/right address information. fig01_asymmetric_functionsComputed Taylor curves. Each entire curve uses terms through z^321; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent. …
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z^{321}
z^4Use in source · 2 passages
Search aliases: z^4
Use in source · Read passage · main.tex, line 491
w^n(Nn)!=E_N,1(-w), D_N(z)=d_N(z^N). Let w_j be the nonzero zeros of d_N, counted with multiplicity. Each corresponds to one rotation orbit of N zeros of D_N. Using Argw∈(-π,π], define Z_N(s)=∑_jw_j^-s, Re s>1/N. fig06_orbit_zerosFor N=4, the four real/imaginary zeros in each orbit share one positive coordinate w=z^4. The plot shows the first six orbits. Multiplicity is retained by the definition; simplicity is not assumed in general. The bound |d_N(w)|≤ e^|w|^1/N and the finite-order zero-count…
Exact mathematical passage
w=z^4
z^5Use in source · 9 passages
Search aliases: z^5
Use in source · Read passage · main.tex, line 95
nctionsComputed Taylor curves. Each entire curve uses terms through z^321; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent. For T_q,r(z)=∑_n≥0c_nz^2n+1, c_0=1, coefficient matching gives c_n=frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1c_ic_n-1-i, n≥1. In particular, T_q,r(z)=z+fracqr3z^3+ fracq^2r^2(q^2+r^2)15z^5+O(z^7). The coefficients are positive and bounded by tangent coefficients, giving a convergent germ. If the …
Exact mathematical passage
T_{q,r}(z)=z+\frac{qr}{3}z^3+ \frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).
z^7Use in source · 9 passages
Search aliases: z^7
Use in source · Read passage · main.tex, line 95
nctionsComputed Taylor curves. Each entire curve uses terms through z^321; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent. For T_q,r(z)=∑_n≥0c_nz^2n+1, c_0=1, coefficient matching gives c_n=frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1c_ic_n-1-i, n≥1. In particular, T_q,r(z)=z+fracqr3z^3+ fracq^2r^2(q^2+r^2)15z^5+O(z^7). The coefficients are positive and bounded by tangent coefficients, giving a convergent germ. If the …
Exact mathematical passage
T_{q,r}(z)=z+\frac{qr}{3}z^3+ \frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).
z^9Use in source · 3 passages
Search aliases: z^9
Use in source · Read passage · main.tex, line 165
claimA worked order-one member T_1/3,2/3(z)=z+frac227z^3+ frac42187z^5+frac23211160261z^7+O(z^9). It has order one, and T_1/3,2/3(1)≈1.0759239897408249. The diagonal point q=r=1/2 gives f_1/2(z)=sqrt2sinh(z/sqrt2). Every off-diagonal member is separated from all rescaled diagonal members by the theorem above.
Exact mathematical passage
T_{1/3,2/3}(z)=z+\frac{2}{27}z^3+ \frac{4}{2187}z^5+\frac{232}{11160261}z^7+O(z^9).
z^bUse in source · 4 passages
Search aliases: z^b
Use in source · Read passage · sequence_full.tex, line 291
X(z)=∑_j=0^b-1z^jX_j(z^b).
Exact mathematical passage
X(z)=\sum_{j=0}^{b-1}z^jX_j(z^b).
z^{bn+j}Use in source · 1 passage
Search aliases: z^bn+j
Use in source · Read passage · sequence_polyphase.tex, line 71
the full sequence For an ordinary generating function G(z)=∑_m≥0a_mz^m and integer b≥1, write G(z)=∑_j=0^b-1z^jG_j(z^b), G_j(t)=∑_n≥0a_bn+jt^n. The root projector is P_b,jG(z)=frac1b∑_k=0^b-1ω^-jkG(ω^kz) =z^jG_j(z^b), ω=e^2π i/b. The same projector works on an EGF, but the section retains its original factorials: ∑_na_bn+jz^bn+j/(bn+j)!. Removing those denominators changes the representation and requires an explicit transport. Root filtering does not do it automatically.
Exact mathematical passage
\sum_na_{bn+j}z^{bn+j}/(bn+j)!
z^jUse in source · 4 passages
Search aliases: z^j
Use in source · Read passage · sequence_full.tex, line 291
X(z)=∑_j=0^b-1z^jX_j(z^b).
Exact mathematical passage
X(z)=\sum_{j=0}^{b-1}z^jX_j(z^b).
z^mUse in source · 8 passages
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Use in source · Read passage · sequence_full.tex, line 265
For an EGF G(z)=∑_m≥0a_m z^m/m!, an integer base b≥1, a residue 0≤ j<b, and ω=e^2π i/b, define
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G(z)=\sum_{m\ge0}a_m z^m/m!
z^NUse in source · 2 passages
Search aliases: z^N
Use in source · Read passage · main.tex, line 480
MonographPartIIIOrbit spectra and exact coordinatesSpectral continuation and the coefficient language of polyphase transforms. Spectral functions of zero orbits. pageheadingFAMILY 4One spectral address for each rotation orbit For integer N≥2, define the classical Mittag--Leffler seed~[source reference] d_N(w)=∑_n≥0frac(-1)^nw^n(Nn)!=E_N,1(-w), D_N(z)=d_N(z^N). Let w_j be the nonzero zeros of d_N, counted with multiplicity. Each corresponds to one rotation orbit of N zeros of D_N. Using Argw∈(-π,…
Exact mathematical passage
d_N(w)=\sum_{n\ge0}\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w), \qquad D_N(z)=d_N(z^N).
z^nUse in source · 3 passages
Search aliases: z^n
Use in source · chapters_sequences.tex, line 39
lessonSEQUENCES / 06Catalan convolution rebuilds the next termS06_catalan_convolution C_n+1=∑_k=0^nC_kC_n-k, C(z)=1+zC(z)^2 The five bars are the five products contributing to C_5. The input indices add to four, and their sum supplies the next coefficient. Here C(z)=∑_n≥0C_nz^n is an ordinary generating function. The contributions are 1·14, 1·5, 2·2, 5·1, and 14·1. Their sum is 14+5+4+5+14=42=C_5. A nonlinear generating-function equation becomes an exact convolution identity among coefficients. …
Exact mathematical passage
C(z)=\sum_{n\ge0}C_nz^n
z^tUse in source · 1 passage
Search aliases: z^t
Use in source · Read passage · sequence_full.tex, line 317
For OGFs this is multiplication by H(z)=∑_th_tz^t: B(z)=H(z)X(z). For EGFs it is a sum of repeated zero-constant antiderivatives:
Exact mathematical passage
H(z)=\sum_th_tz^t
Z_0Use in source · 1 passage
Search aliases: Z_0
Use in source · Read passage · coupled_growth_full.tex, line 293
To bound a seed interval, the coefficient majorant from Theorem F.1 gives F_i(1)≤2. Put R_k=Q^-k, Z_0=2, and
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Z_0=2
z_0Use in source · 3 passages
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Use in source · Read passage · coupled_structure_full.tex, line 66
If (S2) holds, Q_c is constant on every fiber of (u,v)↦ c_1u+c_2v. Fix z_0≠0. Polynomial interpolation gives a scalar polynomial f with f(0)=0 and f(sz_0)=1 at every distinct scale s∈{q,r,a,b}. For inputs u=xf,v=yf, where x,y∈ C,
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z_0\ne0
Z_1Use in source · 1 passage
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Use in source · Read passage · notation.tex, line 41
item[Squarefree Dirichlet sum.] For v>1, Z_Q(v)=ζ(v)ζ(2v)^-1∏_pmid Q(1+p^-v)^-1. The unrestricted notation is D(v)=Z_1(v). Thus A_a,β,Q(0)=Z_Q(2a)-1. item[Prime-sieve measure.] ν_F,λ is the arithmetic atomic measure obtained by multiplying the mass at x_n=(1-n^-2)^2 by ∏_p∈ F(1+λ_p1_pmid n). It is positive exactly when every real strength satisfies λ_p≥-1. item[Hankel order and polynomial degree.] For moments M_j=∫ x^jddν, the order-m matrix is H_m=[M_i+j]_0≤ i,j<m. It tests quadratic forms of r…
Exact mathematical passage
D(v)=Z_1(v)
\mathcal Z_2Use in source · 2 passages
Search aliases: Z_2 mathcal
Use in source · Read passage · main.tex, line 529
claimTwo exact Hurwitz-zeta reductions For N=2, d_2(w)=cossqrt w, whose zeros are w_j=π^2(j+1/2)^2, j≥0. Therefore Z_2(s)=π^-2sζ(2s,1/2). For N=4, D_4(z)=cos(z/sqrt2)cosh(z/sqrt2), and the rotation quotient has one zero w_j=(sqrt2π)^4(j+1/2)^4 per orbit. Hence Z_4(s)=(sqrt2π)^-4sζ(4s,1/2). These anchors are classical reductions. The general spectral-zeta mechanism also has substantial antecedents, including hypergeometric zero zeta functions.
Exact mathematical passage
\mathcal Z_2(s)=\pi^{-2s}\zeta(2s,1/2).
\mathcal Z_4Use in source · 2 passages
Search aliases: Z_4 mathcal
Use in source · Read passage · main.tex, line 538
claimTwo exact Hurwitz-zeta reductions For N=2, d_2(w)=cossqrt w, whose zeros are w_j=π^2(j+1/2)^2, j≥0. Therefore Z_2(s)=π^-2sζ(2s,1/2). For N=4, D_4(z)=cos(z/sqrt2)cosh(z/sqrt2), and the rotation quotient has one zero w_j=(sqrt2π)^4(j+1/2)^4 per orbit. Hence Z_4(s)=(sqrt2π)^-4sζ(4s,1/2). These anchors are classical reductions. The general spectral-zeta mechanism also has substantial antecedents, including hypergeometric zero zeta functions.
Exact mathematical passage
\mathcal Z_4(s)=(\sqrt2\pi)^{-4s}\zeta(4s,1/2).
Z_FUse in source · 1 passage
Search aliases: Z_F
Use in source · Read passage · sieve_full.tex, line 32
Z_F(s;λ) =∑_n≥1μ(n)^2n^-sκ_F(n;λ) =D(s)∏_p∈ F(1+fracλ_pp^s+1).
Exact mathematical passage
Z_F(s;\boldsymbol\lambda) =\sum_{n\ge1}\mu(n)^2n^{-s}\kappa_F(n;\boldsymbol\lambda) =D(s)\prod_{p\in F}\left(1+\frac{\lambda_p}{p^s+1}\right).
Z_kUse in source · 4 passages
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Use in source · Read passage · coupled_growth_full.tex, line 295
Z_k=Q^-k(1+Z_k-1^2).
Exact mathematical passage
Z_k=Q^{-k}(1+Z_{k-1}^2).
Z_{k-1}Use in source · 1 passage
Search aliases: Z_k-1
Use in source · Read passage · coupled_growth_full.tex, line 302
log Z_k≤ klog(1/Q)+log2+2log Z_k-1.
Exact mathematical passage
\log Z_k\le k\log(1/Q)+\log2+2\log Z_{k-1}.
Z_{k-1}^2Use in source · 1 passage
Search aliases: Z_k-1^2
Use in source · Read passage · coupled_growth_full.tex, line 295
Z_k=Q^-k(1+Z_k-1^2).
Exact mathematical passage
Z_k=Q^{-k}(1+Z_{k-1}^2).
Z_NUse in source · 1 passage
Search aliases: Z_N
Use in source · Read passage · coupled_growth_full.tex, line 306
Summing this elementary recurrence gives log Z_N≤ W_N. Therefore H_i(e^t)≤ W_N throughout 1≤ t≤1+Δ, since e^t≤ e/m<3/m≤ R_N.
Exact mathematical passage
\log Z_N\le W_N
\mathcal Z_NUse in source · 28 passages
Search aliases: Z_N mathcal
Use in source · Read passage · main.tex, line 487
4One spectral address for each rotation orbit For integer N≥2, define the classical Mittag--Leffler seed~[source reference] d_N(w)=∑_n≥0frac(-1)^nw^n(Nn)!=E_N,1(-w), D_N(z)=d_N(z^N). Let w_j be the nonzero zeros of d_N, counted with multiplicity. Each corresponds to one rotation orbit of N zeros of D_N. Using Argw∈(-π,π], define Z_N(s)=∑_jw_j^-s, Re s>1/N. fig06_orbit_zerosFor N=4, the four real/imaginary zeros in each orbit share one positive coordinate w=z^4. The plot shows the first six orbit…
Exact mathematical passage
\boxed{\mathcal Z_N(s)=\sum_jw_j^{-s},\qquad\Re s>1/N.}
\mathcal Z_N'Use in source · 8 passages
Search aliases: Z_N' mathcal
Use in source · Read passage · spectral_continuation.tex, line 9
pageheadingORBIT SPECTRA / CONTINUATION THEOREMGlobal meromorphic continuation For every integer N≥2, the orbit zeta has a meromorphic continuation to the whole complex plane with exactly one pole: s=1/N, Res_s=1/N Z_N(s)=fracsin(π/N)Nπ. Its exact absolute-convergence abscissa is 1/N. At nonpositive integers, Z_N(-m)=0 (m≥0), Z_N'(0)=-log N. The zeta-regularized product of the orbit coordinates is consequently exp[- Z_N'(0)]=N. fig15_spectral_polesPole and zero locations follow from the proved c…
Exact mathematical passage
\boxed{\mathcal Z_N(-m)=0\quad(m\ge0),\qquad \mathcal Z_N'(0)=-\log N.}
Z_QUse in source · 10 passages
Search aliases: Z_Q
Use in source · Read passage · main.tex, line 237
(n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At depth zero the corresponding mass-one atom is absent. In particular, A(0)=Z_Q(2a)-1, where Z_Q(v)=∑_substackn≥1
Exact mathematical passage
A(0)=Z_Q(2a)-1
|^{-(2a-1)/\beta}Use in source · 1 passage
Search aliases: |^-(2a-1)/β beta
Use in source · Read passage · arithmetic_inverse.tex, line 113
acter resolution For characters χ,ψ modulo a fixed Q, let η=χoverlineψ. When η is nonprincipal, the elementary squarefree character sum is O_Q(sqrt x), with no density term. The same sector partial-integration argument therefore gives A_χ,ψ(s) =O_a,β,Q,ϵ(|s|^-(2a-1/2)/β), |arg s|≤π/2-ϵ. Compared with the principal |s|^-(2a-1)/β scale, this is faster by |s|^-1/(2β). The modulus is fixed; the constants are not asserted uniform as Q grows. Individual off-diagonal functions can be complex and are no…
Exact mathematical passage
|s|^{-(2a-1)/\beta}
|^{-(2a-1/2)/\beta}Use in source · 3 passages
Search aliases: |^-(2a-1/2)/β beta
Use in source · Read passage · main.tex, line 315
Let α=frac2a-1β, δ_Q=frac1ζ(2)∏_pmid Qfracpp+1. As |s|→∞ in any closed sector |arg s|≤π/2-ϵ with 0<ϵ<π/2, A_a,β,Q(s)=fracδ_QβB(α,2s+1) +O_a,β,Q,ϵ(|s|^-(2a-1/2)/β). In particular, A_a,β,Q(s)∼fracδ_QΓ(α)β(2s)^-α. The principal power is used in the right half-plane. The beta expression is an asymptotic continuous-density model, not an identity with the arithmetic sum.
Exact mathematical passage
\boxed{\A(s)=\frac{\delta_Q}{\beta}B(\alpha,2s+1) +O_{a,\beta,Q,\epsilon}\left(|s|^{-(2a-1/2)/\beta}\right).}
|^{-1/(2\beta)}Use in source · 2 passages
Search aliases: |^-1/(2β) beta
Use in source · Read passage · arithmetic_full.tex, line 441
A_χ,ψ(s) =O_a,β,Q,ε(|s|^-(2a-1/2)/β), fracA_χ,ψ(s)A_a,β,Q(s) =O_a,β,Q,ε(|s|^-1/(2β)).
Exact mathematical passage
A_{\chi,\psi}(s) =O_{a,\beta,Q,\varepsilon}\bigl(|s|^{-(2a-1/2)/\beta}\bigr), \qquad \frac{A_{\chi,\psi}(s)}{A_{a,\beta,Q}(s)} =O_{a,\beta,Q,\varepsilon}\bigl(|s|^{-1/(2\beta)}\bigr).
|^{-\Re s}Use in source · 1 passage
Search aliases: |^-Re s Re
Use in source · Read passage · spectral_continuation.tex, line 56
s have different jobs For 1/N<Re s<1, the genus-zero log derivative and the angular gap give absolute interchange in Z_N(s)=fracsinπ sπ ∫_0^∞ t^-sfracddtlog E_N(t)dd t. Each term uses the principal-branch beta integral ∫_0^∞ t^-s(t+w)^-1dd t=π w^-s/sinπ s. The denominator estimate bounds the absolute integral by a constant times |w|^-Re s, whose sum converges in this strip.
Exact mathematical passage
|w|^{-\Re s}
|^{-\alpha}Use in source · 1 passage
Search aliases: |^-α alpha
Use in source · Read passage · arithmetic_full.tex, line 478
which is O_a,β,ε(|s|^-γ) because u≥ |s|sinε. The principal diagonal has the nonzero leading term δ_QΓ(α)β^-1(2s)^-α, α=(2a-1)/β, by the base sector theorem. Its modulus is bounded below by a positive fixed-sector multiple of |s|^-α for sufficiently large |s|. Division gives the second bound, since γ-α=1/(2β).
Exact mathematical passage
|s|^{-\alpha}
|^{-\gamma}Use in source · 1 passage
Search aliases: |^-γ gamma
Use in source · Read passage · arithmetic_full.tex, line 474
which is O_a,β,ε(|s|^-γ) because u≥ |s|sinε. The principal diagonal has the nonzero leading term δ_QΓ(α)β^-1(2s)^-α, α=(2a-1)/β, by the base sector theorem. Its modulus is bounded below by a positive fixed-sector multiple of |s|^-α for sufficiently large |s|. Division gives the second bound, since γ-α=1/(2β).
Exact mathematical passage
O_{a,\beta,\varepsilon}(|s|^{-\gamma})
|^{-\sigma}Use in source · 2 passages
Search aliases: |^-σ sigma
Use in source · Read passage · spectral_full.tex, line 112
For completeness, the series bound |d_N(w)|≤exp(|w|^1/N) gives order at most 1/N<1. Hadamard factorization therefore has genus zero and a constant exponential factor, which equals one because d_N(0)=1. The usual zero-count consequence gives ∑_j|w_j|^-σ<∞ for every σ>1/N. Along the negative real axis (S3) also shows the order is exactly 1/N.
Exact mathematical passage
\sum_j|w_j|^{-\sigma}<\infty
|^{1/N}Use in source · 2 passages
Search aliases: |^1/N
Use in source · Read passage · main.tex, line 493
one rotation orbit of N zeros of D_N. Using Argw∈(-π,π], define Z_N(s)=∑_jw_j^-s, Re s>1/N. fig06_orbit_zerosFor N=4, the four real/imaginary zeros in each orbit share one positive coordinate w=z^4. The plot shows the first six orbits. Multiplicity is retained by the definition; simplicity is not assumed in general. The bound |d_N(w)|≤ e^|w|^1/N and the finite-order zero-counting theorem give absolute local uniform convergence in the stated half-plane. Since the order is below one and d_N(0)=1, …
Exact mathematical passage
|d_N(w)|\le e^{|w|^{1/N}}
|^2Use in source · 4 passages
Search aliases: |^2
Use in source · REPORT.md, line 28
|∑_n≥ Nc_nz^2n+1| ≤ frac|z|ρ^N1-ρ, |z|^2<2.
Exact mathematical passage
\left|\sum_{n\ge N}c_nz^{2n+1}\right| \le \frac{|z|\rho^N}{1-\rho},\qquad |z|^2<2.
\cdotUse in source · 17 passages
Search aliases: · cdot
Use in source · chapters_sequences.tex, line 5
01Two seeds, one Fibonacci recurrenceS01_fibonacci_lucas F_n+2=F_n+1+F_n, L_n+2=L_n+1+L_n, L_n^2-5F_n^2=4(-1)^n The two families share the same recurrence and have different seeds: (F_0,F_1)=(0,1) and (L_0,L_1)=(2,1). The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude. At n=7, F_7=13 and L_7=29. Thus 29^2-5·13^2=841-845=-4. Comparing terms at different indices would destroy this identity. A recurrence operator and its initial state are se…
Exact mathematical passage
29^2-5\cdot13^2=841-845=-4
\timesUse in source · 33 passages
Search aliases: × times
Use in source · Read passage · arithmetic_inverse.tex, line 83
le squarefree atoms. Their frequencies are at most λ_p and their total weight is the Euler-factor expression on the right. Termwise differentiation proves the bound for every derivative. fig11_prime_instabilityThe proved compact-window envelope shrinks as the excluded prime increases. At a=β=2, Q=1, R=5, excluding p=10007 changes the entire window by at most approximately 1.0749×10^-16. The depth-zero difference is known exactly. The radical differs for every such prime, while all bounded-window…
Exact mathematical passage
1.0749\times10^{-16}
\alphaUse in source · 34 passages
Search aliases: α alpha
Use in source · Read passage · evidence_notes.tex, line 18
The earlier moving-width lower argument used a fixed α<1; it proved orders below ρ and then recovered the order by αuparrow1, but did not establish positive lower type at ρ. This is a limitation of that argument, not a counterexample. The fixed-width comparison closes precisely that missing implication. The evidence archive retains execution failures, their causes and the subsequent corrected runs.
Exact mathematical passage
\alpha<1
\alpha^sUse in source · 1 passage
Search aliases: α^s alpha
Use in source · Read passage · evidence_notes.tex, line 34
The decisive extension is from a scalar positive comparison to a comparison weighted by a positive eigenvector. The fixed-width integral controls the lower type at the exact exponent. A fixed proportional interval [α R,R], with α<1, would only give a smaller exponent through α^sP(s). Letting α tend to one recovers exact order but does not, by itself, yield positive lower type. This is a limitation of that alternative route, not a failure of the theorem.
Exact mathematical passage
\alpha^sP(s)
\alpha_kUse in source · 3 passages
Search aliases: α_k alpha
Use in source · Read passage · spectral_continuation.tex, line 58
Near zero write f(t)=(log E_N)'(t)=∑_k≥1α_kt^k-1, where α_k=(-1)^k+1 Z_N(k). For any integer K≥1, subtract K terms on [0,1]: I_K(s)= · ∫_0^1t^-s[f(t)-∑_k=1^Kα_kt^k-1]dd t +∑_k=1^Kfracα_kk-s
Exact mathematical passage
f(t)=(\log E_N)'(t)=\sum_{k\ge1}\alpha_kt^{k-1}
\betaUse in source · 62 passages
Search aliases: β beta
Use in source · Read passage · main.tex, line 229
MonographPartIIArithmetic depth and signed measuresComplex parameters, positive kernels and the boundaries of positivity. Arithmetic at complex depth. pageheadingFAMILY 2Arithmetic energy at complex depth For real a>1/2, β>0, and integer Q≥1, define A_a,β,Q(s)=∑_substackn≥2
Exact mathematical passage
\beta>0
\beta'Use in source · 1 passage
Search aliases: β' beta
Use in source · Read passage · arithmetic_full.tex, line 48
a=a', β=β', rad(Q)=rad(Q').
Exact mathematical passage
a=a',\qquad \beta=\beta',\qquad \operatorname{rad}(Q)=\operatorname{rad}(Q').
\beta^{-1}Use in source · 1 passage
Search aliases: β^-1 beta
Use in source · Read passage · arithmetic_full.tex, line 476
which is O_a,β,ε(|s|^-γ) because u≥ |s|sinε. The principal diagonal has the nonzero leading term δ_QΓ(α)β^-1(2s)^-α, α=(2a-1)/β, by the base sector theorem. Its modulus is bounded below by a positive fixed-sector multiple of |s|^-α for sufficiently large |s|. Division gives the second bound, since γ-α=1/(2β).
Exact mathematical passage
\delta_Q\Gamma(\alpha)\beta^{-1}(2s)^{-\alpha}
\GammaUse in source · 2 passages
Search aliases: Γ Gamma
Use in source · Read passage · main.tex, line 320
Let α=frac2a-1β, δ_Q=frac1ζ(2)∏_pmid Qfracpp+1. As |s|→∞ in any closed sector |arg s|≤π/2-ϵ with 0<ϵ<π/2, A_a,β,Q(s)=fracδ_QβB(α,2s+1) +O_a,β,Q,ϵ(|s|^-(2a-1/2)/β). In particular, A_a,β,Q(s)∼fracδ_QΓ(α)β(2s)^-α. The principal power is used in the right half-plane. The beta expression is an asymptotic continuous-density model, not an identity with the arithmetic sum.
Exact mathematical passage
\A(s)\sim\frac{\delta_Q\Gamma(\alpha)}{\beta}(2s)^{-\alpha}.
\boldsymbol\gammaUse in source · 1 passage
Search aliases: γ boldsymbol gamma
Use in source · Read passage · sieve_full.tex, line 288
widetildeν_F,γ =∑_n≥1μ(n)^2n^-4 (1+∑_p∈ Fγ_p1_pmid n)δ_x_n.
Exact mathematical passage
\widetilde\nu_{F,\boldsymbol\gamma} =\sum_{n\ge1}\mu(n)^2n^{-4} \left(1+\sum_{p\in F}\gamma_p\mathbf1_{p\mid n}\right)\delta_{x_n}.
\gammaUse in source · 13 passages
Search aliases: γ gamma
Use in source · Read passage · arithmetic_full.tex, line 466
Use partial integration against g_s(x)=x^-2a(1-x^-β)^2s, whose endpoint values vanish when Re s>0. The counting bound reduces the magnitude to a constant times ∫_1^∞sqrt x |g_s'(x)| dx. Writing u=Re s and γ=(2a-1/2)/β>0, direct substitution y=x^-β bounds its two derivative terms by
Exact mathematical passage
\gamma=(2a-1/2)/\beta>0
\gamma_2Use in source · 1 passage
Search aliases: γ_2 gamma
Use in source · Read passage · sieve_full.tex, line 307
ecessity uses the squarefree integer n=∏_p∈ Sp; sufficiency checks every atom's prime subset. Its finite Hankel matrices are affine functions of the strengths, so their positive-semidefinite feasible sets are convex spectrahedra. In contrast, the multiplicative family includes the arithmetic prime-overlap terms and has polynomial parameter dependence. For example, additive strengths γ_2=γ_3=-3/4 leave both individual prime weights positive but make the n=6 weight negative; multiplicative strengt…
Exact mathematical passage
\gamma_2=\gamma_3=-3/4
\gamma_3Use in source · 1 passage
Search aliases: γ_3 gamma
Use in source · Read passage · sieve_full.tex, line 307
ecessity uses the squarefree integer n=∏_p∈ Sp; sufficiency checks every atom's prime subset. Its finite Hankel matrices are affine functions of the strengths, so their positive-semidefinite feasible sets are convex spectrahedra. In contrast, the multiplicative family includes the arithmetic prime-overlap terms and has polynomial parameter dependence. For example, additive strengths γ_2=γ_3=-3/4 leave both individual prime weights positive but make the n=6 weight negative; multiplicative strengt…
Exact mathematical passage
\gamma_2=\gamma_3=-3/4
\gamma_NUse in source · 2 passages
Search aliases: γ_N gamma
Use in source · Read passage · spectral_full.tex, line 122
Re x-Re(ω^j x) ≥ [cosfracπ2N-cosfrac3π2N]|x| =:γ_N|x|>0. tagS5
Exact mathematical passage
\Re x-\Re(\omega^j x) \ge \left[\cos\frac{\pi}{2N}-\cos\frac{3\pi}{2N}\right]|x| =:\gamma_N|x|>0. \tag{S5}
\gamma_pUse in source · 4 passages
Search aliases: γ_p gamma
Use in source · Read passage · multi_prime.tex, line 55
claimA separate additive family has different geometry For weights 1+∑_p∈ Fγ_p1_pmid n, positivity is equivalent to 1+∑_γ_p<0γ_p≥0. Every prime subset appears at its squarefree product, proving necessity. Checking each atom proves sufficiency. Its finite Hankel matrices are affine in γ, so the semidefinite feasible sets are convex spectrahedra. The multiplicative family has genuine overlap terms and polynomial parameter dependence. These are two distinct deformations, with distinct domains.
Exact mathematical passage
1+\sum_{p\in F}\gamma_p\mathbf1_{p\mid n}
\DeltaUse in source · 24 passages
Search aliases: Δ Delta
Use in source · chapters_sequences.tex, line 109
lessonSEQUENCES / 16Sums and differences form a polynomial ladderS16_polynomial_ladder T_n=∑_j=1^nj, H_n=∑_j=1^nT_j, Δ T_n=n, Δ H_n=T_n The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, Δ f_n=f_n-f_n-1. All plotted values use their displayed common index. At n=4, T_4=10 and H_4=20. The differences are 10-6=4 and 20-10=10. Separately, 4^2-3^2=7 and 4^3-3…
Exact mathematical passage
\Delta f_n=f_n-f_{n-1}
\deltaUse in source · 49 passages
Search aliases: δ delta
Use in source · Read passage · main.tex, line 309
Let α=frac2a-1β, δ_Q=frac1ζ(2)∏_pmid Qfracpp+1. As |s|→∞ in any closed sector |arg s|≤π/2-ϵ with 0<ϵ<π/2, A_a,β,Q(s)=fracδ_QβB(α,2s+1) +O_a,β,Q,ϵ(|s|^-(2a-1/2)/β). In particular, A_a,β,Q(s)∼fracδ_QΓ(α)β(2s)^-α. The principal power is used in the right half-plane. The beta expression is an asymptotic continuous-density model, not an identity with the arithmetic sum.
Exact mathematical passage
\alpha=\frac{2a-1}{\beta},\qquad \delta_Q=\frac1{\zeta(2)}\prod_{p\mid Q}\frac{p}{p+1}.
\delta_0Use in source · 1 passage
Search aliases: δ_0 delta
Use in source · Read passage · main.tex, line 442
claimAn endpoint mass is numerically delicate For the undeformed depth-zero measure, set ν_θ=ν+θδ_0, θ≥-1. Write h_j=∫ P_j(x)^2 dν(x) and K_m(x,y)=∑_j=0^mP_j(x)P_j(y)/h_j. The Uvarov formula~[source reference] is P_m^[θ](x)=P_m(x)- fracθ P_m(0)1+θ K_m-1(0,0)K_m-1(x,0), with K_-1=0 for degree zero. The mass at zero is recovered by 1≤frac1K_m(0,0)≤1+(M_0-1)(7/16)^2m. The lower bound comes from the zero atom, whose mass is one. The test polynomial (1-x)^m gives the upper bound since every other nod…
Exact mathematical passage
\nu_\theta=\nu+\theta\delta_0
\delta_1Use in source · 2 passages
Search aliases: δ_1 delta
Use in source · Read passage · main.tex, line 423
For p=2, test with f(x)=x(1-x)^m-2, m≥2: δ_m≤frac409681(frac16M_017-1) (frac272567)^2(m-2), M_0=105π^-4. This follows from x_2=9/16 and x_n≥64/81 at nonzero odd nodes. Exactly δ_1=p^4; the limit is an infinite-resolution phenomenon.
Exact mathematical passage
\delta_1=p^4
\Delta_jUse in source · 7 passages
Search aliases: Δ_j Delta
Use in source · Read passage · arithmetic_full.tex, line 223
Theorem B.2 (AI-2), gap bound. Suppose the first j-1 atoms have already been removed exactly, put Δ_j=λ_j-λ_j+1>0, and fix h>0. Define
Exact mathematical passage
\Delta_j=\lambda_j-\lambda_{j+1}>0
\delta_{jl}Use in source · 1 passage
Search aliases: δ_jl delta
Use in source · Read passage · sequence_full.tex, line 279
∑_j=0^b-1 P_b,j=I, P_b,j P_b,l=δ_jl P_b,j.
Exact mathematical passage
\sum_{j=0}^{b-1}\mathcal P_{b,j}=I,\qquad \mathcal P_{b,j}\mathcal P_{b,l}=\delta_{jl}\mathcal P_{b,j}.
\delta_mUse in source · 8 passages
Search aliases: δ_m delta
Use in source · Read passage · main.tex, line 406
nvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and below -1; possible coefficient poles occur there, although numerator cancellation is possible. fig05_sieve_boundaryThe line is a proved bound for p=2; dots are ordinary high-precision pencil roots. A defect λ=-1.001 eventually violates positivity even though it can pass low-order checks. The nearest root is -1-δ_m, with δ_m=min_substackf∈ R[x]∖{0}
Exact mathematical passage
-1-\delta_m
\delta_{p+m}Use in source · 3 passages
Search aliases: δ_p+m delta
Use in source · Read passage · multi_prime.tex, line 107
deg f<m frac∫ f^2ddν_p-free ∫ f^2ddν_p-divisible, δ_mdownarrow0. The detector yields, for C_κ=∏_q≠ pmax(1,1+λ_q), δ_p+m≤fracpC_κ C_p^23r_p^2m. The denominator retains the atom p; all lower nodes in the numerator are killed. This extends the prime-two bound to every prime and to simultaneous positive or deletion strengths. It does not assert simple roots or global multivariate real stability.
Exact mathematical passage
\delta_{p+m}\le\frac{pC_\kappa C_p^2}{3}r_p^{2m}.
\delta_QUse in source · 13 passages
Search aliases: δ_Q delta
Use in source · Read passage · main.tex, line 309
Let α=frac2a-1β, δ_Q=frac1ζ(2)∏_pmid Qfracpp+1. As |s|→∞ in any closed sector |arg s|≤π/2-ϵ with 0<ϵ<π/2, A_a,β,Q(s)=fracδ_QβB(α,2s+1) +O_a,β,Q,ϵ(|s|^-(2a-1/2)/β). In particular, A_a,β,Q(s)∼fracδ_QΓ(α)β(2s)^-α. The principal power is used in the right half-plane. The beta expression is an asymptotic continuous-density model, not an identity with the arithmetic sum.
Exact mathematical passage
\alpha=\frac{2a-1}{\beta},\qquad \delta_Q=\frac1{\zeta(2)}\prod_{p\mid Q}\frac{p}{p+1}.
\delta_Q^\betaUse in source · 1 passage
Search aliases: δ_Q^β delta beta
Use in source · Read passage · arithmetic_inverse.tex, line 25
The counting law N_Q(x)=δ_Qx+O_Q(sqrt x) gives λ_j=2δ_Q^β j^-β [1+O_Q,β(j^-1/2+j^-β)], β=-lim_j→∞fraclogλ_jlog j. Once β is recovered, n_j=(1-e^-λ_j/2)^-1/β, a=-fraclog w_j2log n_j. Prime membership in this integer support then recovers the excluded prime set. The reconstruction is exact from infinite data; it does not certify that a finite search has found every excluded prime.
Exact mathematical passage
\lambda_j=2\delta_Q^\beta j^{-\beta} [1+O_{Q,\beta}(j^{-1/2}+j^{-\beta})], \quad \beta=-\lim_{j\to\infty}\frac{\log\lambda_j}{\log j}.
\delta_Q^{\beta}Use in source · 1 passage
Search aliases: δ_Q^β delta beta
Use in source · Read passage · arithmetic_full.tex, line 111
n_j=fracjδ_Q(1+O_Q(j^-1/2)), λ_j=2δ_Q^βj^-β (1+O_Q,β(j^-1/2+j^-β)).
Exact mathematical passage
n_j=\frac{j}{\delta_Q}\left(1+O_Q(j^{-1/2})\right), \quad \lambda_j=2\delta_Q^{\beta}j^{-\beta} \left(1+O_{Q,\beta}(j^{-1/2}+j^{-\beta})\right).
\delta_{x_n}Use in source · 4 passages
Search aliases: δ_x_n delta
Use in source · Read passage · main.tex, line 360
Prime sieves and signed measures. pageheadingFAMILY 3An infinite prime shelf changes the packets For a prime p and λ≥-1, let x_n=(1-n^-2)^2, ν_p,λ=∑_n≥1μ(n)^2n^-4 (1+λ1_pmid n)δ_x_n. Define monic orthogonal polynomials P_m^(k;p,λ) for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the or…
Exact mathematical passage
x_n=(1-n^{-2})^2,\qquad \nu_{p,\lambda}=\sum_{n\ge1}\mu(n)^2n^{-4} (1+\lambda\mathbf1_{p\mid n})\delta_{x_n}.
\delta_{\lambda_j}Use in source · 1 passage
Search aliases: δ_λ_j delta lambda
Use in source · Read passage · arithmetic_full.tex, line 498
Uniqueness of a positive measure from its Laplace transform is classical. Schilling, Song and Vondraček~[source reference] give the uniqueness and Bernstein representation results. Here the measure is ∑_jw_jδ_λ_j, the transform variable is real depth s>0, and finiteness follows from ∑_jw_j<∞. The exact recovery of a, β and rad(Q), the rank asymptotic and the large-prime instability estimates are proved above. They are not attributed to that reference.
Exact mathematical passage
\sum_jw_j\delta_{\lambda_j}
\epsilonUse in source · 4 passages
Search aliases: ϵ epsilon
Use in source · Read passage · main.tex, line 314
Let α=frac2a-1β, δ_Q=frac1ζ(2)∏_pmid Qfracpp+1. As |s|→∞ in any closed sector |arg s|≤π/2-ϵ with 0<ϵ<π/2, A_a,β,Q(s)=fracδ_QβB(α,2s+1) +O_a,β,Q,ϵ(|s|^-(2a-1/2)/β). In particular, A_a,β,Q(s)∼fracδ_QΓ(α)β(2s)^-α. The principal power is used in the right half-plane. The beta expression is an asymptotic continuous-density model, not an identity with the arithmetic sum.
Exact mathematical passage
|\arg s|\le\pi/2-\epsilon
\varepsilonUse in source · 55 passages
Search aliases: ε varepsilon
Use in source · Read passage · main.tex, line 207
fracμ_13z^3+fracμ_1μ_215z^5+O(z^7). If q_j^ρ_j+r_j^ρ_j=1, then ρ(F_Θ)=max_j activeρ_j. The positive weights alter local coefficients. The most expansive active channel determines the asymptotic order. A zero-weight channel is removed before taking the maximum. fig07_weak_tapA two-channel example: scales (1/4,1/4) with weight 1-ε, and (3/4,3/4) with weight ε. The function varies continuously on every fixed compact set, while its growth order jumps when the second channel becomes active. For 0<ε≤1…
Exact mathematical passage
1-\varepsilon
\varepsilon^{-\rho/(\rho-1/2)}Use in source · 1 passage
Search aliases: ε^-ρ/(ρ-1/2) varepsilon rho rho
Use in source · Read passage · resolution_full.tex, line 209
The radius column displays the analytic formula approximately. For a certificate use the formula itself or a rigorously rounded upper value. These large numbers reflect the conservatism of the global comparison. They do not mean that observations at all smaller radii fail. The formula grows proportionally to ε^-ρ/(ρ-1/2) as the tap vanishes.
Exact mathematical passage
\varepsilon^{-\rho/(\rho-1/2)}
\etaUse in source · 26 passages
Search aliases: η eta
Use in source · Read passage · main.tex, line 396
ARYFinite positivity can conceal a signed defect Let ν_nonp and ν_p be the restrictions of the undeformed measure ν_p,0 to indices not divisible by p and divisible by p, respectively. Then ν_p,λ=ν_nonp+(1+λ)ν_p. Let A_m,B_m be the order-m Hankel matrices of these two unscaled positive measures. Both are positive definite. Thus τ_m(λ)=det(A_m+(1+λ)B_m) =det A_m∏_i=1^m[1+(1+λ)η_i], where η_i>0 are the eigenvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and below -1; possible coeffic…
Exact mathematical passage
\tau_m(\lambda)=\det(A_m+(1+\lambda)B_m) =\det A_m\prod_{i=1}^{m}[1+(1+\lambda)\eta_i],
\eta^2Use in source · 1 passage
Search aliases: η^2 eta
Use in source · Read passage · arithmetic_full.tex, line 422
A_χ,ψ(s) =∑_j≥0(-1)^jbinom2sj [fracL(2a+jβ,η) L(4a+2jβ,η^2)-1].
Exact mathematical passage
A_{\chi,\psi}(s) =\sum_{j\ge0}(-1)^j\binom{2s}{j} \left[\frac{L(2a+j\beta,\eta)} {L(4a+2j\beta,\eta^2)}-1\right].
\eta_iUse in source · 4 passages
Search aliases: η_i eta
Use in source · Read passage · main.tex, line 396
ARYFinite positivity can conceal a signed defect Let ν_nonp and ν_p be the restrictions of the undeformed measure ν_p,0 to indices not divisible by p and divisible by p, respectively. Then ν_p,λ=ν_nonp+(1+λ)ν_p. Let A_m,B_m be the order-m Hankel matrices of these two unscaled positive measures. Both are positive definite. Thus τ_m(λ)=det(A_m+(1+λ)B_m) =det A_m∏_i=1^m[1+(1+λ)η_i], where η_i>0 are the eigenvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and below -1; possible coeffic…
Exact mathematical passage
\tau_m(\lambda)=\det(A_m+(1+\lambda)B_m) =\det A_m\prod_{i=1}^{m}[1+(1+\lambda)\eta_i],
\eta_NUse in source · 3 passages
Search aliases: η_N eta
Use in source · Read passage · spectral_full.tex, line 137
[0,π]. Thus, once (N-1)e^-γ_N|x|<1, the distinguished term e^x in (S1) is larger in modulus than the sum of all competitors, and d_N(w)≠0. Consequently, sufficiently large zeros avoid the entire closed left half-plane. There are only finitely many remaining zeros in a bounded disk. None lies on the negative ray, since d_N(-t)>0 for t≥0. It follows that some η_N>0 satisfies
Exact mathematical passage
\eta_N>0
\thetaUse in source · 5 passages
Search aliases: θ theta
Use in source · Read passage · main.tex, line 442
claimAn endpoint mass is numerically delicate For the undeformed depth-zero measure, set ν_θ=ν+θδ_0, θ≥-1. Write h_j=∫ P_j(x)^2 dν(x) and K_m(x,y)=∑_j=0^mP_j(x)P_j(y)/h_j. The Uvarov formula~[source reference] is P_m^[θ](x)=P_m(x)- fracθ P_m(0)1+θ K_m-1(0,0)K_m-1(x,0), with K_-1=0 for degree zero. The mass at zero is recovered by 1≤frac1K_m(0,0)≤1+(M_0-1)(7/16)^2m. The lower bound comes from the zero atom, whose mass is one. The test polynomial (1-x)^m gives the upper bound since every other nod…
Exact mathematical passage
\nu_\theta=\nu+\theta\delta_0
\kappaUse in source · 19 passages
Search aliases: κ kappa
Use in source · Read passage · arithmetic_full.tex, line 299
η κ_j(t), κ_j(t)=fracA(-t)R_j(-t) ∼fracw_1w_je^(λ_1-λ_j)t.
Exact mathematical passage
\eta\,\kappa_j(t),\qquad \kappa_j(t)=\frac{A(-t)}{R_j(-t)} \sim\frac{w_1}{w_j}e^{(\lambda_1-\lambda_j)t}.
\kappa_FUse in source · 5 passages
Search aliases: κ_F kappa
Use in source · Read passage · sieve_full.tex, line 16
x_n=(1-n^-2)^2, κ_F(n;λ)=∏_p∈ F(1+λ_p1_pmid n), ν_F,λ =∑_n≥1μ(n)^2n^-4κ_F(n;λ)δ_x_n.
Exact mathematical passage
x_n=(1-n^{-2})^2,\qquad \kappa_F(n;\boldsymbol\lambda)=\prod_{p\in F}(1+\lambda_p\mathbf1_{p\mid n}),
\kappa_jUse in source · 3 passages
Search aliases: κ_j kappa
Use in source · Read passage · arithmetic_full.tex, line 299
η κ_j(t), κ_j(t)=fracA(-t)R_j(-t) ∼fracw_1w_je^(λ_1-λ_j)t.
Exact mathematical passage
\eta\,\kappa_j(t),\qquad \kappa_j(t)=\frac{A(-t)}{R_j(-t)} \sim\frac{w_1}{w_j}e^{(\lambda_1-\lambda_j)t}.
\kappa_NUse in source · 1 passage
Search aliases: κ_N kappa
Use in source · Read passage · spectral_continuation.tex, line 29
pageheadingORBIT SPECTRA / LARGE RADIUSA finite exponential sum controls the tail Put E_N(t)=d_N(-t)=∑_n≥0t^n/(Nn)! and ω=e^2π i/N. For x≥0, E_N(x^N)=frac1N∑_j=0^N-1e^ω^jx. Root filtering proves this identity term by term: only powers divisible by N remain. The positive real exponential is dominant, so log E_N(t)=t^1/N-log N+O(e^-κ_N t^1/N), κ_N=1-cos(2π/N)>0, with corresponding differentiated estimates on the positive ray. fig14_spectral_tailComputed logarithmic derivatives of the finite expone…
Exact mathematical passage
\log E_N(t)=t^{1/N}-\log N+O(e^{-\kappa_N t^{1/N}}), \quad \kappa_N=1-\cos(2\pi/N)>0,
\boldsymbol\lambdaUse in source · 11 passages
Search aliases: λ boldsymbol lambda
Use in source · Read passage · notation.tex, line 43
item[Prime-sieve measure.] ν_F,λ is the arithmetic atomic measure obtained by multiplying the mass at x_n=(1-n^-2)^2 by ∏_p∈ F(1+λ_p1_pmid n). It is positive exactly when every real strength satisfies λ_p≥-1. item[Hankel order and polynomial degree.] For moments M_j=∫ x^jddν, the order-m matrix is H_m=[M_i+j]_0≤ i,j<m. It tests quadratic forms of real polynomials of degree at most m-1. At depth k, the matrix is [M_k+i+j]_0≤ i,j<m. An order-five test therefore covers degree four, not degree five.…
Exact mathematical passage
\nu_{F,\boldsymbol\lambda}
\lambdaUse in source · 124 passages
Search aliases: λ lambda
Use in source · Read passage · main.tex, line 247
-v =fracζ(v)ζ(2v)∏_pmid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit working precision. Right: initial squarefree frequencies and weights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If p_* is the lea…
Exact mathematical passage
\lambda_n=-2\log(1-n^{-\beta})
\lambda_1Use in source · 4 passages
Search aliases: λ_1 lambda
Use in source · Read passage · arithmetic_full.tex, line 28
The logarithm is real. The omitted n=1 atom is not restored in any inverse formula. Let n_1<n_2<⋯ be the support integers, and abbreviate λ_j=λ_n_j, w_j=n_j^-2a. Then λ_1>λ_2>⋯>0, λ_j→0, and ∑_jw_j=M=Z_Q(2a)-1<∞, where
Exact mathematical passage
\lambda_1>\lambda_2>\cdots>0
\lambda_2Use in source · 12 passages
Search aliases: λ_2 lambda
Use in source · Read passage · arithmetic_full.tex, line 28
The logarithm is real. The omitted n=1 atom is not restored in any inverse formula. Let n_1<n_2<⋯ be the support integers, and abbreviate λ_j=λ_n_j, w_j=n_j^-2a. Then λ_1>λ_2>⋯>0, λ_j→0, and ∑_jw_j=M=Z_Q(2a)-1<∞, where
Exact mathematical passage
\lambda_1>\lambda_2>\cdots>0
\lambda_3Use in source · 8 passages
Search aliases: λ_3 lambda
Use in source · Read passage · multi_prime.tex, line 40
pageheadingPRIME SIEVES / FINITE TESTSMoment positivity is a weaker requirement fig12_multi_prime_domainThe exact positive-measure domain and the order-one moment test have different parameter geometry. Regions passing a finite test outside the safe orthant still contain negative atoms. For primes two and three, M_0=105π^-4(1+λ_2/17)(1+λ_3/82). Both (0,0) and (-34,-164) have positive total mass. Their midpoint (-17,-82) has zero mass. Thus the parameter set passing the order-one positive-definit…
Exact mathematical passage
M_0=105\pi^{-4}(1+\lambda_2/17)(1+\lambda_3/82).
\lambda_5Use in source · 1 passage
Search aliases: λ_5 lambda
Use in source · Read passage · sieve_full.tex, line 415
The separate rational certificates (5) do not depend on those approximate roots. The script constructs negative detectors for primes 2,3,5,7, at Hankel orders 8,11,19,40, respectively. These sufficient orders are not asserted to be minimal. In the p=7 example, the other strengths are λ_2=1/2,λ_3=-1,λ_5=2, and λ_7=-1.0001; the exact rational upper bound on the detector quadratic form is approximately -9.74271×10^-9.
Exact mathematical passage
\lambda_2=1/2,\lambda_3=-1,\lambda_5=2
\lambda_7Use in source · 1 passage
Search aliases: λ_7 lambda
Use in source · Read passage · sieve_full.tex, line 415
The separate rational certificates (5) do not depend on those approximate roots. The script constructs negative detectors for primes 2,3,5,7, at Hankel orders 8,11,19,40, respectively. These sufficient orders are not asserted to be minimal. In the p=7 example, the other strengths are λ_2=1/2,λ_3=-1,λ_5=2, and λ_7=-1.0001; the exact rational upper bound on the detector quadratic form is approximately -9.74271×10^-9.
Exact mathematical passage
\lambda_7=-1.0001
\lambda_jUse in source · 28 passages
Search aliases: λ_j lambda
Use in source · Read passage · arithmetic_full.tex, line 27
The logarithm is real. The omitted n=1 atom is not restored in any inverse formula. Let n_1<n_2<⋯ be the support integers, and abbreviate λ_j=λ_n_j, w_j=n_j^-2a. Then λ_1>λ_2>⋯>0, λ_j→0, and ∑_jw_j=M=Z_Q(2a)-1<∞, where
Exact mathematical passage
\lambda_j=\lambda_{n_j}
\lambda_{j+1}Use in source · 4 passages
Search aliases: λ_j+1 lambda
Use in source · Read passage · arithmetic_full.tex, line 75
R_j(-t)=w_je^λ_jt(1+r_j(t)), 0≤ r_j(t)≤fracM_j+1w_j e^-(λ_j-λ_j+1)t.
Exact mathematical passage
R_j(-t)=w_je^{\lambda_jt}(1+r_j(t)),\qquad 0\le r_j(t)\le\frac{M_{j+1}}{w_j} e^{-(\lambda_j-\lambda_{j+1})t}.
\lambda_j^kUse in source · 1 passage
Search aliases: λ_j^k lambda
Use in source · Read passage · arithmetic_full.tex, line 402
The monomial moment matrices also have infinite rank: m_k=(-1)^kA^(k)(0)=∑_jw_jλ_j^k, and for every nonzero polynomial P, ∑_jw_j|P(λ_j)|^2>0, because a nonzero polynomial cannot vanish at infinitely many distinct frequencies. All finite derivative Hankel matrices are strictly positive definite.
Exact mathematical passage
m_k=(-1)^kA^{(k)}(0)=\sum_jw_j\lambda_j^k
\lambda_{L+1}Use in source · 2 passages
Search aliases: λ_L+1 lambda
Use in source · Read passage · arithmetic_full.tex, line 269
E_L(t)=e^tλ_L+1fracL^1-2a2a-1.
Exact mathematical passage
E_L(t)=e^{t\lambda_{L+1}}\frac{L^{1-2a}}{2a-1}.
\lambda_{\lfloor cj\rfloor}Use in source · 1 passage
Search aliases: λ_lfloor cjrfloor lambda lfloor rfloor
Use in source · Read passage · arithmetic_full.tex, line 133
-fraclog(λ_lfloor cjrfloor/λ_j)log c =β+O_Q,β,c(j^-1/2+j^-β).
Exact mathematical passage
-\frac{\log(\lambda_{\lfloor cj\rfloor}/\lambda_j)}{\log c} =\beta+O_{Q,\beta,c}(j^{-1/2}+j^{-\beta}).
\lambda_nUse in source · 5 passages
Search aliases: λ_n lambda
Use in source · Read passage · main.tex, line 247
-v =fracζ(v)ζ(2v)∏_pmid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit working precision. Right: initial squarefree frequencies and weights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If p_* is the lea…
Exact mathematical passage
\lambda_n=-2\log(1-n^{-\beta})
\lambda_{n_j}Use in source · 1 passage
Search aliases: λ_n_j lambda
Use in source · Read passage · arithmetic_full.tex, line 27
The logarithm is real. The omitted n=1 atom is not restored in any inverse formula. Let n_1<n_2<⋯ be the support integers, and abbreviate λ_j=λ_n_j, w_j=n_j^-2a. Then λ_1>λ_2>⋯>0, λ_j→0, and ∑_jw_j=M=Z_Q(2a)-1<∞, where
Exact mathematical passage
\lambda_j=\lambda_{n_j}
\lambda_pUse in source · 49 passages
Search aliases: λ_p lambda
Use in source · Read passage · notation.tex, line 45
item[Prime-sieve measure.] ν_F,λ is the arithmetic atomic measure obtained by multiplying the mass at x_n=(1-n^-2)^2 by ∏_p∈ F(1+λ_p1_pmid n). It is positive exactly when every real strength satisfies λ_p≥-1. item[Hankel order and polynomial degree.] For moments M_j=∫ x^jddν, the order-m matrix is H_m=[M_i+j]_0≤ i,j<m. It tests quadratic forms of real polynomials of degree at most m-1. At depth k, the matrix is [M_k+i+j]_0≤ i,j<m. An order-five test therefore covers degree four, not degree five.…
Exact mathematical passage
\prod_{p\in F}(1+\lambda_p\mathbf1_{p\mid n})
\lambda_p^kUse in source · 4 passages
Search aliases: λ_p^k lambda
Use in source · Read passage · arithmetic_full.tex, line 318
displaystyle sup_|s|≤ R|A_a,β,Q^(k)(s)-A_a,β,Qp^(k)(s)| ≤ λ_p^k e^Rλ_p fracZ_Q(2a)p^2a+1.
Exact mathematical passage
\boxed{\displaystyle \sup_{|s|\le R}|A_{a,\beta,Q}^{(k)}(s)-A_{a,\beta,Qp}^{(k)}(s)| \le \lambda_p^k e^{R\lambda_p} \frac{Z_Q(2a)}{p^{2a}+1}.}
\lambda_{p_*}Use in source · 1 passage
Search aliases: λ_p_* lambda
Use in source · Read passage · main.tex, line 285
claimNo polynomial-coefficient linear ODE Suppose ∑_j=0^rP_j(s)A^(j)(s)=0, where the P_j are polynomials and are not all zero. At s=-t, divide by the exponential of the first support atom. The remaining tail is exponentially smaller, so the polynomial ∑_jP_j(-t)(-λ_p_*)^j tends to zero exponentially and must vanish identically. Remove that atom and repeat. Every support frequency therefore satisfies ∑_jP_j(s)(-λ_n)^j=0 for all s. For fixed s, this polynomial in λ has infinitely many distinct roo…
Exact mathematical passage
\sum_jP_j(-t)(-\lambda_{p_*})^j
\lambda_qUse in source · 8 passages
Search aliases: λ_q lambda
Use in source · Read passage · multi_prime.tex, line 28
The exact coordinate formula is M_j(λ)=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) ∏_p∈ F(1+fracλ_pp^4+2r+1). Coefficients of the monic families lie in Q(π^-2,λ_p:p∈ F). For two primes, the overlap term is visible: M_j=M_j(0,0)+λ_pU_p,j+λ_qU_q,j +λ_pλ_qU_pq,j. Here U_pq,j>0 is the pq-divisible submeasure moment. The family does not reduce to adding separate prime responses without their overlap.
Exact mathematical passage
M_j=M_j(0,0)+\lambda_pU_{p,j}+\lambda_qU_{q,j} +\lambda_p\lambda_qU_{pq,j}.
\muUse in source · 33 passages
Search aliases: μ mu
Use in source · chapters_sequences.tex, line 96
changes the address ruleS14_divisor_convolution ∑_dmid nμ(d)=1_n=1, ∑_dmid nφ(d)=n The figure places the six divisors of twelve in matching columns. Möbius contributions sum to zero; totient contributions sum to twelve. The address rule is divisibility. This differs from the additive index splitting used by ordinary power-series convolution. For n=12, μ(1),μ(2),μ(3),μ(4),μ(6),μ(12) are 1,-1,-1,0,1,0. The totient values are 1,1,2,2,2,4. Their sums are zero and twelve. Choose the convolution law b…
Exact mathematical passage
\mu(1),\mu(2),\mu(3),\mu(4),\mu(6),\mu(12)
\mu_1Use in source · 2 passages
Search aliases: μ_1 mu
Use in source · Read passage · main.tex, line 194
pageheadingFAMILY 1 / TAPSFeedback banks and weak-channel sensitivity Choose finitely many channels j=1,…,J, with h_j>0, ∑_jh_j=1, and 0<q_j,r_j<1. Define F_Θ'(z)=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz), F_Θ(0)=0. These taps act inside the nonlinear definition. With μ_1=∑_jh_jq_jr_j and μ_2=∑_jh_jq_jr_j(q_j^2+r_j^2), F_Θ(z)=z+fracμ_13z^3+fracμ_1μ_215z^5+O(z^7). If q_j^ρ_j+r_j^ρ_j=1, then ρ(F_Θ)=max_j activeρ_j. The positive weights alter local coefficients. The most expansive active channel determines the as…
Exact mathematical passage
\mu_1=\sum_jh_jq_jr_j
\mu_2Use in source · 2 passages
Search aliases: μ_2 mu
Use in source · Read passage · main.tex, line 195
pageheadingFAMILY 1 / TAPSFeedback banks and weak-channel sensitivity Choose finitely many channels j=1,…,J, with h_j>0, ∑_jh_j=1, and 0<q_j,r_j<1. Define F_Θ'(z)=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz), F_Θ(0)=0. These taps act inside the nonlinear definition. With μ_1=∑_jh_jq_jr_j and μ_2=∑_jh_jq_jr_j(q_j^2+r_j^2), F_Θ(z)=z+fracμ_13z^3+fracμ_1μ_215z^5+O(z^7). If q_j^ρ_j+r_j^ρ_j=1, then ρ(F_Θ)=max_j activeρ_j. The positive weights alter local coefficients. The most expansive active channel determines the as…
Exact mathematical passage
\mu_2=\sum_jh_jq_jr_j(q_j^2+r_j^2)
\nuUse in source · 24 passages
Search aliases: ν nu
Use in source · Read passage · main.tex, line 360
Prime sieves and signed measures. pageheadingFAMILY 3An infinite prime shelf changes the packets For a prime p and λ≥-1, let x_n=(1-n^-2)^2, ν_p,λ=∑_n≥1μ(n)^2n^-4 (1+λ1_pmid n)δ_x_n. Define monic orthogonal polynomials P_m^(k;p,λ) for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the or…
Exact mathematical passage
x_n=(1-n^{-2})^2,\qquad \nu_{p,\lambda}=\sum_{n\ge1}\mu(n)^2n^{-4} (1+\lambda\mathbf1_{p\mid n})\delta_{x_n}.
\nu_{F,\boldsymbol\lambda}Use in source · 6 passages
Search aliases: ν_F,λ nu boldsymbol lambda
Use in source · Read passage · notation.tex, line 43
item[Prime-sieve measure.] ν_F,λ is the arithmetic atomic measure obtained by multiplying the mass at x_n=(1-n^-2)^2 by ∏_p∈ F(1+λ_p1_pmid n). It is positive exactly when every real strength satisfies λ_p≥-1. item[Hankel order and polynomial degree.] For moments M_j=∫ x^jddν, the order-m matrix is H_m=[M_i+j]_0≤ i,j<m. It tests quadratic forms of real polynomials of degree at most m-1. At depth k, the matrix is [M_k+i+j]_0≤ i,j<m. An order-five test therefore covers degree four, not degree five.…
Exact mathematical passage
\nu_{F,\boldsymbol\lambda}
\nu_{\mathrm{non}p}Use in source · 3 passages
Search aliases: ν_nonp nu mathrm
Use in source · Read passage · main.tex, line 389
pageheadingFAMILY 3 / BOUNDARYFinite positivity can conceal a signed defect Let ν_nonp and ν_p be the restrictions of the undeformed measure ν_p,0 to indices not divisible by p and divisible by p, respectively. Then ν_p,λ=ν_nonp+(1+λ)ν_p. Let A_m,B_m be the order-m Hankel matrices of these two unscaled positive measures. Both are positive definite. Thus τ_m(λ)=det(A_m+(1+λ)B_m) =det A_m∏_i=1^m[1+(1+λ)η_i], where η_i>0 are the eigenvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and…
Exact mathematical passage
\nu_{\mathrm{non}p}
\nu_pUse in source · 3 passages
Search aliases: ν_p nu
Use in source · Read passage · main.tex, line 389
pageheadingFAMILY 3 / BOUNDARYFinite positivity can conceal a signed defect Let ν_nonp and ν_p be the restrictions of the undeformed measure ν_p,0 to indices not divisible by p and divisible by p, respectively. Then ν_p,λ=ν_nonp+(1+λ)ν_p. Let A_m,B_m be the order-m Hankel matrices of these two unscaled positive measures. Both are positive definite. Thus τ_m(λ)=det(A_m+(1+λ)B_m) =det A_m∏_i=1^m[1+(1+λ)η_i], where η_i>0 are the eigenvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and…
Exact mathematical passage
\nu_p
\nu_{p,0}Use in source · 1 passage
Search aliases: ν_p,0 nu
Use in source · Read passage · main.tex, line 390
pageheadingFAMILY 3 / BOUNDARYFinite positivity can conceal a signed defect Let ν_nonp and ν_p be the restrictions of the undeformed measure ν_p,0 to indices not divisible by p and divisible by p, respectively. Then ν_p,λ=ν_nonp+(1+λ)ν_p. Let A_m,B_m be the order-m Hankel matrices of these two unscaled positive measures. Both are positive definite. Thus τ_m(λ)=det(A_m+(1+λ)B_m) =det A_m∏_i=1^m[1+(1+λ)η_i], where η_i>0 are the eigenvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and…
Exact mathematical passage
\nu_{p,0}
\nu_{p,\lambda}Use in source · 4 passages
Search aliases: ν_p,λ nu lambda
Use in source · Read passage · main.tex, line 360
Prime sieves and signed measures. pageheadingFAMILY 3An infinite prime shelf changes the packets For a prime p and λ≥-1, let x_n=(1-n^-2)^2, ν_p,λ=∑_n≥1μ(n)^2n^-4 (1+λ1_pmid n)δ_x_n. Define monic orthogonal polynomials P_m^(k;p,λ) for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the or…
Exact mathematical passage
x_n=(1-n^{-2})^2,\qquad \nu_{p,\lambda}=\sum_{n\ge1}\mu(n)^2n^{-4} (1+\lambda\mathbf1_{p\mid n})\delta_{x_n}.
\nu_{p\text{-divisible}}Use in source · 2 passages
Search aliases: ν_p-divisible nu text
Use in source · Read passage · multi_prime.tex, line 101
Thus every one-variable determinant fiber has exact degree m and all roots real below -1. Its nearest margin satisfies δ_m=min_substackf∈ R[x]∖{0} deg f<m frac∫ f^2ddν_p-free ∫ f^2ddν_p-divisible, δ_mdownarrow0. The detector yields, for C_κ=∏_q≠ pmax(1,1+λ_q), δ_p+m≤fracpC_κ C_p^23r_p^2m. The denominator retains the atom p; all lower nodes in the numerator are killed. This extends the prime-two bound to every prime and to simultaneous positive or deletion strengths. It does not assert simple roo…
Exact mathematical passage
\delta_m=\min_{\substack{f\in\mathbb R[x]\setminus\{0\}\\\deg f<m}} \frac{\int f^2\dd\nu_{p\text{-free}}} {\int f^2\dd\nu_{p\text{-divisible}}},\qquad\delta_m\downarrow0.
\nu_{p\text{-free}}Use in source · 2 passages
Search aliases: ν_p-free nu text
Use in source · Read passage · multi_prime.tex, line 101
Thus every one-variable determinant fiber has exact degree m and all roots real below -1. Its nearest margin satisfies δ_m=min_substackf∈ R[x]∖{0} deg f<m frac∫ f^2ddν_p-free ∫ f^2ddν_p-divisible, δ_mdownarrow0. The detector yields, for C_κ=∏_q≠ pmax(1,1+λ_q), δ_p+m≤fracpC_κ C_p^23r_p^2m. The denominator retains the atom p; all lower nodes in the numerator are killed. This extends the prime-two bound to every prime and to simultaneous positive or deletion strengths. It does not assert simple roo…
Exact mathematical passage
\delta_m=\min_{\substack{f\in\mathbb R[x]\setminus\{0\}\\\deg f<m}} \frac{\int f^2\dd\nu_{p\text{-free}}} {\int f^2\dd\nu_{p\text{-divisible}}},\qquad\delta_m\downarrow0.
\nu_\thetaUse in source · 1 passage
Search aliases: ν_θ nu theta
Use in source · Read passage · main.tex, line 442
claimAn endpoint mass is numerically delicate For the undeformed depth-zero measure, set ν_θ=ν+θδ_0, θ≥-1. Write h_j=∫ P_j(x)^2 dν(x) and K_m(x,y)=∑_j=0^mP_j(x)P_j(y)/h_j. The Uvarov formula~[source reference] is P_m^[θ](x)=P_m(x)- fracθ P_m(0)1+θ K_m-1(0,0)K_m-1(x,0), with K_-1=0 for degree zero. The mass at zero is recovered by 1≤frac1K_m(0,0)≤1+(M_0-1)(7/16)^2m. The lower bound comes from the zero atom, whose mass is one. The test polynomial (1-x)^m gives the upper bound since every other nod…
Exact mathematical passage
\nu_\theta=\nu+\theta\delta_0
\piUse in source · 57 passages
Search aliases: π pi
Use in source · Read passage · main.tex, line 314
Let α=frac2a-1β, δ_Q=frac1ζ(2)∏_pmid Qfracpp+1. As |s|→∞ in any closed sector |arg s|≤π/2-ϵ with 0<ϵ<π/2, A_a,β,Q(s)=fracδ_QβB(α,2s+1) +O_a,β,Q,ϵ(|s|^-(2a-1/2)/β). In particular, A_a,β,Q(s)∼fracδ_QΓ(α)β(2s)^-α. The principal power is used in the right half-plane. The beta expression is an asymptotic continuous-density model, not an identity with the arithmetic sum.
Exact mathematical passage
|\arg s|\le\pi/2-\epsilon
\pi^{-2}Use in source · 3 passages
Search aliases: π^-2 pi
Use in source · Read passage · main.tex, line 377
two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P_1=x-fracM_1M_0, P_2=x^2-fracM_0M_3-M_1M_2Δx+ fracM_1M_3-M_2^2Δ. The infinitely many positive nodes make every Hankel matrix strictly positi…
Exact mathematical passage
\mathbb Q(\pi^{-2},\lambda)
\pi^{-2s}Use in source · 2 passages
Search aliases: π^-2s pi
Use in source · Read passage · main.tex, line 529
claimTwo exact Hurwitz-zeta reductions For N=2, d_2(w)=cossqrt w, whose zeros are w_j=π^2(j+1/2)^2, j≥0. Therefore Z_2(s)=π^-2sζ(2s,1/2). For N=4, D_4(z)=cos(z/sqrt2)cosh(z/sqrt2), and the rotation quotient has one zero w_j=(sqrt2π)^4(j+1/2)^4 per orbit. Hence Z_4(s)=(sqrt2π)^-4sζ(4s,1/2). These anchors are classical reductions. The general spectral-zeta mechanism also has substantial antecedents, including hypergeometric zero zeta functions.
Exact mathematical passage
\mathcal Z_2(s)=\pi^{-2s}\zeta(2s,1/2).
\pi^{-4}Use in source · 4 passages
Search aliases: π^-4 pi
Use in source · Read passage · main.tex, line 418
For p=2, test with f(x)=x(1-x)^m-2, m≥2: δ_m≤frac409681(frac16M_017-1) (frac272567)^2(m-2), M_0=105π^-4. This follows from x_2=9/16 and x_n≥64/81 at nonzero odd nodes. Exactly δ_1=p^4; the limit is an infinite-resolution phenomenon.
Exact mathematical passage
\delta_m\le\frac{4096}{81}\left(\frac{16M_0}{17}-1\right) \left(\frac{272}{567}\right)^{2(m-2)},\qquad M_0=105\pi^{-4}.
\pi^2Use in source · 1 passage
Search aliases: π^2 pi
Use in source · Read passage · main.tex, line 528
claimTwo exact Hurwitz-zeta reductions For N=2, d_2(w)=cossqrt w, whose zeros are w_j=π^2(j+1/2)^2, j≥0. Therefore Z_2(s)=π^-2sζ(2s,1/2). For N=4, D_4(z)=cos(z/sqrt2)cosh(z/sqrt2), and the rotation quotient has one zero w_j=(sqrt2π)^4(j+1/2)^4 per orbit. Hence Z_4(s)=(sqrt2π)^-4sζ(4s,1/2). These anchors are classical reductions. The general spectral-zeta mechanism also has substantial antecedents, including hypergeometric zero zeta functions.
Exact mathematical passage
w_j=\pi^2(j+1/2)^2
\pi^4Use in source · 3 passages
Search aliases: π^4 pi
Use in source · Read passage · arithmetic_full.tex, line 353
For a=β=2,Q=1, the mass removed by p is exactly 105/[π^4(p^4+1)]. The examples p=101,1009,10007 quantify how quickly large-prime distinctions disappear on |s|≤5.
Exact mathematical passage
105/[\pi^4(p^4+1)]
\rho^NUse in source · 1 passage
Search aliases: ρ^N rho
Use in source · REPORT.md, line 28
|∑_n≥ Nc_nz^2n+1| ≤ frac|z|ρ^N1-ρ, |z|^2<2.
Exact mathematical passage
\left|\sum_{n\ge N}c_nz^{2n+1}\right| \le \frac{|z|\rho^N}{1-\rho},\qquad |z|^2<2.
\rho_jUse in source · 5 passages
Search aliases: ρ_j rho
Use in source · Read passage · main.tex, line 199
vity Choose finitely many channels j=1,…,J, with h_j>0, ∑_jh_j=1, and 0<q_j,r_j<1. Define F_Θ'(z)=1+∑_jh_jF_Θ(q_jz)F_Θ(r_jz), F_Θ(0)=0. These taps act inside the nonlinear definition. With μ_1=∑_jh_jq_jr_j and μ_2=∑_jh_jq_jr_j(q_j^2+r_j^2), F_Θ(z)=z+fracμ_13z^3+fracμ_1μ_215z^5+O(z^7). If q_j^ρ_j+r_j^ρ_j=1, then ρ(F_Θ)=max_j activeρ_j. The positive weights alter local coefficients. The most expansive active channel determines the asymptotic order. A zero-weight channel is removed before taking th…
Exact mathematical passage
q_j^{\rho_j}+r_j^{\rho_j}=1
\rho_\alphaUse in source · 2 passages
Search aliases: ρ_α rho alpha
Use in source · Read passage · main.tex, line 134
claimWhy this equation appears Nonnegative coefficients give M(R)=T_q,r(R). The integral equation implies M(R)≤ R[1+M(qR)M(rR)], M(R)≥(1-α)R M(α qR)M(α rR),quad0<α<1. Taking logarithms turns product feedback into a sum over contracted scales. Comparison with CR^ρ proves the upper bound; comparison with cR^ρ_α, where (α q)^ρ_α+ (α r)^ρ_α=1, proves the lower bound. Letting αuparrow1 gives the same order. The complete interval-induction proof is given in Appendix~[source reference].
Exact mathematical passage
cR^{\rho_\alpha}
\sigmaUse in source · 5 passages
Search aliases: σ sigma
Use in source · Read passage · spectral_full.tex, line 112
For completeness, the series bound |d_N(w)|≤exp(|w|^1/N) gives order at most 1/N<1. Hadamard factorization therefore has genus zero and a constant exponential factor, which equals one because d_N(0)=1. The usual zero-count consequence gives ∑_j|w_j|^-σ<∞ for every σ>1/N. Along the negative real axis (S3) also shows the order is exactly 1/N.
Exact mathematical passage
\sum_j|w_j|^{-\sigma}<\infty
\tau_0^{(k)}Use in source · 1 passage
Search aliases: τ_0^(k) tau
Use in source · Read passage · main.tex, line 433
operations with arithmetic coordinates The depth tilt x^k is a Christoffel transformation. Suppressing p,λ, P_m^(k+1)(x)=fracP_m+1^(k)(x) -[P_m+1^(k)(0)/P_m^(k)(0)]P_m^(k)(x)x. The numerator vanishes at zero. All roots are positive, so the displayed denominator coefficient is nonzero. For shifted Hankel determinants τ_m^(k)=det[M_k+i+j]_i,j<m, with τ_0^(k)=1, τ_m^(k)τ_m^(k+2)-(τ_m^(k+1))^2 =τ_m+1^(k)τ_m-1^(k+2)>0, m≥1. This is the Desnanot--Jacobi identity applied to exact arithmetic moments. Th…
Exact mathematical passage
\tau_0^{(k)}=1
\tau_{m+1}^{(k)}Use in source · 1 passage
Search aliases: τ_m+1^(k) tau
Use in source · Read passage · main.tex, line 434
thmetic coordinates The depth tilt x^k is a Christoffel transformation. Suppressing p,λ, P_m^(k+1)(x)=fracP_m+1^(k)(x) -[P_m+1^(k)(0)/P_m^(k)(0)]P_m^(k)(x)x. The numerator vanishes at zero. All roots are positive, so the displayed denominator coefficient is nonzero. For shifted Hankel determinants τ_m^(k)=det[M_k+i+j]_i,j<m, with τ_0^(k)=1, τ_m^(k)τ_m^(k+2)-(τ_m^(k+1))^2 =τ_m+1^(k)τ_m-1^(k+2)>0, m≥1. This is the Desnanot--Jacobi identity applied to exact arithmetic moments. The construction mech…
Exact mathematical passage
\tau_m^{(k)}\tau_m^{(k+2)}-(\tau_m^{(k+1)})^2 =\tau_{m+1}^{(k)}\tau_{m-1}^{(k+2)}>0,\qquad m\ge1.
\tau_{m-1}^{(k+2)}Use in source · 1 passage
Search aliases: τ_m-1^(k+2) tau
Use in source · Read passage · main.tex, line 434
thmetic coordinates The depth tilt x^k is a Christoffel transformation. Suppressing p,λ, P_m^(k+1)(x)=fracP_m+1^(k)(x) -[P_m+1^(k)(0)/P_m^(k)(0)]P_m^(k)(x)x. The numerator vanishes at zero. All roots are positive, so the displayed denominator coefficient is nonzero. For shifted Hankel determinants τ_m^(k)=det[M_k+i+j]_i,j<m, with τ_0^(k)=1, τ_m^(k)τ_m^(k+2)-(τ_m^(k+1))^2 =τ_m+1^(k)τ_m-1^(k+2)>0, m≥1. This is the Desnanot--Jacobi identity applied to exact arithmetic moments. The construction mech…
Exact mathematical passage
\tau_m^{(k)}\tau_m^{(k+2)}-(\tau_m^{(k+1)})^2 =\tau_{m+1}^{(k)}\tau_{m-1}^{(k+2)}>0,\qquad m\ge1.
\tau_m^{(k)}Use in source · 2 passages
Search aliases: τ_m^(k) tau
Use in source · Read passage · main.tex, line 432
eheadingFAMILY 3 / TRANSPORTClassical operations with arithmetic coordinates The depth tilt x^k is a Christoffel transformation. Suppressing p,λ, P_m^(k+1)(x)=fracP_m+1^(k)(x) -[P_m+1^(k)(0)/P_m^(k)(0)]P_m^(k)(x)x. The numerator vanishes at zero. All roots are positive, so the displayed denominator coefficient is nonzero. For shifted Hankel determinants τ_m^(k)=det[M_k+i+j]_i,j<m, with τ_0^(k)=1, τ_m^(k)τ_m^(k+2)-(τ_m^(k+1))^2 =τ_m+1^(k)τ_m-1^(k+2)>0, m≥1. This is the Desnanot--Jacobi identity a…
Exact mathematical passage
\tau_m^{(k)}=\det[M_{k+i+j}]_{i,j<m}
\tau_m^{(k+1)}Use in source · 1 passage
Search aliases: τ_m^(k+1) tau
Use in source · Read passage · main.tex, line 434
thmetic coordinates The depth tilt x^k is a Christoffel transformation. Suppressing p,λ, P_m^(k+1)(x)=fracP_m+1^(k)(x) -[P_m+1^(k)(0)/P_m^(k)(0)]P_m^(k)(x)x. The numerator vanishes at zero. All roots are positive, so the displayed denominator coefficient is nonzero. For shifted Hankel determinants τ_m^(k)=det[M_k+i+j]_i,j<m, with τ_0^(k)=1, τ_m^(k)τ_m^(k+2)-(τ_m^(k+1))^2 =τ_m+1^(k)τ_m-1^(k+2)>0, m≥1. This is the Desnanot--Jacobi identity applied to exact arithmetic moments. The construction mech…
Exact mathematical passage
\tau_m^{(k)}\tau_m^{(k+2)}-(\tau_m^{(k+1)})^2 =\tau_{m+1}^{(k)}\tau_{m-1}^{(k+2)}>0,\qquad m\ge1.
\tau_m^{(k+2)}Use in source · 1 passage
Search aliases: τ_m^(k+2) tau
Use in source · Read passage · main.tex, line 434
thmetic coordinates The depth tilt x^k is a Christoffel transformation. Suppressing p,λ, P_m^(k+1)(x)=fracP_m+1^(k)(x) -[P_m+1^(k)(0)/P_m^(k)(0)]P_m^(k)(x)x. The numerator vanishes at zero. All roots are positive, so the displayed denominator coefficient is nonzero. For shifted Hankel determinants τ_m^(k)=det[M_k+i+j]_i,j<m, with τ_0^(k)=1, τ_m^(k)τ_m^(k+2)-(τ_m^(k+1))^2 =τ_m+1^(k)τ_m-1^(k+2)>0, m≥1. This is the Desnanot--Jacobi identity applied to exact arithmetic moments. The construction mech…
Exact mathematical passage
\tau_m^{(k)}\tau_m^{(k+2)}-(\tau_m^{(k+1)})^2 =\tau_{m+1}^{(k)}\tau_{m-1}^{(k+2)}>0,\qquad m\ge1.
\phiUse in source · 1 passage
Search aliases: ϕ phi
Use in source · Read passage · arithmetic_inverse.tex, line 92
pageheadingARITHMETIC DEPTH / REALIZATIONAn infinite feature space with arithmetic labels For the ordered support, define ϕ_s(j)=sqrtw_je^-λ_js. The kernel A(s+overline t) is its weighted ℓ^2 Gram kernel. Summability and bounded frequencies make every feature vector well defined.
Exact mathematical passage
\phi_s(j)=\sqrt{w_j}e^{-\lambda_js}
\varphiUse in source · 4 passages
Search aliases: φ varphi
Use in source · Read passage · arithmetic_full.tex, line 189
|N_Q(x)-δ_Qx|≤ C_Qsqrt x+1, C_Q=2^ω(Q)+2fracφ(Q)Q.
Exact mathematical passage
|N_Q(x)-\delta_Qx|\le C_Q\sqrt x+1, \qquad C_Q=2^{\omega(Q)}+2\frac{\varphi(Q)}Q.
\phi_sUse in source · 1 passage
Search aliases: ϕ_s phi
Use in source · Read passage · arithmetic_inverse.tex, line 92
pageheadingARITHMETIC DEPTH / REALIZATIONAn infinite feature space with arithmetic labels For the ordered support, define ϕ_s(j)=sqrtw_je^-λ_js. The kernel A(s+overline t) is its weighted ℓ^2 Gram kernel. Summability and bounded frequencies make every feature vector well defined.
Exact mathematical passage
\phi_s(j)=\sqrt{w_j}e^{-\lambda_js}
\chiUse in source · 10 passages
Search aliases: χ chi
Use in source · Read passage · arithmetic_full.tex, line 411
Let χ,ψ be Dirichlet characters modulo the same fixed integer Q, extended by zero outside the units, and put η=χoverlineψ. The base extension is
Exact mathematical passage
\chi,\psi
\psiUse in source · 8 passages
Search aliases: ψ psi
Use in source · Read passage · arithmetic_full.tex, line 411
Let χ,ψ be Dirichlet characters modulo the same fixed integer Q, extended by zero outside the units, and put η=χoverlineψ. The base extension is
Exact mathematical passage
\chi,\psi
\omegaUse in source · 14 passages
Search aliases: ω omega
Use in source · Read passage · arithmetic_full.tex, line 186
The rank asymptotic can be made effective without an RH input. Write ω(Q)=#{p:pmid Q}. For real x≥4, inclusion and exclusion gives
Exact mathematical passage
\omega(Q)=\#\{p:p\mid Q\}
\omega^{-jk}Use in source · 2 passages
Search aliases: ω^-jk omega
Use in source · Read passage · sequence_full.tex, line 268
P_b,jG(z)=frac1b∑_k=0^b-1ω^-jkG(ω^kz) =∑_m≡ j (b)a_mfracz^mm!.
Exact mathematical passage
\boxed{ \mathcal P_{b,j}G(z)=\frac1b\sum_{k=0}^{b-1}\omega^{-jk}G(\omega^kz) =\sum_{m\equiv j\,(b)}a_m\frac{z^m}{m!}.}
\omega^jUse in source · 4 passages
Search aliases: ω^j omega
Use in source · Read passage · spectral_continuation.tex, line 24
pageheadingORBIT SPECTRA / LARGE RADIUSA finite exponential sum controls the tail Put E_N(t)=d_N(-t)=∑_n≥0t^n/(Nn)! and ω=e^2π i/N. For x≥0, E_N(x^N)=frac1N∑_j=0^N-1e^ω^jx. Root filtering proves this identity term by term: only powers divisible by N remain. The positive real exponential is dominant, so log E_N(t)=t^1/N-log N+O(e^-κ_N t^1/N), κ_N=1-cos(2π/N)>0, with corresponding differentiated estimates on the positive ray. fig14_spectral_tailComputed logarithmic derivatives of the finite expone…
Exact mathematical passage
E_N(x^N)=\frac1N\sum_{j=0}^{N-1}e^{\omega^jx}.
\omega^{jm}Use in source · 1 passage
Search aliases: ω^jm omega
Use in source · Read passage · spectral_full.tex, line 67
This equality follows by expanding the exponentials and using N^-1∑_jω^jm=1 when Nmid m, and zero otherwise. For real x≥0, the power series makes F_N(x)>0. Put
Exact mathematical passage
N^{-1}\sum_j\omega^{jm}=1
\omega^kUse in source · 2 passages
Search aliases: ω^k omega
Use in source · Read passage · sequence_full.tex, line 268
P_b,jG(z)=frac1b∑_k=0^b-1ω^-jkG(ω^kz) =∑_m≡ j (b)a_mfracz^mm!.
Exact mathematical passage
\boxed{ \mathcal P_{b,j}G(z)=\frac1b\sum_{k=0}^{b-1}\omega^{-jk}G(\omega^kz) =\sum_{m\equiv j\,(b)}a_m\frac{z^m}{m!}.}
\omega^{k(m-j)}Use in source · 1 passage
Search aliases: ω^k(m-j) omega
Use in source · Read passage · sequence_full.tex, line 276
This holds as a formal series and wherever the original series converges. Indeed the coefficient multiplier is the geometric sum b^-1∑_k=0^b-1ω^k(m-j), equal to one or zero according to the residue. It proves all the projector identities
Exact mathematical passage
b^{-1}\sum_{k=0}^{b-1}\omega^{k(m-j)}
\ldotsUse in source · 16 passages
Search aliases: … ldots
Use in source · chapters_sequences.tex, line 47
^nn!=exp(e^t-1) Each matrix row contains Stirling numbers of the second kind, counting partitions into k nonempty blocks. Summing across the row gives the Bell number at its right. The generating function is exponential, so its coefficient is B_n/n!. The fifth row contains 0,1,15,25,10,1. Its sum is 52=B_5. Equivalently, apply the Stirling transform to the constant input sequence 1,1,1,…. A triangular transform uses all relevant input positions. It is not automatically a fixed-length tap filter.…
Exact mathematical passage
1,1,1,\ldots
\ellUse in source · 24 passages
Search aliases: ℓ ell
Use in source · Read passage · arithmetic_full.tex, line 364
For c=(c_j)∈ℓ^2, define
Exact mathematical passage
c=(c_j)\in\ell^2
\ell^2Use in source · 6 passages
Search aliases: ℓ^2 ell
Use in source · Read passage · arithmetic_full.tex, line 364
For c=(c_j)∈ℓ^2, define
Exact mathematical passage
c=(c_j)\in\ell^2
\ell^TUse in source · 1 passage
Search aliases: ℓ^T ell
Use in source · Read passage · coupled_growth_full.tex, line 152
ℓ=(a^ρ,1-q^ρ)^T, ℓ^TP(ρ)=ℓ^T.
Exact mathematical passage
\ell=(a^\rho,1-q^\rho)^T,\qquad \ell^TP(\rho)=\ell^T.
\ell_1Use in source · 5 passages
Search aliases: ℓ_1 ell
Use in source · Read passage · coupled_structure_full.tex, line 124
The preceding proof also makes a coordinate invariant obstruction explicit. Under an invertible state change, write the original coordinates as u=ℓ_1(W),v=ℓ_2(W), where ℓ_1,ℓ_2 are independent linear forms. At the pair of distinct nodes (qz,rz), the first quadratic term has bilinear coefficient
Exact mathematical passage
u=\ell_1(W),v=\ell_2(W)
\ell_2Use in source · 5 passages
Search aliases: ℓ_2 ell
Use in source · Read passage · coupled_structure_full.tex, line 124
The preceding proof also makes a coordinate invariant obstruction explicit. Under an invertible state change, write the original coordinates as u=ℓ_1(W),v=ℓ_2(W), where ℓ_1,ℓ_2 are independent linear forms. At the pair of distinct nodes (qz,rz), the first quadratic term has bilinear coefficient
Exact mathematical passage
u=\ell_1(W),v=\ell_2(W)
\ell_iUse in source · 3 passages
Search aliases: ℓ_i ell
Use in source · Read passage · resolution_full.tex, line 399
ℓ_i(x)=∏_j≠ ifracx-y_jy_i-y_j, b_i=∏_j≠ ifracmax(y_j,1-y_j)|y_i-y_j|, K_d=∑_i=0^dn_i^4b_i^2.
Exact mathematical passage
\ell_i(x)=\prod_{j\ne i}\frac{x-y_j}{y_i-y_j}, \quad b_i=\prod_{j\ne i}\frac{\max(y_j,1-y_j)}{|y_i-y_j|}, \quad K_d=\sum_{i=0}^{d}n_i^4b_i^2.
\mapstoUse in source · 2 passages
Search aliases: ↦ mapsto
Use in source · Read passage · sequence_full.tex, line 210
The integrals are path independent on the disk. Linear-ODE uniqueness shows U=T on the previous disk, and there T(qz) equals the newly extended function at qz. Thus the same nonlinear equation holds on the enlarged disk. Iterating R↦ R/q exhausts the plane.
Exact mathematical passage
R\mapsto R/q
\partialUse in source · 1 passage
Search aliases: ∂ partial
Use in source · Read passage · dilation_full.tex, line 233
0≤∂_h_jc_n≤ n d_n(Q).
Exact mathematical passage
0\le\partial_{h_j}c_n\le n d_n(Q).
\partial_{h_j}Use in source · 1 passage
Search aliases: ∂_h_j partial
Use in source · Read passage · dilation_full.tex, line 233
0≤∂_h_jc_n≤ n d_n(Q).
Exact mathematical passage
0\le\partial_{h_j}c_n\le n d_n(Q).
\nablaUse in source · 1 passage
Search aliases: ∇ nabla
Use in source · Read passage · dilation_full.tex, line 242
|v·∇ c_n|≤frac d2 n d_n(Q), |c_n(h)-c_n(g)|≤frac d2 n d_n(Q).
Exact mathematical passage
|v\cdot\nabla c_n|\le\frac d2 n d_n(Q),\qquad |c_n(h)-c_n(g)|\le\frac d2 n d_n(Q).
\notinUse in source · 1 passage
Search aliases: ∉ notin
Use in source · Read passage · arithmetic_full.tex, line 168
displaystyle rad(Q)=∏_substackp prime p∉{n_j:j≥1}p.
Exact mathematical passage
\boxed{\displaystyle \operatorname{rad}(Q)=\prod_{\substack{p\ \mathrm{prime}\\ p\notin\{n_j:j\ge1\}}}p.}
\prodUse in source · 35 passages
Search aliases: ∏ prod
Use in source · Read passage · main.tex, line 238
(n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At depth zero the corresponding mass-one atom is absent. In particular, A(0)=Z_Q(2a)-1, where Z_Q(v)=∑_substackn≥1 (n,Q)=1μ(n)^2n^-v =fracζ(v)ζ(2v)∏_pmid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit wor…
Exact mathematical passage
Z_Q(v)=\sum_{\substack{n\ge1\\(n,Q)=1}}\mu(n)^2n^{-v} =\frac{\zeta(v)}{\zeta(2v)}\prod_{p\mid Q}(1+p^{-v})^{-1},\qquad v>1.
\prod_{i=1}^{m}Use in source · 1 passage
Search aliases: ∏_i=1^m prod
Use in source · Read passage · main.tex, line 396
ARYFinite positivity can conceal a signed defect Let ν_nonp and ν_p be the restrictions of the undeformed measure ν_p,0 to indices not divisible by p and divisible by p, respectively. Then ν_p,λ=ν_nonp+(1+λ)ν_p. Let A_m,B_m be the order-m Hankel matrices of these two unscaled positive measures. Both are positive definite. Thus τ_m(λ)=det(A_m+(1+λ)B_m) =det A_m∏_i=1^m[1+(1+λ)η_i], where η_i>0 are the eigenvalues of A_m^-1/2B_mA_m^-1/2. Every determinant root is real and below -1; possible coeffic…
Exact mathematical passage
\tau_m(\lambda)=\det(A_m+(1+\lambda)B_m) =\det A_m\prod_{i=1}^{m}[1+(1+\lambda)\eta_i],
\prod_jUse in source · 3 passages
Search aliases: ∏_j prod
Use in source · Read passage · main.tex, line 496
bit share one positive coordinate w=z^4. The plot shows the first six orbits. Multiplicity is retained by the definition; simplicity is not assumed in general. The bound |d_N(w)|≤ e^|w|^1/N and the finite-order zero-counting theorem give absolute local uniform convergence in the stated half-plane. Since the order is below one and d_N(0)=1, the Hadamard factorization has genus zero: d_N(w)=∏_j(1-w/w_j). At positive integers, a log derivative gives -wfracd_N'(w)d_N(w)=∑_k≥1 Z_N(k)w^k. If a_n=(-1)^…
Exact mathematical passage
d_N(w)=\prod_j(1-w/w_j).
\prod_j^{\rm reg}Use in source · 1 passage
Search aliases: ∏_j^rm reg prod rm
Use in source · Read passage · spectral_continuation.tex, line 103
The negative-integer values remain zero, and Z_N'(0;a)=-log[N E_N(a)], ∏_j^rm reg(w_j+a)=N E_N,1(a). The finite part -log N and normalization -log E_N(a) give a zeta-regularized product, not an ordinary product. At N=2 it is 2coshsqrt a, the classical half-integer spectral determinant reproduced by the four shifted benchmarks. Functional equations and derivative-weighted zero sums remain open here.
Exact mathematical passage
\mathcal Z_N'(0;a)=-\log[N E_N(a)],\qquad \boxed{\prod_j^{\rm reg}(w_j+a)=N E_{N,1}(a).}
\prod_{j\ne i}Use in source · 2 passages
Search aliases: ∏_j≠ i prod ne
Use in source · Read passage · resolution_full.tex, line 399
ℓ_i(x)=∏_j≠ ifracx-y_jy_i-y_j, b_i=∏_j≠ ifracmax(y_j,1-y_j)|y_i-y_j|, K_d=∑_i=0^dn_i^4b_i^2.
Exact mathematical passage
\ell_i(x)=\prod_{j\ne i}\frac{x-y_j}{y_i-y_j}, \quad b_i=\prod_{j\ne i}\frac{\max(y_j,1-y_j)}{|y_i-y_j|}, \quad K_d=\sum_{i=0}^{d}n_i^4b_i^2.
\prod_{n=1}^{a-1}Use in source · 2 passages
Search aliases: ∏_n=1^a-1 prod
Use in source · Read passage · multi_prime.tex, line 68
pageheadingPRIME SIEVES / CONSTRUCTIVE WITNESSEvery negative prime atom has a finite detector For an integer a≥2, let L_a(x)=∏_n=1^a-1fracx-x_nx_a-x_n, C_a=L_a(1), r_a=frac1-x_a+11-x_a<1, f_a,m(x)=L_a(x)(frac1-x1-x_a)^m. This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects …
Exact mathematical passage
L_a(x)=\prod_{n=1}^{a-1}\frac{x-x_n}{x_a-x_n},\quad C_a=L_a(1),\quad r_a=\frac{1-x_{a+1}}{1-x_a}<1,
\prod_{n\in S}Use in source · 3 passages
Search aliases: ∏_n∈ S prod in
Use in source · Read passage · resolution_full.tex, line 452
P_S,m(x)=∏_n∈ Sfracx-x_nx_2-x_n (frac1-x1-x_2)^m, C_S=∏_n∈ Sfrac1-x_n|x_2-x_n|, r_S=frac1-x_n_+1-x_2<1.
Exact mathematical passage
P_{S,m}(x)=\prod_{n\in S}\frac{x-x_n}{x_2-x_n} \left(\frac{1-x}{1-x_2}\right)^m,
\prod_{p\mid Q}Use in source · 7 passages
Search aliases: ∏_pmid Q prod mid
Use in source · Read passage · main.tex, line 238
(n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At depth zero the corresponding mass-one atom is absent. In particular, A(0)=Z_Q(2a)-1, where Z_Q(v)=∑_substackn≥1 (n,Q)=1μ(n)^2n^-v =fracζ(v)ζ(2v)∏_pmid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit wor…
Exact mathematical passage
Z_Q(v)=\sum_{\substack{n\ge1\\(n,Q)=1}}\mu(n)^2n^{-v} =\frac{\zeta(v)}{\zeta(2v)}\prod_{p\mid Q}(1+p^{-v})^{-1},\qquad v>1.
\prod_{p\in F}Use in source · 6 passages
Search aliases: ∏_p∈ F prod in
Use in source · Read passage · notation.tex, line 45
item[Prime-sieve measure.] ν_F,λ is the arithmetic atomic measure obtained by multiplying the mass at x_n=(1-n^-2)^2 by ∏_p∈ F(1+λ_p1_pmid n). It is positive exactly when every real strength satisfies λ_p≥-1. item[Hankel order and polynomial degree.] For moments M_j=∫ x^jddν, the order-m matrix is H_m=[M_i+j]_0≤ i,j<m. It tests quadratic forms of real polynomials of degree at most m-1. At depth k, the matrix is [M_k+i+j]_0≤ i,j<m. An order-five test therefore covers degree four, not degree five.…
Exact mathematical passage
\prod_{p\in F}(1+\lambda_p\mathbf1_{p\mid n})
\prod_{p\in S}Use in source · 3 passages
Search aliases: ∏_p∈ S prod in
Use in source · Read passage · sieve_full.tex, line 39
Equivalently, expansion in subsets S⊆ F gives the coefficient D(s)/∏_p∈ S(p^s+1) for ∏_p∈ Sλ_p. This is an independently useful coordinate formula for prime overlap.
Exact mathematical passage
D(s)/\prod_{p\in S}(p^s+1)
\prod_{q\in F}Use in source · 2 passages
Search aliases: ∏_q∈ F prod in
Use in source · Read passage · multi_prime.tex, line 80
x_a<1, f_a,m(x)=L_a(x)(frac1-x1-x_a)^m. This polynomial kills every lower node, equals one at x_a, and has absolute value at most C_ar_a^m at every higher node. fig13_signed_detectorA rational detector for p=3, λ_3=-1.1, λ_2=1. The first exact bound proving negativity is m=5, so a Hankel matrix of order eight detects the signed defect. Set C_F=∏_q∈ Fmax(1,|1+λ_q|). The summable tail gives ∫ f_p,m^2ddν ≤frac1+λ_pp^4+ fracC_FC_p^23p^3r_p^2m. Hence the quadratic form is negative whenever pC_FC_p^2r…
Exact mathematical passage
C_F=\prod_{q\in F}\max(1,|1+\lambda_q|)
\prod_{q\in F\setminus\{p\}}Use in source · 1 passage
Search aliases: ∏_q∈ F∖{p} prod in setminus
Use in source · Read passage · sieve_full.tex, line 187
Fix p∈ F, with all other strengths at least -1, and write κ(n)=∏_q∈ F∖{p}(1+λ_q1_qmid n). Separate the positive measure into its p-free and p-divisible parts:
Exact mathematical passage
\kappa(n)=\prod_{q\in F\setminus\{p\}}(1+\lambda_q\mathbf1_{q\mid n})
\prod_{q\ne p}Use in source · 3 passages
Search aliases: ∏_q≠ p prod ne
Use in source · Read passage · multi_prime.tex, line 106
deg f<m frac∫ f^2ddν_p-free ∫ f^2ddν_p-divisible, δ_mdownarrow0. The detector yields, for C_κ=∏_q≠ pmax(1,1+λ_q), δ_p+m≤fracpC_κ C_p^23r_p^2m. The denominator retains the atom p; all lower nodes in the numerator are killed. This extends the prime-two bound to every prime and to simultaneous positive or deletion strengths. It does not assert simple roots or global multivariate real stability.
Exact mathematical passage
C_\kappa=\prod_{q\ne p}\max(1,1+\lambda_q)
\prod_{\substack{p\ \mathrm{prime}\\
p\notin\{n_j:j\ge1\}}}Use in source · 1 passage
Search aliases: ∏_substackp prime p∉{n_j:j≥1} prod substack mathrm notin ge
Use in source · Read passage · arithmetic_full.tex, line 168
displaystyle rad(Q)=∏_substackp prime p∉{n_j:j≥1}p.
Exact mathematical passage
\boxed{\displaystyle \operatorname{rad}(Q)=\prod_{\substack{p\ \mathrm{prime}\\ p\notin\{n_j:j\ge1\}}}p.}
\sum_{(d,Q)=1}Use in source · 1 passage
Search aliases: ∑_(d,Q)=1 sum
Use in source · Read passage · arithmetic_full.tex, line 201
2=∑_d^2mid nμ(d). For (d,Q)=1, the number of m≤ x/d^2 coprime to Q differs from (φ(Q)/Q)x/d^2 by at most 2^ω(Q). Summing over at most sqrt x values of d gives the first part of C_Q. The omitted main-term tail is at most (φ(Q)/Q)x/lfloorsqrt xrfloor≤2(φ(Q)/Q)sqrt x. Removing n=1 adds one. The main-term density is (φ(Q)/Q)∑_(d,Q)=1μ(d)d^-2=δ_Q by its absolutely convergent Euler product.
Exact mathematical passage
(\varphi(Q)/Q)\sum_{(d,Q)=1}\mu(d)d^{-2}=\delta_Q
\sum_{1\le n<p}Use in source · 1 passage
Search aliases: ∑_1≤ n<p sum le
Use in source · Read passage · sieve_full.tex, line 239
A_rm tail=D(4)fracp^4p^4+1 ∏_q≠ p(1+fracλ_qq^4+1) -∑_1≤ n<pμ(n)^2n^-4κ(n)>0,
Exact mathematical passage
A_{\rm tail}=D(4)\frac{p^4}{p^4+1} \prod_{q\ne p}\left(1+\frac{\lambda_q}{q^4+1}\right) -\sum_{1\le n<p}\mu(n)^2n^{-4}\kappa(n)>0,
\sum_{d^2\mid n}Use in source · 1 passage
Search aliases: ∑_d^2mid n sum mid
Use in source · Read passage · arithmetic_full.tex, line 194
To verify the constants, expand μ(n)^2=∑_d^2mid nμ(d). For (d,Q)=1, the number of m≤ x/d^2 coprime to Q differs from (φ(Q)/Q)x/d^2 by at most 2^ω(Q). Summing over at most sqrt x values of d gives the first part of C_Q. The omitted main-term tail is at most (φ(Q)/Q)x/lfloorsqrt xrfloor≤2(φ(Q)/Q)sqrt x. Removing n=1 adds one. The main-term density is (φ(Q)/Q)∑_(d,Q)=1μ(d)d^-2=δ_Q by its absolutely convergent Euler product.
Exact mathematical passage
\mu(n)^2=\sum_{d^2\mid n}\mu(d)
\sum_{d\le\sqrt x}Use in source · 1 passage
Search aliases: ∑_d≤sqrt x sum le sqrt
Use in source · Read passage · arithmetic_full.tex, line 456
∑_n≤ xμ(n)^2η(n) =∑_d≤sqrt xμ(d)η(d)^2 ∑_m≤ x/d^2η(m)=O_Q(sqrt x).
Exact mathematical passage
\sum_{n\le x}\mu(n)^2\eta(n) =\sum_{d\le\sqrt x}\mu(d)\eta(d)^2 \sum_{m\le x/d^2}\eta(m)=O_Q(\sqrt x).
\sum_{i+j=n-1}Use in source · 3 passages
Search aliases: ∑_i+j=n-1 sum
Use in source · Read passage · coupled_lab_full.tex, line 25
A_n=∑_i+j=n-1binom2n2i+1q^2i+1r^2j+1A_iB_j, B_n=∑_i+j=n-1binom2n2i+1a^2i+1b^2j+1A_iB_j,
Exact mathematical passage
A_n=\sum_{i+j=n-1}\binom{2n}{2i+1}q^{2i+1}r^{2j+1}A_iB_j, \qquad B_n=\sum_{i+j=n-1}\binom{2n}{2i+1}a^{2i+1}b^{2j+1}A_iB_j,
\sum_{i=0}^{4}Use in source · 2 passages
Search aliases: ∑_i=0^4 sum
Use in source · Read passage · resolution_visual.tex, line 198
claimInterpolation controls the whole polynomial space Choose the odd squarefree labels n_i=1,3,5,7,11, write y_i=x_n_i, and put b_i=∏_j≠ ifracmax(y_j,1-y_j)|y_i-y_j|, K_4=∑_i=0^4n_i^4b_i^2. Lagrange interpolation and weighted Cauchy--Schwarz show that, for every polynomial f of degree at most four, Q_δ(f)≥ (1-fracδ K_412) ∑_i=0^4n_i^-4f(y_i)^2. Exact rational comparison gives 12/K_4>7.9790×10^-11>10^-12. A nonzero polynomial cannot vanish at all five selected nodes, so the right side is strictl…
Exact mathematical passage
b_i=\prod_{j\ne i}\frac{\max(y_j,1-y_j)}{|y_i-y_j|}, \qquad K_4=\sum_{i=0}^{4}n_i^4b_i^2.
\sum_{i=0}^{d}Use in source · 3 passages
Search aliases: ∑_i=0^d sum
Use in source · Read passage · resolution_full.tex, line 399
ℓ_i(x)=∏_j≠ ifracx-y_jy_i-y_j, b_i=∏_j≠ ifracmax(y_j,1-y_j)|y_i-y_j|, K_d=∑_i=0^dn_i^4b_i^2.
Exact mathematical passage
\ell_i(x)=\prod_{j\ne i}\frac{x-y_j}{y_i-y_j}, \quad b_i=\prod_{j\ne i}\frac{\max(y_j,1-y_j)}{|y_i-y_j|}, \quad K_d=\sum_{i=0}^{d}n_i^4b_i^2.
\sum_{i=0}^{n}Use in source · 1 passage
Search aliases: ∑_i=0^n sum
Use in source · REPORT.md, line 13
u_n+1=frac∑_i=0^nq^2i+1r^2(n-i)+1u_i v_n-i2n+3, v_n+1=frac∑_i=0^na^2i+1b^2(n-i)+1u_i v_n-i2n+3.
Exact mathematical passage
u_{n+1}=\frac{\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\qquad v_{n+1}=\frac{\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.
\sum_{i=0}^{n-1}Use in source · 5 passages
Search aliases: ∑_i=0^n-1 sum
Use in source · Read passage · main.tex, line 90
on is symmetric in q,r, but unequal scales retain left/right address information. fig01_asymmetric_functionsComputed Taylor curves. Each entire curve uses terms through z^321; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent. For T_q,r(z)=∑_n≥0c_nz^2n+1, c_0=1, coefficient matching gives c_n=frac12n+1∑_i=0^n-1 q^2i+1r^2(n-1-i)+1c_ic_n-1-i, n≥1. In particular, T_q,r(z)=z+fracqr3z^3+ fracq^2r^2(q^2+r^2)15z^5+O(z^7).…
Exact mathematical passage
c_n=\frac{1}{2n+1}\sum_{i=0}^{n-1} q^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\qquad n\ge1.
\sum_{j<m}Use in source · 1 passage
Search aliases: ∑_j<m sum
Use in source · Read passage · sieve_full.tex, line 320
If at least two strengths in a positive multiplicative family are nonzero, its entire depth-zero monic polynomial sequence cannot equal any one-prime sequence on the same variable and node system. A direct proof avoids importing a determinacy theorem. Write P_m=x^m+∑_j<mc_mjx^j. The equations ∫ P_m dν=0 recursively determine every normalized moment from the monic sequence. Equal sequences therefore give equal normalized Stieltjes series for |z|>1. Analytic continuation on the connected complemen…
Exact mathematical passage
P_m=x^m+\sum_{j<m}c_{mj}x^j
\sum_{j=0}^{b-1}Use in source · 4 passages
Search aliases: ∑_j=0^b-1 sum
Use in source · Read passage · sequence_full.tex, line 279
∑_j=0^b-1 P_b,j=I, P_b,j P_b,l=δ_jl P_b,j.
Exact mathematical passage
\sum_{j=0}^{b-1}\mathcal P_{b,j}=I,\qquad \mathcal P_{b,j}\mathcal P_{b,l}=\delta_{jl}\mathcal P_{b,j}.
\sum_{j=0}^{m}Use in source · 1 passage
Search aliases: ∑_j=0^m sum
Use in source · Read passage · main.tex, line 444
claimAn endpoint mass is numerically delicate For the undeformed depth-zero measure, set ν_θ=ν+θδ_0, θ≥-1. Write h_j=∫ P_j(x)^2 dν(x) and K_m(x,y)=∑_j=0^mP_j(x)P_j(y)/h_j. The Uvarov formula~[source reference] is P_m^[θ](x)=P_m(x)- fracθ P_m(0)1+θ K_m-1(0,0)K_m-1(x,0), with K_-1=0 for degree zero. The mass at zero is recovered by 1≤frac1K_m(0,0)≤1+(M_0-1)(7/16)^2m. The lower bound comes from the zero atom, whose mass is one. The test polynomial (1-x)^m gives the upper bound since every other nod…
Exact mathematical passage
K_m(x,y)=\sum_{j=0}^{m}P_j(x)P_j(y)/h_j
\sum_{j=0}^{N-1}Use in source · 2 passages
Search aliases: ∑_j=0^N-1 sum
Use in source · Read passage · spectral_continuation.tex, line 24
pageheadingORBIT SPECTRA / LARGE RADIUSA finite exponential sum controls the tail Put E_N(t)=d_N(-t)=∑_n≥0t^n/(Nn)! and ω=e^2π i/N. For x≥0, E_N(x^N)=frac1N∑_j=0^N-1e^ω^jx. Root filtering proves this identity term by term: only powers divisible by N remain. The positive real exponential is dominant, so log E_N(t)=t^1/N-log N+O(e^-κ_N t^1/N), κ_N=1-cos(2π/N)>0, with corresponding differentiated estimates on the positive ray. fig14_spectral_tailComputed logarithmic derivatives of the finite expone…
Exact mathematical passage
E_N(x^N)=\frac1N\sum_{j=0}^{N-1}e^{\omega^jx}.
\sum_{j=0}^{r}Use in source · 1 passage
Search aliases: ∑_j=0^r sum
Use in source · Read passage · main.tex, line 282
claimNo polynomial-coefficient linear ODE Suppose ∑_j=0^rP_j(s)A^(j)(s)=0, where the P_j are polynomials and are not all zero. At s=-t, divide by the exponential of the first support atom. The remaining tail is exponentially smaller, so the polynomial ∑_jP_j(-t)(-λ_p_*)^j tends to zero exponentially and must vanish identically. Remove that atom and repeat. Every support frequency therefore satisfies ∑_jP_j(s)(-λ_n)^j=0 for all s. For fixed s, this polynomial in λ has infinitely many distinct roo…
Exact mathematical passage
\sum_{j=0}^{r}P_j(s)A^{(j)}(s)=0
\sum_{j=1}^{J}Use in source · 1 passage
Search aliases: ∑_j=1^J sum
Use in source · Read passage · dilation_full.tex, line 13
F'(z)=1+∑_j=1^Jh_jF(q_jz)F(r_jz), F(0)=0,
Exact mathematical passage
F'(z)=1+\sum_{j=1}^{J}h_jF(q_jz)F(r_jz),\qquad F(0)=0,
\sum_{j=1}^{k-1}Use in source · 1 passage
Search aliases: ∑_j=1^k-1 sum
Use in source · Read passage · main.tex, line 504
1/N and the finite-order zero-counting theorem give absolute local uniform convergence in the stated half-plane. Since the order is below one and d_N(0)=1, the Hadamard factorization has genus zero: d_N(w)=∏_j(1-w/w_j). At positive integers, a log derivative gives -wfracd_N'(w)d_N(w)=∑_k≥1 Z_N(k)w^k. If a_n=(-1)^n/(Nn)! and b_k= Z_N(k), then b_k=-ka_k-∑_j=1^k-1a_jb_k-j. Thus every positive integer value is rational without numerical zero finding.
Exact mathematical passage
b_k=-ka_k-\sum_{j=1}^{k-1}a_jb_{k-j}.
\sum_{j=1}^{m-1}Use in source · 1 passage
Search aliases: ∑_j=1^m-1 sum
Use in source · Read passage · spectral_full.tex, line 295
b_m:= Z_N(m) =-m c_m-∑_j=1^m-1c_jb_m-j. tagS17
Exact mathematical passage
b_m:=\mathcal Z_N(m) =-m c_m-\sum_{j=1}^{m-1}c_jb_{m-j}. \tag{S17}
\sum_{j=1}^{N-1}Use in source · 1 passage
Search aliases: ∑_j=1^N-1 sum
Use in source · Read passage · spectral_full.tex, line 70
h_N(x)=∑_j=1^N-1e^(ω^j-1)x, c_N=1-cos(2π/N)>0.
Exact mathematical passage
h_N(x)=\sum_{j=1}^{N-1}e^{(\omega^j-1)x}, \qquad c_N=1-\cos(2\pi/N)>0.
\sum_{j\ge0}Use in source · 3 passages
Search aliases: ∑_j≥0 sum ge
Use in source · Read passage · main.tex, line 271
pageheadingFAMILY 2 / CALCULUSClosed shifts, infinite differential complexity The Euler product for Z_Q shows that only rad(Q)=∏_pmid Qp matters. Expansion of the depth factor gives A_a,β,Q(s+1)=A_a,β,Q(s)-2A_a+β/2,β,Q(s) +A_a+β,β,Q(s). The fractional transfer and Newton formulas are A_a,β,Q(s+t)=∑_j≥0(-1)^jbinom2tjA_a+jβ/2,β,Q(s), A_a,β,Q(t)=∑_j≥0(-1)^jbinom2tj[Z_Q(2a+jβ)-1]. They converge absolutely and locally uniformly: binomial coefficients have at most polynomial growth on compact paramete…
Exact mathematical passage
\A(s+t)=\sum_{j\ge0}(-1)^j\binom{2t}{j}A_{a+j\beta/2,\beta,Q}(s),
\sum_{j\ge1}Use in source · 1 passage
Search aliases: ∑_j≥1 sum ge
Use in source · Read passage · arithmetic_full.tex, line 366
(Uc)(s)=∑_j≥1c_jsqrtw_j e^-λ_js.
Exact mathematical passage
(Uc)(s)=\sum_{j\ge1}c_j\sqrt{w_j}\,e^{-\lambda_js}.
\sum_{k=0}^{b-1}Use in source · 3 passages
Search aliases: ∑_k=0^b-1 sum
Use in source · Read passage · sequence_full.tex, line 268
P_b,jG(z)=frac1b∑_k=0^b-1ω^-jkG(ω^kz) =∑_m≡ j (b)a_mfracz^mm!.
Exact mathematical passage
\boxed{ \mathcal P_{b,j}G(z)=\frac1b\sum_{k=0}^{b-1}\omega^{-jk}G(\omega^kz) =\sum_{m\equiv j\,(b)}a_m\frac{z^m}{m!}.}
\sum_{k=0}^{K}Use in source · 1 passage
Search aliases: ∑_k=0^K sum
Use in source · Read passage · spectral_full.tex, line 398
G_N,a(t)=∑_k=0^Kαbinomα-1ka^k t^α-k-1 +O(t^α-K-2). tagT5
Exact mathematical passage
G_{N,a}(t)=\sum_{k=0}^{K}\alpha\binom{\alpha-1}{k}a^k t^{\alpha-k-1} +O(t^{\alpha-K-2}). \tag{T5}
\sum_{k=0}^mUse in source · 1 passage
Search aliases: ∑_k=0^m sum
Use in source · Read passage · sequence_polyphase.tex, line 107
claimThree operations that must be kept distinct For an EGF E_a(z)=∑_ma_mz^m/m!, differentiation transports future coefficients: E_a^(k)=∑_ma_m+kz^m/m!. Causal ordinary taps instead act by repeated zero-constant integration: E_b=∑_ℓ h_ℓ I^ℓ E_a. Multiplication of two EGFs gives Hurwitz convolution, b_m=∑_k=0^mbinom mk a_kh_m-k. It is not ordinary sequence FIR convolution. Intrinsic nonlinear feedback taps alter the defining family and are a fourth operation with their own coefficient recurrence.
Exact mathematical passage
b_m=\sum_{k=0}^m\binom mk a_kh_{m-k}
\sum_{k=0}^{n}Use in source · 1 passage
Search aliases: ∑_k=0^n sum
Use in source · chapters_sequences.tex, line 88
lessonSEQUENCES / 13Apéry uses taps that depend on the indexS13_apery_recurrence (n+1)^3A_n+1=(2n+1)(17n^2+17n+5)A_n-n^3A_n-1 The prefix annotation begins the Apéry sequence defined by A_n=∑_k=0^nbinom nk^2binomn+kk^2. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at n=2. Using A_1=5 and A_2=73 gives 27A_3=535·73-8·5=39015. Therefore A_3=1445. The recurrence applies for n≥1, with A_0=1. This is a three-term recurrenc…
Exact mathematical passage
A_n=\sum_{k=0}^{n}\binom nk^2\binom{n+k}{k}^2
\sum_{k=1}^KUse in source · 1 passage
Search aliases: ∑_k=1^K sum
Use in source · Read passage · spectral_continuation.tex, line 61
Near zero write f(t)=(log E_N)'(t)=∑_k≥1α_kt^k-1, where α_k=(-1)^k+1 Z_N(k). For any integer K≥1, subtract K terms on [0,1]: I_K(s)= · ∫_0^1t^-s[f(t)-∑_k=1^Kα_kt^k-1]dd t +∑_k=1^Kfracα_kk-s · +∫_1^∞ t^-s[f(t)-frac1N t^1/N-1]dd t +frac1N(s-1/N). The first integral is holomorphic for Re s<K+1. The second is entire because its remainder decays exponentially in t^1/N. The factor sinπ s cancels the positive-integer poles. Increasing K covers the plane, leaving only 1/N. At nonpositive integers I_K is…
Exact mathematical passage
\begin{aligned} I_K(s)={}&\int_0^1t^{-s}\left[f(t)-\sum_{k=1}^K\alpha_kt^{k-1}\right]\dd t +\sum_{k=1}^K\frac{\alpha_k}{k-s}\\ &+\int_1^\infty t^{-s}\left[f(t)-\frac1N t^{1/N-1}\right]\dd t +\frac{1}{N(s-1/N)}. \end{aligned}
\sum_{k>j}Use in source · 3 passages
Search aliases: ∑_k>j sum
Use in source · Read passage · arithmetic_full.tex, line 72
Every limit exists and recovers exactly the j-th support frequency and weight. Indeed, if M_j+1=∑_k>jw_k, then for every t≥0,
Exact mathematical passage
M_{j+1}=\sum_{k>j}w_k
\sum_{k\ge0}Use in source · 2 passages
Search aliases: ∑_k≥0 sum ge
Use in source · Read passage · spectral_continuation.tex, line 94
^-s, Re s>1/N. The rightward shift preserves avoidance of the negative branch cut. Normalize the product with E_N(a) and use log E_N(a+t) in the Mellin argument. For a>0 there are genuine simple poles at s=1/N-k, k≥0: Res_s=1/N-k Z_N(s;a) =fracsin(π/N)Nπfrac(1-1/N)_kk!a^k. Indeed the large-radius derivative expands as frac1N(a+t)^1/N-1 =frac1N∑_k≥0frac(-1)^k(1-1/N)_kk!a^kt^1/N-1-k as an asymptotic series. The alternating sign is canceled by sin(π(1/N-k))=(-1)^ksin(π/N). At a=0 all secondary resi…
Exact mathematical passage
\frac1N(a+t)^{1/N-1} =\frac1N\sum_{k\ge0}\frac{(-1)^k(1-1/N)_k}{k!}a^kt^{1/N-1-k}
\sum_{k\ge1}Use in source · 2 passages
Search aliases: ∑_k≥1 sum ge
Use in source · Read passage · main.tex, line 500
ts. Multiplicity is retained by the definition; simplicity is not assumed in general. The bound |d_N(w)|≤ e^|w|^1/N and the finite-order zero-counting theorem give absolute local uniform convergence in the stated half-plane. Since the order is below one and d_N(0)=1, the Hadamard factorization has genus zero: d_N(w)=∏_j(1-w/w_j). At positive integers, a log derivative gives -wfracd_N'(w)d_N(w)=∑_k≥1 Z_N(k)w^k. If a_n=(-1)^n/(Nn)! and b_k= Z_N(k), then b_k=-ka_k-∑_j=1^k-1a_jb_k-j. Thus every posi…
Exact mathematical passage
-w\frac{d_N'(w)}{d_N(w)}=\sum_{k\ge1}\mathcal Z_N(k)w^k.
\sum_{l=0}^{b-1}Use in source · 1 passage
Search aliases: ∑_l=0^b-1 sum
Use in source · Read passage · sequence_full.tex, line 360
Y_j(w)=∑_l=0^b-1H_jl(w)X_l(w), H_jl(w)=∑_substack0≤ t<L t+l≡ j (b) h_t w^(t+l-j)/b.
Exact mathematical passage
\boxed{Y_j(w)=\sum_{l=0}^{b-1}H_{jl}(w)X_l(w),\qquad H_{jl}(w)=\sum_{\substack{0\le t<L\\t+l\equiv j\,(b)}} h_t w^{(t+l-j)/b}.}
\sum_mUse in source · 4 passages
Search aliases: ∑_m sum
Use in source · Read passage · sequence_full.tex, line 288
For an ordinary generating function X(z)=∑_ma_m z^m, define X_j(w)=∑_na_bn+jw^n. Then
Exact mathematical passage
X(z)=\sum_ma_m z^m
\sum_{m=1}^{K}Use in source · 1 passage
Search aliases: ∑_m=1^K sum
Use in source · Read passage · spectral_full.tex, line 226
B_N,K(s)= · ∫_0^1x^-Ns[g_N(x)-∑_m=1^KN a_N,m-1x^Nm-1]dx · +∑_m=1^Kfraca_N,m-1m-s +∫_1^∞ x^-Ns[g_N(x)-1]dx +frac1Ns-1. tagS13
Exact mathematical passage
\begin{aligned} B_{N,K}(s)={}& \int_0^1x^{-Ns}\left[g_N(x)-\sum_{m=1}^{K}N a_{N,m-1}x^{Nm-1}\right]dx\\ &+\sum_{m=1}^{K}\frac{a_{N,m-1}}{m-s} +\int_1^\infty x^{-Ns}\bigl[g_N(x)-1\bigr]dx +\frac1{Ns-1}. \end{aligned} \tag{S13}
\sum_{m\equiv j\,(b)}Use in source · 1 passage
Search aliases: ∑_m≡ j (b) sum equiv
Use in source · Read passage · sequence_full.tex, line 268
P_b,jG(z)=frac1b∑_k=0^b-1ω^-jkG(ω^kz) =∑_m≡ j (b)a_mfracz^mm!.
Exact mathematical passage
\boxed{ \mathcal P_{b,j}G(z)=\frac1b\sum_{k=0}^{b-1}\omega^{-jk}G(\omega^kz) =\sum_{m\equiv j\,(b)}a_m\frac{z^m}{m!}.}
\sum_{m\le x/d^2}Use in source · 1 passage
Search aliases: ∑_m≤ x/d^2 sum le
Use in source · Read passage · arithmetic_full.tex, line 456
∑_n≤ xμ(n)^2η(n) =∑_d≤sqrt xμ(d)η(d)^2 ∑_m≤ x/d^2η(m)=O_Q(sqrt x).
Exact mathematical passage
\sum_{n\le x}\mu(n)^2\eta(n) =\sum_{d\le\sqrt x}\mu(d)\eta(d)^2 \sum_{m\le x/d^2}\eta(m)=O_Q(\sqrt x).
\sum_{m\le y}Use in source · 1 passage
Search aliases: ∑_m≤ y sum le
Use in source · Read passage · arithmetic_full.tex, line 453
Proof. Distinct characters give a nonprincipal η. Its sum over a complete period is zero, so |∑_m≤ yη(m)|≤ Q. Hence
Exact mathematical passage
|\sum_{m\le y}\eta(m)|\le Q
\sum_{m\ge t}Use in source · 1 passage
Search aliases: ∑_m≥ t sum ge
Use in source · Read passage · sequence_full.tex, line 327
Proof. The EGF of the delayed sequence is ∑_m≥ ta_m-tz^m/m!=I^tG(z). Summing taps gives the formula. Conversely, G^(t) has coefficient sequence a_m+t, so derivatives sample future coefficients and do not implement these past delays. square
Exact mathematical passage
\sum_{m\ge t}a_{m-t}z^m/m!=I^tG(z)
\sum_{m\ge0}Use in source · 2 passages
Search aliases: ∑_m≥0 sum ge
Use in source · Read passage · sequence_full.tex, line 265
For an EGF G(z)=∑_m≥0a_m z^m/m!, an integer base b≥1, a residue 0≤ j<b, and ω=e^2π i/b, define
Exact mathematical passage
G(z)=\sum_{m\ge0}a_m z^m/m!
\sum_{m\ge1}Use in source · 1 passage
Search aliases: ∑_m≥1 sum ge
Use in source · Read passage · spectral_full.tex, line 218
G_N(t)=∑_k≥0a_N,kt^k, g_N(x)=∑_m≥1N a_N,m-1x^Nm-1. tagS12
Exact mathematical passage
G_N(t)=\sum_{k\ge0}a_{N,k}t^k, \qquad g_N(x)=\sum_{m\ge1}N a_{N,m-1}x^{Nm-1}. \tag{S12}
\sum_nUse in source · 8 passages
Search aliases: ∑_n sum
Use in source · Read passage · notation.tex, line 64
item[Coefficient conventions.] An ordinary generating function uses ∑_na_nz^n. An exponential generating function uses ∑_na_nz^n/n!. Their product and shift rules differ. item[Polyphase section and finite taps.] Polyphase splits a coefficient sequence by residue class. Finite impulse response (FIR) taps act by finite ordinary convolution. A phase matrix must retain the carry delay when a tap crosses a residue-class boundary. This is not the Hurwitz product of exponential generating functions. it…
Exact mathematical passage
\sum_na_nz^n
\sum_{n\text{ even}}Use in source · 1 passage
Search aliases: ∑_n even sum text
Use in source · Read passage · resolution_full.tex, line 418
The total even mass satisfies ∑_n evenμ(n)^2n^-4<frac116(1+frac13)=1/12. Keeping the selected nodes in the positive part yields the explicit bound
Exact mathematical passage
\sum_{n\text{ even}}\mu(n)^2n^{-4}<\frac1{16}(1+\frac13)=1/12
\sum_{n=0}^{\infty}Use in source · 2 passages
Search aliases: ∑_n=0^∞ sum infty
Use in source · Read passage · spectral_full.tex, line 13
d_N(w)=∑_n=0^∞frac(-1)^n w^n(Nn)! =E_N,1(-w).
Exact mathematical passage
d_N(w)=\sum_{n=0}^{\infty}\frac{(-1)^n w^n}{(Nn)!} =E_{N,1}(-w).
\sum_{n=1}^{N}Use in source · 1 passage
Search aliases: ∑_n=1^N sum
Use in source · Read passage · resolution_full.tex, line 101
0≤ L(R)-∑_n=1^Nn d_nR^2n+1 ≤12(N+1)t^2N+3. tagRA-2
Exact mathematical passage
0\le L(R)-\sum_{n=1}^{N}n d_nR^{2n+1} \le12(N+1)t^{2N+3}. \tag{RA-2}
\sum_{n>a}Use in source · 1 passage
Search aliases: ∑_n>a sum
Use in source · Read passage · sieve_full.tex, line 133
∑_n>a|μ(n)^2n^-4κ_F(n)| ≤fracC_F3a^3. tag3
Exact mathematical passage
\sum_{n>a}|\mu(n)^2n^{-4}\kappa_F(n)| \le\frac{C_F}{3a^3}. \tag{3}
\sum_{n>K}Use in source · 1 passage
Search aliases: ∑_n>K sum
Use in source · Read passage · resolution_full.tex, line 465
This polynomial has degree |S|+m, equals one at x_2, and kills every positive atom through the cutoff. At each remaining positive node, the product of the absolute root factors is at most C_S, and the last factor is at most r_S^m. Dropping all other negative atoms and bounding the positive tail by the integral ∑_n>Kn^-4≤1/(3K^3) gives
Exact mathematical passage
\sum_{n>K}n^{-4}\le1/(3K^3)
\sum_{n>L}Use in source · 1 passage
Search aliases: ∑_n>L sum
Use in source · Read passage · arithmetic_full.tex, line 274
Indeed all omitted frequencies are at most λ_L+1, and ∑_n>Ln^-2a≤∫_L^∞ x^-2a dx. Ignoring squarefreeness and coprimality makes this a valid upper bound. If r_0,r_1 are the two positive truncated residual sums at t,t+h, the true secant belongs to
Exact mathematical passage
\sum_{n>L}n^{-2a}\le\int_L^\infty x^{-2a}\,dx
\sum_{n\in\mathcal S_Q}Use in source · 1 passage
Search aliases: ∑_n∈ S_Q sum in mathcal
Use in source · Read passage · main.tex, line 248
mid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit working precision. Right: initial squarefree frequencies and weights for Q=1. Only finitely many atoms are shown; the proof uses the infinite support. Let S_Q={n≥2:μ(n)^2=1, (n,Q)=1} and λ_n=-2log(1-n^-β). Then A(s)=∑_n∈ S_Qw_ne^-λ_ns, w_n=n^-2a. The positive weights are summable, and the distinct support frequencies decrease to zero. If p_* is the least prime not dividin…
Exact mathematical passage
A(s)=\sum_{n\in\mathcal S_Q}w_ne^{-\lambda_ns},\qquad w_n=n^{-2a}.
\sum_{n\le x}Use in source · 1 passage
Search aliases: ∑_n≤ x sum le
Use in source · Read passage · arithmetic_full.tex, line 456
∑_n≤ xμ(n)^2η(n) =∑_d≤sqrt xμ(d)η(d)^2 ∑_m≤ x/d^2η(m)=O_Q(sqrt x).
Exact mathematical passage
\sum_{n\le x}\mu(n)^2\eta(n) =\sum_{d\le\sqrt x}\mu(d)\eta(d)^2 \sum_{m\le x/d^2}\eta(m)=O_Q(\sqrt x).
\sum_{n\ge N}Use in source · 1 passage
Search aliases: ∑_n≥ N sum ge
Use in source · REPORT.md, line 28
|∑_n≥ Nc_nz^2n+1| ≤ frac|z|ρ^N1-ρ, |z|^2<2.
Exact mathematical passage
\left|\sum_{n\ge N}c_nz^{2n+1}\right| \le \frac{|z|\rho^N}{1-\rho},\qquad |z|^2<2.
\sum_{n\ge1}Use in source · 8 passages
Search aliases: ∑_n≥1 sum ge
Use in source · Read passage · main.tex, line 360
Prime sieves and signed measures. pageheadingFAMILY 3An infinite prime shelf changes the packets For a prime p and λ≥-1, let x_n=(1-n^-2)^2, ν_p,λ=∑_n≥1μ(n)^2n^-4 (1+λ1_pmid n)δ_x_n. Define monic orthogonal polynomials P_m^(k;p,λ) for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the or…
Exact mathematical passage
x_n=(1-n^{-2})^2,\qquad \nu_{p,\lambda}=\sum_{n\ge1}\mu(n)^2n^{-4} (1+\lambda\mathbf1_{p\mid n})\delta_{x_n}.
\sum_{n\ge2}Use in source · 1 passage
Search aliases: ∑_n≥2 sum ge
Use in source · Read passage · arithmetic_full.tex, line 415
A_χ,ψ(s)=∑_n≥2μ(n)^2 η(n)n^-2a(1-n^-β)^2s.
Exact mathematical passage
A_{\chi,\psi}(s)=\sum_{n\ge2}\mu(n)^2 \eta(n)n^{-2a}(1-n^{-\beta})^{2s}.
\sum_{n\ge2,\ (n,Q)=1}Use in source · 1 passage
Search aliases: ∑_n≥2, (n,Q)=1 sum ge
Use in source · Read passage · notation.tex, line 31
item[Arithmetic depth A_a,β,Q.] The entire sum ∑_n≥2, (n,Q)=1μ(n)^2n^-2a(1-n^-β)^2s, with a>1/2, β>0, Q≥1 integer and the real logarithm in the complex power. The atom n=1 is excluded at every depth. item[Möbius function and modulus radical.] μ(1)=1. For a product of k distinct primes, μ(n)=(-1)^k. If a prime square divides n, then μ(n)=0. The radical rad(Q)=∏_pmid Qp records the excluded prime set, with rad(1)=1. item[Squarefree Dirichlet sum.] For v>1, Z_Q(v)=ζ(v)ζ(2v)^-1∏_pmid Q(1+p^-v)^-1. T…
Exact mathematical passage
\sum_{n\ge2,\ (n,Q)=1}\mu(n)^2n^{-2a}(1-n^{-\beta})^{2s}
\sum_{p\mid n}Use in source · 1 passage
Search aliases: ∑_pmid n sum mid
Use in source · Read passage · sieve_full.tex, line 191
A_m=[∑_pnmid nμ(n)^2n^-4κ(n)x_n^i+j]_i,j<m, B_m=[∑_pmid nμ(n)^2n^-4κ(n)x_n^i+j]_i,j<m.
Exact mathematical passage
A_m=[\sum_{p\nmid n}\mu(n)^2n^{-4}\kappa(n)x_n^{i+j}]_{i,j<m}, \quad B_m=[\sum_{p\mid n}\mu(n)^2n^{-4}\kappa(n)x_n^{i+j}]_{i,j<m}.
\sum_{p\nmid n}Use in source · 1 passage
Search aliases: ∑_pnmid n sum nmid
Use in source · Read passage · sieve_full.tex, line 191
A_m=[∑_pnmid nμ(n)^2n^-4κ(n)x_n^i+j]_i,j<m, B_m=[∑_pmid nμ(n)^2n^-4κ(n)x_n^i+j]_i,j<m.
Exact mathematical passage
A_m=[\sum_{p\nmid n}\mu(n)^2n^{-4}\kappa(n)x_n^{i+j}]_{i,j<m}, \quad B_m=[\sum_{p\mid n}\mu(n)^2n^{-4}\kappa(n)x_n^{i+j}]_{i,j<m}.
\sum_{pq\mid n}Use in source · 1 passage
Search aliases: ∑_pqmid n sum mid
Use in source · Read passage · sieve_full.tex, line 67
where U_pq,j=∑_pqmid nμ(n)^2n^-4x_n^j>0. More generally, with all other strengths at least -1, every mixed moment derivative in a set of distinct primes is strictly positive. It is the integral over atoms divisible by those primes, with the remaining nonnegative prime factors. Infinitely many such atoms remain after any finite collection of deletions. Pure second derivatives in any one strength vanish because the moments are separately affine.
Exact mathematical passage
U_{pq,j}=\sum_{pq\mid n}\mu(n)^2n^{-4}x_n^j>0
\sum_{p\in F}Use in source · 2 passages
Search aliases: ∑_p∈ F sum in
Use in source · Read passage · multi_prime.tex, line 55
claimA separate additive family has different geometry For weights 1+∑_p∈ Fγ_p1_pmid n, positivity is equivalent to 1+∑_γ_p<0γ_p≥0. Every prime subset appears at its squarefree product, proving necessity. Checking each atom proves sufficiency. Its finite Hankel matrices are affine in γ, so the semidefinite feasible sets are convex spectrahedra. The multiplicative family has genuine overlap terms and polynomial parameter dependence. These are two distinct deformations, with distinct domains.
Exact mathematical passage
1+\sum_{p\in F}\gamma_p\mathbf1_{p\mid n}
\sum_{p\in S}Use in source · 1 passage
Search aliases: ∑_p∈ S sum in
Use in source · Read passage · sieve_full.tex, line 296
1+∑_p∈ Sγ_p≥0 (S⊆ F), equivalently 1+∑_γ_p<0γ_p≥0.
Exact mathematical passage
1+\sum_{p\in S}\gamma_p\ge0\quad(S\subseteq F), \quad\text{equivalently}\quad 1+\sum_{\gamma_p<0}\gamma_p\ge0.
\sum_{r=0}^{2j}Use in source · 3 passages
Search aliases: ∑_r=0^2j sum
Use in source · Read passage · main.tex, line 372
for the measure x^k dν_p,λ, k≥0 integer. The zero atom has mass one at k=0 and is annihilated when k≥1. fig04_prime_polynomialsLeft: degree-two packets for p=2, depth zero. Right: initial atom masses. Strength -1 deletes every even squarefree atom; strength 0 is the original measure; strength 1 doubles that shelf. The exact moments are finite zeta stencils: M_j^p,λ=∑_r=0^2j(-1)^rbinom2jr fracζ(4+2r)ζ(8+4r) (1+fracλp^4+2r+1). All polynomial coefficients belong to Q(π^-2,λ). With Δ=M_0M_2-M_1^2, P…
Exact mathematical passage
M_j^{p,\lambda}=\sum_{r=0}^{2j}(-1)^r\binom{2j}{r} \frac{\zeta(4+2r)}{\zeta(8+4r)} \left(1+\frac{\lambda}{p^{4+2r}+1}\right).
\sum_{\substack{0\le t<L\\t+l\equiv j\,(b)}}Use in source · 1 passage
Search aliases: ∑_substack0≤ t<L t+l≡ j (b) sum substack le t equiv
Use in source · Read passage · sequence_full.tex, line 360
Y_j(w)=∑_l=0^b-1H_jl(w)X_l(w), H_jl(w)=∑_substack0≤ t<L t+l≡ j (b) h_t w^(t+l-j)/b.
Exact mathematical passage
\boxed{Y_j(w)=\sum_{l=0}^{b-1}H_{jl}(w)X_l(w),\qquad H_{jl}(w)=\sum_{\substack{0\le t<L\\t+l\equiv j\,(b)}} h_t w^{(t+l-j)/b}.}
\sum_{\substack{n\le x\\(n,Q)=1}}Use in source · 1 passage
Search aliases: ∑_substackn≤ x (n,Q)=1 sum substack le
Use in source · Read passage · main.tex, line 328
claimProof mechanism Squarefree inclusion--exclusion gives ∑_substackn≤ x (n,Q)=1μ(n)^2 =δ_Q x+O_Q(sqrtx). Integrate against g_s(x)=x^-2a(1-x^-β)^2s on [1,∞). The density term becomes the beta integral after y=x^-β. Partial integration bounds the error by C_Q∫_1^∞sqrtx|g_s'(x)| d x. The two derivative terms are beta integrals with decay exponent (2a-1/2)/β; |s|/Re s stays bounded in the stated sector. This gives the error term and then the leading asymptotic.
Exact mathematical passage
\sum_{\substack{n\le x\\(n,Q)=1}}\mu(n)^2 =\delta_Q x+O_Q(\sqrt{x}).
\sum_{\substack{n\ge1\\(n,Q)=1}}Use in source · 1 passage
Search aliases: ∑_substackn≥1 (n,Q)=1 sum substack ge
Use in source · Read passage · main.tex, line 238
(n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At depth zero the corresponding mass-one atom is absent. In particular, A(0)=Z_Q(2a)-1, where Z_Q(v)=∑_substackn≥1 (n,Q)=1μ(n)^2n^-v =fracζ(v)ζ(2v)∏_pmid Q(1+p^-v)^-1, v>1. fig03_complex_depthLeft: a=β=2 at real depths, using the exact zeta--Newton representation at 90-digit wor…
Exact mathematical passage
Z_Q(v)=\sum_{\substack{n\ge1\\(n,Q)=1}}\mu(n)^2n^{-v} =\frac{\zeta(v)}{\zeta(2v)}\prod_{p\mid Q}(1+p^{-v})^{-1},\qquad v>1.
\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ even}}}Use in source · 2 passages
Search aliases: ∑_substackn≥1 μ(n)^2=1 n even sum substack ge mu n text
Use in source · Read passage · resolution_full.tex, line 380
Q_δ(f)=∑_substackn≥1 μ(n)^2=1 n oddn^-4f(x_n)^2 -δ∑_substackn≥1 μ(n)^2=1 n evenn^-4f(x_n)^2.
Exact mathematical passage
Q_\delta(f)=\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ odd}}}n^{-4}f(x_n)^2 -\delta\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ even}}}n^{-4}f(x_n)^2.
\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ odd}}}Use in source · 2 passages
Search aliases: ∑_substackn≥1 μ(n)^2=1 n odd sum substack ge mu n text
Use in source · Read passage · resolution_full.tex, line 380
Q_δ(f)=∑_substackn≥1 μ(n)^2=1 n oddn^-4f(x_n)^2 -δ∑_substackn≥1 μ(n)^2=1 n evenn^-4f(x_n)^2.
Exact mathematical passage
Q_\delta(f)=\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ odd}}}n^{-4}f(x_n)^2 -\delta\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ even}}}n^{-4}f(x_n)^2.
\sum_{\substack{n\ge1\\\mu(n)^2=1,(n,Q)=1\\p\mid n}}Use in source · 1 passage
Search aliases: ∑_substackn≥1 μ(n)^2=1,(n,Q)=1 pmid n sum substack ge mu p mid
Use in source · Read passage · arithmetic_full.tex, line 334
∑_substackn≥1 μ(n)^2=1,(n,Q)=1 pmid nn^-2a =fracZ_Q(2a)p^2a+1.
Exact mathematical passage
\sum_{\substack{n\ge1\\\mu(n)^2=1,(n,Q)=1\\p\mid n}}n^{-2a} =\frac{Z_Q(2a)}{p^{2a}+1}.
\sum_{\substack{n\ge2\\(n,Q)=1}}Use in source · 1 passage
Search aliases: ∑_substackn≥2 (n,Q)=1 sum substack ge
Use in source · Read passage · main.tex, line 230
MonographPartIIArithmetic depth and signed measuresComplex parameters, positive kernels and the boundaries of positivity. Arithmetic at complex depth. pageheadingFAMILY 2Arithmetic energy at complex depth For real a>1/2, β>0, and integer Q≥1, define A_a,β,Q(s)=∑_substackn≥2 (n,Q)=1 μ(n)^2n^-2a(1-n^-β)^2s, s∈ C. The complex power uses the real logarithm of 1-n^-β>0. The atom n=1 is excluded because its logarithm is undefined. At positive integer depths its omitted contribution would be zero. At d…
Exact mathematical passage
\boxed{\A(s)=\sum_{\substack{n\ge2\\(n,Q)=1}} \mu(n)^2n^{-2a}(1-n^{-\beta})^{2s},\qquad s\in\mathbb C.}
\sum_{\substack{n\ge2\\ \mu(n)^2=1\\ (n,Q)=1}}Use in source · 1 passage
Search aliases: ∑_substackn≥2 μ(n)^2=1 (n,Q)=1 sum substack ge mu
Use in source · Read passage · arithmetic_full.tex, line 19
A_a,β,Q(s)=∑_substackn≥2 μ(n)^2=1 (n,Q)=1 n^-2aexp(-λ_n s), λ_n=-2log(1-n^-β).
Exact mathematical passage
A_{a,\beta,Q}(s)=\sum_{\substack{n\ge2\\ \mu(n)^2=1\\ (n,Q)=1}} n^{-2a}\exp(-\lambda_n s),\qquad \lambda_n=-2\log(1-n^{-\beta}).
\sum_{\substack{n\ge2\\\mu(n)^2=1\\(n,Q)=1}}Use in source · 2 passages
Search aliases: ∑_substackn≥2 μ(n)^2=1 (n,Q)=1 sum substack ge mu
Use in source · Read passage · resolution_full.tex, line 220
A_Q(s)=∑_substackn≥2 μ(n)^2=1 (n,Q)=1 n^-4(1-n^-2)^2s.
Exact mathematical passage
A_Q(s)=\sum_{\substack{n\ge2\\\mu(n)^2=1\\(n,Q)=1}} n^{-4}(1-n^{-2})^{2s}.
\sum_{\substack{\ell\ge0\\j+\ell\equiv i\pmod b}}Use in source · 1 passage
Search aliases: ∑_substackℓ≥0 j+ℓ≡ ipmod b sum substack ell ge j ell equiv pmod
Use in source · Read passage · sequence_polyphase.tex, line 78
For causal FIR taps h_ℓ, b_m=∑_ℓ h_ℓ a_m-ℓ with zero extension to negative indices, split A(z) and B(z) into b phases. Then B_i(t)=∑_jH_ij(t)A_j(t), where H_ij(t)=∑_substackℓ≥0 j+ℓ≡ ipmod b h_ℓ t^(j+ℓ-i)/b. The exponent is the nonnegative carry to the next phase block. This is an exact coefficient law, with no approximation or loss of terms.
Exact mathematical passage
\boxed{H_{ij}(t)=\sum_{\substack{\ell\ge0\\j+\ell\equiv i\pmod b}} h_\ell t^{(j+\ell-i)/b}.}
\sum_tUse in source · 3 passages
Search aliases: ∑_t sum
Use in source · Read passage · sequence_full.tex, line 91
F'(z)=1+∑_t h_tF(q_tz)F(r_tz), F(0)=0.
Exact mathematical passage
F'(z)=1+\sum_t h_tF(q_tz)F(r_tz),\qquad F(0)=0.
\sum_{t=0}^{L-1}Use in source · 2 passages
Search aliases: ∑_t=0^L-1 sum
Use in source · Read passage · sequence_full.tex, line 313
b_m=∑_t=0^L-1h_ta_m-t.
Exact mathematical passage
b_m=\sum_{t=0}^{L-1}h_ta_{m-t}.
\sum_{t=0}^mUse in source · 1 passage
Search aliases: ∑_t=0^m sum
Use in source · Read passage · sequence_full.tex, line 335
G_a(z)G_h(z)=∑_m[ ∑_t=0^mbinom mt h_ta_m-t]fracz^mm!.
Exact mathematical passage
G_a(z)G_h(z)=\sum_m\left[ \sum_{t=0}^m\binom mt h_ta_{m-t}\right]\frac{z^m}{m!}.
\sum_{\gamma_p<0}Use in source · 2 passages
Search aliases: ∑_γ_p<0 sum gamma
Use in source · Read passage · multi_prime.tex, line 57
claimA separate additive family has different geometry For weights 1+∑_p∈ Fγ_p1_pmid n, positivity is equivalent to 1+∑_γ_p<0γ_p≥0. Every prime subset appears at its squarefree product, proving necessity. Checking each atom proves sufficiency. Its finite Hankel matrices are affine in γ, so the semidefinite feasible sets are convex spectrahedra. The multiplicative family has genuine overlap terms and polynomial parameter dependence. These are two distinct deformations, with distinct domains.
Exact mathematical passage
1+\sum_{\gamma_p<0}\gamma_p\ge0.
\sum_\ellUse in source · 2 passages
Search aliases: ∑_ℓ sum ell
Use in source · Read passage · sequence_polyphase.tex, line 75
For causal FIR taps h_ℓ, b_m=∑_ℓ h_ℓ a_m-ℓ with zero extension to negative indices, split A(z) and B(z) into b phases. Then B_i(t)=∑_jH_ij(t)A_j(t), where H_ij(t)=∑_substackℓ≥0
Exact mathematical passage
b_m=\sum_\ell h_\ell a_{m-\ell}
\setminusUse in source · 3 passages
Search aliases: ∖ setminus
Use in source · Read passage · main.tex, line 407
B_mA_m^-1/2. Every determinant root is real and below -1; possible coefficient poles occur there, although numerator cancellation is possible. fig05_sieve_boundaryThe line is a proved bound for p=2; dots are ordinary high-precision pencil roots. A defect λ=-1.001 eventually violates positivity even though it can pass low-order checks. The nearest root is -1-δ_m, with δ_m=min_substackf∈ R[x]∖{0} deg f<m frac∫ f^2 dν_nonp∫ f^2 dν_p>0, δ_mdownarrow0. To prove the limit, approximate a continuous bum…
Exact mathematical passage
\delta_m=\min_{\substack{f\in\mathbb R[x]\setminus\{0\}\\\deg f<m}} \frac{\int f^2\dd\nu_{\mathrm{non}p}}{\int f^2\dd\nu_p}>0, \qquad\delta_m\downarrow0.
\intUse in source · 31 passages
Search aliases: ∫ int
Use in source · Read passage · main.tex, line 335
(n,Q)=1μ(n)^2 =δ_Q x+O_Q(sqrtx). Integrate against g_s(x)=x^-2a(1-x^-β)^2s on [1,∞). The density term becomes the beta integral after y=x^-β. Partial integration bounds the error by C_Q∫_1^∞sqrtx|g_s'(x)| d x. The two derivative terms are beta integrals with decay exponent (2a-1/2)/β; |s|/Re s stays bounded in the stated sector. This gives the error term and then the leading asymptotic.
Exact mathematical passage
C_Q\int_1^\infty\sqrt{x}|g_s'(x)|\dd x
\int_0^1Use in source · 4 passages
Search aliases: ∫_0^1 int
Use in source · Read passage · spectral_continuation.tex, line 61
Near zero write f(t)=(log E_N)'(t)=∑_k≥1α_kt^k-1, where α_k=(-1)^k+1 Z_N(k). For any integer K≥1, subtract K terms on [0,1]: I_K(s)= · ∫_0^1t^-s[f(t)-∑_k=1^Kα_kt^k-1]dd t +∑_k=1^Kfracα_kk-s · +∫_1^∞ t^-s[f(t)-frac1N t^1/N-1]dd t +frac1N(s-1/N). The first integral is holomorphic for Re s<K+1. The second is entire because its remainder decays exponentially in t^1/N. The factor sinπ s cancels the positive-integer poles. Increasing K covers the plane, leaving only 1/N. At nonpositive integers I_K is…
Exact mathematical passage
\begin{aligned} I_K(s)={}&\int_0^1t^{-s}\left[f(t)-\sum_{k=1}^K\alpha_kt^{k-1}\right]\dd t +\sum_{k=1}^K\frac{\alpha_k}{k-s}\\ &+\int_1^\infty t^{-s}\left[f(t)-\frac1N t^{1/N-1}\right]\dd t +\frac{1}{N(s-1/N)}. \end{aligned}
\int_0^zUse in source · 3 passages
Search aliases: ∫_0^z int
Use in source · Read passage · sequence_full.tex, line 202
U(z)=e^B(z)∫_0^z e^-B(w) dw, B(z)=∫_0^zT(qw) dw.
Exact mathematical passage
U(z)=e^{B(z)}\int_0^z e^{-B(w)}\,dw, \qquad B(z)=\int_0^zT(qw)\,dw.
\int_0^\inftyUse in source · 6 passages
Search aliases: ∫_0^∞ int infty
Use in source · Read passage · spectral_continuation.tex, line 49
pageheadingORBIT SPECTRA / MELLIN PROOFThe two ends have different jobs For 1/N<Re s<1, the genus-zero log derivative and the angular gap give absolute interchange in Z_N(s)=fracsinπ sπ ∫_0^∞ t^-sfracddtlog E_N(t)dd t. Each term uses the principal-branch beta integral ∫_0^∞ t^-s(t+w)^-1dd t=π w^-s/sinπ s. The denominator estimate bounds the absolute integral by a constant times |w|^-Re s, whose sum converges in this strip.
Exact mathematical passage
\mathcal Z_N(s)=\frac{\sin\pi s}{\pi} \int_0^\infty t^{-s}\frac{d}{dt}\log E_N(t)\dd t.
\int_1^\inftyUse in source · 6 passages
Search aliases: ∫_1^∞ int infty
Use in source · Read passage · main.tex, line 335
(n,Q)=1μ(n)^2 =δ_Q x+O_Q(sqrtx). Integrate against g_s(x)=x^-2a(1-x^-β)^2s on [1,∞). The density term becomes the beta integral after y=x^-β. Partial integration bounds the error by C_Q∫_1^∞sqrtx|g_s'(x)| d x. The two derivative terms are beta integrals with decay exponent (2a-1/2)/β; |s|/Re s stays bounded in the stated sector. This gives the error term and then the leading asymptotic.
Exact mathematical passage
C_Q\int_1^\infty\sqrt{x}|g_s'(x)|\dd x
\int_L^\inftyUse in source · 1 passage
Search aliases: ∫_L^∞ int infty
Use in source · Read passage · arithmetic_full.tex, line 274
Indeed all omitted frequencies are at most λ_L+1, and ∑_n>Ln^-2a≤∫_L^∞ x^-2a dx. Ignoring squarefreeness and coprimality makes this a valid upper bound. If r_0,r_1 are the two positive truncated residual sums at t,t+h, the true secant belongs to
Exact mathematical passage
\sum_{n>L}n^{-2a}\le\int_L^\infty x^{-2a}\,dx
\int_{R-1}^RUse in source · 1 passage
Search aliases: ∫_R-1^R int
Use in source · Read passage · coupled_growth_full.tex, line 226
U(R)≥∫_R-1^R U(qt)V(rt) dt ≥ U(q(R-1))V(r(R-1)).
Exact mathematical passage
U(R)\ge\int_{R-1}^R U(qt)V(rt)\,dt \ge U(q(R-1))V(r(R-1)).
\int_{R-L}^{R}Use in source · 2 passages
Search aliases: ∫_R-L^R int
Use in source · Read passage · dilation_full.tex, line 70
M(R) · ≥ h∫_R-L^RM(qt)M(rt) dt · ≥ hL M(q(R-L))M(r(R-L)) · =M(q(R-L))M(r(R-L)).
Exact mathematical passage
\begin{aligned} M(R)&\ge h\int_{R-L}^{R}M(qt)M(rt)\,dt\\ &\ge hL M(q(R-L))M(r(R-L))\\ &=M(q(R-L))M(r(R-L)). \end{aligned}
\simUse in source · 3 passages
Search aliases: ∼ sim
Use in source · Read passage · main.tex, line 320
Let α=frac2a-1β, δ_Q=frac1ζ(2)∏_pmid Qfracpp+1. As |s|→∞ in any closed sector |arg s|≤π/2-ϵ with 0<ϵ<π/2, A_a,β,Q(s)=fracδ_QβB(α,2s+1) +O_a,β,Q,ϵ(|s|^-(2a-1/2)/β). In particular, A_a,β,Q(s)∼fracδ_QΓ(α)β(2s)^-α. The principal power is used in the right half-plane. The beta expression is an asymptotic continuous-density model, not an identity with the arithmetic sum.
Exact mathematical passage
\A(s)\sim\frac{\delta_Q\Gamma(\alpha)}{\beta}(2s)^{-\alpha}.
\approxUse in source · 6 passages
Search aliases: ≈ approx
Use in source · Read passage · main.tex, line 169
claimA worked order-one member T_1/3,2/3(z)=z+frac227z^3+ frac42187z^5+frac23211160261z^7+O(z^9). It has order one, and T_1/3,2/3(1)≈1.0759239897408249. The diagonal point q=r=1/2 gives f_1/2(z)=sqrt2sinh(z/sqrt2). Every off-diagonal member is separated from all rescaled diagonal members by the theorem above.
Exact mathematical passage
T_{1/3,2/3}(1)\approx1.0759239897408249
\neUse in source · 25 passages
Search aliases: ≠ ne
Use in source · Read passage · main.tex, line 142
pageheadingFAMILY 1 / SEPARATIONMore than a rescaled one-scale seed claimNonrescaling theorem If q≠ r, then T_q,r cannot equal A f_s(Bz) for nonzero constants A,B and any real 0<s<1, where f_s'=1+f_s(sz)^2 is the known family.
Exact mathematical passage
q\ne r
\equivUse in source · 5 passages
Search aliases: ≡ equiv
Use in source · Read passage · sequence_full.tex, line 51
u,v≥ n, u≡ v≡ npmod2, u+v≤ n(n+1).
Exact mathematical passage
u,v\ge n,\qquad u\equiv v\equiv n\pmod2,\qquad u+v\le n(n+1).
\leUse in source · 167 passages
Search aliases: ≤ le
Use in source · chapters_sequences.tex, line 21
_n=(-1)^n. What remains is an alternating mode of constant magnitude. Both modes belong to the same second-order recurrence. At n=6, J_6=21 and 2^6-3·21=1. At n=7, J_7=43 and 2^7-3·43=-1. A recurrence can combine growth and oscillation. Splitting the modes explains the parity correction. OEIS A001045, indexed from zero. Direct closed form and exact recurrence agree for 0≤ n≤18.
Exact mathematical passage
0\le n\le18
\subseteqUse in source · 2 passages
Search aliases: ⊆ subseteq
Use in source · Read passage · sieve_full.tex, line 38
Equivalently, expansion in subsets S⊆ F gives the coefficient D(s)/∏_p∈ S(p^s+1) for ∏_p∈ Sλ_p. This is an independently useful coordinate formula for prime overlap.
Exact mathematical passage
S\subseteq F
\otimesUse in source · 3 passages
Search aliases: ⊗ otimes
Use in source · Read passage · coupled_structure_full.tex, line 128
B=ℓ_1⊗ℓ_2.
Exact mathematical passage
B=\ell_1\otimes\ell_2.
\cdotsUse in source · 10 passages
Search aliases: ⋯ cdots
Use in source · Read passage · notation.tex, line 61
item[Rising factorial.] (s)_k=s(s+1)⋯(s+k-1) for k≥1, and (s)_0=1. It occurs in the shifted spectral residues. item[Coefficient conventions.] An ordinary generating function uses ∑_na_nz^n. An exponential generating function uses ∑_na_nz^n/n!. Their product and shift rules differ. item[Polyphase section and finite taps.] Polyphase splits a coefficient sequence by residue class. Finite impulse response (FIR) taps act by finite ordinary convolution. A phase matrix must retain the carry delay when …
Exact mathematical passage
(s)_k=s(s+1)\cdots(s+k-1)
\langleUse in source · 1 passage
Search aliases: ⟨ langle
Use in source · Read passage · arithmetic_full.tex, line 395
Using an inner product linear in the first argument, the coefficient vector of K(·,t) is (sqrtw_je^-λ_joverline t)_j, so ⟨ Uc,K(·,t)⟩=(Uc)(t). This also verifies the sign and conjugation convention directly. Arbitrarily large finite Gram matrices at distinct complex depths are strictly positive, as proved in the base route. Thus the feature space has infinite rank, not a finite-dimensional Prony model.
Exact mathematical passage
\langle Uc,K(\cdot,t)\rangle=(Uc)(t)
\rangleUse in source · 1 passage
Search aliases: ⟩ rangle
Use in source · Read passage · arithmetic_full.tex, line 395
Using an inner product linear in the first argument, the coefficient vector of K(·,t) is (sqrtw_je^-λ_joverline t)_j, so ⟨ Uc,K(·,t)⟩=(Uc)(t). This also verifies the sign and conjugation convention directly. Arbitrarily large finite Gram matrices at distinct complex depths are strictly positive, as proved in the base route. Thus the feature space has infinite rank, not a finite-dimensional Prony model.
Exact mathematical passage
\langle Uc,K(\cdot,t)\rangle=(Uc)(t)
\longrightarrowUse in source · 1 passage
Search aliases: ⟶ longrightarrow
Use in source · Read passage · sequence_full.tex, line 137
c_n^1/(2n+1)≥ q^n/(2n+1)r^n^2/(2n+1)(2n+1)^-n/(2n+1)⟶∞.
Exact mathematical passage
c_n^{1/(2n+1)}\ge q^{n/(2n+1)}r^{n^2/(2n+1)}(2n+1)^{-n/(2n+1)}\longrightarrow\infty.
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- monographs/special-functions/monograph/sections/multi_prime.tex · 56 mathematical passages
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- monographs/special-functions/monograph/sections/sequence_full.tex · 185 mathematical passages
- monographs/special-functions/monograph/sections/sequence_polyphase.tex · 54 mathematical passages
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