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Paper IV: Cyclic Analytic Functions, Reciprocal Families, and Spectral Zero Data
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781 symbol spellings in 2030 mathematical passages across 31 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.
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+Definition passage · 275 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
=Definition passage · 779 passages
Search aliases: =
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
\mathfrak ADefinition passage · 6 passages
Search aliases: A mathfrak
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 762
Let A_N^+= C[ξ]/(ξ^N-1) and A_N^-= C[ξ]/(ξ^N+1). Every element has a unique representative of degree below N.
Exact mathematical passage
\mathfrak A_N^+=\CC[\xi]/(\xi^N-1)
\mathfrak A_N^+Definition passage · 3 passages
Search aliases: A_N^+ mathfrak
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 762
Let A_N^+= C[ξ]/(ξ^N-1) and A_N^-= C[ξ]/(ξ^N+1). Every element has a unique representative of degree below N.
Exact mathematical passage
\mathfrak A_N^+=\CC[\xi]/(\xi^N-1)
\mathfrak A_N^-Definition passage · 3 passages
Search aliases: A_N^- mathfrak
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 766
Let A_N^+= C[ξ]/(ξ^N-1) and A_N^-= C[ξ]/(ξ^N+1). Every element has a unique representative of degree below N.
Exact mathematical passage
\mathfrak A_N^-=\CC[\xi]/(\xi^N+1).
\mathrm{abs}Definition passage · 10 passages
Search aliases: abs mathrm
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2718
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{abs}}(s) = \sum_{\rho\in\Zcal(\Dfun_N)}|\rho|^{-s}
\BcalDefinition passage · 17 passages
Search aliases: Bcal B mathcal
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
\Bcal_{N,j}Definition passage · 10 passages
Search aliases: Bcal_N,j B_N,j Bcal mathcal
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
\bmodDefinition passage · 10 passages
Search aliases: bmod
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1776
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
Exact mathematical passage
\mathfrak T_N = \left\{\frac{2\pi k}{N}:0\leq k<N\right\} \quad\bmod 2\pi.
CDefinition passage · 36 passages
Search aliases: C
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1379
or 0≤ j<N, define Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)!. Then E_N,j(z) = fracz^jj!Φ_N,j(z^N). Define the residue-section reciprocal function C_N,j(z) = Φ_N,j(z^N)^-1 = fracz^j/j! E_N,j(z). For j>0, the last quotient at z=0 is interpreted by its removable continuation, namely the first expression in the display. Write C_N,j(z) = ∑_n≥0C_N,j(n)fracz^Nn(Nn)!.
Exact mathematical passage
\Cfun_{N,j}(z) = \sum_{n\geq0}C_{N,j}(n)\frac{z^{Nn}}{(Nn)!}.
C_{N,j}Definition passage · 9 passages
Search aliases: C_N,j
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1379
or 0≤ j<N, define Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)!. Then E_N,j(z) = fracz^jj!Φ_N,j(z^N). Define the residue-section reciprocal function C_N,j(z) = Φ_N,j(z^N)^-1 = fracz^j/j! E_N,j(z). For j>0, the last quotient at z=0 is interpreted by its removable continuation, namely the first expression in the display. Write C_N,j(z) = ∑_n≥0C_N,j(n)fracz^Nn(Nn)!.
Exact mathematical passage
\Cfun_{N,j}(z) = \sum_{n\geq0}C_{N,j}(n)\frac{z^{Nn}}{(Nn)!}.
C_{N,j,n}Definition passage · 7 passages
Search aliases: C_N,j,n
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1414
For r∈ C, define formal coefficients C_N,j,n(r) by C_N,j(z)^r = ∑_n≥0C_N,j,n(r)fracz^Nn(Nn)!. On a simply connected region containing 0 and avoiding the zeros of Φ_N,j(z^N), interpret this power as e^rL, where L is the holomorphic logarithm of C_N,j normalized by L(0)=0.
Exact mathematical passage
C_{N,j,n}(r)
\CCDefinition passage · 21 passages
Search aliases: CC C mathbb
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 433
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\lambda\in\CC
\CC^NDefinition passage · 1 passage
Search aliases: CC^N C^N CC mathbb
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 799
Let J_N be the signed cyclic shift on the standard basis e_0,…,e_N-1 of C^N, defined by J_Ne_j=e_j+1 (0≤ j<N-1), J_Ne_N-1=-e_0. Then J_N^N=-I.
Exact mathematical passage
\CC^N
\CfunDefinition passage · 15 passages
Search aliases: Cfun C mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1369
For 0≤ j<N, define Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)!. Then E_N,j(z) = fracz^jj!Φ_N,j(z^N). Define the residue-section reciprocal function C_N,j(z) = Φ_N,j(z^N)^-1 = fracz^j/j! E_N,j(z). For j>0, the last quotient at z=0 is interpreted by its removable continuation, namely the first expression in the display. Write C_N,j(z) = ∑_n≥0C_N,j(n)fracz^Nn(Nn)!.
Exact mathematical passage
\Cfun_{N,j}(z) = \Phi_{N,j}(z^N)^{-1} = \frac{z^j/j!}{\Efun_{N,j}(z)}.
\Cfun_{N,j}Definition passage · 11 passages
Search aliases: Cfun_N,j C_N,j Cfun mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1369
For 0≤ j<N, define Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)!. Then E_N,j(z) = fracz^jj!Φ_N,j(z^N). Define the residue-section reciprocal function C_N,j(z) = Φ_N,j(z^N)^-1 = fracz^j/j! E_N,j(z). For j>0, the last quotient at z=0 is interpreted by its removable continuation, namely the first expression in the display. Write C_N,j(z) = ∑_n≥0C_N,j(n)fracz^Nn(Nn)!.
Exact mathematical passage
\Cfun_{N,j}(z) = \Phi_{N,j}(z^N)^{-1} = \frac{z^j/j!}{\Efun_{N,j}(z)}.
\cosDefinition passage · 35 passages
Search aliases: cos
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1769
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
Exact mathematical passage
h_N(\theta) = \max_{0\leq k<N} \cos\left(\theta+\frac{(2k+1)\pi}{N}\right).
\mathrm{der}Definition passage · 2 passages
Search aliases: der mathrm
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2900
Let q be a positive integer. Choose logarithms of the simple zeros of D_N with arguments in a fixed bounded interval. Define Z_N,q^der(s) = ∑_substack D_N(ρ)=0 D_N'(ρ)≠0 frace^-sLogρ D_N'(ρ)^q.
Exact mathematical passage
\mathfrak Z_{N,q}^{\mathrm{der}}(s) = \sum_{\substack{\Dfun_N(\rho)=0\\ \Dfun_N'(\rho)\ne0}} \frac{e^{-s\Log\rho}}{\Dfun_N'(\rho)^q}.
\DfunDefinition passage · 108 passages
Search aliases: Dfun D mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 581
For 0≤ j<N, define D_N,j(z) = ∑_n≥0(-1)^nfracz^Nn+j(Nn+j)!. The distinguished signed denominator is D_N(z)= D_N,0(z).
Exact mathematical passage
\Dfun_{N,j}(z) = \sum_{n\geq0}(-1)^n\frac{z^{Nn+j}}{(Nn+j)!}.
\Dfun_NDefinition passage · 70 passages
Search aliases: Dfun_N D_N Dfun mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 587
For 0≤ j<N, define D_N,j(z) = ∑_n≥0(-1)^nfracz^Nn+j(Nn+j)!. The distinguished signed denominator is D_N(z)= D_N,0(z).
Exact mathematical passage
\Dfun_N(z)=\Dfun_{N,0}(z).
\Dfun_N'Definition passage · 17 passages
Search aliases: Dfun_N' D_N' Dfun mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1237
Define the transfer tangent T_N(z) =-frac D_N'(z) D_N(z). Near the origin, its convergent block expansion has the form T_N(z) = ∑_n≥0T_N(n) fracz^Nn+N-1(Nn+N-1)!.
Exact mathematical passage
\Tfun_N(z) =-\frac{\Dfun_N'(z)}{\Dfun_N(z)}.
\Dfun_{N,0}Definition passage · 4 passages
Search aliases: Dfun_N,0 D_N,0 Dfun mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 587
For 0≤ j<N, define D_N,j(z) = ∑_n≥0(-1)^nfracz^Nn+j(Nn+j)!. The distinguished signed denominator is D_N(z)= D_N,0(z).
Exact mathematical passage
\Dfun_N(z)=\Dfun_{N,0}(z).
\Dfun_{N,j}Definition passage · 8 passages
Search aliases: Dfun_N,j D_N,j Dfun mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 581
For 0≤ j<N, define D_N,j(z) = ∑_n≥0(-1)^nfracz^Nn+j(Nn+j)!. The distinguished signed denominator is D_N(z)= D_N,0(z).
Exact mathematical passage
\Dfun_{N,j}(z) = \sum_{n\geq0}(-1)^n\frac{z^{Nn+j}}{(Nn+j)!}.
EDefinition passage · 41 passages
Search aliases: E
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1458
Define polynomials E_m^(N,j,r)(x) by C_N,j(z)^re^xz = ∑_m≥0E_m^(N,j,r)(x)fracz^mm!.
Exact mathematical passage
E_m^{(N,j,r)}(x)
eDefinition passage · 157 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
e^{-s\log|\lambda|}Definition passage · 1 passage
Search aliases: e^-slog|λ| log lambda
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2729
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
e^{-s\log|\lambda|}
e^{-s\log|\rho|}Definition passage · 1 passage
Search aliases: e^-slog|ρ| log rho
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2729
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
e^{-s\log|\rho|}
e^{-s\Log\lambda}Definition passage · 1 passage
Search aliases: e^-sLogλ Log lambda operatorname lambda
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2882
Choose a logarithm for each member of the finite exceptional part of Λ_N. Use the real logarithm for the eventual positive sequence. Define Z_N^orb(s) = ∑_λ∈Λ_Ne^-sLogλ. This series is initially defined on Re s>1/N.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{orb}}(s) = \sum_{\lambda\in\Lambda_N}e^{-s\Log\lambda}.
e^{-s\Log\rho}Definition passage · 2 passages
Search aliases: e^-sLogρ Log rho operatorname rho
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2900
Let q be a positive integer. Choose logarithms of the simple zeros of D_N with arguments in a fixed bounded interval. Define Z_N,q^der(s) = ∑_substack D_N(ρ)=0 D_N'(ρ)≠0 frace^-sLogρ D_N'(ρ)^q.
Exact mathematical passage
\mathfrak Z_{N,q}^{\mathrm{der}}(s) = \sum_{\substack{\Dfun_N(\rho)=0\\ \Dfun_N'(\rho)\ne0}} \frac{e^{-s\Log\rho}}{\Dfun_N'(\rho)^q}.
e^{rL}Definition passage · 1 passage
Search aliases: e^rL
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1420
For r∈ C, define formal coefficients C_N,j,n(r) by C_N,j(z)^r = ∑_n≥0C_N,j,n(r)fracz^Nn(Nn)!. On a simply connected region containing 0 and avoiding the zeros of Φ_N,j(z^N), interpret this power as e^rL, where L is the holomorphic logarithm of C_N,j normalized by L(0)=0.
Exact mathematical passage
e^{rL}
e^{xz}Definition passage · 6 passages
Search aliases: e^xz
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1459
Define polynomials E_m^(N,j,r)(x) by C_N,j(z)^re^xz = ∑_m≥0E_m^(N,j,r)(x)fracz^mm!.
Exact mathematical passage
\Cfun_{N,j}(z)^re^{xz} = \sum_{m\geq0}E_m^{(N,j,r)}(x)\frac{z^m}{m!}.
e^{\lambda z}Definition passage · 3 passages
Search aliases: e^λ z lambda
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
e_0Definition passage · 15 passages
Search aliases: e_0
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 799
Let J_N be the signed cyclic shift on the standard basis e_0,…,e_N-1 of C^N, defined by J_Ne_j=e_j+1 (0≤ j<N-1), J_Ne_N-1=-e_0. Then J_N^N=-I.
Exact mathematical passage
e_0,\ldots,e_{N-1}
e_jDefinition passage · 1 passage
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 800
Let J_N be the signed cyclic shift on the standard basis e_0,…,e_N-1 of C^N, defined by J_Ne_j=e_j+1 (0≤ j<N-1), J_Ne_N-1=-e_0. Then J_N^N=-I.
Exact mathematical passage
J_Ne_j=e_{j+1} \quad (0\leq j<N-1), \qquad J_Ne_{N-1}=-e_0.
e_{j+1}Definition passage · 1 passage
Search aliases: e_j+1
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 800
Let J_N be the signed cyclic shift on the standard basis e_0,…,e_N-1 of C^N, defined by J_Ne_j=e_j+1 (0≤ j<N-1), J_Ne_N-1=-e_0. Then J_N^N=-I.
Exact mathematical passage
J_Ne_j=e_{j+1} \quad (0\leq j<N-1), \qquad J_Ne_{N-1}=-e_0.
E_m^{(N,j,r)}Definition passage · 9 passages
Search aliases: E_m^(N,j,r)
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1458
Define polynomials E_m^(N,j,r)(x) by C_N,j(z)^re^xz = ∑_m≥0E_m^(N,j,r)(x)fracz^mm!.
Exact mathematical passage
E_m^{(N,j,r)}(x)
e_{N-1}Definition passage · 2 passages
Search aliases: e_N-1
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 799
Let J_N be the signed cyclic shift on the standard basis e_0,…,e_N-1 of C^N, defined by J_Ne_j=e_j+1 (0≤ j<N-1), J_Ne_N-1=-e_0. Then J_N^N=-I.
Exact mathematical passage
e_0,\ldots,e_{N-1}
\EfunDefinition passage · 21 passages
Search aliases: Efun E mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
\Efun_{N,0}Definition passage · 3 passages
Search aliases: Efun_N,0 E_N,0 Efun mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1611
Set H_N(z)= E_N,0(z) = ∑_n≥0fracz^Nn(Nn)!. For u∈ C, define symupΘ_N,n(u) by H_N(uz)^-1H_N(z)^-u = ∑_n≥0symupΘ_N,n(u)fracz^Nn(Nn)!. This is first a formal identity at the origin. Analytically, it is interpreted on a common simply connected zero-free region with the branch of log H_N normalized by log H_N(0)=0.
Exact mathematical passage
H_N(z)=\Efun_{N,0}(z) = \sum_{n\geq0}\frac{z^{Nn}}{(Nn)!}.
\Efun_{N,j}Definition passage · 17 passages
Search aliases: Efun_N,j E_N,j Efun mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
\expDefinition passage · 10 passages
Search aliases: exp
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 933
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
\Sfun_N^{(r)}(z) = \Dfun_N(z)^{-r} = \exp\bigl(-r\log\Dfun_N(z)\bigr).
HDefinition passage · 17 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1611
Set H_N(z)= E_N,0(z) = ∑_n≥0fracz^Nn(Nn)!. For u∈ C, define symupΘ_N,n(u) by H_N(uz)^-1H_N(z)^-u = ∑_n≥0symupΘ_N,n(u)fracz^Nn(Nn)!. This is first a formal identity at the origin. Analytically, it is interpreted on a common simply connected zero-free region with the branch of log H_N normalized by log H_N(0)=0.
Exact mathematical passage
H_N(z)=\Efun_{N,0}(z) = \sum_{n\geq0}\frac{z^{Nn}}{(Nn)!}.
hDefinition passage · 10 passages
Search aliases: h
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1769
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
Exact mathematical passage
h_N(\theta) = \max_{0\leq k<N} \cos\left(\theta+\frac{(2k+1)\pi}{N}\right).
H_NDefinition passage · 17 passages
Search aliases: H_N
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1611
Set H_N(z)= E_N,0(z) = ∑_n≥0fracz^Nn(Nn)!. For u∈ C, define symupΘ_N,n(u) by H_N(uz)^-1H_N(z)^-u = ∑_n≥0symupΘ_N,n(u)fracz^Nn(Nn)!. This is first a formal identity at the origin. Analytically, it is interpreted on a common simply connected zero-free region with the branch of log H_N normalized by log H_N(0)=0.
Exact mathematical passage
H_N(z)=\Efun_{N,0}(z) = \sum_{n\geq0}\frac{z^{Nn}}{(Nn)!}.
h_NDefinition passage · 7 passages
Search aliases: h_N
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1769
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
Exact mathematical passage
h_N(\theta) = \max_{0\leq k<N} \cos\left(\theta+\frac{(2k+1)\pi}{N}\right).
IDefinition passage · 13 passages
Search aliases: I
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 806
Let J_N be the signed cyclic shift on the standard basis e_0,…,e_N-1 of C^N, defined by J_Ne_j=e_j+1 (0≤ j<N-1), J_Ne_N-1=-e_0. Then J_N^N=-I.
Exact mathematical passage
J_N^N=-I
JDefinition passage · 38 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 799
Let J_N be the signed cyclic shift on the standard basis e_0,…,e_N-1 of C^N, defined by J_Ne_j=e_j+1 (0≤ j<N-1), J_Ne_N-1=-e_0. Then J_N^N=-I.
Exact mathematical passage
J_N
jDefinition passage · 190 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 433
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
0\leq j<N
J_NDefinition passage · 17 passages
Search aliases: J_N
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 799
Let J_N be the signed cyclic shift on the standard basis e_0,…,e_N-1 of C^N, defined by J_Ne_j=e_j+1 (0≤ j<N-1), J_Ne_N-1=-e_0. Then J_N^N=-I.
Exact mathematical passage
J_N
J_N^NDefinition passage · 2 passages
Search aliases: J_N^N
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 806
Let J_N be the signed cyclic shift on the standard basis e_0,…,e_N-1 of C^N, defined by J_Ne_j=e_j+1 (0≤ j<N-1), J_Ne_N-1=-e_0. Then J_N^N=-I.
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J_N^N=-I
kDefinition passage · 65 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1769
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
Exact mathematical passage
h_N(\theta) = \max_{0\leq k<N} \cos\left(\theta+\frac{(2k+1)\pi}{N}\right).
LDefinition passage · 24 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1420
For r∈ C, define formal coefficients C_N,j,n(r) by C_N,j(z)^r = ∑_n≥0C_N,j,n(r)fracz^Nn(Nn)!. On a simply connected region containing 0 and avoiding the zeros of Φ_N,j(z^N), interpret this power as e^rL, where L is the holomorphic logarithm of C_N,j normalized by L(0)=0.
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e^{rL}
\LogDefinition passage · 3 passages
Search aliases: Log operatorname
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2882
Choose a logarithm for each member of the finite exceptional part of Λ_N. Use the real logarithm for the eventual positive sequence. Define Z_N^orb(s) = ∑_λ∈Λ_Ne^-sLogλ. This series is initially defined on Re s>1/N.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{orb}}(s) = \sum_{\lambda\in\Lambda_N}e^{-s\Log\lambda}.
\logDefinition passage · 27 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 933
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
\Sfun_N^{(r)}(z) = \Dfun_N(z)^{-r} = \exp\bigl(-r\log\Dfun_N(z)\bigr).
mDefinition passage · 237 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 946
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
\kappa_{N,m}
\maxDefinition passage · 3 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1769
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
Exact mathematical passage
h_N(\theta) = \max_{0\leq k<N} \cos\left(\theta+\frac{(2k+1)\pi}{N}\right).
\max_{0\leq k<N}Definition passage · 1 passage
Search aliases: max_0≤ k<N max leq
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1769
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
Exact mathematical passage
h_N(\theta) = \max_{0\leq k<N} \cos\left(\theta+\frac{(2k+1)\pi}{N}\right).
NDefinition passage · 830 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 433
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
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0\leq j<N
nDefinition passage · 222 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
\mathrm{orb}Definition passage · 4 passages
Search aliases: orb mathrm
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2882
Choose a logarithm for each member of the finite exceptional part of Λ_N. Use the real logarithm for the eventual positive sequence. Define Z_N^orb(s) = ∑_λ∈Λ_Ne^-sLogλ. This series is initially defined on Re s>1/N.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{orb}}(s) = \sum_{\lambda\in\Lambda_N}e^{-s\Log\lambda}.
PDefinition passage · 60 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 941
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
\Sfun_N^{(r)}(z) = \sum_{n\geq0}P_{N,n}(r)\frac{z^{Nn}}{(Nn)!}.
P_{N,n}Definition passage · 29 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 941
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
\Sfun_N^{(r)}(z) = \sum_{n\geq0}P_{N,n}(r)\frac{z^{Nn}}{(Nn)!}.
qDefinition passage · 50 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2899
Let q be a positive integer. Choose logarithms of the simple zeros of D_N with arguments in a fixed bounded interval. Define Z_N,q^der(s) = ∑_substack D_N(ρ)=0
Exact mathematical passage
q
\mathbb RDefinition passage · 1 passage
Search aliases: R mathbb
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1768
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
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\theta\in\mathbb R
rDefinition passage · 215 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 932
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
r\in\CC
\ReDefinition passage · 23 passages
Search aliases: Re
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2887
Choose a logarithm for each member of the finite exceptional part of Λ_N. Use the real logarithm for the eventual positive sequence. Define Z_N^orb(s) = ∑_λ∈Λ_Ne^-sLogλ. This series is initially defined on Re s>1/N.
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\Re s>1/N
\mathrm{red}Definition passage · 12 passages
Search aliases: red mathrm
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2724
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{red}}(s) = \sum_{\lambda\in\Lambda_N}|\lambda|^{-s},
sDefinition passage · 99 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2718
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{abs}}(s) = \sum_{\rho\in\Zcal(\Dfun_N)}|\rho|^{-s}
\SfunDefinition passage · 7 passages
Search aliases: Sfun S mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 933
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
\Sfun_N^{(r)}(z) = \Dfun_N(z)^{-r} = \exp\bigl(-r\log\Dfun_N(z)\bigr).
\Sfun_N^{(r)}Definition passage · 6 passages
Search aliases: Sfun_N^(r) S_N^(r) Sfun mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 933
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
\Sfun_N^{(r)}(z) = \Dfun_N(z)^{-r} = \exp\bigl(-r\log\Dfun_N(z)\bigr).
TDefinition passage · 32 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1242
Define the transfer tangent T_N(z) =-frac D_N'(z) D_N(z). Near the origin, its convergent block expansion has the form T_N(z) = ∑_n≥0T_N(n) fracz^Nn+N-1(Nn+N-1)!.
Exact mathematical passage
\Tfun_N(z) = \sum_{n\geq0}T_N(n) \frac{z^{Nn+N-1}}{(Nn+N-1)!}.
\mathfrak TDefinition passage · 2 passages
Search aliases: T mathfrak
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1776
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
Exact mathematical passage
\mathfrak T_N = \left\{\frac{2\pi k}{N}:0\leq k<N\right\} \quad\bmod 2\pi.
T_NDefinition passage · 19 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1242
Define the transfer tangent T_N(z) =-frac D_N'(z) D_N(z). Near the origin, its convergent block expansion has the form T_N(z) = ∑_n≥0T_N(n) fracz^Nn+N-1(Nn+N-1)!.
Exact mathematical passage
\Tfun_N(z) = \sum_{n\geq0}T_N(n) \frac{z^{Nn+N-1}}{(Nn+N-1)!}.
\mathfrak T_NDefinition passage · 2 passages
Search aliases: T_N mathfrak
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1776
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
Exact mathematical passage
\mathfrak T_N = \left\{\frac{2\pi k}{N}:0\leq k<N\right\} \quad\bmod 2\pi.
\TfunDefinition passage · 13 passages
Search aliases: Tfun T mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1237
Define the transfer tangent T_N(z) =-frac D_N'(z) D_N(z). Near the origin, its convergent block expansion has the form T_N(z) = ∑_n≥0T_N(n) fracz^Nn+N-1(Nn+N-1)!.
Exact mathematical passage
\Tfun_N(z) =-\frac{\Dfun_N'(z)}{\Dfun_N(z)}.
\Tfun_NDefinition passage · 10 passages
Search aliases: Tfun_N T_N Tfun mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1237
Define the transfer tangent T_N(z) =-frac D_N'(z) D_N(z). Near the origin, its convergent block expansion has the form T_N(z) = ∑_n≥0T_N(n) fracz^Nn+N-1(Nn+N-1)!.
Exact mathematical passage
\Tfun_N(z) =-\frac{\Dfun_N'(z)}{\Dfun_N(z)}.
\ThetafunDefinition passage · 10 passages
Search aliases: Thetafun symupΘ symup Theta
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1616
Set H_N(z)= E_N,0(z) = ∑_n≥0fracz^Nn(Nn)!. For u∈ C, define symupΘ_N,n(u) by H_N(uz)^-1H_N(z)^-u = ∑_n≥0symupΘ_N,n(u)fracz^Nn(Nn)!. This is first a formal identity at the origin. Analytically, it is interpreted on a common simply connected zero-free region with the branch of log H_N normalized by log H_N(0)=0.
Exact mathematical passage
\Thetafun_{N,n}(u)
\Thetafun_{N,n}Definition passage · 7 passages
Search aliases: Thetafun_N,n symupΘ_N,n Thetafun symup Theta
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1616
Set H_N(z)= E_N,0(z) = ∑_n≥0fracz^Nn(Nn)!. For u∈ C, define symupΘ_N,n(u) by H_N(uz)^-1H_N(z)^-u = ∑_n≥0symupΘ_N,n(u)fracz^Nn(Nn)!. This is first a formal identity at the origin. Analytically, it is interpreted on a common simply connected zero-free region with the branch of log H_N normalized by log H_N(0)=0.
Exact mathematical passage
\Thetafun_{N,n}(u)
uDefinition passage · 61 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1616
Set H_N(z)= E_N,0(z) = ∑_n≥0fracz^Nn(Nn)!. For u∈ C, define symupΘ_N,n(u) by H_N(uz)^-1H_N(z)^-u = ∑_n≥0symupΘ_N,n(u)fracz^Nn(Nn)!. This is first a formal identity at the origin. Analytically, it is interpreted on a common simply connected zero-free region with the branch of log H_N normalized by log H_N(0)=0.
Exact mathematical passage
u\in\CC
wDefinition passage · 131 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 698
Define the reduced denominator Δ_N(w) = ∑_n≥0(-1)^nfracw^n(Nn)!. Then D_N(z)=Δ_N(z^N).
Exact mathematical passage
\Delta_N(w) = \sum_{n\geq0}(-1)^n\frac{w^n}{(Nn)!}.
w^mDefinition passage · 14 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 947
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
-\log\Delta_N(w) = \sum_{m\geq1}\kappa_{N,m}\frac{w^m}{m}.
w^nDefinition passage · 28 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 698
Define the reduced denominator Δ_N(w) = ∑_n≥0(-1)^nfracw^n(Nn)!. Then D_N(z)=Δ_N(z^N).
Exact mathematical passage
\Delta_N(w) = \sum_{n\geq0}(-1)^n\frac{w^n}{(Nn)!}.
xDefinition passage · 66 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1458
Define polynomials E_m^(N,j,r)(x) by C_N,j(z)^re^xz = ∑_m≥0E_m^(N,j,r)(x)fracz^mm!.
Exact mathematical passage
E_m^{(N,j,r)}(x)
\mathfrak ZDefinition passage · 24 passages
Search aliases: Z mathfrak
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2718
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{abs}}(s) = \sum_{\rho\in\Zcal(\Dfun_N)}|\rho|^{-s}
zDefinition passage · 372 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
z^jDefinition passage · 11 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1363
For 0≤ j<N, define Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)!. Then E_N,j(z) = fracz^jj!Φ_N,j(z^N). Define the residue-section reciprocal function C_N,j(z) = Φ_N,j(z^N)^-1 = fracz^j/j! E_N,j(z). For j>0, the last quotient at z=0 is interpreted by its removable continuation, namely the first expression in the display. Write C_N,j(z) = ∑_n≥0C_N,j(n)fracz^Nn(Nn)!.
Exact mathematical passage
\Efun_{N,j}(z) = \frac{z^j}{j!}\Phi_{N,j}(z^N).
z^mDefinition passage · 13 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1459
Define polynomials E_m^(N,j,r)(x) by C_N,j(z)^re^xz = ∑_m≥0E_m^(N,j,r)(x)fracz^mm!.
Exact mathematical passage
\Cfun_{N,j}(z)^re^{xz} = \sum_{m\geq0}E_m^{(N,j,r)}(x)\frac{z^m}{m!}.
z^NDefinition passage · 41 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 704
Define the reduced denominator Δ_N(w) = ∑_n≥0(-1)^nfracw^n(Nn)!. Then D_N(z)=Δ_N(z^N).
Exact mathematical passage
\Dfun_N(z)=\Delta_N(z^N).
z^{Nn}Definition passage · 12 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 941
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
\Sfun_N^{(r)}(z) = \sum_{n\geq0}P_{N,n}(r)\frac{z^{Nn}}{(Nn)!}.
z^{Nn+j}Definition passage · 6 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 581
For 0≤ j<N, define D_N,j(z) = ∑_n≥0(-1)^nfracz^Nn+j(Nn+j)!. The distinguished signed denominator is D_N(z)= D_N,0(z).
Exact mathematical passage
\Dfun_{N,j}(z) = \sum_{n\geq0}(-1)^n\frac{z^{Nn+j}}{(Nn+j)!}.
z^{Nn+N-1}Definition passage · 2 passages
Search aliases: z^Nn+N-1
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1242
Define the transfer tangent T_N(z) =-frac D_N'(z) D_N(z). Near the origin, its convergent block expansion has the form T_N(z) = ∑_n≥0T_N(n) fracz^Nn+N-1(Nn+N-1)!.
Exact mathematical passage
\Tfun_N(z) = \sum_{n\geq0}T_N(n) \frac{z^{Nn+N-1}}{(Nn+N-1)!}.
\mathfrak Z_{N,q}^{\mathrm{der}}Definition passage · 2 passages
Search aliases: Z_N,q^der mathfrak mathrm
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2900
Let q be a positive integer. Choose logarithms of the simple zeros of D_N with arguments in a fixed bounded interval. Define Z_N,q^der(s) = ∑_substack D_N(ρ)=0 D_N'(ρ)≠0 frace^-sLogρ D_N'(ρ)^q.
Exact mathematical passage
\mathfrak Z_{N,q}^{\mathrm{der}}(s) = \sum_{\substack{\Dfun_N(\rho)=0\\ \Dfun_N'(\rho)\ne0}} \frac{e^{-s\Log\rho}}{\Dfun_N'(\rho)^q}.
\mathfrak Z_N^{\mathrm{abs}}Definition passage · 6 passages
Search aliases: Z_N^abs mathfrak mathrm
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2718
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{abs}}(s) = \sum_{\rho\in\Zcal(\Dfun_N)}|\rho|^{-s}
\mathfrak Z_N^{\mathrm{orb}}Definition passage · 2 passages
Search aliases: Z_N^orb mathfrak mathrm
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2882
Choose a logarithm for each member of the finite exceptional part of Λ_N. Use the real logarithm for the eventual positive sequence. Define Z_N^orb(s) = ∑_λ∈Λ_Ne^-sLogλ. This series is initially defined on Re s>1/N.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{orb}}(s) = \sum_{\lambda\in\Lambda_N}e^{-s\Log\lambda}.
\mathfrak Z_N^{\mathrm{red}}Definition passage · 6 passages
Search aliases: Z_N^red mathfrak mathrm
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2724
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{red}}(s) = \sum_{\lambda\in\Lambda_N}|\lambda|^{-s},
\ZcalDefinition passage · 8 passages
Search aliases: Zcal Z mathcal
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2718
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{abs}}(s) = \sum_{\rho\in\Zcal(\Dfun_N)}|\rho|^{-s}
|Definition passage · 63 passages
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Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2718
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{abs}}(s) = \sum_{\rho\in\Zcal(\Dfun_N)}|\rho|^{-s}
|^{-s}Definition passage · 3 passages
Search aliases: |^-s
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2718
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{abs}}(s) = \sum_{\rho\in\Zcal(\Dfun_N)}|\rho|^{-s}
\DeltaDefinition passage · 84 passages
Search aliases: Δ Delta
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 698
Define the reduced denominator Δ_N(w) = ∑_n≥0(-1)^nfracw^n(Nn)!. Then D_N(z)=Δ_N(z^N).
Exact mathematical passage
\Delta_N(w) = \sum_{n\geq0}(-1)^n\frac{w^n}{(Nn)!}.
\Delta_NDefinition passage · 59 passages
Search aliases: Δ_N Delta
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 698
Define the reduced denominator Δ_N(w) = ∑_n≥0(-1)^nfracw^n(Nn)!. Then D_N(z)=Δ_N(z^N).
Exact mathematical passage
\Delta_N(w) = \sum_{n\geq0}(-1)^n\frac{w^n}{(Nn)!}.
\thetaDefinition passage · 33 passages
Search aliases: θ theta
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1768
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
Exact mathematical passage
\theta\in\mathbb R
\kappaDefinition passage · 56 passages
Search aliases: κ kappa
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 946
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
\kappa_{N,m}
\kappa_{N,m}Definition passage · 23 passages
Search aliases: κ_N,m kappa
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 946
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
\kappa_{N,m}
\LambdaDefinition passage · 30 passages
Search aliases: Λ Lambda
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2413
For m≥1, define σ_N,m = ∑_λ∈Λ_Nλ^-m, with multiplicities.
Exact mathematical passage
\sigma_{N,m} = \sum_{\lambda\in\Lambda_N}\lambda^{-m},
\lambdaDefinition passage · 78 passages
Search aliases: λ lambda
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 433
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\lambda\in\CC
\lambda^{-m}Definition passage · 1 passage
Search aliases: λ^-m lambda
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2413
For m≥1, define σ_N,m = ∑_λ∈Λ_Nλ^-m, with multiplicities.
Exact mathematical passage
\sigma_{N,m} = \sum_{\lambda\in\Lambda_N}\lambda^{-m},
\Lambda_NDefinition passage · 27 passages
Search aliases: Λ_N Lambda
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2413
For m≥1, define σ_N,m = ∑_λ∈Λ_Nλ^-m, with multiplicities.
Exact mathematical passage
\sigma_{N,m} = \sum_{\lambda\in\Lambda_N}\lambda^{-m},
\xiDefinition passage · 36 passages
Search aliases: ξ xi
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 762
Let A_N^+= C[ξ]/(ξ^N-1) and A_N^-= C[ξ]/(ξ^N+1). Every element has a unique representative of degree below N.
Exact mathematical passage
\mathfrak A_N^+=\CC[\xi]/(\xi^N-1)
\xi^NDefinition passage · 2 passages
Search aliases: ξ^N xi
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 762
Let A_N^+= C[ξ]/(ξ^N-1) and A_N^-= C[ξ]/(ξ^N+1). Every element has a unique representative of degree below N.
Exact mathematical passage
\mathfrak A_N^+=\CC[\xi]/(\xi^N-1)
\piDefinition passage · 94 passages
Search aliases: π pi
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1769
For θ∈ R, define h_N(θ) = max_0≤ k<N cos(θ+frac(2k+1)πN). The tie directions are T_N = {frac2π kN:0≤ k<N} bmod 2π.
Exact mathematical passage
h_N(\theta) = \max_{0\leq k<N} \cos\left(\theta+\frac{(2k+1)\pi}{N}\right).
\rhoDefinition passage · 38 passages
Search aliases: ρ rho
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2718
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{abs}}(s) = \sum_{\rho\in\Zcal(\Dfun_N)}|\rho|^{-s}
\sigmaDefinition passage · 32 passages
Search aliases: σ sigma
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2413
For m≥1, define σ_N,m = ∑_λ∈Λ_Nλ^-m, with multiplicities.
Exact mathematical passage
\sigma_{N,m} = \sum_{\lambda\in\Lambda_N}\lambda^{-m},
\sigma_{N,m}Definition passage · 8 passages
Search aliases: σ_N,m sigma
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2413
For m≥1, define σ_N,m = ∑_λ∈Λ_Nλ^-m, with multiplicities.
Exact mathematical passage
\sigma_{N,m} = \sum_{\lambda\in\Lambda_N}\lambda^{-m},
\PhiDefinition passage · 24 passages
Search aliases: Φ Phi
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1357
For 0≤ j<N, define Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)!. Then E_N,j(z) = fracz^jj!Φ_N,j(z^N). Define the residue-section reciprocal function C_N,j(z) = Φ_N,j(z^N)^-1 = fracz^j/j! E_N,j(z). For j>0, the last quotient at z=0 is interpreted by its removable continuation, namely the first expression in the display. Write C_N,j(z) = ∑_n≥0C_N,j(n)fracz^Nn(Nn)!.
Exact mathematical passage
\Phi_{N,j}(w) = j!\sum_{n\geq0}\frac{w^n}{(Nn+j)!}.
\Phi_{N,j}Definition passage · 20 passages
Search aliases: Φ_N,j Phi
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1357
For 0≤ j<N, define Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)!. Then E_N,j(z) = fracz^jj!Φ_N,j(z^N). Define the residue-section reciprocal function C_N,j(z) = Φ_N,j(z^N)^-1 = fracz^j/j! E_N,j(z). For j>0, the last quotient at z=0 is interpreted by its removable continuation, namely the first expression in the display. Write C_N,j(z) = ∑_n≥0C_N,j(n)fracz^Nn(Nn)!.
Exact mathematical passage
\Phi_{N,j}(w) = j!\sum_{n\geq0}\frac{w^n}{(Nn+j)!}.
\ldotsDefinition passage · 23 passages
Search aliases: … ldots
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 799
Let J_N be the signed cyclic shift on the standard basis e_0,…,e_N-1 of C^N, defined by J_Ne_j=e_j+1 (0≤ j<N-1), J_Ne_N-1=-e_0. Then J_N^N=-I.
Exact mathematical passage
e_0,\ldots,e_{N-1}
\inDefinition passage · 43 passages
Search aliases: ∈ in
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 433
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\lambda\in\CC
\sumDefinition passage · 136 passages
Search aliases: ∑ sum
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
\sum_{m\geq0}Definition passage · 3 passages
Search aliases: ∑_m≥0 sum geq
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 1459
Define polynomials E_m^(N,j,r)(x) by C_N,j(z)^re^xz = ∑_m≥0E_m^(N,j,r)(x)fracz^mm!.
Exact mathematical passage
\Cfun_{N,j}(z)^re^{xz} = \sum_{m\geq0}E_m^{(N,j,r)}(x)\frac{z^m}{m!}.
\sum_{m\geq1}Definition passage · 13 passages
Search aliases: ∑_m≥1 sum geq
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 947
For r∈ C, define S_N^(r)(z) = D_N(z)^-r = exp(-rlog D_N(z)). Write S_N^(r)(z) = ∑_n≥0P_N,n(r)fracz^Nn(Nn)!. Define the logarithmic coefficients κ_N,m by -logΔ_N(w) = ∑_m≥1κ_N,mfracw^mm.
Exact mathematical passage
-\log\Delta_N(w) = \sum_{m\geq1}\kappa_{N,m}\frac{w^m}{m}.
\sum_{n\geq0}Definition passage · 19 passages
Search aliases: ∑_n≥0 sum geq
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
\sum_{\substack{\Dfun_N(\rho)=0\\ \Dfun_N'(\rho)\ne0}}Definition passage · 1 passage
Search aliases: ∑_substackDfun_N(ρ)=0 Dfun_N'(ρ)≠0 ∑_substack D_N(ρ)=0 D_N'(ρ)≠0 sum substack Dfun rho Dfun rho ne sum substack mathsf rho mathsf rho ne
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2900
Let q be a positive integer. Choose logarithms of the simple zeros of D_N with arguments in a fixed bounded interval. Define Z_N,q^der(s) = ∑_substack D_N(ρ)=0 D_N'(ρ)≠0 frace^-sLogρ D_N'(ρ)^q.
Exact mathematical passage
\mathfrak Z_{N,q}^{\mathrm{der}}(s) = \sum_{\substack{\Dfun_N(\rho)=0\\ \Dfun_N'(\rho)\ne0}} \frac{e^{-s\Log\rho}}{\Dfun_N'(\rho)^q}.
\sum_{\lambda\in\Lambda_N}Definition passage · 11 passages
Search aliases: ∑_λ∈Λ_N sum lambda in Lambda
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2413
For m≥1, define σ_N,m = ∑_λ∈Λ_Nλ^-m, with multiplicities.
Exact mathematical passage
\sigma_{N,m} = \sum_{\lambda\in\Lambda_N}\lambda^{-m},
\sum_{\rho\in\Zcal(\Dfun_N)}Definition passage · 1 passage
Search aliases: ∑_ρ∈Zcal(Dfun_N) ∑_ρ∈ Z( D_N) sum rho in Zcal Dfun sum rho in mathcal mathsf
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2718
Define Z_N^abs(s) = ∑_ρ∈ Z( D_N)|ρ|^-s and Z_N^red(s) = ∑_λ∈Λ_N|λ|^-s, with multiplicities. Each term means e^-slog|ρ| or e^-slog|λ|.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{abs}}(s) = \sum_{\rho\in\Zcal(\Dfun_N)}|\rho|^{-s}
\neDefinition passage · 13 passages
Search aliases: ≠ ne
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 2900
Let q be a positive integer. Choose logarithms of the simple zeros of D_N with arguments in a fixed bounded interval. Define Z_N,q^der(s) = ∑_substack D_N(ρ)=0 D_N'(ρ)≠0 frace^-sLogρ D_N'(ρ)^q.
Exact mathematical passage
\mathfrak Z_{N,q}^{\mathrm{der}}(s) = \sum_{\substack{\Dfun_N(\rho)=0\\ \Dfun_N'(\rho)\ne0}} \frac{e^{-s\Log\rho}}{\Dfun_N'(\rho)^q}.
\leqDefinition passage · 66 passages
Search aliases: ≤ leq
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 433
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
0\leq j<N
\geqDefinition passage · 106 passages
Search aliases: ≥ geq
Definition passage · Read passage · Paper_IV_Traditional_Final.tex, line 434
For λ∈ C and 0≤ j<N, define E_N,j(λ z) = B_N,j(e^λ z) = ∑_n≥0frac(λ z)^Nn+j(Nn+j)!.
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \Bcal_{N,j}(e^{\lambda z}) = \sum_{n\geq0}\frac{(\lambda z)^{Nn+j}}{(Nn+j)!}.
BIntroduction passage · 18 passages
Search aliases: B
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1596
For (N,j)=(2,0), C_2,0(z)=sech z. For (N,j)=(2,1), C_2,1(z)=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
B_m(x)
bIntroduction passage · 35 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1145
Let D_n denote the displayed determinant, with D_0=1. Expansion along the last row gives D_n = ∑_k=1^n frac(n-1)!(n-k)!p_kD_n-k. Here the cofactor sign is cancelled by the product of the negative superdiagonal entries -(n-k+1),…,-(n-1). Thus D_n/n! satisfies the same recurrence and initial value as b_N,n(r).
Exact mathematical passage
b_{N,n}(r)
B_mIntroduction passage · 5 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1596
For (N,j)=(2,0), C_2,0(z)=sech z. For (N,j)=(2,1), C_2,1(z)=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
B_m(x)
b_{N,n}Introduction passage · 7 passages
Search aliases: b_N,n
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1145
Let D_n denote the displayed determinant, with D_0=1. Expansion along the last row gives D_n = ∑_k=1^n frac(n-1)!(n-k)!p_kD_n-k. Here the cofactor sign is cancelled by the product of the negative superdiagonal entries -(n-k+1),…,-(n-1). Thus D_n/n! satisfies the same recurrence and initial value as b_N,n(r).
Exact mathematical passage
b_{N,n}(r)
\Cfun_{2,0}Introduction passage · 1 passage
Search aliases: Cfun_2,0 C_2,0 Cfun mathsf
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1583
For (N,j)=(2,0), C_2,0(z)=sech z. For (N,j)=(2,1), C_2,1(z)=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
\Cfun_{2,0}(z)=\sech z.
\Cfun_{2,1}Introduction passage · 1 passage
Search aliases: Cfun_2,1 C_2,1 Cfun mathsf
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1587
For (N,j)=(2,0), C_2,0(z)=sech z. For (N,j)=(2,1), C_2,1(z)=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
\Cfun_{2,1}(z)=\frac z{\sinh z}.
DIntroduction passage · 91 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1136
Let D_n denote the displayed determinant, with D_0=1. Expansion along the last row gives D_n = ∑_k=1^n frac(n-1)!(n-k)!p_kD_n-k. Here the cofactor sign is cancelled by the product of the negative superdiagonal entries -(n-k+1),…,-(n-1). Thus D_n/n! satisfies the same recurrence and initial value as b_N,n(r).
Exact mathematical passage
D_n
D_0Introduction passage · 2 passages
Search aliases: D_0
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1136
Let D_n denote the displayed determinant, with D_0=1. Expansion along the last row gives D_n = ∑_k=1^n frac(n-1)!(n-k)!p_kD_n-k. Here the cofactor sign is cancelled by the product of the negative superdiagonal entries -(n-k+1),…,-(n-1). Thus D_n/n! satisfies the same recurrence and initial value as b_N,n(r).
Exact mathematical passage
D_0=1
D_nIntroduction passage · 3 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1136
Let D_n denote the displayed determinant, with D_0=1. Expansion along the last row gives D_n = ∑_k=1^n frac(n-1)!(n-k)!p_kD_n-k. Here the cofactor sign is cancelled by the product of the negative superdiagonal entries -(n-k+1),…,-(n-1). Thus D_n/n! satisfies the same recurrence and initial value as b_N,n(r).
Exact mathematical passage
D_n
D_{n-k}Introduction passage · 1 passage
Search aliases: D_n-k
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1137
Let D_n denote the displayed determinant, with D_0=1. Expansion along the last row gives D_n = ∑_k=1^n frac(n-1)!(n-k)!p_kD_n-k. Here the cofactor sign is cancelled by the product of the negative superdiagonal entries -(n-k+1),…,-(n-1). Thus D_n/n! satisfies the same recurrence and initial value as b_N,n(r).
Exact mathematical passage
D_n = \sum_{k=1}^{n} \frac{(n-1)!}{(n-k)!}p_kD_{n-k}.
\Dfun_N^{(N)}Introduction passage · 1 passage
Search aliases: Dfun_N^(N) D_N^(N) Dfun mathsf
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1338
Write D_N=e^g locally. Then g'=- T_N. Divide the equation D_N^(N)=- D_N by D_N and use the defining identity for Y_N.
Exact mathematical passage
\Dfun_N^{(N)}=-\Dfun_N
\Dfun_N^{-r}Introduction passage · 4 passages
Search aliases: Dfun_N^-r D_N^-r Dfun mathsf
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 2590
For an entire function f, let Z(f) denote its multiset of nonzero zeros. Let r be a positive integer. Then D_N^-r is meromorphic and a simple zero of D_N becomes a pole of order r.
Exact mathematical passage
\Dfun_N^{-r}
e^gIntroduction passage · 5 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1338
Write D_N=e^g locally. Then g'=- T_N. Divide the equation D_N^(N)=- D_N by D_N and use the defining identity for Y_N.
Exact mathematical passage
\Dfun_N=e^g
e^tIntroduction passage · 3 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1604
=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
t e^{yt}/(e^t-1)
e^{yt}Introduction passage · 1 passage
Search aliases: e^yt
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1604
=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
t e^{yt}/(e^t-1)
E_mIntroduction passage · 10 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1590
For (N,j)=(2,0), C_2,0(z)=sech z. For (N,j)=(2,1), C_2,1(z)=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
E_m(x)
E_m^{(2,0,1)}Introduction passage · 1 passage
Search aliases: E_m^(2,0,1)
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1591
For (N,j)=(2,0), C_2,0(z)=sech z. For (N,j)=(2,1), C_2,1(z)=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
E_m^{(2,0,1)}(x) = 2^mE_m\left(\frac{x+1}{2}\right).
E_m^{(2,1,1)}Introduction passage · 1 passage
Search aliases: E_m^(2,1,1)
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1598
daries For (N,j)=(2,0), C_2,0(z)=sech z. For (N,j)=(2,1), C_2,1(z)=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
E_m^{(2,1,1)}(x) = 2^mB_m\left(\frac{x+1}{2}\right),
fIntroduction passage · 6 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 2590
For an entire function f, let Z(f) denote its multiset of nonzero zeros. Let r be a positive integer. Then D_N^-r is meromorphic and a simple zero of D_N becomes a pole of order r.
Exact mathematical passage
f
GIntroduction passage · 20 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 2035
The localization estimate puts every sufficiently large zero in the fundamental sector into one of these disks. The zero sets of G and D_N coincide because their prefactor is nonzero. Each disk is invariant under complex conjugation, and D_N has real coefficients. The conjugate of its unique zero is therefore the same zero. It is real, positive for large m, and simple by the derivative bound above. Denote it by τ_m.
Exact mathematical passage
G
gIntroduction passage · 10 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1338
Write D_N=e^g locally. Then g'=- T_N. Divide the equation D_N^(N)=- D_N by D_N and use the defining identity for Y_N.
Exact mathematical passage
\Dfun_N=e^g
g'Introduction passage · 4 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1338
Write D_N=e^g locally. Then g'=- T_N. Divide the equation D_N^(N)=- D_N by D_N and use the defining identity for Y_N.
Exact mathematical passage
g'=-\Tfun_N
h_nIntroduction passage · 2 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 2346
[Complete symmetric function form] For a positive integer r, let each eigenvalue λ^-1 of K_N be repeated r times. If h_n denotes the complete homogeneous symmetric function of degree n in this absolutely summable multiset, then fracP_N,n(r)(Nn)!=h_n. For arbitrary complex r, the Bell polynomial formula is the polynomial continuation obtained by multiplying every power sum by r.
Exact mathematical passage
h_n
KIntroduction passage · 14 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 2346
[Complete symmetric function form] For a positive integer r, let each eigenvalue λ^-1 of K_N be repeated r times. If h_n denotes the complete homogeneous symmetric function of degree n in this absolutely summable multiset, then fracP_N,n(r)(Nn)!=h_n. For arbitrary complex r, the Bell polynomial formula is the polynomial continuation obtained by multiplying every power sum by r.
Exact mathematical passage
K_N
K_NIntroduction passage · 8 passages
Search aliases: K_N
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 2346
[Complete symmetric function form] For a positive integer r, let each eigenvalue λ^-1 of K_N be repeated r times. If h_n denotes the complete homogeneous symmetric function of degree n in this absolutely summable multiset, then fracP_N,n(r)(Nn)!=h_n. For arbitrary complex r, the Bell polynomial formula is the polynomial continuation obtained by multiplying every power sum by r.
Exact mathematical passage
K_N
pIntroduction passage · 14 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1137
Let D_n denote the displayed determinant, with D_0=1. Expansion along the last row gives D_n = ∑_k=1^n frac(n-1)!(n-k)!p_kD_n-k. Here the cofactor sign is cancelled by the product of the negative superdiagonal entries -(n-k+1),…,-(n-1). Thus D_n/n! satisfies the same recurrence and initial value as b_N,n(r).
Exact mathematical passage
D_n = \sum_{k=1}^{n} \frac{(n-1)!}{(n-k)!}p_kD_{n-k}.
p_kIntroduction passage · 2 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1137
Let D_n denote the displayed determinant, with D_0=1. Expansion along the last row gives D_n = ∑_k=1^n frac(n-1)!(n-k)!p_kD_n-k. Here the cofactor sign is cancelled by the product of the negative superdiagonal entries -(n-k+1),…,-(n-1). Thus D_n/n! satisfies the same recurrence and initial value as b_N,n(r).
Exact mathematical passage
D_n = \sum_{k=1}^{n} \frac{(n-1)!}{(n-k)!}p_kD_{n-k}.
\sechIntroduction passage · 9 passages
Search aliases: sech operatorname
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1583
For (N,j)=(2,0), C_2,0(z)=sech z. For (N,j)=(2,1), C_2,1(z)=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
\Cfun_{2,0}(z)=\sech z.
\sinhIntroduction passage · 8 passages
Search aliases: sinh
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1587
For (N,j)=(2,0), C_2,0(z)=sech z. For (N,j)=(2,1), C_2,1(z)=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
\Cfun_{2,1}(z)=\frac z{\sinh z}.
tIntroduction passage · 64 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1603
n_2,1(z)=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
t=2z
YIntroduction passage · 13 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1338
Write D_N=e^g locally. Then g'=- T_N. Divide the equation D_N^(N)=- D_N by D_N and use the defining identity for Y_N.
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Y_N
yIntroduction passage · 24 passages
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Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1604
=frac zsinh z. The first case is a scaled and translated Euler family. In fact, if E_m(x) denotes the classical Euler polynomial, then E_m^(2,0,1)(x) = 2^mE_m(fracx+12). If B_m(x) denotes the classical Bernoulli polynomial, the second case has the exact identification E_m^(2,1,1)(x) = 2^mB_m(fracx+12), obtained by substituting t=2z in t e^yt/(e^t-1).
Exact mathematical passage
t e^{yt}/(e^t-1)
Y_NIntroduction passage · 4 passages
Search aliases: Y_N
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1338
Write D_N=e^g locally. Then g'=- T_N. Divide the equation D_N^(N)=- D_N by D_N and use the defining identity for Y_N.
Exact mathematical passage
Y_N
\lambda^{-1}Introduction passage · 4 passages
Search aliases: λ^-1 lambda
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 2346
[Complete symmetric function form] For a positive integer r, let each eigenvalue λ^-1 of K_N be repeated r times. If h_n denotes the complete homogeneous symmetric function of degree n in this absolutely summable multiset, then fracP_N,n(r)(Nn)!=h_n. For arbitrary complex r, the Bell polynomial formula is the polynomial continuation obtained by multiplying every power sum by r.
Exact mathematical passage
\lambda^{-1}
\tauIntroduction passage · 32 passages
Search aliases: τ tau
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 2040
sufficiently large zero in the fundamental sector into one of these disks. The zero sets of G and D_N coincide because their prefactor is nonzero. Each disk is invariant under complex conjugation, and D_N has real coefficients. The conjugate of its unique zero is therefore the same zero. It is real, positive for large m, and simple by the derivative bound above. Denote it by τ_m.
Exact mathematical passage
\tau_m
\tau_mIntroduction passage · 23 passages
Search aliases: τ_m tau
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 2040
sufficiently large zero in the fundamental sector into one of these disks. The zero sets of G and D_N coincide because their prefactor is nonzero. Each disk is invariant under complex conjugation, and D_N has real coefficients. The conjugate of its unique zero is therefore the same zero. It is real, positive for large m, and simple by the derivative bound above. Denote it by τ_m.
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\tau_m
\sum_{k=1}^{n}Introduction passage · 8 passages
Search aliases: ∑_k=1^n sum
Introduction passage · Read passage · Paper_IV_Traditional_Final.tex, line 1137
Let D_n denote the displayed determinant, with D_0=1. Expansion along the last row gives D_n = ∑_k=1^n frac(n-1)!(n-k)!p_kD_n-k. Here the cofactor sign is cancelled by the product of the negative superdiagonal entries -(n-k+1),…,-(n-1). Thus D_n/n! satisfies the same recurrence and initial value as b_N,n(r).
Exact mathematical passage
D_n = \sum_{k=1}^{n} \frac{(n-1)!}{(n-k)!}p_kD_{n-k}.
*Use in source · 7 passages
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2606
rest-pole coefficient expansion] Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
R_*
AUse in source · 40 passages
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 528
[Hypergeometric form] Let A_N,j = { fracj+mN mathrel: 1≤ m≤ N and j+m≠ N }. Then E_N,j(λ z) = frac(λ z)^jj! _0F_N-1 ( -
Exact mathematical passage
A_{N,j} = \left\{ \frac{j+m}{N} \mathrel{:} 1\leq m\leq N \text{ and } j+m\ne N \right\}.
aUse in source · 58 passages
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 214
For a power series F(z)=∑_m≥ 0a_mz^m and a positive integer N, the coefficients may be separated according to the residue of m modulo N. Root of unity averaging gives exact projection operators for this separation. This elementary construction is usually called a multisection or a root filter. It appears throughout the classical theory of generating functions and special functions. Background on multisections, hypergeometric functions, and related coefficient systems may be found in [source refe…
Exact mathematical passage
F(z)=\sum_{m\geq 0}a_mz^m
a_0Use in source · 1 passage
Search aliases: a_0
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2436
[Newton sums for the reduced zeros] For |w| below the least zero modulus, with the analytic logarithm normalized to zero at the origin, -logΔ_N(w) = ∑_m≥1σ_N,mfracw^mm. Thus κ_N,m=σ_N,m. If a_m=frac(-1)^m(Nm)!, a_0=1, then σ_N,m = -ma_m- ∑_k=1^m-1a_kσ_N,m-k.
Exact mathematical passage
a_m=\frac{(-1)^m}{(Nm)!}, \qquad a_0=1,
A_2Use in source · 1 passage
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2154
[The case N=2] For N=2, D_2(z)=cos z. Its zeros are ±(m+frac12)π, m≥0, and all are simple. They agree with the formula t_m=A_2(m+1/2).
Exact mathematical passage
t_m=A_2(m+1/2)
A_{3,1}Use in source · 1 passage
Search aliases: A_3,1
Use in source · hypergeometric.tex, line 18
,2.60000) (1)_2=2!; draw[fill=black!0,draw=black!40,line width=.35pt] (0.80000,1.35000) rectangle ++(1.60000,0.70000); node[] at (1.60000,1.70000) frac43; draw[fill=black!0,draw=black!40,line width=.35pt] (2.40000,1.35000) rectangle ++(1.60000,0.70000); node[] at (3.20000,1.70000) frac73; node[anchor=west] at (4.50000,1.70000) (4/3)_2; node[] at (9.30000,3.30000) A_3,1={2/3,4/3}; node[] at (9.30000,2.00000) displaystyle E_3,1(z)=z _0F_2 (;frac23,frac43;fracz^327); node[] at (6.30000,-0.05000) Th…
Exact mathematical passage
A_{3,1}=\{2/3,4/3\}
A_4Use in source · 8 passages
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2183
[Exact fourth-order strings] For N=4, D_4(z) = cos(frac zsqrt2) cosh(frac zsqrt2). The positive zeros are exactly τ_m = sqrt2π(m+frac12) = A_4(m+frac12). Rotation by the fourth roots of unity gives the real and imaginary zero strings. In this case the error term in [source reference] is identically zero.
Exact mathematical passage
\tau_m = \sqrt2\pi\left(m+\frac12\right) = A_4\left(m+\frac12\right).
A_4^{-4s}Use in source · 1 passage
Search aliases: A_4^-4s
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2847
[Exact zeta models in orders two and four] For N=2, Z_2^red(s) = π^-2sζ(2s,frac12) and Z_2^abs(s) = 2π^-sζ(s,frac12). For N=4, put A_4=sqrt2π. Then Z_4^red(s) = A_4^-4sζ(4s,frac12) and Z_4^abs(s) = 4A_4^-sζ(s,frac12). In these two cases the entire correction terms in [source reference] vanish identically.
Exact mathematical passage
\mathfrak Z_4^{\mathrm{red}}(s) = A_4^{-4s}\zeta\left(4s,\frac12\right)
A_4^{-s}Use in source · 1 passage
Search aliases: A_4^-s
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2853
[Exact zeta models in orders two and four] For N=2, Z_2^red(s) = π^-2sζ(2s,frac12) and Z_2^abs(s) = 2π^-sζ(s,frac12). For N=4, put A_4=sqrt2π. Then Z_4^red(s) = A_4^-4sζ(4s,frac12) and Z_4^abs(s) = 4A_4^-sζ(s,frac12). In these two cases the entire correction terms in [source reference] vanish identically.
Exact mathematical passage
\mathfrak Z_4^{\mathrm{abs}}(s) = 4A_4^{-s}\zeta\left(s,\frac12\right).
A_4^4Use in source · 5 passages
Search aliases: A_4^4
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2868
For N=2, the reduced zeros are π^2(m+frac12)^2, m≥0. For N=4, [source reference] gives the reduced zeros A_4^4(m+frac12)^4, m≥0. Summing their negative powers proves the reduced formulas. The absolute formulas follow from Z_N^abs(s) = N Z_N^red(s/N).
Exact mathematical passage
A_4^4\left(m+\frac12\right)^4, \qquad m\geq0.
a_kUse in source · 2 passages
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2441
[Newton sums for the reduced zeros] For |w| below the least zero modulus, with the analytic logarithm normalized to zero at the origin, -logΔ_N(w) = ∑_m≥1σ_N,mfracw^mm. Thus κ_N,m=σ_N,m. If a_m=frac(-1)^m(Nm)!, a_0=1, then σ_N,m = -ma_m- ∑_k=1^m-1a_kσ_N,m-k.
Exact mathematical passage
\sigma_{N,m} = -ma_m- \sum_{k=1}^{m-1}a_k\sigma_{N,m-k}.
a_mUse in source · 9 passages
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 214
For a power series F(z)=∑_m≥ 0a_mz^m and a positive integer N, the coefficients may be separated according to the residue of m modulo N. Root of unity averaging gives exact projection operators for this separation. This elementary construction is usually called a multisection or a root filter. It appears throughout the classical theory of generating functions and special functions. Background on multisections, hypergeometric functions, and related coefficient systems may be found in [source refe…
Exact mathematical passage
F(z)=\sum_{m\geq 0}a_mz^m
A_NUse in source · 4 passages
Search aliases: A_N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1833
The zero strings are governed by two adjacent exponents. Put a_N=fracπ N, s_N=sin a_N, c_N=cos a_N, and A_N=fracπs_N, t_m=A_N(m+frac12). For N≥3, also put γ_N=c_N-cos(3a_N)>0.
Exact mathematical passage
A_N=\frac\pi{s_N}, \qquad t_m=A_N\left(m+\frac12\right).
a_NUse in source · 23 passages
Search aliases: a_N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1825
The zero strings are governed by two adjacent exponents. Put a_N=fracπ N, s_N=sin a_N, c_N=cos a_N, and A_N=fracπs_N, t_m=A_N(m+frac12). For N≥3, also put γ_N=c_N-cos(3a_N)>0.
Exact mathematical passage
a_N=\frac\pi N, \qquad s_N=\sin a_N, \qquad c_N=\cos a_N,
a_nUse in source · 2 passages
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1728
[Order from Taylor coefficients] If f(w)=∑_n≥0a_nw^n is an entire function with infinitely many nonzero coefficients, then ord f = limsup_substackn→∞
Exact mathematical passage
f(w)=\sum_{n\geq0}a_nw^n
A_{N,j}Use in source · 7 passages
Search aliases: A_N,j
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 528
[Hypergeometric form] Let A_N,j = { fracj+mN mathrel: 1≤ m≤ N and j+m≠ N }. Then E_N,j(λ z) = frac(λ z)^jj! _0F_N-1 ( -
Exact mathematical passage
A_{N,j} = \left\{ \frac{j+m}{N} \mathrel{:} 1\leq m\leq N \text{ and } j+m\ne N \right\}.
A_{N,n}Use in source · 3 passages
Search aliases: A_N,n
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1061
[A sequence of binomial type] Define A_N,n(r) = fracn!(Nn)!P_N,n(r). Then A_N,n(r+s) = ∑_q=0^n binom nq A_N,q(r)A_N,n-q(s). Thus (A_N,n)_n≥0 is a polynomial sequence of binomial type.
Exact mathematical passage
A_{N,n}(r) = \frac{n!}{(Nn)!}P_{N,n}(r).
A_{N,n-q}Use in source · 1 passage
Search aliases: A_N,n-q
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1067
[A sequence of binomial type] Define A_N,n(r) = fracn!(Nn)!P_N,n(r). Then A_N,n(r+s) = ∑_q=0^n binom nq A_N,q(r)A_N,n-q(s). Thus (A_N,n)_n≥0 is a polynomial sequence of binomial type.
Exact mathematical passage
A_{N,n}(r+s) = \sum_{q=0}^{n} \binom nq A_{N,q}(r)A_{N,n-q}(s).
A_{N,q}Use in source · 1 passage
Search aliases: A_N,q
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1067
[A sequence of binomial type] Define A_N,n(r) = fracn!(Nn)!P_N,n(r). Then A_N,n(r+s) = ∑_q=0^n binom nq A_N,q(r)A_N,n-q(s). Thus (A_N,n)_n≥0 is a polynomial sequence of binomial type.
Exact mathematical passage
A_{N,n}(r+s) = \sum_{q=0}^{n} \binom nq A_{N,q}(r)A_{N,n-q}(s).
A_N^{-Ns}Use in source · 2 passages
Search aliases: A_N^-Ns
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2754
[Meromorphic continuation of the symmetry-orbit zero zeta] Put A_N=fracπsin(π/N). There is an entire function Q_N^red such that Z_N^red(s) = A_N^-Nsζ(Ns,frac12)+Q_N^red(s) on Re s>1/N. The right hand side gives a meromorphic continuation to C with one simple pole at s=1/N. Its residue is fracsin(π/N)Nπ.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{red}}(s) = A_N^{-Ns}\zeta\left(Ns,\frac12\right)+Q_N^{\mathrm{red}}(s)
A_N^{-s}Use in source · 1 passage
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2766
There is an entire function Q_N^abs such that, initially on Re s>1, Z_N^abs(s) = N A_N^-sζ(s,frac12)+Q_N^abs(s). Thus Z_N^abs continues meromorphically to C with one simple pole at s=1, of residue fracNsin(π/N)π.
Exact mathematical passage
\mathfrak Z_N^{\mathrm{abs}}(s) = N A_N^{-s}\zeta\left(s,\frac12\right)+Q_N^{\mathrm{abs}}(s).
a_qUse in source · 2 passages
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1526
[Formal differential-operator representation on polynomials] Let a formal series A(z)=∑_q≥0a_qz^q/q! act on polynomials by A(∂_x)f(x) = ∑_q≥0fraca_qq!∂_x^qf(x). The power r is applied to the formal symbol before the substitution z=∂_x. Then E_m^(N,j,r)(x) = C_N,j(∂_x)^r x^m. The inverse relation is x^m = Φ_N,j(∂_x^N)^r E_m^(N,j,r)(x). Both sums are finite on every polynomial.
Exact mathematical passage
A(z)=\sum_{q\geq0}a_qz^q/q!
a_{\rho,r}Use in source · 2 passages
Search aliases: a_ρ,r rho
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2620
_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
a_{\rho,r} = (-\rho\Dfun_N'(\rho))^{-r}.
a_{\rho,\ell}Use in source · 2 passages
Search aliases: a_ρ,ℓ rho ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2597
[Nearest-pole coefficient expansion] Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
\Dfun_N(z)^{-r} = \sum_{\ell=1}^{r} a_{\rho,\ell} \left(1-\frac z\rho\right)^{-\ell} + O(1).
a_\ellUse in source · 5 passages
Search aliases: a_ℓ ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 493
At the origin, the derivative matrix is diagonal with entries 1,λ,…,λ^N-1. For an explicit determinant proof, write each basis function as a finite sum of exponentials with exponents a_ℓ=λω^ℓ. Its derivative matrix factors as W(z)=V diag(e^a_ℓ z) C, V_kℓ=a_ℓ^k, where C is the constant matrix of Fourier weights. Thus det W(z)=e^z∑_ℓ a_ℓdet W(0)=det W(0), because N≥2 implies ∑_ℓω^ℓ=0. The product of the diagonal initial values is the stated power of λ.
Exact mathematical passage
a_\ell=\lambda\omega^\ell
a_\ell^kUse in source · 1 passage
Search aliases: a_ℓ^k ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 495
At the origin, the derivative matrix is diagonal with entries 1,λ,…,λ^N-1. For an explicit determinant proof, write each basis function as a finite sum of exponentials with exponents a_ℓ=λω^ℓ. Its derivative matrix factors as W(z)=V diag(e^a_ℓ z) C, V_kℓ=a_ℓ^k, where C is the constant matrix of Fourier weights. Thus det W(z)=e^z∑_ℓ a_ℓdet W(0)=det W(0), because N≥2 implies ∑_ℓω^ℓ=0. The product of the diagonal initial values is the stated power of λ.
Exact mathematical passage
W(z)=V\,\operatorname{diag}(e^{a_\ell z})\,C, \qquad V_{k\ell}=a_\ell^k,
\argUse in source · 2 passages
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1846
If z=x+iy lies in the sector |arg z|≤ a_N, has x≥ R, and is a zero of D_N, then |y|≤ Y.
Exact mathematical passage
|\arg z|\leq a_N
\mathcal BUse in source · 7 passages
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Use in source · atlas-pages.tex, line 46
vpage3A root filter is an exact voteCYCLIC PROJECTIONTraditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5). For N=4, write ω=i. To extract residue j, evaluate the function at four rotated arguments and attach the phase i^-jℓ: ( B_4,jF)(z)=frac14∑_ℓ=0^3i^-jℓF(i^ℓ z). A monomial a_mz^m is multiplied by the finite average of i^(m-j)ℓ. Matching phases all vote 1. Nonmatching phases cancel exactly. vfigroot_filter headAn example that can be checked coefficient by coefficient Take F(z)=∑_m…
Exact mathematical passage
(\mathcal B_{4,j}F)(z)=\frac14\sum_{\ell=0}^{3}i^{-j\ell}F(i^\ell z).
b_0Use in source · 2 passages
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Use in source · atlas-pages.tex, line 199
vpage11Two recurrences reach the same answerNEWTON AND ADDITIONTraditional monograph: Theorem 7.3 (p. 14), Theorem 7.4 (p. 15), Theorem 7.6 (p. 15). One recurrence uses the denominator coefficients. Another uses its logarithm. With b_n(r)=P_N,n(r)/(Nn)!, the shorter logarithmic form is nb_n(r)=r∑_k=1^nκ_N,kb_n-k(r), b_0=1. The negative superdiagonal in the Newton determinant is essential. vfignewton_addition headThe additive parameter law has block weights Using the same logarithm, D_N^-rD_N^-s=…
Exact mathematical passage
nb_n(r)=r\sum_{k=1}^{n}\kappa_{N,k}b_{n-k}(r),\qquad b_0=1.
b_1Use in source · 6 passages
Search aliases: b_1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1226
[title=A coefficient calculation at order two] Write Δ_2(w)^-r=∑_n≥0b_n(r)w^n. Since Δ_2(w)=1-frac w2+fracw^224+O(w^3), -logΔ_2(w)=frac w2+fracw^212+O(w^3), the first logarithmic coefficients are κ_2,1=1/2 and κ_2,2=1/6. Newton's recurrence gives b_1(r)=frac r2, 2b_2(r)=r(frac r4+frac16). Restoring the factorial normalization yields P_2,1(r)=r and P_2,2(r)=r(3r+2). At r=1, the second value is five. No zero-location assumption is needed for this finite calculation. modelfigureparameter_polynomial…
Exact mathematical passage
b_1(r)=\frac r2,\qquad 2b_2(r)=r\left(\frac r4+\frac16\right).
b_2Use in source · 6 passages
Search aliases: b_2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1226
[title=A coefficient calculation at order two] Write Δ_2(w)^-r=∑_n≥0b_n(r)w^n. Since Δ_2(w)=1-frac w2+fracw^224+O(w^3), -logΔ_2(w)=frac w2+fracw^212+O(w^3), the first logarithmic coefficients are κ_2,1=1/2 and κ_2,2=1/6. Newton's recurrence gives b_1(r)=frac r2, 2b_2(r)=r(frac r4+frac16). Restoring the factorial normalization yields P_2,1(r)=r and P_2,2(r)=r(3r+2). At r=1, the second value is five. No zero-location assumption is needed for this finite calculation. modelfigureparameter_polynomial…
Exact mathematical passage
b_1(r)=\frac r2,\qquad 2b_2(r)=r\left(\frac r4+\frac16\right).
B_3Use in source · 3 passages
Search aliases: B_3
Use in source · atlas-pages.tex, line 184
r=exp (r∑_m≥1κ_N,mfracw^mm), the coefficient of w^n is a polynomial of degree n in r. Bell polynomials collect the possible ways logarithmic terms contribute total degree n. vfigbell_partitions For N=2, the first three logarithmic coefficients are κ_1=frac12, κ_2=frac16, κ_3=frac115. The third Bell polynomial is B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3. Substituting x_1=rκ_1, x_2=rκ_2, x_3=2rκ_3 and multiplying by 6!/3! gives P_2,3(r)=15r^3+30r^2+16r. At r=1 this is 61. At r=2 it is 272. At r=0 it van…
Exact mathematical passage
B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3
b_3Use in source · 5 passages
Search aliases: b_3
Use in source · atlas-pages.tex, line 163
FICIENT RECONSTRUCTIONTraditional monograph: Theorem 7.3 (p. 14), Theorem 7.6 (p. 15). Start in the reduced variable with Δ_2(w)=1-frac w2+fracw^224-fracw^3720+⋯. Write 1/Δ_2(w)=∑ b_nw^n. Matching the product with 1 gives a unit lower-triangular system. At each row there is exactly one new unknown. vfigtriangular_inverse The first three equations are b_1-tfrac12=0, b_2-tfrac12b_1+tfrac124=0, b_3-tfrac12b_2+tfrac124b_1-tfrac1720=0. They yield b_1=1/2, b_2=5/24, and b_3=61/720. Multiplication by (…
Exact mathematical passage
b_1-\tfrac12=0,\qquad b_2-\tfrac12b_1+\tfrac1{24}=0, \qquad b_3-\tfrac12b_2+\tfrac1{24}b_1-\tfrac1{720}=0.
\mathcal B_{3,0}Use in source · 1 passage
Search aliases: B_3,0 mathcal
Use in source · cyclic_product.tex, line 26
0,1.25000) rectangle ++(2.65000,0.85000); node[] at (3.72500,1.67500) 6 z^2; draw[fill=black!18,draw=black!40,line width=.35pt] (5.05000,1.25000) rectangle ++(2.65000,0.85000); node[] at (6.37500,1.67500) 3 z^3; draw[fill=black!0,draw=black!40,line width=.35pt] (7.70000,1.25000) rectangle ++(2.65000,0.85000); node[] at (9.02500,1.67500) 3 z^4; node[] at (6.20000,0.25000) B_3,0(FG)=2+7z^3; node[] at (6.20000,-0.40000) shaded cells: (0,0), (1,2), (2,1) all land in residue 0;
Exact mathematical passage
\mathcal B_{3,0}(FG)=2+7z^3
\mathcal B_{4,2}Use in source · 2 passages
Search aliases: B_4,2 mathcal
Use in source · atlas-pages.tex, line 53
r rotated arguments and attach the phase i^-jℓ: ( B_4,jF)(z)=frac14∑_ℓ=0^3i^-jℓF(i^ℓ z). A monomial a_mz^m is multiplied by the finite average of i^(m-j)ℓ. Matching phases all vote 1. Nonmatching phases cancel exactly. vfigroot_filter headAn example that can be checked coefficient by coefficient Take F(z)=∑_m=0^12(m+1)z^m. Then B_4,2F=3z^2+7z^6+11z^10. Applying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the ori…
Exact mathematical passage
\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.
\mathcal B_{4,j}Use in source · 2 passages
Search aliases: B_4,j mathcal
Use in source · atlas-pages.tex, line 46
vpage3A root filter is an exact voteCYCLIC PROJECTIONTraditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5). For N=4, write ω=i. To extract residue j, evaluate the function at four rotated arguments and attach the phase i^-jℓ: ( B_4,jF)(z)=frac14∑_ℓ=0^3i^-jℓF(i^ℓ z). A monomial a_mz^m is multiplied by the finite average of i^(m-j)ℓ. Matching phases all vote 1. Nonmatching phases cancel exactly. vfigroot_filter headAn example that can be checked coefficient by coefficient Take F(z)=∑_m…
Exact mathematical passage
(\mathcal B_{4,j}F)(z)=\frac14\sum_{\ell=0}^{3}i^{-j\ell}F(i^\ell z).
B_nUse in source · 7 passages
Search aliases: B_n
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 954
The polynomial dependence on r can be read directly from the formal logarithm. We use the complete exponential Bell polynomials B_n defined by exp(∑_m≥1x_mfract^mm!) = ∑_n≥0B_n(x_1,…,x_n)fract^nn!.
Exact mathematical passage
B_n
b_nUse in source · 9 passages
Search aliases: b_n
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1220
[title=A coefficient calculation at order two] Write Δ_2(w)^-r=∑_n≥0b_n(r)w^n. Since Δ_2(w)=1-frac w2+fracw^224+O(w^3), -logΔ_2(w)=frac w2+fracw^212+O(w^3), the first logarithmic coefficients are κ_2,1=1/2 and κ_2,2=1/6. Newton's recurrence gives b_1(r)=frac r2, 2b_2(r)=r(frac r4+frac16). Restoring the factorial normalization yields P_2,1(r)=r and P_2,2(r)=r(3r+2). At r=1, the second value is five. No zero-location assumption is needed for this finite calculation. modelfigureparameter_polynomial…
Exact mathematical passage
\Delta_2(w)^{-r}=\sum_{n\geq0}b_n(r)w^n
b_{N,0}Use in source · 1 passage
Search aliases: b_N,0
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1089
[Logarithmic Newton recurrence] Set b_N,n(r)=fracP_N,n(r)(Nn)!. Then b_N,0(r)=1 and for every n≥1, nb_N,n(r) = r∑_k=1^nκ_N,kb_N,n-k(r). Equivalently, P_N,n(r) = frac r n ∑_k=1^n frac(Nn)!(N(n-k))! κ_N,kP_N,n-k(r). If p_k=rκ_N,k, then, for n≥1, b_N,n(r) = frac1n! det p_1 · -1 · 0 · ⋯ · 0
Exact mathematical passage
b_{N,0}(r)=1
\mathcal B_{N,i}Use in source · 2 passages
Search aliases: B_N,i mathcal
Use in source · atlas-pages.tex, line 76
1+2z+3z^2+z^3, G=2+z+z^2. The table displays every product of their three selected sections. vfigcyclic_product The shaded cells contribute to residue zero: 2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3. Keeping only the product of the residue-zero inputs would give 2+2z^3 and lose five units of the z^3 coefficient. The mixed cells are part of the exact result. headThe general rule B_N,k(FG)= ∑_i+j≡ k (N)( B_N,iF)( B_N,jG). Every pair belongs to exactly one output residue. No term is discarded and no coeffi…
Exact mathematical passage
\mathcal B_{N,k}(FG)= \sum_{i+j\equiv k\ (N)}(\mathcal B_{N,i}F)(\mathcal B_{N,j}G).
\mathcal B_{N,j}Use in source · 2 passages
Search aliases: B_N,j mathcal
Use in source · atlas-pages.tex, line 76
1+2z+3z^2+z^3, G=2+z+z^2. The table displays every product of their three selected sections. vfigcyclic_product The shaded cells contribute to residue zero: 2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3. Keeping only the product of the residue-zero inputs would give 2+2z^3 and lose five units of the z^3 coefficient. The mixed cells are part of the exact result. headThe general rule B_N,k(FG)= ∑_i+j≡ k (N)( B_N,iF)( B_N,jG). Every pair belongs to exactly one output residue. No term is discarded and no coeffi…
Exact mathematical passage
\mathcal B_{N,k}(FG)= \sum_{i+j\equiv k\ (N)}(\mathcal B_{N,i}F)(\mathcal B_{N,j}G).
\mathcal B_{N,k}Use in source · 2 passages
Search aliases: B_N,k mathcal
Use in source · atlas-pages.tex, line 76
1+2z+3z^2+z^3, G=2+z+z^2. The table displays every product of their three selected sections. vfigcyclic_product The shaded cells contribute to residue zero: 2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3. Keeping only the product of the residue-zero inputs would give 2+2z^3 and lose five units of the z^3 coefficient. The mixed cells are part of the exact result. headThe general rule B_N,k(FG)= ∑_i+j≡ k (N)( B_N,iF)( B_N,jG). Every pair belongs to exactly one output residue. No term is discarded and no coeffi…
Exact mathematical passage
\mathcal B_{N,k}(FG)= \sum_{i+j\equiv k\ (N)}(\mathcal B_{N,i}F)(\mathcal B_{N,j}G).
b_{N,n-k}Use in source · 1 passage
Search aliases: b_N,n-k
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1091
[Logarithmic Newton recurrence] Set b_N,n(r)=fracP_N,n(r)(Nn)!. Then b_N,0(r)=1 and for every n≥1, nb_N,n(r) = r∑_k=1^nκ_N,kb_N,n-k(r). Equivalently, P_N,n(r) = frac r n ∑_k=1^n frac(Nn)!(N(n-k))! κ_N,kP_N,n-k(r). If p_k=rκ_N,k, then, for n≥1, b_N,n(r) = frac1n! det p_1 · -1 · 0 · ⋯ · 0
Exact mathematical passage
nb_{N,n}(r) = r\sum_{k=1}^{n}\kappa_{N,k}b_{N,n-k}(r).
b_{n-k}Use in source · 2 passages
Search aliases: b_n-k
Use in source · atlas-pages.tex, line 199
vpage11Two recurrences reach the same answerNEWTON AND ADDITIONTraditional monograph: Theorem 7.3 (p. 14), Theorem 7.4 (p. 15), Theorem 7.6 (p. 15). One recurrence uses the denominator coefficients. Another uses its logarithm. With b_n(r)=P_N,n(r)/(Nn)!, the shorter logarithmic form is nb_n(r)=r∑_k=1^nκ_N,kb_n-k(r), b_0=1. The negative superdiagonal in the Newton determinant is essential. vfignewton_addition headThe additive parameter law has block weights Using the same logarithm, D_N^-rD_N^-s=…
Exact mathematical passage
nb_n(r)=r\sum_{k=1}^{n}\kappa_{N,k}b_{n-k}(r),\qquad b_0=1.
B_rUse in source · 2 passages
Search aliases: B_r
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1123
Let B_r(w)=Δ_N(w)^-r = ∑_n≥0b_N,n(r)w^n. Logarithmic differentiation gives fracB_r'(w)B_r(w) = r∑_k≥1κ_N,kw^k-1. Comparison of the coefficient of w^n-1 proves the recurrence.
Exact mathematical passage
B_r(w)=\Delta_N(w)^{-r} = \sum_{n\geq0}b_{N,n}(r)w^n.
B_r'Use in source · 1 passage
Search aliases: B_r'
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1129
Let B_r(w)=Δ_N(w)^-r = ∑_n≥0b_N,n(r)w^n. Logarithmic differentiation gives fracB_r'(w)B_r(w) = r∑_k≥1κ_N,kw^k-1. Comparison of the coefficient of w^n-1 proves the recurrence.
Exact mathematical passage
\frac{B_r'(w)}{B_r(w)} = r\sum_{k\geq1}\kappa_{N,k}w^{k-1}.
\Bcal_{N,i}Use in source · 3 passages
Search aliases: Bcal_N,i B_N,i Bcal mathcal
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 358
[Root filter] If F(z)=∑_m≥0a_mz^m, then B_N,jF(z) = ∑_m≡ jbmod Na_mz^m. The operators satisfy B_N,i B_N,j=δ_ij B_N,j and ∑_j=0^N-1 B_N,j=I.
Exact mathematical passage
\Bcal_{N,i}\Bcal_{N,j}=\delta_{ij}\Bcal_{N,j}
\Bcal_{N,k}Use in source · 7 passages
Search aliases: Bcal_N,k B_N,k Bcal mathcal
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 388
Every monomial selected by B_N,j has the stated covariance. Conversely, write F=∑_k B_N,kF. Replacing z by ω z gives ∑_kω^k B_N,kF = ω^j∑_k B_N,kF. Application of B_N,k gives (ω^k-ω^j) B_N,kF=0. The powers 1,ω,…,ω^N-1 are distinct, so B_N,kF=0 for k≠ j.
Exact mathematical passage
F=\sum_k\Bcal_{N,k}F
cUse in source · 11 passages
Search aliases: c
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1817
At a tie direction exactly two adjacent vertices attain the maximum. After rotation it is enough to take θ=0. With c_N=cos(π/N) and s_N=sin(π/N), the two dominant terms give frac2N e^c_Nrcos(s_Nr) and every other term is exponentially smaller. Along a sequence on which |cos(s_Nr)|≥1/2, the lower indicator bound is c_N=h_N(0). Rotational symmetry treats every tie direction.
Exact mathematical passage
c_N=\cos(\pi/N)
C_1Use in source · 1 passage
Search aliases: C_1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1992
If G(z)=0, then |1+e^2is_Nz| ≤ C e^-γ_NRe z. The real part of 2is_Nz is bounded in the strip. Apply [source reference] with w=2is_Nz. There is a unique integer m such that |z-t_m|≤ C_1e^-γ_NRe z, where t_m=frac(2m+1)π2s_N. In particular, |Re z-t_m|=o(1). For all sufficiently large zeros we have |Re z-t_m|≤1, and hence |z-t_m|≤ C_2e^-γ_Nt_m with a constant independent of m. Because Re z→+∞ in this argument, the corresponding integers m are eventually nonnegative and tend to infinity.
Exact mathematical passage
|z-t_m|\leq C_1e^{-\gamma_N\Re z},
C_2Use in source · 1 passage
Search aliases: C_2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2001
G(z)=0, then |1+e^2is_Nz| ≤ C e^-γ_NRe z. The real part of 2is_Nz is bounded in the strip. Apply [source reference] with w=2is_Nz. There is a unique integer m such that |z-t_m|≤ C_1e^-γ_NRe z, where t_m=frac(2m+1)π2s_N. In particular, |Re z-t_m|=o(1). For all sufficiently large zeros we have |Re z-t_m|≤1, and hence |z-t_m|≤ C_2e^-γ_Nt_m with a constant independent of m. Because Re z→+∞ in this argument, the corresponding integers m are eventually nonnegative and tend to infinity.
Exact mathematical passage
|z-t_m|\leq C_2e^{-\gamma_Nt_m}
C_{2,0}Use in source · 4 passages
Search aliases: C_2,0
Use in source · atlas-pages.tex, line 244
vpage14Three similar reciprocals differNORMALIZED SECTIONSTraditional monograph: Definition 9.1 (p. 19), Theorem 9.2 (p. 19), Example 9.11 (p. 22). The signed denominator and the normalized unsigned sections produce different reciprocal families. In order two, D_2^-1=sec z, C_2,0=sech z, C_2,1=z/sinh z. Their first block-EGF coefficients make the difference visible. vfigdifferent_reciprocals headRemove the forced zero before taking a reciprocal The section E_N,j begins with z^j/j!. Define E_N,j(…
Exact mathematical passage
D_2^{-1}=\sec z,\qquad C_{2,0}=\sech z,\qquad C_{2,1}=z/\sinh z.
C_{2,1}Use in source · 3 passages
Search aliases: C_2,1
Use in source · atlas-pages.tex, line 244
vpage14Three similar reciprocals differNORMALIZED SECTIONSTraditional monograph: Definition 9.1 (p. 19), Theorem 9.2 (p. 19), Example 9.11 (p. 22). The signed denominator and the normalized unsigned sections produce different reciprocal families. In order two, D_2^-1=sec z, C_2,0=sech z, C_2,1=z/sinh z. Their first block-EGF coefficients make the difference visible. vfigdifferent_reciprocals headRemove the forced zero before taking a reciprocal The section E_N,j begins with z^j/j!. Define E_N,j(…
Exact mathematical passage
D_2^{-1}=\sec z,\qquad C_{2,0}=\sech z,\qquad C_{2,1}=z/\sinh z.
C_LUse in source · 2 passages
Search aliases: C_L
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2924
For N≥3, [source reference] gives | D_N'(ρ)|^-1 ≤ C e^-c_N|ρ| for every sufficiently large simple zero. Let L be a compact subset of the s plane. Because the chosen arguments of the zeros lie in a fixed bounded interval, there are constants C_L and M_L such that sup_s∈ L|e^-sLogρ| ≤ C_L|ρ|^M_L for every sufficiently large zero. Hence the absolute value of the corresponding summand is at most C_L'|ρ|^M_Le^-q c_N|ρ|. The linear zero counting estimate makes this majorant summable. The Weierstrass t…
Exact mathematical passage
C_L
C_L'Use in source · 1 passage
Search aliases: C_L'
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2931
-1 ≤ C e^-c_N|ρ| for every sufficiently large simple zero. Let L be a compact subset of the s plane. Because the chosen arguments of the zeros lie in a fixed bounded interval, there are constants C_L and M_L such that sup_s∈ L|e^-sLogρ| ≤ C_L|ρ|^M_L for every sufficiently large zero. Hence the absolute value of the corresponding summand is at most C_L'|ρ|^M_Le^-q c_N|ρ|. The linear zero counting estimate makes this majorant summable. The Weierstrass test gives absolute local uniform convergence …
Exact mathematical passage
C_L'|\rho|^{M_L}e^{-q c_N|\rho|}.
C_MUse in source · 3 passages
Search aliases: C_M
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1911
[Near cancellation of two exponentials] For every M>0, there are constants ε_M>0 and C_M>0 such that the following holds. If |Re w|≤ M and |1+e^w|≤ε_M, then there is a unique integer m with |w-(2m+1)π i| ≤ C_M|1+e^w|.
Exact mathematical passage
C_M>0
c_NUse in source · 11 passages
Search aliases: c_N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1817
At a tie direction exactly two adjacent vertices attain the maximum. After rotation it is enough to take θ=0. With c_N=cos(π/N) and s_N=sin(π/N), the two dominant terms give frac2N e^c_Nrcos(s_Nr) and every other term is exponentially smaller. Along a sequence on which |cos(s_Nr)|≥1/2, the lower indicator bound is c_N=h_N(0). Rotational symmetry treats every tie direction.
Exact mathematical passage
c_N=\cos(\pi/N)
C_{N,j,n-q}Use in source · 1 passage
Search aliases: C_N,j,n-q
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1430
[Parameter-addition law for residue-section reciprocals] For each n≥1, the coefficient C_N,j,n(r) is a polynomial in r of exact degree n. Its leading coefficient is frac(Nn)!n! (-fracj!(N+j)!)^n. It satisfies C_N,j,n(r+s) = ∑_q=0^n binomNnNq C_N,j,q(r)C_N,j,n-q(s). In particular, C_N,j,n(1)=C_N,j(n).
Exact mathematical passage
C_{N,j,n}(r+s) = \sum_{q=0}^{n} \binom{Nn}{Nq} C_{N,j,q}(r)C_{N,j,n-q}(s).
C_{N,j,q}Use in source · 1 passage
Search aliases: C_N,j,q
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1430
[Parameter-addition law for residue-section reciprocals] For each n≥1, the coefficient C_N,j,n(r) is a polynomial in r of exact degree n. Its leading coefficient is frac(Nn)!n! (-fracj!(N+j)!)^n. It satisfies C_N,j,n(r+s) = ∑_q=0^n binomNnNq C_N,j,q(r)C_N,j,n-q(s). In particular, C_N,j,n(1)=C_N,j(n).
Exact mathematical passage
C_{N,j,n}(r+s) = \sum_{q=0}^{n} \binom{Nn}{Nq} C_{N,j,q}(r)C_{N,j,n-q}(s).
\Cfun_{N,j}^rUse in source · 2 passages
Search aliases: Cfun_N,j^r C_N,j^r Cfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1499
Multiply the block series for C_N,j^r by the exponential series and collect the coefficient of total degree m.
Exact mathematical passage
\Cfun_{N,j}^r
\coshUse in source · 10 passages
Search aliases: cosh
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 569
For N=1, the only section is e^λ z. For N=2, E_2,0(z)=cosh z, E_2,1(z)=sinh z. These are classical instances of the same cyclic basis.
Exact mathematical passage
\Efun_{2,0}(z)=\cosh z, \qquad \Efun_{2,1}(z)=\sinh z.
\mathbf DUse in source · 5 passages
Search aliases: D mathbf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 811
[Companion matrix realization] Put D_N(z) = D_N,0(z) D_N,1(z) vdots D_N,N-1(z) . Then D_N(z)=e^zJ_Ne_0 and D_N(z)=frac1Ntr(e^zJ_N). The eigenvalues of J_N are the N roots of -1, each with multiplicity one.
Exact mathematical passage
\mathbf D_N(z) = \begin{pmatrix} \Dfun_{N,0}(z)\\ \Dfun_{N,1}(z)\\ \vdots\\ \Dfun_{N,N-1}(z) \end{pmatrix}.
dUse in source · 20 passages
Search aliases: d
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 445
[Differential cycle and initial data] For 1≤ j<N, fracddz E_N,j(λ z) = λ E_N,j-1(λ z), while fracddz E_N,0(λ z) = λ E_N,N-1(λ z). Thus fracd^Ndz^N E_N,j(λ z) = λ^N E_N,j(λ z). For 0≤ k<N, .fracd^kdz^k E_N,j(λ z)|_z=0 = δ_kjλ^j. If λ≠0, the N functions E_N,j(λ z) form a basis of the solution space of y^(N)=λ^Ny.
Exact mathematical passage
\frac{d}{dz}\Efun_{N,j}(\lambda z) = \lambda\Efun_{N,j-1}(\lambda z),
D'''Use in source · 4 passages
Search aliases: D'''
Use in source · atlas-pages.tex, line 224
retain the reduced logarithmic information: T_N(z)=N∑_m≥1κ_N,mz^Nm-1, T_N(n)=N(Nn+N-1)!κ_N,n+1. For order four, κ_4,1=1/24 and κ_4,2=17/10080. These give T_4(0)=1 and T_4(1)=34, explaining the last line of the plate. headA nonlinear equation with a precise origin If D=e^g locally, then g'=-T. Differentiating e^g three times gives D'''/D=-T''+3TT'-T^3. Setting D'''/D=-1 yields the cubic Riccati identity shown above. No additional nonlinear model has been assumed. The reciprocal family also satisf…
Exact mathematical passage
D'''/D=-T''+3TT'-T^3
D^{-(r+s)}Use in source · 2 passages
Search aliases: D^-(r+s)
Use in source · atlas-pages.tex, line 238
2π i, and the power by e^-2π ir. headThree different kinds of statement A formal identity compares coefficients at the origin. A holomorphic identity for complex powers uses a common specified logarithm. A positive integer reciprocal power is globally meromorphic and needs no branch choice. These scopes cannot be exchanged without a further argument. For example, D^-rD^-s=D^-(r+s) is immediate using one fixed logarithm. Combining values built from unrelated branches can introduce additional phas…
Exact mathematical passage
D^{-r}D^{-s}=D^{-(r+s)}
D^{-r}Use in source · 3 passages
Search aliases: D^-r
Use in source · atlas-pages.tex, line 238
2π i, and the power by e^-2π ir. headThree different kinds of statement A formal identity compares coefficients at the origin. A holomorphic identity for complex powers uses a common specified logarithm. A positive integer reciprocal power is globally meromorphic and needs no branch choice. These scopes cannot be exchanged without a further argument. For example, D^-rD^-s=D^-(r+s) is immediate using one fixed logarithm. Combining values built from unrelated branches can introduce additional phas…
Exact mathematical passage
D^{-r}D^{-s}=D^{-(r+s)}
D^{-s}Use in source · 2 passages
Search aliases: D^-s
Use in source · atlas-pages.tex, line 238
2π i, and the power by e^-2π ir. headThree different kinds of statement A formal identity compares coefficients at the origin. A holomorphic identity for complex powers uses a common specified logarithm. A positive integer reciprocal power is globally meromorphic and needs no branch choice. These scopes cannot be exchanged without a further argument. For example, D^-rD^-s=D^-(r+s) is immediate using one fixed logarithm. Combining values built from unrelated branches can introduce additional phas…
Exact mathematical passage
D^{-r}D^{-s}=D^{-(r+s)}
d^kUse in source · 1 passage
Search aliases: d^k
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 463
[Differential cycle and initial data] For 1≤ j<N, fracddz E_N,j(λ z) = λ E_N,j-1(λ z), while fracddz E_N,0(λ z) = λ E_N,N-1(λ z). Thus fracd^Ndz^N E_N,j(λ z) = λ^N E_N,j(λ z). For 0≤ k<N, .fracd^kdz^k E_N,j(λ z)|_z=0 = δ_kjλ^j. If λ≠0, the N functions E_N,j(λ z) form a basis of the solution space of y^(N)=λ^Ny.
Exact mathematical passage
\left.\frac{d^k}{dz^k}\Efun_{N,j}(\lambda z)\right|_{z=0} = \delta_{kj}\lambda^j.
d^NUse in source · 2 passages
Search aliases: d^N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 457
[Differential cycle and initial data] For 1≤ j<N, fracddz E_N,j(λ z) = λ E_N,j-1(λ z), while fracddz E_N,0(λ z) = λ E_N,N-1(λ z). Thus fracd^Ndz^N E_N,j(λ z) = λ^N E_N,j(λ z). For 0≤ k<N, .fracd^kdz^k E_N,j(λ z)|_z=0 = δ_kjλ^j. If λ≠0, the N functions E_N,j(λ z) form a basis of the solution space of y^(N)=λ^Ny.
Exact mathematical passage
\frac{d^N}{dz^N}\Efun_{N,j}(\lambda z) = \lambda^N\Efun_{N,j}(\lambda z).
D^{\prime}Use in source · 1 passage
Search aliases: D^prime prime
Use in source · branch_domain.tex, line 11
draw[fill=black] (2.80000,2.30000) circle (0.055); node[] at (2.80000,4.65000) fixed-log domain about 0; draw[fill=white,line width=1pt] (4.40000,2.30000) circle (0.09); node[] at (5.10000,1.45000) ρ=π/2; draw[->,line width=1.1pt] (4.88,2.3) arc[start angle=0,end angle=330,radius=.48]; node[] at (2.80000,0.10000) the loop is not inside the disk; node[] at (9.20000,4.10000) D(z)∼ D^prime(ρ)(z-ρ); node[] at (9.20000,2.90000) log D⟼log D+2π i; node[] at (9.20000,1.65000) D^-r⟼ e^-2π i rD^-r; node[]…
Exact mathematical passage
D(z)\sim D^{\prime}(\rho)(z-\rho)
D_{0}Use in source · 3 passages
Search aliases: D_0
Use in source · signed_shift.tex, line 23
pt] (1.40000,0.90000) rectangle ++(0.80000,0.80000); node[] at (1.80000,1.30000) 1; draw[fill=black!0,draw=black!40,line width=.35pt] (2.20000,0.90000) rectangle ++(0.80000,0.80000); node[] at (2.60000,1.30000) 0; node[] at (6.67500,3.78000) e^zJ_3; draw[fill=black!12,draw=black!40,line width=.35pt] (5.10000,2.25000) rectangle ++(1.05000,1.05000); node[] at (5.62500,2.77500) D_0; draw[fill=black!12,draw=black!40,line width=.35pt] (6.15000,2.25000) rectangle ++(1.05000,1.05000); node[] at (6.6750…
Exact mathematical passage
D_{0}
D_1Use in source · 1 passage
Search aliases: D_1
Use in source · signed_shift.tex, line 43
15000) rectangle ++(1.05000,1.05000); node[] at (6.67500,0.67500) D_1; draw[fill=black!12,draw=black!40,line width=.35pt] (7.20000,0.15000) rectangle ++(1.05000,1.05000); node[] at (7.72500,0.67500) D_0; node[] at (10.80000,2.90000) J_3^3=-I; node[] at (10.80000,1.70000) D_3(0)=e_0; node[] at (10.80000,0.60000) det e^zJ_3=1; node[] at (6.10000,-0.80000) D_0↦-D_2, D_1↦ D_0, D_2↦ D_1;
Exact mathematical passage
D_0\mapsto-D_2,\qquad D_1\mapsto D_0,\qquad D_2\mapsto D_1
D_{1}Use in source · 3 passages
Search aliases: D_1
Use in source · signed_shift.tex, line 27
!12,draw=black!40,line width=.35pt] (5.10000,2.25000) rectangle ++(1.05000,1.05000); node[] at (5.62500,2.77500) D_0; draw[fill=black!12,draw=black!40,line width=.35pt] (6.15000,2.25000) rectangle ++(1.05000,1.05000); node[] at (6.67500,2.77500) - D_2; draw[fill=black!12,draw=black!40,line width=.35pt] (7.20000,2.25000) rectangle ++(1.05000,1.05000); node[] at (7.72500,2.77500) - D_1; draw[fill=black!12,draw=black!40,line width=.35pt] (5.10000,1.20000) rectangle ++(1.05000,1.05000); node[] at (5…
Exact mathematical passage
- D_{1}
D_2Use in source · 6 passages
Search aliases: D_2
Use in source · atlas-pages.tex, line 5
p. 13), Section 11 (p. 25), Section 12 (p. 33). This guide follows a concrete analytic family through three descriptions: selected Taylor coefficients, a finite differential system, and the zeros of an entire function. The formulas are linked, but the evidence needed for each description is different. vfigthree_views headKeep three small models in view For N=2, the signed denominator is D_2(z)=cos z. Its reciprocal is the familiar secant series. For N=3, a third exponential perturbs the zeros of…
Exact mathematical passage
D_2(z)=\cos z
D_{2}Use in source · 3 passages
Search aliases: D_2
Use in source · signed_shift.tex, line 25
2.20000,0.90000) rectangle ++(0.80000,0.80000); node[] at (2.60000,1.30000) 0; node[] at (6.67500,3.78000) e^zJ_3; draw[fill=black!12,draw=black!40,line width=.35pt] (5.10000,2.25000) rectangle ++(1.05000,1.05000); node[] at (5.62500,2.77500) D_0; draw[fill=black!12,draw=black!40,line width=.35pt] (6.15000,2.25000) rectangle ++(1.05000,1.05000); node[] at (6.67500,2.77500) - D_2; draw[fill=black!12,draw=black!40,line width=.35pt] (7.20000,2.25000) rectangle ++(1.05000,1.05000); node[] at (7.7250…
Exact mathematical passage
- D_{2}
D_2^{-1}Use in source · 2 passages
Search aliases: D_2^-1
Use in source · atlas-pages.tex, line 244
vpage14Three similar reciprocals differNORMALIZED SECTIONSTraditional monograph: Definition 9.1 (p. 19), Theorem 9.2 (p. 19), Example 9.11 (p. 22). The signed denominator and the normalized unsigned sections produce different reciprocal families. In order two, D_2^-1=sec z, C_2,0=sech z, C_2,1=z/sinh z. Their first block-EGF coefficients make the difference visible. vfigdifferent_reciprocals headRemove the forced zero before taking a reciprocal The section E_N,j begins with z^j/j!. Define E_N,j(…
Exact mathematical passage
D_2^{-1}=\sec z,\qquad C_{2,0}=\sech z,\qquad C_{2,1}=z/\sinh z.
D_2^{\prime\prime}Use in source · 1 passage
Search aliases: D_2^primeprime prime prime
Use in source · riccati.tex, line 4
[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[] at (3.00000,4.50000) LINEAR EQUATION; node[] at (9.30000,4.50000) LOGARITHMIC DERIVATIVE; node[] at (3.00000,3.25000) D_2^primeprime=-D_2; node[] at (9.30000,3.25000) T_2^prime=1+T_2^2; node[] at (3.00000,1.85000) D_3^primeprimeprime=-D_3; node[] at (9.30000,1.85000) T_3^primeprime-3T_3T_3^prime+T_3^3=1; node[] at (6.15000,0.50000) T_N=-D_N^prime/D_N only where D_N≠0; draw[black!25] (6.10000,1.10000) -- (6.10000,4.650…
Exact mathematical passage
D_2^{\prime\prime}=-D_2
D_3Use in source · 6 passages
Search aliases: D_3
Use in source · atlas-pages.tex, line 95
ordinates and columns are input coordinates. The upper-triangular minus signs record crossing the signed edge while moving from the column to the row. They are not arbitrary orientation choices. headTwo checks connect the matrix and the scalar equation The initial derivative matrix is the identity. Each initial jet (y(0),y'(0),y''(0)) therefore determines a unique linear combination. Also, frac13Tr(e^zJ_3)=D_3(z), det(XI-J_3)=X^3+1. The roots of -1 appearing later in the exponential sum are alre…
Exact mathematical passage
\frac13\Tr(e^{zJ_3})=D_3(z),\qquad \det(XI-J_3)=X^3+1.
\mathbf D_3Use in source · 1 passage
Search aliases: D_3 mathbf
Use in source · signed_shift.tex, line 41
; node[] at (5.62500,0.67500) D_2; draw[fill=black!12,draw=black!40,line width=.35pt] (6.15000,0.15000) rectangle ++(1.05000,1.05000); node[] at (6.67500,0.67500) D_1; draw[fill=black!12,draw=black!40,line width=.35pt] (7.20000,0.15000) rectangle ++(1.05000,1.05000); node[] at (7.72500,0.67500) D_0; node[] at (10.80000,2.90000) J_3^3=-I; node[] at (10.80000,1.70000) D_3(0)=e_0; node[] at (10.80000,0.60000) det e^zJ_3=1; node[] at (6.10000,-0.80000) D_0↦-D_2, D_1↦ D_0, D_2↦ D_1;
Exact mathematical passage
\mathbf D_3(0)=e_0
D_3'Use in source · 2 passages
Search aliases: D_3'
Use in source · atlas-pages.tex, line 343
vpage19A root estimate becomes a certificateEXACT RATIONAL BRACKETSTraditional monograph: Theorem 11.7 (p. 28), Theorem 11.8 (p. 30), Corollary 11.9 (p. 31). The root finder is used only to suggest intervals. Each displayed interval is accepted by exact rational arithmetic: opposite endpoint signs prove existence, and an interval bound for D_3' excludes zero throughout the interval, proving uniqueness inside it. vfigcubic_certificates headHow the certificate works Evaluate the Taylor polynomial …
Exact mathematical passage
D_3'
D_{3,0}Use in source · 5 passages
Search aliases: D_3,0
Use in source · atlas-pages.tex, line 88
ed edge determines the cycleDIFFERENTIAL SYSTEMTraditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12). Let J_3e_0=e_1, J_3e_1=e_2, and J_3e_2=-e_0. A complete turn changes sign, so J_3^3=-I. The signed sections are the coordinates of e^zJ_3e_0. vfigsigned_shift Here D_j inside the matrix means D_3,j(z). For example, D_3,0=1-fracz^33!+fracz^66!-⋯, D_3,1=z-fracz^44!+⋯. Rows are output coordinates and columns are input coordinates. The upper-tri…
Exact mathematical passage
D_{3,0}=1-\frac{z^3}{3!}+\frac{z^6}{6!}-\cdots, \qquad D_{3,1}=z-\frac{z^4}{4!}+\cdots.
D_{3,1}Use in source · 5 passages
Search aliases: D_3,1
Use in source · atlas-pages.tex, line 88
ed edge determines the cycleDIFFERENTIAL SYSTEMTraditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12). Let J_3e_0=e_1, J_3e_1=e_2, and J_3e_2=-e_0. A complete turn changes sign, so J_3^3=-I. The signed sections are the coordinates of e^zJ_3e_0. vfigsigned_shift Here D_j inside the matrix means D_3,j(z). For example, D_3,0=1-fracz^33!+fracz^66!-⋯, D_3,1=z-fracz^44!+⋯. Rows are output coordinates and columns are input coordinates. The upper-tri…
Exact mathematical passage
D_{3,0}=1-\frac{z^3}{3!}+\frac{z^6}{6!}-\cdots, \qquad D_{3,1}=z-\frac{z^4}{4!}+\cdots.
D_{3,2}Use in source · 3 passages
Search aliases: D_3,2
Use in source · atlas-pages.tex, line 104
vpage6Addition is quotient multiplicationFINITE ALGEBRATraditional monograph: Proposition 5.5 (p. 8), Theorem 6.2 (p. 10). Work in C[ξ]/(ξ^3+1). Every element has coordinates in 1,ξ,ξ^2. The entire exponential has the finite-coordinate form e^zξ=D_3,0(z)+D_3,1(z)ξ+D_3,2(z)ξ^2. The multiplication table makes the carry sign unavoidable. vfigsigned_addition Multiply the coordinate forms of e^zξ and e^wξ. Their product is e^(z+w)ξ. The coefficient of 1 receives one contribution without a wrap and tw…
Exact mathematical passage
e^{z\xi}=D_{3,0}(z)+D_{3,1}(z)\xi+D_{3,2}(z)\xi^2.
D_{3,j}Use in source · 2 passages
Search aliases: D_3,j
Use in source · atlas-pages.tex, line 87
vpage5One signed edge determines the cycleDIFFERENTIAL SYSTEMTraditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12). Let J_3e_0=e_1, J_3e_1=e_2, and J_3e_2=-e_0. A complete turn changes sign, so J_3^3=-I. The signed sections are the coordinates of e^zJ_3e_0. vfigsigned_shift Here D_j inside the matrix means D_3,j(z). For example, D_3,0=1-fracz^33!+fracz^66!-⋯, D_3,1=z-fracz^44!+⋯. Rows are output coordinates and columns are input coordinates.…
Exact mathematical passage
D_{3,j}(z)
D_3^{\prime\prime\prime}Use in source · 1 passage
Search aliases: D_3^primeprimeprime prime prime prime
Use in source · riccati.tex, line 6
[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[] at (3.00000,4.50000) LINEAR EQUATION; node[] at (9.30000,4.50000) LOGARITHMIC DERIVATIVE; node[] at (3.00000,3.25000) D_2^primeprime=-D_2; node[] at (9.30000,3.25000) T_2^prime=1+T_2^2; node[] at (3.00000,1.85000) D_3^primeprimeprime=-D_3; node[] at (9.30000,1.85000) T_3^primeprime-3T_3T_3^prime+T_3^3=1; node[] at (6.15000,0.50000) T_N=-D_N^prime/D_N only where D_N≠0; draw[black!25] (6.10000,1.10000) -- (6.10000,4.650…
Exact mathematical passage
D_3^{\prime\prime\prime}=-D_3
D_4Use in source · 7 passages
Search aliases: D_4
Use in source · atlas-pages.tex, line 6
n entire function. The formulas are linked, but the evidence needed for each description is different. vfigthree_views headKeep three small models in view For N=2, the signed denominator is D_2(z)=cos z. Its reciprocal is the familiar secant series. For N=3, a third exponential perturbs the zeros of a dominant cosine. For N=4, an exact factorization gives a complete zero picture: D_4(z)=cos(z/sqrt2)cosh(z/sqrt2). The examples move between these orders deliberately. Order two makes the coefficien…
Exact mathematical passage
D_4(z)=\cos(z/\sqrt2)\cosh(z/\sqrt2).
D_jUse in source · 2 passages
Search aliases: D_j
Use in source · atlas-pages.tex, line 87
vpage5One signed edge determines the cycleDIFFERENTIAL SYSTEMTraditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12). Let J_3e_0=e_1, J_3e_1=e_2, and J_3e_2=-e_0. A complete turn changes sign, so J_3^3=-I. The signed sections are the coordinates of e^zJ_3e_0. vfigsigned_shift Here D_j inside the matrix means D_3,j(z). For example, D_3,0=1-fracz^33!+fracz^66!-⋯, D_3,1=z-fracz^44!+⋯. Rows are output coordinates and columns are input coordinates.…
Exact mathematical passage
D_j
D_NUse in source · 25 passages
Search aliases: D_N
Use in source · atlas-pages.tex, line 31
D_N(z) · D_N,0(z), the denominator whose zeros control the signed reciprocal family.
Exact mathematical passage
D_N(z)
\mathbf D_NUse in source · 4 passages
Search aliases: D_N mathbf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 811
[Companion matrix realization] Put D_N(z) = D_N,0(z) D_N,1(z) vdots D_N,N-1(z) . Then D_N(z)=e^zJ_Ne_0 and D_N(z)=frac1Ntr(e^zJ_N). The eigenvalues of J_N are the N roots of -1, each with multiplicity one.
Exact mathematical passage
\mathbf D_N(z) = \begin{pmatrix} \Dfun_{N,0}(z)\\ \Dfun_{N,1}(z)\\ \vdots\\ \Dfun_{N,N-1}(z) \end{pmatrix}.
D_N'Use in source · 4 passages
Search aliases: D_N'
Use in source · atlas-pages.tex, line 35
T_N(z) · -D_N'(z)/D_N(z), a function, not the coefficient T_N(n).
Exact mathematical passage
-D_N'(z)/D_N(z)
D_{N,0}Use in source · 2 passages
Search aliases: D_N,0
Use in source · atlas-pages.tex, line 31
D_N(z) · D_N,0(z), the denominator whose zeros control the signed reciprocal family.
Exact mathematical passage
D_{N,0}(z)
D_{N,j}Use in source · 2 passages
Search aliases: D_N,j
Use in source · atlas-pages.tex, line 30
D_N,j(z) · The same series with (-1)^n. One full derivative cycle contributes a minus sign.
Exact mathematical passage
D_{N,j}(z)
D_N^{-(r+s)}Use in source · 2 passages
Search aliases: D_N^-(r+s)
Use in source · atlas-pages.tex, line 205
Theorem 7.6 (p. 15). One recurrence uses the denominator coefficients. Another uses its logarithm. With b_n(r)=P_N,n(r)/(Nn)!, the shorter logarithmic form is nb_n(r)=r∑_k=1^nκ_N,kb_n-k(r), b_0=1. The negative superdiagonal in the Newton determinant is essential. vfignewton_addition headThe additive parameter law has block weights Using the same logarithm, D_N^-rD_N^-s=D_N^-(r+s). Thus P_N,n(r+s)=∑_q=0^nbinomNnNqP_N,q(r)P_N,n-q(s). For N=2,n=3,r=s=1, the four weights are 1,15,15,1. The two middl…
Exact mathematical passage
D_N^{-r}D_N^{-s}=D_N^{-(r+s)}
D_N^{-r}Use in source · 6 passages
Search aliases: D_N^-r
Use in source · atlas-pages.tex, line 33
P_N,n(r) · Block-EGF coefficients of D_N^-r. They are polynomials in r.
Exact mathematical passage
D_N^{-r}
D_N^{-s}Use in source · 2 passages
Search aliases: D_N^-s
Use in source · atlas-pages.tex, line 205
Theorem 7.6 (p. 15). One recurrence uses the denominator coefficients. Another uses its logarithm. With b_n(r)=P_N,n(r)/(Nn)!, the shorter logarithmic form is nb_n(r)=r∑_k=1^nκ_N,kb_n-k(r), b_0=1. The negative superdiagonal in the Newton determinant is essential. vfignewton_addition headThe additive parameter law has block weights Using the same logarithm, D_N^-rD_N^-s=D_N^-(r+s). Thus P_N,n(r+s)=∑_q=0^nbinomNnNqP_N,q(r)P_N,n-q(s). For N=2,n=3,r=s=1, the four weights are 1,15,15,1. The two middl…
Exact mathematical passage
D_N^{-r}D_N^{-s}=D_N^{-(r+s)}
D_N^{\prime}Use in source · 1 passage
Search aliases: D_N^prime prime
Use in source · riccati.tex, line 8
oin=round] node[] at (3.00000,4.50000) LINEAR EQUATION; node[] at (9.30000,4.50000) LOGARITHMIC DERIVATIVE; node[] at (3.00000,3.25000) D_2^primeprime=-D_2; node[] at (9.30000,3.25000) T_2^prime=1+T_2^2; node[] at (3.00000,1.85000) D_3^primeprimeprime=-D_3; node[] at (9.30000,1.85000) T_3^primeprime-3T_3T_3^prime+T_3^3=1; node[] at (6.15000,0.50000) T_N=-D_N^prime/D_N only where D_N≠0; draw[black!25] (6.10000,1.10000) -- (6.10000,4.65000); node[] at (6.10000,-0.40000) T_4(z)=fracz^33!+34fracz^77…
Exact mathematical passage
T_N=-D_N^{\prime}/D_N\quad\text{only where }D_N\ne0
d_{pq}Use in source · 5 passages
Search aliases: d_pq
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 857
[Entries of the fundamental matrix] For 0≤ p,q<N, let d_pq be the member of {0,…,N-1} congruent to p-q modulo N. Put ε_pq = 1, · p≥ q,
Exact mathematical passage
d_{pq}
\ddotsUse in source · 1 passage
Search aliases: ddots
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1106
[Logarithmic Newton recurrence] Set b_N,n(r)=fracP_N,n(r)(Nn)!. Then b_N,0(r)=1 and for every n≥1, nb_N,n(r) = r∑_k=1^nκ_N,kb_N,n-k(r). Equivalently, P_N,n(r) = frac r n ∑_k=1^n frac(Nn)!(N(n-k))! κ_N,kP_N,n-k(r). If p_k=rκ_N,k, then, for n≥1, b_N,n(r) = frac1n! det p_1 · -1 · 0 · ⋯ · 0 p_2 · p_1 · -2 · ⋯ · 0 p_3 · p_2 · p_1 · ddots · vdots vdots · vdots · ddots · ddots · -(n-1) p_n · p_n-1 · ⋯ · p_2 · p_1 .
Exact mathematical passage
b_{N,n}(r) = \frac1{n!} \det \begin{pmatrix} p_1&-1&0&\cdots&0\\ p_2&p_1&-2&\cdots&0\\ p_3&p_2&p_1&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&-(n-1)\\ p_n&p_{n-1}&\cdots&p_2&p_1 \end{pmatrix}.
\detUse in source · 8 passages
Search aliases: det
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 500
igin, the derivative matrix is diagonal with entries 1,λ,…,λ^N-1. For an explicit determinant proof, write each basis function as a finite sum of exponentials with exponents a_ℓ=λω^ℓ. Its derivative matrix factors as W(z)=V diag(e^a_ℓ z) C, V_kℓ=a_ℓ^k, where C is the constant matrix of Fourier weights. Thus det W(z)=e^z∑_ℓ a_ℓdet W(0)=det W(0), because N≥2 implies ∑_ℓω^ℓ=0. The product of the diagonal initial values is the stated power of λ.
Exact mathematical passage
\det W(z)=e^{z\sum_\ell a_\ell}\det W(0)=\det W(0),
\Dfun_1Use in source · 2 passages
Search aliases: Dfun_1 D_1 Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 727
The first denominators are D_1(z)=e^-z, D_2(z)=cos z. For N=3, D_3(z) = frac13( e^-z +2e^z/2cos(fracsqrt3z2) ). For N=4, D_4(z) = cos(frac zsqrt2) cosh(frac zsqrt2).
Exact mathematical passage
\Dfun_1(z)=e^{-z}, \qquad \Dfun_2(z)=\cos z.
\Dfun_2Use in source · 4 passages
Search aliases: Dfun_2 D_2 Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 727
The first denominators are D_1(z)=e^-z, D_2(z)=cos z. For N=3, D_3(z) = frac13( e^-z +2e^z/2cos(fracsqrt3z2) ). For N=4, D_4(z) = cos(frac zsqrt2) cosh(frac zsqrt2).
Exact mathematical passage
\Dfun_1(z)=e^{-z}, \qquad \Dfun_2(z)=\cos z.
\Dfun_3Use in source · 1 passage
Search aliases: Dfun_3 D_3 Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 733
The first denominators are D_1(z)=e^-z, D_2(z)=cos z. For N=3, D_3(z) = frac13( e^-z +2e^z/2cos(fracsqrt3z2) ). For N=4, D_4(z) = cos(frac zsqrt2) cosh(frac zsqrt2).
Exact mathematical passage
\Dfun_3(z) = \frac13\left( e^{-z} +2e^{z/2}\cos\left(\frac{\sqrt3z}{2}\right) \right).
\Dfun_4Use in source · 2 passages
Search aliases: Dfun_4 D_4 Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 742
The first denominators are D_1(z)=e^-z, D_2(z)=cos z. For N=3, D_3(z) = frac13( e^-z +2e^z/2cos(fracsqrt3z2) ). For N=4, D_4(z) = cos(frac zsqrt2) cosh(frac zsqrt2).
Exact mathematical passage
\Dfun_4(z) = \cos\left(\frac z{\sqrt2}\right) \cosh\left(\frac z{\sqrt2}\right).
\Dfun_{N,0}'Use in source · 1 passage
Search aliases: Dfun_N,0' D_N,0' Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 618
[Signed derivative cycle] For 1≤ j<N, D_N,j'= D_N,j-1, and D_N,0'=- D_N,N-1. Every signed section satisfies D_N,j^(N)=- D_N,j. The N signed sections form the basis of y^(N)=-y determined by D_N,j^(k)(0)=δ_jk for 0≤ j,k<N.
Exact mathematical passage
\Dfun_{N,0}'=-\Dfun_{N,N-1}.
\Dfun_{N,1}Use in source · 1 passage
Search aliases: Dfun_N,1 D_N,1 Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 811
[Companion matrix realization] Put D_N(z) = D_N,0(z) D_N,1(z) vdots D_N,N-1(z) . Then D_N(z)=e^zJ_Ne_0 and D_N(z)=frac1Ntr(e^zJ_N). The eigenvalues of J_N are the N roots of -1, each with multiplicity one.
Exact mathematical passage
\mathbf D_N(z) = \begin{pmatrix} \Dfun_{N,0}(z)\\ \Dfun_{N,1}(z)\\ \vdots\\ \Dfun_{N,N-1}(z) \end{pmatrix}.
\Dfun_{N,d_{pq}}Use in source · 1 passage
Search aliases: Dfun_N,d_pq D_N,d_pq Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 867
-1, · p<q. Then (e^zJ_N)_pq = ε_pq D_N,d_pq(z). Thus every entry of the fundamental matrix is one of the signed sections, with the sign recording whether the cyclic path from q to p crosses the signed edge.
Exact mathematical passage
\bigl(e^{zJ_N}\bigr)_{pq} = \varepsilon_{pq}\Dfun_{N,d_{pq}}(z).
\Dfun_{N,i}Use in source · 1 passage
Search aliases: Dfun_N,i D_N,i Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 654
[Twisted addition law] For 0≤ k<N, D_N,k(z+w) = ∑_substack0≤ i,j<N i+j≡ kbmod N (-1)^lfloor(i+j)/Nrfloor D_N,i(z) D_N,j(w).
Exact mathematical passage
\Dfun_{N,k}(z+w) = \sum_{\substack{0\leq i,j<N\\ i+j\equiv k\bmod N}} (-1)^{\lfloor(i+j)/N\rfloor} \Dfun_{N,i}(z)\Dfun_{N,j}(w).
\Dfun_{N,j}'Use in source · 1 passage
Search aliases: Dfun_N,j' D_N,j' Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 614
[Signed derivative cycle] For 1≤ j<N, D_N,j'= D_N,j-1, and D_N,0'=- D_N,N-1. Every signed section satisfies D_N,j^(N)=- D_N,j. The N signed sections form the basis of y^(N)=-y determined by D_N,j^(k)(0)=δ_jk for 0≤ j,k<N.
Exact mathematical passage
\Dfun_{N,j}'=\Dfun_{N,j-1},
\Dfun_{N,j-1}Use in source · 1 passage
Search aliases: Dfun_N,j-1 D_N,j-1 Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 614
[Signed derivative cycle] For 1≤ j<N, D_N,j'= D_N,j-1, and D_N,0'=- D_N,N-1. Every signed section satisfies D_N,j^(N)=- D_N,j. The N signed sections form the basis of y^(N)=-y determined by D_N,j^(k)(0)=δ_jk for 0≤ j,k<N.
Exact mathematical passage
\Dfun_{N,j}'=\Dfun_{N,j-1},
\Dfun_{N,j}^{(k)}Use in source · 1 passage
Search aliases: Dfun_N,j^(k) D_N,j^(k) Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 626
[Signed derivative cycle] For 1≤ j<N, D_N,j'= D_N,j-1, and D_N,0'=- D_N,N-1. Every signed section satisfies D_N,j^(N)=- D_N,j. The N signed sections form the basis of y^(N)=-y determined by D_N,j^(k)(0)=δ_jk for 0≤ j,k<N.
Exact mathematical passage
\Dfun_{N,j}^{(k)}(0)=\delta_{jk}
\Dfun_{N,j}^{(N)}Use in source · 1 passage
Search aliases: Dfun_N,j^(N) D_N,j^(N) Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 622
[Signed derivative cycle] For 1≤ j<N, D_N,j'= D_N,j-1, and D_N,0'=- D_N,N-1. Every signed section satisfies D_N,j^(N)=- D_N,j. The N signed sections form the basis of y^(N)=-y determined by D_N,j^(k)(0)=δ_jk for 0≤ j,k<N.
Exact mathematical passage
\Dfun_{N,j}^{(N)}=-\Dfun_{N,j}.
\Dfun_{N,k}Use in source · 1 passage
Search aliases: Dfun_N,k D_N,k Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 654
[Twisted addition law] For 0≤ k<N, D_N,k(z+w) = ∑_substack0≤ i,j<N i+j≡ kbmod N (-1)^lfloor(i+j)/Nrfloor D_N,i(z) D_N,j(w).
Exact mathematical passage
\Dfun_{N,k}(z+w) = \sum_{\substack{0\leq i,j<N\\ i+j\equiv k\bmod N}} (-1)^{\lfloor(i+j)/N\rfloor} \Dfun_{N,i}(z)\Dfun_{N,j}(w).
\Dfun_{N,N-1}Use in source · 3 passages
Search aliases: Dfun_N,N-1 D_N,N-1 Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 618
[Signed derivative cycle] For 1≤ j<N, D_N,j'= D_N,j-1, and D_N,0'=- D_N,N-1. Every signed section satisfies D_N,j^(N)=- D_N,j. The N signed sections form the basis of y^(N)=-y determined by D_N,j^(k)(0)=δ_jk for 0≤ j,k<N.
Exact mathematical passage
\Dfun_{N,0}'=-\Dfun_{N,N-1}.
\Dfun_N^{-q}Use in source · 1 passage
Search aliases: Dfun_N^-q D_N^-q Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2943
The derivative weighted sum omits the finitely many zeros that might be multiple. A treatment of those zeros requires the complete principal parts of D_N^-q rather than the single derivative weight.
Exact mathematical passage
\Dfun_N^{-q}
\Dfun_N^{\prime}Use in source · 1 passage
Search aliases: Dfun_N^prime D_N^prime Dfun prime mathsf prime
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 296
ic objects are introduced in the following order. First, E_N,j(z) = ∑_n≥0fracz^Nn+j(Nn+j)! is the residue j part of the exponential. Second, D_N,j(z) = ∑_n≥0(-1)^nfracz^Nn+j(Nn+j)! is the signed basis for y^(N)=-y. The section D_N= D_N,0 is the denominator used in the reciprocal family S_N^(r)= D_N^-r. Set T_N=- D_N^prime/ D_N. Equivalently, T_N is the logarithmic derivative of S_N^(1).
Exact mathematical passage
\Tfun_N=-\Dfun_N^{\prime}/\Dfun_N
\operatorname{diag}Use in source · 1 passage
Search aliases: diag operatorname
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 495
At the origin, the derivative matrix is diagonal with entries 1,λ,…,λ^N-1. For an explicit determinant proof, write each basis function as a finite sum of exponentials with exponents a_ℓ=λω^ℓ. Its derivative matrix factors as W(z)=V diag(e^a_ℓ z) C, V_kℓ=a_ℓ^k, where C is the constant matrix of Fourier weights. Thus det W(z)=e^z∑_ℓ a_ℓdet W(0)=det W(0), because N≥2 implies ∑_ℓω^ℓ=0. The product of the diagonal initial values is the stated power of λ.
Exact mathematical passage
W(z)=V\,\operatorname{diag}(e^{a_\ell z})\,C, \qquad V_{k\ell}=a_\ell^k,
e^{(2k+1)\pi i/N}Use in source · 1 passage
Search aliases: e^(2k+1)π i/N pi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1718
We now assume N≥2. Write η_k=e^(2k+1)π i/N, 0≤ k<N. These are the vertices of a regular polygon and D_N(z)=frac1N∑_k=0^N-1e^η_kz.
Exact mathematical passage
\eta_k=e^{(2k+1)\pi i/N}, \qquad 0\leq k<N.
e^{(z+w)\xi}Use in source · 3 passages
Search aliases: e^(z+w)ξ xi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 792
Differentiation multiplies by λξ in the first algebra and by ξ in the second. Reduction of the product by the defining relation gives the derivative cycles. The identity e^(z+w)ξ=e^zξe^wξ gives the addition laws after comparison of the coefficients of 1,ξ,…,ξ^N-1.
Exact mathematical passage
e^{(z+w)\xi}=e^{z\xi}e^{w\xi}
e^{-2s_Ny}Use in source · 1 passage
Search aliases: e^-2s_Ny
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1888
a_N-θ. Monotonicity of cosine on [0,π], followed by reflection about π, therefore gives Re(η_ℓ z)≤ rcos(a_N+θ). The remaining vertex e^-ia_N has projection rcos(a_N-θ). Their difference is at least 2s_Nrsinθ=2s_Ny. If D_N(z)=0, divide its finite exponential sum by the term with exponent e^-ia_Nz and apply the triangle inequality: 1≤(N-1)e^-2s_Ny. Thus y≤ Y. Conjugation gives the same bound for -y on the lower half sector. Since N≥3 and s_N>0, this Y is positive.
Exact mathematical passage
1\leq(N-1)e^{-2s_Ny}.
e^{-2\gamma_Nt_m}Use in source · 1 passage
Search aliases: e^-2γ_Nt_m gamma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2068
[Leading displacement term and derivative sign] With the notation of [source reference], τ_m = t_m+ fracR_N(t_m)2is_N + O(e^-2γ_Nt_m). The quotient R_N(t_m)/(2is_N) is real. One also has D_N'(τ_m) = (-1)^m+1 frac2s_NN e^c_Nτ_m (1+O(e^-γ_Nτ_m)).
Exact mathematical passage
\tau_m = t_m+ \frac{R_N(t_m)}{2is_N} + O(e^{-2\gamma_Nt_m}).
e^{-2\pi i r}Use in source · 1 passage
Search aliases: e^-2π i r pi
Use in source · branch_domain.tex, line 13
width=1pt] (4.40000,2.30000) circle (0.09); node[] at (5.10000,1.45000) ρ=π/2; draw[->,line width=1.1pt] (4.88,2.3) arc[start angle=0,end angle=330,radius=.48]; node[] at (2.80000,0.10000) the loop is not inside the disk; node[] at (9.20000,4.10000) D(z)∼ D^prime(ρ)(z-ρ); node[] at (9.20000,2.90000) log D⟼log D+2π i; node[] at (9.20000,1.65000) D^-r⟼ e^-2π i rD^-r; node[] at (9.20000,0.40000) r=1/2: -1 r∈ Z: 1;
Exact mathematical passage
D^{-r}\longmapsto e^{-2\pi i r}D^{-r}
e^{-2\pi ir}Use in source · 2 passages
Search aliases: e^-2π ir pi
Use in source · atlas-pages.tex, line 235
gion containing 0 and normalize the logarithm to vanish there. D_N(z)^-r=exp(-rL(z)), e^L(z)=D_N(z), L(0)=0. vfigbranch_domain For D_2=cos z, the open disk |z|<π/2 is an admissible domain. A small loop around the simple zero ρ=π/2 is not contained in that disk. Analytic continuation around the zero changes the local logarithm by 2π i, and the power by e^-2π ir. headThree different kinds of statement A formal identity compares coefficients at the origin. A holomorphic identity for complex powers …
Exact mathematical passage
e^{-2\pi ir}
e^{-3t/2}Use in source · 1 passage
Search aliases: e^-3t/2
Use in source · two_term.tex, line 27
60) (12.05750,1.33419) (12.08675,1.31211) (12.11600,1.29036) (12.14525,1.26895) (12.17450,1.24790) (12.20375,1.22721) (12.23300,1.20689) (12.26225,1.18695) (12.29150,1.16740) (12.32075,1.14826) (12.35000,1.12952); draw[fill=white] (2.77215,2.02182) circle (0.07); draw[fill=white] (7.01644,2.02182) circle (0.07); draw[fill=white] (11.26073,2.02182) circle (0.07); node[] at (6.50000,4.70000) frac32e^-t/2D_3(t)=cos(sqrt3t/2)+frac12e^-3t/2; node[] at (6.50000,-0.25000) solid: full expression; dashed…
Exact mathematical passage
\frac32e^{-t/2}D_3(t)=\cos(\sqrt3t/2)+\frac12e^{-3t/2}
e^{-3t_m}Use in source · 1 passage
Search aliases: e^-3t_m
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2124
[Leading correction for N=3] For N=3, t_m=frac2πsqrt3(m+frac12) and τ_m = t_m+ frac(-1)^msqrt3e^-3t_m/2 + O(e^-3t_m).
Exact mathematical passage
\tau_m = t_m+ \frac{(-1)^m}{\sqrt3}e^{-3t_m/2} + O(e^{-3t_m}).
e^{-3t_m/2}Use in source · 5 passages
Search aliases: e^-3t_m/2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2124
[Leading correction for N=3] For N=3, t_m=frac2πsqrt3(m+frac12) and τ_m = t_m+ frac(-1)^msqrt3e^-3t_m/2 + O(e^-3t_m).
Exact mathematical passage
\tau_m = t_m+ \frac{(-1)^m}{\sqrt3}e^{-3t_m/2} + O(e^{-3t_m}).
e^{-c_N|\rho|}Use in source · 1 passage
Search aliases: e^-c_N|ρ| rho
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2917
For N≥3, [source reference] gives | D_N'(ρ)|^-1 ≤ C e^-c_N|ρ| for every sufficiently large simple zero. Let L be a compact subset of the s plane. Because the chosen arguments of the zeros lie in a fixed bounded interval, there are constants C_L and M_L such that sup_s∈ L|e^-sLogρ| ≤ C_L|ρ|^M_L for every sufficiently large zero. Hence the absolute value of the corresponding summand is at most C_L'|ρ|^M_Le^-q c_N|ρ|. The linear zero counting estimate makes this majorant summable. The Weierstrass t…
Exact mathematical passage
|\Dfun_N'(\rho)|^{-1} \leq C e^{-c_N|\rho|}
e^{-e^{-ia_N}t_m}Use in source · 1 passage
Search aliases: e^-e^-ia_Nt_m
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2103
At the real point t_m, the normalized value N e^-e^-ia_Nt_m D_N(t_m) is purely imaginary. The two dominant terms cancel there, so R_N(t_m) is purely imaginary and the displayed correction is real.
Exact mathematical passage
N e^{-e^{-ia_N}t_m}\Dfun_N(t_m)
e^{-e^{-ia_N}z}Use in source · 2 passages
Search aliases: e^-e^-ia_Nz
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1857
η≠ e^ia_N,e^-ia_N exp((η-e^-ia_N)z). Then N e^-e^-ia_Nz D_N(z) = 1+e^2is_Nz+R_N(z), and, for q=0,1,2, |R_N^(q)(z)| ≤ C e^-γ_Nx.
Exact mathematical passage
N e^{-e^{-ia_N}z}\Dfun_N(z) = 1+e^{2is_Nz}+R_N(z),
e^{-ia_N}Use in source · 11 passages
Search aliases: e^-ia_N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1849
On the strip x≥ R and |y|≤ Y, define R_N(z) = ∑_substackη^N=-1 η≠ e^ia_N,e^-ia_N exp((η-e^-ia_N)z). Then N e^-e^-ia_Nz D_N(z) = 1+e^2is_Nz+R_N(z), and, for q=0,1,2, |R_N^(q)(z)| ≤ C e^-γ_Nx.
Exact mathematical passage
R_N(z) = \sum_{\substack{\eta^N=-1\\ \eta\ne e^{ia_N},e^{-ia_N}}} \exp\bigl((\eta-e^{-ia_N})z\bigr).
e^{-is_N\tau_m}Use in source · 1 passage
Search aliases: e^-is_Nτ_m tau
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2109
For the derivative, use the formula in the proof of [source reference] and e^-is_Nτ_m = (-1)^m(-i) (1+O(e^-γ_Nτ_m)). The stated real sign follows.
Exact mathematical passage
e^{-is_N\tau_m} = (-1)^m(-i) \left(1+O(e^{-\gamma_N\tau_m})\right).
e^{-q c_N|\rho|}Use in source · 1 passage
Search aliases: e^-q c_N|ρ| rho
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2931
-1 ≤ C e^-c_N|ρ| for every sufficiently large simple zero. Let L be a compact subset of the s plane. Because the chosen arguments of the zeros lie in a fixed bounded interval, there are constants C_L and M_L such that sup_s∈ L|e^-sLogρ| ≤ C_L|ρ|^M_L for every sufficiently large zero. Hence the absolute value of the corresponding summand is at most C_L'|ρ|^M_Le^-q c_N|ρ|. The linear zero counting estimate makes this majorant summable. The Weierstrass test gives absolute local uniform convergence …
Exact mathematical passage
C_L'|\rho|^{M_L}e^{-q c_N|\rho|}.
e^{-t}Use in source · 2 passages
Search aliases: e^-t
Use in source · atlas-pages.tex, line 323
vpage18Two large terms nearly cancelTHE CUBIC MODELTraditional monograph: Lemma 11.5 (p. 27), Theorem 11.7 (p. 28), Corollary 11.9 (p. 31). For real t, order three has the exact formula D_3(t)=frace^-t+2e^t/2cos(sqrt3t/2)3. Divide by the nonzero factor 2e^t/2/3. The resulting equation separates the dominant cosine from the small third exponential. vfigtwo_term Ignoring the last term gives model zeros t_m=frac2πsqrt3(m+1/2). At t_m, the full normalized expression equals tfrac12e^-3t_m/2>0, so t_m…
Exact mathematical passage
D_3(t)=\frac{e^{-t}+2e^{t/2}\cos(\sqrt3t/2)}{3}.
e^{-t/2}Use in source · 1 passage
Search aliases: e^-t/2
Use in source · two_term.tex, line 27
60) (12.05750,1.33419) (12.08675,1.31211) (12.11600,1.29036) (12.14525,1.26895) (12.17450,1.24790) (12.20375,1.22721) (12.23300,1.20689) (12.26225,1.18695) (12.29150,1.16740) (12.32075,1.14826) (12.35000,1.12952); draw[fill=white] (2.77215,2.02182) circle (0.07); draw[fill=white] (7.01644,2.02182) circle (0.07); draw[fill=white] (11.26073,2.02182) circle (0.07); node[] at (6.50000,4.70000) frac32e^-t/2D_3(t)=cos(sqrt3t/2)+frac12e^-3t/2; node[] at (6.50000,-0.25000) solid: full expression; dashed…
Exact mathematical passage
\frac32e^{-t/2}D_3(t)=\cos(\sqrt3t/2)+\frac12e^{-3t/2}
e^{-w}Use in source · 2 passages
Search aliases: e^-w
Use in source · atlas-pages.tex, line 398
and integral comparison. Thus truncation has an explicit error mechanism. takeawayThe infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem. platecaptionThe exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: Δ_1(w)=e^-w has no zeros, so its empty product would be 1, not the function.
Exact mathematical passage
\Delta_1(w)=e^{-w}
e^{-z}Use in source · 2 passages
Search aliases: e^-z
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 727
The first denominators are D_1(z)=e^-z, D_2(z)=cos z. For N=3, D_3(z) = frac13( e^-z +2e^z/2cos(fracsqrt3z2) ). For N=4, D_4(z) = cos(frac zsqrt2) cosh(frac zsqrt2).
Exact mathematical passage
\Dfun_1(z)=e^{-z}, \qquad \Dfun_2(z)=\cos z.
e^{-\gamma t_m}Use in source · 3 passages
Search aliases: e^-γ t_m gamma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1946
[Large zero strings] Assume N≥3. There are constants m_0, C, and γ>0 and a sequence of positive real simple zeros (τ_m)_m≥ m_0 of D_N such that τ_m = t_m+O(e^-γ t_m). Every zero of sufficiently large modulus is of the form ω^kτ_m for a unique 0≤ k<N and a unique m≥ m_0.
Exact mathematical passage
\tau_m = t_m+O(e^{-\gamma t_m}).
e^{-\gamma_N\Re z}Use in source · 2 passages
Search aliases: e^-γ_NRe z gamma Re
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1984
If G(z)=0, then |1+e^2is_Nz| ≤ C e^-γ_NRe z. The real part of 2is_Nz is bounded in the strip. Apply [source reference] with w=2is_Nz. There is a unique integer m such that |z-t_m|≤ C_1e^-γ_NRe z, where t_m=frac(2m+1)π2s_N. In particular, |Re z-t_m|=o(1). For all sufficiently large zeros we have |Re z-t_m|≤1, and hence |z-t_m|≤ C_2e^-γ_Nt_m with a constant independent of m. Because Re z→+∞ in this argument, the corresponding integers m are eventually nonnegative and tend to infinity.
Exact mathematical passage
|1+e^{2is_Nz}| \leq C e^{-\gamma_N\Re z}.
e^{-\gamma_Nt_m}Use in source · 4 passages
Search aliases: e^-γ_Nt_m gamma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2001
G(z)=0, then |1+e^2is_Nz| ≤ C e^-γ_NRe z. The real part of 2is_Nz is bounded in the strip. Apply [source reference] with w=2is_Nz. There is a unique integer m such that |z-t_m|≤ C_1e^-γ_NRe z, where t_m=frac(2m+1)π2s_N. In particular, |Re z-t_m|=o(1). For all sufficiently large zeros we have |Re z-t_m|≤1, and hence |z-t_m|≤ C_2e^-γ_Nt_m with a constant independent of m. Because Re z→+∞ in this argument, the corresponding integers m are eventually nonnegative and tend to infinity.
Exact mathematical passage
|z-t_m|\leq C_2e^{-\gamma_Nt_m}
e^{-\gamma_Nx}Use in source · 2 passages
Search aliases: e^-γ_Nx gamma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1863
η≠ e^ia_N,e^-ia_N exp((η-e^-ia_N)z). Then N e^-e^-ia_Nz D_N(z) = 1+e^2is_Nz+R_N(z), and, for q=0,1,2, |R_N^(q)(z)| ≤ C e^-γ_Nx.
Exact mathematical passage
|R_N^{(q)}(z)| \leq C e^{-\gamma_Nx}.
e^{-\gamma_N\tau_m}Use in source · 3 passages
Search aliases: e^-γ_Nτ_m gamma tau
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2050
Differentiate the normalized expression. At z=τ_m, G'(τ_m) = 2is_Ne^2is_Nτ_m+R_N'(τ_m) = -2is_N+O(e^-γ_Nτ_m). Since G(τ_m)=0, D_N'(τ_m) = frac1N e^e^-ia_Nτ_mG'(τ_m). Taking absolute values gives the derivative asymptotic. Differentiating D_N(ω z)= D_N(z) gives the rotation formula.
Exact mathematical passage
G'(\tau_m) = 2is_Ne^{2is_N\tau_m}+R_N'(\tau_m) = -2is_N+O(e^{-\gamma_N\tau_m}).
e^{-\gamma\tau_m}Use in source · 1 passage
Search aliases: e^-γτ_m gamma tau
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1964
The derivatives satisfy D_N'(ω^kτ_m) = ω^-k D_N'(τ_m) and | D_N'(τ_m)| = frac2s_NNe^c_Nτ_m (1+O(e^-γτ_m)).
Exact mathematical passage
|\Dfun_N'(\tau_m)| = \frac{2s_N}{N}e^{c_N\tau_m} \left(1+O(e^{-\gamma\tau_m})\right).
e^{-\delta r}Use in source · 1 passage
Search aliases: e^-δ r delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1793
The indicator of D_N equals h_N, with the limsup taken over radii where D_N(re^iθ)≠0: limsup_r→∞ frac1rlog| D_N(re^iθ)| = h_N(θ). If K is a compact set of directions disjoint from T_N, then there are δ>0 and R>0 such that, for each θ∈ K, one root η(θ) of -1 satisfies D_N(re^iθ) = frac1N exp(η(θ)re^iθ) (1+O(e^-δ r)) uniformly for r≥ R. In particular, no sufficiently large zero can lie in those directions.
Exact mathematical passage
\Dfun_N(re^{i\theta}) = \frac1N \exp\bigl(\eta(\theta)re^{i\theta}\bigr) \left(1+O(e^{-\delta r})\right)
e^{2is_N(t_m+\delta_m)}Use in source · 1 passage
Search aliases: e^2is_N(t_m+δ_m) delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2089
Put δ_m=τ_m-t_m. The large zero theorem gives δ_m=O(e^-γ_Nt_m). Since e^2is_Nt_m=-1, 1+e^2is_N(t_m+δ_m) = -2is_Nδ_m+O(δ_m^2). The remainder estimate through the first derivative gives R_N(t_m+δ_m) = R_N(t_m)+O(e^-γ_Nt_m|δ_m|). Substitution in the zero equation gives the displacement formula.
Exact mathematical passage
1+e^{2is_N(t_m+\delta_m)} = -2is_N\delta_m+O(\delta_m^2).
e^{2is_Nt_m}Use in source · 1 passage
Search aliases: e^2is_Nt_m
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2088
Put δ_m=τ_m-t_m. The large zero theorem gives δ_m=O(e^-γ_Nt_m). Since e^2is_Nt_m=-1, 1+e^2is_N(t_m+δ_m) = -2is_Nδ_m+O(δ_m^2). The remainder estimate through the first derivative gives R_N(t_m+δ_m) = R_N(t_m)+O(e^-γ_Nt_m|δ_m|). Substitution in the zero equation gives the displacement formula.
Exact mathematical passage
e^{2is_Nt_m}=-1
e^{2is_Nz}Use in source · 3 passages
Search aliases: e^2is_Nz
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1857
η≠ e^ia_N,e^-ia_N exp((η-e^-ia_N)z). Then N e^-e^-ia_Nz D_N(z) = 1+e^2is_Nz+R_N(z), and, for q=0,1,2, |R_N^(q)(z)| ≤ C e^-γ_Nx.
Exact mathematical passage
N e^{-e^{-ia_N}z}\Dfun_N(z) = 1+e^{2is_Nz}+R_N(z),
e^{2is_N\tau_m}Use in source · 1 passage
Search aliases: e^2is_Nτ_m tau
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2050
Differentiate the normalized expression. At z=τ_m, G'(τ_m) = 2is_Ne^2is_Nτ_m+R_N'(τ_m) = -2is_N+O(e^-γ_Nτ_m). Since G(τ_m)=0, D_N'(τ_m) = frac1N e^e^-ia_Nτ_mG'(τ_m). Taking absolute values gives the derivative asymptotic. Differentiating D_N(ω z)= D_N(z) gives the rotation formula.
Exact mathematical passage
G'(\tau_m) = 2is_Ne^{2is_N\tau_m}+R_N'(\tau_m) = -2is_N+O(e^{-\gamma_N\tau_m}).
e^{2Y}Use in source · 2 passages
Search aliases: e^2Y
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1904
ng strip, the roots e^ia_N and e^-ia_N have real part c_N, while each remaining root η=u+iv satisfies u≤cos(3a_N)=c_N-γ_N. Hence Re((η-e^-ia_N)z) ≤-γ_Nx+2Y. Division by e^e^-ia_Nz gives the displayed normalization. Each differentiation multiplies a remainder term by η-e^-ia_N, whose modulus is at most two. Thus, for q=0,1,2, |R_N^(q)(z)|≤(N-2)2^qe^2Ye^-γ_Nx. One may take R=1 and C=4(N-2)e^2Y.
Exact mathematical passage
|R_N^{(q)}(z)|\leq(N-2)2^qe^{2Y}e^{-\gamma_Nx}.
e^{2\pi i/N}Use in source · 1 passage
Search aliases: e^2π i/N pi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 334
Fix N≥1 and write ω=e^2π i/N. Let Ω be a disk centered at the origin, with the whole plane allowed, and assume that Ω is invariant under multiplication by ω. For a holomorphic function F on Ω, define ( B_N,jF)(z) = frac1N∑_ℓ=0^N-1ω^-jℓF(ω^ℓ z), 0≤ j<N. The same formula is valid for formal power series.
Exact mathematical passage
\omega=e^{2\pi i/N}.
e^{a_\ell z}Use in source · 1 passage
Search aliases: e^a_ℓ z ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 495
At the origin, the derivative matrix is diagonal with entries 1,λ,…,λ^N-1. For an explicit determinant proof, write each basis function as a finite sum of exponentials with exponents a_ℓ=λω^ℓ. Its derivative matrix factors as W(z)=V diag(e^a_ℓ z) C, V_kℓ=a_ℓ^k, where C is the constant matrix of Fourier weights. Thus det W(z)=e^z∑_ℓ a_ℓdet W(0)=det W(0), because N≥2 implies ∑_ℓω^ℓ=0. The product of the diagonal initial values is the stated power of λ.
Exact mathematical passage
W(z)=V\,\operatorname{diag}(e^{a_\ell z})\,C, \qquad V_{k\ell}=a_\ell^k,
e^{c_Nr}Use in source · 1 passage
Search aliases: e^c_Nr
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1818
At a tie direction exactly two adjacent vertices attain the maximum. After rotation it is enough to take θ=0. With c_N=cos(π/N) and s_N=sin(π/N), the two dominant terms give frac2N e^c_Nrcos(s_Nr) and every other term is exponentially smaller. Along a sequence on which |cos(s_Nr)|≥1/2, the lower indicator bound is c_N=h_N(0). Rotational symmetry treats every tie direction.
Exact mathematical passage
\frac2N e^{c_Nr}\cos(s_Nr)
e^{c_N\tau_m}Use in source · 2 passages
Search aliases: e^c_Nτ_m tau
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1964
The derivatives satisfy D_N'(ω^kτ_m) = ω^-k D_N'(τ_m) and | D_N'(τ_m)| = frac2s_NNe^c_Nτ_m (1+O(e^-γτ_m)).
Exact mathematical passage
|\Dfun_N'(\tau_m)| = \frac{2s_N}{N}e^{c_N\tau_m} \left(1+O(e^{-\gamma\tau_m})\right).
e^{e^{-ia_N}z}Use in source · 1 passage
Search aliases: e^e^-ia_Nz
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1901
On the resulting strip, the roots e^ia_N and e^-ia_N have real part c_N, while each remaining root η=u+iv satisfies u≤cos(3a_N)=c_N-γ_N. Hence Re((η-e^-ia_N)z) ≤-γ_Nx+2Y. Division by e^e^-ia_Nz gives the displayed normalization. Each differentiation multiplies a remainder term by η-e^-ia_N, whose modulus is at most two. Thus, for q=0,1,2, |R_N^(q)(z)|≤(N-2)2^qe^2Ye^-γ_Nx. One may take R=1 and C=4(N-2)e^2Y.
Exact mathematical passage
e^{e^{-ia_N}z}
e^{e^{-ia_N}\tau_m}Use in source · 1 passage
Search aliases: e^e^-ia_Nτ_m tau
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2058
Differentiate the normalized expression. At z=τ_m, G'(τ_m) = 2is_Ne^2is_Nτ_m+R_N'(τ_m) = -2is_N+O(e^-γ_Nτ_m). Since G(τ_m)=0, D_N'(τ_m) = frac1N e^e^-ia_Nτ_mG'(τ_m). Taking absolute values gives the derivative asymptotic. Differentiating D_N(ω z)= D_N(z) gives the rotation formula.
Exact mathematical passage
\Dfun_N'(\tau_m) = \frac1N e^{e^{-ia_N}\tau_m}G'(\tau_m).
e^{g(z)}Use in source · 1 passage
Search aliases: e^g(z)
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1318
For a function g, define the complete differential Bell polynomial Y_N by fracd^Ndz^Ne^g(z) = e^g(z)Y_N(g',g'',…,g^(N)).
Exact mathematical passage
\frac{d^N}{dz^N}e^{g(z)} = e^{g(z)}Y_N\bigl(g',g'',\ldots,g^{(N)}\bigr).
e^{ia_N}Use in source · 2 passages
Search aliases: e^ia_N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1849
On the strip x≥ R and |y|≤ Y, define R_N(z) = ∑_substackη^N=-1 η≠ e^ia_N,e^-ia_N exp((η-e^-ia_N)z). Then N e^-e^-ia_Nz D_N(z) = 1+e^2is_Nz+R_N(z), and, for q=0,1,2, |R_N^(q)(z)| ≤ C e^-γ_Nx.
Exact mathematical passage
R_N(z) = \sum_{\substack{\eta^N=-1\\ \eta\ne e^{ia_N},e^{-ia_N}}} \exp\bigl((\eta-e^{-ia_N})z\bigr).
e^{i\theta}Use in source · 11 passages
Search aliases: e^iθ theta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1785
[Indicator polygon and dominant sectors] The indicator of D_N equals h_N, with the limsup taken over radii where D_N(re^iθ)≠0: limsup_r→∞ frac1rlog| D_N(re^iθ)| = h_N(θ). If K is a compact set of directions disjoint from T_N, then there are δ>0 and R>0 such that, for each θ∈ K, one root η(θ) of -1 satisfies D_N(re^iθ) = frac1N exp(η(θ)re^iθ) (1+O(e^-δ r)) uniformly for r≥ R. In particular, no sufficiently large zero can lie in those directions.
Exact mathematical passage
\Dfun_N(re^{i\theta})\ne0
e^{L(z)}Use in source · 2 passages
Search aliases: e^L(z)
Use in source · atlas-pages.tex, line 231
vpage13A complex power needs a domainBRANCH BOOKKEEPINGTraditional monograph: Section 2 (p. 2), Definition 7.1 (p. 13), Corollary 8.4 (p. 18). A formal unit series has a formal logarithm at the origin. To interpret that germ analytically, choose a simply connected zero-free region containing 0 and normalize the logarithm to vanish there. D_N(z)^-r=exp(-rL(z)), e^L(z)=D_N(z), L(0)=0. vfigbranch_domain For D_2=cos z, the open disk |z|<π/2 is an admissible domain. A small loop around the simple zer…
Exact mathematical passage
D_N(z)^{-r}=\exp(-rL(z)),\qquad e^{L(z)}=D_N(z),\quad L(0)=0.
e^{L+\pi i}Use in source · 1 passage
Search aliases: e^L+π i pi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1932
In fact, the constants ε_M=1/2 and C_M=3/2 work for every complex w, without the strip restriction. Put u=-(1+e^w) and L=log(1+u)=log(-e^w), using the principal logarithm. The elementary logarithm estimate on |u|≤1/2 gives |L|≤(3/2)|u|. Since e^L+π i=e^w, there is an integer m such that w=L+(2m+1)π i. This gives the required distance bound. Distinct centers are separated by at least 2π, whereas two centers satisfying the bound would be separated by at most 3|1+e^w|≤3/2. Thus the integer is uniqu…
Exact mathematical passage
e^{L+\pi i}=e^w
e^{r\Re(\eta e^{i\theta})}Use in source · 2 passages
Search aliases: e^rRe(η e^iθ) Re eta theta
Use in source · atlas-pages.tex, line 308
vpage17Growth comes from a polygonDOMINANT EXPONENTSTraditional monograph: Definition 11.3 (p. 26), Theorem 11.4 (p. 26). Write the signed denominator as a finite exponential sum: D_N(z)=frac1N∑_η^N=-1e^η z. For z=re^iθ, the size of one exponential is e^rRe(η e^iθ). The maximal projection determines the directional exponential growth. vfigindicator The indicator is h_N(θ)=max_0≤ k<Ncos(θ+frac(2k+1)πN). Away from a tie direction, one exponential dominates all the others by a uniform positive proj…
Exact mathematical passage
e^{r\Re(\eta e^{i\theta})}
e^{rz}Use in source · 1 passage
Search aliases: e^rz
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1203
For N=1, one has D_1(z)^-r=e^rz and P_1,n(r)=r^n. For N=2, P_2,1(r)=r, P_2,2(r)=r(3r+2), and P_2,3(r)=r(15r^2+30r+16). At r=1, these are the positive secant coefficients.
Exact mathematical passage
\Dfun_1(z)^{-r}=e^{rz}
e^{t/2}Use in source · 4 passages
Search aliases: e^t/2
Use in source · atlas-pages.tex, line 323
vpage18Two large terms nearly cancelTHE CUBIC MODELTraditional monograph: Lemma 11.5 (p. 27), Theorem 11.7 (p. 28), Corollary 11.9 (p. 31). For real t, order three has the exact formula D_3(t)=frace^-t+2e^t/2cos(sqrt3t/2)3. Divide by the nonzero factor 2e^t/2/3. The resulting equation separates the dominant cosine from the small third exponential. vfigtwo_term Ignoring the last term gives model zeros t_m=frac2πsqrt3(m+1/2). At t_m, the full normalized expression equals tfrac12e^-3t_m/2>0, so t_m…
Exact mathematical passage
D_3(t)=\frac{e^{-t}+2e^{t/2}\cos(\sqrt3t/2)}{3}.
e^wUse in source · 6 passages
Search aliases: e^w
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1916
[Near cancellation of two exponentials] For every M>0, there are constants ε_M>0 and C_M>0 such that the following holds. If |Re w|≤ M and |1+e^w|≤ε_M, then there is a unique integer m with |w-(2m+1)π i| ≤ C_M|1+e^w|.
Exact mathematical passage
|1+e^w|\leq\varepsilon_M,
e^{w\xi}Use in source · 3 passages
Search aliases: e^wξ xi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 792
Differentiation multiplies by λξ in the first algebra and by ξ in the second. Reduction of the product by the defining relation gives the derivative cycles. The identity e^(z+w)ξ=e^zξe^wξ gives the addition laws after comparison of the coefficients of 1,ξ,…,ξ^N-1.
Exact mathematical passage
e^{(z+w)\xi}=e^{z\xi}e^{w\xi}
e^{yz}Use in source · 2 passages
Search aliases: e^yz
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1484
Differentiate the generating function with respect to x for the first identity. Multiply it by e^yz and compare coefficients for the second.
Exact mathematical passage
e^{yz}
e^{z/2}Use in source · 1 passage
Search aliases: e^z/2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 733
The first denominators are D_1(z)=e^-z, D_2(z)=cos z. For N=3, D_3(z) = frac13( e^-z +2e^z/2cos(fracsqrt3z2) ). For N=4, D_4(z) = cos(frac zsqrt2) cosh(frac zsqrt2).
Exact mathematical passage
\Dfun_3(z) = \frac13\left( e^{-z} +2e^{z/2}\cos\left(\frac{\sqrt3z}{2}\right) \right).
e^{zJ_3}Use in source · 6 passages
Search aliases: e^zJ_3
Use in source · atlas-pages.tex, line 85
vpage5One signed edge determines the cycleDIFFERENTIAL SYSTEMTraditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12). Let J_3e_0=e_1, J_3e_1=e_2, and J_3e_2=-e_0. A complete turn changes sign, so J_3^3=-I. The signed sections are the coordinates of e^zJ_3e_0. vfigsigned_shift Here D_j inside the matrix means D_3,j(z). For example, D_3,0=1-fracz^33!+fracz^66!-⋯, D_3,1=z-fracz^44!+⋯. Rows are output coordinates and columns are input coordinates.…
Exact mathematical passage
e^{zJ_3}e_0
e^{zJ_N}Use in source · 9 passages
Search aliases: e^zJ_N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 822
D_N,N-1(z) . Then D_N(z)=e^zJ_Ne_0 and D_N(z)=frac1Ntr(e^zJ_N). The eigenvalues of J_N are the N roots of -1, each with multiplicity one.
Exact mathematical passage
\mathbf D_N(z)=e^{zJ_N}e_0
e^{z\tr J_N}Use in source · 1 passage
Search aliases: e^ztr J_N tr operatorname
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 911
The first two identities are the exponential group law. Since tr J_N=0, det(e^zJ_N)=e^ztr J_N=1. The first column statement is [source reference].
Exact mathematical passage
\det(e^{zJ_N})=e^{z\tr J_N}=1.
e^{z\xi}Use in source · 6 passages
Search aliases: e^zξ xi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 780
[Finite algebra exponential identities] In A_N^+ one has e^λ zξ = ∑_j=0^N-1 E_N,j(λ z)ξ^j. In A_N^- one has e^zξ = ∑_j=0^N-1 D_N,j(z)ξ^j. Differentiation and multiplication of these identities recover the cyclic derivative systems and the two addition laws.
Exact mathematical passage
e^{z\xi} = \sum_{j=0}^{N-1}\Dfun_{N,j}(z)\xi^j.
e^{z\sum_\ell a_\ell}Use in source · 1 passage
Search aliases: e^z∑_ℓ a_ℓ sum ell ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 500
igin, the derivative matrix is diagonal with entries 1,λ,…,λ^N-1. For an explicit determinant proof, write each basis function as a finite sum of exponentials with exponents a_ℓ=λω^ℓ. Its derivative matrix factors as W(z)=V diag(e^a_ℓ z) C, V_kℓ=a_ℓ^k, where C is the constant matrix of Fourier weights. Thus det W(z)=e^z∑_ℓ a_ℓdet W(0)=det W(0), because N≥2 implies ∑_ℓω^ℓ=0. The product of the diagonal initial values is the stated power of λ.
Exact mathematical passage
\det W(z)=e^{z\sum_\ell a_\ell}\det W(0)=\det W(0),
e^{|z|}Use in source · 1 passage
Search aliases: e^|z|
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1754
The finite exponential representation gives | D_N(z)|≤ e^|z|, so the type is at most one. Choose a root η_0 and put z=toverlineη_0 with t>0. The term e^η_0z=e^t has strictly larger real exponent than every other term. Hence D_N(toverlineη_0) = frac1Ne^t(1+o(1)). The type is therefore one.
Exact mathematical passage
|\Dfun_N(z)|\leq e^{|z|},
e^{\eta z}Use in source · 5 passages
Search aliases: e^η z eta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 244
The second half of the paper concerns the zeros of D_N. The finite exponential sum D_N(z)=frac1N∑_η^N=-1e^η z has a regular polygon of exponents. The general relation between an exponent polygon and the zero strips of an exponential sum goes back to P olya and Langer. Later developments include the work of Moreno and the treatment in Levin's monograph [source reference]. Bleher and Mallison give a particularly close general result: for convex-polygon exponential sums they locate all sufficiently…
Exact mathematical passage
\Dfun_N(z)=\frac1N\sum_{\eta^N=-1}e^{\eta z}
e^{\eta_0z}Use in source · 1 passage
Search aliases: e^η_0z eta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1757
The finite exponential representation gives | D_N(z)|≤ e^|z|, so the type is at most one. Choose a root η_0 and put z=toverlineη_0 with t>0. The term e^η_0z=e^t has strictly larger real exponent than every other term. Hence D_N(toverlineη_0) = frac1Ne^t(1+o(1)). The type is therefore one.
Exact mathematical passage
e^{\eta_0z}=e^t
e^{\eta_kz}Use in source · 1 passage
Search aliases: e^η_kz eta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1723
We now assume N≥2. Write η_k=e^(2k+1)π i/N, 0≤ k<N. These are the vertices of a regular polygon and D_N(z)=frac1N∑_k=0^N-1e^η_kz.
Exact mathematical passage
\Dfun_N(z)=\frac1N\sum_{k=0}^{N-1}e^{\eta_kz}.
e^{\lambda w}Use in source · 1 passage
Search aliases: e^λ w lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 518
Expand e^λ(z+w)=e^λ ze^λ w as a bivariate power series. A term z^a w^b contributes to the kth section exactly when a+b≡ kbmod N. Grouping first by i≡ abmod N and j≡ bbmod N gives the displayed sum.
Exact mathematical passage
e^{\lambda(z+w)}=e^{\lambda z}e^{\lambda w}
e^{\lambda z\xi}Use in source · 1 passage
Search aliases: e^λ zξ lambda xi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 774
[Finite algebra exponential identities] In A_N^+ one has e^λ zξ = ∑_j=0^N-1 E_N,j(λ z)ξ^j. In A_N^- one has e^zξ = ∑_j=0^N-1 D_N,j(z)ξ^j. Differentiation and multiplication of these identities recover the cyclic derivative systems and the two addition laws.
Exact mathematical passage
e^{\lambda z\xi} = \sum_{j=0}^{N-1}\Efun_{N,j}(\lambda z)\xi^j.
e^{\lambda(z+w)}Use in source · 1 passage
Search aliases: e^λ(z+w) lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 518
Expand e^λ(z+w)=e^λ ze^λ w as a bivariate power series. A term z^a w^b contributes to the kth section exactly when a+b≡ kbmod N. Grouping first by i≡ abmod N and j≡ bbmod N gives the displayed sum.
Exact mathematical passage
e^{\lambda(z+w)}=e^{\lambda z}e^{\lambda w}
e^{\rho\omega^\ell z}Use in source · 1 passage
Search aliases: e^ρω^ℓ z rho omega ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 683
[Finite exponential representation] Let R_N={η∈ Cmathrel:η^N=-1}. Then D_N(z) = frac1N∑_η∈ R_Ne^η z. More generally, D_N,j(z) = fracρ^-jN ∑_ℓ=0^N-1 ω^-jℓe^ρω^ℓ z.
Exact mathematical passage
\Dfun_{N,j}(z) = \frac{\rho^{-j}}N \sum_{\ell=0}^{N-1} \omega^{-j\ell}e^{\rho\omega^\ell z}.
e_1Use in source · 4 passages
Search aliases: e_1
Use in source · atlas-pages.tex, line 85
vpage5One signed edge determines the cycleDIFFERENTIAL SYSTEMTraditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12). Let J_3e_0=e_1, J_3e_1=e_2, and J_3e_2=-e_0. A complete turn changes sign, so J_3^3=-I. The signed sections are the coordinates of e^zJ_3e_0. vfigsigned_shift Here D_j inside the matrix means D_3,j(z). For example, D_3,0=1-fracz^33!+fracz^66!-⋯, D_3,1=z-fracz^44!+⋯. Rows are output coordinates and columns are input coordinates.…
Exact mathematical passage
J_3e_0=e_1
e_2Use in source · 4 passages
Search aliases: e_2
Use in source · atlas-pages.tex, line 85
vpage5One signed edge determines the cycleDIFFERENTIAL SYSTEMTraditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12). Let J_3e_0=e_1, J_3e_1=e_2, and J_3e_2=-e_0. A complete turn changes sign, so J_3^3=-I. The signed sections are the coordinates of e^zJ_3e_0. vfigsigned_shift Here D_j inside the matrix means D_3,j(z). For example, D_3,0=1-fracz^33!+fracz^66!-⋯, D_3,1=z-fracz^44!+⋯. Rows are output coordinates and columns are input coordinates.…
Exact mathematical passage
J_3e_1=e_2
E_{2,1}Use in source · 1 passage
Search aliases: E_2,1
Use in source · different_reciprocals.tex, line 42
++(2.10000,0.80000); node[] at (6.05000,1.65000) - frac13; draw[fill=black!0,draw=black!40,line width=.35pt] (7.10000,1.25000) rectangle ++(2.10000,0.80000); node[] at (8.15000,1.65000) frac715; draw[fill=black!8,draw=black!40,line width=.35pt] (9.20000,1.25000) rectangle ++(2.10000,0.80000); node[] at (10.25000,1.65000) - frac3121; node[] at (5.65000,-0.15000) E_2,1(z)=sinh z=zΦ_2,1(z^2), C_2,1(0)=1;
Exact mathematical passage
E_{2,1}(z)=\sinh z=z\Phi_{2,1}(z^2),\qquad C_{2,1}(0)=1
E_{3,1}Use in source · 3 passages
Search aliases: E_3,1
Use in source · atlas-pages.tex, line 130
because the hypergeometric convention already supplies (1)_n=n!. vfighypergeometric headCheck the displayed case For N=3,j=1,n=2, the six factors after the initial 1! are 2,3,4,5,6,7. Dividing by three and grouping them gives the three rows of the plate. The middle row supplies 2!. The others supply (2/3)_2 and (4/3)_2. The right-hand side is exactly 5040. Consequently, E_3,1(z)=z(1+fracz^34!+fracz^67!+⋯). The prefactor z retains the forced zero at the origin. The hypergeometric argument z^3/27 …
Exact mathematical passage
E_{3,1}(z)=z\left(1+\frac{z^3}{4!}+\frac{z^6}{7!}+\cdots\right).
E_{3,j}Use in source · 2 passages
Search aliases: E_3,j
Use in source · atlas-pages.tex, line 111
Multiply the coordinate forms of e^zξ and e^wξ. Their product is e^(z+w)ξ. The coefficient of 1 receives one contribution without a wrap and two contributions with a wrap. Reading that coefficient gives the displayed addition law. headCompare the unsigned model In C[ξ]/(ξ^3-1), the same table has no minus signs. Its exponential coordinates are the unsigned sections E_3,j. Thus the difference between the two differential systems is a single algebraic relation, not a change in how coefficients are…
Exact mathematical passage
E_{3,j}
E_4Use in source · 2 passages
Search aliases: E_4
Use in source · appell_operator.tex, line 33
9091) (11.95200,5.46972) (12.01000,5.65433) (12.06800,5.84486) (12.12600,6.04145) (12.18400,6.24420) (12.24200,6.45326) (12.30000,6.66875); draw[fill=black] (1.31232,1.20000) circle (0.055); draw[fill=black] (4.18000,1.20000) circle (0.055); draw[fill=black] (8.82000,1.20000) circle (0.055); draw[fill=black] (11.68768,1.20000) circle (0.055); node[] at (3.20000,4.65000) solid: E_4(x)=x^4-6x^2+5; node[] at (9.80000,4.65000) dashed: x^4; node[] at (6.50000,-0.40000) cosh(∂_x)E_4=E_4+frac12E_4^prim…
Exact mathematical passage
E_4(x)=x^4-6x^2+5
E_4^{(4)}Use in source · 1 passage
Search aliases: E_4^(4)
Use in source · appell_operator.tex, line 35
) (12.24200,6.45326) (12.30000,6.66875); draw[fill=black] (1.31232,1.20000) circle (0.055); draw[fill=black] (4.18000,1.20000) circle (0.055); draw[fill=black] (8.82000,1.20000) circle (0.055); draw[fill=black] (11.68768,1.20000) circle (0.055); node[] at (3.20000,4.65000) solid: E_4(x)=x^4-6x^2+5; node[] at (9.80000,4.65000) dashed: x^4; node[] at (6.50000,-0.40000) cosh(∂_x)E_4=E_4+frac12E_4^primeprime+frac124E_4^(4)=x^4;
Exact mathematical passage
\cosh(\partial_x)E_4=E_4+\frac12E_4^{\prime\prime}+\frac1{24}E_4^{(4)}=x^4
E_4^{\prime\prime}Use in source · 1 passage
Search aliases: E_4^primeprime prime prime
Use in source · appell_operator.tex, line 35
) (12.24200,6.45326) (12.30000,6.66875); draw[fill=black] (1.31232,1.20000) circle (0.055); draw[fill=black] (4.18000,1.20000) circle (0.055); draw[fill=black] (8.82000,1.20000) circle (0.055); draw[fill=black] (11.68768,1.20000) circle (0.055); node[] at (3.20000,4.65000) solid: E_4(x)=x^4-6x^2+5; node[] at (9.80000,4.65000) dashed: x^4; node[] at (6.50000,-0.40000) cosh(∂_x)E_4=E_4+frac12E_4^primeprime+frac124E_4^(4)=x^4;
Exact mathematical passage
\cosh(\partial_x)E_4=E_4+\frac12E_4^{\prime\prime}+\frac1{24}E_4^{(4)}=x^4
E_m'Use in source · 2 passages
Search aliases: E_m'
Use in source · atlas-pages.tex, line 274
The four real roots of this polynomial are ±1,±sqrt5. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output. headAn inverse operation, not a truncation error Applying cosh(∂_x) returns x^4 exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are E_m'=mE_m-1, E_m(x+y)=∑_q=0^mbinom mqE_q(x)y^m-q. Both follow from the generating function. No operator-series convergence as…
Exact mathematical passage
E_m'=mE_{m-1},\qquad E_m(x+y)=\sum_{q=0}^m\binom mqE_q(x)y^{m-q}.
E_{m-1}Use in source · 2 passages
Search aliases: E_m-1
Use in source · atlas-pages.tex, line 274
The four real roots of this polynomial are ±1,±sqrt5. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output. headAn inverse operation, not a truncation error Applying cosh(∂_x) returns x^4 exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are E_m'=mE_m-1, E_m(x+y)=∑_q=0^mbinom mqE_q(x)y^m-q. Both follow from the generating function. No operator-series convergence as…
Exact mathematical passage
E_m'=mE_{m-1},\qquad E_m(x+y)=\sum_{q=0}^m\binom mqE_q(x)y^{m-q}.
E_{m-1}^{(N,j,r)}Use in source · 1 passage
Search aliases: E_m-1^(N,j,r)
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1468
[Appell laws] For m≥1, the polynomials satisfy fracddxE_m^(N,j,r)(x) = mE_m-1^(N,j,r)(x) and, for m≥0, E_m^(N,j,r)(x+y) = ∑_q=0^m binom mq E_q^(N,j,r)(x)y^m-q.
Exact mathematical passage
\frac{d}{dx}E_m^{(N,j,r)}(x) = mE_{m-1}^{(N,j,r)}(x)
E_{m-q}^{(N,j,s)}Use in source · 1 passage
Search aliases: E_m-q^(N,j,s)
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1505
[Joint Appell convolution in the space and order parameters] For all r,s,x,y∈ C and m≥0, E_m^(N,j,r+s)(x+y) = ∑_q=0^m binom mq E_q^(N,j,r)(x) E_m-q^(N,j,s)(y).
Exact mathematical passage
E_m^{(N,j,r+s)}(x+y) = \sum_{q=0}^{m} \binom mq E_q^{(N,j,r)}(x) E_{m-q}^{(N,j,s)}(y).
E_m^{(N,j,r+s)}Use in source · 1 passage
Search aliases: E_m^(N,j,r+s)
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1505
[Joint Appell convolution in the space and order parameters] For all r,s,x,y∈ C and m≥0, E_m^(N,j,r+s)(x+y) = ∑_q=0^m binom mq E_q^(N,j,r)(x) E_m-q^(N,j,s)(y).
Exact mathematical passage
E_m^{(N,j,r+s)}(x+y) = \sum_{q=0}^{m} \binom mq E_q^{(N,j,r)}(x) E_{m-q}^{(N,j,s)}(y).
E_{N,0}Use in source · 2 passages
Search aliases: E_N,0
Use in source · atlas-pages.tex, line 282
vpage16Scale and order enter differentlyTHETA COUPLINGTraditional monograph: Theorem 10.2 (p. 23), Proposition 10.3 (p. 23), Proposition 10.4 (p. 23). Let H_N(z)=E_N,0(z). The theta family uses the same parameter u twice, but in two different roles: H_N(uz)^-1H_N(z)^-u =∑_n≥0Θ_N,n(u)fracz^Nn(Nn)!. Argument scaling contributes u^Nm at logarithmic block m. Reciprocal order contributes u. vfigtheta Their combined logarithm has coefficient (-1)^mκ_N,mfracu^Nm+um. At the first block this gives Θ_N,1(…
Exact mathematical passage
H_N(z)=E_{N,0}(z)
E_{N,j}Use in source · 10 passages
Search aliases: E_N,j
Use in source · atlas-pages.tex, line 29
midrule E_N,j(z) · displaystyle∑_n≥0fracz^Nn+j(Nn+j)! selects one residue class of the exponential.
Exact mathematical passage
E_{N,j}(z)
e_pUse in source · 1 passage
Search aliases: e_p
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 877
For every nonnegative integer n, J_N^Nn+d_pqe_q = (-1)^nε_pqe_p. Only powers whose exponent is congruent to d_pq modulo N contribute to the (p,q) entry of the exponential. Therefore (e^zJ_N)_pq = ε_pq ∑_n≥0(-1)^n fracz^Nn+d_pq(Nn+d_pq)!. This is the stated formula.
Exact mathematical passage
J_N^{Nn+d_{pq}}e_q = (-1)^n\varepsilon_{pq}e_p.
E_qUse in source · 2 passages
Search aliases: E_q
Use in source · atlas-pages.tex, line 274
The four real roots of this polynomial are ±1,±sqrt5. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output. headAn inverse operation, not a truncation error Applying cosh(∂_x) returns x^4 exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are E_m'=mE_m-1, E_m(x+y)=∑_q=0^mbinom mqE_q(x)y^m-q. Both follow from the generating function. No operator-series convergence as…
Exact mathematical passage
E_m'=mE_{m-1},\qquad E_m(x+y)=\sum_{q=0}^m\binom mqE_q(x)y^{m-q}.
e_qUse in source · 1 passage
Search aliases: e_q
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 877
For every nonnegative integer n, J_N^Nn+d_pqe_q = (-1)^nε_pqe_p. Only powers whose exponent is congruent to d_pq modulo N contribute to the (p,q) entry of the exponential. Therefore (e^zJ_N)_pq = ε_pq ∑_n≥0(-1)^n fracz^Nn+d_pq(Nn+d_pq)!. This is the stated formula.
Exact mathematical passage
J_N^{Nn+d_{pq}}e_q = (-1)^n\varepsilon_{pq}e_p.
E_q^{(N,j,r)}Use in source · 2 passages
Search aliases: E_q^(N,j,r)
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1474
[Appell laws] For m≥1, the polynomials satisfy fracddxE_m^(N,j,r)(x) = mE_m-1^(N,j,r)(x) and, for m≥0, E_m^(N,j,r)(x+y) = ∑_q=0^m binom mq E_q^(N,j,r)(x)y^m-q.
Exact mathematical passage
E_m^{(N,j,r)}(x+y) = \sum_{q=0}^{m} \binom mq E_q^{(N,j,r)}(x)y^{m-q}.
e_\lambdaUse in source · 3 passages
Search aliases: e_λ lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2306
[Diagonal Fredholm determinant representation] Let I_N be a multiplicity-sensitive countable index set for Λ_N and put H_N=ℓ^2(I_N). On its standard orthonormal basis, define the diagonal operator K_Ne_λ=λ^-1e_λ. Then K_N is trace class and Δ_N(w)=det_ F(I-wK_N). For every m≥1, κ_N,m=tr(K_N^m).
Exact mathematical passage
K_Ne_\lambda=\lambda^{-1}e_\lambda.
\Efun_{2,0}Use in source · 1 passage
Search aliases: Efun_2,0 E_2,0 Efun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 569
For N=1, the only section is e^λ z. For N=2, E_2,0(z)=cosh z, E_2,1(z)=sinh z. These are classical instances of the same cyclic basis.
Exact mathematical passage
\Efun_{2,0}(z)=\cosh z, \qquad \Efun_{2,1}(z)=\sinh z.
\Efun_{2,1}Use in source · 1 passage
Search aliases: Efun_2,1 E_2,1 Efun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 569
For N=1, the only section is e^λ z. For N=2, E_2,0(z)=cosh z, E_2,1(z)=sinh z. These are classical instances of the same cyclic basis.
Exact mathematical passage
\Efun_{2,0}(z)=\cosh z, \qquad \Efun_{2,1}(z)=\sinh z.
\Efun_{N,i}Use in source · 1 passage
Search aliases: Efun_N,i E_N,i Efun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 509
[Addition law] For 0≤ k<N, E_N,k(λ(z+w)) = ∑_substack0≤ i,j<N i+j≡ kbmod N E_N,i(λ z) E_N,j(λ w).
Exact mathematical passage
\Efun_{N,k}(\lambda(z+w)) = \sum_{\substack{0\leq i,j<N\\ i+j\equiv k\bmod N}} \Efun_{N,i}(\lambda z)\Efun_{N,j}(\lambda w).
\Efun_{N,j-1}Use in source · 1 passage
Search aliases: Efun_N,j-1 E_N,j-1 Efun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 445
[Differential cycle and initial data] For 1≤ j<N, fracddz E_N,j(λ z) = λ E_N,j-1(λ z), while fracddz E_N,0(λ z) = λ E_N,N-1(λ z). Thus fracd^Ndz^N E_N,j(λ z) = λ^N E_N,j(λ z). For 0≤ k<N, .fracd^kdz^k E_N,j(λ z)|_z=0 = δ_kjλ^j. If λ≠0, the N functions E_N,j(λ z) form a basis of the solution space of y^(N)=λ^Ny.
Exact mathematical passage
\frac{d}{dz}\Efun_{N,j}(\lambda z) = \lambda\Efun_{N,j-1}(\lambda z),
\Efun_{N,k}Use in source · 1 passage
Search aliases: Efun_N,k E_N,k Efun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 509
[Addition law] For 0≤ k<N, E_N,k(λ(z+w)) = ∑_substack0≤ i,j<N i+j≡ kbmod N E_N,i(λ z) E_N,j(λ w).
Exact mathematical passage
\Efun_{N,k}(\lambda(z+w)) = \sum_{\substack{0\leq i,j<N\\ i+j\equiv k\bmod N}} \Efun_{N,i}(\lambda z)\Efun_{N,j}(\lambda w).
\Efun_{N,N-1}Use in source · 2 passages
Search aliases: Efun_N,N-1 E_N,N-1 Efun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 451
[Differential cycle and initial data] For 1≤ j<N, fracddz E_N,j(λ z) = λ E_N,j-1(λ z), while fracddz E_N,0(λ z) = λ E_N,N-1(λ z). Thus fracd^Ndz^N E_N,j(λ z) = λ^N E_N,j(λ z). For 0≤ k<N, .fracd^kdz^k E_N,j(λ z)|_z=0 = δ_kjλ^j. If λ≠0, the N functions E_N,j(λ z) form a basis of the solution space of y^(N)=λ^Ny.
Exact mathematical passage
\frac{d}{dz}\Efun_{N,0}(\lambda z) = \lambda\Efun_{N,N-1}(\lambda z).
FUse in source · 39 passages
Search aliases: F
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 214
For a power series F(z)=∑_m≥ 0a_mz^m and a positive integer N, the coefficients may be separated according to the residue of m modulo N. Root of unity averaging gives exact projection operators for this separation. This elementary construction is usually called a multisection or a root filter. It appears throughout the classical theory of generating functions and special functions. Background on multisections, hypergeometric functions, and related coefficient systems may be found in [source refe…
Exact mathematical passage
F(z)=\sum_{m\geq 0}a_mz^m
F_2Use in source · 5 passages
Search aliases: F_2
Use in source · atlas-pages.tex, line 58
vfigroot_filter headAn example that can be checked coefficient by coefficient Take F(z)=∑_m=0^12(m+1)z^m. Then B_4,2F=3z^2+7z^6+11z^10. Applying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original F. headSelection is also a symmetry condition For the selected polynomial F_2, rotation gives F_2(iz)=-F_2(z). More generally, a residue-j section satisfies F_j(ω z)=ω^jF_j(z). The phase label therefore describes …
Exact mathematical passage
F_2
F_iUse in source · 4 passages
Search aliases: F_i
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 417
Write F=∑_iF_i and G=∑_jG_j, where F_i= B_N,iF and G_j= B_N,jG. Their covariance gives (F_iG_j)(ω z)=ω^i+jF_i(z)G_j(z), so F_iG_j belongs to the packet labelled by i+j modulo N. Expanding the finite product and applying B_N,k gives the formula.
Exact mathematical passage
F=\sum_iF_i
F_jUse in source · 2 passages
Search aliases: F_j
Use in source · atlas-pages.tex, line 58
F(z)=∑_m=0^12(m+1)z^m. Then B_4,2F=3z^2+7z^6+11z^10. Applying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original F. headSelection is also a symmetry condition For the selected polynomial F_2, rotation gives F_2(iz)=-F_2(z). More generally, a residue-j section satisfies F_j(ω z)=ω^jF_j(z). The phase label therefore describes both the coefficient addresses and the response to rotation. takeawayThis is not an…
Exact mathematical passage
F_j(\omega z)=\omega^jF_j(z)
F_{N-1}Use in source · 1 passage
Search aliases: F_N-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 540
[Hypergeometric form] Let A_N,j = { fracj+mN mathrel: 1≤ m≤ N and j+m≠ N }. Then E_N,j(λ z) = frac(λ z)^jj! _0F_N-1 ( - A_N,j mathrel, frac(λ z)^NN^N ).
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \frac{(\lambda z)^j}{j!} {}_0F_{N-1} \left( \begin{matrix}-\\ A_{N,j}\end{matrix} \mathrel{,} \frac{(\lambda z)^N}{N^N} \right).
\FdetUse in source · 2 passages
Search aliases: Fdet det_ F operatorname
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2310
[Diagonal Fredholm determinant representation] Let I_N be a multiplicity-sensitive countable index set for Λ_N and put H_N=ℓ^2(I_N). On its standard orthonormal basis, define the diagonal operator K_Ne_λ=λ^-1e_λ. Then K_N is trace class and Δ_N(w)=det_ F(I-wK_N). For every m≥1, κ_N,m=tr(K_N^m).
Exact mathematical passage
\Delta_N(w)=\Fdet(I-wK_N).
G'Use in source · 3 passages
Search aliases: G'
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2026
/2)|u| shows that Φ_m maps B_m into itself. The derivative bound |Φ_m'|≤1/2 makes it a contraction. Banach's fixed-point theorem gives a unique fixed point. On B_m, |a(z-t_m)|≤1<π, so the principal logarithm inverts the exponential and the fixed-point equation is equivalent to G(z)=0. Moreover, |z-t_m|≤frac34s_N|R_N(z)|=O(e^-γ_Nt_m). At this zero, |G'(z)|≥ |a| |1+R_N(z)|-|R_N'(z)|≥ s_N/2>0, so the zero is simple. The disks are mutually disjoint because their diameter 1/s_N is smaller than the ce…
Exact mathematical passage
|G'(z)|\geq |a|\,|1+R_N(z)|-|R_N'(z)|\geq s_N/2>0,
g''Use in source · 1 passage
Search aliases: g''
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1318
For a function g, define the complete differential Bell polynomial Y_N by fracd^Ndz^Ne^g(z) = e^g(z)Y_N(g',g'',…,g^(N)).
Exact mathematical passage
\frac{d^N}{dz^N}e^{g(z)} = e^{g(z)}Y_N\bigl(g',g'',\ldots,g^{(N)}\bigr).
g^{(N)}Use in source · 1 passage
Search aliases: g^(N)
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1318
For a function g, define the complete differential Bell polynomial Y_N by fracd^Ndz^Ne^g(z) = e^g(z)Y_N(g',g'',…,g^(N)).
Exact mathematical passage
\frac{d^N}{dz^N}e^{g(z)} = e^{g(z)}Y_N\bigl(g',g'',\ldots,g^{(N)}\bigr).
G_jUse in source · 4 passages
Search aliases: G_j
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 417
Write F=∑_iF_i and G=∑_jG_j, where F_i= B_N,iF and G_j= B_N,jG. Their covariance gives (F_iG_j)(ω z)=ω^i+jF_i(z)G_j(z), so F_iG_j belongs to the packet labelled by i+j modulo N. Expanding the finite product and applying B_N,k gives the formula.
Exact mathematical passage
G=\sum_jG_j
\mathcal HUse in source · 1 passage
Search aliases: H mathcal
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2304
[Diagonal Fredholm determinant representation] Let I_N be a multiplicity-sensitive countable index set for Λ_N and put H_N=ℓ^2(I_N). On its standard orthonormal basis, define the diagonal operator K_Ne_λ=λ^-1e_λ. Then K_N is trace class and Δ_N(w)=det_ F(I-wK_N). For every m≥1, κ_N,m=tr(K_N^m).
Exact mathematical passage
\mathcal H_N=\ell^2(I_N)
h_4Use in source · 1 passage
Search aliases: h_4
Use in source · indicator.tex, line 28
1615) (11.95250,3.25792) (11.98292,3.19481) (12.01333,3.12685) (12.04375,3.05410) (12.07417,2.97659) (12.10458,2.89439) (12.13500,2.80755) (12.16542,2.71613) (12.19583,2.62019) (12.22625,2.51980) (12.25667,2.41503) (12.28708,2.30595) (12.31750,2.19263) (12.34792,2.07515) (12.37833,1.95359) (12.40875,1.82804) (12.43917,1.69859) (12.46958,1.56531) (12.50000,1.42830); node[] at (8.90000,4.65000) h_4(θ)=frac|cosθ|+|sinθ|sqrt2; node[] at (8.90000,0.05000) minima: two dominant exponentials tie;
Exact mathematical passage
h_4(\theta)=\frac{|\cos\theta|+|\sin\theta|}{\sqrt2}
\mathcal H_NUse in source · 1 passage
Search aliases: H_N mathcal
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2304
[Diagonal Fredholm determinant representation] Let I_N be a multiplicity-sensitive countable index set for Λ_N and put H_N=ℓ^2(I_N). On its standard orthonormal basis, define the diagonal operator K_Ne_λ=λ^-1e_λ. Then K_N is trace class and Δ_N(w)=det_ F(I-wK_N). For every m≥1, κ_N,m=tr(K_N^m).
Exact mathematical passage
\mathcal H_N=\ell^2(I_N)
\hboxUse in source · 2 passages
Search aliases: hbox
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1517
Multiply the generating functions C_N,j(z)^re^xz hboxand C_N,j(z)^se^yz. Their product is the generating function with parameters r+s and x+y. Coefficient comparison gives the formula.
Exact mathematical passage
\Cfun_{N,j}(z)^re^{xz} \quad\hbox{and}\quad \Cfun_{N,j}(z)^se^{yz}.
iUse in source · 91 passages
Search aliases: i
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 334
Fix N≥1 and write ω=e^2π i/N. Let Ω be a disk centered at the origin, with the whole plane allowed, and assume that Ω is invariant under multiplication by ω. For a holomorphic function F on Ω, define ( B_N,jF)(z) = frac1N∑_ℓ=0^N-1ω^-jℓF(ω^ℓ z), 0≤ j<N. The same formula is valid for formal power series.
Exact mathematical passage
\omega=e^{2\pi i/N}.
i^{(m-j)\ell}Use in source · 3 passages
Search aliases: i^(m-j)ℓ ell
Use in source · atlas-pages.tex, line 49
vpage3A root filter is an exact voteCYCLIC PROJECTIONTraditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5). For N=4, write ω=i. To extract residue j, evaluate the function at four rotated arguments and attach the phase i^-jℓ: ( B_4,jF)(z)=frac14∑_ℓ=0^3i^-jℓF(i^ℓ z). A monomial a_mz^m is multiplied by the finite average of i^(m-j)ℓ. Matching phases all vote 1. Nonmatching phases cancel exactly. vfigroot_filter headAn example that can be checked coefficient by coefficient Take F(z)=∑_m…
Exact mathematical passage
i^{(m-j)\ell}
i^{-j\ell}Use in source · 4 passages
Search aliases: i^-jℓ ell
Use in source · atlas-pages.tex, line 45
vpage3A root filter is an exact voteCYCLIC PROJECTIONTraditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5). For N=4, write ω=i. To extract residue j, evaluate the function at four rotated arguments and attach the phase i^-jℓ: ( B_4,jF)(z)=frac14∑_ℓ=0^3i^-jℓF(i^ℓ z). A monomial a_mz^m is multiplied by the finite average of i^(m-j)ℓ. Matching phases all vote 1. Nonmatching phases cancel exactly. vfigroot_filter headAn example that can be checked coefficient by coefficient Take F(z)=∑_m…
Exact mathematical passage
i^{-j\ell}
i^kUse in source · 3 passages
Search aliases: i^k
Use in source · atlas-pages.tex, line 142
vpage8Four zeros become one orbit zeroTHE REDUCED VARIABLETraditional monograph: Definition 5.7 (p. 9), Example 11.12 (p. 31), Corollary 11.13 (p. 31). For N=4 the factorization is exact: D_4(z)=cos(z/sqrt2)cosh(z/sqrt2), D_4(z)=Δ_4(z^4). Set A_4=sqrt2π. All the zeros are A_4(m+1/2)i^k, with m≥0 and 0≤ k<4. Each zero is simple. vfigorbit_reduction The drawing uses u=z/A_4 and v=w/A_4^4, so its marked coordinates are rational. The first four-point orbit has fourth power 1/16. The second has fourt…
Exact mathematical passage
A_4(m+1/2)i^k
i^\ellUse in source · 2 passages
Search aliases: i^ℓ ell
Use in source · atlas-pages.tex, line 46
vpage3A root filter is an exact voteCYCLIC PROJECTIONTraditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5). For N=4, write ω=i. To extract residue j, evaluate the function at four rotated arguments and attach the phase i^-jℓ: ( B_4,jF)(z)=frac14∑_ℓ=0^3i^-jℓF(i^ℓ z). A monomial a_mz^m is multiplied by the finite average of i^(m-j)ℓ. Matching phases all vote 1. Nonmatching phases cancel exactly. vfigroot_filter headAn example that can be checked coefficient by coefficient Take F(z)=∑_m…
Exact mathematical passage
(\mathcal B_{4,j}F)(z)=\frac14\sum_{\ell=0}^{3}i^{-j\ell}F(i^\ell z).
I_NUse in source · 2 passages
Search aliases: I_N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2303
[Diagonal Fredholm determinant representation] Let I_N be a multiplicity-sensitive countable index set for Λ_N and put H_N=ℓ^2(I_N). On its standard orthonormal basis, define the diagonal operator K_Ne_λ=λ^-1e_λ. Then K_N is trace class and Δ_N(w)=det_ F(I-wK_N). For every m≥1, κ_N,m=tr(K_N^m).
Exact mathematical passage
I_N
J_3Use in source · 13 passages
Search aliases: J_3
Use in source · atlas-pages.tex, line 85
vpage5One signed edge determines the cycleDIFFERENTIAL SYSTEMTraditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12). Let J_3e_0=e_1, J_3e_1=e_2, and J_3e_2=-e_0. A complete turn changes sign, so J_3^3=-I. The signed sections are the coordinates of e^zJ_3e_0. vfigsigned_shift Here D_j inside the matrix means D_3,j(z). For example, D_3,0=1-fracz^33!+fracz^66!-⋯, D_3,1=z-fracz^44!+⋯. Rows are output coordinates and columns are input coordinates.…
Exact mathematical passage
J_3e_0=e_1
J_3^3Use in source · 3 passages
Search aliases: J_3^3
Use in source · atlas-pages.tex, line 85
vpage5One signed edge determines the cycleDIFFERENTIAL SYSTEMTraditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12). Let J_3e_0=e_1, J_3e_1=e_2, and J_3e_2=-e_0. A complete turn changes sign, so J_3^3=-I. The signed sections are the coordinates of e^zJ_3e_0. vfigsigned_shift Here D_j inside the matrix means D_3,j(z). For example, D_3,0=1-fracz^33!+fracz^66!-⋯, D_3,1=z-fracz^44!+⋯. Rows are output coordinates and columns are input coordinates.…
Exact mathematical passage
J_3^3=-I
J_4Use in source · 1 passage
Search aliases: J_4
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 755
The signed shift is a finite carrier for the differential cycle. At order four, three basis steps have coefficient one and the final step has coefficient minus one. Four steps return the vector with reversed sign. The matrix exponential packages every finite path length in one analytic expression. modelfiguresigned_cycleThe exact signed shift J_4. Source columns and target rows describe the same action as the cycle on the left. Direct integer multiplication verifies J_4^4=-I.fig:signed-cycle
Exact mathematical passage
J_4
J_4^4Use in source · 1 passage
Search aliases: J_4^4
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 755
steps have coefficient one and the final step has coefficient minus one. Four steps return the vector with reversed sign. The matrix exponential packages every finite path length in one analytic expression. modelfiguresigned_cycleThe exact signed shift J_4. Source columns and target rows describe the same action as the cycle on the left. Direct integer multiplication verifies J_4^4=-I.fig:signed-cycle
Exact mathematical passage
J_4^4=-I
J_N^{N-1}Use in source · 1 passage
Search aliases: J_N^N-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 836
The vectors e_0,J_Ne_0,…,J_N^N-1e_0 are the standard basis. Thus e_0 is cyclic for J_N, so the minimal polynomial of J_N has degree N. Since J_N^N=-I, that minimal polynomial divides x^N+1 and therefore equals x^N+1. The characteristic polynomial has the same degree and is divisible by the minimal polynomial, so it too equals x^N+1. This polynomial has distinct roots. Consequently J_N is diagonalizable and its spectrum consists of all N roots of -1, each once. Hence tr(e^zJ_N) = ∑_η^N=-1e^η z = …
Exact mathematical passage
e_0,J_Ne_0,\ldots,J_N^{N-1}e_0
J_N^{Nn+d_{pq}}Use in source · 1 passage
Search aliases: J_N^Nn+d_pq
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 877
For every nonnegative integer n, J_N^Nn+d_pqe_q = (-1)^nε_pqe_p. Only powers whose exponent is congruent to d_pq modulo N contribute to the (p,q) entry of the exponential. Therefore (e^zJ_N)_pq = ε_pq ∑_n≥0(-1)^n fracz^Nn+d_pq(Nn+d_pq)!. This is the stated formula.
Exact mathematical passage
J_N^{Nn+d_{pq}}e_q = (-1)^n\varepsilon_{pq}e_p.
K_N^mUse in source · 2 passages
Search aliases: K_N^m
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2314
[Diagonal Fredholm determinant representation] Let I_N be a multiplicity-sensitive countable index set for Λ_N and put H_N=ℓ^2(I_N). On its standard orthonormal basis, define the diagonal operator K_Ne_λ=λ^-1e_λ. Then K_N is trace class and Δ_N(w)=det_ F(I-wK_N). For every m≥1, κ_N,m=tr(K_N^m).
Exact mathematical passage
\kappa_{N,m}=\tr(K_N^m).
K_N^qUse in source · 2 passages
Search aliases: K_N^q
Use in source · atlas-pages.tex, line 402
vpage22Finite coefficients give infinite momentsZERO POWER SUMSTraditional monograph: Theorem 12.6 (p. 35), Corollary 12.7 (p. 36), Corollary 12.9 (p. 37), Corollary 12.10 (p. 37). Taking the logarithm of the canonical product identifies κ_N,q=σ_N,q:=∑_λ∈Λ_Nλ^-q =Tr(K_N^q), q≥1. The same exact numbers can be computed from finitely many denominator coefficients by Newton's recurrence. No numerical zero list is needed for that calculation. vfigmoments For order two, σ_2,1=frac12, σ_2,2=frac16, σ_2…
Exact mathematical passage
\kappa_{N,q}=\sigma_{N,q}:=\sum_{\lambda\in\Lambda_N}\lambda^{-q} =\Tr(K_N^q),\qquad q\ge1.
\lfloorUse in source · 4 passages
Search aliases: lfloor
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 654
[Twisted addition law] For 0≤ k<N, D_N,k(z+w) = ∑_substack0≤ i,j<N i+j≡ kbmod N (-1)^lfloor(i+j)/Nrfloor D_N,i(z) D_N,j(w).
Exact mathematical passage
\Dfun_{N,k}(z+w) = \sum_{\substack{0\leq i,j<N\\ i+j\equiv k\bmod N}} (-1)^{\lfloor(i+j)/N\rfloor} \Dfun_{N,i}(z)\Dfun_{N,j}(w).
\limsupUse in source · 2 passages
Search aliases: limsup
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1729
[Order from Taylor coefficients] If f(w)=∑_n≥0a_nw^n is an entire function with infinitely many nonzero coefficients, then ord f = limsup_substackn→∞ a_n≠0 fracnlog nlog(1/|a_n|). Canonical products for entire functions of order below one have genus zero. We use these standard facts in the form given in [source reference].
Exact mathematical passage
\ord f = \limsup_{\substack{n\to\infty\\a_n\ne0}} \frac{n\log n}{\log(1/|a_n|)}.
\limsup_{r\to\infty}Use in source · 1 passage
Search aliases: limsup_r→∞ limsup to infty
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1786
[Indicator polygon and dominant sectors] The indicator of D_N equals h_N, with the limsup taken over radii where D_N(re^iθ)≠0: limsup_r→∞ frac1rlog| D_N(re^iθ)| = h_N(θ). If K is a compact set of directions disjoint from T_N, then there are δ>0 and R>0 such that, for each θ∈ K, one root η(θ) of -1 satisfies D_N(re^iθ) = frac1N exp(η(θ)re^iθ) (1+O(e^-δ r)) uniformly for r≥ R. In particular, no sufficiently large zero can lie in those directions.
Exact mathematical passage
\limsup_{r\to\infty} \frac1r\log|\Dfun_N(re^{i\theta})| = h_N(\theta).
\limsup_{\substack{n\to\infty\\a_n\ne0}}Use in source · 1 passage
Search aliases: limsup_substackn→∞ a_n≠0 limsup substack to infty a ne
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1729
[Order from Taylor coefficients] If f(w)=∑_n≥0a_nw^n is an entire function with infinitely many nonzero coefficients, then ord f = limsup_substackn→∞ a_n≠0 fracnlog nlog(1/|a_n|). Canonical products for entire functions of order below one have genus zero. We use these standard facts in the form given in [source reference].
Exact mathematical passage
\ord f = \limsup_{\substack{n\to\infty\\a_n\ne0}} \frac{n\log n}{\log(1/|a_n|)}.
\log_{10}Use in source · 1 passage
Search aliases: log_10 log
Use in source · pole_asymptotic.tex, line 39
); draw[fill=black] (2.42857,3.23383) circle (0.055); draw[fill=black] (4.05714,2.85954) circle (0.055); draw[fill=black] (5.68571,2.48068) circle (0.055); draw[fill=black] (7.31429,2.10005) circle (0.055); draw[fill=black] (8.94286,1.71875) circle (0.055); draw[fill=black] (10.57143,1.33719) circle (0.055); draw[fill=black] (12.20000,0.95555) circle (0.055); node[] at (6.50000,4.90000) log_10(1-R_n), R_n=fracP_2,n(1)2(2n)!(2/π)^2n+1; node[] at (6.50000,-0.25000) Each ratio is enclosed with a ra…
Exact mathematical passage
\log_{10}(1-R_n),\quad R_n=\frac{P_{2,n}(1)}{2(2n)!(2/\pi)^{2n+1}}
MUse in source · 12 passages
Search aliases: M
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1749
The coefficient of w^n in Δ_N has modulus 1/(Nn)!. Stirling's formula and [source reference] give order 1/N. Since M_f( · ^N)(r)=M_f(r^N) for the maximum modulus, composition with z^N multiplies the order by N and gives order one for D_N.
Exact mathematical passage
M_{f(\,\cdot\,^N)}(r)=M_f(r^N)
m^NUse in source · 2 passages
Search aliases: m^N
Use in source · atlas-pages.tex, line 438
vpage24Convergence is not continuationZERO ZETA FUNCTIONSTraditional monograph: Theorem 14.2 (p. 39), Theorem 14.3 (p. 39), Corollary 14.4 (p. 41). Reduced zeros grow like m^N. Their absolute zero-zeta series therefore has half-plane of absolute convergence Re s>1/N. The absolute zero zeta in the original variable instead requires Re s>1. vfigzeta_plane For N=2 and N=4, the lattices are exact: Z_2^rm red(s)=π^-2sζ(2s,1/2), Z_4^rm red(s)=(sqrt2π)^-4sζ(4s,1/2). Each identity first holds where its …
Exact mathematical passage
m^N
m^{r-2}Use in source · 3 passages
Search aliases: m^r-2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2651
[Leading dominant zero term] Under the hypotheses of [source reference], P_N,n(r) = · m!binomm+r-1r-1 ∑_ρ∈ Z_R fracρ^-m(-ρ D_N'(ρ))^r · +O(m!m^r-2R^-m) +O_R_*(m!R_*^-m). For r=1, the first error term is absent.
Exact mathematical passage
\begin{aligned} P_{N,n}(r) ={}&m!\binom{m+r-1}{r-1} \sum_{\rho\in\Zcal_R} \frac{\rho^{-m}}{(-\rho\Dfun_N'(\rho))^r} \\ &+O\left(m!m^{r-2}R^{-m}\right) +O_{R_*}\left(m!R_*^{-m}\right). \end{aligned}
m_0Use in source · 3 passages
Search aliases: m_0
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1945
[Large zero strings] Assume N≥3. There are constants m_0, C, and γ>0 and a sequence of positive real simple zeros (τ_m)_m≥ m_0 of D_N such that τ_m = t_m+O(e^-γ t_m). Every zero of sufficiently large modulus is of the form ω^kτ_m for a unique 0≤ k<N and a unique m≥ m_0.
Exact mathematical passage
m_0
M_fUse in source · 1 passage
Search aliases: M_f
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1749
The coefficient of w^n in Δ_N has modulus 1/(Nn)!. Stirling's formula and [source reference] give order 1/N. Since M_f( · ^N)(r)=M_f(r^N) for the maximum modulus, composition with z^N multiplies the order by N and gives order one for D_N.
Exact mathematical passage
M_{f(\,\cdot\,^N)}(r)=M_f(r^N)
M_{f(\,\cdot\,^N)}Use in source · 1 passage
Search aliases: M_f( · ^N) cdot
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1749
The coefficient of w^n in Δ_N has modulus 1/(Nn)!. Stirling's formula and [source reference] give order 1/N. Since M_f( · ^N)(r)=M_f(r^N) for the maximum modulus, composition with z^N multiplies the order by N and gives order one for D_N.
Exact mathematical passage
M_{f(\,\cdot\,^N)}(r)=M_f(r^N)
M_LUse in source · 3 passages
Search aliases: M_L
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2924
For N≥3, [source reference] gives | D_N'(ρ)|^-1 ≤ C e^-c_N|ρ| for every sufficiently large simple zero. Let L be a compact subset of the s plane. Because the chosen arguments of the zeros lie in a fixed bounded interval, there are constants C_L and M_L such that sup_s∈ L|e^-sLogρ| ≤ C_L|ρ|^M_L for every sufficiently large zero. Hence the absolute value of the corresponding summand is at most C_L'|ρ|^M_Le^-q c_N|ρ|. The linear zero counting estimate makes this majorant summable. The Weierstrass t…
Exact mathematical passage
M_L
\mathrelUse in source · 3 passages
Search aliases: mathrel
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 528
[Hypergeometric form] Let A_N,j = { fracj+mN mathrel: 1≤ m≤ N and j+m≠ N }. Then E_N,j(λ z) = frac(λ z)^jj! _0F_N-1 ( -
Exact mathematical passage
A_{N,j} = \left\{ \frac{j+m}{N} \mathrel{:} 1\leq m\leq N \text{ and } j+m\ne N \right\}.
\max_{0\le k<N}Use in source · 2 passages
Search aliases: max_0≤ k<N max le
Use in source · atlas-pages.tex, line 311
om a polygonDOMINANT EXPONENTSTraditional monograph: Definition 11.3 (p. 26), Theorem 11.4 (p. 26). Write the signed denominator as a finite exponential sum: D_N(z)=frac1N∑_η^N=-1e^η z. For z=re^iθ, the size of one exponential is e^rRe(η e^iθ). The maximal projection determines the directional exponential growth. vfigindicator The indicator is h_N(θ)=max_0≤ k<Ncos(θ+frac(2k+1)πN). Away from a tie direction, one exponential dominates all the others by a uniform positive projection gap on compact …
Exact mathematical passage
h_N(\theta)=\max_{0\le k<N}\cos\left(\theta+\frac{(2k+1)\pi}{N}\right).
N^NUse in source · 3 passages
Search aliases: N^N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 540
[Hypergeometric form] Let A_N,j = { fracj+mN mathrel: 1≤ m≤ N and j+m≠ N }. Then E_N,j(λ z) = frac(λ z)^jj! _0F_N-1 ( - A_N,j mathrel, frac(λ z)^NN^N ).
Exact mathematical passage
\Efun_{N,j}(\lambda z) = \frac{(\lambda z)^j}{j!} {}_0F_{N-1} \left( \begin{matrix}-\\ A_{N,j}\end{matrix} \mathrel{,} \frac{(\lambda z)^N}{N^N} \right).
N^{Nn}Use in source · 3 passages
Search aliases: N^Nn
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 555
The multiplication formula for the gamma function gives (Nn+j)! = j!N^Nnn! ∏_a∈ A_N,j(a)_n. Indeed, the N Pochhammer factors supplied by the multiplication formula are ((j+m)/N)_n for 1≤ m≤ N; the excluded factor is ((j+N-j)/N)_n=(1)_n=n!. Insertion into the defining series gives the stated hypergeometric expansion.
Exact mathematical passage
(Nn+j)! = j!N^{Nn}n! \prod_{a\in A_{N,j}}(a)_n.
n_{D_4}Use in source · 1 passage
Search aliases: n_D_4
Use in source · counting.tex, line 30
(2.19219,1.00000) circle (0.055); node[] at (2.19219,0.58000) 81/16; node[font=sffamilyfootnotesize] at (2.19219,1.45000) count 2; draw[line width=1pt] (11.54219,0.88000) -- (11.54219,1.12000); draw[fill=black] (11.54219,1.00000) circle (0.055); node[] at (11.54219,0.58000) 625/16; node[font=sffamilyfootnotesize] at (11.54219,1.45000) count 3; node[] at (6.50000,-0.05000) n_D_4(sqrt2π u)=4lfloor u+1/2rfloor, n_Δ_4(4π^4 v)=lfloor v^1/4+1/2rfloor;
Exact mathematical passage
n_{D_4}(\sqrt2\pi u)=4\lfloor u+1/2\rfloor,\quad n_{\Delta_4}(4\pi^4 v)=\lfloor v^{1/4}+1/2\rfloor
n_{D_N}Use in source · 4 passages
Search aliases: n_D_N
Use in source · atlas-pages.tex, line 369
ions are v=(m+1/2)^4. vfigcounting The two rows describe the same zero orbits. The upper row counts four zeros each time a new radius is reached. The lower row counts one reduced zero each time its fourth-power location is reached. Both formulas include zeros on the boundary of the disk. headSeparate radius conversion from orbit multiplicity In unscaled variables the general relation is n_D_N(R^1/N)=Nn_Δ_N(R). There are two operations here: replace the radius by its Nth root and divide the count…
Exact mathematical passage
n_{D_N}(R^{1/N})=Nn_{\Delta_N}(R).
n_{\Dfun_N}Use in source · 3 passages
Search aliases: n_Dfun_N n_ D_N Dfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2219
[Zero counting asymptotics] Let n_ D_N(R) count the zeros of D_N in |z|≤ R, with multiplicity. Let n_Δ_N(R) count the zeros of Δ_N in |w|≤ R. Then, as R→∞, n_ D_N(R) = fracNsin(π/N)πR+O(1) and n_Δ_N(R) = fracsin(π/N)πR^1/N+O(1).
Exact mathematical passage
n_{\Dfun_N}(R)
n_{\Delta_4}Use in source · 1 passage
Search aliases: n_Δ_4 Delta
Use in source · counting.tex, line 30
(2.19219,1.00000) circle (0.055); node[] at (2.19219,0.58000) 81/16; node[font=sffamilyfootnotesize] at (2.19219,1.45000) count 2; draw[line width=1pt] (11.54219,0.88000) -- (11.54219,1.12000); draw[fill=black] (11.54219,1.00000) circle (0.055); node[] at (11.54219,0.58000) 625/16; node[font=sffamilyfootnotesize] at (11.54219,1.45000) count 3; node[] at (6.50000,-0.05000) n_D_4(sqrt2π u)=4lfloor u+1/2rfloor, n_Δ_4(4π^4 v)=lfloor v^1/4+1/2rfloor;
Exact mathematical passage
n_{D_4}(\sqrt2\pi u)=4\lfloor u+1/2\rfloor,\quad n_{\Delta_4}(4\pi^4 v)=\lfloor v^{1/4}+1/2\rfloor
n_{\Delta_N}Use in source · 7 passages
Search aliases: n_Δ_N Delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2219
[Zero counting asymptotics] Let n_ D_N(R) count the zeros of D_N in |z|≤ R, with multiplicity. Let n_Δ_N(R) count the zeros of Δ_N in |w|≤ R. Then, as R→∞, n_ D_N(R) = fracNsin(π/N)πR+O(1) and n_Δ_N(R) = fracsin(π/N)πR^1/N+O(1).
Exact mathematical passage
n_{\Delta_N}(R)
OUse in source · 30 passages
Search aliases: O
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 994
The identity Δ_N(w)^-r = exp( r∑_m≥1κ_N,mfracw^mm ) fits the defining generating function for B_n with x_m=r(m-1)!κ_N,m. The coefficient of w^n is therefore the displayed Bell polynomial divided by n!. Multiplication by (Nn)! gives P_N,n(r). Since Δ_N(w)=1-fracwN!+O(w^2), we have κ_N,1=1/N!. The term x_1^n gives the stated leading coefficient, which is nonzero.
Exact mathematical passage
\Delta_N(w)=1-\frac{w}{N!}+O(w^2),
oUse in source · 3 passages
Search aliases: o
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1758
The finite exponential representation gives | D_N(z)|≤ e^|z|, so the type is at most one. Choose a root η_0 and put z=toverlineη_0 with t>0. The term e^η_0z=e^t has strictly larger real exponent than every other term. Hence D_N(toverlineη_0) = frac1Ne^t(1+o(1)). The type is therefore one.
Exact mathematical passage
\Dfun_N(t\overline{\eta_0}) = \frac1Ne^t(1+o(1)).
O_{R_*}Use in source · 4 passages
Search aliases: O_R_*
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2607
rcle_expansion Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
P_{N,n}(r) = m! \sum_{\rho\in\Zcal_R} \sum_{\ell=1}^{r} a_{\rho,\ell} \binom{m+\ell-1}{\ell-1} \rho^{-m} + O_{R_*}(m!R_*^{-m}).
\ordUse in source · 2 passages
Search aliases: ord operatorname
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1729
[Order from Taylor coefficients] If f(w)=∑_n≥0a_nw^n is an entire function with infinitely many nonzero coefficients, then ord f = limsup_substackn→∞ a_n≠0 fracnlog nlog(1/|a_n|). Canonical products for entire functions of order below one have genus zero. We use these standard facts in the form given in [source reference].
Exact mathematical passage
\ord f = \limsup_{\substack{n\to\infty\\a_n\ne0}} \frac{n\log n}{\log(1/|a_n|)}.
\overline zUse in source · 1 passage
Search aliases: overline z overline
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 715
[Symmetries] The denominator satisfies D_N(ω z)= D_N(z) and D_N(overline z)=overline D_N(z). Its nonzero zeros occur in orbits under multiplication by the Nth roots of unity and under complex conjugation.
Exact mathematical passage
\Dfun_N(\overline z)=\overline{\Dfun_N(z)}.
\overline{\Dfun_N(z)}Use in source · 1 passage
Search aliases: overlineDfun_N(z) overline D_N(z) overline Dfun overline mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 715
[Symmetries] The denominator satisfies D_N(ω z)= D_N(z) and D_N(overline z)=overline D_N(z). Its nonzero zeros occur in orbits under multiplication by the Nth roots of unity and under complex conjugation.
Exact mathematical passage
\Dfun_N(\overline z)=\overline{\Dfun_N(z)}.
\overline{\eta_0}Use in source · 2 passages
Search aliases: overlineη_0 overline eta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1757
The finite exponential representation gives | D_N(z)|≤ e^|z|, so the type is at most one. Choose a root η_0 and put z=toverlineη_0 with t>0. The term e^η_0z=e^t has strictly larger real exponent than every other term. Hence D_N(toverlineη_0) = frac1Ne^t(1+o(1)). The type is therefore one.
Exact mathematical passage
z=t\overline{\eta_0}
p_1Use in source · 1 passage
Search aliases: p_1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1106
[Logarithmic Newton recurrence] Set b_N,n(r)=fracP_N,n(r)(Nn)!. Then b_N,0(r)=1 and for every n≥1, nb_N,n(r) = r∑_k=1^nκ_N,kb_N,n-k(r). Equivalently, P_N,n(r) = frac r n ∑_k=1^n frac(Nn)!(N(n-k))! κ_N,kP_N,n-k(r). If p_k=rκ_N,k, then, for n≥1, b_N,n(r) = frac1n! det p_1 · -1 · 0 · ⋯ · 0 p_2 · p_1 · -2 · ⋯ · 0 p_3 · p_2 · p_1 · ddots · vdots vdots · vdots · ddots · ddots · -(n-1) p_n · p_n-1 · ⋯ · p_2 · p_1 .
Exact mathematical passage
b_{N,n}(r) = \frac1{n!} \det \begin{pmatrix} p_1&-1&0&\cdots&0\\ p_2&p_1&-2&\cdots&0\\ p_3&p_2&p_1&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&-(n-1)\\ p_n&p_{n-1}&\cdots&p_2&p_1 \end{pmatrix}.
P_{1,n}Use in source · 1 passage
Search aliases: P_1,n
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1203
For N=1, one has D_1(z)^-r=e^rz and P_1,n(r)=r^n. For N=2, P_2,1(r)=r, P_2,2(r)=r(3r+2), and P_2,3(r)=r(15r^2+30r+16). At r=1, these are the positive secant coefficients.
Exact mathematical passage
P_{1,n}(r)=r^n
p_2Use in source · 1 passage
Search aliases: p_2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1106
[Logarithmic Newton recurrence] Set b_N,n(r)=fracP_N,n(r)(Nn)!. Then b_N,0(r)=1 and for every n≥1, nb_N,n(r) = r∑_k=1^nκ_N,kb_N,n-k(r). Equivalently, P_N,n(r) = frac r n ∑_k=1^n frac(Nn)!(N(n-k))! κ_N,kP_N,n-k(r). If p_k=rκ_N,k, then, for n≥1, b_N,n(r) = frac1n! det p_1 · -1 · 0 · ⋯ · 0 p_2 · p_1 · -2 · ⋯ · 0 p_3 · p_2 · p_1 · ddots · vdots vdots · vdots · ddots · ddots · -(n-1) p_n · p_n-1 · ⋯ · p_2 · p_1 .
Exact mathematical passage
b_{N,n}(r) = \frac1{n!} \det \begin{pmatrix} p_1&-1&0&\cdots&0\\ p_2&p_1&-2&\cdots&0\\ p_3&p_2&p_1&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&-(n-1)\\ p_n&p_{n-1}&\cdots&p_2&p_1 \end{pmatrix}.
P_{2,1}Use in source · 2 passages
Search aliases: P_2,1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1204
For N=1, one has D_1(z)^-r=e^rz and P_1,n(r)=r^n. For N=2, P_2,1(r)=r, P_2,2(r)=r(3r+2), and P_2,3(r)=r(15r^2+30r+16). At r=1, these are the positive secant coefficients.
Exact mathematical passage
P_{2,1}(r)=r,
P_{2,2}Use in source · 2 passages
Search aliases: P_2,2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1207
For N=1, one has D_1(z)^-r=e^rz and P_1,n(r)=r^n. For N=2, P_2,1(r)=r, P_2,2(r)=r(3r+2), and P_2,3(r)=r(15r^2+30r+16). At r=1, these are the positive secant coefficients.
Exact mathematical passage
P_{2,2}(r)=r(3r+2),
P_{2,3}Use in source · 6 passages
Search aliases: P_2,3
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1211
For N=1, one has D_1(z)^-r=e^rz and P_1,n(r)=r^n. For N=2, P_2,1(r)=r, P_2,2(r)=r(3r+2), and P_2,3(r)=r(15r^2+30r+16). At r=1, these are the positive secant coefficients.
Exact mathematical passage
P_{2,3}(r)=r(15r^2+30r+16).
P_{2,n}Use in source · 5 passages
Search aliases: P_2,n
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2701
[Secant asymptotic] For N=2 and r=1, P_2,n(1) ∼ 2(2n)!(frac2π)^2n+1.
Exact mathematical passage
P_{2,n}(1) \sim 2(2n)!\left(\frac2\pi\right)^{2n+1}.
p_3Use in source · 1 passage
Search aliases: p_3
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1106
[Logarithmic Newton recurrence] Set b_N,n(r)=fracP_N,n(r)(Nn)!. Then b_N,0(r)=1 and for every n≥1, nb_N,n(r) = r∑_k=1^nκ_N,kb_N,n-k(r). Equivalently, P_N,n(r) = frac r n ∑_k=1^n frac(Nn)!(N(n-k))! κ_N,kP_N,n-k(r). If p_k=rκ_N,k, then, for n≥1, b_N,n(r) = frac1n! det p_1 · -1 · 0 · ⋯ · 0 p_2 · p_1 · -2 · ⋯ · 0 p_3 · p_2 · p_1 · ddots · vdots vdots · vdots · ddots · ddots · -(n-1) p_n · p_n-1 · ⋯ · p_2 · p_1 .
Exact mathematical passage
b_{N,n}(r) = \frac1{n!} \det \begin{pmatrix} p_1&-1&0&\cdots&0\\ p_2&p_1&-2&\cdots&0\\ p_3&p_2&p_1&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&-(n-1)\\ p_n&p_{n-1}&\cdots&p_2&p_1 \end{pmatrix}.
P_{4,2}Use in source · 4 passages
Search aliases: P_4,2
Use in source · atlas-pages.tex, line 463
ample 11.12 (p. 31), Theorem 12.6 (p. 35). Use N=4 throughout this page. Its denominator, its exact zero lattice, and the reciprocal recurrences provide two routes to the same coefficient. All symbols in the plate refer to this one order. vfigworked_circuit headRoute one: invert the denominator The w^2 coefficient of Δ_4^-1 is frac124^2-frac18!=frac698!. Thus P_4,2(1)=69. Keeping r variable gives P_4,2(r)=35r^2+34r. headRoute two: use the zero moments The exact reduced zeros are 4π^4(m+1/2)^4. T…
Exact mathematical passage
P_{4,2}(1)=69
P_LUse in source · 5 passages
Search aliases: P_L
Use in source · atlas-pages.tex, line 390
ltiplicity-sensitive genus-zero product is Δ_N(w)=∏_λ∈Λ_N(1-w/λ). Putting K_Ne_λ=λ^-1e_λ on ℓ^2(Λ_N) gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries. vfigproduct headFinite compression is an approximation For N=2, λ_m=π^2(m+1/2)^2. The finite determinant P_L is exactly the displayed finite product, but it is not exactly Δ_2. For 0≤ w≤1 and L≥1, an elementary tail estimate gives 0≤1-fracΔ_2(w)P_L(w) ≤ w∑_m≥ Lλ_m^-1 ≤frac19(L-1/2). The last b…
Exact mathematical passage
P_L
p_nUse in source · 1 passage
Search aliases: p_n
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1106
[Logarithmic Newton recurrence] Set b_N,n(r)=fracP_N,n(r)(Nn)!. Then b_N,0(r)=1 and for every n≥1, nb_N,n(r) = r∑_k=1^nκ_N,kb_N,n-k(r). Equivalently, P_N,n(r) = frac r n ∑_k=1^n frac(Nn)!(N(n-k))! κ_N,kP_N,n-k(r). If p_k=rκ_N,k, then, for n≥1, b_N,n(r) = frac1n! det p_1 · -1 · 0 · ⋯ · 0 p_2 · p_1 · -2 · ⋯ · 0 p_3 · p_2 · p_1 · ddots · vdots vdots · vdots · ddots · ddots · -(n-1) p_n · p_n-1 · ⋯ · p_2 · p_1 .
Exact mathematical passage
b_{N,n}(r) = \frac1{n!} \det \begin{pmatrix} p_1&-1&0&\cdots&0\\ p_2&p_1&-2&\cdots&0\\ p_3&p_2&p_1&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&-(n-1)\\ p_n&p_{n-1}&\cdots&p_2&p_1 \end{pmatrix}.
P_{N,0}Use in source · 1 passage
Search aliases: P_N,0
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1003
[Parameter recurrence] We have P_N,0(r)=1. For n≥1, P_N,n(r) = · - ∑_k=1^n-1 (-1)^k binomNn-1Nk P_N,n-k(r)
Exact mathematical passage
P_{N,0}(r)=1
P_{N,1}Use in source · 1 passage
Search aliases: P_N,1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1150
[The first three reciprocal polynomials] For every N≥1, P_N,1(r)=r, P_N,2(r) = r[ frac12binom2NN(r+1)-1 ], and P_N,3(r) = r[ frac(3N)!6(N!)^3(r+1)(r+2) - binom3NN(r+1) +1 ].
Exact mathematical passage
P_{N,1}(r)=r,
P_{N,2}Use in source · 1 passage
Search aliases: P_N,2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1153
[The first three reciprocal polynomials] For every N≥1, P_N,1(r)=r, P_N,2(r) = r[ frac12binom2NN(r+1)-1 ], and P_N,3(r) = r[ frac(3N)!6(N!)^3(r+1)(r+2) - binom3NN(r+1) +1 ].
Exact mathematical passage
P_{N,2}(r) = r\left[ \frac12\binom{2N}{N}(r+1)-1 \right],
P_{N,3}Use in source · 1 passage
Search aliases: P_N,3
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1161
[The first three reciprocal polynomials] For every N≥1, P_N,1(r)=r, P_N,2(r) = r[ frac12binom2NN(r+1)-1 ], and P_N,3(r) = r[ frac(3N)!6(N!)^3(r+1)(r+2) - binom3NN(r+1) +1 ].
Exact mathematical passage
P_{N,3}(r) = r\left[ \frac{(3N)!}{6(N!)^3}(r+1)(r+2) - \binom{3N}{N}(r+1) +1 \right].
P_{N,n-k}Use in source · 3 passages
Search aliases: P_N,n-k
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1004
[Parameter recurrence] We have P_N,0(r)=1. For n≥1, P_N,n(r) = · - ∑_k=1^n-1 (-1)^k binomNn-1Nk P_N,n-k(r) · -r ∑_k=1^n (-1)^k binomNn-1Nk-1 P_N,n-k(r).
Exact mathematical passage
\begin{aligned} P_{N,n}(r) ={}&- \sum_{k=1}^{n-1} (-1)^k \binom{Nn-1}{Nk} P_{N,n-k}(r) \\ &-r \sum_{k=1}^{n} (-1)^k \binom{Nn-1}{Nk-1} P_{N,n-k}(r). \end{aligned}
P_{N,n-q}Use in source · 4 passages
Search aliases: P_N,n-q
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1046
[Additive parameter law for reciprocal powers] For all r,s∈ C, S_N^(r)(z) S_N^(s)(z) = S_N^(r+s)(z) as a formal identity and on every common zero-free region with a fixed logarithm. Consequently, P_N,n(r+s) = ∑_q=0^n binomNnNq P_N,q(r)P_N,n-q(s).
Exact mathematical passage
P_{N,n}(r+s) = \sum_{q=0}^{n} \binom{Nn}{Nq} P_{N,q}(r)P_{N,n-q}(s).
P_{N,q}Use in source · 4 passages
Search aliases: P_N,q
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1046
[Additive parameter law for reciprocal powers] For all r,s∈ C, S_N^(r)(z) S_N^(s)(z) = S_N^(r+s)(z) as a formal identity and on every common zero-free region with a fixed logarithm. Consequently, P_N,n(r+s) = ∑_q=0^n binomNnNq P_N,q(r)P_N,n-q(s).
Exact mathematical passage
P_{N,n}(r+s) = \sum_{q=0}^{n} \binom{Nn}{Nq} P_{N,q}(r)P_{N,n-q}(s).
p_{n-1}Use in source · 1 passage
Search aliases: p_n-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1106
[Logarithmic Newton recurrence] Set b_N,n(r)=fracP_N,n(r)(Nn)!. Then b_N,0(r)=1 and for every n≥1, nb_N,n(r) = r∑_k=1^nκ_N,kb_N,n-k(r). Equivalently, P_N,n(r) = frac r n ∑_k=1^n frac(Nn)!(N(n-k))! κ_N,kP_N,n-k(r). If p_k=rκ_N,k, then, for n≥1, b_N,n(r) = frac1n! det p_1 · -1 · 0 · ⋯ · 0 p_2 · p_1 · -2 · ⋯ · 0 p_3 · p_2 · p_1 · ddots · vdots vdots · vdots · ddots · ddots · -(n-1) p_n · p_n-1 · ⋯ · p_2 · p_1 .
Exact mathematical passage
b_{N,n}(r) = \frac1{n!} \det \begin{pmatrix} p_1&-1&0&\cdots&0\\ p_2&p_1&-2&\cdots&0\\ p_3&p_2&p_1&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&-(n-1)\\ p_n&p_{n-1}&\cdots&p_2&p_1 \end{pmatrix}.
\pmodUse in source · 3 passages
Search aliases: pmod
Use in source · atlas-pages.tex, line 63
vpage4Products mix the residue lanesCYCLIC CONVOLUTIONTraditional monograph: Theorem 3.3 (p. 5). Selection separates a function additively. It does not make multiplication diagonal. A residue-i term multiplied by a residue-j term lands in residue i+jpmod N. For N=3, use F=1+2z+3z^2+z^3, G=2+z+z^2. The table displays every product of their three selected sections. vfigcyclic_product The shaded cells contribute to residue zero: 2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3. Keeping only the product of the res…
Exact mathematical passage
i+j\pmod N
\primeUse in source · 8 passages
Search aliases: prime
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 296
ic objects are introduced in the following order. First, E_N,j(z) = ∑_n≥0fracz^Nn+j(Nn+j)! is the residue j part of the exponential. Second, D_N,j(z) = ∑_n≥0(-1)^nfracz^Nn+j(Nn+j)! is the signed basis for y^(N)=-y. The section D_N= D_N,0 is the denominator used in the reciprocal family S_N^(r)= D_N^-r. Set T_N=- D_N^prime/ D_N. Equivalently, T_N is the logarithmic derivative of S_N^(1).
Exact mathematical passage
\Tfun_N=-\Dfun_N^{\prime}/\Dfun_N
QUse in source · 7 passages
Search aliases: Q
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2753
[Meromorphic continuation of the symmetry-orbit zero zeta] Put A_N=fracπsin(π/N). There is an entire function Q_N^red such that Z_N^red(s) = A_N^-Nsζ(Ns,frac12)+Q_N^red(s) on Re s>1/N. The right hand side gives a meromorphic continuation to C with one simple pole at s=1/N. Its residue is fracsin(π/N)Nπ.
Exact mathematical passage
Q_N^{\mathrm{red}}
Q_2^{\mathrm{red}}Use in source · 1 passage
Search aliases: Q_2^red mathrm
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2784
For N=2, [source reference] and D_2(z)=Δ_2(z^2) give λ_m=π^2(m+frac12)^2, m≥0. Thus the first formula holds with Q_2^red=0.
Exact mathematical passage
Q_2^{\mathrm{red}}=0
Q_N^{\mathrm{abs}}Use in source · 3 passages
Search aliases: Q_N^abs mathrm
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2764
There is an entire function Q_N^abs such that, initially on Re s>1, Z_N^abs(s) = N A_N^-sζ(s,frac12)+Q_N^abs(s). Thus Z_N^abs continues meromorphically to C with one simple pole at s=1, of residue fracNsin(π/N)π.
Exact mathematical passage
Q_N^{\mathrm{abs}}
Q_N^{\mathrm{red}}Use in source · 4 passages
Search aliases: Q_N^red mathrm
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2753
[Meromorphic continuation of the symmetry-orbit zero zeta] Put A_N=fracπsin(π/N). There is an entire function Q_N^red such that Z_N^red(s) = A_N^-Nsζ(Ns,frac12)+Q_N^red(s) on Re s>1/N. The right hand side gives a meromorphic continuation to C with one simple pole at s=1/N. Its residue is fracsin(π/N)Nπ.
Exact mathematical passage
Q_N^{\mathrm{red}}
RUse in source · 59 passages
Search aliases: R
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1792
[Indicator polygon and dominant sectors] The indicator of D_N equals h_N, with the limsup taken over radii where D_N(re^iθ)≠0: limsup_r→∞ frac1rlog| D_N(re^iθ)| = h_N(θ). If K is a compact set of directions disjoint from T_N, then there are δ>0 and R>0 such that, for each θ∈ K, one root η(θ) of -1 satisfies D_N(re^iθ) = frac1N exp(η(θ)re^iθ) (1+O(e^-δ r)) uniformly for r≥ R. In particular, no sufficiently large zero can lie in those directions.
Exact mathematical passage
R>0
r^{-1}Use in source · 2 passages
Search aliases: r^-1
Use in source · atlas-pages.tex, line 317
o-free. headWhy zeros remain possible on a growing ray At a tie, two adjacent exponents have the same real projection. Their complex phases can be opposite. Their sum can therefore vanish even though each term is exponentially large. The minima in the order-four indicator plot occur precisely at the tie directions θ=kπ/2. The indicator uses a limsup in radius. It does not assert that r^-1log|D_N(re^iθ)| has an ordinary limit through the zeros on a tie ray. takeawayThe polygon predicts where larg…
Exact mathematical passage
r^{-1}\log|D_N(re^{i\theta})|
R^{-m}Use in source · 3 passages
Search aliases: R^-m
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2651
[Leading dominant zero term] Under the hypotheses of [source reference], P_N,n(r) = · m!binomm+r-1r-1 ∑_ρ∈ Z_R fracρ^-m(-ρ D_N'(ρ))^r · +O(m!m^r-2R^-m) +O_R_*(m!R_*^-m). For r=1, the first error term is absent.
Exact mathematical passage
\begin{aligned} P_{N,n}(r) ={}&m!\binom{m+r-1}{r-1} \sum_{\rho\in\Zcal_R} \frac{\rho^{-m}}{(-\rho\Dfun_N'(\rho))^r} \\ &+O\left(m!m^{r-2}R^{-m}\right) +O_{R_*}\left(m!R_*^{-m}\right). \end{aligned}
R^{1/N}Use in source · 7 passages
Search aliases: R^1/N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2226
[Zero counting asymptotics] Let n_ D_N(R) count the zeros of D_N in |z|≤ R, with multiplicity. Let n_Δ_N(R) count the zeros of Δ_N in |w|≤ R. Then, as R→∞, n_ D_N(R) = fracNsin(π/N)πR+O(1) and n_Δ_N(R) = fracsin(π/N)πR^1/N+O(1).
Exact mathematical passage
n_{\Delta_N}(R) = \frac{\sin(\pi/N)}{\pi}R^{1/N}+O(1).
r^2Use in source · 7 passages
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Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1211
For N=1, one has D_1(z)^-r=e^rz and P_1,n(r)=r^n. For N=2, P_2,1(r)=r, P_2,2(r)=r(3r+2), and P_2,3(r)=r(15r^2+30r+16). At r=1, these are the positive secant coefficients.
Exact mathematical passage
P_{2,3}(r)=r(15r^2+30r+16).
r^3Use in source · 4 passages
Search aliases: r^3
Use in source · atlas-pages.tex, line 185
ynomials collect the possible ways logarithmic terms contribute total degree n. vfigbell_partitions For N=2, the first three logarithmic coefficients are κ_1=frac12, κ_2=frac16, κ_3=frac115. The third Bell polynomial is B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3. Substituting x_1=rκ_1, x_2=rκ_2, x_3=2rκ_3 and multiplying by 6!/3! gives P_2,3(r)=15r^3+30r^2+16r. At r=1 this is 61. At r=2 it is 272. At r=0 it vanishes, as every positive-index reciprocal coefficient must. headThe leading term has a simple …
Exact mathematical passage
P_{2,3}(r)=15r^3+30r^2+16r.
r^NUse in source · 1 passage
Search aliases: r^N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1749
The coefficient of w^n in Δ_N has modulus 1/(Nn)!. Stirling's formula and [source reference] give order 1/N. Since M_f( · ^N)(r)=M_f(r^N) for the maximum modulus, composition with z^N multiplies the order by N and gives order one for D_N.
Exact mathematical passage
M_{f(\,\cdot\,^N)}(r)=M_f(r^N)
r^nUse in source · 6 passages
Search aliases: r^n
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1203
For N=1, one has D_1(z)^-r=e^rz and P_1,n(r)=r^n. For N=2, P_2,1(r)=r, P_2,2(r)=r(3r+2), and P_2,3(r)=r(15r^2+30r+16). At r=1, these are the positive secant coefficients.
Exact mathematical passage
P_{1,n}(r)=r^n
R_*Use in source · 7 passages
Search aliases: R_*
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2606
rest-pole coefficient expansion] Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
R_*
R_*^{-m}Use in source · 4 passages
Search aliases: R_*^-m
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2607
rcle_expansion Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
P_{N,n}(r) = m! \sum_{\rho\in\Zcal_R} \sum_{\ell=1}^{r} a_{\rho,\ell} \binom{m+\ell-1}{\ell-1} \rho^{-m} + O_{R_*}(m!R_*^{-m}).
R_2Use in source · 4 passages
Search aliases: R_2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2595
[Nearest-pole coefficient expansion] Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
R_2>R
R_3Use in source · 1 passage
Search aliases: R_3
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2136
The only nondominant exponent is -1. Hence R_3(t) = exp((-frac32+fracsqrt32i)t). At t=t_m, its phase is i(-1)^m. Insert this value into [source reference].
Exact mathematical passage
R_3(t) = \exp\left(\left(-\frac32+\frac{\sqrt3}{2}i\right)t\right).
R_NUse in source · 11 passages
Search aliases: R_N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1849
On the strip x≥ R and |y|≤ Y, define R_N(z) = ∑_substackη^N=-1 η≠ e^ia_N,e^-ia_N exp((η-e^-ia_N)z). Then N e^-e^-ia_Nz D_N(z) = 1+e^2is_Nz+R_N(z), and, for q=0,1,2, |R_N^(q)(z)| ≤ C e^-γ_Nx.
Exact mathematical passage
R_N(z) = \sum_{\substack{\eta^N=-1\\ \eta\ne e^{ia_N},e^{-ia_N}}} \exp\bigl((\eta-e^{-ia_N})z\bigr).
R_nUse in source · 5 passages
Search aliases: R_n
Use in source · atlas-pages.tex, line 431
t poles are ρ=±π/2. They contribute equally to even-degree coefficients. Including both poles gives P_2,n(1)∼2(2n)!(frac2π)^2n+1. The factor two is orbit multiplicity, not an adjustable fitted constant. vfigpole_asymptotic headA finite quantitative check For n=1,…,8, the exact integer coefficients and a rational enclosure for π bound the ratio R_n strictly between zero and one. In this tested range, 1-R_n<3^-(2n+1). The plot displays the logarithm of the enclosed error, not an independently asse…
Exact mathematical passage
R_n
R_N'Use in source · 3 passages
Search aliases: R_N'
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2011
Fix a large m, put a=2is_N, and consider the closed disk B_m={z:|z-t_m|≤ 1/(2s_N)}. The remainder bounds hold uniformly on these fixed-radius disks. For sufficiently large m, |R_N|≤ 1/2 and |R_N'|≤ s_N/2 there. Define Φ_m(z)=t_m+fraclog(1+R_N(z))2is_N, using the principal logarithm. The estimate |log(1+u)|≤(3/2)|u| shows that Φ_m maps B_m into itself. The derivative bound |Φ_m'|≤1/2 makes it a contraction. Banach's fixed-point theorem gives a unique fixed point. On B_m, |a(z-t_m)|≤1<π, so the pr…
Exact mathematical passage
|R_N'|\leq s_N/2
R_N^{(q)}Use in source · 2 passages
Search aliases: R_N^(q)
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1863
η≠ e^ia_N,e^-ia_N exp((η-e^-ia_N)z). Then N e^-e^-ia_Nz D_N(z) = 1+e^2is_Nz+R_N(z), and, for q=0,1,2, |R_N^(q)(z)| ≤ C e^-γ_Nx.
Exact mathematical passage
|R_N^{(q)}(z)| \leq C e^{-\gamma_Nx}.
\RcalUse in source · 2 passages
Search aliases: Rcal R mathcal
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 673
[Finite exponential representation] Let R_N={η∈ Cmathrel:η^N=-1}. Then D_N(z) = frac1N∑_η∈ R_Ne^η z. More generally, D_N,j(z) = fracρ^-jN ∑_ℓ=0^N-1 ω^-jℓe^ρω^ℓ z.
Exact mathematical passage
\Rcal_N=\{\eta\in\CC\mathrel{:}\eta^N=-1\}.
\Rcal_NUse in source · 2 passages
Search aliases: Rcal_N R_N Rcal mathcal
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 673
[Finite exponential representation] Let R_N={η∈ Cmathrel:η^N=-1}. Then D_N(z) = frac1N∑_η∈ R_Ne^η z. More generally, D_N,j(z) = fracρ^-jN ∑_ℓ=0^N-1 ω^-jℓe^ρω^ℓ z.
Exact mathematical passage
\Rcal_N=\{\eta\in\CC\mathrel{:}\eta^N=-1\}.
\rfloorUse in source · 4 passages
Search aliases: rfloor
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 654
[Twisted addition law] For 0≤ k<N, D_N,k(z+w) = ∑_substack0≤ i,j<N i+j≡ kbmod N (-1)^lfloor(i+j)/Nrfloor D_N,i(z) D_N,j(w).
Exact mathematical passage
\Dfun_{N,k}(z+w) = \sum_{\substack{0\leq i,j<N\\ i+j\equiv k\bmod N}} (-1)^{\lfloor(i+j)/N\rfloor} \Dfun_{N,i}(z)\Dfun_{N,j}(w).
\rmUse in source · 6 passages
Search aliases: rm
Use in source · atlas-pages.tex, line 441
is not continuationZERO ZETA FUNCTIONSTraditional monograph: Theorem 14.2 (p. 39), Theorem 14.3 (p. 39), Corollary 14.4 (p. 41). Reduced zeros grow like m^N. Their absolute zero-zeta series therefore has half-plane of absolute convergence Re s>1/N. The absolute zero zeta in the original variable instead requires Re s>1. vfigzeta_plane For N=2 and N=4, the lattices are exact: Z_2^rm red(s)=π^-2sζ(2s,1/2), Z_4^rm red(s)=(sqrt2π)^-4sζ(4s,1/2). Each identity first holds where its defining series con…
Exact mathematical passage
\mathfrak Z_2^{\rm red}(s)=\pi^{-2s}\zeta(2s,1/2),\qquad \mathfrak Z_4^{\rm red}(s)=(\sqrt2\pi)^{-4s}\zeta(4s,1/2).
SUse in source · 4 passages
Search aliases: S
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1023
Set S= D_N^-r. Logarithmic differentiation gives D_NS'+r D_N'S=0. Taking the coefficient of z^Nn-1/(Nn-1)! gives ∑_k=0^n-1(-1)^k binomNn-1NkP_N,n-k(r) +r∑_k=1^n(-1)^k binomNn-1Nk-1P_N,n-k(r)=0. The term with k=0 in the first sum is P_N,n(r), and every other term has a smaller block index. Rearrangement gives the formula.
Exact mathematical passage
S=\Dfun_N^{-r}
S'Use in source · 1 passage
Search aliases: S'
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1024
Set S= D_N^-r. Logarithmic differentiation gives D_NS'+r D_N'S=0. Taking the coefficient of z^Nn-1/(Nn-1)! gives ∑_k=0^n-1(-1)^k binomNn-1NkP_N,n-k(r) +r∑_k=1^n(-1)^k binomNn-1Nk-1P_N,n-k(r)=0. The term with k=0 in the first sum is P_N,n(r), and every other term has a smaller block index. Rearrangement gives the formula.
Exact mathematical passage
\Dfun_NS'+r\Dfun_N'S=0.
S_{90}Use in source · 2 passages
Search aliases: S_90
Use in source · atlas-pages.tex, line 347
Theorem 11.8 (p. 30), Corollary 11.9 (p. 31). The root finder is used only to suggest intervals. Each displayed interval is accepted by exact rational arithmetic: opposite endpoint signs prove existence, and an interval bound for D_3' excludes zero throughout the interval, proving uniqueness inside it. vfigcubic_certificates headHow the certificate works Evaluate the Taylor polynomial S_90(t)=∑_n=0^90frac(-1)^nt^3n(3n)! at the two rational endpoints. From the next term onward, magnitudes decreas…
Exact mathematical passage
S_{90}(t)=\sum_{n=0}^{90}\frac{(-1)^nt^{3n}}{(3n)!}
s_NUse in source · 31 passages
Search aliases: s_N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1817
At a tie direction exactly two adjacent vertices attain the maximum. After rotation it is enough to take θ=0. With c_N=cos(π/N) and s_N=sin(π/N), the two dominant terms give frac2N e^c_Nrcos(s_Nr) and every other term is exponentially smaller. Along a sequence on which |cos(s_Nr)|≥1/2, the lower indicator bound is c_N=h_N(0). Rotational symmetry treats every tie direction.
Exact mathematical passage
s_N=\sin(\pi/N)
\secUse in source · 4 passages
Search aliases: sec
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 923
A unit constant term makes the formal reciprocal unambiguous. To compute one coefficient, only finitely many earlier coefficients are needed. The later analytic theory explains how those exact coefficients relate to the full zero set. modelfigurereciprocal_normalizationAt N=2 and r=1, the reciprocal is sec z. The upper row gives block exponential coefficients and the lower row gives ordinary Taylor coefficients. Both rows encode the same five terms.fig:reciprocal-normalization
Exact mathematical passage
\sec z
\operatorname{sech}Use in source · 1 passage
Search aliases: sech operatorname
Use in source · different_reciprocals.tex, line 23
[fill=black!0,draw=black!40,line width=.35pt] (7.10000,3.05000) rectangle ++(2.10000,0.80000); node[] at (8.15000,3.45000) 5; draw[fill=black!8,draw=black!40,line width=.35pt] (9.20000,3.05000) rectangle ++(2.10000,0.80000); node[] at (10.25000,3.45000) 61; draw[fill=black!0,draw=black!40,line width=.35pt] (0.00000,2.15000) rectangle ++(3.40000,0.80000); node[] at (1.70000,2.55000) sechz; draw[fill=black!0,draw=black!40,line width=.35pt] (3.40000,2.15000) rectangle ++(1.60000,0.80000); node[] at…
Exact mathematical passage
\operatorname{sech}z
\Sfun_N^{(1)}Use in source · 1 passage
Search aliases: Sfun_N^(1) S_N^(1) Sfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 296
acz^Nn+j(Nn+j)! is the residue j part of the exponential. Second, D_N,j(z) = ∑_n≥0(-1)^nfracz^Nn+j(Nn+j)! is the signed basis for y^(N)=-y. The section D_N= D_N,0 is the denominator used in the reciprocal family S_N^(r)= D_N^-r. Set T_N=- D_N^prime/ D_N. Equivalently, T_N is the logarithmic derivative of S_N^(1).
Exact mathematical passage
\Sfun_N^{(1)}
\Sfun_N^{(r+s)}Use in source · 1 passage
Search aliases: Sfun_N^(r+s) S_N^(r+s) Sfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1040
[Additive parameter law for reciprocal powers] For all r,s∈ C, S_N^(r)(z) S_N^(s)(z) = S_N^(r+s)(z) as a formal identity and on every common zero-free region with a fixed logarithm. Consequently, P_N,n(r+s) = ∑_q=0^n binomNnNq P_N,q(r)P_N,n-q(s).
Exact mathematical passage
\Sfun_N^{(r)}(z)\Sfun_N^{(s)}(z) = \Sfun_N^{(r+s)}(z)
\Sfun_N^{(s)}Use in source · 1 passage
Search aliases: Sfun_N^(s) S_N^(s) Sfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1040
[Additive parameter law for reciprocal powers] For all r,s∈ C, S_N^(r)(z) S_N^(s)(z) = S_N^(r+s)(z) as a formal identity and on every common zero-free region with a fixed logarithm. Consequently, P_N,n(r+s) = ∑_q=0^n binomNnNq P_N,q(r)P_N,n-q(s).
Exact mathematical passage
\Sfun_N^{(r)}(z)\Sfun_N^{(s)}(z) = \Sfun_N^{(r+s)}(z)
\sinUse in source · 13 passages
Search aliases: sin
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1817
At a tie direction exactly two adjacent vertices attain the maximum. After rotation it is enough to take θ=0. With c_N=cos(π/N) and s_N=sin(π/N), the two dominant terms give frac2N e^c_Nrcos(s_Nr) and every other term is exponentially smaller. Along a sequence on which |cos(s_Nr)|≥1/2, the lower indicator bound is c_N=h_N(0). Rotational symmetry treats every tie direction.
Exact mathematical passage
s_N=\sin(\pi/N)
\supUse in source · 1 passage
Search aliases: sup
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2925
For N≥3, [source reference] gives | D_N'(ρ)|^-1 ≤ C e^-c_N|ρ| for every sufficiently large simple zero. Let L be a compact subset of the s plane. Because the chosen arguments of the zeros lie in a fixed bounded interval, there are constants C_L and M_L such that sup_s∈ L|e^-sLogρ| ≤ C_L|ρ|^M_L for every sufficiently large zero. Hence the absolute value of the corresponding summand is at most C_L'|ρ|^M_Le^-q c_N|ρ|. The linear zero counting estimate makes this majorant summable. The Weierstrass t…
Exact mathematical passage
\sup_{s\in L}|e^{-s\Log\rho}| \leq C_L|\rho|^{M_L}
\sup_{s\in L}Use in source · 1 passage
Search aliases: sup_s∈ L sup in
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2925
For N≥3, [source reference] gives | D_N'(ρ)|^-1 ≤ C e^-c_N|ρ| for every sufficiently large simple zero. Let L be a compact subset of the s plane. Because the chosen arguments of the zeros lie in a fixed bounded interval, there are constants C_L and M_L such that sup_s∈ L|e^-sLogρ| ≤ C_L|ρ|^M_L for every sufficiently large zero. Hence the absolute value of the corresponding summand is at most C_L'|ρ|^M_Le^-q c_N|ρ|. The linear zero counting estimate makes this majorant summable. The Weierstrass t…
Exact mathematical passage
\sup_{s\in L}|e^{-s\Log\rho}| \leq C_L|\rho|^{M_L}
T'Use in source · 2 passages
Search aliases: T'
Use in source · atlas-pages.tex, line 224
retain the reduced logarithmic information: T_N(z)=N∑_m≥1κ_N,mz^Nm-1, T_N(n)=N(Nn+N-1)!κ_N,n+1. For order four, κ_4,1=1/24 and κ_4,2=17/10080. These give T_4(0)=1 and T_4(1)=34, explaining the last line of the plate. headA nonlinear equation with a precise origin If D=e^g locally, then g'=-T. Differentiating e^g three times gives D'''/D=-T''+3TT'-T^3. Setting D'''/D=-1 yields the cubic Riccati identity shown above. No additional nonlinear model has been assumed. The reciprocal family also satisf…
Exact mathematical passage
D'''/D=-T''+3TT'-T^3
T''Use in source · 2 passages
Search aliases: T''
Use in source · atlas-pages.tex, line 224
retain the reduced logarithmic information: T_N(z)=N∑_m≥1κ_N,mz^Nm-1, T_N(n)=N(Nn+N-1)!κ_N,n+1. For order four, κ_4,1=1/24 and κ_4,2=17/10080. These give T_4(0)=1 and T_4(1)=34, explaining the last line of the plate. headA nonlinear equation with a precise origin If D=e^g locally, then g'=-T. Differentiating e^g three times gives D'''/D=-T''+3TT'-T^3. Setting D'''/D=-1 yields the cubic Riccati identity shown above. No additional nonlinear model has been assumed. The reciprocal family also satisf…
Exact mathematical passage
D'''/D=-T''+3TT'-T^3
T^3Use in source · 2 passages
Search aliases: T^3
Use in source · atlas-pages.tex, line 224
retain the reduced logarithmic information: T_N(z)=N∑_m≥1κ_N,mz^Nm-1, T_N(n)=N(Nn+N-1)!κ_N,n+1. For order four, κ_4,1=1/24 and κ_4,2=17/10080. These give T_4(0)=1 and T_4(1)=34, explaining the last line of the plate. headA nonlinear equation with a precise origin If D=e^g locally, then g'=-T. Differentiating e^g three times gives D'''/D=-T''+3TT'-T^3. Setting D'''/D=-1 yields the cubic Riccati identity shown above. No additional nonlinear model has been assumed. The reciprocal family also satisf…
Exact mathematical passage
D'''/D=-T''+3TT'-T^3
t^3Use in source · 2 passages
Search aliases: t^3
Use in source · atlas-pages.tex, line 351
te endpoint signs prove existence, and an interval bound for D_3' excludes zero throughout the interval, proving uniqueness inside it. vfigcubic_certificates headHow the certificate works Evaluate the Taylor polynomial S_90(t)=∑_n=0^90frac(-1)^nt^3n(3n)! at the two rational endpoints. From the next term onward, magnitudes decrease, because their successive ratio is fract^3(3n+1)(3n+2)(3n+3)<1 there. The alternating-series remainder is bounded by the first omitted term. A separate outward interva…
Exact mathematical passage
\frac{t^3}{(3n+1)(3n+2)(3n+3)}<1
t^{3n}Use in source · 2 passages
Search aliases: t^3n
Use in source · atlas-pages.tex, line 347
Theorem 11.8 (p. 30), Corollary 11.9 (p. 31). The root finder is used only to suggest intervals. Each displayed interval is accepted by exact rational arithmetic: opposite endpoint signs prove existence, and an interval bound for D_3' excludes zero throughout the interval, proving uniqueness inside it. vfigcubic_certificates headHow the certificate works Evaluate the Taylor polynomial S_90(t)=∑_n=0^90frac(-1)^nt^3n(3n)! at the two rational endpoints. From the next term onward, magnitudes decreas…
Exact mathematical passage
S_{90}(t)=\sum_{n=0}^{90}\frac{(-1)^nt^{3n}}{(3n)!}
t^mUse in source · 1 passage
Search aliases: t^m
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 955
The polynomial dependence on r can be read directly from the formal logarithm. We use the complete exponential Bell polynomials B_n defined by exp(∑_m≥1x_mfract^mm!) = ∑_n≥0B_n(x_1,…,x_n)fract^nn!.
Exact mathematical passage
\exp\left(\sum_{m\geq1}x_m\frac{t^m}{m!}\right) = \sum_{n\geq0}B_n(x_1,\ldots,x_n)\frac{t^n}{n!}.
t^nUse in source · 1 passage
Search aliases: t^n
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 955
The polynomial dependence on r can be read directly from the formal logarithm. We use the complete exponential Bell polynomials B_n defined by exp(∑_m≥1x_mfract^mm!) = ∑_n≥0B_n(x_1,…,x_n)fract^nn!.
Exact mathematical passage
\exp\left(\sum_{m\geq1}x_m\frac{t^m}{m!}\right) = \sum_{n\geq0}B_n(x_1,\ldots,x_n)\frac{t^n}{n!}.
T_1Use in source · 2 passages
Search aliases: T_1
Use in source · atlas-pages.tex, line 227
pressions have poles rather than finite values. takeawayThe nonlinear tangent equation is exact where division is legal. It is not a global regular differential equation across the zeros. platecaptionThe tests verify the tangent coefficient recurrence, the exact cubic ODE, and the symbolic Riccati reduction after clearing denominators. The order-one tangent is the separate constant function T_1=1.
Exact mathematical passage
T_1=1
T_2^2Use in source · 1 passage
Search aliases: T_2^2
Use in source · riccati.tex, line 5
[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[] at (3.00000,4.50000) LINEAR EQUATION; node[] at (9.30000,4.50000) LOGARITHMIC DERIVATIVE; node[] at (3.00000,3.25000) D_2^primeprime=-D_2; node[] at (9.30000,3.25000) T_2^prime=1+T_2^2; node[] at (3.00000,1.85000) D_3^primeprimeprime=-D_3; node[] at (9.30000,1.85000) T_3^primeprime-3T_3T_3^prime+T_3^3=1; node[] at (6.15000,0.50000) T_N=-D_N^prime/D_N only where D_N≠0; draw[black!25] (6.10000,1.10000) -- (6.10000,4.650…
Exact mathematical passage
T_2^{\prime}=1+T_2^2
T_2^{\prime}Use in source · 1 passage
Search aliases: T_2^prime prime
Use in source · riccati.tex, line 5
[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[] at (3.00000,4.50000) LINEAR EQUATION; node[] at (9.30000,4.50000) LOGARITHMIC DERIVATIVE; node[] at (3.00000,3.25000) D_2^primeprime=-D_2; node[] at (9.30000,3.25000) T_2^prime=1+T_2^2; node[] at (3.00000,1.85000) D_3^primeprimeprime=-D_3; node[] at (9.30000,1.85000) T_3^primeprime-3T_3T_3^prime+T_3^3=1; node[] at (6.15000,0.50000) T_N=-D_N^prime/D_N only where D_N≠0; draw[black!25] (6.10000,1.10000) -- (6.10000,4.650…
Exact mathematical passage
T_2^{\prime}=1+T_2^2
T_3Use in source · 1 passage
Search aliases: T_3
Use in source · riccati.tex, line 7
intikzpicture[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[] at (3.00000,4.50000) LINEAR EQUATION; node[] at (9.30000,4.50000) LOGARITHMIC DERIVATIVE; node[] at (3.00000,3.25000) D_2^primeprime=-D_2; node[] at (9.30000,3.25000) T_2^prime=1+T_2^2; node[] at (3.00000,1.85000) D_3^primeprimeprime=-D_3; node[] at (9.30000,1.85000) T_3^primeprime-3T_3T_3^prime+T_3^3=1; node[] at (6.15000,0.50000) T_N=-D_N^prime/D_N only where D_N≠0; draw[black!25] (6.10000,1.10000) -- (…
Exact mathematical passage
T_3^{\prime\prime}-3T_3T_3^{\prime}+T_3^3=1
T_3^3Use in source · 1 passage
Search aliases: T_3^3
Use in source · riccati.tex, line 7
intikzpicture[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[] at (3.00000,4.50000) LINEAR EQUATION; node[] at (9.30000,4.50000) LOGARITHMIC DERIVATIVE; node[] at (3.00000,3.25000) D_2^primeprime=-D_2; node[] at (9.30000,3.25000) T_2^prime=1+T_2^2; node[] at (3.00000,1.85000) D_3^primeprimeprime=-D_3; node[] at (9.30000,1.85000) T_3^primeprime-3T_3T_3^prime+T_3^3=1; node[] at (6.15000,0.50000) T_N=-D_N^prime/D_N only where D_N≠0; draw[black!25] (6.10000,1.10000) -- (…
Exact mathematical passage
T_3^{\prime\prime}-3T_3T_3^{\prime}+T_3^3=1
T_3^{\prime}Use in source · 1 passage
Search aliases: T_3^prime prime
Use in source · riccati.tex, line 7
intikzpicture[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[] at (3.00000,4.50000) LINEAR EQUATION; node[] at (9.30000,4.50000) LOGARITHMIC DERIVATIVE; node[] at (3.00000,3.25000) D_2^primeprime=-D_2; node[] at (9.30000,3.25000) T_2^prime=1+T_2^2; node[] at (3.00000,1.85000) D_3^primeprimeprime=-D_3; node[] at (9.30000,1.85000) T_3^primeprime-3T_3T_3^prime+T_3^3=1; node[] at (6.15000,0.50000) T_N=-D_N^prime/D_N only where D_N≠0; draw[black!25] (6.10000,1.10000) -- (…
Exact mathematical passage
T_3^{\prime\prime}-3T_3T_3^{\prime}+T_3^3=1
T_3^{\prime\prime}Use in source · 1 passage
Search aliases: T_3^primeprime prime prime
Use in source · riccati.tex, line 7
intikzpicture[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[] at (3.00000,4.50000) LINEAR EQUATION; node[] at (9.30000,4.50000) LOGARITHMIC DERIVATIVE; node[] at (3.00000,3.25000) D_2^primeprime=-D_2; node[] at (9.30000,3.25000) T_2^prime=1+T_2^2; node[] at (3.00000,1.85000) D_3^primeprimeprime=-D_3; node[] at (9.30000,1.85000) T_3^primeprime-3T_3T_3^prime+T_3^3=1; node[] at (6.15000,0.50000) T_N=-D_N^prime/D_N only where D_N≠0; draw[black!25] (6.10000,1.10000) -- (…
Exact mathematical passage
T_3^{\prime\prime}-3T_3T_3^{\prime}+T_3^3=1
T_4Use in source · 5 passages
Search aliases: T_4
Use in source · atlas-pages.tex, line 222
/D_N. It compresses the linear differential equation into a nonlinear scalar identity. This division is allowed only on the zero-free set of the denominator. vfigriccati The coefficients retain the reduced logarithmic information: T_N(z)=N∑_m≥1κ_N,mz^Nm-1, T_N(n)=N(Nn+N-1)!κ_N,n+1. For order four, κ_4,1=1/24 and κ_4,2=17/10080. These give T_4(0)=1 and T_4(1)=34, explaining the last line of the plate. headA nonlinear equation with a precise origin If D=e^g locally, then g'=-T. Differentiating e^g…
Exact mathematical passage
T_4(0)=1
t_mUse in source · 44 passages
Search aliases: t_m
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1833
The zero strings are governed by two adjacent exponents. Put a_N=fracπ N, s_N=sin a_N, c_N=cos a_N, and A_N=fracπs_N, t_m=A_N(m+frac12). For N≥3, also put γ_N=c_N-cos(3a_N)>0.
Exact mathematical passage
A_N=\frac\pi{s_N}, \qquad t_m=A_N\left(m+\frac12\right).
t_m^{-1}Use in source · 1 passage
Search aliases: t_m^-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2793
For N≥3, [source reference] writes all but finitely many reduced zeros as λ_m = (A_N(m+frac12))^N(1+ε_m) where λ_m>0, 1+ε_m>0, and ε_m = O(t_m^-1e^-γ t_m). On every compact set of s values, (1+ε_m)^-s-1 = O(|ε_m|) uniformly for large m. The difference λ_m^-s - A_N^-Ns(m+frac12)^-Ns therefore has a normally convergent sum on every compact subset of C. Indeed, the model power has at most polynomial growth in m on a compact set, while |ε_m| decays exponentially. The finitely many exceptional terms …
Exact mathematical passage
\varepsilon_m = O\left(t_m^{-1}e^{-\gamma t_m}\right).
t_m^2Use in source · 1 passage
Search aliases: t_m^2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2207
he nonzero zero multiset of Δ_N. All but finitely many members of Λ_N are positive and simple. They can be indexed as λ_m=τ_m^N for all sufficiently large m. If N≥3, the notation may be chosen as in [source reference], and λ_m = t_m^N +O(t_m^N-1e^-γ t_m). For N=2, take τ_m=t_m=π(m+1/2) for m≥0; then λ_m=t_m^2 exactly.
Exact mathematical passage
\lambda_m=t_m^2
t_m^NUse in source · 1 passage
Search aliases: t_m^N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2200
[Reduced zeros] Let Λ_N be the nonzero zero multiset of Δ_N. All but finitely many members of Λ_N are positive and simple. They can be indexed as λ_m=τ_m^N for all sufficiently large m. If N≥3, the notation may be chosen as in [source reference], and λ_m = t_m^N +O(t_m^N-1e^-γ t_m). For N=2, take τ_m=t_m=π(m+1/2) for m≥0; then λ_m=t_m^2 exactly.
Exact mathematical passage
\lambda_m = t_m^N +O\left(t_m^{N-1}e^{-\gamma t_m}\right).
t_m^{N-1}Use in source · 1 passage
Search aliases: t_m^N-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2200
[Reduced zeros] Let Λ_N be the nonzero zero multiset of Δ_N. All but finitely many members of Λ_N are positive and simple. They can be indexed as λ_m=τ_m^N for all sufficiently large m. If N≥3, the notation may be chosen as in [source reference], and λ_m = t_m^N +O(t_m^N-1e^-γ t_m). For N=2, take τ_m=t_m=π(m+1/2) for m≥0; then λ_m=t_m^2 exactly.
Exact mathematical passage
\lambda_m = t_m^N +O\left(t_m^{N-1}e^{-\gamma t_m}\right).
\tanUse in source · 1 passage
Search aliases: tan
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1345
For N=1, T_1=1. For N=2, D_2=cos z and T_2=tan z. The Riccati equation becomes T_2'=1+ T_2^2.
Exact mathematical passage
\Tfun_2=\tan z
\Tfun_1Use in source · 1 passage
Search aliases: Tfun_1 T_1 Tfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1345
For N=1, T_1=1. For N=2, D_2=cos z and T_2=tan z. The Riccati equation becomes T_2'=1+ T_2^2.
Exact mathematical passage
\Tfun_1=1
\Tfun_2Use in source · 1 passage
Search aliases: Tfun_2 T_2 Tfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1345
For N=1, T_1=1. For N=2, D_2=cos z and T_2=tan z. The Riccati equation becomes T_2'=1+ T_2^2.
Exact mathematical passage
\Tfun_2=\tan z
\Tfun_2'Use in source · 1 passage
Search aliases: Tfun_2' T_2' Tfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1346
For N=1, T_1=1. For N=2, D_2=cos z and T_2=tan z. The Riccati equation becomes T_2'=1+ T_2^2.
Exact mathematical passage
\Tfun_2'=1+\Tfun_2^2.
\Tfun_2^2Use in source · 1 passage
Search aliases: Tfun_2^2 T_2^2 Tfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1346
For N=1, T_1=1. For N=2, D_2=cos z and T_2=tan z. The Riccati equation becomes T_2'=1+ T_2^2.
Exact mathematical passage
\Tfun_2'=1+\Tfun_2^2.
\Tfun_N'Use in source · 1 passage
Search aliases: Tfun_N' T_N' Tfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1326
[Scalar higher Riccati identity] At every point of the open set {z: D_N(z)≠0}, Y_N( - T_N, - T_N', …, - T_N^(N-1) ) =-1.
Exact mathematical passage
Y_N\bigl( -\Tfun_N, -\Tfun_N', \ldots, -\Tfun_N^{(N-1)} \bigr) =-1.
\Tfun_N^{(N-1)}Use in source · 1 passage
Search aliases: Tfun_N^(N-1) T_N^(N-1) Tfun mathsf
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1326
[Scalar higher Riccati identity] At every point of the open set {z: D_N(z)≠0}, Y_N( - T_N, - T_N', …, - T_N^(N-1) ) =-1.
Exact mathematical passage
Y_N\bigl( -\Tfun_N, -\Tfun_N', \ldots, -\Tfun_N^{(N-1)} \bigr) =-1.
\Thetafun_{2,1}Use in source · 1 passage
Search aliases: Thetafun_2,1 symupΘ_2,1 Thetafun symup Theta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1696
For N=2, symupΘ_2,1(u)=-u(u+1) and symupΘ_2,2(u) = u(u+1)(5u^2+u+2). The two factors predicted by [source reference] are visible.
Exact mathematical passage
\Thetafun_{2,1}(u)=-u(u+1)
\Thetafun_{2,2}Use in source · 1 passage
Search aliases: Thetafun_2,2 symupΘ_2,2 Thetafun symup Theta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1700
For N=2, symupΘ_2,1(u)=-u(u+1) and symupΘ_2,2(u) = u(u+1)(5u^2+u+2). The two factors predicted by [source reference] are visible.
Exact mathematical passage
\Thetafun_{2,2}(u) = u(u+1)(5u^2+u+2).
\Thetafun_{N,1}Use in source · 1 passage
Search aliases: Thetafun_N,1 symupΘ_N,1 Thetafun symup Theta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1672
[Theta Newton recurrence] Put θ_N,n(u)=fracsymupΘ_N,n(u)(Nn)!. Then θ_N,0(u)=1 and nθ_N,n(u) = ∑_m=1^n (-1)^mκ_N,m (u^Nm+u) θ_N,n-m(u). In particular, symupΘ_N,1(u)=-(u^N+u).
Exact mathematical passage
\Thetafun_{N,1}(u)=-(u^N+u).
\TrUse in source · 4 passages
Search aliases: Tr
Use in source · atlas-pages.tex, line 95
ordinates and columns are input coordinates. The upper-triangular minus signs record crossing the signed edge while moving from the column to the row. They are not arbitrary orientation choices. headTwo checks connect the matrix and the scalar equation The initial derivative matrix is the identity. Each initial jet (y(0),y'(0),y''(0)) therefore determines a unique linear combination. Also, frac13Tr(e^zJ_3)=D_3(z), det(XI-J_3)=X^3+1. The roots of -1 appearing later in the exponential sum are alre…
Exact mathematical passage
\frac13\Tr(e^{zJ_3})=D_3(z),\qquad \det(XI-J_3)=X^3+1.
\trUse in source · 6 passages
Search aliases: tr operatorname
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 826
D_N,N-1(z) . Then D_N(z)=e^zJ_Ne_0 and D_N(z)=frac1Ntr(e^zJ_N). The eigenvalues of J_N are the N roots of -1, each with multiplicity one.
Exact mathematical passage
\Dfun_N(z)=\frac1N\tr(e^{zJ_N}).
UUse in source · 2 passages
Search aliases: U
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 895
[Addition and inversion laws for the companion matrix] Assume N≥2. The matrix U_N(z)=e^zJ_N satisfies U_N(z+w)=U_N(z)U_N(w), U_N(z)^-1=U_N(-z), det U_N(z)=1. Its first column is D_N(z), and all of its entries are given by [source reference].
Exact mathematical passage
U_N(z)=e^{zJ_N}
u^2Use in source · 3 passages
Search aliases: u^2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1700
For N=2, symupΘ_2,1(u)=-u(u+1) and symupΘ_2,2(u) = u(u+1)(5u^2+u+2). The two factors predicted by [source reference] are visible.
Exact mathematical passage
\Thetafun_{2,2}(u) = u(u+1)(5u^2+u+2).
u^{2N}Use in source · 1 passage
Search aliases: u^2N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2505
[Theta Bell polynomial formula] Put widetildeσ_N,m=(-1)^mσ_N,m. Then symupΘ_N,n(u) = frac(Nn)!n! B_n( widetildeσ_N,1(u^N+u), 1!widetildeσ_N,2(u^2N+u), …, (n-1)!widetildeσ_N,n(u^nN+u) ).
Exact mathematical passage
\Thetafun_{N,n}(u) = \frac{(Nn)!}{n!} B_n\bigl( \widetilde{\sigma}_{N,1}(u^N+u), 1!\widetilde{\sigma}_{N,2}(u^{2N}+u), \ldots, (n-1)!\widetilde{\sigma}_{N,n}(u^{nN}+u) \bigr).
u^3Use in source · 1 passage
Search aliases: u^3
Use in source · theta.tex, line 25
) (11.64750,1.00832) (11.72000,0.96637) (11.79250,0.92343) (11.86500,0.87947) (11.93750,0.83449) (12.01000,0.78848) (12.08250,0.74141) (12.15500,0.69327) (12.22750,0.64406) (12.30000,0.59375); draw[fill=black] (2.63333,2.30000) circle (0.08); draw[fill=black] (2.63333,3.00000) circle (0.08); node[] at (3.00000,4.65000) solid: Θ_2,1(u)=-u(u+1); node[] at (9.50000,4.65000) dashed: Θ_3,1(u)=-(u^3+u); node[] at (6.50000,-0.35000) at u=-1: even order cancels, odd order need not;
Exact mathematical passage
\Theta_{3,1}(u)=-(u^3+u)
u^4Use in source · 1 passage
Search aliases: u^4
Use in source · orbit_reduction.tex, line 17
00,2.50000) circle (0.055); draw[black!45] (2.70000,2.50000) -- (2.70000,0.50000); draw[fill=black] (2.70000,1.90000) circle (0.055); draw[fill=black] (2.70000,0.70000) circle (0.055); node[] at (2.70000,5.00000) u=z/(sqrt2π); node[] at (2.70000,-0.30000) u=tfrac12 i^k, tfrac32 i^k; draw[->,line width=.9pt] (5.30000,2.50000) -- (7.00000,2.50000); node[] at (6.15000,3.00000) v=u^4; draw[->] (7.50000,2.50000) -- (12.80000,2.50000); draw[fill=black] (7.65000,2.50000) circle (0.09); node[] at (7.650…
Exact mathematical passage
v=u^4
u^NUse in source · 4 passages
Search aliases: u^N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1672
[Theta Newton recurrence] Put θ_N,n(u)=fracsymupΘ_N,n(u)(Nn)!. Then θ_N,0(u)=1 and nθ_N,n(u) = ∑_m=1^n (-1)^mκ_N,m (u^Nm+u) θ_N,n-m(u). In particular, symupΘ_N,1(u)=-(u^N+u).
Exact mathematical passage
\Thetafun_{N,1}(u)=-(u^N+u).
u^{Nm}Use in source · 7 passages
Search aliases: u^Nm
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1663
[Theta Newton recurrence] Put θ_N,n(u)=fracsymupΘ_N,n(u)(Nn)!. Then θ_N,0(u)=1 and nθ_N,n(u) = ∑_m=1^n (-1)^mκ_N,m (u^Nm+u) θ_N,n-m(u). In particular, symupΘ_N,1(u)=-(u^N+u).
Exact mathematical passage
n\theta_{N,n}(u) = \sum_{m=1}^{n} (-1)^m\kappa_{N,m} \bigl(u^{Nm}+u\bigr) \theta_{N,n-m}(u).
u^{nN}Use in source · 1 passage
Search aliases: u^nN
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2505
[Theta Bell polynomial formula] Put widetildeσ_N,m=(-1)^mσ_N,m. Then symupΘ_N,n(u) = frac(Nn)!n! B_n( widetildeσ_N,1(u^N+u), 1!widetildeσ_N,2(u^2N+u), …, (n-1)!widetildeσ_N,n(u^nN+u) ).
Exact mathematical passage
\Thetafun_{N,n}(u) = \frac{(Nn)!}{n!} B_n\bigl( \widetilde{\sigma}_{N,1}(u^N+u), 1!\widetilde{\sigma}_{N,2}(u^{2N}+u), \ldots, (n-1)!\widetilde{\sigma}_{N,n}(u^{nN}+u) \bigr).
u^{Nq}Use in source · 1 passage
Search aliases: u^Nq
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1629
[Theta reconstruction] For every n≥0, symupΘ_N,n(u) = (-1)^n ∑_q=0^n binomNnNq P_N,q(1)u^NqP_N,n-q(u). Thus symupΘ_N,n is a polynomial of degree at most Nn.
Exact mathematical passage
\Thetafun_{N,n}(u) = (-1)^n \sum_{q=0}^{n} \binom{Nn}{Nq} P_{N,q}(1)u^{Nq}P_{N,n-q}(u).
U_NUse in source · 2 passages
Search aliases: U_N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 895
[Addition and inversion laws for the companion matrix] Assume N≥2. The matrix U_N(z)=e^zJ_N satisfies U_N(z+w)=U_N(z)U_N(w), U_N(z)^-1=U_N(-z), det U_N(z)=1. Its first column is D_N(z), and all of its entries are given by [source reference].
Exact mathematical passage
U_N(z)=e^{zJ_N}
VUse in source · 1 passage
Search aliases: V
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 495
At the origin, the derivative matrix is diagonal with entries 1,λ,…,λ^N-1. For an explicit determinant proof, write each basis function as a finite sum of exponentials with exponents a_ℓ=λω^ℓ. Its derivative matrix factors as W(z)=V diag(e^a_ℓ z) C, V_kℓ=a_ℓ^k, where C is the constant matrix of Fourier weights. Thus det W(z)=e^z∑_ℓ a_ℓdet W(0)=det W(0), because N≥2 implies ∑_ℓω^ℓ=0. The product of the diagonal initial values is the stated power of λ.
Exact mathematical passage
W(z)=V\,\operatorname{diag}(e^{a_\ell z})\,C, \qquad V_{k\ell}=a_\ell^k,
vUse in source · 9 passages
Search aliases: v
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1895
On the resulting strip, the roots e^ia_N and e^-ia_N have real part c_N, while each remaining root η=u+iv satisfies u≤cos(3a_N)=c_N-γ_N. Hence Re((η-e^-ia_N)z) ≤-γ_Nx+2Y. Division by e^e^-ia_Nz gives the displayed normalization. Each differentiation multiplies a remainder term by η-e^-ia_N, whose modulus is at most two. Thus, for q=0,1,2, |R_N^(q)(z)|≤(N-2)2^qe^2Ye^-γ_Nx. One may take R=1 and C=4(N-2)e^2Y.
Exact mathematical passage
\eta=u+iv
v^{1/4}Use in source · 1 passage
Search aliases: v^1/4
Use in source · counting.tex, line 30
(2.19219,1.00000) circle (0.055); node[] at (2.19219,0.58000) 81/16; node[font=sffamilyfootnotesize] at (2.19219,1.45000) count 2; draw[line width=1pt] (11.54219,0.88000) -- (11.54219,1.12000); draw[fill=black] (11.54219,1.00000) circle (0.055); node[] at (11.54219,0.58000) 625/16; node[font=sffamilyfootnotesize] at (11.54219,1.45000) count 3; node[] at (6.50000,-0.05000) n_D_4(sqrt2π u)=4lfloor u+1/2rfloor, n_Δ_4(4π^4 v)=lfloor v^1/4+1/2rfloor;
Exact mathematical passage
n_{D_4}(\sqrt2\pi u)=4\lfloor u+1/2\rfloor,\quad n_{\Delta_4}(4\pi^4 v)=\lfloor v^{1/4}+1/2\rfloor
V_{k\ell}Use in source · 1 passage
Search aliases: V_kℓ ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 495
At the origin, the derivative matrix is diagonal with entries 1,λ,…,λ^N-1. For an explicit determinant proof, write each basis function as a finite sum of exponentials with exponents a_ℓ=λω^ℓ. Its derivative matrix factors as W(z)=V diag(e^a_ℓ z) C, V_kℓ=a_ℓ^k, where C is the constant matrix of Fourier weights. Thus det W(z)=e^z∑_ℓ a_ℓdet W(0)=det W(0), because N≥2 implies ∑_ℓω^ℓ=0. The product of the diagonal initial values is the stated power of λ.
Exact mathematical passage
W(z)=V\,\operatorname{diag}(e^{a_\ell z})\,C, \qquad V_{k\ell}=a_\ell^k,
\vdotsUse in source · 2 passages
Search aliases: vdots
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 811
[Companion matrix realization] Put D_N(z) = D_N,0(z) D_N,1(z) vdots D_N,N-1(z) . Then D_N(z)=e^zJ_Ne_0 and D_N(z)=frac1Ntr(e^zJ_N). The eigenvalues of J_N are the N roots of -1, each with multiplicity one.
Exact mathematical passage
\mathbf D_N(z) = \begin{pmatrix} \Dfun_{N,0}(z)\\ \Dfun_{N,1}(z)\\ \vdots\\ \Dfun_{N,N-1}(z) \end{pmatrix}.
WUse in source · 2 passages
Search aliases: W
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 495
At the origin, the derivative matrix is diagonal with entries 1,λ,…,λ^N-1. For an explicit determinant proof, write each basis function as a finite sum of exponentials with exponents a_ℓ=λω^ℓ. Its derivative matrix factors as W(z)=V diag(e^a_ℓ z) C, V_kℓ=a_ℓ^k, where C is the constant matrix of Fourier weights. Thus det W(z)=e^z∑_ℓ a_ℓdet W(0)=det W(0), because N≥2 implies ∑_ℓω^ℓ=0. The product of the diagonal initial values is the stated power of λ.
Exact mathematical passage
W(z)=V\,\operatorname{diag}(e^{a_\ell z})\,C, \qquad V_{k\ell}=a_\ell^k,
w^2Use in source · 13 passages
Search aliases: w^2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 994
The identity Δ_N(w)^-r = exp( r∑_m≥1κ_N,mfracw^mm ) fits the defining generating function for B_n with x_m=r(m-1)!κ_N,m. The coefficient of w^n is therefore the displayed Bell polynomial divided by n!. Multiplication by (Nn)! gives P_N,n(r). Since Δ_N(w)=1-fracwN!+O(w^2), we have κ_N,1=1/N!. The term x_1^n gives the stated leading coefficient, which is nonzero.
Exact mathematical passage
\Delta_N(w)=1-\frac{w}{N!}+O(w^2),
w^3Use in source · 3 passages
Search aliases: w^3
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1221
[title=A coefficient calculation at order two] Write Δ_2(w)^-r=∑_n≥0b_n(r)w^n. Since Δ_2(w)=1-frac w2+fracw^224+O(w^3), -logΔ_2(w)=frac w2+fracw^212+O(w^3), the first logarithmic coefficients are κ_2,1=1/2 and κ_2,2=1/6. Newton's recurrence gives b_1(r)=frac r2, 2b_2(r)=r(frac r4+frac16). Restoring the factorial normalization yields P_2,1(r)=r and P_2,2(r)=r(3r+2). At r=1, the second value is five. No zero-location assumption is needed for this finite calculation. modelfigureparameter_polynomial…
Exact mathematical passage
\Delta_2(w)=1-\frac w2+\frac{w^2}{24}+O(w^3), \qquad -\log\Delta_2(w)=\frac w2+\frac{w^2}{12}+O(w^3),
w^bUse in source · 1 passage
Search aliases: w^b
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 519
Expand e^λ(z+w)=e^λ ze^λ w as a bivariate power series. A term z^a w^b contributes to the kth section exactly when a+b≡ kbmod N. Grouping first by i≡ abmod N and j≡ bbmod N gives the displayed sum.
Exact mathematical passage
z^a w^b
w^{k-1}Use in source · 1 passage
Search aliases: w^k-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1129
Let B_r(w)=Δ_N(w)^-r = ∑_n≥0b_N,n(r)w^n. Logarithmic differentiation gives fracB_r'(w)B_r(w) = r∑_k≥1κ_N,kw^k-1. Comparison of the coefficient of w^n-1 proves the recurrence.
Exact mathematical passage
\frac{B_r'(w)}{B_r(w)} = r\sum_{k\geq1}\kappa_{N,k}w^{k-1}.
w^{m-1}Use in source · 2 passages
Search aliases: w^m-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2459
Differentiate the product identity in logarithmic form to obtain Δ_N'(w) = -Δ_N(w) ∑_m≥1σ_N,mw^m-1. Comparison of the coefficient of w^m-1 gives ma_m = -∑_k=0^m-1a_kσ_N,m-k. Solving for σ_N,m gives the recurrence.
Exact mathematical passage
\Delta_N'(w) = -\Delta_N(w) \sum_{m\geq1}\sigma_{N,m}w^{m-1}.
w^{n-1}Use in source · 1 passage
Search aliases: w^n-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1134
Let B_r(w)=Δ_N(w)^-r = ∑_n≥0b_N,n(r)w^n. Logarithmic differentiation gives fracB_r'(w)B_r(w) = r∑_k≥1κ_N,kw^k-1. Comparison of the coefficient of w^n-1 proves the recurrence.
Exact mathematical passage
w^{n-1}
\widetilde{\sigma}Use in source · 4 passages
Search aliases: widetildeσ widetilde sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2501
[Theta Bell polynomial formula] Put widetildeσ_N,m=(-1)^mσ_N,m. Then symupΘ_N,n(u) = frac(Nn)!n! B_n( widetildeσ_N,1(u^N+u), 1!widetildeσ_N,2(u^2N+u), …, (n-1)!widetildeσ_N,n(u^nN+u) ).
Exact mathematical passage
\widetilde{\sigma}_{N,m}=(-1)^m\sigma_{N,m}.
\widetilde{\sigma}_{N,1}Use in source · 1 passage
Search aliases: widetildeσ_N,1 widetilde sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2505
[Theta Bell polynomial formula] Put widetildeσ_N,m=(-1)^mσ_N,m. Then symupΘ_N,n(u) = frac(Nn)!n! B_n( widetildeσ_N,1(u^N+u), 1!widetildeσ_N,2(u^2N+u), …, (n-1)!widetildeσ_N,n(u^nN+u) ).
Exact mathematical passage
\Thetafun_{N,n}(u) = \frac{(Nn)!}{n!} B_n\bigl( \widetilde{\sigma}_{N,1}(u^N+u), 1!\widetilde{\sigma}_{N,2}(u^{2N}+u), \ldots, (n-1)!\widetilde{\sigma}_{N,n}(u^{nN}+u) \bigr).
\widetilde{\sigma}_{N,2}Use in source · 1 passage
Search aliases: widetildeσ_N,2 widetilde sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2505
[Theta Bell polynomial formula] Put widetildeσ_N,m=(-1)^mσ_N,m. Then symupΘ_N,n(u) = frac(Nn)!n! B_n( widetildeσ_N,1(u^N+u), 1!widetildeσ_N,2(u^2N+u), …, (n-1)!widetildeσ_N,n(u^nN+u) ).
Exact mathematical passage
\Thetafun_{N,n}(u) = \frac{(Nn)!}{n!} B_n\bigl( \widetilde{\sigma}_{N,1}(u^N+u), 1!\widetilde{\sigma}_{N,2}(u^{2N}+u), \ldots, (n-1)!\widetilde{\sigma}_{N,n}(u^{nN}+u) \bigr).
\widetilde{\sigma}_{N,m}Use in source · 3 passages
Search aliases: widetildeσ_N,m widetilde sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2501
[Theta Bell polynomial formula] Put widetildeσ_N,m=(-1)^mσ_N,m. Then symupΘ_N,n(u) = frac(Nn)!n! B_n( widetildeσ_N,1(u^N+u), 1!widetildeσ_N,2(u^2N+u), …, (n-1)!widetildeσ_N,n(u^nN+u) ).
Exact mathematical passage
\widetilde{\sigma}_{N,m}=(-1)^m\sigma_{N,m}.
\widetilde{\sigma}_{N,n}Use in source · 1 passage
Search aliases: widetildeσ_N,n widetilde sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2505
[Theta Bell polynomial formula] Put widetildeσ_N,m=(-1)^mσ_N,m. Then symupΘ_N,n(u) = frac(Nn)!n! B_n( widetildeσ_N,1(u^N+u), 1!widetildeσ_N,2(u^2N+u), …, (n-1)!widetildeσ_N,n(u^nN+u) ).
Exact mathematical passage
\Thetafun_{N,n}(u) = \frac{(Nn)!}{n!} B_n\bigl( \widetilde{\sigma}_{N,1}(u^N+u), 1!\widetilde{\sigma}_{N,2}(u^{2N}+u), \ldots, (n-1)!\widetilde{\sigma}_{N,n}(u^{nN}+u) \bigr).
XUse in source · 3 passages
Search aliases: X
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 3009
6.1--6.6 · Quotient-algebra power bases and actual exponential-series convergence in the finite-dimensional coefficient topology. Both pointwise section-coordinate identities. Signed-shift powers, all matrix-exponential entries, group laws and trace are checked. The characteristic polynomial X^N+1, its distinct roots, exact spectrum and determinant one are proved.
Exact mathematical passage
X^N+1
x^2Use in source · 3 passages
Search aliases: x^2
Use in source · atlas-pages.tex, line 267
IALSTraditional monograph: Theorem 9.6 (p. 21), Theorem 9.8 (p. 21), Proposition 9.9 (p. 21). For C_2,0(z)=sech z, define E_m(x) by sech z e^xz=∑ E_m(x)z^m/m!. The formal differential-operator formula is E_m(x)=sech(∂_x)x^m. Although the symbol is an infinite series, derivatives above degree m vanish. vfigappell_operator For m=4, only three terms survive: (1-frac∂_x^22!+5frac∂_x^44!)x^4 =x^4-6x^2+5. The four real roots of this polynomial are ±1,±sqrt5. They are not the zeros of the original anal…
Exact mathematical passage
\left(1-\frac{\partial_x^2}{2!}+5\frac{\partial_x^4}{4!}\right)x^4 =x^4-6x^2+5.
X^3Use in source · 2 passages
Search aliases: X^3
Use in source · atlas-pages.tex, line 95
ordinates and columns are input coordinates. The upper-triangular minus signs record crossing the signed edge while moving from the column to the row. They are not arbitrary orientation choices. headTwo checks connect the matrix and the scalar equation The initial derivative matrix is the identity. Each initial jet (y(0),y'(0),y''(0)) therefore determines a unique linear combination. Also, frac13Tr(e^zJ_3)=D_3(z), det(XI-J_3)=X^3+1. The roots of -1 appearing later in the exponential sum are alre…
Exact mathematical passage
\frac13\Tr(e^{zJ_3})=D_3(z),\qquad \det(XI-J_3)=X^3+1.
x^4Use in source · 7 passages
Search aliases: x^4
Use in source · atlas-pages.tex, line 267
IALSTraditional monograph: Theorem 9.6 (p. 21), Theorem 9.8 (p. 21), Proposition 9.9 (p. 21). For C_2,0(z)=sech z, define E_m(x) by sech z e^xz=∑ E_m(x)z^m/m!. The formal differential-operator formula is E_m(x)=sech(∂_x)x^m. Although the symbol is an infinite series, derivatives above degree m vanish. vfigappell_operator For m=4, only three terms survive: (1-frac∂_x^22!+5frac∂_x^44!)x^4 =x^4-6x^2+5. The four real roots of this polynomial are ±1,±sqrt5. They are not the zeros of the original anal…
Exact mathematical passage
\left(1-\frac{\partial_x^2}{2!}+5\frac{\partial_x^4}{4!}\right)x^4 =x^4-6x^2+5.
x^mUse in source · 5 passages
Search aliases: x^m
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1533
[Formal differential-operator representation on polynomials] Let a formal series A(z)=∑_q≥0a_qz^q/q! act on polynomials by A(∂_x)f(x) = ∑_q≥0fraca_qq!∂_x^qf(x). The power r is applied to the formal symbol before the substitution z=∂_x. Then E_m^(N,j,r)(x) = C_N,j(∂_x)^r x^m. The inverse relation is x^m = Φ_N,j(∂_x^N)^r E_m^(N,j,r)(x). Both sums are finite on every polynomial.
Exact mathematical passage
E_m^{(N,j,r)}(x) = \Cfun_{N,j}(\partial_x)^r x^m.
x^{m-Nn}Use in source · 1 passage
Search aliases: x^m-Nn
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1489
[Explicit Appell polynomial] For every m≥0, E_m^(N,j,r)(x) = ∑_0≤ n≤ m/N binommNn C_N,j,n(r)x^m-Nn.
Exact mathematical passage
E_m^{(N,j,r)}(x) = \sum_{0\leq n\leq m/N} \binom{m}{Nn} C_{N,j,n}(r)x^{m-Nn}.
X^NUse in source · 1 passage
Search aliases: X^N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 3009
6.1--6.6 · Quotient-algebra power bases and actual exponential-series convergence in the finite-dimensional coefficient topology. Both pointwise section-coordinate identities. Signed-shift powers, all matrix-exponential entries, group laws and trace are checked. The characteristic polynomial X^N+1, its distinct roots, exact spectrum and determinant one are proved.
Exact mathematical passage
X^N+1
x^NUse in source · 3 passages
Search aliases: x^N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 841
The vectors e_0,J_Ne_0,…,J_N^N-1e_0 are the standard basis. Thus e_0 is cyclic for J_N, so the minimal polynomial of J_N has degree N. Since J_N^N=-I, that minimal polynomial divides x^N+1 and therefore equals x^N+1. The characteristic polynomial has the same degree and is divisible by the minimal polynomial, so it too equals x^N+1. This polynomial has distinct roots. Consequently J_N is diagonalizable and its spectrum consists of all N roots of -1, each once. Hence tr(e^zJ_N) = ∑_η^N=-1e^η z = …
Exact mathematical passage
x^N+1
x_1Use in source · 5 passages
Search aliases: x_1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 955
The polynomial dependence on r can be read directly from the formal logarithm. We use the complete exponential Bell polynomials B_n defined by exp(∑_m≥1x_mfract^mm!) = ∑_n≥0B_n(x_1,…,x_n)fract^nn!.
Exact mathematical passage
\exp\left(\sum_{m\geq1}x_m\frac{t^m}{m!}\right) = \sum_{n\geq0}B_n(x_1,\ldots,x_n)\frac{t^n}{n!}.
x_1^3Use in source · 2 passages
Search aliases: x_1^3
Use in source · atlas-pages.tex, line 184
r=exp (r∑_m≥1κ_N,mfracw^mm), the coefficient of w^n is a polynomial of degree n in r. Bell polynomials collect the possible ways logarithmic terms contribute total degree n. vfigbell_partitions For N=2, the first three logarithmic coefficients are κ_1=frac12, κ_2=frac16, κ_3=frac115. The third Bell polynomial is B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3. Substituting x_1=rκ_1, x_2=rκ_2, x_3=2rκ_3 and multiplying by 6!/3! gives P_2,3(r)=15r^3+30r^2+16r. At r=1 this is 61. At r=2 it is 272. At r=0 it van…
Exact mathematical passage
B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3
x_1^nUse in source · 1 passage
Search aliases: x_1^n
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 997
e identity Δ_N(w)^-r = exp( r∑_m≥1κ_N,mfracw^mm ) fits the defining generating function for B_n with x_m=r(m-1)!κ_N,m. The coefficient of w^n is therefore the displayed Bell polynomial divided by n!. Multiplication by (Nn)! gives P_N,n(r). Since Δ_N(w)=1-fracwN!+O(w^2), we have κ_N,1=1/N!. The term x_1^n gives the stated leading coefficient, which is nonzero.
Exact mathematical passage
x_1^n
x_2Use in source · 4 passages
Search aliases: x_2
Use in source · atlas-pages.tex, line 184
r=exp (r∑_m≥1κ_N,mfracw^mm), the coefficient of w^n is a polynomial of degree n in r. Bell polynomials collect the possible ways logarithmic terms contribute total degree n. vfigbell_partitions For N=2, the first three logarithmic coefficients are κ_1=frac12, κ_2=frac16, κ_3=frac115. The third Bell polynomial is B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3. Substituting x_1=rκ_1, x_2=rκ_2, x_3=2rκ_3 and multiplying by 6!/3! gives P_2,3(r)=15r^3+30r^2+16r. At r=1 this is 61. At r=2 it is 272. At r=0 it van…
Exact mathematical passage
B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3
x_3Use in source · 4 passages
Search aliases: x_3
Use in source · atlas-pages.tex, line 184
r=exp (r∑_m≥1κ_N,mfracw^mm), the coefficient of w^n is a polynomial of degree n in r. Bell polynomials collect the possible ways logarithmic terms contribute total degree n. vfigbell_partitions For N=2, the first three logarithmic coefficients are κ_1=frac12, κ_2=frac16, κ_3=frac115. The third Bell polynomial is B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3. Substituting x_1=rκ_1, x_2=rκ_2, x_3=2rκ_3 and multiplying by 6!/3! gives P_2,3(r)=15r^3+30r^2+16r. At r=1 this is 61. At r=2 it is 272. At r=0 it van…
Exact mathematical passage
B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3
x_mUse in source · 2 passages
Search aliases: x_m
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 955
The polynomial dependence on r can be read directly from the formal logarithm. We use the complete exponential Bell polynomials B_n defined by exp(∑_m≥1x_mfract^mm!) = ∑_n≥0B_n(x_1,…,x_n)fract^nn!.
Exact mathematical passage
\exp\left(\sum_{m\geq1}x_m\frac{t^m}{m!}\right) = \sum_{n\geq0}B_n(x_1,\ldots,x_n)\frac{t^n}{n!}.
x_nUse in source · 1 passage
Search aliases: x_n
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 955
The polynomial dependence on r can be read directly from the formal logarithm. We use the complete exponential Bell polynomials B_n defined by exp(∑_m≥1x_mfract^mm!) = ∑_n≥0B_n(x_1,…,x_n)fract^nn!.
Exact mathematical passage
\exp\left(\sum_{m\geq1}x_m\frac{t^m}{m!}\right) = \sum_{n\geq0}B_n(x_1,\ldots,x_n)\frac{t^n}{n!}.
y'Use in source · 2 passages
Search aliases: y'
Use in source · atlas-pages.tex, line 94
z^66!-⋯, D_3,1=z-fracz^44!+⋯. Rows are output coordinates and columns are input coordinates. The upper-triangular minus signs record crossing the signed edge while moving from the column to the row. They are not arbitrary orientation choices. headTwo checks connect the matrix and the scalar equation The initial derivative matrix is the identity. Each initial jet (y(0),y'(0),y''(0)) therefore determines a unique linear combination. Also, frac13Tr(e^zJ_3)=D_3(z), det(XI-J_3)=X^3+1. The roots of -1…
Exact mathematical passage
(y(0),y'(0),y''(0))
y''Use in source · 2 passages
Search aliases: y''
Use in source · atlas-pages.tex, line 94
z^66!-⋯, D_3,1=z-fracz^44!+⋯. Rows are output coordinates and columns are input coordinates. The upper-triangular minus signs record crossing the signed edge while moving from the column to the row. They are not arbitrary orientation choices. headTwo checks connect the matrix and the scalar equation The initial derivative matrix is the identity. Each initial jet (y(0),y'(0),y''(0)) therefore determines a unique linear combination. Also, frac13Tr(e^zJ_3)=D_3(z), det(XI-J_3)=X^3+1. The roots of -1…
Exact mathematical passage
(y(0),y'(0),y''(0))
y^{(N)}Use in source · 6 passages
Search aliases: y^(N)
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 179
Fix a positive integer N. Root of unity projection of the exponential produces residue-class sections that, for λ≠0, form a basis for y^(N)=λ^N y. The signed sections for y^(N)=-y yield the denominator D_N and its reduced form Δ_N. We derive reciprocal coefficient polynomials, Bell and Newton recurrences, an additive parameter law, and residue section Appell and theta families. For N≥2, Δ_N has order 1/N. We specialize classical two term zero theory for exponential sums to the regular exponent p…
Exact mathematical passage
y^{(N)}=\lambda^N y
y^{m-q}Use in source · 3 passages
Search aliases: y^m-q
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1474
[Appell laws] For m≥1, the polynomials satisfy fracddxE_m^(N,j,r)(x) = mE_m-1^(N,j,r)(x) and, for m≥0, E_m^(N,j,r)(x+y) = ∑_q=0^m binom mq E_q^(N,j,r)(x)y^m-q.
Exact mathematical passage
E_m^{(N,j,r)}(x+y) = \sum_{q=0}^{m} \binom mq E_q^{(N,j,r)}(x)y^{m-q}.
\mathbb ZUse in source · 1 passage
Search aliases: Z mathbb
Use in source · branch_domain.tex, line 14
ρ=π/2; draw[->,line width=1.1pt] (4.88,2.3) arc[start angle=0,end angle=330,radius=.48]; node[] at (2.80000,0.10000) the loop is not inside the disk; node[] at (9.20000,4.10000) D(z)∼ D^prime(ρ)(z-ρ); node[] at (9.20000,2.90000) log D⟼log D+2π i; node[] at (9.20000,1.65000) D^-r⟼ e^-2π i rD^-r; node[] at (9.20000,0.40000) r=1/2: -1 r∈ Z: 1;
Exact mathematical passage
r=1/2:\ -1\qquad r\in\mathbb Z:\ 1
z^{0}Use in source · 2 passages
Search aliases: z^0
Use in source · root_filter.tex, line 14
.00000); node[] at (0.60000,2.00000) -1; draw[->,line width=1pt] (2.50000,2.00000) -- (2.50000,0.55000); node[] at (2.50000,0.10000) -i; node[] at (2.50000,-0.30000) frac14∑_ℓ=0^3 i^(m-j)ℓ; node[anchor=east] at (6.10000,3.70000) j=0; draw[fill=black!0,draw=black!40,line width=.35pt] (6.45000,3.40000) rectangle ++(1.70000,0.60000); node[] at (7.30000,3.70000) z^0; draw[fill=black!0,draw=black!40,line width=.35pt] (8.15000,3.40000) rectangle ++(1.70000,0.60000); node[] at (9.00000,3.70000) z^4; dr…
Exact mathematical passage
z^{0}
z^{1}Use in source · 2 passages
Search aliases: z^1
Use in source · root_filter.tex, line 21
.40000) rectangle ++(1.70000,0.60000); node[] at (9.00000,3.70000) z^4; draw[fill=black!0,draw=black!40,line width=.35pt] (9.85000,3.40000) rectangle ++(1.70000,0.60000); node[] at (10.70000,3.70000) z^8; node[anchor=east] at (6.10000,2.85000) j=1; draw[fill=black!0,draw=black!40,line width=.35pt] (6.45000,2.55000) rectangle ++(1.70000,0.60000); node[] at (7.30000,2.85000) z^1; draw[fill=black!0,draw=black!40,line width=.35pt] (8.15000,2.55000) rectangle ++(1.70000,0.60000); node[] at (9.00000,2…
Exact mathematical passage
z^{1}
z^{10}Use in source · 4 passages
Search aliases: z^10
Use in source · atlas-pages.tex, line 53
r rotated arguments and attach the phase i^-jℓ: ( B_4,jF)(z)=frac14∑_ℓ=0^3i^-jℓF(i^ℓ z). A monomial a_mz^m is multiplied by the finite average of i^(m-j)ℓ. Matching phases all vote 1. Nonmatching phases cancel exactly. vfigroot_filter headAn example that can be checked coefficient by coefficient Take F(z)=∑_m=0^12(m+1)z^m. Then B_4,2F=3z^2+7z^6+11z^10. Applying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the ori…
Exact mathematical passage
\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.
z^{11}Use in source · 2 passages
Search aliases: z^11
Use in source · root_filter.tex, line 39
lack!0,draw=black!40,line width=.35pt] (6.45000,0.85000) rectangle ++(1.70000,0.60000); node[] at (7.30000,1.15000) z^3; draw[fill=black!0,draw=black!40,line width=.35pt] (8.15000,0.85000) rectangle ++(1.70000,0.60000); node[] at (9.00000,1.15000) z^7; draw[fill=black!0,draw=black!40,line width=.35pt] (9.85000,0.85000) rectangle ++(1.70000,0.60000); node[] at (10.70000,1.15000) z^11; node[] at (9.10000,-0.30000) only the matching phase survives;
Exact mathematical passage
z^{11}
z^2Use in source · 13 passages
Search aliases: z^2
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2779
For N=2, [source reference] and D_2(z)=Δ_2(z^2) give λ_m=π^2(m+frac12)^2, m≥0. Thus the first formula holds with Q_2^red=0.
Exact mathematical passage
\Dfun_2(z)=\Delta_2(z^2)
z^{2}Use in source · 7 passages
Search aliases: z^2
Use in source · cyclic_product.tex, line 4
[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[] at (3.72500,4.22000) 2; node[] at (6.37500,4.22000) z; node[] at (9.02500,4.22000) z^2; node[anchor=east] at (2.15000,3.37500) z^3 + 1; draw[fill=black!18,draw=black!40,line width=.35pt] (2.40000,2.95000) rectangle ++(2.65000,0.85000); node[] at (3.72500,3.37500) 2 z^3 + 2; draw[fill=black!0,draw=black!40,line width=.35pt] (5.05000,2.95000) rectangle ++(2.65000,0.85000); node[] at (6.37500,3.37500) z^4 + z; draw[fill=…
Exact mathematical passage
z^{2}
z^3Use in source · 18 passages
Search aliases: z^3
Use in source · atlas-pages.tex, line 65
vpage4Products mix the residue lanesCYCLIC CONVOLUTIONTraditional monograph: Theorem 3.3 (p. 5). Selection separates a function additively. It does not make multiplication diagonal. A residue-i term multiplied by a residue-j term lands in residue i+jpmod N. For N=3, use F=1+2z+3z^2+z^3, G=2+z+z^2. The table displays every product of their three selected sections. vfigcyclic_product The shaded cells contribute to residue zero: 2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3. Keeping only the product of the res…
Exact mathematical passage
F=1+2z+3z^2+z^3,\qquad G=2+z+z^2.
z^{3}Use in source · 6 passages
Search aliases: z^3
Use in source · cyclic_product.tex, line 5
[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[] at (3.72500,4.22000) 2; node[] at (6.37500,4.22000) z; node[] at (9.02500,4.22000) z^2; node[anchor=east] at (2.15000,3.37500) z^3 + 1; draw[fill=black!18,draw=black!40,line width=.35pt] (2.40000,2.95000) rectangle ++(2.65000,0.85000); node[] at (3.72500,3.37500) 2 z^3 + 2; draw[fill=black!0,draw=black!40,line width=.35pt] (5.05000,2.95000) rectangle ++(2.65000,0.85000); node[] at (6.37500,3.37500) z^4 + z; draw[fill=…
Exact mathematical passage
z^{3} + 1
z^4Use in source · 12 passages
Search aliases: z^4
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1714
ion gives a polygon of exponents. Large zeros can occur where two dominant terms nearly cancel. The proof first localizes those zeros and then uses symmetry and uniqueness to place them on the rays. Finite exceptional zeros remain part of the later products and sums. modelfigurezero_orbitsThe exact order-four model. The first three zeros on each of four rays are mapped by w=z^4 to three positive reduced zeros. The reduced axis uses a logarithmic scale. This complete order-four example does not c…
Exact mathematical passage
w=z^4
z^{4}Use in source · 4 passages
Search aliases: z^4
Use in source · cyclic_product.tex, line 9
t (6.37500,4.22000) z; node[] at (9.02500,4.22000) z^2; node[anchor=east] at (2.15000,3.37500) z^3 + 1; draw[fill=black!18,draw=black!40,line width=.35pt] (2.40000,2.95000) rectangle ++(2.65000,0.85000); node[] at (3.72500,3.37500) 2 z^3 + 2; draw[fill=black!0,draw=black!40,line width=.35pt] (5.05000,2.95000) rectangle ++(2.65000,0.85000); node[] at (6.37500,3.37500) z^4 + z; draw[fill=black!0,draw=black!40,line width=.35pt] (7.70000,2.95000) rectangle ++(2.65000,0.85000); node[] at (9.02500,3.3…
Exact mathematical passage
z^{4} + z
z^{5}Use in source · 3 passages
Search aliases: z^5
Use in source · cyclic_product.tex, line 11
aw=black!40,line width=.35pt] (2.40000,2.95000) rectangle ++(2.65000,0.85000); node[] at (3.72500,3.37500) 2 z^3 + 2; draw[fill=black!0,draw=black!40,line width=.35pt] (5.05000,2.95000) rectangle ++(2.65000,0.85000); node[] at (6.37500,3.37500) z^4 + z; draw[fill=black!0,draw=black!40,line width=.35pt] (7.70000,2.95000) rectangle ++(2.65000,0.85000); node[] at (9.02500,3.37500) z^5 + z^2; node[anchor=east] at (2.15000,2.52500) 2 z; draw[fill=black!0,draw=black!40,line width=.35pt] (2.40000,2.100…
Exact mathematical passage
z^{5} + z^{2}
z^6Use in source · 7 passages
Search aliases: z^6
Use in source · atlas-pages.tex, line 53
r rotated arguments and attach the phase i^-jℓ: ( B_4,jF)(z)=frac14∑_ℓ=0^3i^-jℓF(i^ℓ z). A monomial a_mz^m is multiplied by the finite average of i^(m-j)ℓ. Matching phases all vote 1. Nonmatching phases cancel exactly. vfigroot_filter headAn example that can be checked coefficient by coefficient Take F(z)=∑_m=0^12(m+1)z^m. Then B_4,2F=3z^2+7z^6+11z^10. Applying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the ori…
Exact mathematical passage
\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.
z^{6}Use in source · 2 passages
Search aliases: z^6
Use in source · root_filter.tex, line 30
5000) rectangle ++(1.70000,0.60000); node[] at (10.70000,2.85000) z^9; node[anchor=east] at (6.10000,2.00000) j=2; draw[fill=black!18,draw=black!40,line width=.35pt] (6.45000,1.70000) rectangle ++(1.70000,0.60000); node[] at (7.30000,2.00000) z^2; draw[fill=black!18,draw=black!40,line width=.35pt] (8.15000,1.70000) rectangle ++(1.70000,0.60000); node[] at (9.00000,2.00000) z^6; draw[fill=black!18,draw=black!40,line width=.35pt] (9.85000,1.70000) rectangle ++(1.70000,0.60000); node[] at (10.70000…
Exact mathematical passage
z^{6}
z^7Use in source · 2 passages
Search aliases: z^7
Use in source · riccati.tex, line 10
D_2^primeprime=-D_2; node[] at (9.30000,3.25000) T_2^prime=1+T_2^2; node[] at (3.00000,1.85000) D_3^primeprimeprime=-D_3; node[] at (9.30000,1.85000) T_3^primeprime-3T_3T_3^prime+T_3^3=1; node[] at (6.15000,0.50000) T_N=-D_N^prime/D_N only where D_N≠0; draw[black!25] (6.10000,1.10000) -- (6.10000,4.65000); node[] at (6.10000,-0.40000) T_4(z)=fracz^33!+34fracz^77!+⋯;
Exact mathematical passage
T_4(z)=\frac{z^3}{3!}+34\frac{z^7}{7!}+\cdots
z^{7}Use in source · 2 passages
Search aliases: z^7
Use in source · root_filter.tex, line 37
70000) rectangle ++(1.70000,0.60000); node[] at (10.70000,2.00000) z^10; node[anchor=east] at (6.10000,1.15000) j=3; draw[fill=black!0,draw=black!40,line width=.35pt] (6.45000,0.85000) rectangle ++(1.70000,0.60000); node[] at (7.30000,1.15000) z^3; draw[fill=black!0,draw=black!40,line width=.35pt] (8.15000,0.85000) rectangle ++(1.70000,0.60000); node[] at (9.00000,1.15000) z^7; draw[fill=black!0,draw=black!40,line width=.35pt] (9.85000,0.85000) rectangle ++(1.70000,0.60000); node[] at (10.70000,…
Exact mathematical passage
z^{7}
z^{8}Use in source · 2 passages
Search aliases: z^8
Use in source · root_filter.tex, line 18
lack!0,draw=black!40,line width=.35pt] (6.45000,3.40000) rectangle ++(1.70000,0.60000); node[] at (7.30000,3.70000) z^0; draw[fill=black!0,draw=black!40,line width=.35pt] (8.15000,3.40000) rectangle ++(1.70000,0.60000); node[] at (9.00000,3.70000) z^4; draw[fill=black!0,draw=black!40,line width=.35pt] (9.85000,3.40000) rectangle ++(1.70000,0.60000); node[] at (10.70000,3.70000) z^8; node[anchor=east] at (6.10000,2.85000) j=1; draw[fill=black!0,draw=black!40,line width=.35pt] (6.45000,2.55000) re…
Exact mathematical passage
z^{8}
z^{9}Use in source · 2 passages
Search aliases: z^9
Use in source · root_filter.tex, line 25
lack!0,draw=black!40,line width=.35pt] (6.45000,2.55000) rectangle ++(1.70000,0.60000); node[] at (7.30000,2.85000) z^1; draw[fill=black!0,draw=black!40,line width=.35pt] (8.15000,2.55000) rectangle ++(1.70000,0.60000); node[] at (9.00000,2.85000) z^5; draw[fill=black!0,draw=black!40,line width=.35pt] (9.85000,2.55000) rectangle ++(1.70000,0.60000); node[] at (10.70000,2.85000) z^9; node[anchor=east] at (6.10000,2.00000) j=2; draw[fill=black!18,draw=black!40,line width=.35pt] (6.45000,1.70000) r…
Exact mathematical passage
z^{9}
z^aUse in source · 1 passage
Search aliases: z^a
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 519
Expand e^λ(z+w)=e^λ ze^λ w as a bivariate power series. A term z^a w^b contributes to the kth section exactly when a+b≡ kbmod N. Grouping first by i≡ abmod N and j≡ bbmod N gives the displayed sum.
Exact mathematical passage
z^a w^b
z^kUse in source · 1 passage
Search aliases: z^k
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 463
[Differential cycle and initial data] For 1≤ j<N, fracddz E_N,j(λ z) = λ E_N,j-1(λ z), while fracddz E_N,0(λ z) = λ E_N,N-1(λ z). Thus fracd^Ndz^N E_N,j(λ z) = λ^N E_N,j(λ z). For 0≤ k<N, .fracd^kdz^k E_N,j(λ z)|_z=0 = δ_kjλ^j. If λ≠0, the N functions E_N,j(λ z) form a basis of the solution space of y^(N)=λ^Ny.
Exact mathematical passage
\left.\frac{d^k}{dz^k}\Efun_{N,j}(\lambda z)\right|_{z=0} = \delta_{kj}\lambda^j.
z^{N-1}Use in source · 5 passages
Search aliases: z^N-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1267
Since D_N(z)=Δ_N(z^N), -frac D_N'(z) D_N(z) = -Nz^N-1fracΔ_N'(z^N)Δ_N(z^N). Differentiate the defining series for -logΔ_N and compare coefficients.
Exact mathematical passage
-\frac{\Dfun_N'(z)}{\Dfun_N(z)} = -Nz^{N-1}\frac{\Delta_N'(z^N)}{\Delta_N(z^N)}.
z^{Nm-1}Use in source · 3 passages
Search aliases: z^Nm-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1252
[Logarithmic coefficient formula] The transfer tangent satisfies T_N(z) = N∑_m≥1κ_N,mz^Nm-1. Equivalently, T_N(n) = N(Nn+N-1)!κ_N,n+1.
Exact mathematical passage
\Tfun_N(z) = N\sum_{m\geq1}\kappa_{N,m}z^{Nm-1}.
z^{Nn+d_{pq}}Use in source · 1 passage
Search aliases: z^Nn+d_pq
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 883
For every nonnegative integer n, J_N^Nn+d_pqe_q = (-1)^nε_pqe_p. Only powers whose exponent is congruent to d_pq modulo N contribute to the (p,q) entry of the exponential. Therefore (e^zJ_N)_pq = ε_pq ∑_n≥0(-1)^n fracz^Nn+d_pq(Nn+d_pq)!. This is the stated formula.
Exact mathematical passage
\bigl(e^{zJ_N}\bigr)_{pq} = \varepsilon_{pq} \sum_{n\geq0}(-1)^n \frac{z^{Nn+d_{pq}}}{(Nn+d_{pq})!}.
z^{Nn-1}Use in source · 1 passage
Search aliases: z^Nn-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1027
Set S= D_N^-r. Logarithmic differentiation gives D_NS'+r D_N'S=0. Taking the coefficient of z^Nn-1/(Nn-1)! gives ∑_k=0^n-1(-1)^k binomNn-1NkP_N,n-k(r) +r∑_k=1^n(-1)^k binomNn-1Nk-1P_N,n-k(r)=0. The term with k=0 in the first sum is P_N,n(r), and every other term has a smaller block index. Rearrangement gives the formula.
Exact mathematical passage
z^{Nn-1}/(Nn-1)!
z^qUse in source · 1 passage
Search aliases: z^q
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1526
[Formal differential-operator representation on polynomials] Let a formal series A(z)=∑_q≥0a_qz^q/q! act on polynomials by A(∂_x)f(x) = ∑_q≥0fraca_qq!∂_x^qf(x). The power r is applied to the formal symbol before the substitution z=∂_x. Then E_m^(N,j,r)(x) = C_N,j(∂_x)^r x^m. The inverse relation is x^m = Φ_N,j(∂_x^N)^r E_m^(N,j,r)(x). Both sums are finite on every polynomial.
Exact mathematical passage
A(z)=\sum_{q\geq0}a_qz^q/q!
\mathfrak Z_2^{\mathrm{abs}}Use in source · 1 passage
Search aliases: Z_2^abs mathfrak mathrm
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2841
[Exact zeta models in orders two and four] For N=2, Z_2^red(s) = π^-2sζ(2s,frac12) and Z_2^abs(s) = 2π^-sζ(s,frac12). For N=4, put A_4=sqrt2π. Then Z_4^red(s) = A_4^-4sζ(4s,frac12) and Z_4^abs(s) = 4A_4^-sζ(s,frac12). In these two cases the entire correction terms in [source reference] vanish identically.
Exact mathematical passage
\mathfrak Z_2^{\mathrm{abs}}(s) = 2\pi^{-s}\zeta\left(s,\frac12\right).
\mathfrak Z_2^{\mathrm{red}}Use in source · 1 passage
Search aliases: Z_2^red mathfrak mathrm
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2835
[Exact zeta models in orders two and four] For N=2, Z_2^red(s) = π^-2sζ(2s,frac12) and Z_2^abs(s) = 2π^-sζ(s,frac12). For N=4, put A_4=sqrt2π. Then Z_4^red(s) = A_4^-4sζ(4s,frac12) and Z_4^abs(s) = 4A_4^-sζ(s,frac12). In these two cases the entire correction terms in [source reference] vanish identically.
Exact mathematical passage
\mathfrak Z_2^{\mathrm{red}}(s) = \pi^{-2s}\zeta\left(2s,\frac12\right)
\mathfrak Z_2^{\rm red}Use in source · 2 passages
Search aliases: Z_2^rm red mathfrak rm
Use in source · atlas-pages.tex, line 441
is not continuationZERO ZETA FUNCTIONSTraditional monograph: Theorem 14.2 (p. 39), Theorem 14.3 (p. 39), Corollary 14.4 (p. 41). Reduced zeros grow like m^N. Their absolute zero-zeta series therefore has half-plane of absolute convergence Re s>1/N. The absolute zero zeta in the original variable instead requires Re s>1. vfigzeta_plane For N=2 and N=4, the lattices are exact: Z_2^rm red(s)=π^-2sζ(2s,1/2), Z_4^rm red(s)=(sqrt2π)^-4sζ(4s,1/2). Each identity first holds where its defining series con…
Exact mathematical passage
\mathfrak Z_2^{\rm red}(s)=\pi^{-2s}\zeta(2s,1/2),\qquad \mathfrak Z_4^{\rm red}(s)=(\sqrt2\pi)^{-4s}\zeta(4s,1/2).
\mathfrak Z_4^{\mathrm{abs}}Use in source · 1 passage
Search aliases: Z_4^abs mathfrak mathrm
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2853
[Exact zeta models in orders two and four] For N=2, Z_2^red(s) = π^-2sζ(2s,frac12) and Z_2^abs(s) = 2π^-sζ(s,frac12). For N=4, put A_4=sqrt2π. Then Z_4^red(s) = A_4^-4sζ(4s,frac12) and Z_4^abs(s) = 4A_4^-sζ(s,frac12). In these two cases the entire correction terms in [source reference] vanish identically.
Exact mathematical passage
\mathfrak Z_4^{\mathrm{abs}}(s) = 4A_4^{-s}\zeta\left(s,\frac12\right).
\mathfrak Z_4^{\mathrm{red}}Use in source · 1 passage
Search aliases: Z_4^red mathfrak mathrm
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2847
[Exact zeta models in orders two and four] For N=2, Z_2^red(s) = π^-2sζ(2s,frac12) and Z_2^abs(s) = 2π^-sζ(s,frac12). For N=4, put A_4=sqrt2π. Then Z_4^red(s) = A_4^-4sζ(4s,frac12) and Z_4^abs(s) = 4A_4^-sζ(s,frac12). In these two cases the entire correction terms in [source reference] vanish identically.
Exact mathematical passage
\mathfrak Z_4^{\mathrm{red}}(s) = A_4^{-4s}\zeta\left(4s,\frac12\right)
\mathfrak Z_4^{\rm red}Use in source · 2 passages
Search aliases: Z_4^rm red mathfrak rm
Use in source · atlas-pages.tex, line 441
is not continuationZERO ZETA FUNCTIONSTraditional monograph: Theorem 14.2 (p. 39), Theorem 14.3 (p. 39), Corollary 14.4 (p. 41). Reduced zeros grow like m^N. Their absolute zero-zeta series therefore has half-plane of absolute convergence Re s>1/N. The absolute zero zeta in the original variable instead requires Re s>1. vfigzeta_plane For N=2 and N=4, the lattices are exact: Z_2^rm red(s)=π^-2sζ(2s,1/2), Z_4^rm red(s)=(sqrt2π)^-4sζ(4s,1/2). Each identity first holds where its defining series con…
Exact mathematical passage
\mathfrak Z_2^{\rm red}(s)=\pi^{-2s}\zeta(2s,1/2),\qquad \mathfrak Z_4^{\rm red}(s)=(\sqrt2\pi)^{-4s}\zeta(4s,1/2).
\mathfrak Z_N^{\rm abs}Use in source · 2 passages
Search aliases: Z_N^rm abs mathfrak rm
Use in source · atlas-pages.tex, line 448
frak Z_2^rm red(s)=π^-2sζ(2s,1/2), Z_4^rm red(s)=(sqrt2π)^-4sζ(4s,1/2). Each identity first holds where its defining series converges. The right side supplies its meromorphic continuation. There is a single pole, at 1/2 or 1/4, with residue 1/(2π) or 1/(4sqrt2π) respectively. headThe factor N matters twice For every order in the paper, Z_N^rm abs(s)=N Z_N^rm red(s/N). For general N, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero s…
Exact mathematical passage
\mathfrak Z_N^{\rm abs}(s)=N\mathfrak Z_N^{\rm red}(s/N).
\mathfrak Z_N^{\rm red}Use in source · 4 passages
Search aliases: Z_N^rm red mathfrak rm
Use in source · atlas-pages.tex, line 448
frak Z_2^rm red(s)=π^-2sζ(2s,1/2), Z_4^rm red(s)=(sqrt2π)^-4sζ(4s,1/2). Each identity first holds where its defining series converges. The right side supplies its meromorphic continuation. There is a single pole, at 1/2 or 1/4, with residue 1/(2π) or 1/(4sqrt2π) respectively. headThe factor N matters twice For every order in the paper, Z_N^rm abs(s)=N Z_N^rm red(s/N). For general N, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero s…
Exact mathematical passage
\mathfrak Z_N^{\rm abs}(s)=N\mathfrak Z_N^{\rm red}(s/N).
\Zcal_RUse in source · 4 passages
Search aliases: Zcal_R Z_R Zcal mathcal
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2594
[Nearest-pole coefficient expansion] Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
\Zcal_R
\Zcal_R^{\mathrm{orb}}Use in source · 2 passages
Search aliases: Zcal_R^orb Z_R^orb Zcal mathrm mathcal mathrm
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2671
[Orbit reduced nearest circle term] Assume the hypotheses of [source reference]. Choose one representative from each orbit of the nearest zeros under multiplication by ω, and call the resulting set Z_R^orb. With m=Nn, P_N,n(r) = · N m!binomm+r-1r-1 ∑_τ∈ Z_R^orb fracτ^-m(-τ D_N'(τ))^r
Exact mathematical passage
\Zcal_R^{\mathrm{orb}}
\ZZUse in source · 1 passage
Search aliases: ZZ Z mathbb
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2593
[Nearest-pole coefficient expansion] Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
r\in\ZZ_{>0}
\ZZ_{>0}Use in source · 1 passage
Search aliases: ZZ_>0 Z_>0 ZZ mathbb
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2593
[Nearest-pole coefficient expansion] Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
r\in\ZZ_{>0}
|^{-1}Use in source · 6 passages
Search aliases: |^-1
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2292
By [source reference], the order of Δ_N is 1/N<1. By [source reference], it has infinitely many zeros, and [source reference] gives ∑_λ∈Λ_N|λ|^-1<∞. Hadamard factorization therefore gives a canonical product of genus zero. The only possible exponential factor is a nonzero constant because the order is below one. Since Δ_N(0)=1, that constant is one. Substitution of w=z^N gives the second formula.
Exact mathematical passage
\sum_{\lambda\in\Lambda_N}|\lambda|^{-1}<\infty.
|^{-q}Use in source · 2 passages
Search aliases: |^-q
Use in source · atlas-pages.tex, line 421
out added regularizing terms. takeawayA global analytic product theorem connects the infinite zero sum to a finite algebraic recurrence. The recurrence alone does not prove the product theorem. platecaptionDisplayed moments are checked against formal logarithm coefficients and exact order-two/order-four zeta values. Bar widths are numerical shares. General power sums use λ^-q, not |λ|^-q when exceptional nonpositive zeros are present.
Exact mathematical passage
|\lambda|^{-q}
|^{1/N}Use in source · 1 passage
Search aliases: |^1/N
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2820
For every N≥2, each nonzero zero λ of Δ_N gives N zeros of D_N, each of modulus |λ|^1/N. Hence Z_N^abs(s) = N Z_N^red(s/N). Substitution gives the second formula with Q_N^abs(s)=NQ_N^red(s/N), which is entire.
Exact mathematical passage
|\lambda|^{1/N}
|^{M_L}Use in source · 2 passages
Search aliases: |^M_L
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2925
For N≥3, [source reference] gives | D_N'(ρ)|^-1 ≤ C e^-c_N|ρ| for every sufficiently large simple zero. Let L be a compact subset of the s plane. Because the chosen arguments of the zeros lie in a fixed bounded interval, there are constants C_L and M_L such that sup_s∈ L|e^-sLogρ| ≤ C_L|ρ|^M_L for every sufficiently large zero. Hence the absolute value of the corresponding summand is at most C_L'|ρ|^M_Le^-q c_N|ρ|. The linear zero counting estimate makes this majorant summable. The Weierstrass t…
Exact mathematical passage
\sup_{s\in L}|e^{-s\Log\rho}| \leq C_L|\rho|^{M_L}
|_{z=0}Use in source · 1 passage
Search aliases: |_z=0
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 463
[Differential cycle and initial data] For 1≤ j<N, fracddz E_N,j(λ z) = λ E_N,j-1(λ z), while fracddz E_N,0(λ z) = λ E_N,N-1(λ z). Thus fracd^Ndz^N E_N,j(λ z) = λ^N E_N,j(λ z). For 0≤ k<N, .fracd^kdz^k E_N,j(λ z)|_z=0 = δ_kjλ^j. If λ≠0, the N functions E_N,j(λ z) form a basis of the solution space of y^(N)=λ^Ny.
Exact mathematical passage
\left.\frac{d^k}{dz^k}\Efun_{N,j}(\lambda z)\right|_{z=0} = \delta_{kj}\lambda^j.
\pmUse in source · 6 passages
Search aliases: ± pm
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2150
[The case N=2] For N=2, D_2(z)=cos z. Its zeros are ±(m+frac12)π, m≥0, and all are simple. They agree with the formula t_m=A_2(m+1/2).
Exact mathematical passage
\pm\left(m+\frac12\right)\pi, \qquad m\geq0,
\cdotUse in source · 8 passages
Search aliases: · cdot
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1749
The coefficient of w^n in Δ_N has modulus 1/(Nn)!. Stirling's formula and [source reference] give order 1/N. Since M_f( · ^N)(r)=M_f(r^N) for the maximum modulus, composition with z^N multiplies the order by N and gives order one for D_N.
Exact mathematical passage
M_{f(\,\cdot\,^N)}(r)=M_f(r^N)
\timesUse in source · 8 passages
Search aliases: × times
Use in source · cubic_certificates.tex, line 4
[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[anchor=west] at (0.45000,3.65000) m=0; node[] at (3.75000,3.65000) 1.84981279919014347629; node[] at (8.30000,3.65000) +3.60134×10^-2; node[] at (11.50000,3.65000) -5.243×10^-2; node[anchor=west] at (0.45000,2.70000) m=1; node[] at (3.75000,2.70000) 5.44123335502366335540; node[] at (8.30000,2.70000) -1.64738×10^-4; node[] at (11.50000,2.70000) +2.471×10^-4; node[anchor=west] at (0.45000,1.75000) m=2; node[] at (3.75000…
Exact mathematical passage
+3.60134\times10^{-2}
\gammaUse in source · 21 passages
Search aliases: γ gamma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1839
The zero strings are governed by two adjacent exponents. Put a_N=fracπ N, s_N=sin a_N, c_N=cos a_N, and A_N=fracπs_N, t_m=A_N(m+frac12). For N≥3, also put γ_N=c_N-cos(3a_N)>0.
Exact mathematical passage
\gamma_N=c_N-\cos(3a_N)>0.
\gamma_NUse in source · 16 passages
Search aliases: γ_N gamma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1839
The zero strings are governed by two adjacent exponents. Put a_N=fracπ N, s_N=sin a_N, c_N=cos a_N, and A_N=fracπs_N, t_m=A_N(m+frac12). For N≥3, also put γ_N=c_N-cos(3a_N)>0.
Exact mathematical passage
\gamma_N=c_N-\cos(3a_N)>0.
\deltaUse in source · 13 passages
Search aliases: δ delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 358
[Root filter] If F(z)=∑_m≥0a_mz^m, then B_N,jF(z) = ∑_m≡ jbmod Na_mz^m. The operators satisfy B_N,i B_N,j=δ_ij B_N,j and ∑_j=0^N-1 B_N,j=I.
Exact mathematical passage
\Bcal_{N,i}\Bcal_{N,j}=\delta_{ij}\Bcal_{N,j}
\Delta_1Use in source · 2 passages
Search aliases: Δ_1 Delta
Use in source · atlas-pages.tex, line 398
and integral comparison. Thus truncation has an explicit error mechanism. takeawayThe infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem. platecaptionThe exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: Δ_1(w)=e^-w has no zeros, so its empty product would be 1, not the function.
Exact mathematical passage
\Delta_1(w)=e^{-w}
\Delta_2Use in source · 15 passages
Search aliases: Δ_2 Delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1220
[title=A coefficient calculation at order two] Write Δ_2(w)^-r=∑_n≥0b_n(r)w^n. Since Δ_2(w)=1-frac w2+fracw^224+O(w^3), -logΔ_2(w)=frac w2+fracw^212+O(w^3), the first logarithmic coefficients are κ_2,1=1/2 and κ_2,2=1/6. Newton's recurrence gives b_1(r)=frac r2, 2b_2(r)=r(frac r4+frac16). Restoring the factorial normalization yields P_2,1(r)=r and P_2,2(r)=r(3r+2). At r=1, the second value is five. No zero-location assumption is needed for this finite calculation. modelfigureparameter_polynomial…
Exact mathematical passage
\Delta_2(w)^{-r}=\sum_{n\geq0}b_n(r)w^n
\Delta_4Use in source · 5 passages
Search aliases: Δ_4 Delta
Use in source · atlas-pages.tex, line 139
vpage8Four zeros become one orbit zeroTHE REDUCED VARIABLETraditional monograph: Definition 5.7 (p. 9), Example 11.12 (p. 31), Corollary 11.13 (p. 31). For N=4 the factorization is exact: D_4(z)=cos(z/sqrt2)cosh(z/sqrt2), D_4(z)=Δ_4(z^4). Set A_4=sqrt2π. All the zeros are A_4(m+1/2)i^k, with m≥0 and 0≤ k<4. Each zero is simple. vfigorbit_reduction The drawing uses u=z/A_4 and v=w/A_4^4, so its marked coordinates are rational. The first four-point orbit has fourth power 1/16. The second has fourt…
Exact mathematical passage
D_4(z)=\cos(z/\sqrt2)\cosh(z/\sqrt2),\qquad D_4(z)=\Delta_4(z^4).
\Delta_4^{-1}Use in source · 2 passages
Search aliases: Δ_4^-1 Delta
Use in source · atlas-pages.tex, line 459
D ZERO WORKED MODELTraditional monograph: Corollary 7.7 (p. 16), Example 11.12 (p. 31), Theorem 12.6 (p. 35). Use N=4 throughout this page. Its denominator, its exact zero lattice, and the reciprocal recurrences provide two routes to the same coefficient. All symbols in the plate refer to this one order. vfigworked_circuit headRoute one: invert the denominator The w^2 coefficient of Δ_4^-1 is frac124^2-frac18!=frac698!. Thus P_4,2(1)=69. Keeping r variable gives P_4,2(r)=35r^2+34r. headRoute two…
Exact mathematical passage
\Delta_4^{-1}
\delta_{ij}Use in source · 1 passage
Search aliases: δ_ij delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 358
[Root filter] If F(z)=∑_m≥0a_mz^m, then B_N,jF(z) = ∑_m≡ jbmod Na_mz^m. The operators satisfy B_N,i B_N,j=δ_ij B_N,j and ∑_j=0^N-1 B_N,j=I.
Exact mathematical passage
\Bcal_{N,i}\Bcal_{N,j}=\delta_{ij}\Bcal_{N,j}
\delta_{jk}Use in source · 1 passage
Search aliases: δ_jk delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 626
[Signed derivative cycle] For 1≤ j<N, D_N,j'= D_N,j-1, and D_N,0'=- D_N,N-1. Every signed section satisfies D_N,j^(N)=- D_N,j. The N signed sections form the basis of y^(N)=-y determined by D_N,j^(k)(0)=δ_jk for 0≤ j,k<N.
Exact mathematical passage
\Dfun_{N,j}^{(k)}(0)=\delta_{jk}
\delta_{kj}Use in source · 1 passage
Search aliases: δ_kj delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 463
[Differential cycle and initial data] For 1≤ j<N, fracddz E_N,j(λ z) = λ E_N,j-1(λ z), while fracddz E_N,0(λ z) = λ E_N,N-1(λ z). Thus fracd^Ndz^N E_N,j(λ z) = λ^N E_N,j(λ z). For 0≤ k<N, .fracd^kdz^k E_N,j(λ z)|_z=0 = δ_kjλ^j. If λ≠0, the N functions E_N,j(λ z) form a basis of the solution space of y^(N)=λ^Ny.
Exact mathematical passage
\left.\frac{d^k}{dz^k}\Efun_{N,j}(\lambda z)\right|_{z=0} = \delta_{kj}\lambda^j.
\delta_mUse in source · 5 passages
Search aliases: δ_m delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2088
Put δ_m=τ_m-t_m. The large zero theorem gives δ_m=O(e^-γ_Nt_m). Since e^2is_Nt_m=-1, 1+e^2is_N(t_m+δ_m) = -2is_Nδ_m+O(δ_m^2). The remainder estimate through the first derivative gives R_N(t_m+δ_m) = R_N(t_m)+O(e^-γ_Nt_m|δ_m|). Substitution in the zero equation gives the displacement formula.
Exact mathematical passage
\delta_m=\tau_m-t_m
\delta_m^{(1)}Use in source · 3 passages
Search aliases: δ_m^(1) delta
Use in source · atlas-pages.tex, line 333
cos(sqrt3t/2)3. Divide by the nonzero factor 2e^t/2/3. The resulting equation separates the dominant cosine from the small third exponential. vfigtwo_term Ignoring the last term gives model zeros t_m=frac2πsqrt3(m+1/2). At t_m, the full normalized expression equals tfrac12e^-3t_m/2>0, so t_m is not an exact zero. Linearizing the cosine at that point predicts δ_m^(1)=frac(-1)^msqrt3e^-3t_m/2. The alternating slope explains why the true zero is to the right of the first model zero, then to the lef…
Exact mathematical passage
\delta_m^{(1)}=\frac{(-1)^m}{\sqrt3}e^{-3t_m/2}.
\delta_m^2Use in source · 1 passage
Search aliases: δ_m^2 delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2089
Put δ_m=τ_m-t_m. The large zero theorem gives δ_m=O(e^-γ_Nt_m). Since e^2is_Nt_m=-1, 1+e^2is_N(t_m+δ_m) = -2is_Nδ_m+O(δ_m^2). The remainder estimate through the first derivative gives R_N(t_m+δ_m) = R_N(t_m)+O(e^-γ_Nt_m|δ_m|). Substitution in the zero equation gives the displacement formula.
Exact mathematical passage
1+e^{2is_N(t_m+\delta_m)} = -2is_N\delta_m+O(\delta_m^2).
\Delta_N'Use in source · 4 passages
Search aliases: Δ_N' Delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1267
Since D_N(z)=Δ_N(z^N), -frac D_N'(z) D_N(z) = -Nz^N-1fracΔ_N'(z^N)Δ_N(z^N). Differentiate the defining series for -logΔ_N and compare coefficients.
Exact mathematical passage
-\frac{\Dfun_N'(z)}{\Dfun_N(z)} = -Nz^{N-1}\frac{\Delta_N'(z^N)}{\Delta_N(z^N)}.
\delta_{n0}Use in source · 1 passage
Search aliases: δ_n0 delta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1388
[Triangular inverse system] The coefficients are determined by C_N,j(0)=1 and ∑_q=0^n binomNn+jNq C_N,j(q) = δ_n0. For n≥1, C_N,j(n) = - frac displaystyle∑_q=0^n-1 binomNn+jNqC_N,j(q) displaystylebinomNn+jNn .
Exact mathematical passage
\sum_{q=0}^{n} \binom{Nn+j}{Nq} C_{N,j}(q) = \delta_{n0}.
\varepsilonUse in source · 12 passages
Search aliases: ε varepsilon
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 858
[Entries of the fundamental matrix] For 0≤ p,q<N, let d_pq be the member of {0,…,N-1} congruent to p-q modulo N. Put ε_pq = 1, · p≥ q, -1, · p<q. Then (e^zJ_N)_pq = ε_pq D_N,d_pq(z). Thus every entry of the fundamental matrix is one of the signed sections, with the sign recording whether the cyclic path from q to p crosses the signed edge.
Exact mathematical passage
\varepsilon_{pq} = \begin{cases} 1,&p\geq q,\\ -1,&p<q. \end{cases}
\varepsilon_MUse in source · 3 passages
Search aliases: ε_M varepsilon
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1911
[Near cancellation of two exponentials] For every M>0, there are constants ε_M>0 and C_M>0 such that the following holds. If |Re w|≤ M and |1+e^w|≤ε_M, then there is a unique integer m with |w-(2m+1)π i| ≤ C_M|1+e^w|.
Exact mathematical passage
\varepsilon_M>0
\varepsilon_mUse in source · 5 passages
Search aliases: ε_m varepsilon
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2787
For N≥3, [source reference] writes all but finitely many reduced zeros as λ_m = (A_N(m+frac12))^N(1+ε_m) where λ_m>0, 1+ε_m>0, and ε_m = O(t_m^-1e^-γ t_m). On every compact set of s values, (1+ε_m)^-s-1 = O(|ε_m|) uniformly for large m. The difference λ_m^-s - A_N^-Ns(m+frac12)^-Ns therefore has a normally convergent sum on every compact subset of C. Indeed, the model power has at most polynomial growth in m on a compact set, while |ε_m| decays exponentially. The finitely many exceptional terms …
Exact mathematical passage
\lambda_m = \left(A_N\left(m+\frac12\right)\right)^N(1+\varepsilon_m)
\varepsilon_{pq}Use in source · 4 passages
Search aliases: ε_pq varepsilon
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 858
[Entries of the fundamental matrix] For 0≤ p,q<N, let d_pq be the member of {0,…,N-1} congruent to p-q modulo N. Put ε_pq = 1, · p≥ q, -1, · p<q. Then (e^zJ_N)_pq = ε_pq D_N,d_pq(z). Thus every entry of the fundamental matrix is one of the signed sections, with the sign recording whether the cyclic path from q to p crosses the signed edge.
Exact mathematical passage
\varepsilon_{pq} = \begin{cases} 1,&p\geq q,\\ -1,&p<q. \end{cases}
\zetaUse in source · 9 passages
Search aliases: ζ zeta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2580
[The cosine orbit sums] For N=2, Δ_2(w)=cossqrt w. Its zeros are λ_m=π^2(m+frac12)^2, m≥0. Hence σ_2,q = frac2^2q-1π^2qζ(2q). The identity in [source reference] recovers the classical tangent coefficients from these zero power sums.
Exact mathematical passage
\sigma_{2,q} = \frac{2^{2q}-1}{\pi^{2q}}\zeta(2q).
\etaUse in source · 21 passages
Search aliases: η eta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 244
The second half of the paper concerns the zeros of D_N. The finite exponential sum D_N(z)=frac1N∑_η^N=-1e^η z has a regular polygon of exponents. The general relation between an exponent polygon and the zero strips of an exponential sum goes back to P olya and Langer. Later developments include the work of Moreno and the treatment in Levin's monograph [source reference]. Bleher and Mallison give a particularly close general result: for convex-polygon exponential sums they locate all sufficiently…
Exact mathematical passage
\Dfun_N(z)=\frac1N\sum_{\eta^N=-1}e^{\eta z}
\eta^4Use in source · 1 passage
Search aliases: η^4 eta
Use in source · indicator.tex, line 9
934,3.66066) -- (1.28934,1.53934) -- (3.41066,1.53934) -- cycle; draw[->] (0.40000,2.60000) -- (4.30000,2.60000); draw[->] (2.35000,0.65000) -- (2.35000,4.55000); draw[fill=black] (3.41066,3.66066) circle (0.08); draw[fill=black] (1.28934,3.66066) circle (0.08); draw[fill=black] (1.28934,1.53934) circle (0.08); draw[fill=black] (3.41066,1.53934) circle (0.08); node[] at (2.35000,5.00000) η^4=-1; node[] at (2.35000,0.05000) exponent plane; draw[black!13] (5.20000,1.00000) -- (5.20000,4.00000); no…
Exact mathematical passage
\eta^4=-1
\eta^NUse in source · 6 passages
Search aliases: η^N eta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 244
The second half of the paper concerns the zeros of D_N. The finite exponential sum D_N(z)=frac1N∑_η^N=-1e^η z has a regular polygon of exponents. The general relation between an exponent polygon and the zero strips of an exponential sum goes back to P olya and Langer. Later developments include the work of Moreno and the treatment in Levin's monograph [source reference]. Bleher and Mallison give a particularly close general result: for convex-polygon exponential sums they locate all sufficiently…
Exact mathematical passage
\Dfun_N(z)=\frac1N\sum_{\eta^N=-1}e^{\eta z}
\eta_0Use in source · 2 passages
Search aliases: η_0 eta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1757
The finite exponential representation gives | D_N(z)|≤ e^|z|, so the type is at most one. Choose a root η_0 and put z=toverlineη_0 with t>0. The term e^η_0z=e^t has strictly larger real exponent than every other term. Hence D_N(toverlineη_0) = frac1Ne^t(1+o(1)). The type is therefore one.
Exact mathematical passage
\eta_0
\eta_kUse in source · 3 passages
Search aliases: η_k eta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1718
We now assume N≥2. Write η_k=e^(2k+1)π i/N, 0≤ k<N. These are the vertices of a regular polygon and D_N(z)=frac1N∑_k=0^N-1e^η_kz.
Exact mathematical passage
\eta_k=e^{(2k+1)\pi i/N}, \qquad 0\leq k<N.
\eta_\ellUse in source · 1 passage
Search aliases: η_ℓ eta ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1881
ojection estimate gives the stronger bound Y=fraclog(N-1)2s_N for every zero in the sector, without a large-radius assumption. Write z=re^iθ with 0≤θ≤ a_N and r≥0. For 0≤ℓ≤ N-2, the angle θ+(2ℓ+1)a_N lies between a_N+θ and 2π-a_N-θ. Monotonicity of cosine on [0,π], followed by reflection about π, therefore gives Re(η_ℓ z)≤ rcos(a_N+θ). The remaining vertex e^-ia_N has projection rcos(a_N-θ). Their difference is at least 2s_Nrsinθ=2s_Ny. If D_N(z)=0, divide its finite exponential sum by the term …
Exact mathematical passage
\Re(\eta_\ell z)\leq r\cos(a_N+\theta).
\ThetaUse in source · 10 passages
Search aliases: Θ Theta
Use in source · atlas-pages.tex, line 283
vpage16Scale and order enter differentlyTHETA COUPLINGTraditional monograph: Theorem 10.2 (p. 23), Proposition 10.3 (p. 23), Proposition 10.4 (p. 23). Let H_N(z)=E_N,0(z). The theta family uses the same parameter u twice, but in two different roles: H_N(uz)^-1H_N(z)^-u =∑_n≥0Θ_N,n(u)fracz^Nn(Nn)!. Argument scaling contributes u^Nm at logarithmic block m. Reciprocal order contributes u. vfigtheta Their combined logarithm has coefficient (-1)^mκ_N,mfracu^Nm+um. At the first block this gives Θ_N,1(…
Exact mathematical passage
H_N(uz)^{-1}H_N(z)^{-u} =\sum_{n\ge0}\Theta_{N,n}(u)\frac{z^{Nn}}{(Nn)!}.
\Theta_{2,1}Use in source · 1 passage
Search aliases: Θ_2,1 Theta
Use in source · theta.tex, line 24
0,1.16641) (11.43000,1.12831) (11.50250,1.08927) (11.57500,1.04928) (11.64750,1.00832) (11.72000,0.96637) (11.79250,0.92343) (11.86500,0.87947) (11.93750,0.83449) (12.01000,0.78848) (12.08250,0.74141) (12.15500,0.69327) (12.22750,0.64406) (12.30000,0.59375); draw[fill=black] (2.63333,2.30000) circle (0.08); draw[fill=black] (2.63333,3.00000) circle (0.08); node[] at (3.00000,4.65000) solid: Θ_2,1(u)=-u(u+1); node[] at (9.50000,4.65000) dashed: Θ_3,1(u)=-(u^3+u); node[] at (6.50000,-0.35000) at u…
Exact mathematical passage
\Theta_{2,1}(u)=-u(u+1)
\Theta_{2,2}Use in source · 2 passages
Search aliases: Θ_2,2 Theta
Use in source · atlas-pages.tex, line 296
efficient (-1)^mκ_N,mfracu^Nm+um. At the first block this gives Θ_N,1(u)=-(u^N+u). Every positive-index theta polynomial vanishes at u=0. For even N, H_N(-z)=H_N(z), so the defining product is also 1 at u=-1. headThe odd-order boundary is visible For N=3, Θ_3,1(-1)=2, not zero. The factor u+1 is not a uniform law across orders. In contrast, Θ_2,2(u)=u(u+1)(5u^2+u+2). The two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots. takeawayA …
Exact mathematical passage
\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).
\Theta_{3,1}Use in source · 3 passages
Search aliases: Θ_3,1 Theta
Use in source · atlas-pages.tex, line 295
garithmic block m. Reciprocal order contributes u. vfigtheta Their combined logarithm has coefficient (-1)^mκ_N,mfracu^Nm+um. At the first block this gives Θ_N,1(u)=-(u^N+u). Every positive-index theta polynomial vanishes at u=0. For even N, H_N(-z)=H_N(z), so the defining product is also 1 at u=-1. headThe odd-order boundary is visible For N=3, Θ_3,1(-1)=2, not zero. The factor u+1 is not a uniform law across orders. In contrast, Θ_2,2(u)=u(u+1)(5u^2+u+2). The two guaranteed factors follow from…
Exact mathematical passage
\Theta_{3,1}(-1)=2
\theta_{N,0}Use in source · 1 passage
Search aliases: θ_N,0 theta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1662
[Theta Newton recurrence] Put θ_N,n(u)=fracsymupΘ_N,n(u)(Nn)!. Then θ_N,0(u)=1 and nθ_N,n(u) = ∑_m=1^n (-1)^mκ_N,m (u^Nm+u) θ_N,n-m(u). In particular, symupΘ_N,1(u)=-(u^N+u).
Exact mathematical passage
\theta_{N,0}(u)=1
\Theta_{N,1}Use in source · 2 passages
Search aliases: Θ_N,1 Theta
Use in source · atlas-pages.tex, line 293
E_N,0(z). The theta family uses the same parameter u twice, but in two different roles: H_N(uz)^-1H_N(z)^-u =∑_n≥0Θ_N,n(u)fracz^Nn(Nn)!. Argument scaling contributes u^Nm at logarithmic block m. Reciprocal order contributes u. vfigtheta Their combined logarithm has coefficient (-1)^mκ_N,mfracu^Nm+um. At the first block this gives Θ_N,1(u)=-(u^N+u). Every positive-index theta polynomial vanishes at u=0. For even N, H_N(-z)=H_N(z), so the defining product is also 1 at u=-1. headThe odd-order bound…
Exact mathematical passage
\Theta_{N,1}(u)=-(u^N+u)
\Theta_{N,n}Use in source · 2 passages
Search aliases: Θ_N,n Theta
Use in source · atlas-pages.tex, line 283
vpage16Scale and order enter differentlyTHETA COUPLINGTraditional monograph: Theorem 10.2 (p. 23), Proposition 10.3 (p. 23), Proposition 10.4 (p. 23). Let H_N(z)=E_N,0(z). The theta family uses the same parameter u twice, but in two different roles: H_N(uz)^-1H_N(z)^-u =∑_n≥0Θ_N,n(u)fracz^Nn(Nn)!. Argument scaling contributes u^Nm at logarithmic block m. Reciprocal order contributes u. vfigtheta Their combined logarithm has coefficient (-1)^mκ_N,mfracu^Nm+um. At the first block this gives Θ_N,1(…
Exact mathematical passage
H_N(uz)^{-1}H_N(z)^{-u} =\sum_{n\ge0}\Theta_{N,n}(u)\frac{z^{Nn}}{(Nn)!}.
\theta_{N,n}Use in source · 2 passages
Search aliases: θ_N,n theta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1659
[Theta Newton recurrence] Put θ_N,n(u)=fracsymupΘ_N,n(u)(Nn)!. Then θ_N,0(u)=1 and nθ_N,n(u) = ∑_m=1^n (-1)^mκ_N,m (u^Nm+u) θ_N,n-m(u). In particular, symupΘ_N,1(u)=-(u^N+u).
Exact mathematical passage
\theta_{N,n}(u)=\frac{\Thetafun_{N,n}(u)}{(Nn)!}.
\theta_{N,n-m}Use in source · 1 passage
Search aliases: θ_N,n-m theta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1663
[Theta Newton recurrence] Put θ_N,n(u)=fracsymupΘ_N,n(u)(Nn)!. Then θ_N,0(u)=1 and nθ_N,n(u) = ∑_m=1^n (-1)^mκ_N,m (u^Nm+u) θ_N,n-m(u). In particular, symupΘ_N,1(u)=-(u^N+u).
Exact mathematical passage
n\theta_{N,n}(u) = \sum_{m=1}^{n} (-1)^m\kappa_{N,m} \bigl(u^{Nm}+u\bigr) \theta_{N,n-m}(u).
\kappa_1Use in source · 6 passages
Search aliases: κ_1 kappa
Use in source · atlas-pages.tex, line 181
Corollary 7.7 (p. 16). Let r vary instead of fixing the reciprocal power. Since Δ_N(w)^-r=exp (r∑_m≥1κ_N,mfracw^mm), the coefficient of w^n is a polynomial of degree n in r. Bell polynomials collect the possible ways logarithmic terms contribute total degree n. vfigbell_partitions For N=2, the first three logarithmic coefficients are κ_1=frac12, κ_2=frac16, κ_3=frac115. The third Bell polynomial is B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3. Substituting x_1=rκ_1, x_2=rκ_2, x_3=2rκ_3 and multiplying by …
Exact mathematical passage
\kappa_1=\frac12,\qquad\kappa_2=\frac16,\qquad\kappa_3=\frac1{15}.
\kappa_1^3Use in source · 1 passage
Search aliases: κ_1^3 kappa
Use in source · bell_partitions.tex, line 9
,0.75000); draw[fill=black] (0.78000,2.82000) circle (0.085); draw[fill=black!12,draw=black!40,line width=.35pt] (1.27000,2.45000) rectangle ++(0.66000,0.75000); draw[fill=black] (1.60000,2.82000) circle (0.085); draw[fill=black!12,draw=black!40,line width=.35pt] (2.09000,2.45000) rectangle ++(0.66000,0.75000); draw[fill=black] (2.42000,2.82000) circle (0.085); node[] at (2.15000,1.40000) r^3κ_1^3; node[] at (2.15000,0.40000) 15r^3; node[font=sffamilyfootnotesize] at (6.50000,4.00000) one pair, …
Exact mathematical passage
r^3\kappa_1^3
\kappa_2Use in source · 6 passages
Search aliases: κ_2 kappa
Use in source · atlas-pages.tex, line 181
Corollary 7.7 (p. 16). Let r vary instead of fixing the reciprocal power. Since Δ_N(w)^-r=exp (r∑_m≥1κ_N,mfracw^mm), the coefficient of w^n is a polynomial of degree n in r. Bell polynomials collect the possible ways logarithmic terms contribute total degree n. vfigbell_partitions For N=2, the first three logarithmic coefficients are κ_1=frac12, κ_2=frac16, κ_3=frac115. The third Bell polynomial is B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3. Substituting x_1=rκ_1, x_2=rκ_2, x_3=2rκ_3 and multiplying by …
Exact mathematical passage
\kappa_1=\frac12,\qquad\kappa_2=\frac16,\qquad\kappa_3=\frac1{15}.
\kappa_{2,1}Use in source · 1 passage
Search aliases: κ_2,1 kappa
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1225
[title=A coefficient calculation at order two] Write Δ_2(w)^-r=∑_n≥0b_n(r)w^n. Since Δ_2(w)=1-frac w2+fracw^224+O(w^3), -logΔ_2(w)=frac w2+fracw^212+O(w^3), the first logarithmic coefficients are κ_2,1=1/2 and κ_2,2=1/6. Newton's recurrence gives b_1(r)=frac r2, 2b_2(r)=r(frac r4+frac16). Restoring the factorial normalization yields P_2,1(r)=r and P_2,2(r)=r(3r+2). At r=1, the second value is five. No zero-location assumption is needed for this finite calculation. modelfigureparameter_polynomial…
Exact mathematical passage
\kappa_{2,1}=1/2
\kappa_{2,2}Use in source · 1 passage
Search aliases: κ_2,2 kappa
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1225
[title=A coefficient calculation at order two] Write Δ_2(w)^-r=∑_n≥0b_n(r)w^n. Since Δ_2(w)=1-frac w2+fracw^224+O(w^3), -logΔ_2(w)=frac w2+fracw^212+O(w^3), the first logarithmic coefficients are κ_2,1=1/2 and κ_2,2=1/6. Newton's recurrence gives b_1(r)=frac r2, 2b_2(r)=r(frac r4+frac16). Restoring the factorial normalization yields P_2,1(r)=r and P_2,2(r)=r(3r+2). At r=1, the second value is five. No zero-location assumption is needed for this finite calculation. modelfigureparameter_polynomial…
Exact mathematical passage
\kappa_{2,2}=1/6
\kappa_3Use in source · 6 passages
Search aliases: κ_3 kappa
Use in source · atlas-pages.tex, line 181
Corollary 7.7 (p. 16). Let r vary instead of fixing the reciprocal power. Since Δ_N(w)^-r=exp (r∑_m≥1κ_N,mfracw^mm), the coefficient of w^n is a polynomial of degree n in r. Bell polynomials collect the possible ways logarithmic terms contribute total degree n. vfigbell_partitions For N=2, the first three logarithmic coefficients are κ_1=frac12, κ_2=frac16, κ_3=frac115. The third Bell polynomial is B_3(x_1,x_2,x_3)=x_1^3+3x_1x_2+x_3. Substituting x_1=rκ_1, x_2=rκ_2, x_3=2rκ_3 and multiplying by …
Exact mathematical passage
\kappa_1=\frac12,\qquad\kappa_2=\frac16,\qquad\kappa_3=\frac1{15}.
\kappa_{4,1}Use in source · 2 passages
Search aliases: κ_4,1 kappa
Use in source · atlas-pages.tex, line 222
8), Theorem 8.5 (p. 19). The transfer tangent is T_N=-D_N'/D_N. It compresses the linear differential equation into a nonlinear scalar identity. This division is allowed only on the zero-free set of the denominator. vfigriccati The coefficients retain the reduced logarithmic information: T_N(z)=N∑_m≥1κ_N,mz^Nm-1, T_N(n)=N(Nn+N-1)!κ_N,n+1. For order four, κ_4,1=1/24 and κ_4,2=17/10080. These give T_4(0)=1 and T_4(1)=34, explaining the last line of the plate. headA nonlinear equation with a precis…
Exact mathematical passage
\kappa_{4,1}=1/24
\kappa_{4,2}Use in source · 2 passages
Search aliases: κ_4,2 kappa
Use in source · atlas-pages.tex, line 222
The transfer tangent is T_N=-D_N'/D_N. It compresses the linear differential equation into a nonlinear scalar identity. This division is allowed only on the zero-free set of the denominator. vfigriccati The coefficients retain the reduced logarithmic information: T_N(z)=N∑_m≥1κ_N,mz^Nm-1, T_N(n)=N(Nn+N-1)!κ_N,n+1. For order four, κ_4,1=1/24 and κ_4,2=17/10080. These give T_4(0)=1 and T_4(1)=34, explaining the last line of the plate. headA nonlinear equation with a precise origin If D=e^g locally…
Exact mathematical passage
\kappa_{4,2}=17/10080
\kappa_{N,1}Use in source · 4 passages
Search aliases: κ_N,1 kappa
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 963
[Bell polynomial formula] For every n≥0, P_N,n(r) = frac(Nn)!n! B_n( rκ_N,1, r 1! κ_N,2, …, r (n-1)! κ_N,n ). For n≥1, this is a polynomial in r of exact degree n. Its leading coefficient is frac(Nn)!n!(N!)^n.
Exact mathematical passage
P_{N,n}(r) = \frac{(Nn)!}{n!} B_n\bigl( r\kappa_{N,1}, r\,1!\,\kappa_{N,2}, \ldots, r\,(n-1)!\,\kappa_{N,n} \bigr).
\kappa_{N,2}Use in source · 2 passages
Search aliases: κ_N,2 kappa
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 963
[Bell polynomial formula] For every n≥0, P_N,n(r) = frac(Nn)!n! B_n( rκ_N,1, r 1! κ_N,2, …, r (n-1)! κ_N,n ). For n≥1, this is a polynomial in r of exact degree n. Its leading coefficient is frac(Nn)!n!(N!)^n.
Exact mathematical passage
P_{N,n}(r) = \frac{(Nn)!}{n!} B_n\bigl( r\kappa_{N,1}, r\,1!\,\kappa_{N,2}, \ldots, r\,(n-1)!\,\kappa_{N,n} \bigr).
\kappa_{N,3}Use in source · 1 passage
Search aliases: κ_N,3 kappa
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1182
The first logarithmic coefficients are κ_N,1=frac1N!, κ_N,2=frac1(N!)^2-frac2(2N)!, and κ_N,3 = frac1(N!)^3 - frac3N!(2N)! + frac3(3N)!. Insert these values into [source reference] and simplify.
Exact mathematical passage
\kappa_{N,3} = \frac1{(N!)^3} - \frac3{N!(2N)!} + \frac3{(3N)!}.
\kappa_{N,k}Use in source · 6 passages
Search aliases: κ_N,k kappa
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1091
[Logarithmic Newton recurrence] Set b_N,n(r)=fracP_N,n(r)(Nn)!. Then b_N,0(r)=1 and for every n≥1, nb_N,n(r) = r∑_k=1^nκ_N,kb_N,n-k(r). Equivalently, P_N,n(r) = frac r n ∑_k=1^n frac(Nn)!(N(n-k))! κ_N,kP_N,n-k(r). If p_k=rκ_N,k, then, for n≥1, b_N,n(r) = frac1n! det p_1 · -1 · 0 · ⋯ · 0
Exact mathematical passage
nb_{N,n}(r) = r\sum_{k=1}^{n}\kappa_{N,k}b_{N,n-k}(r).
\kappa_{N,n}Use in source · 1 passage
Search aliases: κ_N,n kappa
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 963
[Bell polynomial formula] For every n≥0, P_N,n(r) = frac(Nn)!n! B_n( rκ_N,1, r 1! κ_N,2, …, r (n-1)! κ_N,n ). For n≥1, this is a polynomial in r of exact degree n. Its leading coefficient is frac(Nn)!n!(N!)^n.
Exact mathematical passage
P_{N,n}(r) = \frac{(Nn)!}{n!} B_n\bigl( r\kappa_{N,1}, r\,1!\,\kappa_{N,2}, \ldots, r\,(n-1)!\,\kappa_{N,n} \bigr).
\kappa_{N,n+1}Use in source · 3 passages
Search aliases: κ_N,n+1 kappa
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1258
[Logarithmic coefficient formula] The transfer tangent satisfies T_N(z) = N∑_m≥1κ_N,mz^Nm-1. Equivalently, T_N(n) = N(Nn+N-1)!κ_N,n+1.
Exact mathematical passage
T_N(n) = N(Nn+N-1)!\kappa_{N,n+1}.
\kappa_{N,q}Use in source · 2 passages
Search aliases: κ_N,q kappa
Use in source · atlas-pages.tex, line 402
vpage22Finite coefficients give infinite momentsZERO POWER SUMSTraditional monograph: Theorem 12.6 (p. 35), Corollary 12.7 (p. 36), Corollary 12.9 (p. 37), Corollary 12.10 (p. 37). Taking the logarithm of the canonical product identifies κ_N,q=σ_N,q:=∑_λ∈Λ_Nλ^-q =Tr(K_N^q), q≥1. The same exact numbers can be computed from finitely many denominator coefficients by Newton's recurrence. No numerical zero list is needed for that calculation. vfigmoments For order two, σ_2,1=frac12, σ_2,2=frac16, σ_2…
Exact mathematical passage
\kappa_{N,q}=\sigma_{N,q}:=\sum_{\lambda\in\Lambda_N}\lambda^{-q} =\Tr(K_N^q),\qquad q\ge1.
\lambda^{-q}Use in source · 4 passages
Search aliases: λ^-q lambda
Use in source · atlas-pages.tex, line 402
vpage22Finite coefficients give infinite momentsZERO POWER SUMSTraditional monograph: Theorem 12.6 (p. 35), Corollary 12.7 (p. 36), Corollary 12.9 (p. 37), Corollary 12.10 (p. 37). Taking the logarithm of the canonical product identifies κ_N,q=σ_N,q:=∑_λ∈Λ_Nλ^-q =Tr(K_N^q), q≥1. The same exact numbers can be computed from finitely many denominator coefficients by Newton's recurrence. No numerical zero list is needed for that calculation. vfigmoments For order two, σ_2,1=frac12, σ_2,2=frac16, σ_2…
Exact mathematical passage
\kappa_{N,q}=\sigma_{N,q}:=\sum_{\lambda\in\Lambda_N}\lambda^{-q} =\Tr(K_N^q),\qquad q\ge1.
\lambda^jUse in source · 1 passage
Search aliases: λ^j lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 463
[Differential cycle and initial data] For 1≤ j<N, fracddz E_N,j(λ z) = λ E_N,j-1(λ z), while fracddz E_N,0(λ z) = λ E_N,N-1(λ z). Thus fracd^Ndz^N E_N,j(λ z) = λ^N E_N,j(λ z). For 0≤ k<N, .fracd^kdz^k E_N,j(λ z)|_z=0 = δ_kjλ^j. If λ≠0, the N functions E_N,j(λ z) form a basis of the solution space of y^(N)=λ^Ny.
Exact mathematical passage
\left.\frac{d^k}{dz^k}\Efun_{N,j}(\lambda z)\right|_{z=0} = \delta_{kj}\lambda^j.
\lambda^mUse in source · 1 passage
Search aliases: λ^m lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2451
Take the logarithm of the product in [source reference]. Absolute convergence in a disk below the first zero permits termwise expansion -log(1-frac wλ) = ∑_m≥1fracw^mmλ^m. This gives the logarithmic formula and identifies κ_N,m with σ_N,m.
Exact mathematical passage
-\log\left(1-\frac w\lambda\right) = \sum_{m\geq1}\frac{w^m}{m\lambda^m}.
\lambda^NUse in source · 4 passages
Search aliases: λ^N lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 179
Fix a positive integer N. Root of unity projection of the exponential produces residue-class sections that, for λ≠0, form a basis for y^(N)=λ^N y. The signed sections for y^(N)=-y yield the denominator D_N and its reduced form Δ_N. We derive reciprocal coefficient polynomials, Bell and Newton recurrences, an additive parameter law, and residue section Appell and theta families. For N≥2, Δ_N has order 1/N. We specialize classical two term zero theory for exponential sums to the regular exponent p…
Exact mathematical passage
y^{(N)}=\lambda^N y
\lambda^{N(N-1)/2}Use in source · 1 passage
Search aliases: λ^N(N-1)/2 lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 482
[Wronskian of the transfer basis] Assume N≥2. For λ≠0, the Wronskian of the ordered basis E_N,0(λ z),…, E_N,N-1(λ z) is the constant λ^N(N-1)/2.
Exact mathematical passage
\lambda^{N(N-1)/2}.
\lambda^{N-1}Use in source · 1 passage
Search aliases: λ^N-1 lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 489
At the origin, the derivative matrix is diagonal with entries 1,λ,…,λ^N-1. For an explicit determinant proof, write each basis function as a finite sum of exponentials with exponents a_ℓ=λω^ℓ. Its derivative matrix factors as W(z)=V diag(e^a_ℓ z) C, V_kℓ=a_ℓ^k, where C is the constant matrix of Fourier weights. Thus det W(z)=e^z∑_ℓ a_ℓdet W(0)=det W(0), because N≥2 implies ∑_ℓω^ℓ=0. The product of the diagonal initial values is the stated power of λ.
Exact mathematical passage
1,\lambda,\ldots,\lambda^{N-1}.
\lambda_kUse in source · 1 passage
Search aliases: λ_k lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2266
ct uses every reduced zero with its multiplicity. Its logarithm turns multiplication into addition, so reciprocal-zero power sums become coefficients. This step depends on the global product theorem. A truncated zero list provides a numerical partial sum, while the identity uses the entire multiset. modelfigurezero_momentsOrder-two moment sums over the first forty reduced zeros λ_k=π^2(k+1/2)^2. The curves are finite numerical partial sums. The dashed limits 1/2 and 1/6 follow from the exact log…
Exact mathematical passage
\lambda_k=\pi^2(k+1/2)^2
\lambda_mUse in source · 14 passages
Search aliases: λ_m lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2195
[Reduced zeros] Let Λ_N be the nonzero zero multiset of Δ_N. All but finitely many members of Λ_N are positive and simple. They can be indexed as λ_m=τ_m^N for all sufficiently large m. If N≥3, the notation may be chosen as in [source reference], and λ_m = t_m^N +O(t_m^N-1e^-γ t_m). For N=2, take τ_m=t_m=π(m+1/2) for m≥0; then λ_m=t_m^2 exactly.
Exact mathematical passage
\lambda_m=\tau_m^N
\lambda_m^{-1}Use in source · 2 passages
Search aliases: λ_m^-1 lambda
Use in source · atlas-pages.tex, line 391
on ℓ^2(Λ_N) gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries. vfigproduct headFinite compression is an approximation For N=2, λ_m=π^2(m+1/2)^2. The finite determinant P_L is exactly the displayed finite product, but it is not exactly Δ_2. For 0≤ w≤1 and L≥1, an elementary tail estimate gives 0≤1-fracΔ_2(w)P_L(w) ≤ w∑_m≥ Lλ_m^-1 ≤frac19(L-1/2). The last bound uses π>3 and integral comparison. Thus truncation has an explicit error mechanism. t…
Exact mathematical passage
0\le1-\frac{\Delta_2(w)}{P_L(w)} \le w\sum_{m\ge L}\lambda_m^{-1} \le\frac1{9(L-1/2)}.
\lambda_m^{-s}Use in source · 1 passage
Search aliases: λ_m^-s lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2805
For N≥3, [source reference] writes all but finitely many reduced zeros as λ_m = (A_N(m+frac12))^N(1+ε_m) where λ_m>0, 1+ε_m>0, and ε_m = O(t_m^-1e^-γ t_m). On every compact set of s values, (1+ε_m)^-s-1 = O(|ε_m|) uniformly for large m. The difference λ_m^-s - A_N^-Ns(m+frac12)^-Ns therefore has a normally convergent sum on every compact subset of C. Indeed, the model power has at most polynomial growth in m on a compact set, while |ε_m| decays exponentially. The finitely many exceptional terms …
Exact mathematical passage
\lambda_m^{-s} - A_N^{-Ns}\left(m+\frac12\right)^{-Ns}
\Lambda_{N,j}Use in source · 3 passages
Search aliases: Λ_N,j Lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2372
[Products for the normalized sections] For every 0≤ j<N, the entire function Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)! has order 1/N. If N≥2 and Λ_N,j is its nonzero zero multiset, then Φ_N,j(w) = ∏_μ∈Λ_N,j (1-frac wμ). For m≥1, put σ_N,j,m = ∑_μ∈Λ_N,jμ^-m. Then C_N,j,n(r) = frac(Nn)!n! B_n( rσ_N,j,1, r1!σ_N,j,2, …, r(n-1)!σ_N,j,n ).
Exact mathematical passage
\Lambda_{N,j}
\muUse in source · 2 passages
Search aliases: μ mu
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2373
[Products for the normalized sections] For every 0≤ j<N, the entire function Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)! has order 1/N. If N≥2 and Λ_N,j is its nonzero zero multiset, then Φ_N,j(w) = ∏_μ∈Λ_N,j (1-frac wμ). For m≥1, put σ_N,j,m = ∑_μ∈Λ_N,jμ^-m. Then C_N,j,n(r) = frac(Nn)!n! B_n( rσ_N,j,1, r1!σ_N,j,2, …, r(n-1)!σ_N,j,n ).
Exact mathematical passage
\Phi_{N,j}(w) = \prod_{\mu\in\Lambda_{N,j}} \left(1-\frac w\mu\right).
\mu^{-m}Use in source · 1 passage
Search aliases: μ^-m mu
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2380
[Products for the normalized sections] For every 0≤ j<N, the entire function Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)! has order 1/N. If N≥2 and Λ_N,j is its nonzero zero multiset, then Φ_N,j(w) = ∏_μ∈Λ_N,j (1-frac wμ). For m≥1, put σ_N,j,m = ∑_μ∈Λ_N,jμ^-m. Then C_N,j,n(r) = frac(Nn)!n! B_n( rσ_N,j,1, r1!σ_N,j,2, …, r(n-1)!σ_N,j,n ).
Exact mathematical passage
\sigma_{N,j,m} = \sum_{\mu\in\Lambda_{N,j}}\mu^{-m}.
\xi^2Use in source · 10 passages
Search aliases: ξ^2 xi
Use in source · atlas-pages.tex, line 103
vpage6Addition is quotient multiplicationFINITE ALGEBRATraditional monograph: Proposition 5.5 (p. 8), Theorem 6.2 (p. 10). Work in C[ξ]/(ξ^3+1). Every element has coordinates in 1,ξ,ξ^2. The entire exponential has the finite-coordinate form e^zξ=D_3,0(z)+D_3,1(z)ξ+D_3,2(z)ξ^2. The multiplication table makes the carry sign unavoidable. vfigsigned_addition Multiply the coordinate forms of e^zξ and e^wξ. Their product is e^(z+w)ξ. The coefficient of 1 receives one contribution without a wrap and tw…
Exact mathematical passage
1,\xi,\xi^2
\xi^3Use in source · 5 passages
Search aliases: ξ^3 xi
Use in source · atlas-pages.tex, line 103
vpage6Addition is quotient multiplicationFINITE ALGEBRATraditional monograph: Proposition 5.5 (p. 8), Theorem 6.2 (p. 10). Work in C[ξ]/(ξ^3+1). Every element has coordinates in 1,ξ,ξ^2. The entire exponential has the finite-coordinate form e^zξ=D_3,0(z)+D_3,1(z)ξ+D_3,2(z)ξ^2. The multiplication table makes the carry sign unavoidable. vfigsigned_addition Multiply the coordinate forms of e^zξ and e^wξ. Their product is e^(z+w)ξ. The coefficient of 1 receives one contribution without a wrap and tw…
Exact mathematical passage
\CC[\xi]/(\xi^3+1)
\xi^jUse in source · 4 passages
Search aliases: ξ^j xi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 774
[Finite algebra exponential identities] In A_N^+ one has e^λ zξ = ∑_j=0^N-1 E_N,j(λ z)ξ^j. In A_N^- one has e^zξ = ∑_j=0^N-1 D_N,j(z)ξ^j. Differentiation and multiplication of these identities recover the cyclic derivative systems and the two addition laws.
Exact mathematical passage
e^{\lambda z\xi} = \sum_{j=0}^{N-1}\Efun_{N,j}(\lambda z)\xi^j.
\xi^{N-1}Use in source · 1 passage
Search aliases: ξ^N-1 xi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 795
Differentiation multiplies by λξ in the first algebra and by ξ in the second. Reduction of the product by the defining relation gives the derivative cycles. The identity e^(z+w)ξ=e^zξe^wξ gives the addition laws after comparison of the coefficients of 1,ξ,…,ξ^N-1.
Exact mathematical passage
1,\xi,\ldots,\xi^{N-1}
\xi^{Nn+j}Use in source · 2 passages
Search aliases: ξ^Nn+j xi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 789
Expand the exponential and write every exponent uniquely as Nn+j. In A_N^+, the relation ξ^Nn+j=ξ^j gives the transfer sections. In A_N^-, the relation ξ^Nn+j=(-1)^nξ^j gives the signed sections.
Exact mathematical passage
\xi^{Nn+j}=\xi^j
\pi^{-2s}Use in source · 3 passages
Search aliases: π^-2s pi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2835
[Exact zeta models in orders two and four] For N=2, Z_2^red(s) = π^-2sζ(2s,frac12) and Z_2^abs(s) = 2π^-sζ(s,frac12). For N=4, put A_4=sqrt2π. Then Z_4^red(s) = A_4^-4sζ(4s,frac12) and Z_4^abs(s) = 4A_4^-sζ(s,frac12). In these two cases the entire correction terms in [source reference] vanish identically.
Exact mathematical passage
\mathfrak Z_2^{\mathrm{red}}(s) = \pi^{-2s}\zeta\left(2s,\frac12\right)
\pi^{-s}Use in source · 1 passage
Search aliases: π^-s pi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2841
[Exact zeta models in orders two and four] For N=2, Z_2^red(s) = π^-2sζ(2s,frac12) and Z_2^abs(s) = 2π^-sζ(s,frac12). For N=4, put A_4=sqrt2π. Then Z_4^red(s) = A_4^-4sζ(4s,frac12) and Z_4^abs(s) = 4A_4^-sζ(s,frac12). In these two cases the entire correction terms in [source reference] vanish identically.
Exact mathematical passage
\mathfrak Z_2^{\mathrm{abs}}(s) = 2\pi^{-s}\zeta\left(s,\frac12\right).
\pi^2Use in source · 6 passages
Search aliases: π^2 pi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2266
ct uses every reduced zero with its multiplicity. Its logarithm turns multiplication into addition, so reciprocal-zero power sums become coefficients. This step depends on the global product theorem. A truncated zero list provides a numerical partial sum, while the identity uses the entire multiset. modelfigurezero_momentsOrder-two moment sums over the first forty reduced zeros λ_k=π^2(k+1/2)^2. The curves are finite numerical partial sums. The dashed limits 1/2 and 1/6 follow from the exact log…
Exact mathematical passage
\lambda_k=\pi^2(k+1/2)^2
\pi^{2q}Use in source · 1 passage
Search aliases: π^2q pi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2580
[The cosine orbit sums] For N=2, Δ_2(w)=cossqrt w. Its zeros are λ_m=π^2(m+frac12)^2, m≥0. Hence σ_2,q = frac2^2q-1π^2qζ(2q). The identity in [source reference] recovers the classical tangent coefficients from these zero power sums.
Exact mathematical passage
\sigma_{2,q} = \frac{2^{2q}-1}{\pi^{2q}}\zeta(2q).
\pi^4Use in source · 7 passages
Search aliases: π^4 pi
Use in source · atlas-pages.tex, line 148
st four-point orbit has fourth power 1/16. The second has fourth power 81/16. headMultiplicity is not multiplied in the reduced function At a nonzero point, z↦ z^4 has nonzero derivative. It preserves the local multiplicity of a zero. Four distinct simple zeros of D_4 therefore correspond to one simple zero of Δ_4, not one zero of multiplicity four. The reduced zeros are λ_m=A_4^4(m+1/2)^4=4π^4(m+1/2)^4. This is why an orbit factor in the z plane is 1-z^4/λ_m. It is the product of the four corre…
Exact mathematical passage
\lambda_m=A_4^4(m+1/2)^4=4\pi^4(m+1/2)^4.
\rho^{-j}Use in source · 4 passages
Search aliases: ρ^-j rho
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 596
[Relation with the transfer exponential] For every j, D_N,j(z) = ρ^-j E_N,j(ρ z).
Exact mathematical passage
\Dfun_{N,j}(z) = \rho^{-j}\Efun_{N,j}(\rho z).
\rho^{-m}Use in source · 4 passages
Search aliases: ρ^-m rho
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2607
rcle_expansion Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
P_{N,n}(r) = m! \sum_{\rho\in\Zcal_R} \sum_{\ell=1}^{r} a_{\rho,\ell} \binom{m+\ell-1}{\ell-1} \rho^{-m} + O_{R_*}(m!R_*^{-m}).
\rho^{i+j-k}Use in source · 1 passage
Search aliases: ρ^i+j-k rho
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 665
Use [source reference]. The factor introduced by a pair (i,j) is ρ^i+j-k. Since 0≤ i+j≤2N-2 and i+j≡ kbmod N, this factor is one when i+j=k and -1 when i+j=k+N.
Exact mathematical passage
\rho^{i+j-k}.
\rho^NUse in source · 2 passages
Search aliases: ρ^N rho
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 592
Fix any ρ∈ C with ρ^N=-1.
Exact mathematical passage
\rho^N=-1
\rho^{Nn+j}Use in source · 1 passage
Search aliases: ρ^Nn+j rho
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 605
The coefficient of z^Nn+j on the right is ρ^-jρ^Nn+jfrac1(Nn+j)! = (-1)^nfrac1(Nn+j)!.
Exact mathematical passage
\rho^{-j}\rho^{Nn+j}\frac1{(Nn+j)!} = (-1)^n\frac1{(Nn+j)!}.
\sigma_1Use in source · 3 passages
Search aliases: σ_1 sigma
Use in source · atlas-pages.tex, line 465
cient. All symbols in the plate refer to this one order. vfigworked_circuit headRoute one: invert the denominator The w^2 coefficient of Δ_4^-1 is frac124^2-frac18!=frac698!. Thus P_4,2(1)=69. Keeping r variable gives P_4,2(r)=35r^2+34r. headRoute two: use the zero moments The exact reduced zeros are 4π^4(m+1/2)^4. Their first two power sums are σ_1=1/24 and σ_2=17/10080. The Bell/Newton identity gives the same reciprocal coefficient (σ_1^2+σ_2)/2. The tangent independently reads off the second …
Exact mathematical passage
\sigma_1=1/24
\sigma_1^2Use in source · 3 passages
Search aliases: σ_1^2 sigma
Use in source · atlas-pages.tex, line 465
he denominator The w^2 coefficient of Δ_4^-1 is frac124^2-frac18!=frac698!. Thus P_4,2(1)=69. Keeping r variable gives P_4,2(r)=35r^2+34r. headRoute two: use the zero moments The exact reduced zeros are 4π^4(m+1/2)^4. Their first two power sums are σ_1=1/24 and σ_2=17/10080. The Bell/Newton identity gives the same reciprocal coefficient (σ_1^2+σ_2)/2. The tangent independently reads off the second moment as 4·7!σ_2=34. This links coefficients, geometry, and derivatives without a fitted parameter…
Exact mathematical passage
(\sigma_1^2+\sigma_2)/2
\sigma_2Use in source · 8 passages
Search aliases: σ_2 sigma
Use in source · atlas-pages.tex, line 465
n the plate refer to this one order. vfigworked_circuit headRoute one: invert the denominator The w^2 coefficient of Δ_4^-1 is frac124^2-frac18!=frac698!. Thus P_4,2(1)=69. Keeping r variable gives P_4,2(r)=35r^2+34r. headRoute two: use the zero moments The exact reduced zeros are 4π^4(m+1/2)^4. Their first two power sums are σ_1=1/24 and σ_2=17/10080. The Bell/Newton identity gives the same reciprocal coefficient (σ_1^2+σ_2)/2. The tangent independently reads off the second moment as 4·7!σ_2=34…
Exact mathematical passage
\sigma_2=17/10080
\sigma_{2,1}Use in source · 3 passages
Search aliases: σ_2,1 sigma
Use in source · atlas-pages.tex, line 409
Corollary 12.9 (p. 37), Corollary 12.10 (p. 37). Taking the logarithm of the canonical product identifies κ_N,q=σ_N,q:=∑_λ∈Λ_Nλ^-q =Tr(K_N^q), q≥1. The same exact numbers can be computed from finitely many denominator coefficients by Newton's recurrence. No numerical zero list is needed for that calculation. vfigmoments For order two, σ_2,1=frac12, σ_2,2=frac16, σ_2,3=frac115, σ_2,4=frac17630. The plate shows how successive inverse powers emphasize the nearest zero. Its shaded pieces are contrib…
Exact mathematical passage
\sigma_{2,1}=\frac12,\quad\sigma_{2,2}=\frac16,\quad \sigma_{2,3}=\frac1{15},\quad\sigma_{2,4}=\frac{17}{630}.
\sigma_{2,2}Use in source · 3 passages
Search aliases: σ_2,2 sigma
Use in source · atlas-pages.tex, line 409
Corollary 12.9 (p. 37), Corollary 12.10 (p. 37). Taking the logarithm of the canonical product identifies κ_N,q=σ_N,q:=∑_λ∈Λ_Nλ^-q =Tr(K_N^q), q≥1. The same exact numbers can be computed from finitely many denominator coefficients by Newton's recurrence. No numerical zero list is needed for that calculation. vfigmoments For order two, σ_2,1=frac12, σ_2,2=frac16, σ_2,3=frac115, σ_2,4=frac17630. The plate shows how successive inverse powers emphasize the nearest zero. Its shaded pieces are contrib…
Exact mathematical passage
\sigma_{2,1}=\frac12,\quad\sigma_{2,2}=\frac16,\quad \sigma_{2,3}=\frac1{15},\quad\sigma_{2,4}=\frac{17}{630}.
\sigma_{2,3}Use in source · 3 passages
Search aliases: σ_2,3 sigma
Use in source · atlas-pages.tex, line 409
Corollary 12.9 (p. 37), Corollary 12.10 (p. 37). Taking the logarithm of the canonical product identifies κ_N,q=σ_N,q:=∑_λ∈Λ_Nλ^-q =Tr(K_N^q), q≥1. The same exact numbers can be computed from finitely many denominator coefficients by Newton's recurrence. No numerical zero list is needed for that calculation. vfigmoments For order two, σ_2,1=frac12, σ_2,2=frac16, σ_2,3=frac115, σ_2,4=frac17630. The plate shows how successive inverse powers emphasize the nearest zero. Its shaded pieces are contrib…
Exact mathematical passage
\sigma_{2,1}=\frac12,\quad\sigma_{2,2}=\frac16,\quad \sigma_{2,3}=\frac1{15},\quad\sigma_{2,4}=\frac{17}{630}.
\sigma_{2,4}Use in source · 2 passages
Search aliases: σ_2,4 sigma
Use in source · atlas-pages.tex, line 409
Corollary 12.9 (p. 37), Corollary 12.10 (p. 37). Taking the logarithm of the canonical product identifies κ_N,q=σ_N,q:=∑_λ∈Λ_Nλ^-q =Tr(K_N^q), q≥1. The same exact numbers can be computed from finitely many denominator coefficients by Newton's recurrence. No numerical zero list is needed for that calculation. vfigmoments For order two, σ_2,1=frac12, σ_2,2=frac16, σ_2,3=frac115, σ_2,4=frac17630. The plate shows how successive inverse powers emphasize the nearest zero. Its shaded pieces are contrib…
Exact mathematical passage
\sigma_{2,1}=\frac12,\quad\sigma_{2,2}=\frac16,\quad \sigma_{2,3}=\frac1{15},\quad\sigma_{2,4}=\frac{17}{630}.
\sigma_{2,q}Use in source · 1 passage
Search aliases: σ_2,q sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2580
[The cosine orbit sums] For N=2, Δ_2(w)=cossqrt w. Its zeros are λ_m=π^2(m+frac12)^2, m≥0. Hence σ_2,q = frac2^2q-1π^2qζ(2q). The identity in [source reference] recovers the classical tangent coefficients from these zero power sums.
Exact mathematical passage
\sigma_{2,q} = \frac{2^{2q}-1}{\pi^{2q}}\zeta(2q).
\sigma_{N,1}Use in source · 1 passage
Search aliases: σ_N,1 sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2552
[Reciprocal coefficients from zero data] For every n≥0, P_N,n(r) = frac(Nn)!n! B_n( rσ_N,1, r1!σ_N,2, …, r(n-1)!σ_N,n ).
Exact mathematical passage
P_{N,n}(r) = \frac{(Nn)!}{n!} B_n\bigl( r\sigma_{N,1}, r1!\sigma_{N,2}, \ldots, r(n-1)!\sigma_{N,n} \bigr).
\sigma_{N,2}Use in source · 1 passage
Search aliases: σ_N,2 sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2552
[Reciprocal coefficients from zero data] For every n≥0, P_N,n(r) = frac(Nn)!n! B_n( rσ_N,1, r1!σ_N,2, …, r(n-1)!σ_N,n ).
Exact mathematical passage
P_{N,n}(r) = \frac{(Nn)!}{n!} B_n\bigl( r\sigma_{N,1}, r1!\sigma_{N,2}, \ldots, r(n-1)!\sigma_{N,n} \bigr).
\sigma_{N,j,1}Use in source · 1 passage
Search aliases: σ_N,j,1 sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2386
ized sections] For every 0≤ j<N, the entire function Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)! has order 1/N. If N≥2 and Λ_N,j is its nonzero zero multiset, then Φ_N,j(w) = ∏_μ∈Λ_N,j (1-frac wμ). For m≥1, put σ_N,j,m = ∑_μ∈Λ_N,jμ^-m. Then C_N,j,n(r) = frac(Nn)!n! B_n( rσ_N,j,1, r1!σ_N,j,2, …, r(n-1)!σ_N,j,n ).
Exact mathematical passage
C_{N,j,n}(r) = \frac{(Nn)!}{n!} B_n\bigl( r\sigma_{N,j,1}, r1!\sigma_{N,j,2}, \ldots, r(n-1)!\sigma_{N,j,n} \bigr).
\sigma_{N,j,2}Use in source · 1 passage
Search aliases: σ_N,j,2 sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2386
ized sections] For every 0≤ j<N, the entire function Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)! has order 1/N. If N≥2 and Λ_N,j is its nonzero zero multiset, then Φ_N,j(w) = ∏_μ∈Λ_N,j (1-frac wμ). For m≥1, put σ_N,j,m = ∑_μ∈Λ_N,jμ^-m. Then C_N,j,n(r) = frac(Nn)!n! B_n( rσ_N,j,1, r1!σ_N,j,2, …, r(n-1)!σ_N,j,n ).
Exact mathematical passage
C_{N,j,n}(r) = \frac{(Nn)!}{n!} B_n\bigl( r\sigma_{N,j,1}, r1!\sigma_{N,j,2}, \ldots, r(n-1)!\sigma_{N,j,n} \bigr).
\sigma_{N,j,m}Use in source · 2 passages
Search aliases: σ_N,j,m sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2380
[Products for the normalized sections] For every 0≤ j<N, the entire function Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)! has order 1/N. If N≥2 and Λ_N,j is its nonzero zero multiset, then Φ_N,j(w) = ∏_μ∈Λ_N,j (1-frac wμ). For m≥1, put σ_N,j,m = ∑_μ∈Λ_N,jμ^-m. Then C_N,j,n(r) = frac(Nn)!n! B_n( rσ_N,j,1, r1!σ_N,j,2, …, r(n-1)!σ_N,j,n ).
Exact mathematical passage
\sigma_{N,j,m} = \sum_{\mu\in\Lambda_{N,j}}\mu^{-m}.
\sigma_{N,j,n}Use in source · 1 passage
Search aliases: σ_N,j,n sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2386
ized sections] For every 0≤ j<N, the entire function Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)! has order 1/N. If N≥2 and Λ_N,j is its nonzero zero multiset, then Φ_N,j(w) = ∏_μ∈Λ_N,j (1-frac wμ). For m≥1, put σ_N,j,m = ∑_μ∈Λ_N,jμ^-m. Then C_N,j,n(r) = frac(Nn)!n! B_n( rσ_N,j,1, r1!σ_N,j,2, …, r(n-1)!σ_N,j,n ).
Exact mathematical passage
C_{N,j,n}(r) = \frac{(Nn)!}{n!} B_n\bigl( r\sigma_{N,j,1}, r1!\sigma_{N,j,2}, \ldots, r(n-1)!\sigma_{N,j,n} \bigr).
\sigma_{N,m-k}Use in source · 2 passages
Search aliases: σ_N,m-k sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2441
[Newton sums for the reduced zeros] For |w| below the least zero modulus, with the analytic logarithm normalized to zero at the origin, -logΔ_N(w) = ∑_m≥1σ_N,mfracw^mm. Thus κ_N,m=σ_N,m. If a_m=frac(-1)^m(Nm)!, a_0=1, then σ_N,m = -ma_m- ∑_k=1^m-1a_kσ_N,m-k.
Exact mathematical passage
\sigma_{N,m} = -ma_m- \sum_{k=1}^{m-1}a_k\sigma_{N,m-k}.
\sigma_{N,n}Use in source · 1 passage
Search aliases: σ_N,n sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2552
[Reciprocal coefficients from zero data] For every n≥0, P_N,n(r) = frac(Nn)!n! B_n( rσ_N,1, r1!σ_N,2, …, r(n-1)!σ_N,n ).
Exact mathematical passage
P_{N,n}(r) = \frac{(Nn)!}{n!} B_n\bigl( r\sigma_{N,1}, r1!\sigma_{N,2}, \ldots, r(n-1)!\sigma_{N,n} \bigr).
\sigma_{N,n+1}Use in source · 1 passage
Search aliases: σ_N,n+1 sigma
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2539
[Tangent coefficients as zero sums] For every n≥0, T_N(n) = N(Nn+N-1)!σ_N,n+1.
Exact mathematical passage
T_N(n) = N(Nn+N-1)!\sigma_{N,n+1}.
\sigma_{N,q}Use in source · 2 passages
Search aliases: σ_N,q sigma
Use in source · atlas-pages.tex, line 402
vpage22Finite coefficients give infinite momentsZERO POWER SUMSTraditional monograph: Theorem 12.6 (p. 35), Corollary 12.7 (p. 36), Corollary 12.9 (p. 37), Corollary 12.10 (p. 37). Taking the logarithm of the canonical product identifies κ_N,q=σ_N,q:=∑_λ∈Λ_Nλ^-q =Tr(K_N^q), q≥1. The same exact numbers can be computed from finitely many denominator coefficients by Newton's recurrence. No numerical zero list is needed for that calculation. vfigmoments For order two, σ_2,1=frac12, σ_2,2=frac16, σ_2…
Exact mathematical passage
\kappa_{N,q}=\sigma_{N,q}:=\sum_{\lambda\in\Lambda_N}\lambda^{-q} =\Tr(K_N^q),\qquad q\ge1.
\tau^{-m}Use in source · 2 passages
Search aliases: τ^-m tau
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2672
[Orbit reduced nearest circle term] Assume the hypotheses of [source reference]. Choose one representative from each orbit of the nearest zeros under multiplication by ω, and call the resulting set Z_R^orb. With m=Nn, P_N,n(r) = · N m!binomm+r-1r-1 ∑_τ∈ Z_R^orb fracτ^-m(-τ D_N'(τ))^r · +O(m!m^r-2R^-m) +O_R_*(m!R_*^-m). For r=1, the first error term is absent.
Exact mathematical passage
\begin{aligned} P_{N,n}(r) ={}&N m!\binom{m+r-1}{r-1} \sum_{\tau\in\Zcal_R^{\mathrm{orb}}} \frac{\tau^{-m}}{(-\tau\Dfun_N'(\tau))^r} \\ &+O\left(m!m^{r-2}R^{-m}\right) +O_{R_*}\left(m!R_*^{-m}\right). \end{aligned}
\tau_0Use in source · 2 passages
Search aliases: τ_0 tau
Use in source · atlas-pages.tex, line 356
1)^nt^3n(3n)! at the two rational endpoints. From the next term onward, magnitudes decrease, because their successive ratio is fract^3(3n+1)(3n+2)(3n+3)<1 there. The alternating-series remainder is bounded by the first omitted term. A separate outward interval evaluation of the differentiated polynomial, plus its tail bound, certifies the derivative sign. The first bracket is 1.84981279919014347629<τ_0<1.84981279919014347630. Its displacement differs from the leading asymptotic prediction by abo…
Exact mathematical passage
1.84981279919014347629<\tau_0<1.84981279919014347630.
\tau_m^NUse in source · 2 passages
Search aliases: τ_m^N tau
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2195
[Reduced zeros] Let Λ_N be the nonzero zero multiset of Δ_N. All but finitely many members of Λ_N are positive and simple. They can be indexed as λ_m=τ_m^N for all sufficiently large m. If N≥3, the notation may be chosen as in [source reference], and λ_m = t_m^N +O(t_m^N-1e^-γ t_m). For N=2, take τ_m=t_m=π(m+1/2) for m≥0; then λ_m=t_m^2 exactly.
Exact mathematical passage
\lambda_m=\tau_m^N
\Phi_{2,1}Use in source · 1 passage
Search aliases: Φ_2,1 Phi
Use in source · different_reciprocals.tex, line 42
++(2.10000,0.80000); node[] at (6.05000,1.65000) - frac13; draw[fill=black!0,draw=black!40,line width=.35pt] (7.10000,1.25000) rectangle ++(2.10000,0.80000); node[] at (8.15000,1.65000) frac715; draw[fill=black!8,draw=black!40,line width=.35pt] (9.20000,1.25000) rectangle ++(2.10000,0.80000); node[] at (10.25000,1.65000) - frac3121; node[] at (5.65000,-0.15000) E_2,1(z)=sinh z=zΦ_2,1(z^2), C_2,1(0)=1;
Exact mathematical passage
E_{2,1}(z)=\sinh z=z\Phi_{2,1}(z^2),\qquad C_{2,1}(0)=1
\Phi_mUse in source · 2 passages
Search aliases: Φ_m Phi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2012
Fix a large m, put a=2is_N, and consider the closed disk B_m={z:|z-t_m|≤ 1/(2s_N)}. The remainder bounds hold uniformly on these fixed-radius disks. For sufficiently large m, |R_N|≤ 1/2 and |R_N'|≤ s_N/2 there. Define Φ_m(z)=t_m+fraclog(1+R_N(z))2is_N, using the principal logarithm. The estimate |log(1+u)|≤(3/2)|u| shows that Φ_m maps B_m into itself. The derivative bound |Φ_m'|≤1/2 makes it a contraction. Banach's fixed-point theorem gives a unique fixed point. On B_m, |a(z-t_m)|≤1<π, so the pr…
Exact mathematical passage
\Phi_m(z)=t_m+\frac{\log(1+R_N(z))}{2is_N},
\Phi_m'Use in source · 1 passage
Search aliases: Φ_m' Phi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2017
_N, and consider the closed disk B_m={z:|z-t_m|≤ 1/(2s_N)}. The remainder bounds hold uniformly on these fixed-radius disks. For sufficiently large m, |R_N|≤ 1/2 and |R_N'|≤ s_N/2 there. Define Φ_m(z)=t_m+fraclog(1+R_N(z))2is_N, using the principal logarithm. The estimate |log(1+u)|≤(3/2)|u| shows that Φ_m maps B_m into itself. The derivative bound |Φ_m'|≤1/2 makes it a contraction. Banach's fixed-point theorem gives a unique fixed point. On B_m, |a(z-t_m)|≤1<π, so the principal logarithm invert…
Exact mathematical passage
|\Phi_m'|\leq1/2
\PsiUse in source · 2 passages
Search aliases: Ψ Psi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2520
The reduced unsigned denominator is Ψ_N(w)=∑_n≥0fracw^n(Nn)!=Δ_N(-w). By [source reference], -logΨ_N(w) = ∑_m≥1widetildeσ_N,mfracw^mm. With w=z^N, the logarithm of the theta generating function is ∑_m≥1 widetildeσ_N,m(u^Nm+u)fracw^mm. The complete Bell polynomial formula follows.
Exact mathematical passage
\Psi_N(w)=\sum_{n\geq0}\frac{w^n}{(Nn)!}=\Delta_N(-w).
\Psi_NUse in source · 2 passages
Search aliases: Ψ_N Psi
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2520
The reduced unsigned denominator is Ψ_N(w)=∑_n≥0fracw^n(Nn)!=Δ_N(-w). By [source reference], -logΨ_N(w) = ∑_m≥1widetildeσ_N,mfracw^mm. With w=z^N, the logarithm of the theta generating function is ∑_m≥1 widetildeσ_N,m(u^Nm+u)fracw^mm. The complete Bell polynomial formula follows.
Exact mathematical passage
\Psi_N(w)=\sum_{n\geq0}\frac{w^n}{(Nn)!}=\Delta_N(-w).
\OmegaUse in source · 3 passages
Search aliases: Ω Omega
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 337
Fix N≥1 and write ω=e^2π i/N. Let Ω be a disk centered at the origin, with the whole plane allowed, and assume that Ω is invariant under multiplication by ω. For a holomorphic function F on Ω, define ( B_N,jF)(z) = frac1N∑_ℓ=0^N-1ω^-jℓF(ω^ℓ z), 0≤ j<N. The same formula is valid for formal power series.
Exact mathematical passage
\Omega
\omegaUse in source · 30 passages
Search aliases: ω omega
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 334
Fix N≥1 and write ω=e^2π i/N. Let Ω be a disk centered at the origin, with the whole plane allowed, and assume that Ω is invariant under multiplication by ω. For a holomorphic function F on Ω, define ( B_N,jF)(z) = frac1N∑_ℓ=0^N-1ω^-jℓF(ω^ℓ z), 0≤ j<N. The same formula is valid for formal power series.
Exact mathematical passage
\omega=e^{2\pi i/N}.
\omega^{(m-j)\ell}Use in source · 1 passage
Search aliases: ω^(m-j)ℓ omega ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 369
Substitution of the power series gives B_N,jF(z) = ∑_m≥0a_mz^m ( frac1N∑_ℓ=0^N-1ω^(m-j)ℓ ). The inner average is one when m≡ jbmod N and zero otherwise. The operator identities follow from the disjoint residue classes.
Exact mathematical passage
\Bcal_{N,j}F(z) = \sum_{m\geq0}a_mz^m \left( \frac1N\sum_{\ell=0}^{N-1}\omega^{(m-j)\ell} \right).
\omega^{-j\ell}Use in source · 2 passages
Search aliases: ω^-jℓ omega ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 338
Fix N≥1 and write ω=e^2π i/N. Let Ω be a disk centered at the origin, with the whole plane allowed, and assume that Ω is invariant under multiplication by ω. For a holomorphic function F on Ω, define ( B_N,jF)(z) = frac1N∑_ℓ=0^N-1ω^-jℓF(ω^ℓ z), 0≤ j<N. The same formula is valid for formal power series.
Exact mathematical passage
(\Bcal_{N,j}F)(z) = \frac1N\sum_{\ell=0}^{N-1}\omega^{-j\ell}F(\omega^\ell z), \qquad 0\leq j<N.
\omega^{-k}Use in source · 2 passages
Search aliases: ω^-k omega
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1958
The derivatives satisfy D_N'(ω^kτ_m) = ω^-k D_N'(τ_m) and | D_N'(τ_m)| = frac2s_NNe^c_Nτ_m (1+O(e^-γτ_m)).
Exact mathematical passage
\Dfun_N'(\omega^k\tau_m) = \omega^{-k}\Dfun_N'(\tau_m)
\omega^{i+j}Use in source · 1 passage
Search aliases: ω^i+j omega
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 419
Write F=∑_iF_i and G=∑_jG_j, where F_i= B_N,iF and G_j= B_N,jG. Their covariance gives (F_iG_j)(ω z)=ω^i+jF_i(z)G_j(z), so F_iG_j belongs to the packet labelled by i+j modulo N. Expanding the finite product and applying B_N,k gives the formula.
Exact mathematical passage
(F_iG_j)(\omega z)=\omega^{i+j}F_i(z)G_j(z),
\omega^jUse in source · 5 passages
Search aliases: ω^j omega
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 382
[Covariance characterization] A holomorphic function F belongs to the range of B_N,j if and only if F(ω z)=ω^jF(z).
Exact mathematical passage
F(\omega z)=\omega^jF(z).
\omega^kUse in source · 7 passages
Search aliases: ω^k omega
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 389
Every monomial selected by B_N,j has the stated covariance. Conversely, write F=∑_k B_N,kF. Replacing z by ω z gives ∑_kω^k B_N,kF = ω^j∑_k B_N,kF. Application of B_N,k gives (ω^k-ω^j) B_N,kF=0. The powers 1,ω,…,ω^N-1 are distinct, so B_N,kF=0 for k≠ j.
Exact mathematical passage
\sum_k\omega^k\Bcal_{N,k}F = \omega^j\sum_k\Bcal_{N,k}F.
\omega^mUse in source · 2 passages
Search aliases: ω^m omega
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1570
[Appell symmetry] Each polynomial E_m^(N,j,r) is monic of degree m, and E_m^(N,j,r)(ω x) = ω^mE_m^(N,j,r)(x).
Exact mathematical passage
E_m^{(N,j,r)}(\omega x) = \omega^mE_m^{(N,j,r)}(x).
\omega^{N-1}Use in source · 1 passage
Search aliases: ω^N-1 omega
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 398
Every monomial selected by B_N,j has the stated covariance. Conversely, write F=∑_k B_N,kF. Replacing z by ω z gives ∑_kω^k B_N,kF = ω^j∑_k B_N,kF. Application of B_N,k gives (ω^k-ω^j) B_N,kF=0. The powers 1,ω,…,ω^N-1 are distinct, so B_N,kF=0 for k≠ j.
Exact mathematical passage
1,\omega,\ldots,\omega^{N-1}
\omega^\ellUse in source · 5 passages
Search aliases: ω^ℓ omega ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 338
Fix N≥1 and write ω=e^2π i/N. Let Ω be a disk centered at the origin, with the whole plane allowed, and assume that Ω is invariant under multiplication by ω. For a holomorphic function F on Ω, define ( B_N,jF)(z) = frac1N∑_ℓ=0^N-1ω^-jℓF(ω^ℓ z), 0≤ j<N. The same formula is valid for formal power series.
Exact mathematical passage
(\Bcal_{N,j}F)(z) = \frac1N\sum_{\ell=0}^{N-1}\omega^{-j\ell}F(\omega^\ell z), \qquad 0\leq j<N.
\ellUse in source · 28 passages
Search aliases: ℓ ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 338
Fix N≥1 and write ω=e^2π i/N. Let Ω be a disk centered at the origin, with the whole plane allowed, and assume that Ω is invariant under multiplication by ω. For a holomorphic function F on Ω, define ( B_N,jF)(z) = frac1N∑_ℓ=0^N-1ω^-jℓF(ω^ℓ z), 0≤ j<N. The same formula is valid for formal power series.
Exact mathematical passage
(\Bcal_{N,j}F)(z) = \frac1N\sum_{\ell=0}^{N-1}\omega^{-j\ell}F(\omega^\ell z), \qquad 0\leq j<N.
\ell^2Use in source · 3 passages
Search aliases: ℓ^2 ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2304
[Diagonal Fredholm determinant representation] Let I_N be a multiplicity-sensitive countable index set for Λ_N and put H_N=ℓ^2(I_N). On its standard orthonormal basis, define the diagonal operator K_Ne_λ=λ^-1e_λ. Then K_N is trace class and Δ_N(w)=det_ F(I-wK_N). For every m≥1, κ_N,m=tr(K_N^m).
Exact mathematical passage
\mathcal H_N=\ell^2(I_N)
\toUse in source · 5 passages
Search aliases: → to
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1729
[Order from Taylor coefficients] If f(w)=∑_n≥0a_nw^n is an entire function with infinitely many nonzero coefficients, then ord f = limsup_substackn→∞ a_n≠0 fracnlog nlog(1/|a_n|). Canonical products for entire functions of order below one have genus zero. We use these standard facts in the form given in [source reference].
Exact mathematical passage
\ord f = \limsup_{\substack{n\to\infty\\a_n\ne0}} \frac{n\log n}{\log(1/|a_n|)}.
\mapstoUse in source · 3 passages
Search aliases: ↦ mapsto
Use in source · atlas-pages.tex, line 146
lta_4(z^4). Set A_4=sqrt2π. All the zeros are A_4(m+1/2)i^k, with m≥0 and 0≤ k<4. Each zero is simple. vfigorbit_reduction The drawing uses u=z/A_4 and v=w/A_4^4, so its marked coordinates are rational. The first four-point orbit has fourth power 1/16. The second has fourth power 81/16. headMultiplicity is not multiplied in the reduced function At a nonzero point, z↦ z^4 has nonzero derivative. It preserves the local multiplicity of a zero. Four distinct simple zeros of D_4 therefore correspond …
Exact mathematical passage
z\mapsto z^4
\partialUse in source · 16 passages
Search aliases: ∂ partial
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1527
[Formal differential-operator representation on polynomials] Let a formal series A(z)=∑_q≥0a_qz^q/q! act on polynomials by A(∂_x)f(x) = ∑_q≥0fraca_qq!∂_x^qf(x). The power r is applied to the formal symbol before the substitution z=∂_x. Then E_m^(N,j,r)(x) = C_N,j(∂_x)^r x^m. The inverse relation is x^m = Φ_N,j(∂_x^N)^r E_m^(N,j,r)(x). Both sums are finite on every polynomial.
Exact mathematical passage
A(\partial_x)f(x) = \sum_{q\geq0}\frac{a_q}{q!}\partial_x^qf(x).
\partial_xUse in source · 12 passages
Search aliases: ∂_x partial
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1527
[Formal differential-operator representation on polynomials] Let a formal series A(z)=∑_q≥0a_qz^q/q! act on polynomials by A(∂_x)f(x) = ∑_q≥0fraca_qq!∂_x^qf(x). The power r is applied to the formal symbol before the substitution z=∂_x. Then E_m^(N,j,r)(x) = C_N,j(∂_x)^r x^m. The inverse relation is x^m = Φ_N,j(∂_x^N)^r E_m^(N,j,r)(x). Both sums are finite on every polynomial.
Exact mathematical passage
A(\partial_x)f(x) = \sum_{q\geq0}\frac{a_q}{q!}\partial_x^qf(x).
\partial_x^2Use in source · 2 passages
Search aliases: ∂_x^2 partial
Use in source · atlas-pages.tex, line 267
IALSTraditional monograph: Theorem 9.6 (p. 21), Theorem 9.8 (p. 21), Proposition 9.9 (p. 21). For C_2,0(z)=sech z, define E_m(x) by sech z e^xz=∑ E_m(x)z^m/m!. The formal differential-operator formula is E_m(x)=sech(∂_x)x^m. Although the symbol is an infinite series, derivatives above degree m vanish. vfigappell_operator For m=4, only three terms survive: (1-frac∂_x^22!+5frac∂_x^44!)x^4 =x^4-6x^2+5. The four real roots of this polynomial are ±1,±sqrt5. They are not the zeros of the original anal…
Exact mathematical passage
\left(1-\frac{\partial_x^2}{2!}+5\frac{\partial_x^4}{4!}\right)x^4 =x^4-6x^2+5.
\partial_x^4Use in source · 2 passages
Search aliases: ∂_x^4 partial
Use in source · atlas-pages.tex, line 267
IALSTraditional monograph: Theorem 9.6 (p. 21), Theorem 9.8 (p. 21), Proposition 9.9 (p. 21). For C_2,0(z)=sech z, define E_m(x) by sech z e^xz=∑ E_m(x)z^m/m!. The formal differential-operator formula is E_m(x)=sech(∂_x)x^m. Although the symbol is an infinite series, derivatives above degree m vanish. vfigappell_operator For m=4, only three terms survive: (1-frac∂_x^22!+5frac∂_x^44!)x^4 =x^4-6x^2+5. The four real roots of this polynomial are ±1,±sqrt5. They are not the zeros of the original anal…
Exact mathematical passage
\left(1-\frac{\partial_x^2}{2!}+5\frac{\partial_x^4}{4!}\right)x^4 =x^4-6x^2+5.
\partial_x^NUse in source · 2 passages
Search aliases: ∂_x^N partial
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1539
ferential-operator representation on polynomials] Let a formal series A(z)=∑_q≥0a_qz^q/q! act on polynomials by A(∂_x)f(x) = ∑_q≥0fraca_qq!∂_x^qf(x). The power r is applied to the formal symbol before the substitution z=∂_x. Then E_m^(N,j,r)(x) = C_N,j(∂_x)^r x^m. The inverse relation is x^m = Φ_N,j(∂_x^N)^r E_m^(N,j,r)(x). Both sums are finite on every polynomial.
Exact mathematical passage
x^m = \Phi_{N,j}(\partial_x^N)^r E_m^{(N,j,r)}(x).
\partial_x^qUse in source · 1 passage
Search aliases: ∂_x^q partial
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1527
[Formal differential-operator representation on polynomials] Let a formal series A(z)=∑_q≥0a_qz^q/q! act on polynomials by A(∂_x)f(x) = ∑_q≥0fraca_qq!∂_x^qf(x). The power r is applied to the formal symbol before the substitution z=∂_x. Then E_m^(N,j,r)(x) = C_N,j(∂_x)^r x^m. The inverse relation is x^m = Φ_N,j(∂_x^N)^r E_m^(N,j,r)(x). Both sums are finite on every polynomial.
Exact mathematical passage
A(\partial_x)f(x) = \sum_{q\geq0}\frac{a_q}{q!}\partial_x^qf(x).
\prodUse in source · 11 passages
Search aliases: ∏ prod
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 555
The multiplication formula for the gamma function gives (Nn+j)! = j!N^Nnn! ∏_a∈ A_N,j(a)_n. Indeed, the N Pochhammer factors supplied by the multiplication formula are ((j+m)/N)_n for 1≤ m≤ N; the excluded factor is ((j+N-j)/N)_n=(1)_n=n!. Insertion into the defining series gives the stated hypergeometric expansion.
Exact mathematical passage
(Nn+j)! = j!N^{Nn}n! \prod_{a\in A_{N,j}}(a)_n.
\prod_{a\in A_{N,j}}Use in source · 3 passages
Search aliases: ∏_a∈ A_N,j prod in
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 555
The multiplication formula for the gamma function gives (Nn+j)! = j!N^Nnn! ∏_a∈ A_N,j(a)_n. Indeed, the N Pochhammer factors supplied by the multiplication formula are ((j+m)/N)_n for 1≤ m≤ N; the excluded factor is ((j+N-j)/N)_n=(1)_n=n!. Insertion into the defining series gives the stated hypergeometric expansion.
Exact mathematical passage
(Nn+j)! = j!N^{Nn}n! \prod_{a\in A_{N,j}}(a)_n.
\prod_{m=0}^{L-1}Use in source · 1 passage
Search aliases: ∏_m=0^L-1 prod
Use in source · product.tex, line 28
4835) (10.90625,1.03091) (10.97813,1.01351) (11.05000,0.99615) (11.12187,0.97883) (11.19375,0.96154) (11.26562,0.94429) (11.33750,0.92707) (11.40937,0.90990) (11.48125,0.89276) (11.55312,0.87566) (11.62500,0.85859) (11.69688,0.84156) (11.76875,0.82457) (11.84062,0.80762) (11.91250,0.79070) (11.98437,0.77382) (12.05625,0.75698) (12.12812,0.74017) (12.20000,0.72340); node[] at (6.50000,4.60000) Δ_2(w)=cossqrt w, P_L(w)=∏_m=0^L-1(1-w/λ_m); node[] at (6.50000,-0.35000) thick: entire function; dotted…
Exact mathematical passage
\Delta_2(w)=\cos\sqrt w,\qquad P_L(w)=\prod_{m=0}^{L-1}(1-w/\lambda_m)
\prod_{\lambda\in\Lambda_N}Use in source · 6 passages
Search aliases: ∏_λ∈Λ_N prod lambda in Lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2273
[Genus zero product] For N≥2, let Λ_N be the nonzero zero multiset of Δ_N. Then Δ_N(w) = ∏_λ∈Λ_N (1-frac wλ), where the product converges locally uniformly. Consequently, D_N(z) = ∏_λ∈Λ_N (1-fracz^Nλ).
Exact mathematical passage
\Delta_N(w) = \prod_{\lambda\in\Lambda_N} \left(1-\frac w\lambda\right),
\prod_{\mu\in\Lambda_{N,j}}Use in source · 1 passage
Search aliases: ∏_μ∈Λ_N,j prod mu in Lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2373
[Products for the normalized sections] For every 0≤ j<N, the entire function Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)! has order 1/N. If N≥2 and Λ_N,j is its nonzero zero multiset, then Φ_N,j(w) = ∏_μ∈Λ_N,j (1-frac wμ). For m≥1, put σ_N,j,m = ∑_μ∈Λ_N,jμ^-m. Then C_N,j,n(r) = frac(Nn)!n! B_n( rσ_N,j,1, r1!σ_N,j,2, …, r(n-1)!σ_N,j,n ).
Exact mathematical passage
\Phi_{N,j}(w) = \prod_{\mu\in\Lambda_{N,j}} \left(1-\frac w\mu\right).
\sum_{0\leq n\leq m/N}Use in source · 1 passage
Search aliases: ∑_0≤ n≤ m/N sum leq leq
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1489
[Explicit Appell polynomial] For every m≥0, E_m^(N,j,r)(x) = ∑_0≤ n≤ m/N binommNn C_N,j,n(r)x^m-Nn.
Exact mathematical passage
E_m^{(N,j,r)}(x) = \sum_{0\leq n\leq m/N} \binom{m}{Nn} C_{N,j,n}(r)x^{m-Nn}.
\sum_iUse in source · 1 passage
Search aliases: ∑_i sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 417
Write F=∑_iF_i and G=∑_jG_j, where F_i= B_N,iF and G_j= B_N,jG. Their covariance gives (F_iG_j)(ω z)=ω^i+jF_i(z)G_j(z), so F_iG_j belongs to the packet labelled by i+j modulo N. Expanding the finite product and applying B_N,k gives the formula.
Exact mathematical passage
F=\sum_iF_i
\sum_{i+j\equiv k\ (N)}Use in source · 2 passages
Search aliases: ∑_i+j≡ k (N) sum equiv
Use in source · atlas-pages.tex, line 76
1+2z+3z^2+z^3, G=2+z+z^2. The table displays every product of their three selected sections. vfigcyclic_product The shaded cells contribute to residue zero: 2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3. Keeping only the product of the residue-zero inputs would give 2+2z^3 and lose five units of the z^3 coefficient. The mixed cells are part of the exact result. headThe general rule B_N,k(FG)= ∑_i+j≡ k (N)( B_N,iF)( B_N,jG). Every pair belongs to exactly one output residue. No term is discarded and no coeffi…
Exact mathematical passage
\mathcal B_{N,k}(FG)= \sum_{i+j\equiv k\ (N)}(\mathcal B_{N,i}F)(\mathcal B_{N,j}G).
\sum_jUse in source · 1 passage
Search aliases: ∑_j sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 417
Write F=∑_iF_i and G=∑_jG_j, where F_i= B_N,iF and G_j= B_N,jG. Their covariance gives (F_iG_j)(ω z)=ω^i+jF_i(z)G_j(z), so F_iG_j belongs to the packet labelled by i+j modulo N. Expanding the finite product and applying B_N,k gives the formula.
Exact mathematical passage
G=\sum_jG_j
\sum_{j=0}^{N-1}Use in source · 3 passages
Search aliases: ∑_j=0^N-1 sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 362
[Root filter] If F(z)=∑_m≥0a_mz^m, then B_N,jF(z) = ∑_m≡ jbmod Na_mz^m. The operators satisfy B_N,i B_N,j=δ_ij B_N,j and ∑_j=0^N-1 B_N,j=I.
Exact mathematical passage
\sum_{j=0}^{N-1}\Bcal_{N,j}=I.
\sum_kUse in source · 2 passages
Search aliases: ∑_k sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 388
Every monomial selected by B_N,j has the stated covariance. Conversely, write F=∑_k B_N,kF. Replacing z by ω z gives ∑_kω^k B_N,kF = ω^j∑_k B_N,kF. Application of B_N,k gives (ω^k-ω^j) B_N,kF=0. The powers 1,ω,…,ω^N-1 are distinct, so B_N,kF=0 for k≠ j.
Exact mathematical passage
F=\sum_k\Bcal_{N,k}F
\sum_{k=0}^{m-1}Use in source · 1 passage
Search aliases: ∑_k=0^m-1 sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2466
Differentiate the product identity in logarithmic form to obtain Δ_N'(w) = -Δ_N(w) ∑_m≥1σ_N,mw^m-1. Comparison of the coefficient of w^m-1 gives ma_m = -∑_k=0^m-1a_kσ_N,m-k. Solving for σ_N,m gives the recurrence.
Exact mathematical passage
ma_m = -\sum_{k=0}^{m-1}a_k\sigma_{N,m-k}.
\sum_{k=0}^{n}Use in source · 1 passage
Search aliases: ∑_k=0^n sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1295
The defining equation is D_N T_N=- D_N'. Extracting the coefficient of z^Nn+N-1/(Nn+N-1)! gives ∑_k=0^n(-1)^k binomNn+N-1NkT_N(n-k)=(-1)^n. Isolating the k=0 term gives the recurrence.
Exact mathematical passage
\sum_{k=0}^{n}(-1)^k \binom{Nn+N-1}{Nk}T_N(n-k)=(-1)^n.
\sum_{k=0}^{N-1}Use in source · 1 passage
Search aliases: ∑_k=0^N-1 sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1723
We now assume N≥2. Write η_k=e^(2k+1)π i/N, 0≤ k<N. These are the vertices of a regular polygon and D_N(z)=frac1N∑_k=0^N-1e^η_kz.
Exact mathematical passage
\Dfun_N(z)=\frac1N\sum_{k=0}^{N-1}e^{\eta_kz}.
\sum_{k=0}^{n-1}Use in source · 1 passage
Search aliases: ∑_k=0^n-1 sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1028
Set S= D_N^-r. Logarithmic differentiation gives D_NS'+r D_N'S=0. Taking the coefficient of z^Nn-1/(Nn-1)! gives ∑_k=0^n-1(-1)^k binomNn-1NkP_N,n-k(r) +r∑_k=1^n(-1)^k binomNn-1Nk-1P_N,n-k(r)=0. The term with k=0 in the first sum is P_N,n(r), and every other term has a smaller block index. Rearrangement gives the formula.
Exact mathematical passage
\sum_{k=0}^{n-1}(-1)^k \binom{Nn-1}{Nk}P_{N,n-k}(r) +r\sum_{k=1}^{n}(-1)^k \binom{Nn-1}{Nk-1}P_{N,n-k}(r)=0.
\sum_{k=1}^{m-1}Use in source · 1 passage
Search aliases: ∑_k=1^m-1 sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2441
[Newton sums for the reduced zeros] For |w| below the least zero modulus, with the analytic logarithm normalized to zero at the origin, -logΔ_N(w) = ∑_m≥1σ_N,mfracw^mm. Thus κ_N,m=σ_N,m. If a_m=frac(-1)^m(Nm)!, a_0=1, then σ_N,m = -ma_m- ∑_k=1^m-1a_kσ_N,m-k.
Exact mathematical passage
\sigma_{N,m} = -ma_m- \sum_{k=1}^{m-1}a_k\sigma_{N,m-k}.
\sum_{k=1}^{n-1}Use in source · 1 passage
Search aliases: ∑_k=1^n-1 sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1004
[Parameter recurrence] We have P_N,0(r)=1. For n≥1, P_N,n(r) = · - ∑_k=1^n-1 (-1)^k binomNn-1Nk P_N,n-k(r) · -r ∑_k=1^n (-1)^k binomNn-1Nk-1 P_N,n-k(r).
Exact mathematical passage
\begin{aligned} P_{N,n}(r) ={}&- \sum_{k=1}^{n-1} (-1)^k \binom{Nn-1}{Nk} P_{N,n-k}(r) \\ &-r \sum_{k=1}^{n} (-1)^k \binom{Nn-1}{Nk-1} P_{N,n-k}(r). \end{aligned}
\sum_{k\geq1}Use in source · 1 passage
Search aliases: ∑_k≥1 sum geq
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1129
Let B_r(w)=Δ_N(w)^-r = ∑_n≥0b_N,n(r)w^n. Logarithmic differentiation gives fracB_r'(w)B_r(w) = r∑_k≥1κ_N,kw^k-1. Comparison of the coefficient of w^n-1 proves the recurrence.
Exact mathematical passage
\frac{B_r'(w)}{B_r(w)} = r\sum_{k\geq1}\kappa_{N,k}w^{k-1}.
\sum_{m=0}^{12}Use in source · 2 passages
Search aliases: ∑_m=0^12 sum
Use in source · atlas-pages.tex, line 52
idue j, evaluate the function at four rotated arguments and attach the phase i^-jℓ: ( B_4,jF)(z)=frac14∑_ℓ=0^3i^-jℓF(i^ℓ z). A monomial a_mz^m is multiplied by the finite average of i^(m-j)ℓ. Matching phases all vote 1. Nonmatching phases cancel exactly. vfigroot_filter headAn example that can be checked coefficient by coefficient Take F(z)=∑_m=0^12(m+1)z^m. Then B_4,2F=3z^2+7z^6+11z^10. Applying the same selector again changes nothing. Applying a different selector gives zero. Summing the four …
Exact mathematical passage
F(z)=\sum_{m=0}^{12}(m+1)z^m
\sum_{m=1}^{n}Use in source · 1 passage
Search aliases: ∑_m=1^n sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1663
[Theta Newton recurrence] Put θ_N,n(u)=fracsymupΘ_N,n(u)(Nn)!. Then θ_N,0(u)=1 and nθ_N,n(u) = ∑_m=1^n (-1)^mκ_N,m (u^Nm+u) θ_N,n-m(u). In particular, symupΘ_N,1(u)=-(u^N+u).
Exact mathematical passage
n\theta_{N,n}(u) = \sum_{m=1}^{n} (-1)^m\kappa_{N,m} \bigl(u^{Nm}+u\bigr) \theta_{N,n-m}(u).
\sum_{m\equiv j\bmod N}Use in source · 1 passage
Search aliases: ∑_m≡ jbmod N sum equiv bmod
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 352
[Root filter] If F(z)=∑_m≥0a_mz^m, then B_N,jF(z) = ∑_m≡ jbmod Na_mz^m. The operators satisfy B_N,i B_N,j=δ_ij B_N,j and ∑_j=0^N-1 B_N,j=I.
Exact mathematical passage
\Bcal_{N,j}F(z) = \sum_{m\equiv j\bmod N}a_mz^m.
\sum_{m\geq 0}Use in source · 1 passage
Search aliases: ∑_m≥ 0 sum geq
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 214
For a power series F(z)=∑_m≥ 0a_mz^m and a positive integer N, the coefficients may be separated according to the residue of m modulo N. Root of unity averaging gives exact projection operators for this separation. This elementary construction is usually called a multisection or a root filter. It appears throughout the classical theory of generating functions and special functions. Background on multisections, hypergeometric functions, and related coefficient systems may be found in [source refe…
Exact mathematical passage
F(z)=\sum_{m\geq 0}a_mz^m
\sum_{m\ge L}Use in source · 2 passages
Search aliases: ∑_m≥ L sum ge
Use in source · atlas-pages.tex, line 391
on ℓ^2(Λ_N) gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries. vfigproduct headFinite compression is an approximation For N=2, λ_m=π^2(m+1/2)^2. The finite determinant P_L is exactly the displayed finite product, but it is not exactly Δ_2. For 0≤ w≤1 and L≥1, an elementary tail estimate gives 0≤1-fracΔ_2(w)P_L(w) ≤ w∑_m≥ Lλ_m^-1 ≤frac19(L-1/2). The last bound uses π>3 and integral comparison. Thus truncation has an explicit error mechanism. t…
Exact mathematical passage
0\le1-\frac{\Delta_2(w)}{P_L(w)} \le w\sum_{m\ge L}\lambda_m^{-1} \le\frac1{9(L-1/2)}.
\sum_{m\ge1}Use in source · 6 passages
Search aliases: ∑_m≥1 sum ge
Use in source · atlas-pages.tex, line 34
κ_N,m · displaystyle-logΔ_N(w)=∑_m≥1κ_N,mw^m/m.
Exact mathematical passage
\displaystyle-\log\Delta_N(w)=\sum_{m\ge1}\kappa_{N,m}w^m/m
\sum_{n=0}^{90}Use in source · 2 passages
Search aliases: ∑_n=0^90 sum
Use in source · atlas-pages.tex, line 347
Theorem 11.8 (p. 30), Corollary 11.9 (p. 31). The root finder is used only to suggest intervals. Each displayed interval is accepted by exact rational arithmetic: opposite endpoint signs prove existence, and an interval bound for D_3' excludes zero throughout the interval, proving uniqueness inside it. vfigcubic_certificates headHow the certificate works Evaluate the Taylor polynomial S_90(t)=∑_n=0^90frac(-1)^nt^3n(3n)! at the two rational endpoints. From the next term onward, magnitudes decreas…
Exact mathematical passage
S_{90}(t)=\sum_{n=0}^{90}\frac{(-1)^nt^{3n}}{(3n)!}
\sum_{n\geq 0}Use in source · 2 passages
Search aliases: ∑_n≥ 0 sum geq
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 231
This volume follows the consequences of applying the cyclic projections to the exponential function. The projected exponentials form a finite differential system. After one sign is inserted around the derivative cycle, they form a basis for y^(N)=-y. The section with initial value one is D_N(z)=∑_n≥ 0(-1)^nfracz^Nn(Nn)!. Its reciprocal powers produce polynomial coefficient families. Its logarithmic derivative produces a scalar higher Riccati identity. The same reciprocal construction applied to …
Exact mathematical passage
\Dfun_N(z)=\sum_{n\geq 0}(-1)^n\frac{z^{Nn}}{(Nn)!}.
\sum_{n\ge0}Use in source · 8 passages
Search aliases: ∑_n≥0 sum ge
Use in source · atlas-pages.tex, line 29
midrule E_N,j(z) · displaystyle∑_n≥0fracz^Nn+j(Nn+j)! selects one residue class of the exponential.
Exact mathematical passage
\displaystyle\sum_{n\ge0}\frac{z^{Nn+j}}{(Nn+j)!}
\sum_{q=0}^mUse in source · 2 passages
Search aliases: ∑_q=0^m sum
Use in source · atlas-pages.tex, line 274
The four real roots of this polynomial are ±1,±sqrt5. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output. headAn inverse operation, not a truncation error Applying cosh(∂_x) returns x^4 exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are E_m'=mE_m-1, E_m(x+y)=∑_q=0^mbinom mqE_q(x)y^m-q. Both follow from the generating function. No operator-series convergence as…
Exact mathematical passage
E_m'=mE_{m-1},\qquad E_m(x+y)=\sum_{q=0}^m\binom mqE_q(x)y^{m-q}.
\sum_{q=0}^{m}Use in source · 2 passages
Search aliases: ∑_q=0^m sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1474
[Appell laws] For m≥1, the polynomials satisfy fracddxE_m^(N,j,r)(x) = mE_m-1^(N,j,r)(x) and, for m≥0, E_m^(N,j,r)(x+y) = ∑_q=0^m binom mq E_q^(N,j,r)(x)y^m-q.
Exact mathematical passage
E_m^{(N,j,r)}(x+y) = \sum_{q=0}^{m} \binom mq E_q^{(N,j,r)}(x)y^{m-q}.
\sum_{q=0}^{n}Use in source · 7 passages
Search aliases: ∑_q=0^n sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1046
[Additive parameter law for reciprocal powers] For all r,s∈ C, S_N^(r)(z) S_N^(s)(z) = S_N^(r+s)(z) as a formal identity and on every common zero-free region with a fixed logarithm. Consequently, P_N,n(r+s) = ∑_q=0^n binomNnNq P_N,q(r)P_N,n-q(s).
Exact mathematical passage
P_{N,n}(r+s) = \sum_{q=0}^{n} \binom{Nn}{Nq} P_{N,q}(r)P_{N,n-q}(s).
\sum_{q=0}^{n-1}Use in source · 1 passage
Search aliases: ∑_q=0^n-1 sum
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1396
[Triangular inverse system] The coefficients are determined by C_N,j(0)=1 and ∑_q=0^n binomNn+jNq C_N,j(q) = δ_n0. For n≥1, C_N,j(n) = - frac displaystyle∑_q=0^n-1 binomNn+jNqC_N,j(q) displaystylebinomNn+jNn .
Exact mathematical passage
C_{N,j}(n) = - \frac{ \displaystyle\sum_{q=0}^{n-1} \binom{Nn+j}{Nq}C_{N,j}(q) }{ \displaystyle\binom{Nn+j}{Nn} }.
\sum_{q\geq0}Use in source · 2 passages
Search aliases: ∑_q≥0 sum geq
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1526
[Formal differential-operator representation on polynomials] Let a formal series A(z)=∑_q≥0a_qz^q/q! act on polynomials by A(∂_x)f(x) = ∑_q≥0fraca_qq!∂_x^qf(x). The power r is applied to the formal symbol before the substitution z=∂_x. Then E_m^(N,j,r)(x) = C_N,j(∂_x)^r x^m. The inverse relation is x^m = Φ_N,j(∂_x^N)^r E_m^(N,j,r)(x). Both sums are finite on every polynomial.
Exact mathematical passage
A(z)=\sum_{q\geq0}a_qz^q/q!
\sum_{\substack{0\leq i,j<N\\ i+j\equiv k\bmod N}}Use in source · 3 passages
Search aliases: ∑_substack0≤ i,j<N i+j≡ kbmod N sum substack leq equiv bmod
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 408
[Cyclic product law] If F and G are holomorphic on the same rotation-invariant domain, then B_N,k(FG) = ∑_substack0≤ i,j<N i+j≡ kbmod N ( B_N,iF)( B_N,jG).
Exact mathematical passage
\Bcal_{N,k}(FG) = \sum_{\substack{0\leq i,j<N\\ i+j\equiv k\bmod N}} (\Bcal_{N,i}F)(\Bcal_{N,j}G).
\sum_{\substack{\eta^N=-1\\
\eta\ne e^{ia_N},e^{-ia_N}}}Use in source · 1 passage
Search aliases: ∑_substackη^N=-1 η≠ e^ia_N,e^-ia_N sum substack eta eta ne
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1849
On the strip x≥ R and |y|≤ Y, define R_N(z) = ∑_substackη^N=-1 η≠ e^ia_N,e^-ia_N exp((η-e^-ia_N)z). Then N e^-e^-ia_Nz D_N(z) = 1+e^2is_Nz+R_N(z), and, for q=0,1,2, |R_N^(q)(z)| ≤ C e^-γ_Nx.
Exact mathematical passage
R_N(z) = \sum_{\substack{\eta^N=-1\\ \eta\ne e^{ia_N},e^{-ia_N}}} \exp\bigl((\eta-e^{-ia_N})z\bigr).
\sum_{\eta^N=-1}Use in source · 4 passages
Search aliases: ∑_η^N=-1 sum eta
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 244
The second half of the paper concerns the zeros of D_N. The finite exponential sum D_N(z)=frac1N∑_η^N=-1e^η z has a regular polygon of exponents. The general relation between an exponent polygon and the zero strips of an exponential sum goes back to P olya and Langer. Later developments include the work of Moreno and the treatment in Levin's monograph [source reference]. Bleher and Mallison give a particularly close general result: for convex-polygon exponential sums they locate all sufficiently…
Exact mathematical passage
\Dfun_N(z)=\frac1N\sum_{\eta^N=-1}e^{\eta z}
\sum_{\eta\in\Rcal_N}Use in source · 1 passage
Search aliases: ∑_η∈Rcal_N ∑_η∈ R_N sum eta in Rcal sum eta in mathcal
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 677
[Finite exponential representation] Let R_N={η∈ Cmathrel:η^N=-1}. Then D_N(z) = frac1N∑_η∈ R_Ne^η z. More generally, D_N,j(z) = fracρ^-jN ∑_ℓ=0^N-1 ω^-jℓe^ρω^ℓ z.
Exact mathematical passage
\Dfun_N(z) = \frac1N\sum_{\eta\in\Rcal_N}e^{\eta z}.
\sum_{\lambda}Use in source · 1 passage
Search aliases: ∑_λ sum lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2496
Logarithmic differentiation of the genus zero product gives -fracΔ_N'(w)Δ_N(w) = ∑_λ∈Λ_Nfrac1λ-w. The series is locally uniform because ∑_λ|λ|^-1<∞. Substitute w=z^N and use the chain rule.
Exact mathematical passage
\sum_{\lambda}|\lambda|^{-1}<\infty
\sum_{\mu\in\Lambda_{N,j}}Use in source · 1 passage
Search aliases: ∑_μ∈Λ_N,j sum mu in Lambda
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2380
[Products for the normalized sections] For every 0≤ j<N, the entire function Φ_N,j(w) = j!∑_n≥0fracw^n(Nn+j)! has order 1/N. If N≥2 and Λ_N,j is its nonzero zero multiset, then Φ_N,j(w) = ∏_μ∈Λ_N,j (1-frac wμ). For m≥1, put σ_N,j,m = ∑_μ∈Λ_N,jμ^-m. Then C_N,j,n(r) = frac(Nn)!n! B_n( rσ_N,j,1, r1!σ_N,j,2, …, r(n-1)!σ_N,j,n ).
Exact mathematical passage
\sigma_{N,j,m} = \sum_{\mu\in\Lambda_{N,j}}\mu^{-m}.
\sum_{\rho\in\Zcal_R}Use in source · 2 passages
Search aliases: ∑_ρ∈Zcal_R ∑_ρ∈ Z_R sum rho in Zcal sum rho in mathcal
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2607
rcle_expansion Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
P_{N,n}(r) = m! \sum_{\rho\in\Zcal_R} \sum_{\ell=1}^{r} a_{\rho,\ell} \binom{m+\ell-1}{\ell-1} \rho^{-m} + O_{R_*}(m!R_*^{-m}).
\sum_{\tau\in\Zcal_R^{\mathrm{orb}}}Use in source · 1 passage
Search aliases: ∑_τ∈Zcal_R^orb ∑_τ∈ Z_R^orb sum tau in Zcal mathrm sum tau in mathcal mathrm
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2672
[Orbit reduced nearest circle term] Assume the hypotheses of [source reference]. Choose one representative from each orbit of the nearest zeros under multiplication by ω, and call the resulting set Z_R^orb. With m=Nn, P_N,n(r) = · N m!binomm+r-1r-1 ∑_τ∈ Z_R^orb fracτ^-m(-τ D_N'(τ))^r · +O(m!m^r-2R^-m) +O_R_*(m!R_*^-m). For r=1, the first error term is absent.
Exact mathematical passage
\begin{aligned} P_{N,n}(r) ={}&N m!\binom{m+r-1}{r-1} \sum_{\tau\in\Zcal_R^{\mathrm{orb}}} \frac{\tau^{-m}}{(-\tau\Dfun_N'(\tau))^r} \\ &+O\left(m!m^{r-2}R^{-m}\right) +O_{R_*}\left(m!R_*^{-m}\right). \end{aligned}
\sum_\ellUse in source · 3 passages
Search aliases: ∑_ℓ sum ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 500
igin, the derivative matrix is diagonal with entries 1,λ,…,λ^N-1. For an explicit determinant proof, write each basis function as a finite sum of exponentials with exponents a_ℓ=λω^ℓ. Its derivative matrix factors as W(z)=V diag(e^a_ℓ z) C, V_kℓ=a_ℓ^k, where C is the constant matrix of Fourier weights. Thus det W(z)=e^z∑_ℓ a_ℓdet W(0)=det W(0), because N≥2 implies ∑_ℓω^ℓ=0. The product of the diagonal initial values is the stated power of λ.
Exact mathematical passage
\det W(z)=e^{z\sum_\ell a_\ell}\det W(0)=\det W(0),
\sum_{\ell=0}^3Use in source · 1 passage
Search aliases: ∑_ℓ=0^3 sum ell
Use in source · root_filter.tex, line 11
000) -- (3.95000,2.00000); node[] at (4.40000,2.00000) 1; draw[->,line width=1pt] (2.50000,2.00000) -- (2.50000,3.45000); node[] at (2.50000,3.90000) i; draw[->,line width=1pt] (2.50000,2.00000) -- (1.05000,2.00000); node[] at (0.60000,2.00000) -1; draw[->,line width=1pt] (2.50000,2.00000) -- (2.50000,0.55000); node[] at (2.50000,0.10000) -i; node[] at (2.50000,-0.30000) frac14∑_ℓ=0^3 i^(m-j)ℓ; node[anchor=east] at (6.10000,3.70000) j=0; draw[fill=black!0,draw=black!40,line width=.35pt] (6.45000…
Exact mathematical passage
\frac14\sum_{\ell=0}^3 i^{(m-j)\ell}
\sum_{\ell=0}^{3}Use in source · 2 passages
Search aliases: ∑_ℓ=0^3 sum ell
Use in source · atlas-pages.tex, line 46
vpage3A root filter is an exact voteCYCLIC PROJECTIONTraditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5). For N=4, write ω=i. To extract residue j, evaluate the function at four rotated arguments and attach the phase i^-jℓ: ( B_4,jF)(z)=frac14∑_ℓ=0^3i^-jℓF(i^ℓ z). A monomial a_mz^m is multiplied by the finite average of i^(m-j)ℓ. Matching phases all vote 1. Nonmatching phases cancel exactly. vfigroot_filter headAn example that can be checked coefficient by coefficient Take F(z)=∑_m…
Exact mathematical passage
(\mathcal B_{4,j}F)(z)=\frac14\sum_{\ell=0}^{3}i^{-j\ell}F(i^\ell z).
\sum_{\ell=0}^{N-1}Use in source · 3 passages
Search aliases: ∑_ℓ=0^N-1 sum ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 338
Fix N≥1 and write ω=e^2π i/N. Let Ω be a disk centered at the origin, with the whole plane allowed, and assume that Ω is invariant under multiplication by ω. For a holomorphic function F on Ω, define ( B_N,jF)(z) = frac1N∑_ℓ=0^N-1ω^-jℓF(ω^ℓ z), 0≤ j<N. The same formula is valid for formal power series.
Exact mathematical passage
(\Bcal_{N,j}F)(z) = \frac1N\sum_{\ell=0}^{N-1}\omega^{-j\ell}F(\omega^\ell z), \qquad 0\leq j<N.
\sum_{\ell=1}^{r}Use in source · 2 passages
Search aliases: ∑_ℓ=1^r sum ell
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2597
[Nearest-pole coefficient expansion] Let r∈ Z_>0. Suppose the zeros of D_N of least modulus R are simple and form the finite set Z_R. Suppose the next zero modulus is R_2>R. At each ρ∈ Z_R, write the principal part D_N(z)^-r = ∑_ℓ=1^r a_ρ,ℓ (1-frac zρ)^-ℓ + O(1). Then, for every R_* with R<R_*<R_2 and m=Nn, as n→∞, P_N,n(r) = m! ∑_ρ∈ Z_R ∑_ℓ=1^r a_ρ,ℓ binomm+ℓ-1ℓ-1 ρ^-m + O_R_*(m!R_*^-m). The leading Laurent coefficient is a_ρ,r = (-ρ D_N'(ρ))^-r.
Exact mathematical passage
\Dfun_N(z)^{-r} = \sum_{\ell=1}^{r} a_{\rho,\ell} \left(1-\frac z\rho\right)^{-\ell} + O(1).
\inftyUse in source · 9 passages
Search aliases: ∞ infty
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1729
[Order from Taylor coefficients] If f(w)=∑_n≥0a_nw^n is an entire function with infinitely many nonzero coefficients, then ord f = limsup_substackn→∞ a_n≠0 fracnlog nlog(1/|a_n|). Canonical products for entire functions of order below one have genus zero. We use these standard facts in the form given in [source reference].
Exact mathematical passage
\ord f = \limsup_{\substack{n\to\infty\\a_n\ne0}} \frac{n\log n}{\log(1/|a_n|)}.
\simUse in source · 4 passages
Search aliases: ∼ sim
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 2701
[Secant asymptotic] For N=2 and r=1, P_2,n(1) ∼ 2(2n)!(frac2π)^2n+1.
Exact mathematical passage
P_{2,n}(1) \sim 2(2n)!\left(\frac2\pi\right)^{2n+1}.
\equivUse in source · 12 passages
Search aliases: ≡ equiv
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 352
[Root filter] If F(z)=∑_m≥0a_mz^m, then B_N,jF(z) = ∑_m≡ jbmod Na_mz^m. The operators satisfy B_N,i B_N,j=δ_ij B_N,j and ∑_j=0^N-1 B_N,j=I.
Exact mathematical passage
\Bcal_{N,j}F(z) = \sum_{m\equiv j\bmod N}a_mz^m.
\leUse in source · 10 passages
Search aliases: ≤ le
Use in source · atlas-pages.tex, line 113
a wrap. Reading that coefficient gives the displayed addition law. headCompare the unsigned model In C[ξ]/(ξ^3-1), the same table has no minus signs. Its exponential coordinates are the unsigned sections E_3,j. Thus the difference between the two differential systems is a single algebraic relation, not a change in how coefficients are grouped. For general N, the carry factor is (-1)^lfloor(i+j)/Nrfloor, 0≤ i,j<N. Only zero or one wrap is possible in a product of two coordinate basis elements. ta…
Exact mathematical passage
(-1)^{\lfloor(i+j)/N\rfloor},\qquad 0\le i,j<N.
\geUse in source · 26 passages
Search aliases: ≥ ge
Use in source · atlas-pages.tex, line 29
midrule E_N,j(z) · displaystyle∑_n≥0fracz^Nn+j(Nn+j)! selects one residue class of the exponential.
Exact mathematical passage
\displaystyle\sum_{n\ge0}\frac{z^{Nn+j}}{(Nn+j)!}
\cdotsUse in source · 12 passages
Search aliases: ⋯ cdots
Use in source · Read passage · Paper_IV_Traditional_Final.tex, line 1106
[Logarithmic Newton recurrence] Set b_N,n(r)=fracP_N,n(r)(Nn)!. Then b_N,0(r)=1 and for every n≥1, nb_N,n(r) = r∑_k=1^nκ_N,kb_N,n-k(r). Equivalently, P_N,n(r) = frac r n ∑_k=1^n frac(Nn)!(N(n-k))! κ_N,kP_N,n-k(r). If p_k=rκ_N,k, then, for n≥1, b_N,n(r) = frac1n! det p_1 · -1 · 0 · ⋯ · 0 p_2 · p_1 · -2 · ⋯ · 0 p_3 · p_2 · p_1 · ddots · vdots vdots · vdots · ddots · ddots · -(n-1) p_n · p_n-1 · ⋯ · p_2 · p_1 .
Exact mathematical passage
b_{N,n}(r) = \frac1{n!} \det \begin{pmatrix} p_1&-1&0&\cdots&0\\ p_2&p_1&-2&\cdots&0\\ p_3&p_2&p_1&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&-(n-1)\\ p_n&p_{n-1}&\cdots&p_2&p_1 \end{pmatrix}.
\longmapstoUse in source · 2 passages
Search aliases: ⟼ longmapsto
Use in source · branch_domain.tex, line 12
000,4.65000) fixed-log domain about 0; draw[fill=white,line width=1pt] (4.40000,2.30000) circle (0.09); node[] at (5.10000,1.45000) ρ=π/2; draw[->,line width=1.1pt] (4.88,2.3) arc[start angle=0,end angle=330,radius=.48]; node[] at (2.80000,0.10000) the loop is not inside the disk; node[] at (9.20000,4.10000) D(z)∼ D^prime(ρ)(z-ρ); node[] at (9.20000,2.90000) log D⟼log D+2π i; node[] at (9.20000,1.65000) D^-r⟼ e^-2π i rD^-r; node[] at (9.20000,0.40000) r=1/2: -1 r∈ Z: 1;
Exact mathematical passage
\log D\longmapsto\log D+2\pi i
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- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/pole_asymptotic.tex · 15 mathematical passages
- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/product.tex · 14 mathematical passages
- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/riccati.tex · 6 mathematical passages
- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/root_filter.tex · 22 mathematical passages
- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/signed_addition.tex · 18 mathematical passages
- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/signed_shift.tex · 24 mathematical passages
- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/theta.tex · 11 mathematical passages
- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/three_views.tex · 24 mathematical passages
- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/triangular_inverse.tex · 30 mathematical passages
- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/two_term.tex · 10 mathematical passages
- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/worked_circuit.tex · 6 mathematical passages
- papers/paper-iv/current/Paper_IV_Verification_Closeout_v1.1.0/publication_baseline/editable/traditional/visual_graphics/zeta_plane.tex · 8 mathematical passages
- papers/paper-iv/current/README.md · 0 mathematical passages
- publication/paper-iv-visual-reconstruction/pages.tex · 327 mathematical passages
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