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0}","kind":"use","occurrences":[9]},{"id":"symbol-6b8e5ecd78509ed8","tex":"\\sum_{m\\ge L}","aliases":"∑_m≥ L sum ge","display":"\\sum_{m\\ge L}","kind":"use","occurrences":[1320,1982]},{"id":"symbol-a0a9cffbc383b69e","tex":"\\sum_{m\\ge1}","aliases":"∑_m≥1 sum ge","display":"\\sum_{m\\ge1}","kind":"use","occurrences":[1066,1184,1217,1728,1846,1879]},{"id":"symbol-4d609c0fc3f9c1f9","tex":"\\sum_{n=0}^{90}","aliases":"∑_n=0^90 sum","display":"\\sum_{n=0}^{90}","kind":"use","occurrences":[1296,1958]},{"id":"symbol-657057157b7882f4","tex":"\\sum_{n\\geq 0}","aliases":"∑_n≥ 0 sum geq","display":"\\sum_{n\\geq 0}","kind":"use","occurrences":[14,19]},{"id":"symbol-f6ef3ac4ec7b2013","tex":"\\sum_{n\\ge0}","aliases":"∑_n≥0 sum ge","display":"\\sum_{n\\ge0}","kind":"use","occurrences":[1054,1060,1071,1266,1716,1722,1733,1928]},{"id":"symbol-1f8f0dbb2884f06c","tex":"\\sum_{q=0}^m","aliases":"∑_q=0^m sum","display":"\\sum_{q=0}^m","kind":"use","occurrences":[1263,1925]},{"id":"symbol-5be06f71bbc1552b","tex":"\\sum_{q=0}^{m}","aliases":"∑_q=0^m sum","display":"\\sum_{q=0}^{m}","kind":"use","occurrences":[433,442]},{"id":"symbol-113d2a04b2350877","tex":"\\sum_{q=0}^{n}","aliases":"∑_q=0^n sum","display":"\\sum_{q=0}^{n}","kind":"use","occurrences":[298,300,392,414,486,1208,1870]},{"id":"symbol-2636c510c23e69cc","tex":"\\sum_{q=0}^{n-1}","aliases":"∑_q=0^n-1 sum","display":"\\sum_{q=0}^{n-1}","kind":"use","occurrences":[394]},{"id":"symbol-eb07f8d7b26ea999","tex":"\\sum_{q\\geq0}","aliases":"∑_q≥0 sum geq","display":"\\sum_{q\\geq0}","kind":"use","occurrences":[446,447]},{"id":"symbol-d4cedeb30578efb5","tex":"\\sum_{\\substack{0\\leq i,j0$, the last quotient at $z=0$ is interpreted by its removable\ncontinuation, namely the first expression in the display.\nWrite\n\\[\n\\Cfun_{N,j}(z)\n=\n\\sum_{n\\geq0}C_{N,j}(n)\\frac{z^{Nn}}{(Nn)!}.\n\\]\n\\end{definition}","or $0\\leq j0$, the last quotient at $z=0$ is interpreted by its removable\ncontinuation, namely the first expression in the display.\nWrite\n\\[\n\\Cfun_{N,j}(z)\n=\n\\sum_{n\\geq0}C_{N,j}(n)\\frac{z^{Nn}}{(Nn)!}.\n\\]\n\\end{definition}","\\begin{theorem}[Triangular inverse system]\\label{thm:triangular_inverse}\nThe coefficients are determined by $C_{N,j}(0)=1$ and\n\\[\n\\sum_{q=0}^{n}\n\\binom{Nn+j}{Nq}\nC_{N,j}(q)\n=\n\\delta_{n0}.\n\\]\nFor $n\\geq1$,\n\\[\nC_{N,j}(n)\n=\n-\n\\frac{\n\\displaystyle\\sum_{q=0}^{n-1}\n\\binom{Nn+j}{Nq}C_{N,j}(q)\n}{\n\\displaystyle\\binom{Nn+j}{Nn}\n}.\n\\]\n\\end{theorem}","\\begin{proof}\nMultiply the defining series for $\\Cfun_{N,j}$ and $\\Efun_{N,j}$. Their product is $z^j/j!$. At degree $Nn+j$, multiplication by $(Nn+j)!$ gives the stated binomial coefficient. The diagonal coefficient is nonzero, so the system is triangular.\n\\end{proof}","\\begin{definition}\\label{def:residue_section_power}\nFor $r\\in\\CC$, define formal coefficients $C_{N,j,n}(r)$ by\n\\[\n\\Cfun_{N,j}(z)^r\n=\n\\sum_{n\\geq0}C_{N,j,n}(r)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nOn a simply connected region containing $0$ and avoiding the zeros of $\\Phi_{N,j}(z^N)$, interpret this power as $e^{rL}$, where $L$ is the holomorphic logarithm of $\\Cfun_{N,j}$ normalized by $L(0)=0$.\n\\end{definition}","\\begin{theorem}[Parameter-addition law for residue-section reciprocals]\\label{thm:residue_section_parameter_addition}\nFor each $n\\geq1$, the coefficient $C_{N,j,n}(r)$ is a polynomial in $r$ of exact degree $n$. Its leading coefficient is\n\\[\n\\frac{(Nn)!}{n!}\n\\left(-\\frac{j!}{(N+j)!}\\right)^n.\n\\]\nIt satisfies\n\\[\nC_{N,j,n}(r+s)\n=\n\\sum_{q=0}^{n}\n\\binom{Nn}{Nq}\nC_{N,j,q}(r)C_{N,j,n-q}(s).\n\\]\nIn particular,\n\\[\nC_{N,j,n}(1)=C_{N,j}(n).\n\\]\n\\end{theorem}","eorem}[Parameter-addition law for residue-section reciprocals]\\label{thm:residue_section_parameter_addition}\nFor each $n\\geq1$, the coefficient $C_{N,j,n}(r)$ is a polynomial in $r$ of exact degree $n$. Its leading coefficient is\n\\[\n\\frac{(Nn)!}{n!}\n\\left(-\\frac{j!}{(N+j)!}\\right)^n.\n\\]\nIt satisfies\n\\[\nC_{N,j,n}(r+s)\n=\n\\sum_{q=0}^{n}\n\\binom{Nn}{Nq}\nC_{N,j,q}(r)C_{N,j,n-q}(s).\n\\]\nIn particular,\n\\[\nC_{N,j,n}(1)=C_{N,j}(n).\n\\]\n\\end{theorem}","\\begin{proof}\nThe unit series $\\Phi_{N,j}(w)^{-1}$ has a formal logarithm with zero constant term. Its linear term is\n\\[\n-\\frac{j!}{(N+j)!}w.\n\\]\nExponentiation after multiplication by $r$ shows that the coefficient of $w^n$ is a polynomial of degree at most $n$. The $n$th power of the linear logarithmic term gives the displayed nonzero leading coefficient, so the degree is exactly $n$. Multiplication of the powers with parameters $r$ and $s$ gives the coefficient identity.\nMore explicitly, before the factorial normalization the coefficient of\n$r^nw^n$ is\n\\[\n\\frac1{n!}\\left(-\\frac{j!}{(N+j)!}\\right)^n;\n\\]\nthus multiplication by $(Nn)!$ gives the displayed leading coefficient.\n\\end{proof}",",j}(w)^{-1}$ has a formal logarithm with zero constant term. Its linear term is\n\\[\n-\\frac{j!}{(N+j)!}w.\n\\]\nExponentiation after multiplication by $r$ shows that the coefficient of $w^n$ is a polynomial of degree at most $n$. The $n$th power of the linear logarithmic term gives the displayed nonzero leading coefficient, so the degree is exactly $n$. Multiplication of the powers with parameters $r$ and $s$ gives the coefficient identity.\nMore explicitly, before the factorial normalization the coefficient of\n$r^nw^n$ is\n\\[\n\\frac1{n!}\\left(-\\frac{j!}{(N+j)!}\\right)^n;\n\\]\nthus multiplication by $(Nn)!$ gives the displayed leading coefficient.\n\\end{proof}","-1}$ has a formal logarithm with zero constant term. Its linear term is\n\\[\n-\\frac{j!}{(N+j)!}w.\n\\]\nExponentiation after multiplication by $r$ shows that the coefficient of $w^n$ is a polynomial of degree at most $n$. The $n$th power of the linear logarithmic term gives the displayed nonzero leading coefficient, so the degree is exactly $n$. Multiplication of the powers with parameters $r$ and $s$ gives the coefficient identity.\nMore explicitly, before the factorial normalization the coefficient of\n$r^nw^n$ is\n\\[\n\\frac1{n!}\\left(-\\frac{j!}{(N+j)!}\\right)^n;\n\\]\nthus multiplication by $(Nn)!$ gives the displayed leading coefficient.\n\\end{proof}","tiation after multiplication by $r$ shows that the coefficient of $w^n$ is a polynomial of degree at most $n$. The $n$th power of the linear logarithmic term gives the displayed nonzero leading coefficient, so the degree is exactly $n$. Multiplication of the powers with parameters $r$ and $s$ gives the coefficient identity.\nMore explicitly, before the factorial normalization the coefficient of\n$r^nw^n$ is\n\\[\n\\frac1{n!}\\left(-\\frac{j!}{(N+j)!}\\right)^n;\n\\]\nthus multiplication by $(Nn)!$ gives the displayed leading coefficient.\n\\end{proof}","multiplication by $r$ shows that the coefficient of $w^n$ is a polynomial of degree at most $n$. The $n$th power of the linear logarithmic term gives the displayed nonzero leading coefficient, so the degree is exactly $n$. Multiplication of the powers with parameters $r$ and $s$ gives the coefficient identity.\nMore explicitly, before the factorial normalization the coefficient of\n$r^nw^n$ is\n\\[\n\\frac1{n!}\\left(-\\frac{j!}{(N+j)!}\\right)^n;\n\\]\nthus multiplication by $(Nn)!$ gives the displayed leading coefficient.\n\\end{proof}","l of degree at most $n$. The $n$th power of the linear logarithmic term gives the displayed nonzero leading coefficient, so the degree is exactly $n$. Multiplication of the powers with parameters $r$ and $s$ gives the coefficient identity.\nMore explicitly, before the factorial normalization the coefficient of\n$r^nw^n$ is\n\\[\n\\frac1{n!}\\left(-\\frac{j!}{(N+j)!}\\right)^n;\n\\]\nthus multiplication by $(Nn)!$ gives the displayed leading coefficient.\n\\end{proof}","\\begin{definition}\\label{def:block_appell}\nDefine polynomials $E_m^{(N,j,r)}(x)$ by\n\\[\n\\Cfun_{N,j}(z)^re^{xz}\n=\n\\sum_{m\\geq0}E_m^{(N,j,r)}(x)\\frac{z^m}{m!}.\n\\]\n\\end{definition}","\\begin{theorem}[Appell laws]\\label{thm:appell_laws}\nFor $m\\geq1$, the polynomials satisfy\n\\[\n\\frac{d}{dx}E_m^{(N,j,r)}(x)\n=\nmE_{m-1}^{(N,j,r)}(x)\n\\]\nand, for $m\\geq0$,\n\\[\nE_m^{(N,j,r)}(x+y)\n=\n\\sum_{q=0}^{m}\n\\binom mq\nE_q^{(N,j,r)}(x)y^{m-q}.\n\\]\n\\end{theorem}","\\begin{proof}\nDifferentiate the generating function with respect to $x$ for the first identity. Multiply it by $e^{yz}$ and compare coefficients for the second.\n\\end{proof}","\\begin{proposition}[Explicit Appell polynomial]\\label{prop:explicit_appell}\nFor every $m\\geq0$,\n\\[\nE_m^{(N,j,r)}(x)\n=\n\\sum_{0\\leq n\\leq m/N}\n\\binom{m}{Nn}\nC_{N,j,n}(r)x^{m-Nn}.\n\\]\n\\end{proposition}","\\begin{proof}\nMultiply the block series for $\\Cfun_{N,j}^r$ by the exponential series and collect the coefficient of total degree $m$.\n\\end{proof}","\\begin{theorem}[Joint Appell convolution in the space and order parameters]\\label{thm:appell_two_parameter_convolution}\nFor all $r,s,x,y\\in\\CC$ and $m\\geq0$,\n\\[\nE_m^{(N,j,r+s)}(x+y)\n=\n\\sum_{q=0}^{m}\n\\binom mq\nE_q^{(N,j,r)}(x)\nE_{m-q}^{(N,j,s)}(y).\n\\]\n\\end{theorem}","\\begin{proof}\nMultiply the generating functions\n\\[\n\\Cfun_{N,j}(z)^re^{xz}\n\\quad\\hbox{and}\\quad\n\\Cfun_{N,j}(z)^se^{yz}.\n\\]\nTheir product is the generating function with parameters $r+s$ and $x+y$. Coefficient comparison gives the formula.\n\\end{proof}","\\begin{proposition}[Formal differential-operator representation on polynomials]\\label{prop:appell_operator_form}\nLet a formal series $A(z)=\\sum_{q\\geq0}a_qz^q/q!$ act on polynomials by\n\\[\nA(\\partial_x)f(x)\n=\n\\sum_{q\\geq0}\\frac{a_q}{q!}\\partial_x^qf(x).\n\\]\nThe power $r$ is applied to the formal symbol before the substitution $z=\\partial_x$. Then\n\\[\nE_m^{(N,j,r)}(x)\n=\n\\Cfun_{N,j}(\\partial_x)^r x^m.\n\\]\nThe inverse relation is\n\\[\nx^m\n=\n\\Phi_{N,j}(\\partial_x^N)^r\nE_m^{(N,j,r)}(x).\n\\]\nBoth sums are finite on every polynomial.\n\\end{proposition}","ferential-operator representation on polynomials]\\label{prop:appell_operator_form}\nLet a formal series $A(z)=\\sum_{q\\geq0}a_qz^q/q!$ act on polynomials by\n\\[\nA(\\partial_x)f(x)\n=\n\\sum_{q\\geq0}\\frac{a_q}{q!}\\partial_x^qf(x).\n\\]\nThe power $r$ is applied to the formal symbol before the substitution $z=\\partial_x$. Then\n\\[\nE_m^{(N,j,r)}(x)\n=\n\\Cfun_{N,j}(\\partial_x)^r x^m.\n\\]\nThe inverse relation is\n\\[\nx^m\n=\n\\Phi_{N,j}(\\partial_x^N)^r\nE_m^{(N,j,r)}(x).\n\\]\nBoth sums are finite on every polynomial.\n\\end{proposition}","\\begin{proof}\nApplying $A(\\partial_x)$ to $e^{xz}$ multiplies it by $A(z)$. Taking\n$A=\\Cfun_{N,j}^r$ gives the first identity by the defining generating\nfunction. Constant coefficient operators on polynomials satisfy\n\\[\n(AB)(\\partial_x)=A(\\partial_x)B(\\partial_x),\n\\]\nbecause all derivative sums are finite and powers of $\\partial_x$\ncommute. The two formal symbols\n\\[\n\\Cfun_{N,j}(z)^r=\\Phi_{N,j}(z^N)^{-r}\n\\quad\\hbox{and}\\quad\n\\Phi_{N,j}(z^N)^r\n\\]\nmultiply to one. Their substitutions at $z=\\partial_x$ are therefore\ninverse constant coefficient operators on polynomials, which proves\nthe second identity.\n\\end{proof}","N,j}^r$ gives the first identity by the defining generating\nfunction. Constant coefficient operators on polynomials satisfy\n\\[\n(AB)(\\partial_x)=A(\\partial_x)B(\\partial_x),\n\\]\nbecause all derivative sums are finite and powers of $\\partial_x$\ncommute. The two formal symbols\n\\[\n\\Cfun_{N,j}(z)^r=\\Phi_{N,j}(z^N)^{-r}\n\\quad\\hbox{and}\\quad\n\\Phi_{N,j}(z^N)^r\n\\]\nmultiply to one. Their substitutions at $z=\\partial_x$ are therefore\ninverse constant coefficient operators on polynomials, which proves\nthe second identity.\n\\end{proof}","\\begin{corollary}[Appell symmetry]\\label{cor:appell_symmetry}\nEach polynomial $E_m^{(N,j,r)}$ is monic of degree $m$, and\n\\[\nE_m^{(N,j,r)}(\\omega x)\n=\n\\omega^mE_m^{(N,j,r)}(x).\n\\]\n\\end{corollary}","\\begin{proof}\nThe term $n=0$ in \\cref{prop:explicit_appell} is $x^m$. Every other exponent is $m-Nn$, so replacement of $x$ by $\\omega x$ multiplies every term by $\\omega^m$.\n\\end{proof}","\\begin{example}\\label{ex:euler_boundaries}\nFor $(N,j)=(2,0)$,\n\\[\n\\Cfun_{2,0}(z)=\\sech z.\n\\]\nFor $(N,j)=(2,1)$,\n\\[\n\\Cfun_{2,1}(z)=\\frac z{\\sinh z}.\n\\]\nThe first case is a scaled and translated Euler family. In fact, if $E_m(x)$ denotes the classical Euler polynomial, then\n\\[\nE_m^{(2,0,1)}(x)\n=\n2^mE_m\\left(\\frac{x+1}{2}\\right).\n\\]\nIf $B_m(x)$ denotes the classical Bernoulli polynomial, the second case has\nthe exact identification\n\\[\nE_m^{(2,1,1)}(x)\n=\n2^mB_m\\left(\\frac{x+1}{2}\\right),\n\\]\nobtained by substituting $t=2z$ in\n$t e^{yt}/(e^t-1)$.\n\\end{example}","daries}\nFor $(N,j)=(2,0)$,\n\\[\n\\Cfun_{2,0}(z)=\\sech z.\n\\]\nFor $(N,j)=(2,1)$,\n\\[\n\\Cfun_{2,1}(z)=\\frac z{\\sinh z}.\n\\]\nThe first case is a scaled and translated Euler family. In fact, if $E_m(x)$ denotes the classical Euler polynomial, then\n\\[\nE_m^{(2,0,1)}(x)\n=\n2^mE_m\\left(\\frac{x+1}{2}\\right).\n\\]\nIf $B_m(x)$ denotes the classical Bernoulli polynomial, the second case has\nthe exact identification\n\\[\nE_m^{(2,1,1)}(x)\n=\n2^mB_m\\left(\\frac{x+1}{2}\\right),\n\\]\nobtained by substituting $t=2z$ in\n$t e^{yt}/(e^t-1)$.\n\\end{example}","n_{2,1}(z)=\\frac z{\\sinh z}.\n\\]\nThe first case is a scaled and translated Euler family. In fact, if $E_m(x)$ denotes the classical Euler polynomial, then\n\\[\nE_m^{(2,0,1)}(x)\n=\n2^mE_m\\left(\\frac{x+1}{2}\\right).\n\\]\nIf $B_m(x)$ denotes the classical Bernoulli polynomial, the second case has\nthe exact identification\n\\[\nE_m^{(2,1,1)}(x)\n=\n2^mB_m\\left(\\frac{x+1}{2}\\right),\n\\]\nobtained by substituting $t=2z$ in\n$t e^{yt}/(e^t-1)$.\n\\end{example}","=\\frac z{\\sinh z}.\n\\]\nThe first case is a scaled and translated Euler family. In fact, if $E_m(x)$ denotes the classical Euler polynomial, then\n\\[\nE_m^{(2,0,1)}(x)\n=\n2^mE_m\\left(\\frac{x+1}{2}\\right).\n\\]\nIf $B_m(x)$ denotes the classical Bernoulli polynomial, the second case has\nthe exact identification\n\\[\nE_m^{(2,1,1)}(x)\n=\n2^mB_m\\left(\\frac{x+1}{2}\\right),\n\\]\nobtained by substituting $t=2z$ in\n$t e^{yt}/(e^t-1)$.\n\\end{example}","\\begin{definition}\\label{def:unsigned_denominator}\nSet\n\\[\nH_N(z)=\\Efun_{N,0}(z)\n=\n\\sum_{n\\geq0}\\frac{z^{Nn}}{(Nn)!}.\n\\]\nFor $u\\in\\CC$, define $\\Thetafun_{N,n}(u)$ by\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\n\\sum_{n\\geq0}\\Thetafun_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nThis is first a formal identity at the origin. Analytically, it is interpreted\non a common simply connected zero-free region with the branch of\n$\\log H_N$ normalized by $\\log H_N(0)=0$.\n\\end{definition}","el{def:unsigned_denominator}\nSet\n\\[\nH_N(z)=\\Efun_{N,0}(z)\n=\n\\sum_{n\\geq0}\\frac{z^{Nn}}{(Nn)!}.\n\\]\nFor $u\\in\\CC$, define $\\Thetafun_{N,n}(u)$ by\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\n\\sum_{n\\geq0}\\Thetafun_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nThis is first a formal identity at the origin. Analytically, it is interpreted\non a common simply connected zero-free region with the branch of\n$\\log H_N$ normalized by $\\log H_N(0)=0$.\n\\end{definition}","\\begin{theorem}[Theta reconstruction]\\label{thm:theta_reconstruction}\nFor every $n\\geq0$,\n\\[\n\\Thetafun_{N,n}(u)\n=\n(-1)^n\n\\sum_{q=0}^{n}\n\\binom{Nn}{Nq}\nP_{N,q}(1)u^{Nq}P_{N,n-q}(u).\n\\]\nThus $\\Thetafun_{N,n}$ is a polynomial of degree at most $Nn$.\n\\end{theorem}","\\begin{proof}\nChoose $\\rho$ with $\\rho^N=-1$. Then\n\\[\nH_N(z)=\\Dfun_N(\\rho z).\n\\]\nThe coefficient of $z^{Nn}$ in $H_N(z)^{-r}$ is therefore $(-1)^nP_{N,n}(r)/(Nn)!$. Apply this to $H_N(uz)^{-1}$ and $H_N(z)^{-u}$, then multiply the two block series.\n\\end{proof}","\\begin{proposition}[Trivial parameter factors]\\label{prop:theta_factors}\nFor $n\\geq1$, the polynomial $\\Thetafun_{N,n}(u)$ is divisible by $u$. If $N$ is even, it is also divisible by $u+1$.\n\\end{proposition}","\\begin{proof}\nAt $u=0$, the defining product is one. Hence every positive-index coefficient vanishes. If $N$ is even, then $H_N(-z)=H_N(z)$. At $u=-1$, the product is again one, so every positive-index coefficient vanishes there as well.\n\\end{proof}","\\begin{proposition}[Theta Newton recurrence]\\label{prop:theta_newton_recurrence}\nPut\n\\[\n\\theta_{N,n}(u)=\\frac{\\Thetafun_{N,n}(u)}{(Nn)!}.\n\\]\nThen $\\theta_{N,0}(u)=1$ and\n\\[\nn\\theta_{N,n}(u)\n=\n\\sum_{m=1}^{n}\n(-1)^m\\kappa_{N,m}\n\\bigl(u^{Nm}+u\\bigr)\n\\theta_{N,n-m}(u).\n\\]\nIn particular,\n\\[\n\\Thetafun_{N,1}(u)=-(u^N+u).\n\\]\n\\end{proposition}","\\begin{proof}\nWrite $w=z^N$. Since $H_N(z)=\\Delta_N(-w)$,\n\\[\n-\\log H_N(z)\n=\n\\sum_{m\\geq1}(-1)^m\\kappa_{N,m}\\frac{w^m}{m}.\n\\]\nThe logarithm of the theta generating function is therefore\n\\[\n\\sum_{m\\geq1}\n(-1)^m\\kappa_{N,m}\n\\bigl(u^{Nm}+u\\bigr)\n\\frac{w^m}{m}.\n\\]\nLogarithmic differentiation and coefficient comparison give the recurrence. The case $n=1$ uses $\\kappa_{N,1}=1/N!$.\n\\end{proof}","\\begin{example}\\label{ex:theta_even}\nFor $N=2$,\n\\[\n\\Thetafun_{2,1}(u)=-u(u+1)\n\\]\nand\n\\[\n\\Thetafun_{2,2}(u)\n=\nu(u+1)(5u^2+u+2).\n\\]\nThe two factors predicted by \\cref{prop:theta_factors} are visible.\n\\end{example}","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.\n\\end{readerroadmap}\n\\modelfigure{zero_orbits}{The 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 classify the finite exceptional set at other orders.}{fig:zero-orbits}","We now assume $N\\geq2$. Write\n\\[\n\\eta_k=e^{(2k+1)\\pi i/N},\n\\qquad 0\\leq k0$. The term $e^{\\eta_0z}=e^t$ has strictly larger real exponent than every other term. Hence\n\\[\n\\Dfun_N(t\\overline{\\eta_0})\n=\n\\frac1Ne^t(1+o(1)).\n\\]\nThe type is therefore one.\n\\end{proof}","\\begin{definition}\\label{def:indicator_polygon}\nFor $\\theta\\in\\mathbb R$, define\n\\[\nh_N(\\theta)\n=\n\\max_{0\\leq k0$ and $R>0$ such that, for each $\\theta\\in K$, one root $\\eta(\\theta)$ of $-1$ satisfies\n\\[\n\\Dfun_N(re^{i\\theta})\n=\n\\frac1N\n\\exp\\bigl(\\eta(\\theta)re^{i\\theta}\\bigr)\n\\left(1+O(e^{-\\delta r})\\right)\n\\]\nuniformly for $r\\geq R$. In particular, no sufficiently large zero can lie in those directions.\n\\end{theorem}","ator polygon and dominant sectors]\\label{thm:indicator_polygon}\nThe indicator of $\\Dfun_N$ equals $h_N$, with the limsup taken over radii where $\\Dfun_N(re^{i\\theta})\\ne0$:\n\\[\n\\limsup_{r\\to\\infty}\n\\frac1r\\log|\\Dfun_N(re^{i\\theta})|\n=\nh_N(\\theta).\n\\]\nIf $K$ is a compact set of directions disjoint from $\\mathfrak T_N$, then there are $\\delta>0$ and $R>0$ such that, for each $\\theta\\in K$, one root $\\eta(\\theta)$ of $-1$ satisfies\n\\[\n\\Dfun_N(re^{i\\theta})\n=\n\\frac1N\n\\exp\\bigl(\\eta(\\theta)re^{i\\theta}\\bigr)\n\\left(1+O(e^{-\\delta r})\\right)\n\\]\nuniformly for $r\\geq R$. In particular, no sufficiently large zero can lie in those directions.\n\\end{theorem}","ominant sectors]\\label{thm:indicator_polygon}\nThe indicator of $\\Dfun_N$ equals $h_N$, with the limsup taken over radii where $\\Dfun_N(re^{i\\theta})\\ne0$:\n\\[\n\\limsup_{r\\to\\infty}\n\\frac1r\\log|\\Dfun_N(re^{i\\theta})|\n=\nh_N(\\theta).\n\\]\nIf $K$ is a compact set of directions disjoint from $\\mathfrak T_N$, then there are $\\delta>0$ and $R>0$ such that, for each $\\theta\\in K$, one root $\\eta(\\theta)$ of $-1$ satisfies\n\\[\n\\Dfun_N(re^{i\\theta})\n=\n\\frac1N\n\\exp\\bigl(\\eta(\\theta)re^{i\\theta}\\bigr)\n\\left(1+O(e^{-\\delta r})\\right)\n\\]\nuniformly for $r\\geq R$. In particular, no sufficiently large zero can lie in those directions.\n\\end{theorem}","\\label{thm:indicator_polygon}\nThe indicator of $\\Dfun_N$ equals $h_N$, with the limsup taken over radii where $\\Dfun_N(re^{i\\theta})\\ne0$:\n\\[\n\\limsup_{r\\to\\infty}\n\\frac1r\\log|\\Dfun_N(re^{i\\theta})|\n=\nh_N(\\theta).\n\\]\nIf $K$ is a compact set of directions disjoint from $\\mathfrak T_N$, then there are $\\delta>0$ and $R>0$ such that, for each $\\theta\\in K$, one root $\\eta(\\theta)$ of $-1$ satisfies\n\\[\n\\Dfun_N(re^{i\\theta})\n=\n\\frac1N\n\\exp\\bigl(\\eta(\\theta)re^{i\\theta}\\bigr)\n\\left(1+O(e^{-\\delta r})\\right)\n\\]\nuniformly for $r\\geq R$. In particular, no sufficiently large zero can lie in those directions.\n\\end{theorem}","\\theta})\\ne0$:\n\\[\n\\limsup_{r\\to\\infty}\n\\frac1r\\log|\\Dfun_N(re^{i\\theta})|\n=\nh_N(\\theta).\n\\]\nIf $K$ is a compact set of directions disjoint from $\\mathfrak T_N$, then there are $\\delta>0$ and $R>0$ such that, for each $\\theta\\in K$, one root $\\eta(\\theta)$ of $-1$ satisfies\n\\[\n\\Dfun_N(re^{i\\theta})\n=\n\\frac1N\n\\exp\\bigl(\\eta(\\theta)re^{i\\theta}\\bigr)\n\\left(1+O(e^{-\\delta r})\\right)\n\\]\nuniformly for $r\\geq R$. In particular, no sufficiently large zero can lie in those directions.\n\\end{theorem}","\\begin{proof}\nFor $z=re^{i\\theta}$, the real part of the exponent $\\eta_kz$ is\n\\[\nr\\cos\\left(\\theta+\\frac{(2k+1)\\pi}{N}\\right).\n\\]\nThe triangle inequality gives the upper indicator bound $h_N(\\theta)$.","At a tie direction exactly two adjacent vertices attain the maximum. After rotation it is enough to take $\\theta=0$. With $c_N=\\cos(\\pi/N)$ and $s_N=\\sin(\\pi/N)$, the two dominant terms give\n\\[\n\\frac2N e^{c_Nr}\\cos(s_Nr)\n\\]\nand every other term is exponentially smaller. Along a sequence on which $|\\cos(s_Nr)|\\geq1/2$, the lower indicator bound is $c_N=h_N(0)$. Rotational symmetry treats every tie direction.\n\\end{proof}","The zero strings are governed by two adjacent exponents. Put\n\\[\na_N=\\frac\\pi N,\n\\qquad\ns_N=\\sin a_N,\n\\qquad\nc_N=\\cos a_N,\n\\]\nand\n\\[\nA_N=\\frac\\pi{s_N},\n\\qquad\n t_m=A_N\\left(m+\\frac12\\right).\n\\]\nFor $N\\geq3$, also put\n\\[\n\\gamma_N=c_N-\\cos(3a_N)>0.\n\\]","\\begin{lemma}[Two term dominance]\\label{lem:two_term_dominance}\nAssume $N\\geq3$. There are positive constants $Y$, $R$, and $C$ with the following properties.","If $z=x+iy$ lies in the sector $|\\arg z|\\leq a_N$, has $x\\geq R$, and is a zero of $\\Dfun_N$, then $|y|\\leq Y$.","On the strip $x\\geq R$ and $|y|\\leq Y$, define\n\\[\nR_N(z)\n=\n\\sum_{\\substack{\\eta^N=-1","On the strip $x\\geq R$ and $|y|\\leq Y$, define\n\\[\nR_N(z)\n=\n\\sum_{\\substack{\\eta^N=-1\\\\\n\\eta\\ne e^{ia_N},e^{-ia_N}}}\n\\exp\\bigl((\\eta-e^{-ia_N})z\\bigr).\n\\]\nThen\n\\[\nN e^{-e^{-ia_N}z}\\Dfun_N(z)\n=\n1+e^{2is_Nz}+R_N(z),\n\\]\nand, for $q=0,1,2$,\n\\[\n|R_N^{(q)}(z)|\n\\leq\nC e^{-\\gamma_Nx}.\n\\]\n\\end{lemma}","\\eta\\ne e^{ia_N},e^{-ia_N}}}\n\\exp\\bigl((\\eta-e^{-ia_N})z\\bigr).\n\\]\nThen\n\\[\nN e^{-e^{-ia_N}z}\\Dfun_N(z)\n=\n1+e^{2is_Nz}+R_N(z),\n\\]\nand, for $q=0,1,2$,\n\\[\n|R_N^{(q)}(z)|\n\\leq\nC e^{-\\gamma_Nx}.\n\\]\n\\end{lemma}","\\begin{proof}\nA direct projection estimate gives the stronger bound\n\\[\nY=\\frac{\\log(N-1)}{2s_N}\n\\]\nfor every zero in the sector, without a large-radius assumption.\nWrite $z=re^{i\\theta}$ with $0\\leq\\theta\\leq a_N$ and $r\\geq0$.\nFor $0\\leq\\ell\\leq N-2$, the angle\n$\\theta+(2\\ell+1)a_N$ lies between $a_N+\\theta$ and\n$2\\pi-a_N-\\theta$. Monotonicity of cosine on $[0,\\pi]$, followed by\nreflection about $\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound","\\begin{proof}\nA direct projection estimate gives the stronger bound\n\\[\nY=\\frac{\\log(N-1)}{2s_N}\n\\]\nfor every zero in the sector, without a large-radius assumption.\nWrite $z=re^{i\\theta}$ with $0\\leq\\theta\\leq a_N$ and $r\\geq0$.\nFor $0\\leq\\ell\\leq N-2$, the angle\n$\\theta+(2\\ell+1)a_N$ lies between $a_N+\\theta$ and\n$2\\pi-a_N-\\theta$. Monotonicity of cosine on $[0,\\pi]$, followed by\nreflection about $\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","begin{proof}\nA direct projection estimate gives the stronger bound\n\\[\nY=\\frac{\\log(N-1)}{2s_N}\n\\]\nfor every zero in the sector, without a large-radius assumption.\nWrite $z=re^{i\\theta}$ with $0\\leq\\theta\\leq a_N$ and $r\\geq0$.\nFor $0\\leq\\ell\\leq N-2$, the angle\n$\\theta+(2\\ell+1)a_N$ lies between $a_N+\\theta$ and\n$2\\pi-a_N-\\theta$. Monotonicity of cosine on $[0,\\pi]$, followed by\nreflection about $\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","ojection estimate gives the stronger bound\n\\[\nY=\\frac{\\log(N-1)}{2s_N}\n\\]\nfor every zero in the sector, without a large-radius assumption.\nWrite $z=re^{i\\theta}$ with $0\\leq\\theta\\leq a_N$ and $r\\geq0$.\nFor $0\\leq\\ell\\leq N-2$, the angle\n$\\theta+(2\\ell+1)a_N$ lies between $a_N+\\theta$ and\n$2\\pi-a_N-\\theta$. Monotonicity of cosine on $[0,\\pi]$, followed by\nreflection about $\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","s_N}\n\\]\nfor every zero in the sector, without a large-radius assumption.\nWrite $z=re^{i\\theta}$ with $0\\leq\\theta\\leq a_N$ and $r\\geq0$.\nFor $0\\leq\\ell\\leq N-2$, the angle\n$\\theta+(2\\ell+1)a_N$ lies between $a_N+\\theta$ and\n$2\\pi-a_N-\\theta$. Monotonicity of cosine on $[0,\\pi]$, followed by\nreflection about $\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","he sector, without a large-radius assumption.\nWrite $z=re^{i\\theta}$ with $0\\leq\\theta\\leq a_N$ and $r\\geq0$.\nFor $0\\leq\\ell\\leq N-2$, the angle\n$\\theta+(2\\ell+1)a_N$ lies between $a_N+\\theta$ and\n$2\\pi-a_N-\\theta$. Monotonicity of cosine on $[0,\\pi]$, followed by\nreflection about $\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","e $z=re^{i\\theta}$ with $0\\leq\\theta\\leq a_N$ and $r\\geq0$.\nFor $0\\leq\\ell\\leq N-2$, the angle\n$\\theta+(2\\ell+1)a_N$ lies between $a_N+\\theta$ and\n$2\\pi-a_N-\\theta$. Monotonicity of cosine on $[0,\\pi]$, followed by\nreflection about $\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","eq\\theta\\leq a_N$ and $r\\geq0$.\nFor $0\\leq\\ell\\leq N-2$, the angle\n$\\theta+(2\\ell+1)a_N$ lies between $a_N+\\theta$ and\n$2\\pi-a_N-\\theta$. Monotonicity of cosine on $[0,\\pi]$, followed by\nreflection about $\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","2\\ell+1)a_N$ lies between $a_N+\\theta$ and\n$2\\pi-a_N-\\theta$. Monotonicity of cosine on $[0,\\pi]$, followed by\nreflection about $\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","a_N-\\theta$. Monotonicity of cosine on $[0,\\pi]$, followed by\nreflection about $\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","ine on $[0,\\pi]$, followed by\nreflection about $\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","\\pi$, therefore gives\n\\[\n\\Re(\\eta_\\ell z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","l z)\\leq r\\cos(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","s(a_N+\\theta).\n\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","\\]\nThe remaining vertex $e^{-ia_N}$ has projection $r\\cos(a_N-\\theta)$.\nTheir difference is at least $2s_Nr\\sin\\theta=2s_Ny$.\nIf $\\Dfun_N(z)=0$, divide its finite exponential sum by the term\nwith exponent $e^{-ia_N}z$ and apply the triangle inequality:\n\\[\n1\\leq(N-1)e^{-2s_Ny}.\n\\]\nThus $y\\leq Y$. Conjugation gives the same bound for $-y$ on the lower\nhalf sector. Since $N\\geq3$ and $s_N>0$, this $Y$ is positive.","On the resulting strip, the roots $e^{ia_N}$ and $e^{-ia_N}$ have\nreal part $c_N$, while each remaining root $\\eta=u+iv$ satisfies\n$u\\leq\\cos(3a_N)=c_N-\\gamma_N$. Hence\n\\[\n\\Re\\bigl((\\eta-e^{-ia_N})z\\bigr)\n\\leq-\\gamma_Nx+2Y.\n\\]\nDivision by $e^{e^{-ia_N}z}$ gives the displayed normalization.\nEach differentiation multiplies a remainder term by\n$\\eta-e^{-ia_N}$, whose modulus is at most two. Thus, for $q=0,1,2$,\n\\[\n|R_N^{(q)}(z)|\\leq(N-2)2^qe^{2Y}e^{-\\gamma_Nx}.\n\\]\nOne may take $R=1$ and $C=4(N-2)e^{2Y}$.\n\\end{proof}","the resulting strip, the roots $e^{ia_N}$ and $e^{-ia_N}$ have\nreal part $c_N$, while each remaining root $\\eta=u+iv$ satisfies\n$u\\leq\\cos(3a_N)=c_N-\\gamma_N$. Hence\n\\[\n\\Re\\bigl((\\eta-e^{-ia_N})z\\bigr)\n\\leq-\\gamma_Nx+2Y.\n\\]\nDivision by $e^{e^{-ia_N}z}$ gives the displayed normalization.\nEach differentiation multiplies a remainder term by\n$\\eta-e^{-ia_N}$, whose modulus is at most two. Thus, for $q=0,1,2$,\n\\[\n|R_N^{(q)}(z)|\\leq(N-2)2^qe^{2Y}e^{-\\gamma_Nx}.\n\\]\nOne may take $R=1$ and $C=4(N-2)e^{2Y}$.\n\\end{proof}","ng strip, the roots $e^{ia_N}$ and $e^{-ia_N}$ have\nreal part $c_N$, while each remaining root $\\eta=u+iv$ satisfies\n$u\\leq\\cos(3a_N)=c_N-\\gamma_N$. Hence\n\\[\n\\Re\\bigl((\\eta-e^{-ia_N})z\\bigr)\n\\leq-\\gamma_Nx+2Y.\n\\]\nDivision by $e^{e^{-ia_N}z}$ gives the displayed normalization.\nEach differentiation multiplies a remainder term by\n$\\eta-e^{-ia_N}$, whose modulus is at most two. Thus, for $q=0,1,2$,\n\\[\n|R_N^{(q)}(z)|\\leq(N-2)2^qe^{2Y}e^{-\\gamma_Nx}.\n\\]\nOne may take $R=1$ and $C=4(N-2)e^{2Y}$.\n\\end{proof}","$, while each remaining root $\\eta=u+iv$ satisfies\n$u\\leq\\cos(3a_N)=c_N-\\gamma_N$. Hence\n\\[\n\\Re\\bigl((\\eta-e^{-ia_N})z\\bigr)\n\\leq-\\gamma_Nx+2Y.\n\\]\nDivision by $e^{e^{-ia_N}z}$ gives the displayed normalization.\nEach differentiation multiplies a remainder term by\n$\\eta-e^{-ia_N}$, whose modulus is at most two. Thus, for $q=0,1,2$,\n\\[\n|R_N^{(q)}(z)|\\leq(N-2)2^qe^{2Y}e^{-\\gamma_Nx}.\n\\]\nOne may take $R=1$ and $C=4(N-2)e^{2Y}$.\n\\end{proof}","ach remaining root $\\eta=u+iv$ satisfies\n$u\\leq\\cos(3a_N)=c_N-\\gamma_N$. Hence\n\\[\n\\Re\\bigl((\\eta-e^{-ia_N})z\\bigr)\n\\leq-\\gamma_Nx+2Y.\n\\]\nDivision by $e^{e^{-ia_N}z}$ gives the displayed normalization.\nEach differentiation multiplies a remainder term by\n$\\eta-e^{-ia_N}$, whose modulus is at most two. Thus, for $q=0,1,2$,\n\\[\n|R_N^{(q)}(z)|\\leq(N-2)2^qe^{2Y}e^{-\\gamma_Nx}.\n\\]\nOne may take $R=1$ and $C=4(N-2)e^{2Y}$.\n\\end{proof}","\\begin{lemma}[Near cancellation of two exponentials]\\label{lem:near_cancellation}\nFor every $M>0$, there are constants $\\varepsilon_M>0$ and $C_M>0$ such that the following holds. If\n\\[\n|\\Re w|\\leq M\n\\]\nand\n\\[\n|1+e^w|\\leq\\varepsilon_M,\n\\]\nthen there is a unique integer $m$ with\n\\[\n|w-(2m+1)\\pi i|\n\\leq\nC_M|1+e^w|.\n\\]\n\\end{lemma}","\\begin{proof}\nIn fact, the constants $\\varepsilon_M=1/2$ and $C_M=3/2$ work\nfor every complex $w$, without the strip restriction. Put\n$u=-(1+e^w)$ and $L=\\log(1+u)=\\log(-e^w)$, using the principal\nlogarithm. The elementary logarithm estimate on $|u|\\leq1/2$ gives\n$|L|\\leq(3/2)|u|$. Since $e^{L+\\pi i}=e^w$, there is an integer\n$m$ such that\n\\[\nw=L+(2m+1)\\pi i.\n\\]\nThis gives the required distance bound. Distinct centers are separated\nby at least $2\\pi$, whereas two centers satisfying the bound would\nbe separated by at most $3|1+e^w|\\leq3/2$. Thus the integer is unique.\nThe formal proof uses the convergent logarithm estimate and the exact\nperiodicity theorem for the complex exponential.\n\\end{proof}","_M=1/2$ and $C_M=3/2$ work\nfor every complex $w$, without the strip restriction. Put\n$u=-(1+e^w)$ and $L=\\log(1+u)=\\log(-e^w)$, using the principal\nlogarithm. The elementary logarithm estimate on $|u|\\leq1/2$ gives\n$|L|\\leq(3/2)|u|$. Since $e^{L+\\pi i}=e^w$, there is an integer\n$m$ such that\n\\[\nw=L+(2m+1)\\pi i.\n\\]\nThis gives the required distance bound. Distinct centers are separated\nby at least $2\\pi$, whereas two centers satisfying the bound would\nbe separated by at most $3|1+e^w|\\leq3/2$. Thus the integer is unique.\nThe formal proof uses the convergent logarithm estimate and the exact\nperiodicity theorem for the complex exponential.\n\\end{proof}",". Put\n$u=-(1+e^w)$ and $L=\\log(1+u)=\\log(-e^w)$, using the principal\nlogarithm. The elementary logarithm estimate on $|u|\\leq1/2$ gives\n$|L|\\leq(3/2)|u|$. Since $e^{L+\\pi i}=e^w$, there is an integer\n$m$ such that\n\\[\nw=L+(2m+1)\\pi i.\n\\]\nThis gives the required distance bound. Distinct centers are separated\nby at least $2\\pi$, whereas two centers satisfying the bound would\nbe separated by at most $3|1+e^w|\\leq3/2$. Thus the integer is unique.\nThe formal proof uses the convergent logarithm estimate and the exact\nperiodicity theorem for the complex exponential.\n\\end{proof}","\\begin{theorem}[Large zero strings]\\label{thm:large_zero_strings}\nAssume $N\\geq3$. There are constants $m_0$, $C$, and $\\gamma>0$ and a sequence of positive real simple zeros $(\\tau_m)_{m\\geq m_0}$ of $\\Dfun_N$ such that\n\\[\n\\tau_m\n=\n t_m+O(e^{-\\gamma t_m}).\n\\]\nEvery zero of sufficiently large modulus is of the form\n\\[\n\\omega^k\\tau_m\n\\]\nfor a unique $0\\leq k0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter","Fix a large $m$, put $a=2is_N$, and consider the closed disk\n$B_m=\\{z:|z-t_m|\\leq 1/(2s_N)\\}$. The remainder bounds hold\nuniformly on these fixed-radius disks. For sufficiently large $m$,\n$|R_N|\\leq 1/2$ and $|R_N'|\\leq s_N/2$ there. Define\n\\[\n\\Phi_m(z)=t_m+\\frac{\\log(1+R_N(z))}{2is_N},\n\\]\nusing the principal logarithm. The estimate\n$|\\log(1+u)|\\leq(3/2)|u|$ shows that $\\Phi_m$ maps $B_m$ into\nitself. The derivative bound $|\\Phi_m'|\\leq1/2$ makes it a contraction.\nBanach's fixed-point theorem gives a unique fixed point.\nOn $B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is small","Fix a large $m$, put $a=2is_N$, and consider the closed disk\n$B_m=\\{z:|z-t_m|\\leq 1/(2s_N)\\}$. The remainder bounds hold\nuniformly on these fixed-radius disks. For sufficiently large $m$,\n$|R_N|\\leq 1/2$ and $|R_N'|\\leq s_N/2$ there. Define\n\\[\n\\Phi_m(z)=t_m+\\frac{\\log(1+R_N(z))}{2is_N},\n\\]\nusing the principal logarithm. The estimate\n$|\\log(1+u)|\\leq(3/2)|u|$ shows that $\\Phi_m$ maps $B_m$ into\nitself. The derivative bound $|\\Phi_m'|\\leq1/2$ makes it a contraction.\nBanach's fixed-point theorem gives a unique fixed point.\nOn $B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spac","Fix a large $m$, put $a=2is_N$, and consider the closed disk\n$B_m=\\{z:|z-t_m|\\leq 1/(2s_N)\\}$. The remainder bounds hold\nuniformly on these fixed-radius disks. For sufficiently large $m$,\n$|R_N|\\leq 1/2$ and $|R_N'|\\leq s_N/2$ there. Define\n\\[\n\\Phi_m(z)=t_m+\\frac{\\log(1+R_N(z))}{2is_N},\n\\]\nusing the principal logarithm. The estimate\n$|\\log(1+u)|\\leq(3/2)|u|$ shows that $\\Phi_m$ maps $B_m$ into\nitself. The derivative bound $|\\Phi_m'|\\leq1/2$ makes it a contraction.\nBanach's fixed-point theorem gives a unique fixed point.\nOn $B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spacing $\\pi/s_N$.\nEnlarging the strip if necessary, these disks li","Fix a large $m$, put $a=2is_N$, and consider the closed disk\n$B_m=\\{z:|z-t_m|\\leq 1/(2s_N)\\}$. The remainder bounds hold\nuniformly on these fixed-radius disks. For sufficiently large $m$,\n$|R_N|\\leq 1/2$ and $|R_N'|\\leq s_N/2$ there. Define\n\\[\n\\Phi_m(z)=t_m+\\frac{\\log(1+R_N(z))}{2is_N},\n\\]\nusing the principal logarithm. The estimate\n$|\\log(1+u)|\\leq(3/2)|u|$ shows that $\\Phi_m$ maps $B_m$ into\nitself. The derivative bound $|\\Phi_m'|\\leq1/2$ makes it a contraction.\nBanach's fixed-point theorem gives a unique fixed point.\nOn $B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spacing $\\pi/s_N$.\nEnlarging the strip if necessary, these disks lie inside the fundamental\nsector and the fixed strip for large $m$.","_N$, and consider the closed disk\n$B_m=\\{z:|z-t_m|\\leq 1/(2s_N)\\}$. The remainder bounds hold\nuniformly on these fixed-radius disks. For sufficiently large $m$,\n$|R_N|\\leq 1/2$ and $|R_N'|\\leq s_N/2$ there. Define\n\\[\n\\Phi_m(z)=t_m+\\frac{\\log(1+R_N(z))}{2is_N},\n\\]\nusing the principal logarithm. The estimate\n$|\\log(1+u)|\\leq(3/2)|u|$ shows that $\\Phi_m$ maps $B_m$ into\nitself. The derivative bound $|\\Phi_m'|\\leq1/2$ makes it a contraction.\nBanach's fixed-point theorem gives a unique fixed point.\nOn $B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spacing $\\pi/s_N$.\nEnlarging the strip if necessary, these disks lie inside the fundamental\nsector and the fixed strip for large $m$.","on these fixed-radius disks. For sufficiently large $m$,\n$|R_N|\\leq 1/2$ and $|R_N'|\\leq s_N/2$ there. Define\n\\[\n\\Phi_m(z)=t_m+\\frac{\\log(1+R_N(z))}{2is_N},\n\\]\nusing the principal logarithm. The estimate\n$|\\log(1+u)|\\leq(3/2)|u|$ shows that $\\Phi_m$ maps $B_m$ into\nitself. The derivative bound $|\\Phi_m'|\\leq1/2$ makes it a contraction.\nBanach's fixed-point theorem gives a unique fixed point.\nOn $B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spacing $\\pi/s_N$.\nEnlarging the strip if necessary, these disks lie inside the fundamental\nsector and the fixed strip for large $m$.","se fixed-radius disks. For sufficiently large $m$,\n$|R_N|\\leq 1/2$ and $|R_N'|\\leq s_N/2$ there. Define\n\\[\n\\Phi_m(z)=t_m+\\frac{\\log(1+R_N(z))}{2is_N},\n\\]\nusing the principal logarithm. The estimate\n$|\\log(1+u)|\\leq(3/2)|u|$ shows that $\\Phi_m$ maps $B_m$ into\nitself. The derivative bound $|\\Phi_m'|\\leq1/2$ makes it a contraction.\nBanach's fixed-point theorem gives a unique fixed point.\nOn $B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spacing $\\pi/s_N$.\nEnlarging the strip if necessary, these disks lie inside the fundamental\nsector and the fixed strip for large $m$.","+\\frac{\\log(1+R_N(z))}{2is_N},\n\\]\nusing the principal logarithm. The estimate\n$|\\log(1+u)|\\leq(3/2)|u|$ shows that $\\Phi_m$ maps $B_m$ into\nitself. The derivative bound $|\\Phi_m'|\\leq1/2$ makes it a contraction.\nBanach's fixed-point theorem gives a unique fixed point.\nOn $B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spacing $\\pi/s_N$.\nEnlarging the strip if necessary, these disks lie inside the fundamental\nsector and the fixed strip for large $m$.","}{2is_N},\n\\]\nusing the principal logarithm. The estimate\n$|\\log(1+u)|\\leq(3/2)|u|$ shows that $\\Phi_m$ maps $B_m$ into\nitself. The derivative bound $|\\Phi_m'|\\leq1/2$ makes it a contraction.\nBanach's fixed-point theorem gives a unique fixed point.\nOn $B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spacing $\\pi/s_N$.\nEnlarging the strip if necessary, these disks lie inside the fundamental\nsector and the fixed strip for large $m$.","/2)|u|$ shows that $\\Phi_m$ maps $B_m$ into\nitself. The derivative bound $|\\Phi_m'|\\leq1/2$ makes it a contraction.\nBanach's fixed-point theorem gives a unique fixed point.\nOn $B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spacing $\\pi/s_N$.\nEnlarging the strip if necessary, these disks lie inside the fundamental\nsector and the fixed strip for large $m$.","nt theorem gives a unique fixed point.\nOn $B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spacing $\\pi/s_N$.\nEnlarging the strip if necessary, these disks lie inside the fundamental\nsector and the fixed strip for large $m$.","B_m$, $|a(z-t_m)|\\leq1<\\pi$, so the principal logarithm inverts\nthe exponential and the fixed-point equation is equivalent to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spacing $\\pi/s_N$.\nEnlarging the strip if necessary, these disks lie inside the fundamental\nsector and the fixed strip for large $m$.","t to $G(z)=0$.\nMoreover,\n\\[\n|z-t_m|\\leq\\frac{3}{4s_N}|R_N(z)|=O(e^{-\\gamma_Nt_m}).\n\\]\nAt this zero,\n\\[\n|G'(z)|\\geq |a|\\,|1+R_N(z)|-|R_N'(z)|\\geq s_N/2>0,\n\\]\nso the zero is simple. The disks are mutually disjoint because their\ndiameter $1/s_N$ is smaller than the center spacing $\\pi/s_N$.\nEnlarging the strip if necessary, these disks lie inside the fundamental\nsector and the fixed strip for large $m$.","The localization estimate puts every sufficiently large zero in the\nfundamental sector into one of these disks. The zero sets of $G$ and\n$\\Dfun_N$ coincide because their prefactor is nonzero. Each disk is\ninvariant under complex conjugation, and $\\Dfun_N$ has real\ncoefficients. The conjugate of its unique zero is therefore the same\nzero. It is real, positive for large $m$, and simple by the\nderivative bound above. Denote it by $\\tau_m$.","sufficiently large zero in the\nfundamental sector into one of these disks. The zero sets of $G$ and\n$\\Dfun_N$ coincide because their prefactor is nonzero. Each disk is\ninvariant under complex conjugation, and $\\Dfun_N$ has real\ncoefficients. The conjugate of its unique zero is therefore the same\nzero. It is real, positive for large $m$, and simple by the\nderivative bound above. Denote it by $\\tau_m$.","The parts of the fundamental sector away from the positive tie ray,\nincluding its boundary rays, were already excluded uniformly.\nRotational invariance now gives every remaining large zero as\n$\\omega^k\\tau_m$. Distinct rays and the disjoint disks give uniqueness\nof $k$ and $m$. Taking $\\gamma=\\gamma_N$ proves the zero location in\nthe statement.","Differentiate the normalized expression. At $z=\\tau_m$,\n\\[\nG'(\\tau_m)\n=\n2is_Ne^{2is_N\\tau_m}+R_N'(\\tau_m)\n=\n-2is_N+O(e^{-\\gamma_N\\tau_m}).\n\\]\nSince $G(\\tau_m)=0$,\n\\[\n\\Dfun_N'(\\tau_m)\n=\n\\frac1N e^{e^{-ia_N}\\tau_m}G'(\\tau_m).\n\\]\nTaking absolute values gives the derivative asymptotic. Differentiating $\\Dfun_N(\\omega z)=\\Dfun_N(z)$ gives the rotation formula.\n\\end{proof}","\\begin{theorem}[Leading displacement term and derivative sign]\\label{thm:first_zero_correction}\nWith the notation of \\cref{lem:two_term_dominance,thm:large_zero_strings},\n\\[\n\\tau_m\n=\nt_m+\n\\frac{R_N(t_m)}{2is_N}\n+\nO(e^{-2\\gamma_Nt_m}).\n\\]\nThe quotient $R_N(t_m)/(2is_N)$ is real. One also has\n\\[\n\\Dfun_N'(\\tau_m)\n=\n(-1)^{m+1}\n\\frac{2s_N}{N}\ne^{c_N\\tau_m}\n\\left(1+O(e^{-\\gamma_N\\tau_m})\\right).\n\\]\n\\end{theorem}","\\begin{proof}\nPut $\\delta_m=\\tau_m-t_m$. The large zero theorem gives $\\delta_m=O(e^{-\\gamma_Nt_m})$. Since $e^{2is_Nt_m}=-1$,\n\\[\n1+e^{2is_N(t_m+\\delta_m)}\n=\n-2is_N\\delta_m+O(\\delta_m^2).\n\\]\nThe remainder estimate through the first derivative gives\n\\[\nR_N(t_m+\\delta_m)\n=\nR_N(t_m)+O(e^{-\\gamma_Nt_m}|\\delta_m|).\n\\]\nSubstitution in the zero equation gives the displacement formula.","At the real point $t_m$, the normalized value\n\\[\nN e^{-e^{-ia_N}t_m}\\Dfun_N(t_m)\n\\]\nis purely imaginary. The two dominant terms cancel there, so $R_N(t_m)$ is purely imaginary and the displayed correction is real.","For the derivative, use the formula in the proof of \\cref{thm:large_zero_strings} and\n\\[\ne^{-is_N\\tau_m}\n=\n(-1)^m(-i)\n\\left(1+O(e^{-\\gamma_N\\tau_m})\\right).\n\\]\nThe stated real sign follows.\n\\end{proof}","\\begin{corollary}[Leading correction for $N=3$]\\label{cor:N3_zero_correction}\nFor $N=3$,\n\\[\nt_m=\\frac{2\\pi}{\\sqrt3}\\left(m+\\frac12\\right)\n\\]\nand\n\\[\n\\tau_m\n=\nt_m+\n\\frac{(-1)^m}{\\sqrt3}e^{-3t_m/2}\n+\nO(e^{-3t_m}).\n\\]\n\\end{corollary}","\\begin{proof}\nThe only nondominant exponent is $-1$. Hence\n\\[\nR_3(t)\n=\n\\exp\\left(\\left(-\\frac32+\\frac{\\sqrt3}{2}i\\right)t\\right).\n\\]\nAt $t=t_m$, its phase is $i(-1)^m$. Insert this value into \\cref{thm:first_zero_correction}.\n\\end{proof}","\\begin{corollary}[The case $N=2$]\\label{cor:case_two_zeros}\nFor $N=2$,\n\\[\n\\Dfun_2(z)=\\cos z.\n\\]\nIts zeros are\n\\[\n\\pm\\left(m+\\frac12\\right)\\pi,\n\\qquad m\\geq0,\n\\]\nand all are simple. They agree with the formula $t_m=A_2(m+1/2)$.\n\\end{corollary}","\\begin{corollary}[Eventual simplicity and exact rays]\\label{cor:eventual_simplicity}\nFor every $N\\geq2$, all but finitely many zeros of $\\Dfun_N$ are simple. Every sufficiently large zero lies on one of the rays\n\\[\n\\omega^k(0,\\infty),\n\\qquad 0\\leq k0}$. Suppose the zeros of $\\Dfun_N$ of least modulus\n$R$ are simple and form the finite set $\\Zcal_R$. Suppose the next\nzero modulus is $R_2>R$. At each $\\rho\\in\\Zcal_R$, write the principal\npart\n\\[\n\\Dfun_N(z)^{-r}\n=\n\\sum_{\\ell=1}^{r}\n a_{\\rho,\\ell}\n\\left(1-\\frac z\\rho\\right)^{-\\ell}\n+\nO(1).\n\\]\nThen, for every $R_*$ with $R0}$. Suppose the zeros of $\\Dfun_N$ of least modulus\n$R$ are simple and form the finite set $\\Zcal_R$. Suppose the next\nzero modulus is $R_2>R$. At each $\\rho\\in\\Zcal_R$, write the principal\npart\n\\[\n\\Dfun_N(z)^{-r}\n=\n\\sum_{\\ell=1}^{r}\n a_{\\rho,\\ell}\n\\left(1-\\frac z\\rho\\right)^{-\\ell}\n+\nO(1).\n\\]\nThen, for every $R_*$ with $R0}$. Suppose the zeros of $\\Dfun_N$ of least modulus\n$R$ are simple and form the finite set $\\Zcal_R$. Suppose the next\nzero modulus is $R_2>R$. At each $\\rho\\in\\Zcal_R$, write the principal\npart\n\\[\n\\Dfun_N(z)^{-r}\n=\n\\sum_{\\ell=1}^{r}\n a_{\\rho,\\ell}\n\\left(1-\\frac z\\rho\\right)^{-\\ell}\n+\nO(1).\n\\]\nThen, for every $R_*$ with $R0}$. Suppose the zeros of $\\Dfun_N$ of least modulus\n$R$ are simple and form the finite set $\\Zcal_R$. Suppose the next\nzero modulus is $R_2>R$. At each $\\rho\\in\\Zcal_R$, write the principal\npart\n\\[\n\\Dfun_N(z)^{-r}\n=\n\\sum_{\\ell=1}^{r}\n a_{\\rho,\\ell}\n\\left(1-\\frac z\\rho\\right)^{-\\ell}\n+\nO(1).\n\\]\nThen, for every $R_*$ with $R0}$. Suppose the zeros of $\\Dfun_N$ of least modulus\n$R$ are simple and form the finite set $\\Zcal_R$. Suppose the next\nzero modulus is $R_2>R$. At each $\\rho\\in\\Zcal_R$, write the principal\npart\n\\[\n\\Dfun_N(z)^{-r}\n=\n\\sum_{\\ell=1}^{r}\n a_{\\rho,\\ell}\n\\left(1-\\frac z\\rho\\right)^{-\\ell}\n+\nO(1).\n\\]\nThen, for every $R_*$ with $R0}$. Suppose the zeros of $\\Dfun_N$ of least modulus\n$R$ are simple and form the finite set $\\Zcal_R$. Suppose the next\nzero modulus is $R_2>R$. At each $\\rho\\in\\Zcal_R$, write the principal\npart\n\\[\n\\Dfun_N(z)^{-r}\n=\n\\sum_{\\ell=1}^{r}\n a_{\\rho,\\ell}\n\\left(1-\\frac z\\rho\\right)^{-\\ell}\n+\nO(1).\n\\]\nThen, for every $R_*$ with $RR$. At each $\\rho\\in\\Zcal_R$, write the principal\npart\n\\[\n\\Dfun_N(z)^{-r}\n=\n\\sum_{\\ell=1}^{r}\n a_{\\rho,\\ell}\n\\left(1-\\frac z\\rho\\right)^{-\\ell}\n+\nO(1).\n\\]\nThen, for every $R_*$ with $R1.\n\\]\nIt fails to converge absolutely when $\\Re s\\leq1$. The series $\\mathfrak Z_N^{\\mathrm{red}}$ converges absolutely and locally uniformly when\n\\[\n\\Re s>\\frac1N.\n\\]\nIt fails to converge absolutely when $\\Re s\\leq1/N$. Both functions are holomorphic on their open half planes of absolute convergence.\n\\end{theorem}","\\begin{proof}\nBy \\cref{thm:zero_counting}, the zeros of $\\Dfun_N$ have linear counting density, while the reduced zeros have counting order $R^{1/N}$. Partial summation gives absolute convergence in the stated open half planes. The positive terms obtained after taking absolute values diverge on the boundary and to its left. Local uniform convergence follows by the Weierstrass test on compact subsets.\n\\end{proof}","\\begin{theorem}[Meromorphic continuation of the symmetry-orbit zero zeta]\\label{thm:zeta_continuation}\nPut\n\\[\nA_N=\\frac\\pi{\\sin(\\pi/N)}.\n\\]\nThere is an entire function $Q_N^{\\mathrm{red}}$ such that\n\\[\n\\mathfrak Z_N^{\\mathrm{red}}(s)\n=\nA_N^{-Ns}\\zeta\\left(Ns,\\frac12\\right)+Q_N^{\\mathrm{red}}(s)\n\\]\non $\\Re s>1/N$. The right hand side gives a meromorphic continuation to $\\CC$ with one simple pole at $s=1/N$. Its residue is\n\\[\n\\frac{\\sin(\\pi/N)}{N\\pi}.\n\\]","gin{theorem}[Meromorphic continuation of the symmetry-orbit zero zeta]\\label{thm:zeta_continuation}\nPut\n\\[\nA_N=\\frac\\pi{\\sin(\\pi/N)}.\n\\]\nThere is an entire function $Q_N^{\\mathrm{red}}$ such that\n\\[\n\\mathfrak Z_N^{\\mathrm{red}}(s)\n=\nA_N^{-Ns}\\zeta\\left(Ns,\\frac12\\right)+Q_N^{\\mathrm{red}}(s)\n\\]\non $\\Re s>1/N$. The right hand side gives a meromorphic continuation to $\\CC$ with one simple pole at $s=1/N$. Its residue is\n\\[\n\\frac{\\sin(\\pi/N)}{N\\pi}.\n\\]","ontinuation of the symmetry-orbit zero zeta]\\label{thm:zeta_continuation}\nPut\n\\[\nA_N=\\frac\\pi{\\sin(\\pi/N)}.\n\\]\nThere is an entire function $Q_N^{\\mathrm{red}}$ such that\n\\[\n\\mathfrak Z_N^{\\mathrm{red}}(s)\n=\nA_N^{-Ns}\\zeta\\left(Ns,\\frac12\\right)+Q_N^{\\mathrm{red}}(s)\n\\]\non $\\Re s>1/N$. The right hand side gives a meromorphic continuation to $\\CC$ with one simple pole at $s=1/N$. Its residue is\n\\[\n\\frac{\\sin(\\pi/N)}{N\\pi}.\n\\]","There is an entire function $Q_N^{\\mathrm{abs}}$ such that, initially on\n$\\Re s>1$,\n\\[\n\\mathfrak Z_N^{\\mathrm{abs}}(s)\n=\nN A_N^{-s}\\zeta\\left(s,\\frac12\\right)+Q_N^{\\mathrm{abs}}(s).\n\\]\nThus $\\mathfrak Z_N^{\\mathrm{abs}}$ continues meromorphically to $\\CC$ with one simple pole at $s=1$, of residue\n\\[\n\\frac{N\\sin(\\pi/N)}\\pi.\n\\]\n\\end{theorem}","\\begin{proof}\nFor $N=2$, \\cref{cor:case_two_zeros} and\n$\\Dfun_2(z)=\\Delta_2(z^2)$ give\n\\[\n\\lambda_m=\\pi^2\\left(m+\\frac12\\right)^2,\n\\qquad m\\geq0.\n\\]\nThus the first formula holds with $Q_2^{\\mathrm{red}}=0$.","For $N\\geq3$, \\cref{cor:reduced_zeros} writes all but finitely many reduced zeros as\n\\[\n\\lambda_m\n=\n\\left(A_N\\left(m+\\frac12\\right)\\right)^N(1+\\varepsilon_m)\n\\]\nwhere $\\lambda_m>0$, $1+\\varepsilon_m>0$, and\n\\[\n\\varepsilon_m\n=\nO\\left(t_m^{-1}e^{-\\gamma t_m}\\right).\n\\]\nOn every compact set of $s$ values,\n\\[\n(1+\\varepsilon_m)^{-s}-1\n=\nO(|\\varepsilon_m|)\n\\]\nuniformly for large $m$. The difference\n\\[\n\\lambda_m^{-s}\n-\nA_N^{-Ns}\\left(m+\\frac12\\right)^{-Ns}\n\\]\ntherefore has a normally convergent sum on every compact subset of\n$\\CC$. Indeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$","For $N\\geq3$, \\cref{cor:reduced_zeros} writes all but finitely many reduced zeros as\n\\[\n\\lambda_m\n=\n\\left(A_N\\left(m+\\frac12\\right)\\right)^N(1+\\varepsilon_m)\n\\]\nwhere $\\lambda_m>0$, $1+\\varepsilon_m>0$, and\n\\[\n\\varepsilon_m\n=\nO\\left(t_m^{-1}e^{-\\gamma t_m}\\right).\n\\]\nOn every compact set of $s$ values,\n\\[\n(1+\\varepsilon_m)^{-s}-1\n=\nO(|\\varepsilon_m|)\n\\]\nuniformly for large $m$. The difference\n\\[\n\\lambda_m^{-s}\n-\nA_N^{-Ns}\\left(m+\\frac12\\right)^{-Ns}\n\\]\ntherefore has a normally convergent sum on every compact subset of\n$\\CC$. Indeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$ and the finitely many\nomitted model terms are entire functions of $s$. Their addition to the\nnormally convergent tail defines $Q_N^{\\mathrm{red}}","For $N\\geq3$, \\cref{cor:reduced_zeros} writes all but finitely many reduced zeros as\n\\[\n\\lambda_m\n=\n\\left(A_N\\left(m+\\frac12\\right)\\right)^N(1+\\varepsilon_m)\n\\]\nwhere $\\lambda_m>0$, $1+\\varepsilon_m>0$, and\n\\[\n\\varepsilon_m\n=\nO\\left(t_m^{-1}e^{-\\gamma t_m}\\right).\n\\]\nOn every compact set of $s$ values,\n\\[\n(1+\\varepsilon_m)^{-s}-1\n=\nO(|\\varepsilon_m|)\n\\]\nuniformly for large $m$. The difference\n\\[\n\\lambda_m^{-s}\n-\nA_N^{-Ns}\\left(m+\\frac12\\right)^{-Ns}\n\\]\ntherefore has a normally convergent sum on every compact subset of\n$\\CC$. Indeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$ and the finitely many\nomitted model terms are entire functions of $s$. Their addition to the\nnormally convergent tail defines $Q_N^{\\mathrm{red}}$ and proves the\nfirs","For $N\\geq3$, \\cref{cor:reduced_zeros} writes all but finitely many reduced zeros as\n\\[\n\\lambda_m\n=\n\\left(A_N\\left(m+\\frac12\\right)\\right)^N(1+\\varepsilon_m)\n\\]\nwhere $\\lambda_m>0$, $1+\\varepsilon_m>0$, and\n\\[\n\\varepsilon_m\n=\nO\\left(t_m^{-1}e^{-\\gamma t_m}\\right).\n\\]\nOn every compact set of $s$ values,\n\\[\n(1+\\varepsilon_m)^{-s}-1\n=\nO(|\\varepsilon_m|)\n\\]\nuniformly for large $m$. The difference\n\\[\n\\lambda_m^{-s}\n-\nA_N^{-Ns}\\left(m+\\frac12\\right)^{-Ns}\n\\]\ntherefore has a normally convergent sum on every compact subset of\n$\\CC$. Indeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$ and the finitely many\nomitted model terms are entire functions of $s$. Their addition to the\nnormally convergent tail defines $Q_N^{\\mathrm{red}}$ and proves the\nfirst formula. Thus that","For $N\\geq3$, \\cref{cor:reduced_zeros} writes all but finitely many reduced zeros as\n\\[\n\\lambda_m\n=\n\\left(A_N\\left(m+\\frac12\\right)\\right)^N(1+\\varepsilon_m)\n\\]\nwhere $\\lambda_m>0$, $1+\\varepsilon_m>0$, and\n\\[\n\\varepsilon_m\n=\nO\\left(t_m^{-1}e^{-\\gamma t_m}\\right).\n\\]\nOn every compact set of $s$ values,\n\\[\n(1+\\varepsilon_m)^{-s}-1\n=\nO(|\\varepsilon_m|)\n\\]\nuniformly for large $m$. The difference\n\\[\n\\lambda_m^{-s}\n-\nA_N^{-Ns}\\left(m+\\frac12\\right)^{-Ns}\n\\]\ntherefore has a normally convergent sum on every compact subset of\n$\\CC$. Indeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$ and the finitely many\nomitted model terms are entire functions of $s$. Their addition to the\nnormally convergent tail defines $Q_N^{\\mathrm{red}}$ and proves the\nfirst formula. Thus that formula is established for every $N\\geq2$.","ight)\\right)^N(1+\\varepsilon_m)\n\\]\nwhere $\\lambda_m>0$, $1+\\varepsilon_m>0$, and\n\\[\n\\varepsilon_m\n=\nO\\left(t_m^{-1}e^{-\\gamma t_m}\\right).\n\\]\nOn every compact set of $s$ values,\n\\[\n(1+\\varepsilon_m)^{-s}-1\n=\nO(|\\varepsilon_m|)\n\\]\nuniformly for large $m$. The difference\n\\[\n\\lambda_m^{-s}\n-\nA_N^{-Ns}\\left(m+\\frac12\\right)^{-Ns}\n\\]\ntherefore has a normally convergent sum on every compact subset of\n$\\CC$. Indeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$ and the finitely many\nomitted model terms are entire functions of $s$. Their addition to the\nnormally convergent tail defines $Q_N^{\\mathrm{red}}$ and proves the\nfirst formula. Thus that formula is established for every $N\\geq2$.","silon_m>0$, and\n\\[\n\\varepsilon_m\n=\nO\\left(t_m^{-1}e^{-\\gamma t_m}\\right).\n\\]\nOn every compact set of $s$ values,\n\\[\n(1+\\varepsilon_m)^{-s}-1\n=\nO(|\\varepsilon_m|)\n\\]\nuniformly for large $m$. The difference\n\\[\n\\lambda_m^{-s}\n-\nA_N^{-Ns}\\left(m+\\frac12\\right)^{-Ns}\n\\]\ntherefore has a normally convergent sum on every compact subset of\n$\\CC$. Indeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$ and the finitely many\nomitted model terms are entire functions of $s$. Their addition to the\nnormally convergent tail defines $Q_N^{\\mathrm{red}}$ and proves the\nfirst formula. Thus that formula is established for every $N\\geq2$.","on_m\n=\nO\\left(t_m^{-1}e^{-\\gamma t_m}\\right).\n\\]\nOn every compact set of $s$ values,\n\\[\n(1+\\varepsilon_m)^{-s}-1\n=\nO(|\\varepsilon_m|)\n\\]\nuniformly for large $m$. The difference\n\\[\n\\lambda_m^{-s}\n-\nA_N^{-Ns}\\left(m+\\frac12\\right)^{-Ns}\n\\]\ntherefore has a normally convergent sum on every compact subset of\n$\\CC$. Indeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$ and the finitely many\nomitted model terms are entire functions of $s$. Their addition to the\nnormally convergent tail defines $Q_N^{\\mathrm{red}}$ and proves the\nfirst formula. Thus that formula is established for every $N\\geq2$.","values,\n\\[\n(1+\\varepsilon_m)^{-s}-1\n=\nO(|\\varepsilon_m|)\n\\]\nuniformly for large $m$. The difference\n\\[\n\\lambda_m^{-s}\n-\nA_N^{-Ns}\\left(m+\\frac12\\right)^{-Ns}\n\\]\ntherefore has a normally convergent sum on every compact subset of\n$\\CC$. Indeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$ and the finitely many\nomitted model terms are entire functions of $s$. Their addition to the\nnormally convergent tail defines $Q_N^{\\mathrm{red}}$ and proves the\nfirst formula. Thus that formula is established for every $N\\geq2$.",". The difference\n\\[\n\\lambda_m^{-s}\n-\nA_N^{-Ns}\\left(m+\\frac12\\right)^{-Ns}\n\\]\ntherefore has a normally convergent sum on every compact subset of\n$\\CC$. Indeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$ and the finitely many\nomitted model terms are entire functions of $s$. Their addition to the\nnormally convergent tail defines $Q_N^{\\mathrm{red}}$ and proves the\nfirst formula. Thus that formula is established for every $N\\geq2$.","2\\right)^{-Ns}\n\\]\ntherefore has a normally convergent sum on every compact subset of\n$\\CC$. Indeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$ and the finitely many\nomitted model terms are entire functions of $s$. Their addition to the\nnormally convergent tail defines $Q_N^{\\mathrm{red}}$ and proves the\nfirst formula. Thus that formula is established for every $N\\geq2$.","ndeed, the model power has at most polynomial growth in $m$\non a compact set, while $|\\varepsilon_m|$ decays exponentially. The\nfinitely many exceptional terms $|\\lambda|^{-s}$ and the finitely many\nomitted model terms are entire functions of $s$. Their addition to the\nnormally convergent tail defines $Q_N^{\\mathrm{red}}$ and proves the\nfirst formula. Thus that formula is established for every $N\\geq2$.","In either case, the Hurwitz zeta function has a meromorphic continuation with one simple pole at its first argument equal to one and residue one. Composition with $Ns$ gives the stated pole and residue.","For every $N\\geq2$, each nonzero zero $\\lambda$ of $\\Delta_N$ gives $N$ zeros of $\\Dfun_N$, each of modulus $|\\lambda|^{1/N}$. Hence\n\\[\n\\mathfrak Z_N^{\\mathrm{abs}}(s)\n=\nN\\mathfrak Z_N^{\\mathrm{red}}(s/N).\n\\]\nSubstitution gives the second formula with\n\\[\nQ_N^{\\mathrm{abs}}(s)=NQ_N^{\\mathrm{red}}(s/N),\n\\]\nwhich is entire.\n\\end{proof}","\\begin{corollary}[Exact zeta models in orders two and four]\\label{cor:exact_zeta_models}\nFor $N=2$,\n\\[\n\\mathfrak Z_2^{\\mathrm{red}}(s)\n=\n\\pi^{-2s}\\zeta\\left(2s,\\frac12\\right)\n\\]\nand\n\\[\n\\mathfrak Z_2^{\\mathrm{abs}}(s)\n=\n2\\pi^{-s}\\zeta\\left(s,\\frac12\\right).\n\\]\nFor $N=4$, put $A_4=\\sqrt2\\pi$. Then\n\\[\n\\mathfrak Z_4^{\\mathrm{red}}(s)\n=\nA_4^{-4s}\\zeta\\left(4s,\\frac12\\right)\n\\]\nand\n\\[\n\\mathfrak Z_4^{\\mathrm{abs}}(s)\n=\n4A_4^{-s}\\zeta\\left(s,\\frac12\\right).\n\\]\nIn these two cases the entire correction terms in \\cref{thm:zeta_continuation} vanish identically.\n\\end{corollary}","\\begin{proof}\nFor $N=2$, the reduced zeros are\n\\[\n\\pi^2\\left(m+\\frac12\\right)^2,\n\\qquad m\\geq0.\n\\]\nFor $N=4$, \\cref{ex:fourth_order_zeros} gives the reduced zeros\n\\[\nA_4^4\\left(m+\\frac12\\right)^4,\n\\qquad m\\geq0.\n\\]\nSumming their negative powers proves the reduced formulas. The absolute formulas follow from\n\\[\n\\mathfrak Z_N^{\\mathrm{abs}}(s)\n=\nN\\mathfrak Z_N^{\\mathrm{red}}(s/N).\n\\]\n\\end{proof}","\\begin{definition}\\label{def:branch_orbit_zeta}\nChoose a logarithm for each member of the finite exceptional part of $\\Lambda_N$. Use the real logarithm for the eventual positive sequence. Define\n\\[\n\\mathfrak Z_N^{\\mathrm{orb}}(s)\n=\n\\sum_{\\lambda\\in\\Lambda_N}e^{-s\\Log\\lambda}.\n\\]\nThis series is initially defined on $\\Re s>1/N$.\n\\end{definition}","\\begin{corollary}[Branch-dependent symmetry-orbit continuation]\\label{cor:branch_grouped_continuation}\nThe branch-dependent orbit zeta $\\mathfrak Z_N^{\\mathrm{orb}}$ continues meromorphically to $\\CC$. It has the same single pole and residue as $\\mathfrak Z_N^{\\mathrm{red}}$. Changing the logarithms of the finite exceptional zeros changes the continuation by an entire function.\n\\end{corollary}","\\begin{definition}\\label{def:derivative_weighted_zeta}\nLet $q$ be a positive integer. Choose logarithms of the simple zeros of $\\Dfun_N$ with arguments in a fixed bounded interval. Define\n\\[\n\\mathfrak Z_{N,q}^{\\mathrm{der}}(s)\n=\n\\sum_{\\substack{\\Dfun_N(\\rho)=0","\\begin{definition}\\label{def:derivative_weighted_zeta}\nLet $q$ be a positive integer. Choose logarithms of the simple zeros of $\\Dfun_N$ with arguments in a fixed bounded interval. Define\n\\[\n\\mathfrak Z_{N,q}^{\\mathrm{der}}(s)\n=\n\\sum_{\\substack{\\Dfun_N(\\rho)=0\\\\ \\Dfun_N'(\\rho)\\ne0}}\n\\frac{e^{-s\\Log\\rho}}{\\Dfun_N'(\\rho)^q}.\n\\]\n\\end{definition}","\\begin{theorem}[Derivative-weighted convergence]\\label{thm:derivative_weighted_convergence}\nIf $N\\geq3$ and $q\\geq1$, the series $\\mathfrak Z_{N,q}^{\\mathrm{der}}(s)$ converges locally uniformly for every $s\\in\\CC$. It therefore defines an entire function.","For $N=2$, the series converges absolutely and locally uniformly on\n$\\Re s>1$, where it defines a holomorphic function.\n\\end{theorem}","\\begin{proof}\nFor $N\\geq3$, \\cref{thm:large_zero_strings} gives\n\\[\n|\\Dfun_N'(\\rho)|^{-1}\n\\leq\nC e^{-c_N|\\rho|}\n\\]\nfor every sufficiently large simple zero. Let $L$ be a compact subset\nof the $s$ plane. Because the chosen arguments of the zeros lie in a\nfixed bounded interval, there are constants $C_L$ and $M_L$ such that\n\\[\n\\sup_{s\\in L}|e^{-s\\Log\\rho}|\n\\leq C_L|\\rho|^{M_L}\n\\]\nfor every sufficiently large zero. Hence the absolute value of the\ncorresponding summand is at most\n\\[\nC_L'|\\rho|^{M_L}e^{-q c_N|\\rho|}.\n\\]\nThe linear zero counting estimate makes this majorant summable. The\nWeierstrass test gives absolute local uniform convergence of the\nseries of entire summands, so its sum is entire. The finite excepti","\\begin{proof}\nFor $N\\geq3$, \\cref{thm:large_zero_strings} gives\n\\[\n|\\Dfun_N'(\\rho)|^{-1}\n\\leq\nC e^{-c_N|\\rho|}\n\\]\nfor every sufficiently large simple zero. Let $L$ be a compact subset\nof the $s$ plane. Because the chosen arguments of the zeros lie in a\nfixed bounded interval, there are constants $C_L$ and $M_L$ such that\n\\[\n\\sup_{s\\in L}|e^{-s\\Log\\rho}|\n\\leq C_L|\\rho|^{M_L}\n\\]\nfor every sufficiently large zero. Hence the absolute value of the\ncorresponding summand is at most\n\\[\nC_L'|\\rho|^{M_L}e^{-q c_N|\\rho|}.\n\\]\nThe linear zero counting estimate makes this majorant summable. The\nWeierstrass test gives absolute local uniform convergence of the\nseries of entire summands, so its sum is entire. The finite exceptional\nset causes no difficulty.","-1}\n\\leq\nC e^{-c_N|\\rho|}\n\\]\nfor every sufficiently large simple zero. Let $L$ be a compact subset\nof the $s$ plane. Because the chosen arguments of the zeros lie in a\nfixed bounded interval, there are constants $C_L$ and $M_L$ such that\n\\[\n\\sup_{s\\in L}|e^{-s\\Log\\rho}|\n\\leq C_L|\\rho|^{M_L}\n\\]\nfor every sufficiently large zero. Hence the absolute value of the\ncorresponding summand is at most\n\\[\nC_L'|\\rho|^{M_L}e^{-q c_N|\\rho|}.\n\\]\nThe linear zero counting estimate makes this majorant summable. The\nWeierstrass test gives absolute local uniform convergence of the\nseries of entire summands, so its sum is entire. The finite exceptional\nset causes no difficulty.","For $N=2$, the derivative of the cosine has modulus one at every zero. The assertion then follows from the linear zero spacing.\n\\end{proof}","\\begin{remark}\\label{rem:multiple_zeros}\nThe derivative weighted sum omits the finitely many zeros that might be multiple. A treatment of those zeros requires the complete principal parts of $\\Dfun_N^{-q}$ rather than the single derivative weight.\n\\end{remark}","The large-zero theorem settles eventual simplicity but does not classify the\nfinite exceptional set. A remaining question is whether every zero of\n$\\Dfun_N$ is simple and every reduced zero of $\\Delta_N$ is positive. Both\nstatements hold for $N=2$ and $N=4$. The argument here proves them only outside\na compact set for general $N$.","The leading correction in \\cref{thm:first_zero_correction} may admit iteration\nto a full exponential expansion for $\\tau_m$. Such an expansion is an open\nproblem, not a result of this paper. It could refine the entire correction in\nthe zero-zeta comparison and lead to explicit special values.","The symmetry-orbit zero zeta has a meromorphic continuation with one pole. No\nfunctional equation is obtained. A branch dependent sum over the $N$ roots of\neach reduced zero must specify logarithms on the rays.","The coefficient families in\n\\cref{sec:reciprocal_powers,sec:residue_section,sec:theta} have exact formal\nparameter-addition laws. Their arithmetic behavior, including odd-prime\ncongruences, dyadic weighting, and $p$-adic or Amice--Mahler interpolation,\nbelongs to Paper IX. General bounded-operator semigroup, resolvent, and\nFeshbach theory belongs to Paper VI. Paper III owns general nonlinear\noperation trees, while abstract completion and occupancy questions belong to\nPaper X. None of those theories is inferred from the scalar complex-analytic\nresults proved here.","6.1--6.6 & Quotient-algebra power bases and actual exponential-series\nconvergence in the finite-dimensional coefficient topology. Both pointwise\nsection-coordinate identities. Signed-shift powers, all matrix-exponential\nentries, group laws and trace are checked. The characteristic polynomial\n$X^N+1$, its distinct roots, exact spectrum and determinant one are proved.","8.1--8.5 & The full tangent coefficient series has the exact kappa and\nfactorial normalization, including $N=1$. Reciprocal differentiation holds on\nevery stated fixed logarithm domain. The actual higher Riccati identity holds\nat every point where the denominator is nonzero.","9.1--9.10 & Arbitrary-parameter residue reciprocals and their analytic germs,\npolynomial degree and leading term, sparse Appell laws, joint convolution,\nmonicity and covariance. The Hasse and derivative-truncation encodings are\nproved equal. The named Appell polynomial satisfies both the forward action\nand the exact $\\Phi_{N,j}(\\partial_x^N)^r$ inverse formula.","10.1--10.4 & Exact formal theta reconstruction, polynomial lift and degree,\nliteral divisibility by $u$ and, for even $N$, by $u+1$, and the theta Newton\nrecurrence with its first coefficient.","11.2, 11.4--11.11 & Two-sided maximum-modulus bounds prove exact order\nand type. The indicator limsup formula holds in every direction, including ties.\nDominant-exponential errors and zero freeness are uniform on compact sets\navoiding ties. Near cancellation and the order-two zero set are checked.\nFor $N\\geq3$, gap and remainder bounds, sector confinement, global exhaustion\nand multiplicity identities give the full large-zero-string theorem.\nThe real leading correction, quadratic exponential error, signed derivative\nasymptotic, eventual alternating sign and cubic correction are checked.","11.13--11.15 & The reduced series is entire, and\n$\\Dfun_N(z)=\\Delta_N(z^N)$ preserves multiplicity away from zero. Finite\nexceptional sets, eventual positive simple reduced zeros, and their stated\ndisplacement are checked, including $N=2$. Both full disk-counting laws\nretain all exceptional multiplicities. The full-zero reciprocal sums have\nexact thresholds $s>1$ and $s>1/N$, including boundary divergence.","12.1 & For every $N\\geq2$, the complete multiplicity-indexed product\nconverges uniformly on compact sets and equals $\\Delta_N$. Jensen and Poisson\nestimates prove rigidity of the actual quotient from derived growth bounds.\nNo growth certificate remains. The original-variable product for $\\Dfun_N$\nis also checked.","12.2--12.4 & The actual diagonal operator is trace class, with a\nbasis-independent trace and determinant given by limits of actual finite\ncompressions. Infinite complete homogeneous coefficients are limits of finite\nmonomial sums. Every normalized section has exact order $1/N$. For $N\\geq2$\nits full zero product and zero-moment Bell formulas are proved. The actual\ncomplex-power germ at zero has the named residue Taylor coefficients.","12.6--12.10 & The normalized logarithm on each zero free disk has the\nconvergent zero-moment series. Its coefficients equal $\\kappa_{N,m}$ and\nsatisfy Newton's recurrence. The Mittag--Leffler tangent expansion converges\nabsolutely and locally uniformly away from poles. Tangent coefficients and\nthe named reciprocal and theta families have their exact zero-moment Bell\nformulas. The full tangent series is identified with the analytic function.","13.1--13.4 & Analytic division constructs each simple-zero principal part.\nIts leading coefficient is uniquely determined. Removing the principal parts\ngives a holomorphic remainder and the exact factorial-scaled Cauchy estimate.\nBoth leading-term errors are bounded, with the first absent for $r=1$.\nRotation gives the orbit-reduced main term. The secant formula holds as a\nratio limit, with an exponentially vanishing relative error.","14.8 & The derivative-weighted series uses every actual simple zero and genuine\nlogarithms with bounded arguments. Absolute and local uniform convergence and\nan entire sum are proved for $N\\geq3$. For $N=2$, absolute/local uniform\nconvergence and holomorphy hold on $\\Re s>1$.","Cyclic projection turns the exponential into a finite differential basis, and\nthe signed cycle produces the distinguished denominator $\\Dfun_N$. Its\nreciprocal powers yield Bell and Newton recurrences, an additive parameter\nlaw, Appell systems, and theta polynomials. Reduction by $w=z^N$ then connects\nthe reciprocal coefficients to a genus zero product and reduced-zero power\nsums.","ph: Section 2 (p. 2), Section 7 (p. 13), Section 11 (p. 25), Section 12 (p. 33).}\nThis 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.\n\\vfig{three_views}\n\\head{Keep three small models in view}\nFor $N=2$, 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:\n\\[\nD_4(z)=\\cos(z/\\sqrt2)\\cosh(z/\\sqrt2).\n\\]\nThe examples move between these orders deliberately. Order two makes the coefficient arithmetic transparent. Order three shows a genuine displacement. Order four makes orbit reduction completely explicit.\n\\head{Three routes through the guide}\n\\par\\noindent\\begin{tabularx}{\\linewidth}{@{}lY@{}}\\toprule\nRoute & Pages and question","p. 13), Section 11 (p. 25), Section 12 (p. 33).}\nThis 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.\n\\vfig{three_views}\n\\head{Keep three small models in view}\nFor $N=2$, 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:\n\\[\nD_4(z)=\\cos(z/\\sqrt2)\\cosh(z/\\sqrt2).\n\\]\nThe examples move between these orders deliberately. Order two makes the coefficient arithmetic transparent. Order three shows a genuine displacement. Order four makes orbit reduction completely explicit.\n\\head{Three routes through the guide}\n\\par\\noindent\\begin{tabularx}{\\linewidth}{@{}lY@{}}\\toprule\nRoute & Pages and question","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.\n\\vfig{three_views}\n\\head{Keep three small models in view}\nFor $N=2$, 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:\n\\[\nD_4(z)=\\cos(z/\\sqrt2)\\cosh(z/\\sqrt2).\n\\]\nThe examples move between these orders deliberately. Order two makes the coefficient arithmetic transparent. Order three shows a genuine displacement. Order four makes orbit reduction completely explicit.\n\\head{Three routes through the guide}\n\\par\\noindent\\begin{tabularx}{\\linewidth}{@{}lY@{}}\\toprule\nRoute & Pages and question","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.\n\\vfig{three_views}\n\\head{Keep three small models in view}\nFor $N=2$, 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:\n\\[\nD_4(z)=\\cos(z/\\sqrt2)\\cosh(z/\\sqrt2).\n\\]\nThe examples move between these orders deliberately. Order two makes the coefficient arithmetic transparent. Order three shows a genuine displacement. Order four makes orbit reduction completely explicit.\n\\head{Three routes through the guide}\n\\par\\noindent\\begin{tabularx}{\\linewidth}{@{}lY@{}}\\toprule\nRoute & Pages and question","n entire function. The formulas are linked, but the evidence needed for each description is different.\n\\vfig{three_views}\n\\head{Keep three small models in view}\nFor $N=2$, 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:\n\\[\nD_4(z)=\\cos(z/\\sqrt2)\\cosh(z/\\sqrt2).\n\\]\nThe examples move between these orders deliberately. Order two makes the coefficient arithmetic transparent. Order three shows a genuine displacement. Order four makes orbit reduction completely explicit.\n\\head{Three routes through the guide}\n\\par\\noindent\\begin{tabularx}{\\linewidth}{@{}lY@{}}\\toprule\nRoute & Pages and question","One full calculation & \\pagerefV{25}: derive the same number, $69$, from denominator coefficients and from zero moments.","\\vpage{2}{Two variables, several normalizations}{OBJECT ATLAS}{Traditional monograph: Section 2 (p. 2), Definition 5.7 (p. 9), Definition 7.1 (p. 13), Definition 9.1 (p. 19).}\nThe most important coordinate change is $w=z^N$. A term of degree $n$ in $w$ has degree $Nn$ in $z$. Its coefficient does not change, but an exponential-generating-function label does.\n\\[\n\\underbrace{b_{N,n}(r)\\,w^n}_{\\text{ordinary coefficient}}\n=\\underbrace{P_{N,n}(r)\\frac{z^{Nn}}{(Nn)!}}_{\\text{block exponential coefficient}},\n\\qquad P_{N,n}(r)=(Nn)!b_{N,n}(r).\n\\]\n\\par\\noindent\\begin{tabularx}{\\linewidth}{@{}lY@{}}\\toprule\nObject & Definition and role","midrule\n$E_{N,j}(z)$ & $\\displaystyle\\sum_{n\\ge0}\\frac{z^{Nn+j}}{(Nn+j)!}$ selects one residue class of the exponential.","$D_{N,j}(z)$ & The same series with $(-1)^n$. One full derivative cycle contributes a minus sign.","$D_N(z)$ & $D_{N,0}(z)$, the denominator whose zeros control the signed reciprocal family.","$\\Delta_N(w)$ & $\\displaystyle\\sum_{n\\ge0}\\frac{(-1)^nw^n}{(Nn)!}$. Thus $D_N(z)=\\Delta_N(z^N)$.","$P_{N,n}(r)$ & Block-EGF coefficients of $D_N^{-r}$. They are polynomials in $r$.","$\\kappa_{N,m}$ & $\\displaystyle-\\log\\Delta_N(w)=\\sum_{m\\ge1}\\kappa_{N,m}w^m/m$.","$T_N(z)$ & $-D_N'(z)/D_N(z)$, a function, not the coefficient $T_N(n)$.","$\\Phi_{N,j}(w)$ & $\\displaystyle j!\\sum_{n\\ge0}\\frac{w^n}{(Nn+j)!}$ removes the forced zero of $E_{N,j}$.","$C_{N,j}(z)$ & $1/\\Phi_{N,j}(z^N)$. This is generally not $1/D_N(z)$.","$\\Lambda_N$ & Zeros of $\\Delta_N$, counted with multiplicity, one per rotational orbit of zeros of $D_N$.","bottomrule\n\\end{tabularx}\n\\head{A normalization test}\nThe coefficient $5/24$ of $w^2$ in $\\Delta_2(w)^{-1}$ is the same term as $5z^4/4!$. It is not $5w^2/2!$. Keeping the two factorials separate prevents errors in Bell formulas, convolution weights, and zero-moment reconstruction.\n\\takeaway{Use $z$ for differentiation and rotational geometry. Use $w$ for reciprocal recurrences, canonical products, and orbit-zero sums.}","\\vpage{3}{A root filter is an exact vote}{CYCLIC PROJECTION}{Traditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5).}\nFor $N=4$, write $\\omega=i$. To extract residue $j$, evaluate the function at four rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a sym","\\vpage{3}{A root filter is an exact vote}{CYCLIC PROJECTION}{Traditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5).}\nFor $N=4$, write $\\omega=i$. To extract residue $j$, evaluate the function at four rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nF","\\vpage{3}{A root filter is an exact vote}{CYCLIC PROJECTION}{Traditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5).}\nFor $N=4$, write $\\omega=i$. To extract residue $j$, evaluate the function at four rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomi","\\vpage{3}{A root filter is an exact vote}{CYCLIC PROJECTION}{Traditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5).}\nFor $N=4$, write $\\omega=i$. To extract residue $j$, evaluate the function at four rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section s","\\vpage{3}{A root filter is an exact vote}{CYCLIC PROJECTION}{Traditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5).}\nFor $N=4$, write $\\omega=i$. To extract residue $j$, evaluate the function at four rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes","\\vpage{3}{A root filter is an exact vote}{CYCLIC PROJECTION}{Traditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5).}\nFor $N=4$, write $\\omega=i$. To extract residue $j$, evaluate the function at four rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient","vpage{3}{A root filter is an exact vote}{CYCLIC PROJECTION}{Traditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5).}\nFor $N=4$, write $\\omega=i$. To extract residue $j$, evaluate the function at four rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\…","YCLIC PROJECTION}{Traditional monograph: Theorem 3.1 (p. 4), Proposition 3.2 (p. 5).}\nFor $N=4$, write $\\omega=i$. To extract residue $j$, evaluate the function at four rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourie…","idue $j$, evaluate the function at four rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\platecaption{The checker compares the actual complex ave…","r rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\platecaption{The checker compares the actual complex average with independent coefficient sele…","(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\platecaption{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\platecaption{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","d{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\platecaption{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\platecaption{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\platecaption{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\platecaption{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\platecaption{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","\\vpage{4}{Products mix the residue lanes}{CYCLIC CONVOLUTION}{Traditional monograph: Theorem 3.3 (p. 5).}\nSelection separates a function additively. It does not make multiplication diagonal. A residue-$i$ term multiplied by a residue-$j$ term lands in residue $i+j\\pmod N$.\nFor $N=3$, use\n\\[\nF=1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cycl","\\vpage{4}{Products mix the residue lanes}{CYCLIC CONVOLUTION}{Traditional monograph: Theorem 3.3 (p. 5).}\nSelection separates a function additively. It does not make multiplication diagonal. A residue-$i$ term multiplied by a residue-$j$ term lands in residue $i+j\\pmod N$.\nFor $N=3$, use\n\\[\nF=1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra beh","\\vpage{4}{Products mix the residue lanes}{CYCLIC CONVOLUTION}{Traditional monograph: Theorem 3.3 (p. 5).}\nSelection separates a function additively. It does not make multiplication diagonal. A residue-$i$ term multiplied by a residue-$j$ term lands in residue $i+j\\pmod N$.\nFor $N=3$, use\n\\[\nF=1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra behind the later addition formulas and","\\vpage{4}{Products mix the residue lanes}{CYCLIC CONVOLUTION}{Traditional monograph: Theorem 3.3 (p. 5).}\nSelection separates a function additively. It does not make multiplication diagonal. A residue-$i$ term multiplied by a residue-$j$ term lands in residue $i+j\\pmod N$.\nFor $N=3$, use\n\\[\nF=1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra behind the later addition formulas and block-coef","\\vpage{4}{Products mix the residue lanes}{CYCLIC CONVOLUTION}{Traditional monograph: Theorem 3.3 (p. 5).}\nSelection separates a function additively. It does not make multiplication diagonal. A residue-$i$ term multiplied by a residue-$j$ term lands in residue $i+j\\pmod N$.\nFor $N=3$, use\n\\[\nF=1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra behind the later addition formulas and block-coefficient recurrences.\n\\takeaway{A packet labe","ditional monograph: Theorem 3.3 (p. 5).}\nSelection separates a function additively. It does not make multiplication diagonal. A residue-$i$ term multiplied by a residue-$j$ term lands in residue $i+j\\pmod N$.\nFor $N=3$, use\n\\[\nF=1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra behind the later addition formulas and block-coefficient recurrences.\n\\takeaway{A packet label tells you where an interaction lands, not that interactions with other packets vanish.}\n\\platecaption{All nine table entries and all thre…","ltiplication diagonal. A residue-$i$ term multiplied by a residue-$j$ term lands in residue $i+j\\pmod N$.\nFor $N=3$, use\n\\[\nF=1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra behind the later addition formulas and block-coefficient recurrences.\n\\takeaway{A packet label tells you where an interaction lands, not that interactions with other packets vanish.}\n\\platecaption{All nine table entries and all three output sections are checked as exact polynomial identities. This is the scalar cyclic specialization …","term multiplied by a residue-$j$ term lands in residue $i+j\\pmod N$.\nFor $N=3$, use\n\\[\nF=1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra behind the later addition formulas and block-coefficient recurrences.\n\\takeaway{A packet label tells you where an interaction lands, not that interactions with other packets vanish.}\n\\platecaption{All nine table entries and all three output sections are checked as exact polynomial identities. This is the scalar cyclic specialization used by the paper.}","1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra behind the later addition formulas and block-coefficient recurrences.\n\\takeaway{A packet label tells you where an interaction lands, not that interactions with other packets vanish.}\n\\platecaption{All nine table entries and all three output sections are checked as exact polynomial identities. This is the scalar cyclic specialization used by the paper.}","\\vpage{5}{One signed edge determines the cycle}{DIFFERENTIAL SYSTEM}{Traditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12).}\nLet $J_3e_0=e_1$, $J_3e_1=e_2$, and $J_3e_2=-e_0$. A complete turn changes sign, so $J_3^3=-I$. The signed sections are the coordinates of $e^{zJ_3}e_0$.\n\\vfig{signed_shift}\nHere $D_j$ inside the matrix means $D_{3,j}(z)$. For example,\n\\[\nD_{3,0}=1-\\frac{z^3}{3!}+\\frac{z^6}{6!}-\\cdots,\n\\qquad D_{3,1}=z-\\frac{z^4}{4!}+\\cdots.\n\\]\nRows 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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a un","\\vpage{5}{One signed edge determines the cycle}{DIFFERENTIAL SYSTEM}{Traditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12).}\nLet $J_3e_0=e_1$, $J_3e_1=e_2$, and $J_3e_2=-e_0$. A complete turn changes sign, so $J_3^3=-I$. The signed sections are the coordinates of $e^{zJ_3}e_0$.\n\\vfig{signed_shift}\nHere $D_j$ inside the matrix means $D_{3,j}(z)$. For example,\n\\[\nD_{3,0}=1-\\frac{z^3}{3!}+\\frac{z^6}{6!}-\\cdots,\n\\qquad D_{3,1}=z-\\frac{z^4}{4!}+\\cdots.\n\\]\nRows 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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear co","\\vpage{5}{One signed edge determines the cycle}{DIFFERENTIAL SYSTEM}{Traditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12).}\nLet $J_3e_0=e_1$, $J_3e_1=e_2$, and $J_3e_2=-e_0$. A complete turn changes sign, so $J_3^3=-I$. The signed sections are the coordinates of $e^{zJ_3}e_0$.\n\\vfig{signed_shift}\nHere $D_j$ inside the matrix means $D_{3,j}(z)$. For example,\n\\[\nD_{3,0}=1-\\frac{z^3}{3!}+\\frac{z^6}{6!}-\\cdots,\n\\qquad D_{3,1}=z-\\frac{z^4}{4!}+\\cdots.\n\\]\nRows 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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[","\\vpage{5}{One signed edge determines the cycle}{DIFFERENTIAL SYSTEM}{Traditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12).}\nLet $J_3e_0=e_1$, $J_3e_1=e_2$, and $J_3e_2=-e_0$. A complete turn changes sign, so $J_3^3=-I$. The signed sections are the coordinates of $e^{zJ_3}e_0$.\n\\vfig{signed_shift}\nHere $D_j$ inside the matrix means $D_{3,j}(z)$. For example,\n\\[\nD_{3,0}=1-\\frac{z^3}{3!}+\\frac{z^6}{6!}-\\cdots,\n\\qquad D_{3,1}=z-\\frac{z^4}{4!}+\\cdots.\n\\]\nRows 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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J","\\vpage{5}{One signed edge determines the cycle}{DIFFERENTIAL SYSTEM}{Traditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12).}\nLet $J_3e_0=e_1$, $J_3e_1=e_2$, and $J_3e_2=-e_0$. A complete turn changes sign, so $J_3^3=-I$. The signed sections are the coordinates of $e^{zJ_3}e_0$.\n\\vfig{signed_shift}\nHere $D_j$ inside the matrix means $D_{3,j}(z)$. For example,\n\\[\nD_{3,0}=1-\\frac{z^3}{3!}+\\frac{z^6}{6!}-\\cdots,\n\\qquad D_{3,1}=z-\\frac{z^4}{4!}+\\cdots.\n\\]\nRows 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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exp","\\vpage{5}{One signed edge determines the cycle}{DIFFERENTIAL SYSTEM}{Traditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12).}\nLet $J_3e_0=e_1$, $J_3e_1=e_2$, and $J_3e_2=-e_0$. A complete turn changes sign, so $J_3^3=-I$. The signed sections are the coordinates of $e^{zJ_3}e_0$.\n\\vfig{signed_shift}\nHere $D_j$ inside the matrix means $D_{3,j}(z)$. For example,\n\\[\nD_{3,0}=1-\\frac{z^3}{3!}+\\frac{z^6}{6!}-\\cdots,\n\\qquad D_{3,1}=z-\\frac{z^4}{4!}+\\cdots.\n\\]\nRows 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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spe","\\vpage{5}{One signed edge determines the cycle}{DIFFERENTIAL SYSTEM}{Traditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12).}\nLet $J_3e_0=e_1$, $J_3e_1=e_2$, and $J_3e_2=-e_0$. A complete turn changes sign, so $J_3^3=-I$. The signed sections are the coordinates of $e^{zJ_3}e_0$.\n\\vfig{signed_shift}\nHere $D_j$ inside the matrix means $D_{3,j}(z)$. For example,\n\\[\nD_{3,0}=1-\\frac{z^3}{3!}+\\frac{z^6}{6!}-\\cdots,\n\\qquad D_{3,1}=z-\\frac{z^4}{4!}+\\cdots.\n\\]\nRows 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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takea","ed edge determines the cycle}{DIFFERENTIAL SYSTEM}{Traditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12).}\nLet $J_3e_0=e_1$, $J_3e_1=e_2$, and $J_3e_2=-e_0$. A complete turn changes sign, so $J_3^3=-I$. The signed sections are the coordinates of $e^{zJ_3}e_0$.\n\\vfig{signed_shift}\nHere $D_j$ inside the matrix means $D_{3,j}(z)$. For example,\n\\[\nD_{3,0}=1-\\frac{z^3}{3!}+\\frac{z^6}{6!}-\\cdots,\n\\qquad D_{3,1}=z-\\frac{z^4}{4!}+\\cdots.\n\\]\nRows 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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions,…","{z^6}{6!}-\\cdots,\n\\qquad D_{3,1}=z-\\frac{z^4}{4!}+\\cdots.\n\\]\nRows 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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\platecaption{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\platecaption{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","igns record crossing the signed edge while moving from the column to the row. They are not arbitrary orientation choices.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\platecaption{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","ombination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\platecaption{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\platecaption{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\platecaption{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","\\vpage{6}{Addition is quotient multiplication}{FINITE ALGEBRA}{Traditional monograph: Proposition 5.5 (p. 8), Theorem 6.2 (p. 10).}\nWork in $\\CC[\\xi]/(\\xi^3+1)$. Every element has coordinates in $1,\\xi,\\xi^2$. The entire exponential has the finite-coordinate form\n\\[\ne^{z\\xi}=D_{3,0}(z)+D_{3,1}(z)\\xi+D_{3,2}(z)\\xi^2.\n\\]\nThe multiplication table makes the carry sign unavoidable.\n\\vfig{signed_addition}\nMultiply the coordinate forms of $e^{z\\xi}$ and $e^{w\\xi}$. Their product is $e^{(z+w)\\xi}$. The coefficient of $1$ receives one contribution without a wrap and two contributions with a wrap. Reading that coefficient gives the displayed addition law.\n\\head{Compare the unsigned model}\nIn $\\CC[\\xi]/(\\xi^3-1)$, the same table has no minus signs. Its exponential coordinates are the unsigned sections $E_{3,j}$. Thus the difference between the two differenti","\\vpage{6}{Addition is quotient multiplication}{FINITE ALGEBRA}{Traditional monograph: Proposition 5.5 (p. 8), Theorem 6.2 (p. 10).}\nWork in $\\CC[\\xi]/(\\xi^3+1)$. Every element has coordinates in $1,\\xi,\\xi^2$. The entire exponential has the finite-coordinate form\n\\[\ne^{z\\xi}=D_{3,0}(z)+D_{3,1}(z)\\xi+D_{3,2}(z)\\xi^2.\n\\]\nThe multiplication table makes the carry sign unavoidable.\n\\vfig{signed_addition}\nMultiply the coordinate forms of $e^{z\\xi}$ and $e^{w\\xi}$. Their product is $e^{(z+w)\\xi}$. The coefficient of $1$ receives one contribution without a wrap and two contributions with a wrap. Reading that coefficient gives the displayed addition law.\n\\head{Compare the unsigned model}\nIn $\\CC[\\xi]/(\\xi^3-1)$, 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","\\vpage{6}{Addition is quotient multiplication}{FINITE ALGEBRA}{Traditional monograph: Proposition 5.5 (p. 8), Theorem 6.2 (p. 10).}\nWork in $\\CC[\\xi]/(\\xi^3+1)$. Every element has coordinates in $1,\\xi,\\xi^2$. The entire exponential has the finite-coordinate form\n\\[\ne^{z\\xi}=D_{3,0}(z)+D_{3,1}(z)\\xi+D_{3,2}(z)\\xi^2.\n\\]\nThe multiplication table makes the carry sign unavoidable.\n\\vfig{signed_addition}\nMultiply the coordinate forms of $e^{z\\xi}$ and $e^{w\\xi}$. Their product is $e^{(z+w)\\xi}$. The coefficient of $1$ receives one contribution without a wrap and two contributions with a wrap. Reading that coefficient gives the displayed addition law.\n\\head{Compare the unsigned model}\nIn $\\CC[\\xi]/(\\xi^3-1)$, 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.\nFor general $N$, the carry factor is\n\\[\n(-1)^{\\lfloor(i+j)/N\\rfloor},\\","lication}{FINITE ALGEBRA}{Traditional monograph: Proposition 5.5 (p. 8), Theorem 6.2 (p. 10).}\nWork in $\\CC[\\xi]/(\\xi^3+1)$. Every element has coordinates in $1,\\xi,\\xi^2$. The entire exponential has the finite-coordinate form\n\\[\ne^{z\\xi}=D_{3,0}(z)+D_{3,1}(z)\\xi+D_{3,2}(z)\\xi^2.\n\\]\nThe multiplication table makes the carry sign unavoidable.\n\\vfig{signed_addition}\nMultiply the coordinate forms of $e^{z\\xi}$ and $e^{w\\xi}$. Their product is $e^{(z+w)\\xi}$. The coefficient of $1$ receives one contribution without a wrap and two contributions with a wrap. Reading that coefficient gives the displayed addition law.\n\\head{Compare the unsigned model}\nIn $\\CC[\\xi]/(\\xi^3-1)$, 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.\nFor general $N$, the carry factor is\n\\[\n(-1)^{\\lfloor(i+j)/N\\rfloor},\\qquad 0\\le i,j0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator se","(p. 19), Theorem 9.2 (p. 19), Example 9.11 (p. 22).}\nThe signed denominator and the normalized unsigned sections produce different reciprocal families. In order two,\n\\[\nD_2^{-1}=\\sec z,\\qquad C_{2,0}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are ex…","p. 19), Example 9.11 (p. 22).}\nThe signed denominator and the normalized unsigned sections produce different reciprocal families. In order two,\n\\[\nD_2^{-1}=\\sec z,\\qquad C_{2,0}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed…","11 (p. 22).}\nThe signed denominator and the normalized unsigned sections produce different reciprocal families. In order two,\n\\[\nD_2^{-1}=\\sec z,\\qquad C_{2,0}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent …","roduce different reciprocal families. In order two,\n\\[\nD_2^{-1}=\\sec z,\\qquad C_{2,0}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary…","order two,\n\\[\nD_2^{-1}=\\sec z,\\qquad C_{2,0}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by","\\sec z,\\qquad C_{2,0}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factoria…","}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","ble.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","eciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","l}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","E_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\platecaption{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","\\vpage{15}{An infinite symbol acts finitely}{APPELL POLYNOMIALS}{Traditional monograph: Theorem 9.6 (p. 21), Theorem 9.8 (p. 21), Proposition 9.9 (p. 21).}\nFor $C_{2,0}(z)=\\sech z$, define $E_m(x)$ by $\\sech z\\,e^{xz}=\\sum E_m(x)z^m/m!$. The formal differential-operator formula is\n\\[\nE_m(x)=\\sech(\\partial_x)x^m.\n\\]\nAlthough the symbol is an infinite series, derivatives above degree $m$ vanish.\n\\vfig{appell_operator}\nFor $m=4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciproca","\\vpage{15}{An infinite symbol acts finitely}{APPELL POLYNOMIALS}{Traditional monograph: Theorem 9.6 (p. 21), Theorem 9.8 (p. 21), Proposition 9.9 (p. 21).}\nFor $C_{2,0}(z)=\\sech z$, define $E_m(x)$ by $\\sech z\\,e^{xz}=\\sum E_m(x)z^m/m!$. The formal differential-operator formula is\n\\[\nE_m(x)=\\sech(\\partial_x)x^m.\n\\]\nAlthough the symbol is an infinite series, derivatives above degree $m$ vanish.\n\\vfig{appell_operator}\nFor $m=4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol.","\\vpage{15}{An infinite symbol acts finitely}{APPELL POLYNOMIALS}{Traditional monograph: Theorem 9.6 (p. 21), Theorem 9.8 (p. 21), Proposition 9.9 (p. 21).}\nFor $C_{2,0}(z)=\\sech z$, define $E_m(x)$ by $\\sech z\\,e^{xz}=\\sum E_m(x)z^m/m!$. The formal differential-operator formula is\n\\[\nE_m(x)=\\sech(\\partial_x)x^m.\n\\]\nAlthough the symbol is an infinite series, derivatives above degree $m$ vanish.\n\\vfig{appell_operator}\nFor $m=4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities","\\vpage{15}{An infinite symbol acts finitely}{APPELL POLYNOMIALS}{Traditional monograph: Theorem 9.6 (p. 21), Theorem 9.8 (p. 21), Proposition 9.9 (p. 21).}\nFor $C_{2,0}(z)=\\sech z$, define $E_m(x)$ by $\\sech z\\,e^{xz}=\\sum E_m(x)z^m/m!$. The formal differential-operator formula is\n\\[\nE_m(x)=\\sech(\\partial_x)x^m.\n\\]\nAlthough the symbol is an infinite series, derivatives above degree $m$ vanish.\n\\vfig{appell_operator}\nFor $m=4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are\n\\[\nE_m'=mE_{m-1},\\qquad E_m(x+y)=\\sum_{q=0}^m\\binom mqE_q(x)y^{m-q}.\n\\]\nBo","\\vpage{15}{An infinite symbol acts finitely}{APPELL POLYNOMIALS}{Traditional monograph: Theorem 9.6 (p. 21), Theorem 9.8 (p. 21), Proposition 9.9 (p. 21).}\nFor $C_{2,0}(z)=\\sech z$, define $E_m(x)$ by $\\sech z\\,e^{xz}=\\sum E_m(x)z^m/m!$. The formal differential-operator formula is\n\\[\nE_m(x)=\\sech(\\partial_x)x^m.\n\\]\nAlthough the symbol is an infinite series, derivatives above degree $m$ vanish.\n\\vfig{appell_operator}\nFor $m=4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are\n\\[\nE_m'=mE_{m-1},\\qquad E_m(x+y)=\\sum_{q=0}^m\\binom mqE_q(x)y^{m-q}.\n\\]\nBoth follow from the generating function. No operator-series convergence as","mbol acts finitely}{APPELL POLYNOMIALS}{Traditional monograph: Theorem 9.6 (p. 21), Theorem 9.8 (p. 21), Proposition 9.9 (p. 21).}\nFor $C_{2,0}(z)=\\sech z$, define $E_m(x)$ by $\\sech z\\,e^{xz}=\\sum E_m(x)z^m/m!$. The formal differential-operator formula is\n\\[\nE_m(x)=\\sech(\\partial_x)x^m.\n\\]\nAlthough the symbol is an infinite series, derivatives above degree $m$ vanish.\n\\vfig{appell_operator}\nFor $m=4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are\n\\[\nE_m'=mE_{m-1},\\qquad E_m(x+y)=\\sum_{q=0}^m\\binom mqE_q(x)y^{m-q}.\n\\]\nBoth follow from the generating function. No operator-series convergence assumption on an infinite-dimensional fu…","IALS}{Traditional monograph: Theorem 9.6 (p. 21), Theorem 9.8 (p. 21), Proposition 9.9 (p. 21).}\nFor $C_{2,0}(z)=\\sech z$, define $E_m(x)$ by $\\sech z\\,e^{xz}=\\sum E_m(x)z^m/m!$. The formal differential-operator formula is\n\\[\nE_m(x)=\\sech(\\partial_x)x^m.\n\\]\nAlthough the symbol is an infinite series, derivatives above degree $m$ vanish.\n\\vfig{appell_operator}\nFor $m=4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are\n\\[\nE_m'=mE_{m-1},\\qquad E_m(x+y)=\\sum_{q=0}^m\\binom mqE_q(x)y^{m-q}.\n\\]\nBoth follow from the generating function. No operator-series convergence assumption on an infinite-dimensional function space is needed for this po…","e $E_m(x)$ by $\\sech z\\,e^{xz}=\\sum E_m(x)z^m/m!$. The formal differential-operator formula is\n\\[\nE_m(x)=\\sech(\\partial_x)x^m.\n\\]\nAlthough the symbol is an infinite series, derivatives above degree $m$ vanish.\n\\vfig{appell_operator}\nFor $m=4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are\n\\[\nE_m'=mE_{m-1},\\qquad E_m(x+y)=\\sum_{q=0}^m\\binom mqE_q(x)y^{m-q}.\n\\]\nBoth follow from the generating function. No operator-series convergence assumption on an infinite-dimensional function space is needed for this polynomial calculation.\n\\takeaway{Formal substitution into the derivative is safe here because every polynomial sees only finitely…","fig{appell_operator}\nFor $m=4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are\n\\[\nE_m'=mE_{m-1},\\qquad E_m(x+y)=\\sum_{q=0}^m\\binom mqE_q(x)y^{m-q}.\n\\]\nBoth follow from the generating function. No operator-series convergence assumption on an infinite-dimensional function space is needed for this polynomial calculation.\n\\takeaway{Formal substitution into the derivative is safe here because every polynomial sees only finitely many derivatives. That is a domain statement.}\n\\platecaption{The polynomial, its inverse image, and the Appell identities are checked symbolically through degree eight in four normalized-section families. The di…","4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are\n\\[\nE_m'=mE_{m-1},\\qquad E_m(x+y)=\\sum_{q=0}^m\\binom mqE_q(x)y^{m-q}.\n\\]\nBoth follow from the generating function. No operator-series convergence assumption on an infinite-dimensional function space is needed for this polynomial calculation.\n\\takeaway{Formal substitution into the derivative is safe here because every polynomial sees only finitely many derivatives. That is a domain statement.}\n\\platecaption{The polynomial, its inverse image, and the Appell identities are checked symbolically through degree eight in four normalized-section families. The displayed curves are samples o…","The four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are\n\\[\nE_m'=mE_{m-1},\\qquad E_m(x+y)=\\sum_{q=0}^m\\binom mqE_q(x)y^{m-q}.\n\\]\nBoth follow from the generating function. No operator-series convergence assumption on an infinite-dimensional function space is needed for this polynomial calculation.\n\\takeaway{Formal substitution into the derivative is safe here because every polynomial sees only finitely many derivatives. That is a domain statement.}\n\\platecaption{The polynomial, its inverse image, and the Appell identities are checked symbolically through degree eight in four normalized-section families. The displayed curves are samples of exact polynomials.}","\\vpage{16}{Scale and order enter differently}{THETA COUPLING}{Traditional monograph: Theorem 10.2 (p. 23), Proposition 10.3 (p. 23), Proposition 10.4 (p. 23).}\nLet $H_N(z)=E_{N,0}(z)$. The theta family uses the same parameter $u$ twice, but in two different roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contras","\\vpage{16}{Scale and order enter differently}{THETA COUPLING}{Traditional monograph: Theorem 10.2 (p. 23), Proposition 10.3 (p. 23), Proposition 10.4 (p. 23).}\nLet $H_N(z)=E_{N,0}(z)$. The theta family uses the same parameter $u$ twice, but in two different roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe","\\vpage{16}{Scale and order enter differently}{THETA COUPLING}{Traditional monograph: Theorem 10.2 (p. 23), Proposition 10.3 (p. 23), Proposition 10.4 (p. 23).}\nLet $H_N(z)=E_{N,0}(z)$. The theta family uses the same parameter $u$ twice, but in two different roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial root","\\vpage{16}{Scale and order enter differently}{THETA COUPLING}{Traditional monograph: Theorem 10.2 (p. 23), Proposition 10.3 (p. 23), Proposition 10.4 (p. 23).}\nLet $H_N(z)=E_{N,0}(z)$. The theta family uses the same parameter $u$ twice, but in two different roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does no","e{16}{Scale and order enter differently}{THETA COUPLING}{Traditional monograph: Theorem 10.2 (p. 23), Proposition 10.3 (p. 23), Proposition 10.4 (p. 23).}\nLet $H_N(z)=E_{N,0}(z)$. The theta family uses the same parameter $u$ twice, but in two different roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling …","ently}{THETA COUPLING}{Traditional monograph: Theorem 10.2 (p. 23), Proposition 10.3 (p. 23), Proposition 10.4 (p. 23).}\nLet $H_N(z)=E_{N,0}(z)$. The theta family uses the same parameter $u$ twice, but in two different roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeab…","p. 23), Proposition 10.3 (p. 23), Proposition 10.4 (p. 23).}\nLet $H_N(z)=E_{N,0}(z)$. The theta family uses the same parameter $u$ twice, but in two different roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\platecaption{…","E_{N,0}(z)$. The theta family uses the same parameter $u$ twice, but in two different roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\platecaption{The exact checker compares theta convolution with its logarithmic Newton …","fferent roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\platecaption{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly …","[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\platecaption{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an i…","(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\platecaption{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invali…","frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\platecaption{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\platecaption{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\platecaption{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","garithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\platecaption{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","ibutes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\platecaption{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","efficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\platecaption{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","\\vpage{17}{Growth comes from a polygon}{DOMINANT EXPONENTS}{Traditional monograph: Definition 11.3 (p. 26), Theorem 11.4 (p. 26).}\nWrite the signed denominator as a finite exponential sum:\n\\[\nD_N(z)=\\frac1N\\sum_{\\eta^N=-1}e^{\\eta z}.\n\\]\nFor $z=re^{i\\theta}$, the size of one exponential is $e^{r\\Re(\\eta e^{i\\theta})}$. The maximal projection determines the directional exponential growth.\n\\vfig{indicator}\nThe indicator is\n\\[\nh_N(\\theta)=\\max_{0\\le k0$, so $t_m$ is not an exact zero. Linearizing the cosine at that point predicts\n\\[\n\\delta_m^{(1)}=\\frac{(-1)^m}{\\sqrt3}e^{-3t_m/2}.\n\\]\nThe alternating slope explains why the true zero is to the right of the first model zero, then to the left of the second, and so on in the eventual regime.\n\\head{Symmetry needs u","\\vpage{18}{Two large terms nearly cancel}{THE CUBIC MODEL}{Traditional monograph: Lemma 11.5 (p. 27), Theorem 11.7 (p. 28), Corollary 11.9 (p. 31).}\nFor real $t$, order three has the exact formula\n\\[\nD_3(t)=\\frac{e^{-t}+2e^{t/2}\\cos(\\sqrt3t/2)}{3}.\n\\]\nDivide by the nonzero factor $2e^{t/2}/3$. The resulting equation separates the dominant cosine from the small third exponential.\n\\vfig{two_term}\nIgnoring the last term gives model zeros\n\\[\nt_m=\\frac{2\\pi}{\\sqrt3}(m+1/2).\n\\]\nAt $t_m$, 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\n\\[\n\\delta_m^{(1)}=\\frac{(-1)^m}{\\sqrt3}e^{-3t_m/2}.\n\\]\nThe alternating slope explains why the true zero is to the right of the first model zero, then to the left of the second, and so on in the eventual regime.\n\\head{Symmetry needs uniqueness}\nReal coefficients alone allow nonreal conjugate pairs. The manuscript first pr","\\vpage{18}{Two large terms nearly cancel}{THE CUBIC MODEL}{Traditional monograph: Lemma 11.5 (p. 27), Theorem 11.7 (p. 28), Corollary 11.9 (p. 31).}\nFor real $t$, order three has the exact formula\n\\[\nD_3(t)=\\frac{e^{-t}+2e^{t/2}\\cos(\\sqrt3t/2)}{3}.\n\\]\nDivide by the nonzero factor $2e^{t/2}/3$. The resulting equation separates the dominant cosine from the small third exponential.\n\\vfig{two_term}\nIgnoring the last term gives model zeros\n\\[\nt_m=\\frac{2\\pi}{\\sqrt3}(m+1/2).\n\\]\nAt $t_m$, 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\n\\[\n\\delta_m^{(1)}=\\frac{(-1)^m}{\\sqrt3}e^{-3t_m/2}.\n\\]\nThe alternating slope explains why the true zero is to the right of the first model zero, then to the left of the second, and so on in the eventual regime.\n\\head{Symmetry needs uniqueness}\nReal coefficients alone allow nonreal conjugate pairs. The manuscript first proves that a near-model disk contains exactl","{THE CUBIC MODEL}{Traditional monograph: Lemma 11.5 (p. 27), Theorem 11.7 (p. 28), Corollary 11.9 (p. 31).}\nFor real $t$, order three has the exact formula\n\\[\nD_3(t)=\\frac{e^{-t}+2e^{t/2}\\cos(\\sqrt3t/2)}{3}.\n\\]\nDivide by the nonzero factor $2e^{t/2}/3$. The resulting equation separates the dominant cosine from the small third exponential.\n\\vfig{two_term}\nIgnoring the last term gives model zeros\n\\[\nt_m=\\frac{2\\pi}{\\sqrt3}(m+1/2).\n\\]\nAt $t_m$, 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\n\\[\n\\delta_m^{(1)}=\\frac{(-1)^m}{\\sqrt3}e^{-3t_m/2}.\n\\]\nThe alternating slope explains why the true zero is to the right of the first model zero, then to the left of the second, and so on in the eventual regime.\n\\head{Symmetry needs uniqueness}\nReal coefficients alone allow nonreal conjugate pairs. The manuscript first proves that a near-model disk contains exactly one zero. Conjugation preserves that disk, so the unique zero must equal its conjugate. This additional uniqueness step, not the symmetry alone, pl…","Lemma 11.5 (p. 27), Theorem 11.7 (p. 28), Corollary 11.9 (p. 31).}\nFor real $t$, order three has the exact formula\n\\[\nD_3(t)=\\frac{e^{-t}+2e^{t/2}\\cos(\\sqrt3t/2)}{3}.\n\\]\nDivide by the nonzero factor $2e^{t/2}/3$. The resulting equation separates the dominant cosine from the small third exponential.\n\\vfig{two_term}\nIgnoring the last term gives model zeros\n\\[\nt_m=\\frac{2\\pi}{\\sqrt3}(m+1/2).\n\\]\nAt $t_m$, 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\n\\[\n\\delta_m^{(1)}=\\frac{(-1)^m}{\\sqrt3}e^{-3t_m/2}.\n\\]\nThe alternating slope explains why the true zero is to the right of the first model zero, then to the left of the second, and so on in the eventual regime.\n\\head{Symmetry needs uniqueness}\nReal coefficients alone allow nonreal conjugate pairs. The manuscript first proves that a near-model disk contains exactly one zero. Conjugation preserves that disk, so the unique zero must equal its conjugate. This additional uniqueness step, not the symmetry alone, places it exactly on the ray.\n\\takeaway{The…","rollary 11.9 (p. 31).}\nFor real $t$, order three has the exact formula\n\\[\nD_3(t)=\\frac{e^{-t}+2e^{t/2}\\cos(\\sqrt3t/2)}{3}.\n\\]\nDivide by the nonzero factor $2e^{t/2}/3$. The resulting equation separates the dominant cosine from the small third exponential.\n\\vfig{two_term}\nIgnoring the last term gives model zeros\n\\[\nt_m=\\frac{2\\pi}{\\sqrt3}(m+1/2).\n\\]\nAt $t_m$, 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\n\\[\n\\delta_m^{(1)}=\\frac{(-1)^m}{\\sqrt3}e^{-3t_m/2}.\n\\]\nThe alternating slope explains why the true zero is to the right of the first model zero, then to the left of the second, and so on in the eventual regime.\n\\head{Symmetry needs uniqueness}\nReal coefficients alone allow nonreal conjugate pairs. The manuscript first proves that a near-model disk contains exactly one zero. Conjugation preserves that disk, so the unique zero must equal its conjugate. This additional uniqueness step, not the symmetry alone, places it exactly on the ray.\n\\takeaway{The correction is caused by a real, nonzero rem…","eal $t$, order three has the exact formula\n\\[\nD_3(t)=\\frac{e^{-t}+2e^{t/2}\\cos(\\sqrt3t/2)}{3}.\n\\]\nDivide by the nonzero factor $2e^{t/2}/3$. The resulting equation separates the dominant cosine from the small third exponential.\n\\vfig{two_term}\nIgnoring the last term gives model zeros\n\\[\nt_m=\\frac{2\\pi}{\\sqrt3}(m+1/2).\n\\]\nAt $t_m$, 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\n\\[\n\\delta_m^{(1)}=\\frac{(-1)^m}{\\sqrt3}e^{-3t_m/2}.\n\\]\nThe alternating slope explains why the true zero is to the right of the first model zero, then to the left of the second, and so on in the eventual regime.\n\\head{Symmetry needs uniqueness}\nReal coefficients alone allow nonreal conjugate pairs. The manuscript first proves that a near-model disk contains exactly one zero. Conjugation preserves that disk, so the unique zero must equal its conjugate. This additional uniqueness step, not the symmetry alone, places it exactly on the ray.\n\\takeaway{The correction is caused by a real, nonzero remainder. A good cosine approx…","cos(\\sqrt3t/2)}{3}.\n\\]\nDivide by the nonzero factor $2e^{t/2}/3$. The resulting equation separates the dominant cosine from the small third exponential.\n\\vfig{two_term}\nIgnoring the last term gives model zeros\n\\[\nt_m=\\frac{2\\pi}{\\sqrt3}(m+1/2).\n\\]\nAt $t_m$, 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\n\\[\n\\delta_m^{(1)}=\\frac{(-1)^m}{\\sqrt3}e^{-3t_m/2}.\n\\]\nThe alternating slope explains why the true zero is to the right of the first model zero, then to the left of the second, and so on in the eventual regime.\n\\head{Symmetry needs uniqueness}\nReal coefficients alone allow nonreal conjugate pairs. The manuscript first proves that a near-model disk contains exactly one zero. Conjugation preserves that disk, so the unique zero must equal its conjugate. This additional uniqueness step, not the symmetry alone, places it exactly on the ray.\n\\takeaway{The correction is caused by a real, nonzero remainder. A good cosine approximation does not make its model zeros exact.}\n\\platecaption{The curves are …","\\vpage{19}{A root estimate becomes a certificate}{EXACT RATIONAL BRACKETS}{Traditional monograph: Theorem 11.7 (p. 28), Theorem 11.8 (p. 30), Corollary 11.9 (p. 31).}\nThe 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.\n\\vfig{cubic_certificates}\n\\head{How the certificate works}\nEvaluate the Taylor polynomial\n\\[\nS_{90}(t)=\\sum_{n=0}^{90}\\frac{(-1)^nt^{3n}}{(3n)!}\n\\]\nat the two rational endpoints. From the next term onward, magnitudes decrease, because their successive ratio is\n\\[\n\\frac{t^3}{(3n+1)(3n+2)(3n+3)}<1\n\\]\nthere. 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.\nThe first bracket is\n\\[\n1.84981279919014347629<\\tau_0<1.84981279919014347630.\n\\]\nIts displacement differs from the leading asy","Theorem 11.8 (p. 30), Corollary 11.9 (p. 31).}\nThe 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.\n\\vfig{cubic_certificates}\n\\head{How the certificate works}\nEvaluate the Taylor polynomial\n\\[\nS_{90}(t)=\\sum_{n=0}^{90}\\frac{(-1)^nt^{3n}}{(3n)!}\n\\]\nat the two rational endpoints. From the next term onward, magnitudes decrease, because their successive ratio is\n\\[\n\\frac{t^3}{(3n+1)(3n+2)(3n+3)}<1\n\\]\nthere. 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.\nThe first bracket is\n\\[\n1.84981279919014347629<\\tau_0<1.84981279919014347630.\n\\]\nIts displacement differs from the leading asymptotic prediction by about five percent. The later examples have much smaller relative discrepancies, as the plate records.\n\\takeaway{A small residual is not a roo…","te endpoint signs prove existence, and an interval bound for $D_3'$ excludes zero throughout the interval, proving uniqueness inside it.\n\\vfig{cubic_certificates}\n\\head{How the certificate works}\nEvaluate the Taylor polynomial\n\\[\nS_{90}(t)=\\sum_{n=0}^{90}\\frac{(-1)^nt^{3n}}{(3n)!}\n\\]\nat the two rational endpoints. From the next term onward, magnitudes decrease, because their successive ratio is\n\\[\n\\frac{t^3}{(3n+1)(3n+2)(3n+3)}<1\n\\]\nthere. 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.\nThe first bracket is\n\\[\n1.84981279919014347629<\\tau_0<1.84981279919014347630.\n\\]\nIts displacement differs from the leading asymptotic prediction by about five percent. The later examples have much smaller relative discrepancies, as the plate records.\n\\takeaway{A small residual is not a root certificate. Endpoint signs plus a derivative bound give a checkable local existence-and-uniqueness statement.}\n\\platecaption{Four certified intervals do not establish …","1)^nt^{3n}}{(3n)!}\n\\]\nat the two rational endpoints. From the next term onward, magnitudes decrease, because their successive ratio is\n\\[\n\\frac{t^3}{(3n+1)(3n+2)(3n+3)}<1\n\\]\nthere. 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.\nThe first bracket is\n\\[\n1.84981279919014347629<\\tau_0<1.84981279919014347630.\n\\]\nIts displacement differs from the leading asymptotic prediction by about five percent. The later examples have much smaller relative discrepancies, as the plate records.\n\\takeaway{A small residual is not a root certificate. Endpoint signs plus a derivative bound give a checkable local existence-and-uniqueness statement.}\n\\platecaption{Four certified intervals do not establish that no other real or complex zeros exist. The displacement columns are high-precision diagnostics, not a finite verification of a global big-$O$ theorem. All rational enclosures are included in the checks file.}","r relative discrepancies, as the plate records.\n\\takeaway{A small residual is not a root certificate. Endpoint signs plus a derivative bound give a checkable local existence-and-uniqueness statement.}\n\\platecaption{Four certified intervals do not establish that no other real or complex zeros exist. The displacement columns are high-precision diagnostics, not a finite verification of a global big-$O$ theorem. All rational enclosures are included in the checks file.}","\\vpage{20}{Counting changes under the power map}{EXACT FOURTH-ORDER STRINGS}{Traditional monograph: Example 11.12 (p. 31), Theorem 11.14 (p. 32).}\nOrder four gives a case with no asymptotic displacement at all. In normalized coordinates the positive radial locations are $u=m+1/2$, while the reduced-zero locations are $v=(m+1/2)^4$.\n\\vfig{counting}\nThe 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.\n\\head{Separate radius conversion from orbit multiplicity}\nIn unscaled variables the general relation is\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+","\\vpage{20}{Counting changes under the power map}{EXACT FOURTH-ORDER STRINGS}{Traditional monograph: Example 11.12 (p. 31), Theorem 11.14 (p. 32).}\nOrder four gives a case with no asymptotic displacement at all. In normalized coordinates the positive radial locations are $u=m+1/2$, while the reduced-zero locations are $v=(m+1/2)^4$.\n\\vfig{counting}\nThe 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.\n\\head{Separate radius conversion from orbit multiplicity}\nIn unscaled variables the general relation is\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^","ions are $v=(m+1/2)^4$.\n\\vfig{counting}\nThe 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.\n\\head{Separate radius conversion from orbit multiplicity}\nIn unscaled variables the general relation is\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\platecaption{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem fo…","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.\n\\head{Separate radius conversion from orbit multiplicity}\nIn unscaled variables the general relation is\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\platecaption{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify d","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.\n\\head{Separate radius conversion from orbit multiplicity}\nIn unscaled variables the general relation is\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\platecaption{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify discarding that boundary elsewhere.…","hed. Both formulas include zeros on the boundary of the disk.\n\\head{Separate radius conversion from orbit multiplicity}\nIn unscaled variables the general relation is\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\platecaption{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify discarding that boundary elsewhere.}","s\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\platecaption{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify discarding that boundary elsewhere.}","D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\platecaption{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify discarding that boundary elsewhere.}","/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\platecaption{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify discarding that boundary elsewhere.}","\\vpage{21}{A product and an operator share zeros}{CANONICAL PRODUCT}{Traditional monograph: Theorem 12.1 (p. 33), Theorem 12.2 (p. 34).}\nFor $N\\ge2$, the reduced entire function has order $1/N<1$. Its multiplicity-sensitive genus-zero product is\n\\[\n\\Delta_N(w)=\\prod_{\\lambda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe las","\\vpage{21}{A product and an operator share zeros}{CANONICAL PRODUCT}{Traditional monograph: Theorem 12.1 (p. 33), Theorem 12.2 (p. 34).}\nFor $N\\ge2$, the reduced entire function has order $1/N<1$. Its multiplicity-sensitive genus-zero product is\n\\[\n\\Delta_N(w)=\\prod_{\\lambda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. T","\\vpage{21}{A product and an operator share zeros}{CANONICAL PRODUCT}{Traditional monograph: Theorem 12.1 (p. 33), Theorem 12.2 (p. 34).}\nFor $N\\ge2$, the reduced entire function has order $1/N<1$. Its multiplicity-sensitive genus-zero product is\n\\[\n\\Delta_N(w)=\\prod_{\\lambda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic t","\\vpage{21}{A product and an operator share zeros}{CANONICAL PRODUCT}{Traditional monograph: Theorem 12.1 (p. 33), Theorem 12.2 (p. 34).}\nFor $N\\ge2$, the reduced entire function has order $1/N<1$. Its multiplicity-sensitive genus-zero product is\n\\[\n\\Delta_N(w)=\\prod_{\\lambda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled","\\vpage{21}{A product and an operator share zeros}{CANONICAL PRODUCT}{Traditional monograph: Theorem 12.1 (p. 33), Theorem 12.2 (p. 34).}\nFor $N\\ge2$, the reduced entire function has order $1/N<1$. Its multiplicity-sensitive genus-zero product is\n\\[\n\\Delta_N(w)=\\prod_{\\lambda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not su","e2$, the reduced entire function has order $1/N<1$. Its multiplicity-sensitive genus-zero product is\n\\[\n\\Delta_N(w)=\\prod_{\\lambda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\platecaption{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Ord…","e reduced entire function has order $1/N<1$. Its multiplicity-sensitive genus-zero product is\n\\[\n\\Delta_N(w)=\\prod_{\\lambda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\platecaption{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one …","ltiplicity-sensitive genus-zero product is\n\\[\n\\Delta_N(w)=\\prod_{\\lambda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\platecaption{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so …","bda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\platecaption{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\platecaption{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","Putting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\platecaption{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\platecaption{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\platecaption{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\platecaption{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\platecaption{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","\\vpage{22}{Finite coefficients give infinite moments}{ZERO POWER SUMS}{Traditional monograph: Theorem 12.6 (p. 35), Corollary 12.7 (p. 36), Corollary 12.9 (p. 37), Corollary 12.10 (p. 37).}\nTaking the logarithm of the canonical product identifies\n\\[\n\\kappa_{N,q}=\\sigma_{N,q}:=\\sum_{\\lambda\\in\\Lambda_N}\\lambda^{-q}\n=\\Tr(K_N^q),\\qquad q\\ge1.\n\\]\nThe same exact numbers can be computed from finitely many denominator coefficients by Newton's recurrence. No numerical zero list is needed for that calculation.\n\\vfig{moments}\nFor order two,\n\\[\n\\sigma_{2,1}=\\frac12,\\quad\\sigma_{2,2}=\\frac16,\\quad\n\\sigma_{2,3}=\\frac1{15},\\quad\\sigma_{2,4}=\\frac{17}{630}.\n\\]\nThe plate shows how successive inverse powers emphasize the nearest zero. Its shaded pieces are contributions from the first five positive reduced zeros, followed by the remaining tail. The total in each row is the exact rational moment.\n\\head{The tangent is a sum of poles}\nLogarithmic differentiation also gives\n\\[\nT_N(z)=Nz^{N-1}\\sum_{\\lambda\\in\\Lambda_N}\\frac1{\\lambda-z^N}\n\\]\naway fr","Corollary 12.9 (p. 37), Corollary 12.10 (p. 37).}\nTaking the logarithm of the canonical product identifies\n\\[\n\\kappa_{N,q}=\\sigma_{N,q}:=\\sum_{\\lambda\\in\\Lambda_N}\\lambda^{-q}\n=\\Tr(K_N^q),\\qquad q\\ge1.\n\\]\nThe same exact numbers can be computed from finitely many denominator coefficients by Newton's recurrence. No numerical zero list is needed for that calculation.\n\\vfig{moments}\nFor order two,\n\\[\n\\sigma_{2,1}=\\frac12,\\quad\\sigma_{2,2}=\\frac16,\\quad\n\\sigma_{2,3}=\\frac1{15},\\quad\\sigma_{2,4}=\\frac{17}{630}.\n\\]\nThe plate shows how successive inverse powers emphasize the nearest zero. Its shaded pieces are contributions from the first five positive reduced zeros, followed by the remaining tail. The total in each row is the exact rational moment.\n\\head{The tangent is a sum of poles}\nLogarithmic differentiation also gives\n\\[\nT_N(z)=Nz^{N-1}\\sum_{\\lambda\\in\\Lambda_N}\\frac1{\\lambda-z^N}\n\\]\naway from poles. Summability of $|\\lambda|^{-1}$ permits the locally uniform partial-fraction expansion without added regularizing terms.\n\\takeaway{A global analytic product theorem connects the infinite z…","ma_{2,2}=\\frac16,\\quad\n\\sigma_{2,3}=\\frac1{15},\\quad\\sigma_{2,4}=\\frac{17}{630}.\n\\]\nThe plate shows how successive inverse powers emphasize the nearest zero. Its shaded pieces are contributions from the first five positive reduced zeros, followed by the remaining tail. The total in each row is the exact rational moment.\n\\head{The tangent is a sum of poles}\nLogarithmic differentiation also gives\n\\[\nT_N(z)=Nz^{N-1}\\sum_{\\lambda\\in\\Lambda_N}\\frac1{\\lambda-z^N}\n\\]\naway from poles. Summability of $|\\lambda|^{-1}$ permits the locally uniform partial-fraction expansion without added regularizing terms.\n\\takeaway{A global analytic product theorem connects the infinite zero sum to a finite algebraic recurrence. The recurrence alone does not prove the product theorem.}\n\\platecaption{Displayed moments are checked against formal logarithm coefficients and exact order-two/order-four zeta values. Bar widths are numerical shares. General power sums use $\\lambda^{-q}$, not $|\\lambda|^{-q}$ when exceptional nonpositive zeros are present.}","s how successive inverse powers emphasize the nearest zero. Its shaded pieces are contributions from the first five positive reduced zeros, followed by the remaining tail. The total in each row is the exact rational moment.\n\\head{The tangent is a sum of poles}\nLogarithmic differentiation also gives\n\\[\nT_N(z)=Nz^{N-1}\\sum_{\\lambda\\in\\Lambda_N}\\frac1{\\lambda-z^N}\n\\]\naway from poles. Summability of $|\\lambda|^{-1}$ permits the locally uniform partial-fraction expansion without added regularizing terms.\n\\takeaway{A global analytic product theorem connects the infinite zero sum to a finite algebraic recurrence. The recurrence alone does not prove the product theorem.}\n\\platecaption{Displayed moments are checked against formal logarithm coefficients and exact order-two/order-four zeta values. Bar widths are numerical shares. General power sums use $\\lambda^{-q}$, not $|\\lambda|^{-q}$ when exceptional nonpositive zeros are present.}","ction expansion without added regularizing terms.\n\\takeaway{A global analytic product theorem connects the infinite zero sum to a finite algebraic recurrence. The recurrence alone does not prove the product theorem.}\n\\platecaption{Displayed moments are checked against formal logarithm coefficients and exact order-two/order-four zeta values. Bar widths are numerical shares. General power sums use $\\lambda^{-q}$, not $|\\lambda|^{-q}$ when exceptional nonpositive zeros are present.}","out added regularizing terms.\n\\takeaway{A global analytic product theorem connects the infinite zero sum to a finite algebraic recurrence. The recurrence alone does not prove the product theorem.}\n\\platecaption{Displayed moments are checked against formal logarithm coefficients and exact order-two/order-four zeta values. Bar widths are numerical shares. General power sums use $\\lambda^{-q}$, not $|\\lambda|^{-q}$ when exceptional nonpositive zeros are present.}","\\vpage{23}{Nearest poles set coefficient growth}{SINGULARITY ANALYSIS}{Traditional monograph: Theorem 13.1 (p. 37), Corollary 13.2 (p. 38), Corollary 13.3 (p. 38), Corollary 13.4 (p. 39).}\nFor $1/\\cos z$, the nearest poles are $\\rho=\\pm\\pi/2$. They contribute equally to even-degree coefficients. Including both poles gives\n\\[\nP_{2,n}(1)\\sim2(2n)!\\left(\\frac2\\pi\\right)^{2n+1}.\n\\]\nThe factor two is orbit multiplicity, not an adjustable fitted constant.\n\\vfig{pole_asymptotic}\n\\head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator","\\vpage{23}{Nearest poles set coefficient growth}{SINGULARITY ANALYSIS}{Traditional monograph: Theorem 13.1 (p. 37), Corollary 13.2 (p. 38), Corollary 13.3 (p. 38), Corollary 13.4 (p. 39).}\nFor $1/\\cos z$, the nearest poles are $\\rho=\\pm\\pi/2$. They contribute equally to even-degree coefficients. Including both poles gives\n\\[\nP_{2,n}(1)\\sim2(2n)!\\left(\\frac2\\pi\\right)^{2n+1}.\n\\]\nThe factor two is orbit multiplicity, not an adjustable fitted constant.\n\\vfig{pole_asymptotic}\n\\head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The","\\vpage{23}{Nearest poles set coefficient growth}{SINGULARITY ANALYSIS}{Traditional monograph: Theorem 13.1 (p. 37), Corollary 13.2 (p. 38), Corollary 13.3 (p. 38), Corollary 13.4 (p. 39).}\nFor $1/\\cos z$, the nearest poles are $\\rho=\\pm\\pi/2$. They contribute equally to even-degree coefficients. Including both poles gives\n\\[\nP_{2,n}(1)\\sim2(2n)!\\left(\\frac2\\pi\\right)^{2n+1}.\n\\]\nThe factor two is orbit multiplicity, not an adjustable fitted constant.\n\\vfig{pole_asymptotic}\n\\head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. S","orollary 13.2 (p. 38), Corollary 13.3 (p. 38), Corollary 13.4 (p. 39).}\nFor $1/\\cos z$, the nearest poles are $\\rho=\\pm\\pi/2$. They contribute equally to even-degree coefficients. Including both poles gives\n\\[\nP_{2,n}(1)\\sim2(2n)!\\left(\\frac2\\pi\\right)^{2n+1}.\n\\]\nThe factor two is orbit multiplicity, not an adjustable fitted constant.\n\\vfig{pole_asymptotic}\n\\head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geo…","$1/\\cos z$, the nearest poles are $\\rho=\\pm\\pi/2$. They contribute equally to even-degree coefficients. Including both poles gives\n\\[\nP_{2,n}(1)\\sim2(2n)!\\left(\\frac2\\pi\\right)^{2n+1}.\n\\]\nThe factor two is orbit multiplicity, not an adjustable fitted constant.\n\\vfig{pole_asymptotic}\n\\head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal …","t poles are $\\rho=\\pm\\pi/2$. They contribute equally to even-degree coefficients. Including both poles gives\n\\[\nP_{2,n}(1)\\sim2(2n)!\\left(\\frac2\\pi\\right)^{2n+1}.\n\\]\nThe factor two is orbit multiplicity, not an adjustable fitted constant.\n\\vfig{pole_asymptotic}\n\\head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower princ…","n-degree coefficients. Including both poles gives\n\\[\nP_{2,n}(1)\\sim2(2n)!\\left(\\frac2\\pi\\right)^{2n+1}.\n\\]\nThe factor two is orbit multiplicity, not an adjustable fitted constant.\n\\vfig{pole_asymptotic}\n\\head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\platecapt…","factor two is orbit multiplicity, not an adjustable fitted constant.\n\\vfig{pole_asymptotic}\n\\head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\platecaption{The ratio certificates use a rational Machin-formula enclosure for $\\pi$. The inherited general theorem ret","head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\platecaption{The ratio certificates use a rational Machin-formula enclosure for $\\pi$. The inherited general theorem retains simple nearest zeros, a gap to the next modulus, and positive integral reciprocal order.","he exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\platecaption{The ratio certificates use a rational Machin-formula enclosure for $\\pi$. The inherited general theorem retains simple nearest zeros, a gap to the next modulus, and positive integral reciprocal order. The plot does not remove those hypotheses.}","ure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\platecaption{The ratio certificates use a rational Machin-formula enclosure for $\\pi$. The inherited general theorem retains simple nearest zeros, a gap to the next modulus, and positive integral reciprocal order. The plot does not remove those hypotheses.}","on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\platecaption{The ratio certificates use a rational Machin-formula enclosure for $\\pi$. The inherited general theorem retains simple nearest zeros, a gap to the next modulus, and positive integral reciprocal order. The plot does not remove those hypotheses.}","\\vpage{24}{Convergence is not continuation}{ZERO ZETA FUNCTIONS}{Traditional monograph: Theorem 14.2 (p. 39), Theorem 14.3 (p. 39), Corollary 14.4 (p. 41).}\nReduced 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 $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N","\\vpage{24}{Convergence is not continuation}{ZERO ZETA FUNCTIONS}{Traditional monograph: Theorem 14.2 (p. 39), Theorem 14.3 (p. 39), Corollary 14.4 (p. 41).}\nReduced 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 $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the","\\vpage{24}{Convergence is not continuation}{ZERO ZETA FUNCTIONS}{Traditional monograph: Theorem 14.2 (p. 39), Theorem 14.3 (p. 39), Corollary 14.4 (p. 41).}\nReduced 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 $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms.","\\vpage{24}{Convergence is not continuation}{ZERO ZETA FUNCTIONS}{Traditional monograph: Theorem 14.2 (p. 39), Theorem 14.3 (p. 39), Corollary 14.4 (p. 41).}\nReduced 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 $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes c","\\vpage{24}{Convergence is not continuation}{ZERO ZETA FUNCTIONS}{Traditional monograph: Theorem 14.2 (p. 39), Theorem 14.3 (p. 39), Corollary 14.4 (p. 41).}\nReduced 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 $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuatio","is not continuation}{ZERO ZETA FUNCTIONS}{Traditional monograph: Theorem 14.2 (p. 39), Theorem 14.3 (p. 39), Corollary 14.4 (p. 41).}\nReduced 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 $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equa…","solute zero zeta in the original variable instead requires $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\platecaption{The shaded regions show convergence, and the open circles mark poles of the continue…","ro zeta in the original variable instead requires $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\platecaption{The shaded regions show convergence, and the open circles mark poles of the continued functio…","nal variable instead requires $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\platecaption{The shaded regions show convergence, and the open circles mark poles of the continued functions. They are not zer…","nstead requires $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\platecaption{The shaded regions show convergence, and the open circles mark poles of the continued functions. They are not zeros of the zeta…","$N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\platecaption{The shaded regions show convergence, and the open circles mark poles of the continued functions. They are not zeros of the zeta functions. Branch-dependent orbit zetas and deri","frak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\platecaption{The shaded regions show convergence, and the open circles mark poles of the continued functions. They are not zeros of the zeta functions. Branch-dependent orbit zetas and derivative-weighted zetas retain their separately sta…","(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\platecaption{The shaded regions show convergence, and the open circles mark poles of the continued functions. They are not zeros of the zeta functions. Branch-dependent orbit zetas and derivative-weighted zetas retain their separately stated conventions.}","\\vpage{25}{One closed calculation: the number 69}{COEFFICIENT AND ZERO WORKED MODEL}{Traditional monograph: Corollary 7.7 (p. 16), Example 11.12 (p. 31), Theorem 12.6 (p. 35).}\nUse $N=4$ 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.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off t","er 69}{COEFFICIENT AND ZERO WORKED MODEL}{Traditional monograph: Corollary 7.7 (p. 16), Example 11.12 (p. 31), Theorem 12.6 (p. 35).}\nUse $N=4$ 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.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrat…","D ZERO WORKED MODEL}{Traditional monograph: Corollary 7.7 (p. 16), Example 11.12 (p. 31), Theorem 12.6 (p. 35).}\nUse $N=4$ 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.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridge…","{Traditional monograph: Corollary 7.7 (p. 16), Example 11.12 (p. 31), Theorem 12.6 (p. 35).}\nUse $N=4$ 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.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof…","ample 11.12 (p. 31), Theorem 12.6 (p. 35).}\nUse $N=4$ 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.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\platecaption…","rem 12.6 (p. 35).}\nUse $N=4$ 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.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\platecaption{Every rational value and","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.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\platecaption{Every rational value and polynomial identit…","recurrences provide two routes to the same coefficient. All symbols in the plate refer to this one order.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\platecaption{Every rational value and polynomial identity on this page is checked exactly. The infinite zero-sum interpretation uses the inherited c…","cient. All symbols in the plate refer to this one order.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\platecaption{Every rational value and polynomial identity on this page is checked exactly. The infinite zero-sum interpretation uses the inherited canonical-product theorem and the exact order-four…","n the plate refer to this one order.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\platecaption{Every rational value and polynomial identity on this page is checked exactly. The infinite zero-sum interpretation uses the inherited canonical-product theorem and the exact order-four zero formula.}","he denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\platecaption{Every rational value and polynomial identity on this page is checked exactly. The infinite zero-sum interpretation uses the inherited canonical-product theorem and the exact order-four zero formula.}","!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\platecaption{Every rational value and polynomial identity on this page is checked exactly. The infinite zero-sum interpretation uses the inherited canonical-product theorem and the exact order-four zero formula.}","Certified zero brackets & Four rational intervals of width $10^{-20}$ establish local existence and uniqueness only. The fresh full suite reproduced all four rational brackets, opposite endpoint signs, and derivative exclusion throughout each interval. These are local certificates, not a global zero census.","\\begin{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\draw[black!13] (1.86000,0.50000) -- (1.86000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (1.86000,0.19000) {$-2$};\n\\draw[black!13] (4.18000,0.50000) -- (4.18000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (4.18000,0.19000) {$-1$};\n\\draw[black!13] (6.50000,0.50000) -- (6.50000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (6.50000,0.19000) {$0$};\n\\draw[black!13] (8.82000,0.50000) -- (8.82000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (8.82000,0.19000) {$1$};\n\\draw[black!13] (11.14000,0.50000) -- (11.14000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (11.14000,0.19000) {$2$};\n\\draw[black!13]","\\begin{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\draw[black!13] (1.86000,0.50000) -- (1.86000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at 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1$};\n\\end{tikzpicture}","width=1pt] (4.40000,2.30000) circle (0.09);\n\\node[] at (5.10000,1.45000) {$\\rho=\\pi/2$};\n\\draw[->,line width=1.1pt] (4.88,2.3) arc[start angle=0,end angle=330,radius=.48];\n\\node[] at (2.80000,0.10000) {the loop is not inside the disk};\n\\node[] at (9.20000,4.10000) {$D(z)\\sim D^{\\prime}(\\rho)(z-\\rho)$};\n\\node[] at (9.20000,2.90000) {$\\log D\\longmapsto\\log D+2\\pi i$};\n\\node[] at (9.20000,1.65000) {$D^{-r}\\longmapsto e^{-2\\pi i r}D^{-r}$};\n\\node[] at (9.20000,0.40000) {$r=1/2:\\ -1\\qquad r\\in\\mathbb Z:\\ 1$};\n\\end{tikzpicture}","{$\\rho=\\pi/2$};\n\\draw[->,line width=1.1pt] (4.88,2.3) arc[start angle=0,end angle=330,radius=.48];\n\\node[] at (2.80000,0.10000) {the loop is not inside the disk};\n\\node[] at (9.20000,4.10000) {$D(z)\\sim D^{\\prime}(\\rho)(z-\\rho)$};\n\\node[] at (9.20000,2.90000) {$\\log D\\longmapsto\\log D+2\\pi i$};\n\\node[] at (9.20000,1.65000) {$D^{-r}\\longmapsto e^{-2\\pi i r}D^{-r}$};\n\\node[] at 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{$1/16$};\n\\node[font=\\sffamily\\footnotesize] at (0.81719,1.45000) {count 1};\n\\draw[line width=1pt] (2.19219,0.88000) -- (2.19219,1.12000);\n\\draw[fill=black] (2.19219,1.00000) circle (0.055);\n\\node[] at (2.19219,0.58000) {$81/16$};\n\\node[font=\\sffamily\\footnotesize] at (2.19219,1.45000) {count 2};\n\\draw[line width=1pt] (11.54219,0.88000) -- (11.54219,1.12000);\n\\draw[fill=black] (11.54219,1.00000) circle (0.055);\n\\node[] at (11.54219,0.58000) {$625/16$};\n\\node[font=\\sffamily\\footnotesize] at (11.54219,1","00,1.00000) -- (12.30000,1.00000);\n\\draw[line width=1pt] (0.81719,0.88000) -- (0.81719,1.12000);\n\\draw[fill=black] (0.81719,1.00000) circle (0.055);\n\\node[] at (0.81719,0.58000) {$1/16$};\n\\node[font=\\sffamily\\footnotesize] at (0.81719,1.45000) {count 1};\n\\draw[line width=1pt] (2.19219,0.88000) -- (2.19219,1.12000);\n\\draw[fill=black] (2.19219,1.00000) circle (0.055);\n\\node[] at (2.19219,0.58000) {$81/16$};\n\\node[font=\\sffamily\\footnotesize] at 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{$\\frac{2}{3}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (2.40000,3.15000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,3.50000) {$\\frac{5}{3}$};\n\\node[anchor=west] at (4.50000,3.50000) {$(2/3)_2$};\n\\draw[fill=black!20,draw=black!40,line width=.35pt] (0.80000,2.25000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.60000,2.60000) {$1$};\n\\draw[fill=black!20,draw=black!40,line width=.35pt] (2.40000,2.25000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,2.60000) {$2$};\n\\node[anchor=west] at (4.50000,2.60000) {$(1)_2=2!$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (0.80000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.6",".50000) {$N=3,\\ j=1,\\ n=2:\\quad 7!=1!\\,3^6\\,2!\\,(2/3)_2(4/3)_2$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (0.80000,3.15000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.60000,3.50000) {$\\frac{2}{3}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (2.40000,3.15000) rectangle 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{$2$};\n\\node[anchor=west] at (4.50000,2.60000) {$(1)_2=2!$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (0.80000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.60000,1.70000) {$\\frac{4}{3}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (2.40000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,1.70000) {$\\frac{7}{3}$};\n\\node[anchor=west] at (4.50000,1.70000) {$(4/3)_2$};\n\\node[] at (9.30000,3.30000) {$A_{3,1}=\\{2/3,4/3\\}$};\n\\node[] at (9.30000,2.00000) {$\\displaystyl","0000,3.50000) {$\\frac{5}{3}$};\n\\node[anchor=west] at (4.50000,3.50000) {$(2/3)_2$};\n\\draw[fill=black!20,draw=black!40,line width=.35pt] (0.80000,2.25000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.60000,2.60000) {$1$};\n\\draw[fill=black!20,draw=black!40,line width=.35pt] (2.40000,2.25000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,2.60000) {$2$};\n\\node[anchor=west] at (4.50000,2.60000) {$(1)_2=2!$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (0.80000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.60000,1.70000) {$\\frac{4}{3}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (2.40000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,1.70000) {$\\frac{7}{3}$};\n\\node[anchor=west] at (4.50000,1.70000) {$(4/3)_2$};\n\\node[] at (9.30000,3.30000) {$A_{3,1}=\\{2/3,4/3\\}$};\n\\node[] at (9.30000,2.00000) {$\\displaystyle E_{3,1}(z)=z\\,{}_0F_2\\!\\left(;\\frac23,\\frac43;\\frac{","0,2.25000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.60000,2.60000) {$1$};\n\\draw[fill=black!20,draw=black!40,line width=.35pt] (2.40000,2.25000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,2.60000) {$2$};\n\\node[anchor=west] at (4.50000,2.60000) {$(1)_2=2!$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (0.80000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.60000,1.70000) {$\\frac{4}{3}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (2.40000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,1.70000) {$\\frac{7}{3}$};\n\\node[anchor=west] at (4.50000,1.70000) {$(4/3)_2$};\n\\node[] at (9.30000,3.30000) {$A_{3,1}=\\{2/3,4/3\\}$};\n\\node[] at (9.30000,2.00000) {$\\displaystyle E_{3,1}(z)=z\\,{}_0F_2\\!\\left(;\\frac23,\\frac43;\\frac{z^3}{27}\\right)$};\n\\node[] at (6.30000,-0.05000) {The omitted unit parameter is already the hypergeometric $n!$.};\n\\end{tikzpicture}",") rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,2.60000) {$2$};\n\\node[anchor=west] at (4.50000,2.60000) {$(1)_2=2!$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (0.80000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.60000,1.70000) {$\\frac{4}{3}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (2.40000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,1.70000) {$\\frac{7}{3}$};\n\\node[anchor=west] at (4.50000,1.70000) {$(4/3)_2$};\n\\node[] at (9.30000,3.30000) {$A_{3,1}=\\{2/3,4/3\\}$};\n\\node[] at (9.30000,2.00000) {$\\displaystyle E_{3,1}(z)=z\\,{}_0F_2\\!\\left(;\\frac23,\\frac43;\\frac{z^3}{27}\\right)$};\n\\node[] at (6.30000,-0.05000) {The omitted unit parameter is already the hypergeometric $n!$.};\n\\end{tikzpicture}","000) {$2$};\n\\node[anchor=west] at (4.50000,2.60000) {$(1)_2=2!$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (0.80000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.60000,1.70000) {$\\frac{4}{3}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (2.40000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,1.70000) {$\\frac{7}{3}$};\n\\node[anchor=west] at (4.50000,1.70000) {$(4/3)_2$};\n\\node[] at (9.30000,3.30000) {$A_{3,1}=\\{2/3,4/3\\}$};\n\\node[] at (9.30000,2.00000) {$\\displaystyle E_{3,1}(z)=z\\,{}_0F_2\\!\\left(;\\frac23,\\frac43;\\frac{z^3}{27}\\right)$};\n\\node[] at (6.30000,-0.05000) {The omitted unit parameter is already the hypergeometric $n!$.};\n\\end{tikzpicture}",",2.60000) {$(1)_2=2!$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (0.80000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.60000,1.70000) {$\\frac{4}{3}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (2.40000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,1.70000) {$\\frac{7}{3}$};\n\\node[anchor=west] at (4.50000,1.70000) {$(4/3)_2$};\n\\node[] at (9.30000,3.30000) {$A_{3,1}=\\{2/3,4/3\\}$};\n\\node[] at (9.30000,2.00000) {$\\displaystyle E_{3,1}(z)=z\\,{}_0F_2\\!\\left(;\\frac23,\\frac43;\\frac{z^3}{27}\\right)$};\n\\node[] at (6.30000,-0.05000) {The omitted unit parameter is already the hypergeometric $n!$.};\n\\end{tikzpicture}","40,line width=.35pt] (0.80000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (1.60000,1.70000) {$\\frac{4}{3}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (2.40000,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,1.70000) {$\\frac{7}{3}$};\n\\node[anchor=west] at (4.50000,1.70000) {$(4/3)_2$};\n\\node[] at (9.30000,3.30000) {$A_{3,1}=\\{2/3,4/3\\}$};\n\\node[] at (9.30000,2.00000) {$\\displaystyle E_{3,1}(z)=z\\,{}_0F_2\\!\\left(;\\frac23,\\frac43;\\frac{z^3}{27}\\right)$};\n\\node[] at (6.30000,-0.05000) {The omitted unit parameter is already the hypergeometric $n!$.};\n\\end{tikzpicture}","0,1.35000) rectangle ++(1.60000,0.70000);\n\\node[] at (3.20000,1.70000) {$\\frac{7}{3}$};\n\\node[anchor=west] at (4.50000,1.70000) {$(4/3)_2$};\n\\node[] at (9.30000,3.30000) {$A_{3,1}=\\{2/3,4/3\\}$};\n\\node[] at (9.30000,2.00000) {$\\displaystyle E_{3,1}(z)=z\\,{}_0F_2\\!\\left(;\\frac23,\\frac43;\\frac{z^3}{27}\\right)$};\n\\node[] at (6.30000,-0.05000) {The omitted unit parameter is already the hypergeometric $n!$.};\n\\end{tikzpicture}","934,3.66066) -- (1.28934,1.53934) -- (3.41066,1.53934) -- cycle;\n\\draw[->] (0.40000,2.60000) -- (4.30000,2.60000);\n\\draw[->] (2.35000,0.65000) -- (2.35000,4.55000);\n\\draw[fill=black] (3.41066,3.66066) circle 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{$0$};\n\\draw[black!13] (7.02500,1.00000) -- (7.02500,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (7.02500,0.69000) {$\\pi/2$};\n\\draw[black!13] (8.85000,1.00000) -- (8.85000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (8.85000,0.69000) {$\\pi$};\n\\draw[black!13] (10.67500,1.00000) -- (10.67500,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (10.67500,0.69000) {$3\\pi/2$};\n\\draw[black!13] (12.50000,1.00000) -- (12.50000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (12.50000,0.69000) {$2\\pi$};\n\\draw[black!13] (5.20000,1.42830) -- (12.50000,1.42830);\n\\node[anchor=east,font=\\sffamily\\footnotesize] at (5.04000,1.42830","de[] at (2.35000,0.05000) {exponent plane};\n\\draw[black!13] (5.20000,1.00000) -- (5.20000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (5.20000,0.69000) {$0$};\n\\draw[black!13] (7.02500,1.00000) -- (7.02500,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (7.02500,0.69000) {$\\pi/2$};\n\\draw[black!13] (8.85000,1.00000) -- (8.85000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (8.85000,0.69000) {$\\pi$};\n\\draw[black!13] (10.67500,1.00000) -- (10.67500,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (10.67500,0.69000) {$3\\pi/2$};\n\\draw[black!13] (12.50000,1.00000) -- (12.50000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (12.50000,0.69000) {$2\\pi$};\n\\draw[black!13] (5.20000,1.42830) -- (12.50000,1.42830);\n\\node[anchor=east,font=\\sffamily\\footnotesize] at (5.04000,1.42830) {$1/\\sqrt2$};\n\\draw[black!13] (5.20000,3.62500) -- (12.50000,3.62500);\n\\node[anchor=east,font=\\sffamily\\footnotesize] a","tnotesize] at (5.20000,0.69000) {$0$};\n\\draw[black!13] (7.02500,1.00000) -- (7.02500,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (7.02500,0.69000) {$\\pi/2$};\n\\draw[black!13] (8.85000,1.00000) -- (8.85000,4.00000);\n\\node[font=\\sffamily\\footnotesize] at (8.85000,0.69000) {$\\pi$};\n\\draw[black!13] (10.67500,1.00000) -- 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(6.50000,4.60000) {$\\Delta_2(w)=\\cos\\sqrt w,\\qquad P_L(w)=\\prod_{m=0}^{L-1}(1-w/\\lambda_m)$};\n\\node[] at (6.50000,-0.35000) {thick: entire function; dotted $L=1$; dashed $L=3$; dash-dot $L=10$};\n\\end{tikzpicture}",") (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)};\n\\node[] at (6.50000,4.60000) {$\\Delta_2(w)=\\cos\\sqrt w,\\qquad P_L(w)=\\prod_{m=0}^{L-1}(1-w/\\lambda_m)$};\n\\node[] at (6.50000,-0.35000) {thick: entire function; dotted $L=1$; dashed $L=3$; dash-dot $L=10$};\n\\end{tikzpicture}","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)};\n\\node[] at (6.50000,4.60000) {$\\Delta_2(w)=\\cos\\sqrt w,\\qquad P_L(w)=\\prod_{m=0}^{L-1}(1-w/\\lambda_m)$};\n\\node[] at (6.50000,-0.35000) {thick: entire function; dotted $L=1$; dashed $L=3$; dash-dot $L=10$};\n\\end{tikzpicture}",",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)};\n\\node[] at (6.50000,4.60000) {$\\Delta_2(w)=\\cos\\sqrt w,\\qquad P_L(w)=\\prod_{m=0}^{L-1}(1-w/\\lambda_m)$};\n\\node[] at (6.50000,-0.35000) {thick: entire function; dotted $L=1$; dashed $L=3$; dash-dot $L=10$};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\node[] at (3.00000,4.50000) {LINEAR EQUATION};\n\\node[] at (9.30000,4.50000) {LOGARITHMIC DERIVATIVE};\n\\node[] at (3.00000,3.25000) {$D_2^{\\prime\\prime}=-D_2$};\n\\node[] at (9.30000,3.25000) {$T_2^{\\prime}=1+T_2^2$};\n\\node[] at (3.00000,1.85000) {$D_3^{\\prime\\prime\\prime}=-D_3$};\n\\node[] at (9.30000,1.85000) {$T_3^{\\prime\\prime}-3T_3T_3^{\\prime}+T_3^3=1$};\n\\node[] at (6.15000,0.50000) {$T_N=-D_N^{\\prime}/D_N\\quad\\text{only where }D_N\\ne0$};\n\\draw[black!25] (6.10000,1.10000) -- (6.10000,4.65000);\n\\node[] at (6.10000,-0.40000) {$T_4(z)=\\frac{z^3}{3!}+34\\frac{z^7}{7!}+\\cdots$};\n\\end{tikzpicture}","in{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\node[] at (3.00000,4.50000) {LINEAR EQUATION};\n\\node[] at (9.30000,4.50000) {LOGARITHMIC DERIVATIVE};\n\\node[] at (3.00000,3.25000) {$D_2^{\\prime\\prime}=-D_2$};\n\\node[] at (9.30000,3.25000) {$T_2^{\\prime}=1+T_2^2$};\n\\node[] at (3.00000,1.85000) {$D_3^{\\prime\\prime\\prime}=-D_3$};\n\\node[] at (9.30000,1.85000) {$T_3^{\\prime\\prime}-3T_3T_3^{\\prime}+T_3^3=1$};\n\\node[] at (6.15000,0.50000) {$T_N=-D_N^{\\prime}/D_N\\quad\\text{only where }D_N\\ne0$};\n\\draw[black!25] (6.10000,1.10000) -- (6.10000,4.65000);\n\\node[] at (6.10000,-0.40000) {$T_4(z)=\\frac{z^3}{3!}+34\\frac{z^7}{7!}+\\cdots$};\n\\end{tikzpicture}","oin=round]\n\\node[] at (3.00000,4.50000) {LINEAR EQUATION};\n\\node[] at (9.30000,4.50000) {LOGARITHMIC DERIVATIVE};\n\\node[] at (3.00000,3.25000) {$D_2^{\\prime\\prime}=-D_2$};\n\\node[] at (9.30000,3.25000) {$T_2^{\\prime}=1+T_2^2$};\n\\node[] at (3.00000,1.85000) {$D_3^{\\prime\\prime\\prime}=-D_3$};\n\\node[] at (9.30000,1.85000) {$T_3^{\\prime\\prime}-3T_3T_3^{\\prime}+T_3^3=1$};\n\\node[] at (6.15000,0.50000) {$T_N=-D_N^{\\prime}/D_N\\quad\\text{only where }D_N\\ne0$};\n\\draw[black!25] (6.10000,1.10000) -- (6.10000,4.65000);\n\\node[] at (6.10000,-0.40000) {$T_4(z)=\\frac{z^3}{3!}+34\\frac{z^7}{7!}+\\cdots$};\n\\end{tikzpicture}","{$D_2^{\\prime\\prime}=-D_2$};\n\\node[] at (9.30000,3.25000) {$T_2^{\\prime}=1+T_2^2$};\n\\node[] at (3.00000,1.85000) {$D_3^{\\prime\\prime\\prime}=-D_3$};\n\\node[] at (9.30000,1.85000) {$T_3^{\\prime\\prime}-3T_3T_3^{\\prime}+T_3^3=1$};\n\\node[] at (6.15000,0.50000) {$T_N=-D_N^{\\prime}/D_N\\quad\\text{only where }D_N\\ne0$};\n\\draw[black!25] (6.10000,1.10000) -- (6.10000,4.65000);\n\\node[] at (6.10000,-0.40000) {$T_4(z)=\\frac{z^3}{3!}+34\\frac{z^7}{7!}+\\cdots$};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\node[] at (2.50000,4.40000) {Four weighted votes for $z^m$};\n\\draw[->,line width=1pt] (2.50000,2.00000) -- (3.95000,2.00000);\n\\node[] at (4.40000,2.00000) {$1$};\n\\draw[->,line width=1pt] (2.50000,2.00000) -- (2.50000,3.45000);\n\\node[] at (2.50000,3.90000) {$i$};\n\\draw[->,line width=1pt] (2.50000,2.00000) -- (1.05000,2.00000);\n\\node[] at (0.60000,2.00000) {$-1$};\n\\draw[->,line width=1pt] (2.50000,2.00000) -- (2.50000,0.55000);\n\\node[] at (2.50000,0.10000) {$-i$};\n\\node[] at (2.50000,-0.30000) {$\\frac14\\sum_{\\ell=0}^3 i^{(m-j)\\ell}$};\n\\node[anchor=east]","\\begin{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\node[] at (2.50000,4.40000) {Four weighted votes for $z^m$};\n\\draw[->,line width=1pt] (2.50000,2.00000) -- (3.95000,2.00000);\n\\node[] at (4.40000,2.00000) {$1$};\n\\draw[->,line width=1pt] (2.50000,2.00000) -- (2.50000,3.45000);\n\\node[] at (2.50000,3.90000) {$i$};\n\\draw[->,line width=1pt] (2.50000,2.00000) -- (1.05000,2.00000);\n\\node[] at (0.60000,2.00000) {$-1$};\n\\draw[->,line width=1pt] (2.50000,2.00000) -- (2.50000,0.55000);\n\\node[] at (2.50000,0.10000) {$-i$};\n\\node[] at (2.50000,-0.30000) {$\\frac14\\sum_{\\ell=0}^3 i^{(m-j)\\ell}$};\n\\node[anchor=east] at (6.10000,3.70000) {$j=0$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (6.45000,3.40000) r","\\begin{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\node[] at (2.50000,4.40000) {Four weighted votes for 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{$z^{0}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (8.15000,3.40000) rectangle ++(1.70000,0.60000);\n\\node[] at (9.00000,3.70000) {$z^{4}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (9.85000,3.40000) rectangle ++(1.70000,0.60000);\n\\node[] at (10.70000,3.70000) {$z^{8}$};\n\\node[anchor=east] at (6.10000,2.85000) {$j=1$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (6.45000,2.55000) rectangle ++(1.70000,0.60000);\n\\node[] at (7.30000,2.85000) {$z^{1}$};\n\\draw[fill=black!0,draw=bl","{$-i$};\n\\node[] at (2.50000,-0.30000) {$\\frac14\\sum_{\\ell=0}^3 i^{(m-j)\\ell}$};\n\\node[anchor=east] at (6.10000,3.70000) {$j=0$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (6.45000,3.40000) rectangle ++(1.70000,0.60000);\n\\node[] at (7.30000,3.70000) {$z^{0}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (8.15000,3.40000) rectangle ++(1.70000,0.60000);\n\\node[] at (9.00000,3.70000) {$z^{4}$};\n\\draw[fill=black!0,draw=black!40,line 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(1.17650,3.48023) (1.20575,3.45","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)};\n\\draw[fill=white] (2.77215,2.02182) circle (0.07);\n\\draw[fill=white] (7.01644,2.02182) circle (0.07);\n\\draw[fill=white] (11.26073,2.02182) circle (0.07);\n\\node[] at (6.50000,4.70000) {$\\frac32e^{-t/2}D_3(t)=\\cos(\\sqrt3t/2)+\\frac12e^{-3t/2}$};\n\\node[] at (6.50000,-0.25000) {solid: full expression; dashed: two-term model; circles: model zeros};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\node[anchor=west] at (0.35000,4.30000) {denominator};\n\\node[] at (8.00000,4.30000) {$1-w/24+w^2/40320+\\cdots$};\n\\draw[black!15] (0.35000,3.87000) -- (12.30000,3.87000);\n\\node[anchor=west] at (0.35000,3.35000) {negative logarithm};\n\\node[] at (8.00000,3.35000) {$w/24+(17/20160)w^2+\\cdots$};\n\\draw[black!15] (0.35000,2.92000) -- (12.30000,2.92000);\n\\node[anchor=west] at (0.35000,2.40000) {zero moments};\n\\node[] at (8.00000,2.40000) {$\\sigma_1=1/24,\\quad\\sigma_2=17/10080$};\n\\draw[black!15] (0.35000,1.97000) -- (12.30000,1.97000);\n\\node[anchor=west] at (0.35000,1.45000) {reciprocal};\n\\node[] at (8.00000,1.","\\begin{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\node[anchor=west] at (0.35000,4.30000) {denominator};\n\\node[] at (8.00000,4.30000) {$1-w/24+w^2/40320+\\cdots$};\n\\draw[black!15] (0.35000,3.87000) -- (12.30000,3.87000);\n\\node[anchor=west] at (0.35000,3.35000) {negative logarithm};\n\\node[] at (8.00000,3.35000) {$w/24+(17/20160)w^2+\\cdots$};\n\\draw[black!15] (0.35000,2.92000) -- (12.30000,2.92000);\n\\node[anchor=west] at (0.35000,2.40000) {zero moments};\n\\node[] at (8.00000,2.40000) {$\\sigma_1=1/24,\\quad\\sigma_2=17/10080$};\n\\draw[black!15] (0.35000,1.97000) -- (12.30000,1.97000);\n\\node[anchor=west] at (0.35000,1.45000) {reciprocal};\n\\node[] at (8.00000,1.45000) {$1+w/24+69w^2/40320+\\cdots$};\n\\draw[black!15] (0.35000,1.02000) -- (12.30000,1.02000);\n\\node[anchor=west] at (0.35000,0.50000) {tangent};\n\\node[] at (8.00000,0.50000) {$z^","00) {denominator};\n\\node[] at (8.00000,4.30000) {$1-w/24+w^2/40320+\\cdots$};\n\\draw[black!15] (0.35000,3.87000) -- (12.30000,3.87000);\n\\node[anchor=west] at (0.35000,3.35000) {negative logarithm};\n\\node[] at (8.00000,3.35000) {$w/24+(17/20160)w^2+\\cdots$};\n\\draw[black!15] (0.35000,2.92000) -- (12.30000,2.92000);\n\\node[anchor=west] at (0.35000,2.40000) {zero moments};\n\\node[] at (8.00000,2.40000) {$\\sigma_1=1/24,\\quad\\sigma_2=17/10080$};\n\\draw[black!15] (0.35000,1.97000) -- (12.30000,1.97000);\n\\node[anchor=west] at (0.35000,1.45000) {reciprocal};\n\\node[] at (8.00000,1.45000) {$1+w/24+69w^2/40320+\\cdots$};\n\\draw[black!15] (0.35000,1.02000) -- (12.30000,1.02000);\n\\node[anchor=west] at (0.35000,0.50000) {tangent};\n\\node[] at (8.00000,0.50000) {$z^3/3!+34z^7/7!+\\cdots$};\n\\node[] at (6.50000,-0.55000) {$69=8!\\,\\frac{\\sigma_1^2+\\sigma_2}{2},\\qquad34=4\\cdot7!\\,\\sigma_2$};\n\\end{tikzpicture}","e logarithm};\n\\node[] at (8.00000,3.35000) {$w/24+(17/20160)w^2+\\cdots$};\n\\draw[black!15] (0.35000,2.92000) -- (12.30000,2.92000);\n\\node[anchor=west] at (0.35000,2.40000) {zero moments};\n\\node[] at (8.00000,2.40000) {$\\sigma_1=1/24,\\quad\\sigma_2=17/10080$};\n\\draw[black!15] (0.35000,1.97000) -- (12.30000,1.97000);\n\\node[anchor=west] at (0.35000,1.45000) {reciprocal};\n\\node[] at (8.00000,1.45000) {$1+w/24+69w^2/40320+\\cdots$};\n\\draw[black!15] (0.35000,1.02000) -- (12.30000,1.02000);\n\\node[anchor=west] at (0.35000,0.50000) {tangent};\n\\node[] at (8.00000,0.50000) {$z^3/3!+34z^7/7!+\\cdots$};\n\\node[] at (6.50000,-0.55000) {$69=8!\\,\\frac{\\sigma_1^2+\\sigma_2}{2},\\qquad34=4\\cdot7!\\,\\sigma_2$};\n\\end{tikzpicture}","0) {zero moments};\n\\node[] at (8.00000,2.40000) {$\\sigma_1=1/24,\\quad\\sigma_2=17/10080$};\n\\draw[black!15] (0.35000,1.97000) -- (12.30000,1.97000);\n\\node[anchor=west] at (0.35000,1.45000) {reciprocal};\n\\node[] at (8.00000,1.45000) {$1+w/24+69w^2/40320+\\cdots$};\n\\draw[black!15] (0.35000,1.02000) -- (12.30000,1.02000);\n\\node[anchor=west] at (0.35000,0.50000) {tangent};\n\\node[] at (8.00000,0.50000) {$z^3/3!+34z^7/7!+\\cdots$};\n\\node[] at (6.50000,-0.55000) {$69=8!\\,\\frac{\\sigma_1^2+\\sigma_2}{2},\\qquad34=4\\cdot7!\\,\\sigma_2$};\n\\end{tikzpicture}","=1/24,\\quad\\sigma_2=17/10080$};\n\\draw[black!15] (0.35000,1.97000) -- (12.30000,1.97000);\n\\node[anchor=west] at (0.35000,1.45000) {reciprocal};\n\\node[] at (8.00000,1.45000) {$1+w/24+69w^2/40320+\\cdots$};\n\\draw[black!15] (0.35000,1.02000) -- (12.30000,1.02000);\n\\node[anchor=west] at (0.35000,0.50000) {tangent};\n\\node[] at (8.00000,0.50000) {$z^3/3!+34z^7/7!+\\cdots$};\n\\node[] at (6.50000,-0.55000) {$69=8!\\,\\frac{\\sigma_1^2+\\sigma_2}{2},\\qquad34=4\\cdot7!\\,\\sigma_2$};\n\\end{tikzpicture}","\\sffamily\\small,line cap=round,line join=round]\n\\draw[fill=black!10,draw=black!40,line width=.35pt] (2.57692,0.70000) rectangle ++(3.32308,3.50000);\n\\draw[->] (0.50000,2.45000) -- (6.10000,2.45000);\n\\draw[->] (0.90000,0.70000) -- (0.90000,4.40000);\n\\draw[dashed] (2.57692,0.70000) -- (2.57692,4.20000);\n\\draw[fill=white,line width=1pt] (2.57692,2.45000) circle (0.09);\n\\node[] at (2.57692,2.05000) {$1/2$};\n\\node[font=\\sffamily\\footnotesize] at (4.28000,3.22000) {series converges};\n\\node[] at (3.20000,4.85000) {$N=2$: $\\mathfrak Z_N^{\\rm red}(s)$};\n\\node[] at (3.20000,0.00000) {single continued pole at $s=1/2$};\n\\draw[fill=black!10,draw=black!40,line width=.35pt] (8.23846,0.70000) rectangle ++(4.36154,3.50000);\n\\draw[->] (7.20000,2.45000) -- (12.80000,2.45000);\n\\draw[->] (7.60000,0.70000) -- (7.60000,4.40000);\n\\draw[dashed] (8.23846,0.70000) -- (8.23846,4.20000);\n\\draw[fill=white,line width=1pt","00) rectangle ++(3.32308,3.50000);\n\\draw[->] (0.50000,2.45000) -- (6.10000,2.45000);\n\\draw[->] (0.90000,0.70000) -- (0.90000,4.40000);\n\\draw[dashed] (2.57692,0.70000) -- (2.57692,4.20000);\n\\draw[fill=white,line width=1pt] (2.57692,2.45000) circle (0.09);\n\\node[] at (2.57692,2.05000) {$1/2$};\n\\node[font=\\sffamily\\footnotesize] at (4.28000,3.22000) {series converges};\n\\node[] at (3.20000,4.85000) {$N=2$: $\\mathfrak Z_N^{\\rm red}(s)$};\n\\node[] at (3.20000,0.00000) {single continued pole at $s=1/2$};\n\\draw[fill=black!10,draw=black!40,line width=.35pt] (8.23846,0.70000) rectangle ++(4.36154,3.50000);\n\\draw[->] (7.20000,2.45000) -- (12.80000,2.45000);\n\\draw[->] (7.60000,0.70000) -- (7.60000,4.40000);\n\\draw[dashed] (8.23846,0.70000) -- (8.23846,4.20000);\n\\draw[fill=white,line width=1pt] (8.23846,2.45000) circle (0.09);\n\\node[] at (8.23846,2.05000) {$1/4$};\n\\node[font=\\sffamily\\footnotesize] at (10","tangle ++(3.32308,3.50000);\n\\draw[->] (0.50000,2.45000) -- (6.10000,2.45000);\n\\draw[->] (0.90000,0.70000) -- (0.90000,4.40000);\n\\draw[dashed] (2.57692,0.70000) -- (2.57692,4.20000);\n\\draw[fill=white,line width=1pt] (2.57692,2.45000) circle (0.09);\n\\node[] at (2.57692,2.05000) {$1/2$};\n\\node[font=\\sffamily\\footnotesize] at (4.28000,3.22000) {series converges};\n\\node[] at (3.20000,4.85000) {$N=2$: $\\mathfrak Z_N^{\\rm red}(s)$};\n\\node[] at (3.20000,0.00000) {single continued pole at $s=1/2$};\n\\draw[fill=black!10,draw=black!40,line width=.35pt] (8.23846,0.70000) rectangle ++(4.36154,3.50000);\n\\draw[->] (7.20000,2.45000) -- (12.80000,2.45000);\n\\draw[->] (7.60000,0.70000) -- (7.60000,4.40000);\n\\draw[dashed] (8.23846,0.70000) -- (8.23846,4.20000);\n\\draw[fill=white,line width=1pt] (8.23846,2.45000) circle (0.09);\n\\node[] at (8.23846,2.05000) {$1/4$};\n\\node[font=\\sffamily\\footnotesize] at (10.98000,3.22000) {series conver","] (0.90000,0.70000) -- (0.90000,4.40000);\n\\draw[dashed] (2.57692,0.70000) -- (2.57692,4.20000);\n\\draw[fill=white,line width=1pt] (2.57692,2.45000) circle (0.09);\n\\node[] at (2.57692,2.05000) {$1/2$};\n\\node[font=\\sffamily\\footnotesize] at (4.28000,3.22000) {series converges};\n\\node[] at (3.20000,4.85000) {$N=2$: $\\mathfrak Z_N^{\\rm red}(s)$};\n\\node[] at (3.20000,0.00000) {single continued pole at $s=1/2$};\n\\draw[fill=black!10,draw=black!40,line width=.35pt] (8.23846,0.70000) rectangle ++(4.36154,3.50000);\n\\draw[->] (7.20000,2.45000) -- (12.80000,2.45000);\n\\draw[->] (7.60000,0.70000) -- (7.60000,4.40000);\n\\draw[dashed] (8.23846,0.70000) -- (8.23846,4.20000);\n\\draw[fill=white,line width=1pt] (8.23846,2.45000) circle (0.09);\n\\node[] at (8.23846,2.05000) {$1/4$};\n\\node[font=\\sffamily\\footnotesize] at (10.98000,3.22000) {series converges};\n\\node[] at (9.90000,4.85000) {$N=4$: $\\mathfrak Z_N^{\\rm re","0,0.00000) {single continued pole at $s=1/2$};\n\\draw[fill=black!10,draw=black!40,line width=.35pt] (8.23846,0.70000) rectangle ++(4.36154,3.50000);\n\\draw[->] (7.20000,2.45000) -- (12.80000,2.45000);\n\\draw[->] (7.60000,0.70000) -- (7.60000,4.40000);\n\\draw[dashed] (8.23846,0.70000) -- (8.23846,4.20000);\n\\draw[fill=white,line width=1pt] (8.23846,2.45000) circle (0.09);\n\\node[] at (8.23846,2.05000) {$1/4$};\n\\node[font=\\sffamily\\footnotesize] at (10.98000,3.22000) {series converges};\n\\node[] at (9.90000,4.85000) {$N=4$: $\\mathfrak Z_N^{\\rm red}(s)$};\n\\node[] at (9.90000,0.00000) {single continued pole at $s=1/4$};\n\\end{tikzpicture}",") rectangle ++(4.36154,3.50000);\n\\draw[->] (7.20000,2.45000) -- (12.80000,2.45000);\n\\draw[->] (7.60000,0.70000) -- (7.60000,4.40000);\n\\draw[dashed] (8.23846,0.70000) -- (8.23846,4.20000);\n\\draw[fill=white,line width=1pt] (8.23846,2.45000) circle (0.09);\n\\node[] at (8.23846,2.05000) {$1/4$};\n\\node[font=\\sffamily\\footnotesize] at (10.98000,3.22000) {series converges};\n\\node[] at (9.90000,4.85000) {$N=4$: $\\mathfrak Z_N^{\\rm red}(s)$};\n\\node[] at (9.90000,0.00000) {single continued pole at $s=1/4$};\n\\end{tikzpicture}","ngle ++(4.36154,3.50000);\n\\draw[->] (7.20000,2.45000) -- (12.80000,2.45000);\n\\draw[->] (7.60000,0.70000) -- (7.60000,4.40000);\n\\draw[dashed] (8.23846,0.70000) -- (8.23846,4.20000);\n\\draw[fill=white,line width=1pt] (8.23846,2.45000) circle (0.09);\n\\node[] at (8.23846,2.05000) {$1/4$};\n\\node[font=\\sffamily\\footnotesize] at (10.98000,3.22000) {series converges};\n\\node[] at (9.90000,4.85000) {$N=4$: $\\mathfrak Z_N^{\\rm red}(s)$};\n\\node[] at (9.90000,0.00000) {single continued pole at $s=1/4$};\n\\end{tikzpicture}","(7.60000,0.70000) -- (7.60000,4.40000);\n\\draw[dashed] (8.23846,0.70000) -- (8.23846,4.20000);\n\\draw[fill=white,line width=1pt] (8.23846,2.45000) circle (0.09);\n\\node[] at (8.23846,2.05000) {$1/4$};\n\\node[font=\\sffamily\\footnotesize] at (10.98000,3.22000) {series converges};\n\\node[] at (9.90000,4.85000) {$N=4$: $\\mathfrak Z_N^{\\rm red}(s)$};\n\\node[] at (9.90000,0.00000) {single continued pole at $s=1/4$};\n\\end{tikzpicture}","idue $j$, evaluate the function at four rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\captiontext{The checker compares the actual complex aver…","r rotated arguments and attach the phase $i^{-j\\ell}$:\n\\[\n(\\mathcal B_{4,j}F)(z)=\\frac14\\sum_{\\ell=0}^{3}i^{-j\\ell}F(i^\\ell z).\n\\]\nA monomial $a_mz^m$ is multiplied by the finite average of $i^{(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\captiontext{The checker compares the actual complex average with independent coefficient selec…","(m-j)\\ell}$. Matching phases all vote $1$. Nonmatching phases cancel exactly.\n\\vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\captiontext{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","vfig{root_filter}\n\\head{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\captiontext{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","d{An example that can be checked coefficient by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\captiontext{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","by coefficient}\nTake $F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\captiontext{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","F(z)=\\sum_{m=0}^{12}(m+1)z^m$. Then\n\\[\n\\mathcal B_{4,2}F=3z^2+7z^6+11z^{10}.\n\\]\nApplying the same selector again changes nothing. Applying a different selector gives zero. Summing the four selected polynomials returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\captiontext{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","returns the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\captiontext{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","the original $F$.\n\\head{Selection is also a symmetry condition}\nFor the selected polynomial $F_2$, rotation gives $F_2(iz)=-F_2(z)$. More generally, a residue-$j$ section satisfies $F_j(\\omega z)=\\omega^jF_j(z)$. The phase label therefore describes both the coefficient addresses and the response to rotation.\n\\takeaway{This is not an approximate Fourier fit. The finite root sum is exactly $0$ or $1$ on every coefficient.}\n\\captiontext{The checker compares the actual complex average with independent coefficient selection for this polynomial. The drawing shows the four complex phases, not a numerical convergence argument.}","ditional monograph: Theorem 3.3 (p. 5).}\nSelection separates a function additively. It does not make multiplication diagonal. A residue-$i$ term multiplied by a residue-$j$ term lands in residue $i+j\\pmod N$.\nFor $N=3$, use\n\\[\nF=1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra behind the later addition formulas and block-coefficient recurrences.\n\\takeaway{A packet label tells you where an interaction lands, not that interactions with other packets vanish.}\n\\captiontext{All nine table entries and all three…","ltiplication diagonal. A residue-$i$ term multiplied by a residue-$j$ term lands in residue $i+j\\pmod N$.\nFor $N=3$, use\n\\[\nF=1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra behind the later addition formulas and block-coefficient recurrences.\n\\takeaway{A packet label tells you where an interaction lands, not that interactions with other packets vanish.}\n\\captiontext{All nine table entries and all three output sections are checked as exact polynomial identities. This is the scalar cyclic specialization u…","term multiplied by a residue-$j$ term lands in residue $i+j\\pmod N$.\nFor $N=3$, use\n\\[\nF=1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra behind the later addition formulas and block-coefficient recurrences.\n\\takeaway{A packet label tells you where an interaction lands, not that interactions with other packets vanish.}\n\\captiontext{All nine table entries and all three output sections are checked as exact polynomial identities. This is the scalar cyclic specialization used by the paper.}","1+2z+3z^2+z^3,\\qquad G=2+z+z^2.\n\\]\nThe table displays every product of their three selected sections.\n\\vfig{cyclic_product}\nThe shaded cells contribute to residue zero:\n\\[\n2(1+z^3)+(2z)(z^2)+(3z^2)z=2+7z^3.\n\\]\nKeeping 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.\n\\head{The general rule}\n\\[\n\\mathcal B_{N,k}(FG)=\n\\sum_{i+j\\equiv k\\ (N)}(\\mathcal B_{N,i}F)(\\mathcal B_{N,j}G).\n\\]\nEvery pair belongs to exactly one output residue. No term is discarded and no coefficient is estimated. This finite cyclic convolution is the algebra behind the later addition formulas and block-coefficient recurrences.\n\\takeaway{A packet label tells you where an interaction lands, not that interactions with other packets vanish.}\n\\captiontext{All nine table entries and all three output sections are checked as exact polynomial identities. This is the scalar cyclic specialization used by the paper.}","{z^6}{6!}-\\cdots,\n\\qquad D_{3,1}=z-\\frac{z^4}{4!}+\\cdots.\n\\]\nRows 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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\captiontext{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","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.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\captiontext{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","igns record crossing the signed edge while moving from the column to the row. They are not arbitrary orientation choices.\n\\head{Two checks connect the matrix and the scalar equation}\nThe initial derivative matrix is the identity. Each initial jet $(y(0),y'(0),y''(0))$ therefore determines a unique linear combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\captiontext{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","combination. Also,\n\\[\n\\frac13\\Tr(e^{zJ_3})=D_3(z),\\qquad \\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\captiontext{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","\\det(XI-J_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\captiontext{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","_3)=X^3+1.\n\\]\nThe roots of $-1$ appearing later in the exponential sum are already the spectrum of this concrete matrix.\n\\takeaway{The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.}\n\\captiontext{The exact tests check every matrix-entry Taylor coefficient through degree 14 for $N=1,\\ldots,5$. The displayed determinant-one law is for $N\\ge2$. $N=1$ is a separate boundary case.}","1}(z)\\xi+D_{3,2}(z)\\xi^2.\n\\]\nThe multiplication table makes the carry sign unavoidable.\n\\vfig{signed_addition}\nMultiply the coordinate forms of $e^{z\\xi}$ and $e^{w\\xi}$. Their product is $e^{(z+w)\\xi}$. The coefficient of $1$ receives one contribution without a wrap and two contributions with a wrap. Reading that coefficient gives the displayed addition law.\n\\head{Compare the unsigned model}\nIn $\\CC[\\xi]/(\\xi^3-1)$, 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.\nFor general $N$, the carry factor is\n\\[\n(-1)^{\\lfloor(i+j)/N\\rfloor},\\qquad 0\\le i,j0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exa…","p. 19), Example 9.11 (p. 22).}\nThe signed denominator and the normalized unsigned sections produce different reciprocal families. In order two,\n\\[\nD_2^{-1}=\\sec z,\\qquad C_{2,0}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed …","11 (p. 22).}\nThe signed denominator and the normalized unsigned sections produce different reciprocal families. In order two,\n\\[\nD_2^{-1}=\\sec z,\\qquad C_{2,0}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent t…","roduce different reciprocal families. In order two,\n\\[\nD_2^{-1}=\\sec z,\\qquad C_{2,0}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary …","order two,\n\\[\nD_2^{-1}=\\sec z,\\qquad C_{2,0}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by …","\\sec z,\\qquad C_{2,0}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorial…","}=\\sech z,\\qquad C_{2,1}=z/\\sinh z.\n\\]\nTheir first block-EGF coefficients make the difference visible.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","ble.\n\\vfig{different_reciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","eciprocals}\n\\head{Remove the forced zero before taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","taking a reciprocal}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","l}\nThe section $E_{N,j}$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","$ begins with $z^j/j!$. Define\n\\[\nE_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","E_{N,j}(z)=\\frac{z^j}{j!}\\Phi_{N,j}(z^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","^N),\\qquad\\Phi_{N,j}(0)=1.\n\\]\nThen $C_{N,j}(z)=1/\\Phi_{N,j}(z^N)$ has value $1$ at the origin. For $j>0$, the notation $(z^j/j!)/E_{N,j}(z)$ means this removable continuation. It does not mean evaluating an undefined $0/0$ expression.\nFor $N=2,j=1$, the first nontrivial coefficient is $-1/3$ in front of $z^2/2!$, equivalently $-z^2/6$. Dropping the numerator $z$ would instead produce the pole of $1/\\sinh z$.\n\\takeaway{A minus sign in the denominator series and removal of an initial zero are different operations. Their reciprocal coefficients should not be interchanged.}\n\\captiontext{All entries in the table are exact rationals computed from independent triangular inverses. The table explicitly uses block-EGF columns. Ordinary coefficients are obtained by dividing by the displayed factorials.}","fig{appell_operator}\nFor $m=4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are\n\\[\nE_m'=mE_{m-1},\\qquad E_m(x+y)=\\sum_{q=0}^m\\binom mqE_q(x)y^{m-q}.\n\\]\nBoth follow from the generating function. No operator-series convergence assumption on an infinite-dimensional function space is needed for this polynomial calculation.\n\\takeaway{Formal substitution into the derivative is safe here because every polynomial sees only finitely many derivatives. That is a domain statement.}\n\\captiontext{The polynomial, its inverse image, and the Appell identities are checked symbolically through degree eight in four normalized-section families. The dis…","4$, only three terms survive:\n\\[\n\\left(1-\\frac{\\partial_x^2}{2!}+5\\frac{\\partial_x^4}{4!}\\right)x^4\n=x^4-6x^2+5.\n\\]\nThe four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are\n\\[\nE_m'=mE_{m-1},\\qquad E_m(x+y)=\\sum_{q=0}^m\\binom mqE_q(x)y^{m-q}.\n\\]\nBoth follow from the generating function. No operator-series convergence assumption on an infinite-dimensional function space is needed for this polynomial calculation.\n\\takeaway{Formal substitution into the derivative is safe here because every polynomial sees only finitely many derivatives. That is a domain statement.}\n\\captiontext{The polynomial, its inverse image, and the Appell identities are checked symbolically through degree eight in four normalized-section families. The displayed curves are samples of…","The four real roots of this polynomial are $\\pm1,\\pm\\sqrt5$. They are not the zeros of the original analytic denominator. The plot distinguishes the input polynomial and its transformed output.\n\\head{An inverse operation, not a truncation error}\nApplying $\\cosh(\\partial_x)$ returns $x^4$ exactly. The general inverse uses the reciprocal formal symbol. The lowering and translation identities are\n\\[\nE_m'=mE_{m-1},\\qquad E_m(x+y)=\\sum_{q=0}^m\\binom mqE_q(x)y^{m-q}.\n\\]\nBoth follow from the generating function. No operator-series convergence assumption on an infinite-dimensional function space is needed for this polynomial calculation.\n\\takeaway{Formal substitution into the derivative is safe here because every polynomial sees only finitely many derivatives. That is a domain statement.}\n\\captiontext{The polynomial, its inverse image, and the Appell identities are checked symbolically through degree eight in four normalized-section families. The displayed curves are samples of exact polynomials.}","p. 23), Proposition 10.3 (p. 23), Proposition 10.4 (p. 23).}\nLet $H_N(z)=E_{N,0}(z)$. The theta family uses the same parameter $u$ twice, but in two different roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\captiontext{T…","E_{N,0}(z)$. The theta family uses the same parameter $u$ twice, but in two different roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\captiontext{The exact checker compares theta convolution with its logarithmic Newton r…","fferent roles:\n\\[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\captiontext{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly r…","[\nH_N(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\captiontext{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an in…","(uz)^{-1}H_N(z)^{-u}\n=\\sum_{n\\ge0}\\Theta_{N,n}(u)\\frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\captiontext{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid…","frac{z^{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\captiontext{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","{Nn}}{(Nn)!}.\n\\]\nArgument scaling contributes $u^{Nm}$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\captiontext{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","$ at logarithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\captiontext{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","garithmic block $m$. Reciprocal order contributes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\captiontext{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","ibutes $u$.\n\\vfig{theta}\nTheir combined logarithm has coefficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\captiontext{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","efficient\n\\[\n(-1)^m\\kappa_{N,m}\\frac{u^{Nm}+u}{m}.\n\\]\nAt the first block this gives $\\Theta_{N,1}(u)=-(u^N+u)$. 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 $u=-1$.\n\\head{The odd-order boundary is visible}\nFor $N=3$, $\\Theta_{3,1}(-1)=2$, not zero. The factor $u+1$ is not a uniform law across orders. In contrast,\n\\[\n\\Theta_{2,2}(u)=u(u+1)(5u^2+u+2).\n\\]\nThe two guaranteed factors follow from the whole generating function, not from guessing the first few polynomial roots.\n\\takeaway{A common parameter does not make argument scaling and reciprocal order interchangeable. Their powers must be tracked separately.}\n\\captiontext{The exact checker compares theta convolution with its logarithmic Newton recurrence through block index four. The odd-order factor claim is explicitly rejected as an invalid control.}","y large radius, smaller terms cannot cancel it. Those sectors are eventually zero-free.\n\\head{Why zeros remain possible on a growing ray}\nAt 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 $\\theta=k\\pi/2$.\nThe indicator uses a limsup in radius. It does not assert that $r^{-1}\\log|D_N(re^{i\\theta})|$ has an ordinary limit through the zeros on a tie ray.\n\\takeaway{The polygon predicts where large cancellation is possible. It does not, by itself, enumerate zeros or prove their simplicity.}\n\\captiontext{The order-four polygon and support-function formula are explicit. The curve is a sampled display of that formula. Eventual localization and uniqueness require the separate zero-analysis arguments.}","o-free.\n\\head{Why zeros remain possible on a growing ray}\nAt 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 $\\theta=k\\pi/2$.\nThe indicator uses a limsup in radius. It does not assert that $r^{-1}\\log|D_N(re^{i\\theta})|$ has an ordinary limit through the zeros on a tie ray.\n\\takeaway{The polygon predicts where large cancellation is possible. It does not, by itself, enumerate zeros or prove their simplicity.}\n\\captiontext{The order-four polygon and support-function formula are explicit. The curve is a sampled display of that formula. Eventual localization and uniqueness require the separate zero-analysis arguments.}","cos(\\sqrt3t/2)}{3}.\n\\]\nDivide by the nonzero factor $2e^{t/2}/3$. The resulting equation separates the dominant cosine from the small third exponential.\n\\vfig{two_term}\nIgnoring the last term gives model zeros\n\\[\nt_m=\\frac{2\\pi}{\\sqrt3}(m+1/2).\n\\]\nAt $t_m$, 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\n\\[\n\\delta_m^{(1)}=\\frac{(-1)^m}{\\sqrt3}e^{-3t_m/2}.\n\\]\nThe alternating slope explains why the true zero is to the right of the first model zero, then to the left of the second, and so on in the eventual regime.\n\\head{Symmetry needs uniqueness}\nReal coefficients alone allow nonreal conjugate pairs. The manuscript first proves that a near-model disk contains exactly one zero. Conjugation preserves that disk, so the unique zero must equal its conjugate. This additional uniqueness step, not the symmetry alone, places it exactly on the ray.\n\\takeaway{The correction is caused by a real, nonzero remainder. A good cosine approximation does not make its model zeros exact.}\n\\captiontext{The curves are s…","te endpoint signs prove existence, and an interval bound for $D_3'$ excludes zero throughout the interval, proving uniqueness inside it.\n\\vfig{cubic_certificates}\n\\head{How the certificate works}\nEvaluate the Taylor polynomial\n\\[\nS_{90}(t)=\\sum_{n=0}^{90}\\frac{(-1)^nt^{3n}}{(3n)!}\n\\]\nat the two rational endpoints. From the next term onward, magnitudes decrease, because their successive ratio is\n\\[\n\\frac{t^3}{(3n+1)(3n+2)(3n+3)}<1\n\\]\nthere. 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.\nThe first bracket is\n\\[\n1.84981279919014347629<\\tau_0<1.84981279919014347630.\n\\]\nIts displacement differs from the leading asymptotic prediction by about five percent. The later examples have much smaller relative discrepancies, as the plate records.\n\\takeaway{A small residual is not a root certificate. Endpoint signs plus a derivative bound give a checkable local existence-and-uniqueness statement.}\n\\captiontext{Four certified intervals do not establish t…","1)^nt^{3n}}{(3n)!}\n\\]\nat the two rational endpoints. From the next term onward, magnitudes decrease, because their successive ratio is\n\\[\n\\frac{t^3}{(3n+1)(3n+2)(3n+3)}<1\n\\]\nthere. 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.\nThe first bracket is\n\\[\n1.84981279919014347629<\\tau_0<1.84981279919014347630.\n\\]\nIts displacement differs from the leading asymptotic prediction by about five percent. The later examples have much smaller relative discrepancies, as the plate records.\n\\takeaway{A small residual is not a root certificate. Endpoint signs plus a derivative bound give a checkable local existence-and-uniqueness statement.}\n\\captiontext{Four certified intervals do not establish that no other real or complex zeros exist. The displacement columns are high-precision diagnostics, not a finite verification of a global big-$O$ theorem. All rational enclosures are included in the checks file.}","er relative discrepancies, as the plate records.\n\\takeaway{A small residual is not a root certificate. Endpoint signs plus a derivative bound give a checkable local existence-and-uniqueness statement.}\n\\captiontext{Four certified intervals do not establish that no other real or complex zeros exist. The displacement columns are high-precision diagnostics, not a finite verification of a global big-$O$ theorem. All rational enclosures are included in the checks file.}","ions are $v=(m+1/2)^4$.\n\\vfig{counting}\nThe 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.\n\\head{Separate radius conversion from orbit multiplicity}\nIn unscaled variables the general relation is\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\captiontext{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for…","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.\n\\head{Separate radius conversion from orbit multiplicity}\nIn unscaled variables the general relation is\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\captiontext{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify di","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.\n\\head{Separate radius conversion from orbit multiplicity}\nIn unscaled variables the general relation is\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\captiontext{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify discarding that boundary elsewhere.}","hed. Both formulas include zeros on the boundary of the disk.\n\\head{Separate radius conversion from orbit multiplicity}\nIn unscaled variables the general relation is\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\captiontext{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify discarding that boundary elsewhere.}","s\n\\[\nn_{D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\captiontext{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify discarding that boundary elsewhere.}","D_N}(R^{1/N})=Nn_{\\Delta_N}(R).\n\\]\nThere are two operations here: replace the radius by its $N$th root and divide the count by $N$. Forgetting either operation changes the leading constant.\nFor this family the asymptotics are\n\\[\nn_{D_N}(R)=\\frac{N\\sin(\\pi/N)}\\pi R+O(1),\\qquad\nn_{\\Delta_N}(R)=\\frac{\\sin(\\pi/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\captiontext{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify discarding that boundary elsewhere.}","i/N)}\\pi R^{1/N}+O(1).\n\\]\nThe different powers agree with the entire-function orders $1$ and $1/N$.\n\\takeaway{The reduced function has fewer repeated zeros and a different radius scale. Its counting law is not obtained by dividing a same-radius count alone.}\n\\captiontext{The exact order-four staircase formulas are tested at rational radii including jump points. The large-zero theorem for general $N$ permits finitely many exceptions. These exact order-four identities do not justify discarding that boundary elsewhere.}","e2$, the reduced entire function has order $1/N<1$. Its multiplicity-sensitive genus-zero product is\n\\[\n\\Delta_N(w)=\\prod_{\\lambda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\captiontext{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Orde…","e reduced entire function has order $1/N<1$. Its multiplicity-sensitive genus-zero product is\n\\[\n\\Delta_N(w)=\\prod_{\\lambda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\captiontext{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one i…","ltiplicity-sensitive genus-zero product is\n\\[\n\\Delta_N(w)=\\prod_{\\lambda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\captiontext{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so i…","bda\\in\\Lambda_N}(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\captiontext{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","(1-w/\\lambda).\n\\]\nPutting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\captiontext{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","Putting $K_Ne_\\lambda=\\lambda^{-1}e_\\lambda$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\captiontext{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","$ on $\\ell^2(\\Lambda_N)$ gives the corresponding diagonal Fredholm determinant. Repeated zeros require repeated diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\captiontext{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","diagonal entries.\n\\vfig{product}\n\\head{Finite compression is an approximation}\nFor $N=2$, $\\lambda_m=\\pi^2(m+1/2)^2$. The finite determinant $P_L$ is exactly the displayed finite product, but it is not exactly $\\Delta_2$. For $0\\le w\\le1$ and $L\\ge1$, an elementary tail estimate gives\n\\[\n0\\le1-\\frac{\\Delta_2(w)}{P_L(w)}\n\\le w\\sum_{m\\ge L}\\lambda_m^{-1}\n\\le\\frac1{9(L-1/2)}.\n\\]\nThe last bound uses $\\pi>3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\captiontext{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","3$ and integral comparison. Thus truncation has an explicit error mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\captiontext{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","mechanism.\n\\takeaway{The infinite product is justified by the analytic theorem. Finite determinants provide controlled approximations, not substitutes for that theorem.}\n\\captiontext{The exact finite determinant test includes a repeated eigenvalue and rejects deduplication. The curves use known cosine zeros. Order one is excluded: $\\Delta_1(w)=e^{-w}$ has no zeros, so its empty product would be $1$, not the function.}","ma_{2,2}=\\frac16,\\quad\n\\sigma_{2,3}=\\frac1{15},\\quad\\sigma_{2,4}=\\frac{17}{630}.\n\\]\nThe plate shows how successive inverse powers emphasize the nearest zero. Its shaded pieces are contributions from the first five positive reduced zeros, followed by the remaining tail. The total in each row is the exact rational moment.\n\\head{The tangent is a sum of poles}\nLogarithmic differentiation also gives\n\\[\nT_N(z)=Nz^{N-1}\\sum_{\\lambda\\in\\Lambda_N}\\frac1{\\lambda-z^N}\n\\]\naway from poles. Summability of $|\\lambda|^{-1}$ permits the locally uniform partial-fraction expansion without added regularizing terms.\n\\takeaway{A global analytic product theorem connects the infinite zero sum to a finite algebraic recurrence. The recurrence alone does not prove the product theorem.}\n\\captiontext{Displayed moments are checked against formal logarithm coefficients and exact order-two/order-four zeta values. Bar widths are numerical shares. General power sums use $\\lambda^{-q}$, not $|\\lambda|^{-q}$ when exceptional nonpositive zeros are present.}","s how successive inverse powers emphasize the nearest zero. Its shaded pieces are contributions from the first five positive reduced zeros, followed by the remaining tail. The total in each row is the exact rational moment.\n\\head{The tangent is a sum of poles}\nLogarithmic differentiation also gives\n\\[\nT_N(z)=Nz^{N-1}\\sum_{\\lambda\\in\\Lambda_N}\\frac1{\\lambda-z^N}\n\\]\naway from poles. Summability of $|\\lambda|^{-1}$ permits the locally uniform partial-fraction expansion without added regularizing terms.\n\\takeaway{A global analytic product theorem connects the infinite zero sum to a finite algebraic recurrence. The recurrence alone does not prove the product theorem.}\n\\captiontext{Displayed moments are checked against formal logarithm coefficients and exact order-two/order-four zeta values. Bar widths are numerical shares. General power sums use $\\lambda^{-q}$, not $|\\lambda|^{-q}$ when exceptional nonpositive zeros are present.}","action expansion without added regularizing terms.\n\\takeaway{A global analytic product theorem connects the infinite zero sum to a finite algebraic recurrence. The recurrence alone does not prove the product theorem.}\n\\captiontext{Displayed moments are checked against formal logarithm coefficients and exact order-two/order-four zeta values. Bar widths are numerical shares. General power sums use $\\lambda^{-q}$, not $|\\lambda|^{-q}$ when exceptional nonpositive zeros are present.}","hout added regularizing terms.\n\\takeaway{A global analytic product theorem connects the infinite zero sum to a finite algebraic recurrence. The recurrence alone does not prove the product theorem.}\n\\captiontext{Displayed moments are checked against formal logarithm coefficients and exact order-two/order-four zeta values. Bar widths are numerical shares. General power sums use $\\lambda^{-q}$, not $|\\lambda|^{-q}$ when exceptional nonpositive zeros are present.}","n-degree coefficients. Including both poles gives\n\\[\nP_{2,n}(1)\\sim2(2n)!\\left(\\frac2\\pi\\right)^{2n+1}.\n\\]\nThe factor two is orbit multiplicity, not an adjustable fitted constant.\n\\vfig{pole_asymptotic}\n\\head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\captionte…","factor two is orbit multiplicity, not an adjustable fitted constant.\n\\vfig{pole_asymptotic}\n\\head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\captiontext{The ratio certificates use a rational Machin-formula enclosure for $\\pi$. The inherited general theorem reta","head{A finite quantitative check}\nFor $n=1,\\ldots,8$, the exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\captiontext{The ratio certificates use a rational Machin-formula enclosure for $\\pi$. The inherited general theorem retains simple nearest zeros, a gap to the next modulus, and positive integral reciprocal order. …","he exact integer coefficients and a rational enclosure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\captiontext{The ratio certificates use a rational Machin-formula enclosure for $\\pi$. The inherited general theorem retains simple nearest zeros, a gap to the next modulus, and positive integral reciprocal order. The plot does not remove those hypotheses.}","ure for $\\pi$ 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 asserted all-$n$ bound.\n\\head{Higher reciprocal powers need more Laurent data}\nFor positive integral order $r$, a simple denominator zero becomes a pole of order $r$. The highest Laurent term contributes a factor $\\binom{m+r-1}{r-1}$, but lower Laurent terms still contribute on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\captiontext{The ratio certificates use a rational Machin-formula enclosure for $\\pi$. The inherited general theorem retains simple nearest zeros, a gap to the next modulus, and positive integral reciprocal order. The plot does not remove those hypotheses.}","on the same exponential scale. Subtracting \\emph{all} principal parts is what leaves a holomorphic remainder controlled by the next zero radius.\n\\takeaway{Nearest-zero geometry explains factorial-times-exponential growth. Multiplicity, reciprocal order, and lower principal parts determine its precise normalization.}\n\\captiontext{The ratio certificates use a rational Machin-formula enclosure for $\\pi$. The inherited general theorem retains simple nearest zeros, a gap to the next modulus, and positive integral reciprocal order. The plot does not remove those hypotheses.}","solute zero zeta in the original variable instead requires $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\captiontext{The shaded regions show convergence, and the open circles mark poles of the continued…","ro zeta in the original variable instead requires $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\captiontext{The shaded regions show convergence, and the open circles mark poles of the continued function…","nal variable instead requires $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\captiontext{The shaded regions show convergence, and the open circles mark poles of the continued functions. They are not zero…","nstead requires $\\Re s>1$.\n\\vfig{zeta_plane}\nFor $N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\captiontext{The shaded regions show convergence, and the open circles mark poles of the continued functions. They are not zeros of the zeta …","$N=2$ and $N=4$, the lattices are exact:\n\\[\n\\mathfrak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\captiontext{The shaded regions show convergence, and the open circles mark poles of the continued functions. They are not zeros of the zeta functions. Branch-dependent orbit zetas and deriv","frak Z_2^{\\rm red}(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathfrak Z_4^{\\rm red}(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\captiontext{The shaded regions show convergence, and the open circles mark poles of the continued functions. They are not zeros of the zeta functions. Branch-dependent orbit zetas and derivative-weighted zetas retain their separately stat…","(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nEach 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 $1/(2\\pi)$ or $1/(4\\sqrt2\\pi)$ respectively.\n\\head{The factor $N$ matters twice}\nFor every order in the paper,\n\\[\n\\mathfrak Z_N^{\\rm abs}(s)=N\\mathfrak Z_N^{\\rm red}(s/N).\n\\]\nFor general $N$, exponentially accurate zero locations leave an entire correction to the Hurwitz model. A finite exceptional zero set also contributes entire terms. This reasoning establishes continuation, not a zeta functional equation.\n\\takeaway{An analytic continuation can exist at points where the original infinite sum diverges. It must not be evaluated there by naively summing that series.}\n\\captiontext{The shaded regions show convergence, and the open circles mark poles of the continued functions. They are not zeros of the zeta functions. Branch-dependent orbit zetas and derivative-weighted zetas retain their separately stated conventions.}","ample 11.12 (p. 31), Theorem 12.6 (p. 35).}\nUse $N=4$ 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.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\captiontext{…","rem 12.6 (p. 35).}\nUse $N=4$ 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.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\captiontext{Every rational value and …","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.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\captiontext{Every rational value and polynomial identity…","recurrences provide two routes to the same coefficient. All symbols in the plate refer to this one order.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\captiontext{Every rational value and polynomial identity on this page is checked exactly. The infinite zero-sum interpretation uses the inherited ca…","cient. All symbols in the plate refer to this one order.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\captiontext{Every rational value and polynomial identity on this page is checked exactly. The infinite zero-sum interpretation uses the inherited canonical-product theorem and the exact order-four …","n the plate refer to this one order.\n\\vfig{worked_circuit}\n\\head{Route one: invert the denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\captiontext{Every rational value and polynomial identity on this page is checked exactly. The infinite zero-sum interpretation uses the inherited canonical-product theorem and the exact order-four zero formula.}","he denominator}\nThe $w^2$ coefficient of $\\Delta_4^{-1}$ is\n\\[\n\\frac1{24^2}-\\frac1{8!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\captiontext{Every rational value and polynomial identity on this page is checked exactly. The infinite zero-sum interpretation uses the inherited canonical-product theorem and the exact order-four zero formula.}","!}=\\frac{69}{8!}.\n\\]\nThus $P_{4,2}(1)=69$. Keeping $r$ variable gives $P_{4,2}(r)=35r^2+34r$.\n\\head{Route two: use the zero moments}\nThe exact reduced zeros are $4\\pi^4(m+1/2)^4$. Their first two power sums are $\\sigma_1=1/24$ and $\\sigma_2=17/10080$. The Bell/Newton identity gives the same reciprocal coefficient $(\\sigma_1^2+\\sigma_2)/2$. The tangent independently reads off the second moment as $4\\cdot7!\\sigma_2=34$.\nThis links coefficients, geometry, and derivatives without a fitted parameter or an inferred zero list.\n\\takeaway{The content is the agreement of precisely normalized descriptions. A matching small example illustrates the general bridges. It is not a proof of all their analytic hypotheses.}\n\\captiontext{Every rational value and polynomial identity on this page is checked exactly. The infinite zero-sum interpretation uses the inherited canonical-product theorem and the exact order-four zero formula.}","Historical certificates & The retained visual run records 147 checks, five rejected controls, and four rational zero brackets of width $10^{-20}$. The brackets certify local existence and uniqueness. The symbolic suites were hash-validated, not rerun in this edition."]}