1246 lines
35 KiB
TeX
1246 lines
35 KiB
TeX
\documentclass[11pt]{article}
|
|
|
|
\usepackage[T1]{fontenc}
|
|
\usepackage{lmodern}
|
|
\usepackage{amsmath,amssymb,amsthm,mathtools}
|
|
\usepackage{array,booktabs}
|
|
\usepackage{enumitem}
|
|
\usepackage[margin=1in]{geometry}
|
|
\usepackage{microtype}
|
|
\usepackage{xcolor}
|
|
\usepackage[hidelinks]{hyperref}
|
|
\usepackage{listings}
|
|
|
|
\definecolor{codegray}{RGB}{245,245,245}
|
|
\lstset{
|
|
basicstyle=\ttfamily\small,
|
|
backgroundcolor=\color{codegray},
|
|
frame=single,
|
|
breaklines=true,
|
|
columns=fullflexible,
|
|
keepspaces=true
|
|
}
|
|
|
|
\newtheorem{theorem}{Theorem}
|
|
\newtheorem{lemma}{Lemma}
|
|
\newtheorem{proposition}{Proposition}
|
|
\newtheorem{corollary}{Corollary}
|
|
\theoremstyle{definition}
|
|
\newtheorem{definition}{Definition}
|
|
\theoremstyle{remark}
|
|
\newtheorem{remark}{Remark}
|
|
|
|
\newcommand{\F}[2]{{}_{#1}F_{#2}}
|
|
\newcommand{\Q}{\mathbb{Q}}
|
|
\newcommand{\e}{\mathbf e}
|
|
\newcommand{\diag}{\operatorname{diag}}
|
|
\newcommand{\ord}{\operatorname{ord}}
|
|
|
|
\title{An Exact Hypergeometric Tail Certificate for\\
|
|
Ramanujan Challenge Problem 2.8}
|
|
\author{Problem 2.8 submission}
|
|
\date{July 2026}
|
|
|
|
\begin{document}
|
|
\maketitle
|
|
|
|
\begin{abstract}
|
|
Let \(G_N=M_0M_1\cdots M_{N-1}\) be the \(4\times4\) transfer
|
|
product in Ramanujan Challenge Problem 2.8, and let \(P_{N,j}\) and
|
|
\(Q_{N,j}\) be the two official seeded rows evaluated in column \(j\).
|
|
We prove
|
|
\[
|
|
\lim_{N\to\infty}\frac{P_{N,j}}{Q_{N,j}}
|
|
=\frac{\sqrt{10005}}{\pi}
|
|
\qquad (j=1,2,3,4).
|
|
\]
|
|
Equivalently, \(Q_{N,j}/P_{N,j}\to\pi/\sqrt{10005}\).
|
|
|
|
The missing connection constant is fixed by an exact rank-three
|
|
hypergeometric tail. A nonterminating \(\F43\) Euler jet is carried
|
|
backward by the displayed matrix, while the first denominator is a
|
|
terminating adjoint \(\F43\). Four denominator-cleared Ore identities
|
|
and three coefficient identities give an all-\(N\) Pad\'e divisibility
|
|
theorem. Positivity at the negative CM point, a Cauchy estimate, and an
|
|
explicit stable-graph contraction prove convergence for all four columns.
|
|
Every algebraic identity used below is reproduced by a dependency-free
|
|
rational-polynomial checker. The classical Chudnovsky formula, cited
|
|
precisely in Section~2, is the sole imported theorem.
|
|
\end{abstract}
|
|
|
|
\enlargethispage{2\baselineskip}
|
|
\tableofcontents
|
|
|
|
\section{Statement and compact form of the seeds}
|
|
|
|
Put
|
|
\[
|
|
R=151931373056001=53360^3+1,\qquad
|
|
x_0=\frac1R,\qquad
|
|
z=-\frac{x}{1-x}.
|
|
\]
|
|
Thus the official CM point is
|
|
\[
|
|
z_0=-\frac1{R-1}=-\frac1{53360^3}.
|
|
\]
|
|
Let
|
|
\[
|
|
G_N=M_0M_1\cdots M_{N-1},\qquad G_0=I_4,
|
|
\]
|
|
where the rational family is defined as follows. Put \(r=x^{-1}\),
|
|
\(\omega=u(3u-2)(3u+2)\), and
|
|
\begin{align*}
|
|
a_1={}&r(144u^5-288u^4+144u^3)
|
|
-99u^5+333u^4-229u^3-114u^2+40u+64,\\
|
|
a_2={}&r(432u^4-864u^3+432u^2)
|
|
-243u^4+909u^3-868u^2-80u+272,\\
|
|
a_3={}&r(432u^3-864u^2+432u)
|
|
-153u^3+648u^2-860u+360,\\
|
|
a_4={}&144r(u-1)^2,\\
|
|
b_1={}&-144ru^3+9u^4+63u^3+158u^2+168u+64,\\
|
|
b_2={}&216ru^2+36u^3-189u^2-316u-168,\\
|
|
b_3={}&108ru+54u^2-189u-158,\\
|
|
c_1={}&-288r^2u^3+
|
|
r(54u^4+378u^3+948u^2+1008u+384)\\
|
|
&\hspace{2.2em}+18u^5+45u^4-251u^3-1086u^2-1384u-576,\\
|
|
c_2={}&-432r^2u^2+
|
|
r(153u^4-657u^3+1292u^2+2064u+1072)\\
|
|
&\hspace{2.2em}-72u^4+702u^3-1069u^2-2508u-1512,\\
|
|
c_3={}&-216r^2u+
|
|
r(180u^3-891u^2+1450u+1116)\\
|
|
&\hspace{2.2em}-108u^3+864u^2-1385u-1422,\\
|
|
c_4={}&-4r^2+r(6u^2-33u+\tfrac{536}{9})
|
|
-4u^2+32u-63.
|
|
\end{align*}
|
|
Define
|
|
\begin{equation}\label{eq:deformed-matrix}
|
|
\mathcal M(u,x)=
|
|
\begin{pmatrix}
|
|
a_1/\omega&a_2/\omega&a_3/\omega&a_4/\omega\\
|
|
-u^3&-3u^2&-3u&-1\\
|
|
xb_1/144&-xb_2/72&-xb_3/36&
|
|
x(-2r-(2u-7))/2\\
|
|
x^2c_1/288&x^2c_2/144&x^2c_3/72&x^2c_4/4
|
|
\end{pmatrix},
|
|
\qquad M_N(x)=\mathcal M(2N+3,x).
|
|
\end{equation}
|
|
At \(x=x_0\), the only deformed challenge coefficient is restored by
|
|
\[
|
|
236337691420383=\frac{14R-567}{9}.
|
|
\]
|
|
Thus every use of Cauchy's theorem below concerns the explicitly displayed
|
|
rational \(x\)-family, not an unspecified continuation.
|
|
|
|
Define four Pascal rows
|
|
\[
|
|
\begin{aligned}
|
|
b_0&=(1,0,0,0),&
|
|
b_1&=(1,1,0,0),\\
|
|
b_2&=(1,2,1,0),&
|
|
b_3&=(1,3,3,1)
|
|
\end{aligned}
|
|
\]
|
|
and let \(\mathcal P\) be the matrix with rows \(b_0,b_1,b_2,b_3\).
|
|
The compact denominator row is
|
|
\[
|
|
C(x)=
|
|
\left(
|
|
\frac{18}{x}+\frac{159}{4},\
|
|
\frac{54}{x}+\frac{131}{2},\
|
|
\frac{54}{x}+27,\
|
|
\frac{18}{x}
|
|
\right).
|
|
\]
|
|
Equivalently,
|
|
\[
|
|
C=\frac{18}{x}b_3+\frac54b_0+\frac{23}{2}b_1+27b_2.
|
|
\]
|
|
Set
|
|
\[
|
|
A=13591409,\qquad B=545140134,\qquad S=426880.
|
|
\]
|
|
The two official initial rows have the exact form
|
|
\begin{equation}\label{eq:seed-identities}
|
|
A_1=SC,\qquad
|
|
A_0=AC-\frac54H_0,\qquad
|
|
H_0=Ab_0+Bb_1=(A+B,B,0,0).
|
|
\end{equation}
|
|
At \(x=x_0\), these identities reproduce the official integer rows
|
|
entry by entry:
|
|
\[
|
|
\begin{aligned}
|
|
A_0={}&(37169305760442252761441,\,
|
|
111507917281327441564208,\\
|
|
&\hspace{4.7em}111507917281327599720129,\,
|
|
37169305760442410917362),\\
|
|
A_1={}&(1167416361542639692320,\,
|
|
3502249084627896132160,\\
|
|
&\hspace{4.7em}3502249084627879697280,\,
|
|
1167416361542622723840).
|
|
\end{aligned}
|
|
\]
|
|
|
|
For \(j=1,\ldots,4\), write
|
|
\[
|
|
P_{N,j}=A_0G_N\e_j,\qquad
|
|
Q_{N,j}=A_1G_N\e_j.
|
|
\]
|
|
We first identify the first-column limit and then invoke the exact cyclic
|
|
frame to cover all four columns.
|
|
|
|
\section{The CM function and its exact value}
|
|
|
|
Let
|
|
\[
|
|
y(z)=\F32\left(
|
|
\begin{matrix}\frac16,\frac12,\frac56\\1,1\end{matrix};z
|
|
\right),
|
|
\qquad
|
|
\theta=z\frac{d}{dz}=(1-x)x\frac{d}{dx},
|
|
\]
|
|
and define
|
|
\begin{equation}\label{eq:phi}
|
|
\Phi(x)=\frac{Ay(z)+B\theta y(z)}{S}.
|
|
\end{equation}
|
|
|
|
We use one external theorem: the classical Chudnovsky identity. In the
|
|
normalization used here it is Theorem~0.1 of Milla's equation-by-equation
|
|
derivation \([2]\), whose modular and CM proof is completed in
|
|
Theorem~9.7 and Chapter~10:
|
|
\[
|
|
\frac1\pi=
|
|
\frac{12}{640320^{3/2}}
|
|
\sum_{k=0}^{\infty}
|
|
\frac{(6k)!}{(3k)!(k!)^3}
|
|
(A+Bk)(-640320^{-3})^k.
|
|
\]
|
|
The elementary coefficient identity
|
|
\[
|
|
\frac{(6k)!}{(3k)!(k!)^3}
|
|
=1728^k
|
|
\frac{(\frac16)_k(\frac12)_k(\frac56)_k}{(k!)^3}
|
|
\]
|
|
and \(640320=12\cdot53360\) give
|
|
\[
|
|
Ay(z_0)+B\theta y(z_0)
|
|
=\frac{640320^{3/2}}{12\pi}
|
|
=\frac{426880\sqrt{10005}}{\pi}.
|
|
\]
|
|
Consequently,
|
|
\begin{equation}\label{eq:CM-value}
|
|
\boxed{\Phi(x_0)=\frac{\sqrt{10005}}{\pi}.}
|
|
\end{equation}
|
|
|
|
\section{The nonterminating adjoint tail}
|
|
|
|
Put \(n=N+1\) and \(\delta_N=\theta-n\). Define
|
|
\begin{equation}\label{eq:tail}
|
|
F_N(z)=\kappa_Nz^n
|
|
\F43\left(
|
|
\begin{matrix}
|
|
n,n+\frac16,n+\frac12,n+\frac56\\
|
|
2n,2n,2n
|
|
\end{matrix};z\right),
|
|
\end{equation}
|
|
where
|
|
\[
|
|
\kappa_0=\frac5{72},\qquad
|
|
\frac{\kappa_{N+1}}{\kappa_N}
|
|
=-\frac{(6N+7)(6N+11)}
|
|
{576(N+1)^2(2N+3)^2}.
|
|
\]
|
|
Its Euler jet is
|
|
\[
|
|
k_N=\left(F_N,\delta_NF_N,\delta_N^2F_N,\delta_N^3F_N\right)^T.
|
|
\]
|
|
|
|
\begin{proposition}[Exact tail contiguity]\label{prop:tail-contiguity}
|
|
For every \(N\ge0\),
|
|
\begin{equation}\label{eq:tail-contiguity}
|
|
\boxed{M_Nk_{N+1}=k_N.}
|
|
\end{equation}
|
|
\end{proposition}
|
|
|
|
\begin{proof}
|
|
Put \(m=(u-1)/2=N+1\). For the \(r\)-th row of
|
|
\(\mathcal M(u,x)\), define
|
|
\[
|
|
P_r(t)=\sum_{s=0}^3\mathcal M(u,x)_{r+1,s+1}t^s,\qquad
|
|
\mathcal TQ=(1-x)x\partial_xQ+(t+1)Q.
|
|
\]
|
|
The shifted tail satisfies
|
|
\[
|
|
L_+(t)F_{N+1}=0,
|
|
\]
|
|
where
|
|
\[
|
|
L_+(t)=(1-x)t(t+u)^3+
|
|
x(t+m+1)(t+m+\tfrac76)(t+m+\tfrac32)(t+m+\tfrac{11}{6}).
|
|
\]
|
|
There is no division step: direct expansion gives the following four
|
|
denominator-cleared polynomial identities:
|
|
\begin{align}
|
|
u(3u-2)(3u+2)x(\mathcal TP_0-P_1)
|
|
& =144(u-1)^2L_+,\label{eq:ore0}\\
|
|
\mathcal TP_1-P_2&=-L_+,\label{eq:ore1}\\
|
|
2(\mathcal TP_2-P_3)&=(-2+7x-2ux)L_+,\label{eq:ore2}\\
|
|
36(\mathcal TP_3+\ell_3P_3+\ell_2P_2+\ell_1P_1+\ell_0P_0)
|
|
&=q_3L_+,\label{eq:ore3}
|
|
\end{align}
|
|
with
|
|
\begin{align*}
|
|
\ell_0={}&(u-1)u(3u-2)(3u+2)x/144,\\
|
|
\ell_1={}&[-576+864u-432u^2+72u^3
|
|
+(580-872u+405u^2-36u^3)x]/72,\\
|
|
\ell_2={}&[432-432u+108u^2
|
|
+(-436+405u-54u^2)x]/36,\\
|
|
\ell_3={}&(-12+6u+11x-2ux)/2,\\
|
|
q_3={}&-36+(536-297u+54u^2)x
|
|
+(-567+288u-36u^2)x^2.
|
|
\end{align*}
|
|
|
|
For clarity, the coefficient calculation producing the first row is also
|
|
written without a special-function routine. Set
|
|
\[
|
|
\begin{aligned}
|
|
A(t)&=\frac{144(u-1)^2(t+u)^3}{u(3u-2)(3u+2)},&
|
|
B(t)&=P_0(t)-\frac{A(t)}x,\\
|
|
\varrho&=-\frac{(3u-2)(3u+2)}{144(u-1)^2u^2},\\
|
|
\chi_j&=
|
|
\frac{(m+j)(m+\frac16+j)(m+\frac12+j)(m+\frac56+j)}
|
|
{m(m+\frac16)(m+\frac12)(m+\frac56)}\\
|
|
&\quad{}\times
|
|
\left(\frac{2m(2m+1)}{(2m+j)(2m+j+1)}\right)^3,\\
|
|
\psi_j&=
|
|
\frac{(m+j)(m+\frac16+j)(m+\frac12+j)(m+\frac56+j)}
|
|
{(2m+j)^3(j+1)} .
|
|
\end{aligned}
|
|
\]
|
|
The constant and generic coefficient equations are
|
|
\begin{equation}\label{eq:tail-coefficients}
|
|
\varrho[-A(0)]=1,\qquad
|
|
\varrho\left[-A(j)\chi_j+
|
|
\frac{(A(j-1)+B(j-1))\chi_{j-1}}{\psi_{j-1}}\right]=1
|
|
\quad(j\ge1).
|
|
\end{equation}
|
|
Clearing the displayed nonzero factors turns
|
|
\eqref{eq:ore0}--\eqref{eq:tail-coefficients} into polynomial
|
|
equalities with every coefficient zero. The independent script
|
|
\texttt{p28\_standalone\_equations.py} expands precisely these equalities
|
|
using only rational addition and multiplication. Equation
|
|
\eqref{eq:tail-coefficients} gives the first row of
|
|
\eqref{eq:tail-contiguity}; applying
|
|
\eqref{eq:ore0}--\eqref{eq:ore3} successively gives the other three
|
|
Euler-jet rows.
|
|
\end{proof}
|
|
|
|
For \(N=0\), write \(y(z)=1+\sum_{k\ge1}c_kz^k\). The coefficient of
|
|
\(z\) is
|
|
\[
|
|
c_1=\frac{(1/6)(1/2)(5/6)}{1^3}=\frac5{72},
|
|
\]
|
|
and, after shifting \(k\mapsto k+1\), both sides below have initial
|
|
coefficient \(5/72\) and consecutive-coefficient ratio
|
|
\[
|
|
\frac{(k+\frac16)(k+\frac12)(k+\frac56)}{(k+1)^3}.
|
|
\]
|
|
Hence coefficient equality, rather than a named ascension rule, gives
|
|
\begin{equation}\label{eq:ascension}
|
|
F_0=y-1
|
|
=\frac5{72}z
|
|
\F43\left(
|
|
\begin{matrix}1,\frac76,\frac32,\frac{11}{6}\\2,2,2\end{matrix};z
|
|
\right).
|
|
\end{equation}
|
|
Because \(\mathcal P\) is the Pascal matrix,
|
|
\[
|
|
\mathcal Pk_0=
|
|
(y-1,\theta y,\theta^2y,\theta^3y)^T.
|
|
\]
|
|
|
|
\section{The rank-three error carrier}
|
|
|
|
Set
|
|
\[
|
|
f=(y-1,\theta y,\theta^2y,\theta^3y)^T,\qquad
|
|
\mathcal E_0=\frac54\mathcal P+fC,\qquad
|
|
\mathcal E_N=\mathcal E_0G_N.
|
|
\]
|
|
The transformed hypergeometric equation is
|
|
\begin{equation}\label{eq:transformed-ode}
|
|
72\theta^3y+108x\theta^2y+46x\theta y+5xy=0.
|
|
\end{equation}
|
|
Indeed the coefficient ratio of \(y\) gives
|
|
\[
|
|
\left[\theta^3-
|
|
z(\theta+\tfrac16)(\theta+\tfrac12)(\theta+\tfrac56)\right]y=0.
|
|
\]
|
|
Substituting \(z=-x/(1-x)\), using
|
|
\(\theta=(1-x)x\partial_x\), and multiplying by \(72(1-x)\)
|
|
expands to \eqref{eq:transformed-ode}.
|
|
Using the displayed decomposition of \(C\), equations
|
|
\eqref{eq:ascension}--\eqref{eq:transformed-ode} give
|
|
\[
|
|
Ck_0=-\frac54.
|
|
\]
|
|
It follows that
|
|
\[
|
|
\mathcal E_0k_0=\frac54f+f(Ck_0)=0.
|
|
\]
|
|
Proposition~\ref{prop:tail-contiguity} therefore implies
|
|
\begin{equation}\label{eq:annihilation}
|
|
\boxed{\mathcal E_Nk_N=0\qquad(N\ge0).}
|
|
\end{equation}
|
|
The same differential equation gives the exact row relation
|
|
\begin{equation}\label{eq:row-relation}
|
|
72(\mathcal E_N)_{3,*}
|
|
+108x(\mathcal E_N)_{2,*}
|
|
+46x(\mathcal E_N)_{1,*}
|
|
+5x(\mathcal E_N)_{0,*}=0.
|
|
\end{equation}
|
|
|
|
\section{A discrete valuation lemma}
|
|
|
|
Let
|
|
\[
|
|
H=\diag(x,1,1,1),\qquad J_N=HM_N.
|
|
\]
|
|
Every entry of \(J_N\) is regular at \(x=0\). If \(u=2N+3\) and
|
|
\[
|
|
a_N=\frac{144(u-1)^2}{u(3u-2)(3u+2)},\qquad
|
|
V_N=(u^3,3u^2,3u,1),
|
|
\]
|
|
direct substitution in the authoritative matrix gives
|
|
\begin{equation}\label{eq:rank-one}
|
|
J_N(0)=
|
|
\begin{pmatrix}a_N\\-1\\-1\\-1\end{pmatrix}V_N.
|
|
\end{equation}
|
|
Thus \(J_N(0)\) has rank one.
|
|
|
|
The first nonconstant coefficient of the \(\F43\) in
|
|
\eqref{eq:tail} equals
|
|
\[
|
|
c_N=\frac{u(3u-2)(3u+2)}{144(u-1)^2}=a_N^{-1}.
|
|
\]
|
|
Since \(z=-x+O(x^2)\),
|
|
\begin{equation}\label{eq:tail-direction}
|
|
Hk_N=x^{N+2}\eta_N
|
|
\left[
|
|
\begin{pmatrix}1\\-c_N\\-c_N\\-c_N\end{pmatrix}
|
|
+O(x)
|
|
\right],\qquad \eta_N\ne0.
|
|
\end{equation}
|
|
The leading vector in \eqref{eq:tail-direction} is precisely the image
|
|
direction in \eqref{eq:rank-one}.
|
|
The other expansion required below is equally direct. Since
|
|
\[
|
|
F_{N+1}=\kappa_{N+1}z^{N+2}(1+O(z)),\qquad
|
|
\delta_{N+1}=\theta-(N+2),
|
|
\]
|
|
the shifted derivatives kill the leading monomial, and therefore
|
|
\begin{equation}\label{eq:next-tail-direction}
|
|
k_{N+1}=x^{N+2}\widetilde\eta_N
|
|
\left(\e_1+xs_N+O(x^2)\right),
|
|
\qquad\widetilde\eta_N\ne0.
|
|
\end{equation}
|
|
Moreover, \eqref{eq:rank-one} gives
|
|
\[
|
|
J_N(0)\e_1=u^3(a_N,-1,-1,-1)^T\ne0.
|
|
\]
|
|
|
|
\begin{lemma}[DVR step, including the extra first-column zero]
|
|
\label{lem:dvr}
|
|
Let \(R_0=\Q[[x]]\), \(H=\diag(x,1,1,1)\), and suppose
|
|
\[
|
|
E=x^NLH,\qquad L\in\operatorname{Mat}_4(R_0),\qquad Ek=0.
|
|
\]
|
|
Assume \(J=HM\in\operatorname{Mat}_4(R_0)\), \(Mk^+=k\), and
|
|
\[
|
|
\begin{aligned}
|
|
k^+&=x^r\alpha(\e_1+xs+O(x^2)),\\
|
|
Hk&=x^r\beta(v_0+xv_1+O(x^2)),
|
|
\end{aligned}
|
|
\]
|
|
with \(\alpha\beta\ne0\). If \(J(0)\) has rank one,
|
|
\(\operatorname{im}J(0)=\Q v_0\), and \(J(0)\e_1\ne0\), then
|
|
\[
|
|
EM=x^{N+1}L^+H
|
|
\]
|
|
for some \(L^+\in\operatorname{Mat}_4(R_0)\).
|
|
\end{lemma}
|
|
|
|
\begin{proof}
|
|
Absorb \(\beta/\alpha\) into \(v_0,v_1\). From \(Jk^+=Hk\),
|
|
\[
|
|
J(\e_1+xs+O(x^2))=v_0+xv_1+O(x^2).
|
|
\]
|
|
Write \(L=L_0+xL_1+\cdots\). The equation \(L(Hk)=0\) gives
|
|
\[
|
|
L_0v_0=0,\qquad L_0v_1+L_1v_0=0.
|
|
\]
|
|
Since \(J(0)\) has image \(\Q v_0\), \(L_0J(0)=0\), so \(LJ\) is
|
|
entrywise divisible by \(x\). The coefficient of \(x\) in its first
|
|
column is
|
|
\[
|
|
L_0(v_1-J(0)s)+L_1v_0=-L_0J(0)s=0.
|
|
\]
|
|
Hence that column is divisible by \(x^2\). Therefore
|
|
\[
|
|
L^+=x^{-1}LJH^{-1}
|
|
\]
|
|
is regular and \(EM=x^{N+1}L^+H\).
|
|
\end{proof}
|
|
|
|
\begin{proposition}[All-\(N\) Pad\'e divisibility]\label{prop:divisibility}
|
|
For every \(N\ge0\), there is
|
|
\(L_N\in\operatorname{Mat}_4(\Q[[x]])\) such that
|
|
\begin{equation}\label{eq:divisibility}
|
|
\boxed{\mathcal E_N=x^NL_NH.}
|
|
\end{equation}
|
|
Thus every row of \(\mathcal E_N\) has componentwise valuations at least
|
|
\[
|
|
(N+1,N,N,N).
|
|
\]
|
|
The last row has the stronger valuations
|
|
\[
|
|
(N+2,N+1,N+1,N+1).
|
|
\]
|
|
\end{proposition}
|
|
|
|
\begin{proof}
|
|
Every component of \(f\) is \(O(x)\), while \(C\) has only a simple pole.
|
|
All four components of \(f\) have leading term \(-5x/72\).
|
|
Consequently the constant term in the first column of \(fC\) is
|
|
\(-5/4\), cancelling the first component of every row of
|
|
\((5/4)\mathcal P\). Hence \(\mathcal E_0=L_0H\).
|
|
|
|
Apply Lemma~\ref{lem:dvr} inductively, using
|
|
\eqref{eq:annihilation}, \eqref{eq:rank-one},
|
|
\eqref{eq:tail-direction}, \eqref{eq:next-tail-direction}, the displayed
|
|
nonzero first column of \(J_N(0)\), and \(M_Nk_{N+1}=k_N\).
|
|
The stronger last-row assertion follows from
|
|
\eqref{eq:row-relation}.
|
|
\end{proof}
|
|
|
|
\section{The terminating denominator}
|
|
|
|
For the first column put
|
|
\[
|
|
q_N(x)=CG_N\e_1,\qquad n=N+1,\qquad Q_N(x)=x^nq_N(x),
|
|
\]
|
|
and define
|
|
\[
|
|
\widehat Q_N(z)
|
|
=(1-z)^nQ_N\!\left(-\frac{z}{1-z}\right).
|
|
\]
|
|
Since \(x=-z/(1-z)\), this is also the first component of
|
|
\((-z)^nCG_N\).
|
|
|
|
\begin{proposition}[Exact terminating denominator]
|
|
\label{prop:terminating}
|
|
For every \(N\ge0\),
|
|
\begin{equation}\label{eq:qhat}
|
|
\frac{\widehat Q_N(z)}{\alpha_n}
|
|
=\F43\left(
|
|
\begin{matrix}
|
|
-n,-n-\frac16,-n-\frac12,-n-\frac56\\
|
|
1-2n,1-2n,1-2n
|
|
\end{matrix};z
|
|
\right),
|
|
\end{equation}
|
|
where
|
|
\begin{equation}\label{eq:normalization}
|
|
\alpha_1=18,\qquad
|
|
\frac{\alpha_{n+1}}{\alpha_n}
|
|
=\frac{576n^2(2n+1)^2}{(6n+1)(6n+5)}.
|
|
\end{equation}
|
|
Here and below the hypergeometric expression denotes the unambiguous finite
|
|
sum over \(0\le k\le n\); it terminates before any lower Pochhammer symbol
|
|
can vanish.
|
|
\end{proposition}
|
|
|
|
\begin{proof}
|
|
Write \(p_n=\widehat Q_{n-1}\) for \(n\ge1\). We use coefficient
|
|
induction, because normalization at \(z=0\) alone
|
|
would not select a unique solution of the fourth-order equation. To expose
|
|
the matrix-to-scalar bridge, write
|
|
\[
|
|
\mathcal L_n(t)=t(t+2n-1)^3
|
|
-z(t+n)(t+n+\tfrac16)(t+n+\tfrac12)(t+n+\tfrac56)
|
|
=\sum_{j=0}^4c_jt^j
|
|
\]
|
|
and define, for operator polynomials with coefficients on the left,
|
|
\[
|
|
\Theta Q=z\partial_zQ+tQ.
|
|
\]
|
|
The four horizontal components are reconstructed from the first by
|
|
\[
|
|
\pi_0=1,\qquad
|
|
\pi_3=\frac{c_4}{c_0}t,\qquad
|
|
\pi_2=\frac{c_3}{c_4}\pi_3-\Theta\pi_3,\qquad
|
|
\pi_1=\frac{c_2}{c_4}\pi_3-\Theta\pi_2.
|
|
\]
|
|
The remaining horizontal equation is the direct factorization
|
|
\[
|
|
\begin{aligned}
|
|
\Theta\pi_1+1-\frac{c_1}{c_4}\pi_3
|
|
={}&-\frac{72}{n(2n+1)(6n+1)(6n+5)z}\bigl[
|
|
t(t-2n)^3\\
|
|
&\hspace{4em}
|
|
-z(t-n)(t-n-\tfrac16)
|
|
(t-n-\tfrac12)(t-n-\tfrac56)\bigr].
|
|
\end{aligned}
|
|
\]
|
|
Let
|
|
\[
|
|
\mathcal C_n(z)=
|
|
\begin{pmatrix}
|
|
0&1&0&0\\
|
|
0&0&1&0\\
|
|
0&0&0&1\\
|
|
-c_0/c_4&-c_1/c_4&-c_2/c_4&-c_3/c_4
|
|
\end{pmatrix}.
|
|
\]
|
|
The propagation of the reconstructed row is the following sixteen-entry
|
|
identity:
|
|
\begin{equation}\label{eq:full-gauge}
|
|
\mathcal C_n[-z\mathcal M(2n+1,-z/(1-z))]
|
|
-\theta[-z\mathcal M(2n+1,-z/(1-z))]
|
|
-[-z\mathcal M(2n+1,-z/(1-z))]\mathcal C_{n+1}=0.
|
|
\end{equation}
|
|
Finally, direct contraction with the displayed matrix gives
|
|
\begin{equation}\label{eq:matrix-scalar-bridge}
|
|
\sum_{r=0}^3\pi_r(t)
|
|
\left[-z\mathcal M(2n+1,-z/(1-z))\right]_{r+1,1}
|
|
=d_0(t)+zd_1(t).
|
|
\end{equation}
|
|
These are identities in \(\Q(n,z)(t)\). The factors depending on the
|
|
integer \(n\) are nonzero for \(n\ge1\); no value is obtained by dividing
|
|
at \(z=0\), because the verifier cross-multiplies first and the reconstructed
|
|
base functions have removable limits there. The mandatory
|
|
sparse-polynomial verifier checks every
|
|
reconstruction equation, the closing factorization, and
|
|
\eqref{eq:full-gauge}--\eqref{eq:matrix-scalar-bridge} by cross
|
|
multiplication. For any reconstructed horizontal row these identities give
|
|
\begin{equation}\label{eq:terminating-step}
|
|
p_{n+1}
|
|
=\bigl(d_0(\theta)+zd_1(\theta)\bigr)p_n,
|
|
\end{equation}
|
|
where
|
|
\[
|
|
d_0(t)=
|
|
\frac{72(2n+1)^2(2n-t)^3}{n(6n+1)(6n+5)}
|
|
\]
|
|
and
|
|
\[
|
|
d_1(t)=-\frac{P(n,t)}{n(2n+1)(6n+1)(6n+5)}
|
|
\]
|
|
with
|
|
\begin{align*}
|
|
P(n,t)={}&-5n-76n^2+1404n^3+4360n^4+4320n^5+1440n^6\\
|
|
&+(5+127n-1760n^2-6536n^3-7632n^4-3024n^5)t\\
|
|
&+(-51+659n+3086n^2+4500n^3+2232n^4)t^2\\
|
|
&+(-72-432n-864n^2-576n^3)t^3.
|
|
\end{align*}
|
|
Thus \eqref{eq:terminating-step} is the scalar form of the displayed
|
|
matrix recurrence; no differential-equation uniqueness is used below.
|
|
|
|
Let
|
|
\[
|
|
h_{n,k}=
|
|
\frac{(-n)_k(-n-\frac16)_k(-n-\frac12)_k(-n-\frac56)_k}
|
|
{(1-2n)_k^3k!}
|
|
\quad(0\le k\le n),
|
|
\qquad
|
|
\nu_n=\frac{576n^2(2n+1)^2}{(6n+1)(6n+5)}.
|
|
\]
|
|
For \(n\ge1\), clearing the displayed nonzero denominators gives exactly
|
|
\begin{align}
|
|
d_0(0)&=\nu_n,\label{eq:term-constant}\\
|
|
d_0(k)+d_1(k-1)\frac{h_{n,k-1}}{h_{n,k}}
|
|
&=\nu_n\frac{h_{n+1,k}}{h_{n,k}}
|
|
\qquad(1\le k\le n),\label{eq:term-generic}\\
|
|
d_1(n)&=-\nu_n
|
|
\frac{(n+\frac76)(n+\frac32)(n+\frac{11}{6})}
|
|
{8(2n+1)^3}.\label{eq:term-top}
|
|
\end{align}
|
|
The base row is not assumed: from the first component of \(C\),
|
|
\[
|
|
q_0(x)=\frac{18}{x}+\frac{159}{4},\qquad
|
|
Q_0(x)=xq_0(x)=18+\frac{159}{4}x,
|
|
\]
|
|
and therefore
|
|
\[
|
|
p_1(z)=(1-z)Q_0\!\left(-\frac{z}{1-z}\right)
|
|
=18\left(1-\frac{77}{24}z\right).
|
|
\]
|
|
Direct application of the four displayed \(\pi_r\) operators gives the
|
|
four base identities
|
|
\begin{equation}\label{eq:base-horizontal-row}
|
|
-zC\!\left(-\frac{z}{1-z}\right)
|
|
=\left.(\pi_0(\theta)p_1,\pi_1(\theta)p_1,
|
|
\pi_2(\theta)p_1,\pi_3(\theta)p_1)\right|_{n=1}.
|
|
\end{equation}
|
|
The fourth base equation is also direct:
|
|
\[
|
|
\left[
|
|
\theta(\theta-2)^3
|
|
-z(\theta-1)(\theta-\tfrac76)
|
|
(\theta-\tfrac32)(\theta-\tfrac{11}{6})
|
|
\right]p_1=0.
|
|
\]
|
|
Equivalently, if \(H_1=-zC(-z/(1-z))\), then all four components of
|
|
\[
|
|
\theta H_1+H_1\mathcal C_1=0
|
|
\]
|
|
vanish. The mandatory verifier checks both forms independently.
|
|
Moreover,
|
|
\[
|
|
\deg((d_0(\theta)+zd_1(\theta))p_n)\le n+1.
|
|
\]
|
|
Equations \eqref{eq:term-constant}--\eqref{eq:term-top} therefore prove,
|
|
coefficient by coefficient, that the right side of
|
|
\eqref{eq:terminating-step} is
|
|
\(\alpha_{n+1}\sum_{k=0}^{n+1}h_{n+1,k}z^k\).
|
|
Starting from \eqref{eq:base-horizontal-row},
|
|
\eqref{eq:full-gauge} propagates the horizontal form at every step.
|
|
This proves \eqref{eq:qhat} and \eqref{eq:normalization} for the actual
|
|
row \(CG_N\), not merely for a scalar surrogate.
|
|
The mandatory standalone checker independently expands the cleared
|
|
identities and rejects any nonzero coefficient.
|
|
\end{proof}
|
|
|
|
\begin{corollary}[Positivity at the CM point]\label{cor:positivity}
|
|
At \(z_0=-1/53360^3\),
|
|
\[
|
|
\widehat Q_N(z_0)\ge\alpha_n>0.
|
|
\]
|
|
Moreover,
|
|
\[
|
|
\alpha_n\ge18\cdot29^N(N!)^2.
|
|
\]
|
|
\end{corollary}
|
|
|
|
\begin{proof}
|
|
For \(0\le k\le n\), the coefficient of \(z^k\) in
|
|
\eqref{eq:qhat} has sign \((-1)^k\). Since \(z_0<0\), every summand is
|
|
nonnegative. Also
|
|
\[
|
|
\frac{576n^2(2n+1)^2}{(6n+1)(6n+5)}-29n^2
|
|
=\frac{n^2(1260n^2+1260n+431)}
|
|
{(6n+1)(6n+5)}>0.
|
|
\]
|
|
Iterating \eqref{eq:normalization} proves the lower bound.
|
|
\end{proof}
|
|
|
|
\section{From formal contact to convergence at
|
|
\texorpdfstring{\(x_0\)}{x0}}
|
|
|
|
This step is included to rule out a beyond-all-orders ambiguity.
|
|
For \(r=0,1\), let
|
|
\[
|
|
E_{N,r}(x)=(\mathcal E_N)_{r,1},\qquad
|
|
\mathcal R_{N,r}(x)=x^nE_{N,r}(x).
|
|
\]
|
|
Proposition~\ref{prop:divisibility} says that
|
|
\(\mathcal R_{N,r}\) has a zero of order at least \(2n\).
|
|
|
|
Choose \(r_0=1/4\). On \(|x|=r_0\), \(|z|\le1/3\). The coefficients of
|
|
\(y\) have modulus at most one, so
|
|
\[
|
|
|y-1|\le\frac12,\qquad
|
|
|\theta^jy|\le\sum_{k\ge1}k^3(1/3)^k=\frac{33}{8}<5
|
|
\quad(1\le j\le3).
|
|
\]
|
|
Termwise estimates give \(\|\mathcal E_0\|_\infty<6000\).
|
|
|
|
For \(m\ge1\), put
|
|
\[
|
|
D(m)=\diag(1,m,m^2,m^3),\qquad
|
|
\mathcal B_m=D(m)^{-1}M_mD(m+1)/(m+1)^2.
|
|
\]
|
|
On \(|x|=1/4\), direct estimates of the authoritative entries give
|
|
\[
|
|
|(M_m)_{ij}|\le10^4u^{\,i+2-j},\qquad u=2m+3,
|
|
\]
|
|
and therefore
|
|
\[
|
|
|(\mathcal B_m)_{ij}|
|
|
\le10^4\left(\frac um\right)^{i-1}
|
|
\left(\frac u{m+1}\right)^{3-j}.
|
|
\]
|
|
For \(m\ge1\), \(u/m\le5\) and \(u/(m+1)\le5/2\). Because the exponent
|
|
\(3-j\) is negative when \(j=4\), we use the exact case split
|
|
\[
|
|
\left(\frac{u}{m+1}\right)^{3-j}\le
|
|
\begin{cases}
|
|
(5/2)^{3-j}\le25/4,&j\le3,\\[1mm]
|
|
1/2,&j=4,
|
|
\end{cases}
|
|
\]
|
|
where the second line follows from \(u/(m+1)=2+1/(m+1)\ge2\).
|
|
Thus every entry is at most \(7{,}812{,}500\), and summing each row gives
|
|
\[
|
|
\|\mathcal B_m\|_\infty\le31{,}250{,}000<4\cdot10^8,\qquad
|
|
\|M_0\|_\infty<10^7.
|
|
\]
|
|
The balancing telescopes:
|
|
\[
|
|
G_N=M_0(N!)^2\mathcal B_1\cdots
|
|
\mathcal B_{N-1}D(N)^{-1}.
|
|
\]
|
|
It follows that, for \(N\ge1\),
|
|
\begin{equation}\label{eq:circle-bound}
|
|
\max_{|x|=1/4}|\mathcal R_{N,r}(x)|
|
|
\le6\cdot10^{10}(N!)^2(4\cdot10^8)^{N-1}(1/4)^n.
|
|
\end{equation}
|
|
We now verify the analytic hypothesis behind the next step. On
|
|
\(|x|\le1/4\),
|
|
\[
|
|
\left|-\frac{x}{1-x}\right|\le\frac13<1,
|
|
\]
|
|
so the series defining \(y\) and its first three Euler derivatives are
|
|
holomorphic on a neighborhood of the closed disk. Because \(f=O(x)\),
|
|
the simple pole of \(C\) cancels in \(fC\), so \(\mathcal E_0\) is
|
|
holomorphic there. Inspection of \eqref{eq:deformed-matrix} shows that
|
|
each \(M_m\) is holomorphic off \(x=0\) in this disk and has at most a
|
|
simple pole at \(0\). Hence
|
|
\(\mathcal R_{N,r}=x^n(\mathcal E_N)_{r,1}\), with \(n=N+1\), is
|
|
holomorphic on the disk after removing its possible singularity at zero.
|
|
Proposition~\ref{prop:divisibility} strengthens its order there to at
|
|
least \(2n\). Therefore
|
|
\(\mathcal R_{N,r}(x)/x^{2n}\) has a removable singularity at zero and
|
|
is holomorphic on a neighborhood of the closed disk.
|
|
Applying the maximum principle to
|
|
\(\mathcal R_{N,r}(x)/x^{2n}\) gives
|
|
\begin{equation}\label{eq:cauchy}
|
|
|\mathcal R_{N,r}(x_0)|
|
|
\le6\cdot10^{10}(N!)^2(4\cdot10^8)^{N-1}(1/4)^n
|
|
(4x_0)^{2n}.
|
|
\end{equation}
|
|
|
|
Because \(1-z=1/(1-x)\),
|
|
\[
|
|
Q_N(x_0)=(1-x_0)^n\widehat Q_N(z_0).
|
|
\]
|
|
Corollary~\ref{cor:positivity} yields
|
|
\[
|
|
Q_N(x_0)\ge
|
|
18\cdot29^N(N!)^2(1-x_0)^n.
|
|
\]
|
|
Combining this with \eqref{eq:cauchy}, we obtain
|
|
\begin{equation}\label{eq:geometric-error}
|
|
\left|\frac{E_{N,r}(x_0)}{q_N(x_0)}\right|
|
|
=\left|\frac{\mathcal R_{N,r}(x_0)}{Q_N(x_0)}\right|
|
|
\le K_0\,\beta(x_0)^N,
|
|
\end{equation}
|
|
where \(K_0<\infty\) is independent of \(N\) and
|
|
\[
|
|
\beta(x_0)=
|
|
\frac{4\cdot10^8}{29}
|
|
\frac{x_0^2}{(1/4)(1-x_0)}
|
|
=\frac{3125}{1307443596565949700399927}
|
|
<4\cdot10^{-19}<1.
|
|
\]
|
|
Therefore
|
|
\begin{equation}\label{eq:error-vanish}
|
|
\frac{E_{N,0}(x_0)}{q_N(x_0)}\longrightarrow0,\qquad
|
|
\frac{E_{N,1}(x_0)}{q_N(x_0)}\longrightarrow0.
|
|
\end{equation}
|
|
|
|
\section{Identification of the first-column limit}
|
|
|
|
By \eqref{eq:seed-identities} and the definition of \(\Phi\),
|
|
\[
|
|
A_0-\Phi A_1
|
|
=-A(\mathcal E_0)_{0,*}-B(\mathcal E_0)_{1,*}.
|
|
\]
|
|
Multiplying by \(G_N\e_1\), dividing by
|
|
\(A_1G_N\e_1=S q_N\), and using
|
|
\eqref{eq:error-vanish}, we get
|
|
\[
|
|
\lim_{N\to\infty}
|
|
\frac{A_0G_N\e_1}{A_1G_N\e_1}
|
|
=\Phi(x_0).
|
|
\]
|
|
Equation \eqref{eq:CM-value} therefore proves
|
|
\begin{equation}\label{eq:first-column}
|
|
\boxed{
|
|
\lim_{N\to\infty}\frac{P_{N,1}}{Q_{N,1}}
|
|
=\frac{\sqrt{10005}}{\pi}.}
|
|
\end{equation}
|
|
|
|
\section{The other three official columns}
|
|
|
|
All matrices in this section are evaluated at \(x=x_0=1/R\).
|
|
For \(m\ge1\), retain
|
|
\[
|
|
D_m=\diag(1,m,m^2,m^3),\qquad
|
|
\mathcal B_m=D_m^{-1}M_mD_{m+1}/(m+1)^2,
|
|
\]
|
|
and, for any row \(a\), put
|
|
\[
|
|
Z_m(a)=\frac{aG_mD_m}{(m!)^2}.
|
|
\]
|
|
Then the balancing gives the exact recurrence
|
|
\begin{equation}\label{eq:balanced-row-recurrence}
|
|
Z_{m+1}(a)=Z_m(a)\mathcal B_m.
|
|
\end{equation}
|
|
Substitution in the displayed matrix shows, entry by entry, that
|
|
\(\mathcal B_m-\mathcal S=O(m^{-1})\), where
|
|
\[
|
|
\mathcal S=
|
|
\begin{pmatrix}
|
|
64R-44&96R-54&48R-17&8R\\
|
|
-8&-12&-6&-1\\
|
|
R^{-1}&-4R^{-1}&-6R^{-1}&-2R^{-1}\\
|
|
2R^{-2}&(17R-8)R^{-2}&4(5R-3)R^{-2}&(6R-4)R^{-2}
|
|
\end{pmatrix}.
|
|
\]
|
|
Its characteristic polynomial is \(Q_R(t)/R^2\), where
|
|
\[
|
|
\begin{aligned}
|
|
Q_R(t)={}&R^2t^4-(64R^3-56R^2-4)t^3\\
|
|
&+(48R^2-262R+220)t^2-(12R-8)t+1.
|
|
\end{aligned}
|
|
\]
|
|
On \(|t|=1\), the cubic coefficient strictly dominates the sum of the
|
|
other four coefficient magnitudes, because
|
|
\[
|
|
(64R^3-56R^2-4)-(49R^2-250R+213)
|
|
=64R^3-105R^2+250R-217>0.
|
|
\]
|
|
To count the roots without an irreducibility or root-finder call, consider
|
|
\[
|
|
H_s(t)=-(64R^3-56R^2-4)t^3+
|
|
s\{R^2t^4+(48R^2-262R+220)t^2-(12R-8)t+1\}.
|
|
\]
|
|
The strict inequality above gives \(H_s(t)\ne0\) for
|
|
\(|t|=1\), \(0\le s\le1\). Thus the winding number of
|
|
\(H_s(e^{i\vartheta})\) about zero cannot change with \(s\); at \(s=0\)
|
|
it is \(3\). Hence \(Q_R=H_1\) has three zeros in \(|t|<1\) and one,
|
|
counted with multiplicity, in \(|t|>1\). Moreover
|
|
\[
|
|
Q_R(1)=-(64R^3-105R^2+274R-233)<0,\qquad
|
|
\lim_{t\to+\infty}Q_R(t)=+\infty.
|
|
\]
|
|
Consequently the unique exterior zero is a simple real number
|
|
\(\rho>1\).
|
|
|
|
\begin{lemma}[Explicit dominant-product dichotomy]
|
|
\label{lem:dominant-product}
|
|
Fix \(\tau\) with
|
|
\[
|
|
\max_{\lambda\ne\rho}|\lambda|<\tau<1,
|
|
\]
|
|
Then there exist a late index \(m_0\), a linear functional \(\Lambda\) on
|
|
rows, nonzero scalars \(L_m\) independent of the row, and a left
|
|
\(\rho\)-eigenvector \(w\) of \(\mathcal S\) such that the following
|
|
alternatives hold:
|
|
\begin{align}
|
|
\Lambda(a)=0&\quad\Longrightarrow\quad
|
|
\|Z_m(a)\|\le K_a\tau^{m-m_0},\label{eq:exceptional-decay}\\
|
|
\Lambda(a)\ne0&\quad\Longrightarrow\quad
|
|
Z_m(a)=\Lambda(a)L_m\bigl(w+o_a(1)\bigr).
|
|
\label{eq:dominant-asymptotic}
|
|
\end{align}
|
|
\end{lemma}
|
|
|
|
\begin{proof}
|
|
Choose an invertible \(P\) with
|
|
\[
|
|
P^{-1}\mathcal SP=
|
|
\begin{pmatrix}\rho&0\\0&A\end{pmatrix},
|
|
\qquad\operatorname{spr}(A)<1.
|
|
\]
|
|
Choose \(\operatorname{spr}(A)<\theta<\tau\). For stable rows define
|
|
\[
|
|
\|\beta\|_\theta=\sum_{k=0}^{\infty}
|
|
\theta^{-k}\|\beta A^k\|_0.
|
|
\]
|
|
The finite Jordan identity
|
|
\[
|
|
J_\lambda^k=\sum_{\ell=0}^{s-1}
|
|
\binom{k}{\ell}\lambda^{k-\ell}N^\ell
|
|
\]
|
|
gives, for \(\operatorname{spr}(A)<\eta<\theta\),
|
|
\(\|A^k\|_0\le Ck^2\eta^k\); hence the series converges and
|
|
\[
|
|
\|\beta A\|_\theta
|
|
=\theta\sum_{k=1}^{\infty}\theta^{-k}\|\beta A^k\|_0
|
|
\le\theta\|\beta\|_\theta.
|
|
\]
|
|
Use the dual norm for stable columns.
|
|
|
|
Write
|
|
\[
|
|
T_m=P^{-1}\mathcal B_mP=
|
|
\begin{pmatrix}a_m&b_m\\c_m&E_m\end{pmatrix}.
|
|
\]
|
|
Then \(a_m\to\rho\), \(b_m,c_m\to0\), and \(E_m\to A\).
|
|
Choose
|
|
\[
|
|
\theta<d_*<\tau<1<a_*<\rho.
|
|
\]
|
|
For sufficiently large \(m_0\), a positive \(\epsilon\) makes, for
|
|
\(m\ge m_0\),
|
|
\begin{gather}
|
|
|a_m|\ge a_*,\quad\|E_m\|\le d_*,
|
|
\quad\|b_m\|,\|c_m\|\le\epsilon,\label{eq:block-bounds}\\
|
|
d_*+\epsilon<\tau,\quad a_*-\epsilon>1,\quad
|
|
d_*+\epsilon<a_*-\epsilon,\label{eq:block-separation}\\
|
|
q:=\frac{d_*}{a_*-\epsilon}
|
|
+\frac{(d_*+\epsilon)\epsilon}{(a_*-\epsilon)^2}<1.
|
|
\label{eq:graph-contraction-constant}
|
|
\end{gather}
|
|
For stable columns \(\|h\|\le1\), set
|
|
\[
|
|
\Psi_m(h)=\frac{E_mh-c_m}{a_m-b_mh}.
|
|
\]
|
|
Equations \eqref{eq:block-bounds}--\eqref{eq:block-separation} give
|
|
\[
|
|
\|\Psi_m(h)\|
|
|
\le\frac{d_*+\epsilon}{a_*-\epsilon}<1.
|
|
\]
|
|
For two such columns,
|
|
\[
|
|
\Psi_m(h)-\Psi_m(k)
|
|
=\frac{E_m(h-k)}{a_m-b_mh}
|
|
+\frac{(E_mk-c_m)b_m(h-k)}
|
|
{(a_m-b_mh)(a_m-b_mk)},
|
|
\]
|
|
so \eqref{eq:graph-contraction-constant} gives
|
|
\[
|
|
\|\Psi_m(h)-\Psi_m(k)\|\le q\|h-k\|.
|
|
\]
|
|
|
|
For \(M>m\), set \(h_M^{(M)}=0\) and recurse backward by
|
|
\(h_j^{(M)}=\Psi_j(h_{j+1}^{(M)})\). Then, for \(M'>M\),
|
|
\[
|
|
\|h_m^{(M')}-h_m^{(M)}\|\le2q^{M-m}.
|
|
\]
|
|
Thus \(h_m=\lim_{M\to\infty}h_m^{(M)}\) exists and satisfies
|
|
\begin{equation}\label{eq:graph-invariance}
|
|
h_ma_m+c_m=(h_mb_m+E_m)h_{m+1}.
|
|
\end{equation}
|
|
The defining recurrence gives the explicit bound
|
|
\[
|
|
\|h_m\|\le
|
|
\frac{\|E_m\|\|h_{m+1}\|+\|c_m\|}
|
|
{|a_m|-\|b_m\|}.
|
|
\]
|
|
Together with \(\|A\|_\theta/\rho<1\), this gives
|
|
\[
|
|
\limsup_{m\to\infty}\|h_m\|
|
|
\le\frac{\theta}{\rho}\limsup_{m\to\infty}\|h_m\|,
|
|
\qquad\text{hence}\qquad h_m\to0.
|
|
\]
|
|
|
|
Write
|
|
\[
|
|
U_m(a)=Z_m(a)P=(\alpha_m,\beta_m),\qquad
|
|
\xi_m=\alpha_m-\beta_mh_m,\qquad
|
|
d_m=a_m-b_mh_{m+1}.
|
|
\]
|
|
Using \eqref{eq:graph-invariance} in
|
|
\(U_{m+1}=U_mT_m\) gives the exact scalar equation
|
|
\[
|
|
\xi_{m+1}=d_m\xi_m.
|
|
\]
|
|
Define the composed seed functional and product
|
|
\[
|
|
\Lambda(a)=\alpha_{m_0}(a)-\beta_{m_0}(a)h_{m_0},\qquad
|
|
L_m=\prod_{\ell=m_0}^{m-1}d_\ell.
|
|
\]
|
|
Both are now explicit, \(\Lambda\) is linear, and
|
|
\(\xi_m=\Lambda(a)L_m\). Equations
|
|
\eqref{eq:block-bounds}--\eqref{eq:block-separation} ensure
|
|
\(d_m\ne0\).
|
|
|
|
If \(\Lambda(a)=0\), then \(\alpha_m=\beta_mh_m\) and
|
|
\[
|
|
\beta_{m+1}=\beta_m(E_m+h_mb_m),\qquad
|
|
\|\beta_{m+1}\|<(d_*+\epsilon)\|\beta_m\|
|
|
<\tau\|\beta_m\|,
|
|
\]
|
|
which proves \eqref{eq:exceptional-decay}.
|
|
|
|
If \(\Lambda(a)\ne0\), put \(r_m=\beta_m/\xi_m\). Exact substitution
|
|
gives
|
|
\[
|
|
r_{m+1}
|
|
=\frac{b_m+r_m(E_m+h_mb_m)}{d_m}.
|
|
\]
|
|
Here \(b_m/d_m\to0\) and
|
|
\((E_m+h_mb_m)/d_m\to A/\rho\), whose norm is below one.
|
|
Enlarge \(m_0\) once more, redefining \(\Lambda\) and \(L_m\) from this
|
|
new index, so that for some \(q_1<1\),
|
|
\[
|
|
\left\|\frac{E_m+h_mb_m}{d_m}\right\|\le q_1
|
|
\qquad(m\ge m_0).
|
|
\]
|
|
Then, for every \(m\ge m_0\),
|
|
\[
|
|
\|r_m\|\le q_1^{m-m_0}\|r_{m_0}\|
|
|
+\sum_{\ell=m_0}^{m-1}q_1^{m-1-\ell}
|
|
\left\|\frac{b_\ell}{d_\ell}\right\|\longrightarrow0.
|
|
\]
|
|
Since \(\alpha_m/\xi_m=1+r_mh_m\to1\),
|
|
\[
|
|
U_m(a)=\Lambda(a)L_m\bigl((1,0)+o_a(1)\bigr).
|
|
\]
|
|
Multiplying by \(P^{-1}\) proves
|
|
\eqref{eq:dominant-asymptotic} with
|
|
\[
|
|
w=(1,0)P^{-1},\qquad w\mathcal S=\rho w.
|
|
\]
|
|
\end{proof}
|
|
|
|
No GCD or irreducibility decision is needed to show that all four
|
|
coordinates of \(w\) are nonzero. Define the polynomial row \(w(t)\) by
|
|
\[
|
|
\begin{aligned}
|
|
w_1(t)={}&Rt(Rt^2-7)+12R^2t^2+4t^2+44t+10,\\
|
|
\frac{w_2(t)}2={}&R^2t((48R-27)t-28)
|
|
+(194R-108)t+40R-23,\\
|
|
w_3(t)={}&R^2t((48R-17)t-8)
|
|
+(198R-68)t+71R-32,\\
|
|
w_4(t)={}&2R(8+(17+3R)t+4R^2t^2).
|
|
\end{aligned}
|
|
\]
|
|
Direct multiplication gives the exact polynomial identity
|
|
\[
|
|
w(t)(tI-\mathcal S)=(Q_R(t),0,0,0).
|
|
\]
|
|
At \(t=\rho\), this is a left \(\rho\)-eigenvector. The exterior
|
|
eigenspace is one-dimensional, so the vector in
|
|
Lemma~\ref{lem:dominant-product} may be rescaled to \(w(\rho)\), with the
|
|
inverse rescaling absorbed into \(L_m\). Since \(R>7\) and \(\rho>1\),
|
|
every displayed grouping is positive. Thus \(w_j(\rho)>0\) for
|
|
\(j=1,2,3,4\); below we abbreviate \(w_j=w_j(\rho)\).
|
|
|
|
Undoing the balancing in \eqref{eq:dominant-asymptotic} gives, whenever
|
|
\(\Lambda(a)\ne0\),
|
|
\begin{equation}\label{eq:all-column-asymptotic}
|
|
aG_m\e_j=(m!)^2m^{-(j-1)}
|
|
\Lambda(a)L_m\bigl(w_j+o_a(1)\bigr).
|
|
\end{equation}
|
|
It remains to verify that the two official rows are not exceptional.
|
|
The positivity estimate and \(Q_m=x_0^{m+1}q_m\) give
|
|
\begin{equation}\label{eq:q-lower}
|
|
\frac{q_m(x_0)}{(m!)^2}
|
|
\ge18(R-1)\,[29(R-1)]^m.
|
|
\end{equation}
|
|
If \(\Lambda(C)=0\), the first coordinate of
|
|
\eqref{eq:exceptional-decay} would contradict \eqref{eq:q-lower}.
|
|
Therefore
|
|
\[
|
|
\Lambda(A_1)=S\Lambda(C)\ne0.
|
|
\]
|
|
If \(\Lambda(A_0)=0\), then
|
|
\eqref{eq:exceptional-decay}, \eqref{eq:dominant-asymptotic}, and
|
|
\(|d_m|>1\) for large \(m\) would make the first-column quotient tend
|
|
to zero, contradicting \eqref{eq:first-column}. Hence
|
|
\(\Lambda(A_0)\ne0\) as well.
|
|
|
|
Because \(w_j>0\), equation \eqref{eq:all-column-asymptotic} first proves
|
|
that every \(Q_{m,j}\) is nonzero for all sufficiently large \(m\), and
|
|
only then permits division:
|
|
\[
|
|
\lim_{m\to\infty}\frac{P_{m,j}}{Q_{m,j}}
|
|
=\frac{\Lambda(A_0)}{\Lambda(A_1)}
|
|
=\lim_{m\to\infty}\frac{P_{m,1}}{Q_{m,1}}
|
|
=\frac{\sqrt{10005}}{\pi}.
|
|
\]
|
|
We conclude:
|
|
|
|
\begin{theorem}[Ramanujan Challenge Problem 2.8]\label{thm:main}
|
|
For every official column \(j=1,2,3,4\),
|
|
\[
|
|
\boxed{
|
|
\lim_{N\to\infty}\frac{P_{N,j}}{Q_{N,j}}
|
|
=\frac{\sqrt{10005}}{\pi}.}
|
|
\]
|
|
Equivalently,
|
|
\[
|
|
\boxed{
|
|
\lim_{N\to\infty}\frac{Q_{N,j}}{P_{N,j}}
|
|
=\frac{\pi}{\sqrt{10005}}.}
|
|
\]
|
|
\end{theorem}
|
|
|
|
\section{Reproducibility map}
|
|
|
|
The proof package contains the following certificates.
|
|
|
|
\begin{center}
|
|
\begin{tabular}{
|
|
>{\raggedright\arraybackslash}p{0.41\textwidth}
|
|
p{0.49\textwidth}}
|
|
\toprule
|
|
File & Exact obligation\\
|
|
\midrule
|
|
\path{p28_full_closure_certificate.wl}
|
|
& Optional independent Wolfram cross-check of the differential gauge and
|
|
hypergeometric closure.\\
|
|
\path{p28_standalone_equations.py}
|
|
& Mandatory dependency-free expansion of the four cleared Ore
|
|
factorizations, the tail coefficient equations, the terminating base,
|
|
generic, and top identities, ascension, and the \(\F32\) equation.\\
|
|
\path{p28_dominant_product_algebra.py}
|
|
& Mandatory dependency-free verification of
|
|
\(\mathcal B_m=\mathcal S+O(m^{-1})\), the characteristic polynomial,
|
|
the root-separation inequalities, the left-eigenvector identity, and
|
|
the four positive coordinate rewrites.\\
|
|
\path{p28_kernel_contiguity_certificate.sage}
|
|
& Optional independent Sage check of the four displayed
|
|
factorizations; it performs no Ore division.\\
|
|
\path{p28_lattice_hypotheses_certificate.sage}
|
|
& Optional exact cross-check of the rank-one factorization, tail
|
|
direction, and transformed ODE identities.\\
|
|
\path{p28_convergence_constants.py}
|
|
& Exact rational verification of the coefficient bounds,
|
|
\(\alpha_{n+1}/\alpha_n\ge29n^2\), and \(\beta(x_0)<1\).\\
|
|
\path{all_four_columns_certificate.sage}
|
|
& Optional Sage cross-check of the balanced limit and exterior-root
|
|
algebra; the analytic contraction is proved in Lemma~\ref{lem:dominant-product}.\\
|
|
\path{p28_parametric_pade_probe.py}
|
|
& Diagnostic finite exact regression of the predicted valuations; it is
|
|
not used as proof of an all-\(N\) statement.\\
|
|
\bottomrule
|
|
\end{tabular}
|
|
\end{center}
|
|
|
|
The two mandatory algebra checkers use only the standard library's
|
|
\texttt{fractions.Fraction}, sparse coefficient dictionaries, and explicit
|
|
addition, multiplication, differentiation, and determinant expansion.
|
|
They invoke no division algorithm, factorizer, root finder, special-function
|
|
library, or numerical sample. Wolfram Language and SageMath are optional
|
|
independent cross-checks, not a trust requirement.
|
|
|
|
\section*{References}
|
|
\addcontentsline{toc}{section}{References}
|
|
|
|
\begin{enumerate}[label={[\arabic*]}]
|
|
\item D. V. Chudnovsky and G. V. Chudnovsky,
|
|
``Approximations and complex multiplication according to Ramanujan,''
|
|
in \emph{Ramanujan Revisited}, Academic Press, 1988, pp.~375--472.
|
|
\item L. Milla,
|
|
``A detailed proof of the Chudnovsky formula with means of basic
|
|
complex analysis,'' arXiv:1809.00533v6, 2021,
|
|
\href{https://arxiv.org/abs/1809.00533}{arXiv:1809.00533}.
|
|
\item J. L. Fields,
|
|
``Rational approximations to generalized hypergeometric functions,''
|
|
\emph{Mathematics of Computation} \textbf{19} (1965), 606--624,
|
|
\href{https://doi.org/10.1090/S0025-5718-1965-0194620-7}
|
|
{doi:10.1090/S0025-5718-1965-0194620-7}.
|
|
\item Yu. V. Nesterenko,
|
|
``Hermite--Pad\'e approximants of generalized hypergeometric
|
|
functions,'' \emph{Russian Acad. Sci. Sb. Math.}
|
|
\textbf{83} (1995), 189--219.
|
|
\item The Ramanujan Machine,
|
|
\href{https://www.ramanujanmachine.com/ramanujan-challenge/}
|
|
{Ramanujan Challenge}, Problem 2.8.
|
|
\end{enumerate}
|
|
|
|
\end{document}
|