437b439c6c
Portable proof handoff intended for verified mirror base 1229ab9e61bee936cb1a29c0693ee56922d2d908.
778 lines
21 KiB
TeX
778 lines
21 KiB
TeX
\documentclass[11pt]{article}
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\usepackage[T1]{fontenc}
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\usepackage{lmodern}
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\usepackage{amsmath,amssymb,amsthm,mathtools}
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\usepackage{array,booktabs}
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\usepackage{enumitem}
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\usepackage[margin=1in]{geometry}
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\usepackage{microtype}
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\usepackage{xcolor}
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\usepackage[hidelinks]{hyperref}
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\usepackage{listings}
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\definecolor{codegray}{RGB}{245,245,245}
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\lstset{
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basicstyle=\ttfamily\small,
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backgroundcolor=\color{codegray},
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frame=single,
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breaklines=true,
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columns=fullflexible,
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keepspaces=true
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}
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\newtheorem{theorem}{Theorem}
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\newtheorem{lemma}{Lemma}
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\newtheorem{proposition}{Proposition}
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\newtheorem{corollary}{Corollary}
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\theoremstyle{definition}
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\newtheorem{definition}{Definition}
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\theoremstyle{remark}
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\newtheorem{remark}{Remark}
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\newcommand{\F}[2]{{}_{#1}F_{#2}}
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\newcommand{\Q}{\mathbb{Q}}
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\newcommand{\e}{\mathbf e}
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\newcommand{\diag}{\operatorname{diag}}
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\newcommand{\ord}{\operatorname{ord}}
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\title{An Exact Hypergeometric Tail Certificate for\\
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Ramanujan Challenge Problem 2.8}
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\author{Problem 2.8 submission}
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\date{July 2026}
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\begin{document}
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\maketitle
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\begin{abstract}
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Let \(G_N=M_0M_1\cdots M_{N-1}\) be the \(4\times4\) transfer
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product in Ramanujan Challenge Problem 2.8, and let \(P_{N,j}\) and
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\(Q_{N,j}\) be the two official seeded rows evaluated in column \(j\).
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We prove
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\[
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\lim_{N\to\infty}\frac{P_{N,j}}{Q_{N,j}}
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=\frac{\sqrt{10005}}{\pi}
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\qquad (j=1,2,3,4).
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\]
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Equivalently, \(Q_{N,j}/P_{N,j}\to\pi/\sqrt{10005}\).
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The missing connection constant is fixed by an exact rank-three
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hypergeometric tail. A nonterminating \(\F43\) Euler jet is carried
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backward by the authoritative matrix, while the first denominator is a
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terminating adjoint \(\F43\). Their common differential gauge gives an
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all-\(N\) Pad\'e divisibility theorem. Positivity of the terminating
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denominator at the negative CM point, together with a balanced-transfer
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Cauchy estimate, turns that formal divisibility into a direct fixed-point
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convergence proof. The symbolic contiguity and adjoint identities are
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included as reproducible Wolfram Language and SageMath certificates.
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\end{abstract}
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\tableofcontents
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\section{Statement and compact form of the seeds}
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Put
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\[
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R=151931373056001=53360^3+1,\qquad
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x_0=\frac1R,\qquad
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z=-\frac{x}{1-x}.
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\]
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Thus the official CM point is
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\[
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z_0=-\frac1{R-1}=-\frac1{53360^3}.
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\]
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Let
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\[
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G_N=M_0M_1\cdots M_{N-1},\qquad G_0=I_4,
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\]
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where \(M_N=M(N,x)\) is the authoritative transfer matrix in the analytic
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deformation
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\[
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236337691420383\ \longmapsto\ \frac{14/x-567}{9}.
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\]
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At \(x=x_0\), this is the exact identity
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\(236337691420383=(14R-567)/9\). Thus every later use of Cauchy's theorem
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concerns this explicitly defined rational \(x\)-family.
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The complete entries of \(M(N,x)\) appear verbatim in the accompanying
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CAS certificates.
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Define four Pascal rows
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\[
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\begin{aligned}
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b_0&=(1,0,0,0),&
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b_1&=(1,1,0,0),\\
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b_2&=(1,2,1,0),&
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b_3&=(1,3,3,1)
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\end{aligned}
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\]
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and let \(\mathcal P\) be the matrix with rows \(b_0,b_1,b_2,b_3\).
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The compact denominator row is
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\[
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C(x)=
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\left(
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\frac{18}{x}+\frac{159}{4},\
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\frac{54}{x}+\frac{131}{2},\
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\frac{54}{x}+27,\
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\frac{18}{x}
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\right).
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\]
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Equivalently,
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\[
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C=\frac{18}{x}b_3+\frac54b_0+\frac{23}{2}b_1+27b_2.
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\]
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Set
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\[
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A=13591409,\qquad B=545140134,\qquad S=426880.
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\]
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The two official initial rows have the exact form
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\begin{equation}\label{eq:seed-identities}
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A_1=SC,\qquad
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A_0=AC-\frac54H_0,\qquad
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H_0=Ab_0+Bb_1=(A+B,B,0,0).
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\end{equation}
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At \(x=x_0\), these identities reproduce the official integer rows
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entry by entry.
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For \(j=1,\ldots,4\), write
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\[
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P_{N,j}=A_0G_N\e_j,\qquad
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Q_{N,j}=A_1G_N\e_j.
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\]
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We first identify the first-column limit and then invoke the exact cyclic
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frame to cover all four columns.
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\section{The CM function and its exact value}
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Let
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\[
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y(z)=\F32\left(
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\begin{matrix}\frac16,\frac12,\frac56\\1,1\end{matrix};z
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\right),
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\qquad
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\theta=z\frac{d}{dz}=(1-x)x\frac{d}{dx},
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\]
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and define
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\begin{equation}\label{eq:phi}
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\Phi(x)=\frac{Ay(z)+B\theta y(z)}{S}.
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\end{equation}
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The classical Chudnovsky identity is
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\[
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\frac1\pi=
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\frac{12}{640320^{3/2}}
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\sum_{k=0}^{\infty}
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\frac{(6k)!}{(3k)!(k!)^3}
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(A+Bk)(-640320^{-3})^k.
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\]
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The elementary coefficient identity
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\[
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\frac{(6k)!}{(3k)!(k!)^3}
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=1728^k
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\frac{(\frac16)_k(\frac12)_k(\frac56)_k}{(k!)^3}
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\]
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and \(640320=12\cdot53360\) give
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\[
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Ay(z_0)+B\theta y(z_0)
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=\frac{640320^{3/2}}{12\pi}
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=\frac{426880\sqrt{10005}}{\pi}.
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\]
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Consequently,
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\begin{equation}\label{eq:CM-value}
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\boxed{\Phi(x_0)=\frac{\sqrt{10005}}{\pi}.}
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\end{equation}
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\section{The nonterminating adjoint tail}
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Put \(n=N+1\) and \(\delta_N=\theta-n\). Define
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\begin{equation}\label{eq:tail}
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F_N(z)=\kappa_Nz^n
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\F43\left(
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\begin{matrix}
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n,n+\frac16,n+\frac12,n+\frac56\\
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2n,2n,2n
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\end{matrix};z\right),
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\end{equation}
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where
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\[
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\kappa_0=\frac5{72},\qquad
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\frac{\kappa_{N+1}}{\kappa_N}
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=-\frac{(6N+7)(6N+11)}
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{576(N+1)^2(2N+3)^2}.
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\]
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Its Euler jet is
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\[
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k_N=\left(F_N,\delta_NF_N,\delta_N^2F_N,\delta_N^3F_N\right)^T.
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\]
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\begin{proposition}[Exact tail contiguity]\label{prop:tail-contiguity}
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For every \(N\ge0\),
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\begin{equation}\label{eq:tail-contiguity}
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\boxed{M_Nk_{N+1}=k_N.}
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\end{equation}
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\end{proposition}
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\begin{proof}
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The first row is verified coefficientwise from the ratio of consecutive
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\(\F43\) coefficients. For the other rows, let \(t=\delta_{N+1}\).
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The shifted tail satisfies
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\[
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\left[
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(1-x)t(t+u)^3+
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x(t+n+1)(t+n+\tfrac76)(t+n+\tfrac32)(t+n+\tfrac{11}{6})
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\right]F_{N+1}=0,
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\]
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where \(u=2N+3\). Each of the remaining three row differences is divided
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by this degree-four Ore polynomial; its remainder is identically zero in
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\(\Q(N,x)[t]\). The exact coefficient identity, the three Ore divisions,
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and the normalization ratio are checked in
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\texttt{p28\_full\_closure\_certificate.wl} and
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\texttt{p28\_kernel\_contiguity\_certificate.sage}.
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\end{proof}
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For \(N=0\), the standard ascension identity gives
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\begin{equation}\label{eq:ascension}
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F_0=y-1
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=\frac5{72}z
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\F43\left(
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\begin{matrix}1,\frac76,\frac32,\frac{11}{6}\\2,2,2\end{matrix};z
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\right).
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\end{equation}
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Because \(\mathcal P\) is the Pascal matrix,
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\[
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\mathcal Pk_0=
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(y-1,\theta y,\theta^2y,\theta^3y)^T.
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\]
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\section{The rank-three error carrier}
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Set
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\[
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f=(y-1,\theta y,\theta^2y,\theta^3y)^T,\qquad
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\mathcal E_0=\frac54\mathcal P+fC,\qquad
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\mathcal E_N=\mathcal E_0G_N.
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\]
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The transformed hypergeometric equation is
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\begin{equation}\label{eq:transformed-ode}
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72\theta^3y+108x\theta^2y+46x\theta y+5xy=0.
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\end{equation}
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Using the displayed decomposition of \(C\), equations
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\eqref{eq:ascension}--\eqref{eq:transformed-ode} give
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\[
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Ck_0=-\frac54.
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\]
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It follows that
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\[
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\mathcal E_0k_0=\frac54f+f(Ck_0)=0.
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\]
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Proposition~\ref{prop:tail-contiguity} therefore implies
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\begin{equation}\label{eq:annihilation}
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\boxed{\mathcal E_Nk_N=0\qquad(N\ge0).}
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\end{equation}
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The same differential equation gives the exact row relation
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\begin{equation}\label{eq:row-relation}
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72(\mathcal E_N)_{3,*}
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+108x(\mathcal E_N)_{2,*}
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+46x(\mathcal E_N)_{1,*}
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+5x(\mathcal E_N)_{0,*}=0.
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\end{equation}
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\section{A discrete valuation lemma}
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Let
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\[
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H=\diag(x,1,1,1),\qquad J_N=HM_N.
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\]
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Every entry of \(J_N\) is regular at \(x=0\). If \(u=2N+3\) and
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\[
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a_N=\frac{144(u-1)^2}{u(3u-2)(3u+2)},\qquad
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V_N=(u^3,3u^2,3u,1),
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\]
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direct substitution in the authoritative matrix gives
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\begin{equation}\label{eq:rank-one}
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J_N(0)=
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\begin{pmatrix}a_N\\-1\\-1\\-1\end{pmatrix}V_N.
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\end{equation}
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Thus \(J_N(0)\) has rank one.
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The first nonconstant coefficient of the \(\F43\) in
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\eqref{eq:tail} equals
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\[
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c_N=\frac{u(3u-2)(3u+2)}{144(u-1)^2}=a_N^{-1}.
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\]
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Since \(z=-x+O(x^2)\),
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\begin{equation}\label{eq:tail-direction}
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Hk_N=x^{N+2}\eta_N
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\left[
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\begin{pmatrix}1\\-c_N\\-c_N\\-c_N\end{pmatrix}
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+O(x)
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\right],\qquad \eta_N\ne0.
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\end{equation}
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The leading vector in \eqref{eq:tail-direction} is precisely the image
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direction in \eqref{eq:rank-one}.
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\begin{lemma}[DVR step, including the extra first-column zero]
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\label{lem:dvr}
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Let \(R_0=\Q[[x]]\), \(H=\diag(x,1,1,1)\), and suppose
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\[
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E=x^NLH,\qquad L\in\operatorname{Mat}_4(R_0),\qquad Ek=0.
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\]
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Assume \(J=HM\in\operatorname{Mat}_4(R_0)\), \(Mk^+=k\), and
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\[
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\begin{aligned}
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k^+&=x^r\alpha(\e_1+xs+O(x^2)),\\
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Hk&=x^r\beta(v_0+xv_1+O(x^2)),
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\end{aligned}
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\]
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with \(\alpha\beta\ne0\). If \(J(0)\) has rank one,
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\(\operatorname{im}J(0)=\Q v_0\), and \(J(0)\e_1\ne0\), then
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\[
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EM=x^{N+1}L^+H
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\]
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for some \(L^+\in\operatorname{Mat}_4(R_0)\).
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\end{lemma}
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\begin{proof}
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Absorb \(\beta/\alpha\) into \(v_0,v_1\). From \(Jk^+=Hk\),
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\[
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J(\e_1+xs+O(x^2))=v_0+xv_1+O(x^2).
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\]
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Write \(L=L_0+xL_1+\cdots\). The equation \(L(Hk)=0\) gives
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\[
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L_0v_0=0,\qquad L_0v_1+L_1v_0=0.
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\]
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Since \(J(0)\) has image \(\Q v_0\), \(L_0J(0)=0\), so \(LJ\) is
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entrywise divisible by \(x\). The coefficient of \(x\) in its first
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column is
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\[
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L_0(v_1-J(0)s)+L_1v_0=-L_0J(0)s=0.
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\]
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Hence that column is divisible by \(x^2\). Therefore
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\[
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L^+=x^{-1}LJH^{-1}
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\]
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is regular and \(EM=x^{N+1}L^+H\).
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\end{proof}
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\begin{proposition}[All-\(N\) Pad\'e divisibility]\label{prop:divisibility}
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For every \(N\ge0\), there is
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\(L_N\in\operatorname{Mat}_4(\Q[[x]])\) such that
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\begin{equation}\label{eq:divisibility}
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\boxed{\mathcal E_N=x^NL_NH.}
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\end{equation}
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Thus every row of \(\mathcal E_N\) has componentwise valuations at least
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\[
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(N+1,N,N,N).
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\]
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The last row has the stronger valuations
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\[
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(N+2,N+1,N+1,N+1).
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\]
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\end{proposition}
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\begin{proof}
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Every component of \(f\) is \(O(x)\), while \(C\) has only a simple pole.
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All four components of \(f\) have leading term \(-5x/72\).
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Consequently the constant term in the first column of \(fC\) is
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\(-5/4\), cancelling the first component of every row of
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\((5/4)\mathcal P\). Hence \(\mathcal E_0=L_0H\).
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Apply Lemma~\ref{lem:dvr} inductively, using
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\eqref{eq:annihilation}, \eqref{eq:rank-one},
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\eqref{eq:tail-direction}, and \(M_Nk_{N+1}=k_N\).
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The stronger last-row assertion follows from
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\eqref{eq:row-relation}.
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\end{proof}
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\section{The terminating denominator}
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For the first column put
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\[
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q_N(x)=CG_N\e_1,\qquad n=N+1,\qquad Q_N(x)=x^nq_N(x),
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\]
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and define
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\[
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\widehat Q_N(z)
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=(1-z)^nQ_N\!\left(-\frac{z}{1-z}\right).
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\]
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Since \(x=-z/(1-z)\), this is also the first component of
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\((-z)^nCG_N\).
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\begin{proposition}[Exact terminating denominator]
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\label{prop:terminating}
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For every \(N\ge0\),
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\begin{equation}\label{eq:qhat}
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\frac{\widehat Q_N(z)}{\alpha_n}
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=\F43\left(
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\begin{matrix}
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-n,-n-\frac16,-n-\frac12,-n-\frac56\\
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1-2n,1-2n,1-2n
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\end{matrix};z
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\right),
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\end{equation}
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where
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\begin{equation}\label{eq:normalization}
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\alpha_1=18,\qquad
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\frac{\alpha_{n+1}}{\alpha_n}
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=\frac{576n^2(2n+1)^2}{(6n+1)(6n+5)}.
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\end{equation}
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Here and below the hypergeometric expression denotes the unambiguous finite
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sum over \(0\le k\le n\); it terminates before any lower Pochhammer symbol
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can vanish.
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\end{proposition}
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\begin{proof}
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The proof is an exact differential-gauge calculation in
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\(\Q(n,z)\). The nonterminating tail
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\[
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\F43\left(
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\begin{matrix}n,n+\frac16,n+\frac12,n+\frac56\\
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2n,2n,2n
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\end{matrix};z\right)
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\]
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has a \(4\times4\) Euler companion system. Direct simplification gives
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\[
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\mathcal C_n(z)\,[-zM(2n+1,-z/(1-z))]
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-\theta[-zM(2n+1,-z/(1-z))]
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-[-zM(2n+1,-z/(1-z))]\mathcal C_{n+1}(z)=0.
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\]
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The transformed seed \(-zC(-z/(1-z))\) is a horizontal adjoint row.
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Eliminating its other three coordinates from the horizontal equation
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produces exactly
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\[
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\left[
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\theta(\theta-2n)^3
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-z(\theta-n)(\theta-n-\tfrac16)
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(\theta-n-\tfrac12)(\theta-n-\tfrac56)
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\right]\widehat Q_N=0.
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\]
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The analytic solution normalized at \(z=0\) is the terminating
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\(\F43\) in \eqref{eq:qhat}.
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For completeness, the CAS certificate does not rely only on this
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differential equation. It computes the actual one-step scalar operator
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and verifies its generic coefficient identity, its \(k=0\) normalization,
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and the separate top boundary \(k=n+1\). Every remainder simplifies
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identically to zero. This proves the statement for all \(n\), not merely
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for sampled values.
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\end{proof}
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\begin{corollary}[Positivity at the CM point]\label{cor:positivity}
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At \(z_0=-1/53360^3\),
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\[
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\widehat Q_N(z_0)\ge\alpha_n>0.
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\]
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Moreover,
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\[
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\alpha_n\ge18\cdot29^N(N!)^2.
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\]
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\end{corollary}
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|
|
\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 \(u/(m+1)\le5/2\); summing four entries in
|
|
each row gives the deliberately loose uniform bound
|
|
\[
|
|
\|\mathcal B_m\|_\infty\le4\cdot10^8,\qquad
|
|
\|M_0\|_\infty<10^7.
|
|
\]
|
|
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}
|
|
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 C(x_0)\,\beta(x_0)^N,
|
|
\end{equation}
|
|
where \(C(x_0)<\infty\) 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}
|
|
|
|
For completeness, we recall the exact finite-frame reduction already used
|
|
to establish convergence of the recurrence. The balanced transfer tends
|
|
to
|
|
\[
|
|
\mathcal S=
|
|
\begin{pmatrix}
|
|
64R-44&96R-54&48R-17&8R\\
|
|
-8&-12&-6&-1\\
|
|
R^{-1}&-4R^{-1}&-6R^{-1}&-2R^{-1}\\
|
|
2R^{-2}&(17R-8)R^{-2}&4(5R-3)R^{-2}&(6R-4)R^{-2}
|
|
\end{pmatrix}.
|
|
\]
|
|
Its characteristic polynomial is \(Q_R(t)/R^2\), where
|
|
\[
|
|
\begin{aligned}
|
|
Q_R(t)={}&R^2t^4-(64R^3-56R^2-4)t^3\\
|
|
&+(48R^2-262R+220)t^2-(12R-8)t+1.
|
|
\end{aligned}
|
|
\]
|
|
The quartic is irreducible. Its spectral separation is also exact: on
|
|
\(|t|=1\), the absolute value of its cubic coefficient exceeds the sum of
|
|
the other coefficient magnitudes, because
|
|
\[
|
|
(64R^3-56R^2-4)-(49R^2-250R+213)
|
|
=64R^3-105R^2+250R-217>0.
|
|
\]
|
|
Rouch\'e's theorem therefore places exactly three roots in \(|t|<1\) and
|
|
the remaining root \(\rho\) in \(|t|>1\). Hence \(\rho\) is the unique
|
|
root of maximal modulus.
|
|
|
|
We next remove any possible nonvanishing assumption about the denominator.
|
|
The positivity estimate above and \(Q_N=x_0^nq_N\) give
|
|
\begin{equation}\label{eq:q-lower}
|
|
q_N(x_0)\ge
|
|
18\cdot29^N(N!)^2
|
|
\left(\frac{1-x_0}{x_0}\right)^{N+1}.
|
|
\end{equation}
|
|
The scalar recurrence obtained from the first cyclic coordinate is of
|
|
Poincar\'e type after the \((N!)^2\) balancing. The discrete
|
|
Birkhoff--Poincar\'e theorem \([4,\text{ Chapters 3 and 5}]\) applies because
|
|
the balanced coefficients are rational in \(N\), have full expansions in
|
|
\(N^{-1}\), and the limiting spectrum is simple. If the coefficient of the
|
|
\(\rho\)-mode in \(q_N\) were zero, the three-root separation just proved
|
|
would give, for some \(\tau<1\),
|
|
\[
|
|
|q_N(x_0)|\le K_\tau (N!)^2\tau^N.
|
|
\]
|
|
This contradicts \eqref{eq:q-lower}. Thus the dominant denominator
|
|
coefficient is nonzero by a wholly exact argument.
|
|
|
|
It remains to transfer the first-column result to the other columns. For
|
|
\(r\ge1\), put
|
|
\[
|
|
\begin{aligned}
|
|
F_r&=[\,\e_1,M_r\e_1,M_rM_{r+1}\e_1,
|
|
M_rM_{r+1}M_{r+2}\e_1\,],\\
|
|
\gamma_{r,k}&=\prod_{\ell=1}^{k}(r+\ell)^2,\\
|
|
C_r&=[\,\e_1,\mathcal B_r\e_1,
|
|
\mathcal B_r\mathcal B_{r+1}\e_1,
|
|
\mathcal B_r\mathcal B_{r+1}\mathcal B_{r+2}\e_1\,].
|
|
\end{aligned}
|
|
\]
|
|
The balancing telescopes exactly:
|
|
\[
|
|
F_r=D(r)C_r\diag(\gamma_{r,0},\ldots,\gamma_{r,3}),
|
|
\qquad
|
|
C_r\longrightarrow
|
|
C=[\,\e_1,\mathcal S\e_1,\mathcal S^2\e_1,\mathcal S^3\e_1\,].
|
|
\]
|
|
The limiting cyclic frame is nonsingular:
|
|
\[
|
|
\det C
|
|
=-\frac{4(27R-11)(128R^2-149R-43)}{R^6}\ne0.
|
|
\]
|
|
Thus \(F_r\) is invertible for all sufficiently large \(r\). If
|
|
\(y_r(a)=aG_r\e_1\), exact inversion of this frame gives
|
|
\[
|
|
aG_r\e_j=r^{-(j-1)}
|
|
\sum_{k=0}^{3}(C_r^{-1})_{k+1,j}
|
|
\frac{y_{r+k}(a)}{\gamma_{r,k}}.
|
|
\]
|
|
The same Birkhoff--Poincar\'e theorem supplies a linear dominant functional
|
|
\(\Lambda\) and an exponent \(\sigma\) such that, for fixed \(k\),
|
|
\[
|
|
\frac{y_{r+k}(a)}
|
|
{(r!)^2\rho^r r^\sigma\gamma_{r,k}}
|
|
\longrightarrow\Lambda(a)\rho^k.
|
|
\]
|
|
Consequently
|
|
\[
|
|
\frac{aG_r\e_j}
|
|
{(r!)^2\rho^r r^{\sigma-(j-1)}}
|
|
\longrightarrow\Lambda(a)\,\widetilde w_j,\qquad
|
|
\widetilde w=[1,\rho,\rho^2,\rho^3]C^{-1}.
|
|
\]
|
|
An explicit left eigenvector is obtained from the first row of
|
|
\(R^2\operatorname{adj}(tI-\mathcal S)\). Each of its four coordinate
|
|
polynomials is coprime to \(Q_R\); hence no coordinate vanishes at \(\rho\).
|
|
It is a nonzero multiple of \(\widetilde w\), so
|
|
\(\widetilde w_j\ne0\) for every \(j\). Applying the last limit to
|
|
\(a=A_0,A_1\), using the exact denominator nonvanishing above, gives
|
|
\[
|
|
\lim_{N\to\infty}\frac{P_{N,j}}{Q_{N,j}}
|
|
=\frac{\Lambda(A_0)}{\Lambda(A_1)}
|
|
\qquad(j=1,2,3,4).
|
|
\]
|
|
Equation \eqref{eq:first-column} evaluates this common ratio. 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}
|
|
& Authoritative differential gauge; nonterminating tail contiguity;
|
|
terminating adjoint equation; coefficientwise \(n\)-contiguity;
|
|
normalization and top boundary; exact spectral and cyclic-frame closure.\\
|
|
\path{p28_kernel_contiguity_certificate.sage}
|
|
& Independent coefficient/Ore proof of \(M_Nk_{N+1}=k_N\).\\
|
|
\path{p28_lattice_hypotheses_certificate.sage}
|
|
& 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}
|
|
& Balanced limit, Rouch\'e separation, nonzero eigenvector coordinates,
|
|
and invertible cyclic frame.\\
|
|
\path{p28_parametric_pade_probe.py}
|
|
& Dependency-free finite exact regression of the predicted valuations.\\
|
|
\bottomrule
|
|
\end{tabular}
|
|
\end{center}
|
|
|
|
The Wolfram certificate performs symbolic identities over
|
|
\(\Q(n,z)\); it uses no numerical samples. The Python constants check uses
|
|
only the standard library's \texttt{fractions.Fraction}. The SageMath
|
|
files are independent exact cross-checks.
|
|
|
|
\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 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 S. Bodine and D. A. Lutz,
|
|
\emph{Asymptotic Integration of Differential and Difference Equations},
|
|
Lecture Notes in Mathematics 2129, Springer, 2015, Chapters 3 and 5.
|
|
\item The Ramanujan Machine,
|
|
\href{https://www.ramanujanmachine.com/ramanujan-challenge/}
|
|
{Ramanujan Challenge}, Problem 2.8.
|
|
\end{enumerate}
|
|
|
|
\end{document}
|