docs(p28): post-submission lint polish (chktex W2/W24) on the audited release line

Whitespace-only polish of solution.tex: 46 chktex warnings fixed
- W2 (43x): 'word \eqref{...}' -> 'word~\eqref{...}' and continuation-line
  joins so references stay glued to their prose
- W24 (3x): \label glued to the \begin{...} line
- W8 kept (5x, all in the DOI identifier 10.1090/S0025-5718-1965-0194620-7:
  single hyphens are correct there, not prose dashes)

Verified: chktex W2+W24 = 0 (total 128, all W3/W25 brace suggestions + W8
DOI false positives); pdflatex 3-pass clean build (0 errors, 0 warnings,
0 overfull, 17 pages, 0 '??'); token-level PDF text diff vs the shipped
e85d7bf9 PDF shows only glyph-extraction/wrap artifacts (content identical,
whitespace-only tex diff).

NOTE: this is the parallel audited line (492c8ab provenance), NOT the
authority manuscript (completed-submissions 1cbb5982, hand-maintained,
must not be regenerated).
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allaun 2026-08-08 03:34:11 -05:00
parent cd4ae757d9
commit 2f4fd8d9ee
2 changed files with 37 additions and 58 deletions

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@ -323,14 +323,10 @@ The constant and generic coefficient equations are
\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
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
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}
@ -377,9 +373,8 @@ 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
expands to~\eqref{eq:transformed-ode}.
Using the displayed decomposition of \(C\), equations~\eqref{eq:ascension}--\eqref{eq:transformed-ode} give
\[
Ck_0=-\frac54.
\]
@ -417,8 +412,7 @@ direct substitution in the authoritative matrix gives
\end{equation}
Thus \(J_N(0)\) has rank one.
The first nonconstant coefficient of the \(\F43\) in
\eqref{eq:tail} equals
The first nonconstant coefficient of the \(\F43\) in~\eqref{eq:tail} equals
\[
c_N=\frac{u(3u-2)(3u+2)}{144(u-1)^2}=a_N^{-1}.
\]
@ -430,8 +424,8 @@ Since \(z=-x+O(x^2)\),
+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 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
@ -443,13 +437,12 @@ 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
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}
\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.
@ -514,12 +507,9 @@ 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
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}.
The stronger last-row assertion follows from~\eqref{eq:row-relation}.
\end{proof}
\section{The terminating denominator}
@ -536,8 +526,7 @@ and define
Since \(x=-z/(1-z)\), this is also the first component of
\((-z)^nCG_N\).
\begin{proposition}[Exact terminating denominator]
\label{prop:terminating}
\begin{proposition}[Exact terminating denominator]\label{prop:terminating}
For every \(N\ge0\),
\begin{equation}\label{eq:qhat}
\frac{\widehat Q_N(z)}{\alpha_n}
@ -619,8 +608,7 @@ 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
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}
@ -642,7 +630,7 @@ P(n,t)={}&-5n-76n^2+1404n^3+4360n^4+4320n^5+1440n^6\\
&+(-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
Thus~\eqref{eq:terminating-step} is the scalar form of the displayed
matrix recurrence; no differential-equation uniqueness is used below.
Let
@ -698,13 +686,11 @@ 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
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
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.
@ -722,15 +708,14 @@ Moreover,
\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
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.
Iterating~\eqref{eq:normalization} proves the lower bound.
\end{proof}
\section{From formal contact to convergence at
@ -802,7 +787,7 @@ We now verify the analytic hypothesis behind the next step. On
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
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
@ -828,7 +813,7 @@ 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
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|
@ -850,20 +835,19 @@ Therefore
\section{Identification of the first-column limit}
By \eqref{eq:seed-identities} and the definition of \(\Phi\),
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
\(A_1G_N\e_1=S q_N\), 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
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}}
@ -927,8 +911,7 @@ Q_R(1)=-(64R^3-105R^2+274R-233)<0,\qquad
Consequently the unique exterior zero is a simple real number
\(\rho>1\).
\begin{lemma}[Explicit dominant-product dichotomy]
\label{lem:dominant-product}
\begin{lemma}[Explicit dominant-product dichotomy]\label{lem:dominant-product}
Fix \(\tau\) with
\[
\max_{\lambda\ne\rho}|\lambda|<\tau<1,
@ -997,7 +980,7 @@ 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
Equations~\eqref{eq:block-bounds}--\eqref{eq:block-separation} give
\[
\|\Psi_m(h)\|
\le\frac{d_*+\epsilon}{a_*-\epsilon}<1.
@ -1009,7 +992,7 @@ For two such columns,
+\frac{(E_mk-c_m)b_m(h-k)}
{(a_m-b_mh)(a_m-b_mk)},
\]
so \eqref{eq:graph-contraction-constant} gives
so~\eqref{eq:graph-contraction-constant} gives
\[
\|\Psi_m(h)-\Psi_m(k)\|\le q\|h-k\|.
\]
@ -1042,7 +1025,7 @@ 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
Using~\eqref{eq:graph-invariance} in
\(U_{m+1}=U_mT_m\) gives the exact scalar equation
\[
\xi_{m+1}=d_m\xi_m.
@ -1053,8 +1036,7 @@ Define the composed seed functional and product
L_m=\prod_{\ell=m_0}^{m-1}d_\ell.
\]
Both are now explicit, \(\Lambda\) is linear, and
\(\xi_m=\Lambda(a)L_m\). Equations
\eqref{eq:block-bounds}--\eqref{eq:block-separation} ensure
\(\xi_m=\Lambda(a)L_m\). Equations~\eqref{eq:block-bounds}--\eqref{eq:block-separation} ensure
\(d_m\ne0\).
If \(\Lambda(a)=0\), then \(\alpha_m=\beta_mh_m\) and
@ -1063,7 +1045,7 @@ If \(\Lambda(a)=0\), then \(\alpha_m=\beta_mh_m\) and
\|\beta_{m+1}\|<(d_*+\epsilon)\|\beta_m\|
<\tau\|\beta_m\|,
\]
which proves \eqref{eq:exceptional-decay}.
which proves~\eqref{eq:exceptional-decay}.
If \(\Lambda(a)\ne0\), put \(r_m=\beta_m/\xi_m\). Exact substitution
gives
@ -1089,8 +1071,7 @@ 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
Multiplying by \(P^{-1}\) proves~\eqref{eq:dominant-asymptotic} with
\[
w=(1,0)P^{-1},\qquad w\mathcal S=\rho w.
\]
@ -1119,7 +1100,7 @@ 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
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)}
@ -1131,19 +1112,17 @@ The positivity estimate and \(Q_m=x_0^{m+1}q_m\) give
\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}.
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
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
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
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:
\[