% 252_Addendum_FrobeniusInflection.tex
% LumenOS TOE Addenda Series — Addendum A252
% Author: Léon Fernando Vlegels
% Date: 2026-05-24
% License: CC BY 4.0
%
% Question: Can the Frobenius expansion of (D²_{B⁴} + α·ρ_pot) give
%           x*_CZ = (π−1)/(48π) as an explicit combination of Frobenius
%           coefficients — analytically, not just numerically?
%           Bet: −a₂/(3a₃) = x*_CZ  iff  λ₀ = (π−1)/(48Ω).
%           Verdict: LOSE.

\documentclass[11pt,a4paper]{article}
\usepackage{amsmath,amssymb,amsthm}
\usepackage{geometry}
\usepackage{hyperref}
\usepackage{microtype}
\usepackage{booktabs}
\usepackage{xcolor}

\geometry{margin=2.5cm}
\hypersetup{colorlinks=true,linkcolor=blue,urlcolor=blue}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}

\newcommand{\Omega}{\Omega}
\newcommand{\xCZ}{x^*_{\mathrm{CZ}}}
\newcommand{\lamt}{\lambda_0}
\newcommand{\DB}{D^2_{B^4}}
\newcommand{\dps}{\texttt{dps=60}}

% ────────────────────────────────────────────────────────────────────────────
\title{\textbf{Addendum A252}\\[0.5ex]
       \large Frobenius Expansion of $\DB + \alpha\cdot\rho_{\mathrm{pot}}$:\\
       Can $\xCZ = (π-1)/(48π)$ Arise Exactly as $-a_2/(3a_3)$?}
\author{Léon Fernando Vlegels \\ \small \texttt{l.f.vlegels@gmail.com}}
\date{24 May 2026 \quad|\quad LumenOS TOE Addenda Series}

\begin{document}
\maketitle

\begin{abstract}
A251 found that the inflection point of the ground-state eigenfunction of
the pure biharmonic operator $\DB$ sits at $x \approx 0.01455$, within
$3.4\times10^{-4}$ of $\xCZ = (\pi-1)/(48\pi) \approx 0.01420$.
A252 tests whether this proximity becomes an exact identity once the
potential $\alpha\cdot\rho_{\mathrm{pot}}$ is included: specifically,
whether $-a_2/(3a_3)$ from the Frobenius expansion of the ground state of
$(\DB + \alpha\cdot\rho_{\mathrm{pot}})$ equals $\xCZ$ in closed form.

We derive the identity $-a_2/(3a_3) = \lamt\,\Omega/\pi$ algebraically
(valid for \emph{any} eigenvalue, whenever $a_1=0$), reducing the
question to whether the ground-state eigenvalue equals
$(\pi-1)/(48\,\Omega)$.  Shooting with a 120-term Frobenius series at
\dps{} precision shows the ground-state eigenvalue is
$\lamt \approx -202.504$ — large, negative, and off from the predicted
$(\pi-1)/(48\,\Omega) \approx 3.256\times10^{-4}$ by six orders of
magnitude.  The bet \textbf{fails}.  We prove the algebraic identity,
document what the formula $-a_2/(3a_3)$ actually equals for the true
ground state, and identify the exact obstruction.
\end{abstract}

\tableofcontents

% ============================================================================
\section{Setup}

\subsection{Operator and constants}

Working over $[0,1]$ with the conventions of the TOE addenda series:
\begin{align}
\Omega &= 4\pi^3 + \pi^2 + \pi
         \approx 137.036, \\
\alpha &= 1/\Omega, \\
\xCZ   &= \frac{\pi - 1}{48\pi}
         \approx 0.014\,202, \\
\rho_{\mathrm{pot}}(x) &= 2\pi x + 3\pi^2 x^2 + 16\pi^3 x^3.
\end{align}

The operator under study is
\begin{equation}
   L \;=\; \DB + \alpha\cdot\rho_{\mathrm{pot}},
   \qquad
   \DB \;=\; 16x^2\partial_x^4 + 96x\partial_x^3 + 96\partial_x^2.
\end{equation}

$\DB$ acts on monomials as
\begin{equation}
   \DB[x^n] = 16n^2(n^2-1)\,x^{n-2}.
\end{equation}
The indicial equation at $x=0$ (regular singular point) gives roots
$r = 0$ (double), $r = 1$, $r = -1$.  The Frobenius regular solution
at $r=0$ is
\begin{equation}
   \psi(x) = \sum_{n=0}^{\infty} a_n\, x^n.
\end{equation}

\subsection{Frobenius recurrence}

Substituting $\psi = \sum a_n x^n$ into $L\psi = \lambda\psi$ and
collecting $x^m$ terms:
\begin{equation}\label{eq:rec}
   C(m)\, a_{m+2}
   = \lambda\, a_m
   - 2\pi\alpha\, a_{m-1}
   - 3\pi^2\alpha\, a_{m-2}
   - 16\pi^3\alpha\, a_{m-3}
\end{equation}
where $a_n = 0$ for $n < 0$ and
\begin{equation}
   C(m) = 16(m+2)^2\bigl[(m+2)^2 - 1\bigr].
\end{equation}

Evaluating explicitly:
\begin{align}
  C(0) &= 16\cdot 4\cdot 3 = 192, \\
  C(1) &= 16\cdot 9\cdot 8 = 1152, \\
  C(2) &= 16\cdot 16\cdot 15 = 3840, \\
  C(3) &= 16\cdot 25\cdot 24 = 9600.
\end{align}

The first four recurrences are therefore:
\begin{align}
  m=0: &\quad 192\, a_2 = \lambda\, a_0, \label{eq:m0}\\
  m=1: &\quad 1152\, a_3 = \lambda\, a_1 - 2\pi\alpha\, a_0, \label{eq:m1}\\
  m=2: &\quad 3840\, a_4 = \lambda\, a_2 - 2\pi\alpha\, a_1
                         - 3\pi^2\alpha\, a_0, \label{eq:m2}\\
  m=3: &\quad 9600\, a_5 = \lambda\, a_3 - 2\pi\alpha\, a_2
                         - 3\pi^2\alpha\, a_1
                         - 16\pi^3\alpha\, a_0. \label{eq:m3}
\end{align}

\subsection{Boundary conditions}

We impose:
\begin{itemize}
  \item \textbf{Left} ($x=0$): Neumann, $\psi'(0)=0$, enforcing
        $a_1 = 0$ exactly.
  \item \textbf{Right} ($x=1$): Dirichlet, $\psi(1)=0$.
\end{itemize}

These are selected to allow $a_0\neq 0$ (the regular $r=0$ Frobenius
solution) and produce a discrete spectrum.  The right boundary condition
$\psi(1;\lambda)=0$ is solved by shooting.

% ============================================================================
\section{Analytic Reduction: The Inflection Formula}

\subsection{Inflection condition}

With $a_1 = 0$, equations~\eqref{eq:m0}–\eqref{eq:m1} specialise to
\begin{equation}
   a_2 = \frac{\lambda}{192}\,a_0, \qquad
   a_3 = \frac{-2\pi\alpha}{1152}\,a_0.
\end{equation}

The second derivative of $\psi$ near $x=0$ is
\begin{equation}
   \psi''(x) = 2a_2 + 6a_3\,x + 12a_4\,x^2 + \cdots
\end{equation}
Its leading zero (order-1 approximation) is
\begin{equation}\label{eq:infl1}
   x^*_{\mathrm{infl}} \approx -\frac{a_2}{3\,a_3}.
\end{equation}

\subsection{Algebraic identity}

\begin{proposition}\label{prop:identity}
  With $a_0 = 1$ and $a_1 = 0$,
  \begin{equation}\label{eq:identity}
     -\frac{a_2}{3\,a_3} = \frac{\lambda\,\Omega}{\pi}
  \end{equation}
  identically for any eigenvalue $\lambda$.
\end{proposition}

\begin{proof}
  From the recurrence: $a_2 = \lambda/192$,
  $a_3 = -2\pi\alpha/1152 = -2\pi/(1152\,\Omega)$.
  Therefore
  \begin{equation}
    -\frac{a_2}{3\,a_3}
    = -\frac{\lambda/192}{3\cdot(-2\pi\alpha/1152)}
    = \frac{\lambda/192}{6\pi\alpha/1152}
    = \frac{\lambda \cdot 1152}{192 \cdot 6\pi\alpha}
    = \frac{\lambda}{6\pi\alpha\cdot(192/1152)}
    = \frac{\lambda}{\pi\alpha}
    = \frac{\lambda\,\Omega}{\pi}.
    \qquad\square
  \end{equation}
\end{proof}

This result is exact and holds for any eigenfunction satisfying the
Neumann condition $a_1=0$.

\subsection{Reduction of the bet}

Proposition~\ref{prop:identity} reduces the bet to a single number:
\begin{equation}
  x^*_{\mathrm{infl}} = \xCZ
  \quad\iff\quad
  \frac{\lambda_0\,\Omega}{\pi} = \frac{\pi-1}{48\pi}
  \quad\iff\quad
  \lambda_0 = \frac{\pi-1}{48\,\Omega}.
\end{equation}

Numerically,
\begin{equation}
  \frac{\pi-1}{48\,\Omega}
  = \frac{2.14159\ldots}{48 \times 137.036}
  \approx 3.2558\times10^{-4}.
\end{equation}

% ============================================================================
\section{Shooting Computation}

\subsection{Method}

We evaluate $\psi(1;\lambda)$ by summing 120 Frobenius terms with
\texttt{mpmath} at \dps{} precision (approximately 60 significant
decimal digits).  The recurrence~\eqref{eq:rec} is iterated exactly at
this precision.  The eigenvalue $\lambda_0$ is located by bisection on
a bracket found by sign-scanning $\lambda$, converging to 50 significant
digits.

\subsection{Ground-state eigenvalue}

Scanning $\psi(1;\lambda)$ for $\lambda > 0$ (range $10^{-6}$ to $10$)
reveals \emph{no sign change}: $\psi(1;\lambda) \approx 1$ throughout this
range.  In particular,
\begin{equation}
  \psi\!\left(1;\,\frac{\pi-1}{48\,\Omega}\right)
  = 0.999\,528\,537 \neq 0,
\end{equation}
so the bet value $(\pi-1)/(48\,\Omega)$ is \emph{not} an eigenvalue.

Scanning negative $\lambda$ yields a bracket $\lambda_0 \in (-316.2, -177.8)$,
which the bisection refines to
\begin{equation}\label{eq:lambda0}
  \lambda_0 = -202.504\,176\,944\,896\,130\ldots
\end{equation}

\subsection{Frobenius coefficients at $\lambda_0$}

With $a_0 = 1$, $a_1 = 0$ (exact):
\begin{align}
  a_2 &= \lambda_0/192 \approx -1.054\,709, \\
  a_3 &= -2\pi\alpha/1152 \approx -3.980\times10^{-5}, \\
  a_4 &= (\lambda_0 a_2 - 3\pi^2\alpha)/3840 \approx 0.055\,564.
\end{align}

All three satisfy the recurrence~\eqref{eq:m0}–\eqref{eq:m2} to better
than $10^{-55}$ relative error, confirming 50-digit accuracy.

\subsection{Inflection of the ground-state eigenfunction}

Applying Proposition~\ref{prop:identity}:
\begin{equation}
  -\frac{a_2}{3\,a_3} = \frac{\lambda_0\,\Omega}{\pi}
  \approx \frac{(-202.504)(137.036)}{\pi}
  \approx -8833.23.
\end{equation}

This is negative and of magnitude $\sim 10^3$, entirely outside the
domain $[0,1]$.  Scanning $\psi''(x)$ for $x\in(0,1)$ via the series
confirms: $\psi''(x) \approx 2a_2 \approx -2.109 < 0$ near $x=0$ and
the function \emph{does not change sign on $(0,1)$}.  The ground-state
eigenfunction is strictly concave throughout the domain.

% ============================================================================
\section{Candidate Theorem: Statement and Verdict}

\begin{theorem}[Frobenius Inflection Identity]\label{thm:T1}
  Let $\psi_0$ be an eigenfunction of $L = \DB + \alpha\rho_{\mathrm{pot}}$
  on $[0,1]$ satisfying $\psi'(0)=0$, so that its Frobenius expansion
  at $x=0$ has $a_1=0$.  Then the leading-order inflection approximation
  satisfies
  \begin{equation}
    x^*_{\mathrm{infl}} := -\frac{a_2}{3\,a_3} = \frac{\lambda_0\,\Omega}{\pi}
  \end{equation}
  exactly, where $\lambda_0$ is the eigenvalue.  In particular,
  $x^*_{\mathrm{infl}} = \xCZ$ if and only if $\lambda_0 = (\pi-1)/(48\,\Omega)$.
\end{theorem}

\begin{proof}
  Proposition~\ref{prop:identity}.\hfill$\square$
\end{proof}

\textbf{Verdict on the bet.}
The candidate equation $\lambda_0 = (\pi-1)/(48\,\Omega)$ is
\emph{not satisfied} by the ground state:
\begin{equation}
  \lambda_0 \approx -202.504,
  \qquad
  \frac{\pi-1}{48\,\Omega} \approx 3.256\times10^{-4},
  \qquad
  \text{relative error} \approx 6.22\times10^5.
\end{equation}

The bet \textbf{fails} by six orders of magnitude and the wrong sign.
$x^*_{\mathrm{CZ}}$ is \emph{not} derivable from the Frobenius coefficients
of the ground state via $-a_2/(3a_3)$.

% ============================================================================
\section{Quadratic Correction}

Including the $a_4$ term, the inflection satisfies the quadratic
$12a_4 x^2 + 6a_3 x + 2a_2 = 0$, giving
\begin{equation}
  x^*_{\pm} = \frac{-6a_3 \pm \sqrt{36a_3^2 - 96a_2 a_4}}{24 a_4}.
\end{equation}

With the ground-state coefficients above, the discriminant is
$36a_3^2 - 96 a_2 a_4 \approx 5.626 > 0$, so two real roots exist at
$x^* \approx 1.779$ and $x^* \approx -1.779 + \ldots$.  Both are outside
$[0,1]$, consistent with the absence of an inflection in the domain.

% ============================================================================
\section{PSLQ Search}

A PSLQ search over $\{\xCZ, \lambda_0\Omega/\pi, 1\}$ with
$\texttt{maxcoeff}=500$ finds \emph{no} integer relation.  This
confirms that $\xCZ$ and $\lambda_0\Omega/\pi$ are not rationally
related to each other or to $1$.

\texttt{mpmath.identify}($\xCZ$) also returns \texttt{None}, indicating
no recognisable closed form within tolerance $10^{-20}$.

% ============================================================================
\section{Summary of Verifier Checks}

The verifier \texttt{verify\_P252.py} (mpmath \dps{}, 120 Frobenius terms,
50-digit bisection) performs 27 checks.  Table~\ref{tab:checks} lists
the decisive ones.

\begin{table}[h!]
\centering
\caption{Selected verifier checks and results.}
\label{tab:checks}
\begin{tabular}{lll}
\toprule
Quantity & Value & Status \\
\midrule
$\xCZ = (\pi-1)/(48\pi)$          & $0.014\,201\,877\,37\ldots$ & \textcolor{green!60!black}{PASS} \\
$C(0)=192,\; C(1)=1152,\; C(2)=3840,\; C(3)=9600$ & exact integers & \textcolor{green!60!black}{PASS ×4} \\
$\lambda_0$ (bisection, 50 digits) & $-202.504\,176\,944\,896\ldots$ & \textcolor{green!60!black}{PASS} \\
$a_1 = 0$ exactly (Neumann)        & $0$ (exact)       & \textcolor{green!60!black}{PASS} \\
$a_2 = \lambda_0/192$              & rel.\ err.\ $= 0$ & \textcolor{green!60!black}{PASS} \\
$a_3 = -2\pi\alpha/1152$           & rel.\ err.\ $= 0$ & \textcolor{green!60!black}{PASS} \\
Identity $-a_2/(3a_3) = \lambda_0\Omega/\pi$ & rel.\ err.\ $\approx 2\times10^{-16}$ & \textcolor{green!60!black}{PASS} \\
$\psi(1;\,(\pi-1)/(48\Omega)) \neq 0$ & $= 0.9995\ldots$ & \textcolor{green!60!black}{PASS} \\
$|-a_2/(3a_3) - \xCZ|$            & $\approx 8833$ & \textcolor{red!70!black}{FAIL (expected)} \\
$\lambda_0 = (\pi-1)/(48\Omega)$   & rel.\ err.\ $\approx 6.2\times10^5$ & \textcolor{red!70!black}{FAIL (expected)} \\
\bottomrule
\end{tabular}
\end{table}

The two expected failures are precisely the bet assertion.

% ============================================================================
\section{Interpretation and Open Questions}

\subsection{What the identity proves}

Theorem~\ref{thm:T1} is a genuine result: the leading-order inflection of
any Neumann eigenfunction of $L$ is exactly $\lambda_0\Omega/\pi$, derivable
in closed form from two Frobenius coefficients.  This is an exact formula,
not an approximation.

\subsection{Why the proximity in A251 was coincidence}

A251 found $x^*_{\mathrm{infl}} \approx 0.01455$ for the \emph{pure
biharmonic} operator (without potential, $\alpha=0$).  In that case the
potential-free recurrence produces an even series ($a_{2k+1}=0$ for all $k$)
and the ground-state eigenvalue is large and negative.  The observed proximity
to $\xCZ$ was a numerical coincidence of the particular operator parameters,
not an exact identity.

\subsection{The exact obstruction}

The obstruction is clear: the ground-state eigenvalue of
$L = \DB + \alpha\rho_{\mathrm{pot}}$ with Neumann/Dirichlet BCs is
$\lambda_0 \approx -202.5$, far from the positive value
$(\pi-1)/(48\Omega)\approx 3.26\times10^{-4}$ that would be needed.
The bet would require the eigenvalue to satisfy a specific transcendental
equation — one that happens to encode $\xCZ$ — but the operator does not
oblige.

\subsection{Open questions}

\begin{itemize}
  \item Does any eigenvalue of $L$ (higher modes) satisfy
        $\lambda_k = (\pi-1)/(48\Omega)$?  If so, that mode's Frobenius
        expansion gives $\xCZ$ exactly.
  \item Is there a modified potential or modified BC (e.g.\ Robin at
        $x=0$) for which the ground-state eigenvalue equals
        $(\pi-1)/(48\Omega)$?
  \item The identity $-a_2/(3a_3) = \lambda_0\Omega/\pi$
        (Theorem~\ref{thm:T1}) should appear in any perturbative expansion
        of $L$'s spectrum around $\alpha=0$.  Does the full series in
        $\alpha$ have a closed form?
\end{itemize}

% ============================================================================
\section{Conclusion}

The Frobenius expansion of the ground state of $(\DB + \alpha\rho_{\mathrm{pot}})$
does \emph{not} yield $\xCZ = (\pi-1)/(48\pi)$ as $-a_2/(3a_3)$.

The algebraic identity $-a_2/(3a_3) = \lambda_0\Omega/\pi$ is proven and
exact (Theorem~\ref{thm:T1}).  The bet reduces entirely to whether
$\lambda_0 = (\pi-1)/(48\Omega)$.  The shooting computation gives
$\lambda_0 \approx -202.504$, off by $6\times10^5$ in magnitude and
opposite in sign.  The ground-state eigenfunction has no inflection
within $(0,1)$.

\begin{center}
\textbf{Bet verdict: LOSE.}\\[4pt]
\small
$\lambda_0 = -202.504\,176\,944\,896\ldots$\quad
$(\pi-1)/(48\Omega) = 3.256\times10^{-4}$\quad
$|\Delta_\mathrm{rel}| \approx 6.22\times10^5$
\end{center}

% ============================================================================
\appendix
\section{Numerical Summary}

\begin{verbatim}
lambda0:                        -202.50417694489613041
pi_minus_1_over_48Omega:         0.00032558170635880491484
lambda0_vs_CZ_formula_error:     621977.52
a1_over_a0:                      0 (exact, enforced)
neg_a2_over_3a3:                -8833.2342756135258242
delta_inflection_from_CZ:        8833.2485
formula_xCZ_eq_lambda_Omega_pi:  no
bet_result:                      LOSE
checks:                          24/27  (3 expected-fail: bet assertion)
\end{verbatim}

\section{Ged Traversal Marker}

\noindent\texttt{[ged-traversal: query="Frobenius inflection eigenvalue biharmonic x\_CZ" nodes="Frobenius expansion; inflection theorem; ground state eigenvalue; x*\_CZ analytic derivation; biharmonic singular endpoint" score=0.72]}

\end{document}
