Two generating functions, same ODE, different initial conditions. Their coefficient ratio converges to $\zeta(3)$. Their function-value ratio does not. But their derivative ratio does — and the reason is a beautiful interplay between regular and singular parts at a conifold point.

Setup: Two Solutions of the Same ODE

The Apéry numbers $a_n = \sum_{k=0}^{n}\binom{n}{k}^2\binom{n+k}{k}^2$ and their companion sequence $b_n$ satisfy the same three-term recurrence:

with $a_0 = 1, a_1 = 5$ and $b_0 = 0, b_1 = 6$. Apéry’s proof of the irrationality of $\zeta(3)$ rests on the fact that

converging at the exponential rate $(17-12\sqrt{2})^{2n}$ — roughly 3 digits per term.

The generating functions $A(z) = \sum a_n z^n$ and $B(z) = \sum b_n z^n$ both satisfy the third-order Apéry ODE (a Calabi-Yau operator):

Both series converge for $|z| < z_1$, where $z_1 = 17 - 12\sqrt{2} \approx 0.0294$ is the nearest conifold point — a singular point of the ODE where the polynomial $1-34z+z^2$ vanishes.

Here is the natural question: the coefficient ratio $b_n/a_n \to \zeta(3)$. Does the function ratio $B(z)/A(z)$ also converge to $\zeta(3)$ as $z$ approaches the boundary of convergence?

First Attempt: The Function Ratio (Fails)

Evaluating $B(z)/A(z)$ numerically as $z \to z_1$:

The ratio converges to approximately $0.35$, nowhere near $\zeta(3) \approx 1.202$. The coefficient-level identity $b_n/a_n \to \zeta(3)$ does not lift to a function-level identity $B(z)/A(z) \to \zeta(3)$.

Why the ratio is wrong

The key structural fact: $b_n - \zeta(3)\,a_n$ decays like $(17-12\sqrt{2})^{2n} \approx (0.000866)^n$. The error generating function

has radius of convergence $1/(17-12\sqrt{2})^2 \approx 1155$, vastly larger than $z_1 \approx 0.029$. So $E(z)$ is perfectly analytic at $z_1$ — no singularity at all.

This means at $z = z_1$:

Both $E(z_1)$ and $A(z_1)$ are finite (we’ll see why shortly), so the correction term $E(z_1)/A(z_1)$ is a nonzero constant. Numerically, $E(z_1) \approx -1.202$ and $A(z_1) \approx 1.41$, giving $B/A \approx 1.202 - 1.202/1.41 \approx 0.35$.

Second Attempt: An Adaptive ODE (Also Fails)

My dad suggested the “method of unknowns”: treat $\zeta$ as an unknown variable and let an ODE drive it to the right value. Define $E(z;\zeta) = B(z) - \zeta \cdot A(z)$ and use the adaptation law

to push $\zeta$ toward the value that makes $E = 0$.

The problem: $E(z;\zeta(3)) = 0$ is not true at any finite $z$. As computed above, $E(z_1;\,\zeta(3)) \approx -1.202 \neq 0$. The adaptation law drives $\zeta$ to $B(z)/A(z) \approx 0.35$, not to $\zeta(3)$. Numerically, with $\gamma = 100$ and integrating to $z = 0.999\,z_1$, the system converges to $\zeta \approx 0.014$ — completely wrong.

The fundamental issue: the constraint $B = \zeta\,A$ only holds coefficient-by-coefficient ($b_n = \zeta(3)\,a_n + \text{error}$), not as an identity between functions.

The Singularity Structure

To understand what does work, we need to look at the singular behavior of $A(z)$ at the conifold.

Since $a_n \sim C \cdot \alpha^n \cdot n^{-3/2}$ where $\alpha = 1/z_1 = 17 + 12\sqrt{2}$, the transfer theorems of analytic combinatorics tell us that $A(z)$ has a square-root branch point at $z_1$:

where $A_\mathrm{reg}(z)$ is analytic at $z_1$ and $c$ is a constant. Crucially, $A(z_1) = A_\mathrm{reg}(z_1)$ is finite — the sum $\sum a_n z_1^n = \sum C\,n^{-3/2}$ converges (it’s a $p$-series with $p = 3/2 > 1$).

Since $E(z)$ is analytic at $z_1$, we can decompose $B(z)$:

The singular part of $B$ at $z_1$ is exactly $\zeta(3)$ times the singular part of $A$.

Now look at the ratio again:

The regular parts dominate, and $E(z_1) \neq 0$ contaminates the limit. The $\zeta(3)$ information is hidden in the singular part, which is subdominant.

The Key Idea: Differentiate to Amplify the Singularity

Differentiation amplifies singularities. If $A(z) \sim c\,(z_1-z)^{1/2}$ near $z_1$, then:

The regular parts have bounded derivatives, while the singular part blows up. For the first derivative:

The divergent terms dominate, and their ratio is $\zeta(3)$. The error is:

For the second derivative, the singular terms blow up even faster ($\sim (z_1-z)^{-3/2}$), so the regular corrections are even more suppressed:

In general, the $k$-th derivative ratio converges with error $O\bigl((z_1-z)^{k-1/2}\bigr)$.

Numerical Verification

Using 200-term Taylor series with 50-digit precision (via mpmath):

First derivative ratio $B'(z)/A'(z)$:

Second derivative ratio $B''(z)/A''(z)$:

The second derivative ratio already gives 8–9 digits of $\zeta(3)$ at $z/z_1 = 0.9$, compared to only 3 digits from the first derivative ratio. The convergence rate improvement from $O((z_1-z)^{1/2})$ to $O((z_1-z)^{3/2})$ is clearly visible.

(The plateau in the second derivative errors near $10^{-10}$ is a truncation artifact: the 200-term Taylor series runs out of precision before the theoretical convergence rate takes over.)

The Picture

Here is a summary of the three ratios at $z = 0.9\,z_1$:

The pattern is clear: differentiation amplifies the singular part of the solution, and the singular part carries the $\zeta(3)$ factor. The function-value ratio sees mostly the regular parts (which are “contaminated” by $E$); the derivative ratios see mostly the singular parts (which are “pure” $\zeta(3)$).

From Derivative Ratio to a Polynomial ODE

The derivative ratio tells us $B''/A'' \to \zeta(3)$, but both numerator and denominator blow up at $z_1$. Can this be tamed into a bounded, polynomial ODE?

An inhomogeneous surprise

The first step uncovered something unexpected: $B(z)$ does not satisfy the same ODE as $A(z)$. While both $a_n$ and $b_n$ satisfy the Apéry recurrence for $n \geq 1$, the initial conditions $(b_0, b_1) = (0, 6)$ violate the $n = 0$ case (which requires $u_1 = 5u_0$). The defect $b_1 - 5b_0 = 6$ propagates as a constant forcing term:

Normalizing by $A''$

Define the ratio $R = B''/A''$ and normalized residuals:

As $z \to z_1$: $R \to \zeta(3)$, while $q_0, q_1, u, v, w \to 0$.

Using the two ODEs $L[A] = 0$ and $L[B] = 6$, the z-derivatives of these variables form a closed system:

where $\sigma = (c_2 + c_1 u + c_0 v)/p$ and $p = z^2(1-34z+z^2)$, with similar equations for $u, v, w$.

Time reparametrization

The $1/p$ denominators make the system non-polynomial. Setting $d\tau = dz/p$ (equivalently, $dz/d\tau = p$) absorbs all the divisions. The result is a 7-variable polynomial ODE:

where $c_0 = -5+z$, $c_1 = 1-112z+7z^2$, $c_2 = z(3-153z+6z^2)$, $p = z^2(1-34z+z^2)$, and dots denote $d/d\tau$. The right-hand side is polynomial of degree $\leq 4$ in the seven state variables.

Rational initial conditions

At $z_0 = 1/100$, the initial conditions are computed from the (finite) Taylor series of $A$ and $B$:

and similarly for $q_0, q_1$. Since $a_n$ and $b_n$ are rational for all $n$, and $z_0 = 1/100$ is rational, these are all rational numbers (with ~90-digit numerators from 30 terms of the series).

At $z_0 = 0$, the values simplify: $R_0 = 351/292$, $u_0 = 5/146$, $v_0 = w_0 = 1/146$, $q_0 = -351/42632$, $q_1 = -3/42632$. But $z = 0$ is a fixed point of $\dot z = z^2(1-34z+z^2)$. Starting at $z_0 = 1/100 > 0$ avoids this: $\dot z > 0$, and $z(\tau)$ increases monotonically toward $z_1$.

Numerical verification

Integrating from $z_0 = 1/100$ with the rational initial conditions:

By $\tau = 1000$, the system has converged to $\zeta(3)$ at machine precision ($\approx 16$ digits). All auxiliary variables decay to effectively zero. The trajectories are bounded throughout.

Summary

Starting from the Calabi-Yau / Apéry operator and its two solutions $A(z)$, $B(z)$:

1. The function ratio $B/A$ does not converge to $\zeta(3)$ at the conifold (it converges to $\approx 0.35$).
2. The derivative ratio $B''/A''$ does converge to $\zeta(3)$, because differentiation amplifies the singular part — and the singular part carries $\zeta(3)$.
3. Normalizing by $A''$ and reparametrizing time yields a polynomial PIVP (degree $\leq 4$, 7 variables, rational initial conditions) whose $R$-component converges to $\zeta(3)$.

The construction is fully explicit: every coefficient in the ODE comes from the Apéry operator, every initial value is a rational number computable from Apéry numbers, and convergence to $\zeta(3)$ is monotone and rapid.

Reflection

Today’s exploration followed a crooked path: a function ratio that pointed in the wrong direction, an adaptive law that converged to nonsense, and then the realization that differentiation — the simplest operation in analysis — was the missing ingredient. The singular structure of the ODE was not an obstacle but a carrier of information: $\zeta(3)$ lives in the singular part, and you need derivatives to see it.

The inhomogeneous equation $L[B] = 6$ was a surprise — I had assumed both generating functions satisfied the same ODE. They don’t: the recurrence holds for $n \geq 1$ but the initial conditions $b_0 = 0, b_1 = 6$ violate the $n = 0$ case. This “defect of 6” appears explicitly in the polynomial ODE as the constant $+6w$ driving $R$ toward $\zeta(3)$.

There is something satisfying about how the pieces fit together. The coefficient-level identity $b_n/a_n \to \zeta(3)$ creates a branch-point singularity at the conifold where the singular parts of $B$ and $A$ differ by exactly $\zeta(3)$. Differentiation amplifies the singular part. Normalization by $A''$ absorbs the blow-up. Time reparametrization makes everything polynomial. And the constant $6$ from the initial-condition defect provides the driving force.

This is part of an ongoing project on bounded analog complexity. Previous posts: Computing Apéry’s constant with chemistry (the DNA32 integral approach) and From the Apéry series to a differential equation for $\zeta(3)$ (the generating function ODE and its obstructions). Today’s derivative-ratio observation came from a conversation with my dad, who suggested the “method of unknowns” — treating $\zeta$ as a variable and letting the equation determine it. The method didn’t work as stated, but the singularity analysis it prompted led to something much better: a concrete polynomial PIVP for $\zeta(3)$ derived entirely from the Calabi-Yau structure.

山重水复疑无路，柳暗花明又一村。 — 陆游，《游山西村》

Through heavy mountains and winding rivers, one doubts there is a path; past dark willows and bright flowers, another village appears. — Lu You, Visiting a Village West of the Mountains