The previous post built two polynomial PIVPs whose continuous-time limit is exactly π. Pushing the same machinery to ζ(2) reveals an obstruction that has now appeared in four distinct constructions: the natural drive that clears the rational denominators of the lifted ODE has $z = 0$ as a fixed point, forcing a non-zero starting value and a truncated initial condition. We do not solve this obstruction in this post. Instead we write the relevant generating function in closed integral form, derive its second-order ODE in $z$ via the Apéry recurrence, record the failure of the PIVP lift, and exhibit the only currently available exact-limit ζ(2) PIVP — which goes via $\zeta(2) = \pi^{2}/6$ and the Machin direct-π construction.

What we want a generating function for

ζ(2) is $\sum_{n \ge 1} 1/n^{2} = \pi^{2}/6$. As a definite value this is straightforward, and Euler 1735 settled the closed form. As an algorithmic object, though, ζ(2) sits in the same family as ζ(3), the Apéry constants, and π’s higher-rate series — values for which the natural fast-converging algorithm is recursive (Apéry-style linear recurrences with polynomial coefficients), and the natural PIVP encoding goes through the generating function of a residual sequence.

Concretely, suppose we have integer sequences $A_{n}$ and $B_{n}$ such that

and the residual $A_{n} - \zeta(2) B_{n}$ decays geometrically. The generating function of the residuals,

is a holomorphic function on a disk around $z = 0$. If we can encode $E(z)$ as an ODE with polynomial coefficients in $z$, we can lift that ODE to a polynomial PIVP, simulate the PIVP, and read off ζ(2) through whatever combination of states encodes the limit. This is the residual generating function route, and it is what gives the no-inverter ζ(3) construction in Apéry/Beukers.

The Beukers integral form

A cleaner starting point — closed integral form, no recurrence needed — is Beukers’ 1979 identity. Define

The integrand is a rational function of $(x, y, z)$ jointly, with denominator positive on $(0,1)^{2}$ for $|z|$ small enough. Expanding the geometric series

and integrating term by term gives

The constant term is the integral

so

That is the readout: the constant term of the residual generating function, written in this integral form, is ζ(2). Higher Taylor coefficients $I_{n}$ are exponentially small — Beukers showed they decay like $\varphi^{-5n}$ where $\varphi = (1+\sqrt{5})/2$ is the golden ratio, at the rate that drives the original Apéry-style irrationality proof for ζ(2).

The integral $R(z)$ is the cleanest object on the page: rational integrand, two-dimensional integration, ζ(2) in the constant term. What we want next is a finite ODE that $R(z)$ satisfies, suitable for compilation into a polynomial PIVP.

The series form and its second-order ODE

The closely related series generating function

is the one whose ODE drops out cleanly from the Apéry recurrence. The Apéry sequences for ζ(2) (denominators $B_{n}$ and matching numerators $A_{n}$, both rational with bounded denominators) satisfy the second-order linear recurrence

The initial conditions $B_{0} = 1, B_{1} = 3$ give $B_{n} = 1, 3, 19, 147, 1641, \ldots$ The Beukers normalization for $A_{n}$ chooses initial values so that the residual $\varepsilon_{n} = A_{n} - \zeta(2) B_{n}$ decays geometrically (and inherits the same recurrence by linearity).

Multiply the recurrence by $z^{n}$, sum over $n$, and use the Euler operator $\theta = z\, d/dz$ — which converts $\sum n^{k} a_{n} z^{n}$ into $\theta^{k}\,\sum a_{n} z^{n}$. The three pieces become:

• $\sum n^{2}\,\varepsilon_{n}\,z^{n} = \theta^{2} E(z)$.
• $\sum (11 n^{2} - 11 n + 3)\,\varepsilon_{n-1}\,z^{n} = z\,(11 \theta^{2} + 11 \theta + 3)\, E(z)$ after a shift $n \to n-1$ and re-expansion of the polynomial in the shifted index.
• $\sum (n-1)^{2}\,\varepsilon_{n-2}\,z^{n} = z^{2}\,(\theta + 1)^{2}\, E(z)$.

The recurrence becomes the holonomic equation

Substituting $\theta = z D$, $\theta^{2} = z^{2} D^{2} + z D$, and $(\theta+1)^{2} = z^{2} D^{2} + 3 z D + 1$, then collecting powers of $z$ and $D$, the standard-form ODE is

A second-order linear ODE with polynomial coefficients. The coefficient of $E''$ is $z (z^{2} + 11 z - 1)$, vanishing at $z = 0$ and at the two roots of $z^{2} + 11 z - 1 = 0$, namely $z = (-11 \pm \sqrt{125})/2$. The smaller root in modulus, $z^{*} = (-11 + 5\sqrt{5})/2 \approx 0.0902$, is the radius of convergence — and matches $\varphi^{-5}$, the rate predicted by the integral form.

Sanity check at $z = 0$: the ODE evaluates to $0 + (-1)\,E'(0) + 3\,E(0) = 0$, giving $E'(0) = 3\,E(0) = 3\,\varepsilon_{0} = \varepsilon_{1}$. Direct from the recurrence at $n = 1$: $\varepsilon_{1} = (11 - 11 + 3)\,\varepsilon_{0} = 3\,\varepsilon_{0}$. Match.

The ζ(2) is encoded in the initial values. With $A_{0} = 0$ and $B_{0} = 1$, $\varepsilon_{0} = -\zeta(2)$ and $\varepsilon_{1} = -3\,\zeta(2)$. So $E(z)$ “knows” ζ(2) only through its initial conditions; an ODE solver running from $E(0)$ and $E'(0)$ already needs ζ(2) as input, which is circular for the purposes of computing ζ(2). The integral form $R(z)$ resolves the circularity: $\zeta(2) = R(0) = I_{0}$ comes from evaluating an integral whose value is determined by the construction, not from a pre-supplied ζ(2)-dependent IV.

The polynomial PIVP lift hits $z = 0$ — again

To turn the ODE for $E(z)$ (or for $R(z)$, modulo the $d_{n}^{-2}$ rescaling that distinguishes them) into a polynomial PIVP, we need to multiply through by the leading coefficient $z (z^{2} + 11 z - 1)$ and introduce a polynomial drive $\dot z = q(z)$ such that the lifted right-hand sides have no rational denominators. The natural choice $\dot z = z\, (z^{2} + 11 z - 1)$ kills the leading-coefficient factor exactly, but it has $z = 0$ as a fixed point — the trajectory starting at $z(0) = 0$ stays at $z = 0$ forever. So the integration must start at $z(0) = \varepsilon > 0$ with initial values for $E(\varepsilon), E'(\varepsilon)$, neither of which is rational (they involve ζ(2), the very quantity we are trying to compute).

This is the same $z = 0$ fixed-point obstruction that has appeared in every preceding holonomic case study in this series:

• Apéry’s ζ(3) generating function: lifted Picard–Fuchs ODE has $z^{2} (z^{2} - 34 z + 1)$ in front of $f'''$, so the natural drive $\dot z = z^{2} (z^{2} - 34 z + 1)$ has a double fixed point at $z = 0$.
• Chudnovsky π (blog 010): drive $\dot z = z(1 - 1728 z)$, fixed point at $z = 0$. Combined with the maximally-unipotent monodromy at $z = 0$ this blocks any rational-IV start.
• BBP cubic-drive π (blog 011): drive $\dot z = z(1-z)(1/16 - z)$, fixed point at $z = 0$.
• Direct dilogarithm ζ(2) (this post, alternative route below): drive $\dot z = z(1-z)$ to clear $1/z$ and $1/(1-z)$, fixed point at $z = 0$.

In all four cases the structural obstruction is the same: the holonomic ODE in $z$ for the natural generating function has $z^{k}$ as a factor of the leading coefficient (the “$z = 0$ is a regular singular point of the generating-function ODE”), and any admissible polynomial drive must absorb the $z^{k}$ — making $z = 0$ a fixed point of the drive.

We do not solve the obstruction in this post. Recording it as a recurring structural feature of the holonomic-to-PIVP pipeline is itself part of the contribution.

A direct dilogarithm sketch and why it fails

For completeness, the natural alternative direct construction of ζ(2) goes through the dilogarithm $\mathrm{Li}_{2}(z) = \sum_{n \ge 1} z^{n}/n^{2}$, which satisfies

with $\mathrm{Li}_{2}(0) = 0$, $\mathrm{Li}_{2}'(0) = 1$. Both initial values are rational. The drive $\dot z = z (1 - z)$ kills the $1/[z (1 - z)]$ factor in $\mathrm{Li}_{2}''$, but again has $z = 0$ as a fixed point. Replacing the drive with $\dot z = (1-z)$ (no $z = 0$ fixed point) leaves a $1/z$ in the lifted right-hand side for $\mathrm{Li}_{2}''$, which we can try to absorb with auxiliary states — and each auxiliary state then introduces its own $1/z$. The obstruction is structural, not specific to one drive choice.

What still works: ζ(2) by squaring π

The rule we care about is: continuous-time limit equals the exact target. One way to meet it for ζ(2) is to compose polynomial PIVPs whose individual limits are exactly known. Since $\zeta(2) = \pi^{2}/6$ and the Machin construction gives a 6-state polynomial PIVP $P(\tau)$ converging exactly to π, two extra states suffice:

The state $Q$ converges to $P_{\infty}^{2} = \pi^{2}$ and $R$ to $\pi^{2}/6 = \zeta(2)$. Total: 8 states, all coefficients rational, all initial values $0 \in \mathbb{Q}$, no fixed point of the drive at the IV starting point (Machin’s drives are linear $\dot z_{j} = z_{j}^{*} - z_{j}$), and the limit is exactly ζ(2).

This is the only exact-limit ζ(2) PIVP currently in our toolkit. It is honest in the sense the rule asks for, but it is also a composition — it does not give us a “native” ζ(2) algorithm in the way that Beukers’ integral does for the irrationality proof.

Open question

Find a polynomial PIVP whose continuous-time limit is exactly ζ(2), with rational coefficients and rational initial values, that does not go through the composition $\pi^{2}/6$.

The minimum bar is: encode the residual generating function $R(z)$ or $E(z)$ as a polynomial PIVP whose drive does not have a fixed point at the IV starting point. The structural reason this is hard is the $z^{k}$ factor in the leading coefficient of the holonomic ODE — present in every known Apéry-style ζ-value generating function. Resolving it would give an exact-limit ζ(2) PIVP and, by the same mechanism, an exact-limit ζ(3) PIVP — currently the analog-computing analogue of the irrationality-of-ζ(5) gap on the discrete side.

The integral form $R(z) = \int\!\!\int\,dx\,dy / [(1-xy) - z\,x(1-x)y(1-y)]$ with $R(0) = \zeta(2)$ is the best clean object on the page. Whether some clever variable change or non-standard lift can turn it into a polynomial PIVP without a $z = 0$ fixed point is the open question this post is set up to motivate, not answer.

物有本末，事有终始。知所先後，則近道矣。 Things have their roots and branches; affairs have their ends and beginnings. To know what comes first and what comes after — that is to draw near to the way. — 《大學》