The Polylog and the Polygamma

Two more members of the zeta family, and both of them are the Hurwitz zeta or the Lerch transcendent wearing a different hat:

$$\operatorname{Li}_s(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^s} = z\,\Phi(z, s, 1)$$$$\psi^{(m)}(z) = \frac{d^{m+1}}{dz^{m+1}}\ln\Gamma(z) = (-1)^{m+1} m!\,\zeta(m+1, z)$$

The polylogarithm is the Lerch transcendent along $a = 1$; the polygamma is the Hurwitz zeta with its two arguments read the other way round — the order $m$ fixes the first slot, and the argument $z$ takes the second. So a single Euler–Maclaurin kernel, already written for $\zeta(s,a)$, computes the entire polygamma family, and the Lerch series computes the polylogarithm.

compute-engine ships both heads natively, so this is an extension rather than an addition. PolyLog is native at integer order (including the continuation past the unit disk); @enumeratio/analytic adds non-integer and complex $s$. PolyGamma is native at real argument; @enumeratio/analytic adds complex $z$, and the GPU kernel neither had.

#Reductions

In$\operatorname{Li}_{1}(z)$Out$-\ln(1-z)$In$\operatorname{Li}_{2}(1)$Out$\frac{π^2}{6}$In$\operatorname{Li}_{-1}(z)$Out$\frac{z}{(1-z)^2}$

source · editable

<notatio-cell value="PolyLog(1, z)" data-allow-mismatch="children" /> <notatio-cell value="PolyLog(2, 1)" data-allow-mismatch="children" /> <notatio-cell value="PolyLog(-1, z)" data-allow-mismatch="children" />

$\operatorname{Li}_1(z) = -\ln(1-z)$, $\operatorname{Li}_2(1) = \zeta(2) = \pi^2/6$, and every negative integer order is a rational function — $\operatorname{Li}_{-1}(z) = z/(1-z)^2$.

#Polygamma as a zeta value

In$N(\mathrm{PolyGamma}(1, 1))$Out$1.644\,934\,066\,848\,226\,436\,47$In$N(\mathrm{PolyGamma}(2, 1))$Out$-2.404\,113\,806\,319\,188\,570\,8$In$N(\mathrm{PolyGamma}(1, \frac{1}{2}))$Out$4.934\,802\,200\,544\,679\,309\,42$

source · editable

<notatio-cell value="N(PolyGamma(1, 1))" data-allow-mismatch="children" /> <notatio-cell value="N(PolyGamma(2, 1))" data-allow-mismatch="children" /> <notatio-cell value="N(PolyGamma(1, 1 / 2))" data-allow-mismatch="children" />

At $z = 1$ every order is a zeta value: $\psi'(1) = \zeta(2) = \pi^2/6$ and $\psi''(1) = -2\zeta(3)$. At the half-integers the alternating zeta shows up instead — $\psi'(\tfrac12) = \pi^2/2$.

Two very different pictures

Colour the complex $z$-plane by hue = argument, brightness = magnitude, and the two functions could hardly look less alike — which is exactly the difference between their kernels. Both are the same <notatio-complex-plot> element, handed a different expression; every constant in them is a slider.

Liₛ(z) is the Lerch series, so it only converges inside $|z| < 1$: a lit disk, dark outside, with the branch point at $z = 1$ sitting on the rim. Raise $\operatorname{Im}(s)$ and the disk fills with spirals, the same winding the Lerch portrait shows; push $\operatorname{Re}(s)$ up and the $n = 1$ term $z$ dominates, leaving a near-linear ramp of hue.

ψ⁽ᵐ⁾(z) is a Hurwitz zeta, which converges over the whole plane — so the picture fills out, and the poles appear: one at each of $z = 0, -1, -2, \ldots$, of order $m+1$, strung along the negative real axis with the full colour wheel turning around each. Step $m$ up and the hue wheels round more times per pole — a pole of order $m+1$ winds the argument $m+1$ times.

(f32 on the GPU, so the rim of the disk and the immediate neighbourhood of a pole are approximate — the CPU kernel is the source of truth. The colouring runs on $\ln|\psi^{(m)}|$ rather than the value, which is what keeps high orders in frame: $\psi^{(m)}(1) \approx m!$ leaves f32's range past $m \approx 34$, but its logarithm never does.)

Where the pieces come from

function	computed as	new here
$\operatorname{Li}_s(z)$, integer $s$	compute-engine's own	—
$\operatorname{Li}_s(z)$, other $s$	$z\,\Phi(z, s, 1)$, the Lerch series	$\lvert z\rvert \leq 1$
$\operatorname{Li}_s(1)$	$\zeta(s)$, exactly	non-integer $s$
$\psi^{(m)}(z)$, real $z$	compute-engine's own	—
$\psi^{(m)}(z)$, complex $z$	$(-1)^{m+1} m!\,\zeta(m+1, z)$	the whole complex plane

The one place we stop short: a non-integer order outside the unit disk. The series does not reach there and the continuation is only implemented for integer $s$, so $\operatorname{Li}_{5/2}(2)$ is returned unevaluated rather than guessed at.