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The polylogarithm can be defined using the power series $$ \operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s}. $$ Contiguous polylogs have the ladder operators $$ \operatorname{Li}_{s+1}(z) = \int_0^z \frac {\operatorname{Li}_s(t)}{t}\,\mathrm{d}t\,, \qquad \operatorname{Li}_{s-1}(z) = z \,{\partial \operatorname{Li}_s(z) \over \partial z}\ , $$ and the sequence can be started with either $$ \operatorname{Li}_{1}(z) = -\ln(1-z)\,,\qquad \operatorname{Li}_{0}(z) = {z \over 1-z} \ . $$

Both $\operatorname{Li}_0$ and $\operatorname{Li}_1$ have inverse functions (up to a choice of branchcut) $$ \operatorname{Li}_0^{-1}(z)=\frac{z}{z+1}\,,\quad \operatorname{Li}_1^{-1}(z)=1-e^{-z}\,, $$ $$ \operatorname{Li}_0\left(\frac{z}{z+1}\right) =z= \operatorname{Li}_1\left(1-e^{-z}\right) + 2 n \pi i\,,\quad n\in\mathbb{Z} $$

Is there a nice/useful inverse function for the dilog ($\operatorname{Li}_2(z)$) and higher polylogs?

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  • $\begingroup$ As $Li_s'(0) \neq 0,$ an inverse exists as a formal powerseries. I would start there. $\endgroup$ – jspecter Oct 18 '11 at 1:25
  • $\begingroup$ @jspecter: $\text{Li}_n^{-1}(z) \approx z-2^{-n} z^2+(2^{1-2 n}-3^{-n}) z^3 + \dots$... but is the general coefficient known in closed form? Can it be summed as, e.g., a hypergeometric? $\endgroup$ – Simon Oct 18 '11 at 1:27
  • $\begingroup$ Certainly, one can use Lagrangian inversion to derive series for the inverse polylogarithms. I haven't encountered any situation where the inverses are needed,though. $\endgroup$ – J. M. isn't a mathematician Oct 18 '11 at 1:29
  • $\begingroup$ @Simon If the inverse polylogarithm is indeed hypergeometric function, it should satisfy a differential equation, and this is unlikely for non-integer $n$. $\endgroup$ – Sasha Oct 18 '11 at 4:34
  • $\begingroup$ @Sasha: You're probably (almost certainly) right about non-integer $n$, but more often than not, the integer case is the one that occurs. In particular, my question asked about the inverse dilog separately from the general polylog. $\endgroup$ – Simon Jan 16 '12 at 22:01
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In astrophysics, specifically in partially degenerated matter, are used what is called Fermi-Dirac Integrals, which are written in terms of polylogaritms, and the z-value is a degeneracy parameter. In some papers I found that in fact they need the inverse of the Fermi-Dirac Integrals, that is, the inverse of the Polylogarithm.

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    $\begingroup$ Thanks Michael, can you give some specifics? $\endgroup$ – Simon Nov 17 '11 at 4:30
  • $\begingroup$ Credible sources would also be much appreciated. $\endgroup$ – Void Apr 7 '20 at 3:38
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I wrote this in a related question and thought I might as well put it here even though the question is quite old. It gives something like a formal representation of the inverse as a series.

It is more a demonstration of how to group the terms in the messy reversion into references to documented (but complicated) sequences.

For the Polylogarithm we have the series representation $$ \mathrm{Li}_s(z) = \sum_{k=1}^\infty \frac{z^k}{k^s} $$ if we perform a series reversion on this (term by term) we end up with an expansion for the inverse function $$ \mathrm{Li}^{-1}_s(z) = \sum_{k=1}^\infty a_k z^k $$ the first few coefficients are $$ a_1 = 1 \\ a_2 = -2^{-s} \\ a_3 = 2^{1-2s} - 3^{-s} \\ a_4 = 5 6^{-s} - 8^{-s}(5+2^s) \\ \cdots $$ there may be a pattern in there somewhere, but the terms seem to grow quite large and complicated rather quickly. For some reason I considered looking at the inverse Mellin transform of these coefficients, multipled by a gamma function, we can denote these functions $e_k(x)$ $$ e_k(x) = \mathcal{M}^{-1}[\Gamma[s]a_k(s)](s) $$ these begin \begin{equation} e_1(x) = e^{-x} \\ e_2(x) = -e^{-2x} \\ e_3(x) = 2 e^{-4x} - e^{-3 x} \\ e_4(x) = -5 e^{-8 x} + 5 e^{-6x} - e^{-4 x} \\ \cdots \end{equation} in each term $e_k(x)$ there are $P(k-1)$ exponential functions, where $P(k)$ from $k=0$ goes like $1,1,2,3,5,7,\cdots$ and are the partition numbers A000041. The coefficients in this grid of exponentials goes like: $$ \alpha_1=[ +1]\\ \alpha_2=[ -1]\\ \alpha_3=[ -1, 2]\\ \alpha_4=[ -1, 5, -5]\\ \alpha_5=[ -1, 6, 3, -21, 14]\\ \alpha_6=[ -1, 7, 7, -28, -28, 84, -42]\\ \alpha_7=[ -1, 8, 8, 4, -36, -72, -12, 120, 180, -330, 132] $$ , and appear to be given by A111785. Interestingly, the powers of the exponentials also appear to have a sequence, the terms go like A074139, which is titled "Number of divisors of A036035(n).". A036035 is titled "Least integer of each prime signature, in graded (reflected or not) colexicographic order of exponents."

We can recreate a coefficient by performing the Mellin transform and dividing through by $\Gamma(s)$ $$ a_k(s) = \frac{1}{\Gamma(s)}\mathcal{M}[e_k(x)](s) = \frac{1}{\Gamma(s)}\int_0^\infty x^{s-1}e_k(x) \; dx $$ Then we can write $$ e_k(x) = \sum_{l=1}^{P(k-1)} \alpha_{kl}e^{-\beta_{kl} x} $$ where $\beta_k$ are the anaolgous rows of A074139 $$ \beta_1=[1]\\ \beta_2=[2]\\ \beta_3=[3, 4]\\ \beta_4=[4, 6, 8]\\ \beta_5=[5, 8, 9, 12, 16]\\ \beta_6=[6, 10, 12, 16, 18, 24, 32]\\ $$ we we know then the Mellin transform of $a e^{-bx}=a b^{-s}\Gamma(s)$, and it's as simple as, grouping the terms which are now understood $$ a_k(s) = \frac{1}{\Gamma(s)}\int_0^\infty x^{s-1}\sum_{l=1}^{P(k-1)} \alpha_{kl}e^{-\beta_{kl} x} \; dx $$ $$ a_k(s) = \sum_{l=1}^{P(k-1)}\frac{1}{\Gamma(s)}\int_0^\infty x^{s-1} \alpha_{kl}e^{-\beta_{kl} x} \; dx $$ $$ a_k(s) = \sum_{l=1}^{P(k-1)}\frac{1}{\Gamma(s)}\alpha_{kl}\beta_{kl}^{-s}\Gamma(s) $$ $$ a_k(s) = \sum_{l=1}^{P(k-1)}\alpha_{kl}\beta_{kl}^{-s} $$ $$ \mathrm{Li}^{-1}_s(z) = \sum_{k=1}^\infty a_k(s)z^{k} = \sum_{k=1}^\infty \sum_{l=1}^{P(k-1)}\alpha_{kl}\beta_{kl}^{-s}z^k $$ the limitation here is understanding where the terms in the sequences referenced come from, and finding whether any tractable forms exist for $\alpha$ and $\beta$. We already know that $\beta_{kl} = \sigma_0(\gamma_{kl})$ for the $\gamma$ in A036035, and divisor counting function $\sigma$. $$ \mathrm{Li}^{-1}_s(z) = \sum_{k=1}^\infty \sum_{l=1}^{P(k-1)}\frac{\alpha_{kl}z^k}{\sigma(\gamma_{kl})^s} $$

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I just ran into a need to solve this very problem. I was able to derive a solution using the inversion of a series.

Starting with

$Li_{s}(z) = \sum_{k=1}^{\infty} k^{-s} z^{k}$

we want the inverse function to be of the form

$Li^{-1}_{s} = \sum_{k=1}^{\infty} b_{k} z^{k}$.

Using the inversion of a series technique (see Morse and Feshbach pg 411) I find

$b_k = \frac{1}{k!}\left[\frac{d^{k-1}}{dx^{k-1}}\left(1 + \sum_{n=1}^{\infty} (n+1)^{-s} x^{n}\right)^{-k}\right]_{x=0}$

These coefficients can be fairly easily calculated by hand or using your favorite software suite. As for a deeper pattern, the one interesting thing I found is that the sum in the coefficient is the Lerch transcendent function

$\sum_{n=1}^{\infty} (n+1)^{-s} x^{n} = x \Phi(x,s,2)$

where

$\Phi(x,s,a) = \sum_{n=0}^{\infty}(n+a)^{-s} x^{n}.$

So the coefficients of the inverse polylog are

$b_k = \frac{1}{k!}\left[\frac{d^{k-1}}{dx^{k-1}}\left(1 + x \Phi(x,s,2)\right)^{-k}\right]_{x=0}$

At this point I just calculated the coefficients using Mathematica and got the following for the first few coefficients

$b_1 = 1$

$b_2 = -2^{-s}$

$b_3 = 2^{1-2s} - 3^{-s}$

$b_4 = -24^{-s} (5\times(3^{s}-4^{s}) + 6^{s})$

They quickly get more complex from there.

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    $\begingroup$ Unfortunately I am not sure if anything nice can be said for $Li_2^{-1}$. My focus is mainly on the fractional orders (for a physics application). I think we are just stuck with the inverse of the polylog being its own special function. In that vein maybe call it the polyexponential? $\endgroup$ – Jeffrey Hersh Mar 11 '20 at 12:41

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