# How to do the series expansion of terms with PolyLog faster?

I have the following expression that I need to series expand, around t=0

(PolyLog[3, 1 + Tanh[J t]] - PolyLog[3, 1 - Tanh[J t]])Tanh[J t]


The amount of time to calculate one extra term is kind of just blowing up exponentially. I need at least 30 terms but for 20 terms only my time exceeded more than 30mins. To further speed up the process, I tried to use Assumptions command, with t>0, but it didn't help either. Infact I further tried to simply and do the series expansion of following expression around t=0

(PolyLog[3, 1 + t] - PolyLog[3, 1 - t])


and found that it is time taking too. I want to know if it is possible to simplify somehow the expression that I have in order to speed up the process? Or if there is some other way to find the Series expansion faster and efficiently?

Edit: I found with a little bit of experimentation that, the expansion of the Polylog is fast. In case instead of PolyLog[3, 1 + Tanh[J t]] if I had  PolyLog[3, Tanh[J t]] then the series expansion around t=0 is pretty quick even up to 50 terms.

• Series expansion of what parameter t or J ? About what point ? Give details ? I don't have time to guess. Aug 29, 2022 at 9:45
• I have edited and specified around t=0. I was also wondering if doing it on a different version of Mathematica will change things. I tried it and found that it is bit faster in Mathematica 12 Aug 29, 2022 at 10:56
• I would do the expansion with J=1 first and substitute t->J t afterwards Aug 29, 2022 at 11:05
• I'm interested in why you need such a high order expansion. Might numerical interpolation give a better result? Aug 29, 2022 at 11:46
• I need such a high-order expansion as the expansion coefficients are "moments", and there are specific "differences"(in comparison with other results that I have) I will get in these moments, which are enhanced when I take the higher-order ones. At low order, these "differences" are not so prominent. So one or the other way I need to calculate those "coefficients"! Aug 29, 2022 at 13:13

Not a full answer. Some hints.

Since function is symmetric, treat even and odd SeriesCoefficients differently.

f[t_, J_] = (PolyLog[3, 1 + Tanh[J t]] -
PolyLog[3, 1 - Tanh[J t]]) Tanh[J t];

f[x, J] == f[-x, J] // Simplify      (*   True   *)

(tab = Table[{2 n,
SeriesCoefficient[f[t, J], {t, 0, 2 n},
Assumptions -> t > 0 && J > 0]}, {n, 1, 7}] //
Expand) // TableForm

(tab = Table[{2 n - 1,
SeriesCoefficient[f[t, J], {t, 0, 2 n - 1},
Assumptions -> t > 0 && J > 0]}, {n, 1, 7}] //
Expand) // TableForm


Leave it to you to find a systematic dependence on 2 n and 2 n-1. This is only valid for t>0 && J>0.