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I was thinking about how the standard proofs that $\pi$ is irrational use variations on $sin(\pi) =0$, in contrast to Apéry's proofs that $\zeta(2),\zeta(3)$ are irrational (the first of course also proving that $\pi$ is irrational.)

My question is as follows:

Let $$Li_2(x)=\sum_{n\ge1} \frac{x^n}{n^2},$$

and let $f(x)$ denote its compositional inverse.

What are the complex roots of $f$?

I didn't find anything looking around, and playing with sage makes it look like maybe there are no zeros.

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  • $\begingroup$ I have no idea whether it helps, but it is at least fun - the compositional inverse satisfies a relatively nice differential equation: $x^{3} f'\left(x\right)^{3} + {\left(x + 1\right)} f\left(x\right)^{3} + 3 \, x f\left(x\right)^{2} f'\left(x\right) - {\left(x^{2} - {\left(x^{3} + 3 \, x^{2}\right)} f\left(x\right)\right)} f'\left(x\right)^{2} - f\left(x\right)^{2} - {\left(x^{3} f\left(x\right)^{2} - x^{2} f\left(x\right)\right)} f''\left(x\right) = 0$ $\endgroup$
    – Martin Rubey
    Commented Oct 11, 2022 at 20:47
  • $\begingroup$ what do you mean by compositional inverse and also by $Li_2$ (in the sense that $Li_2$ is a multivalued function with branch points at $1, \infty$ so do you think of $Li_2$ as defined on say the complex plane minus $[1,\infty]$ as the "main branch" or something else? $\endgroup$
    – Conrad
    Commented Oct 12, 2022 at 0:03
  • $\begingroup$ I think that the answer is: the integer multiples of $4\pi^2$. $\endgroup$
    – Tom Goodwillie
    Commented Oct 12, 2022 at 1:03
  • $\begingroup$ That is, the numbers that arise as values of the multi-valued function $Li_2(x)$ at $x=0$ are precisely the numbewrs $4\pi^2k$, $k\in\mathbb Z$. $\endgroup$
    – Tom Goodwillie
    Commented Oct 12, 2022 at 1:04
  • $\begingroup$ @Conrad : the compositional inverse can be defined from the power series, and I don't think it's a multivalued function. This is similar to log being multivalued but exp is not. $\endgroup$
    – Dror Speiser
    Commented Oct 12, 2022 at 6:47

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