I'm given a generating function $$G(x) = \sum\limits_{k=0}^{\infty}a_k x^k$$ for a sequence $(a_0, a_1, a_2, \ldots)$.

I know, that I can express generating functions for sequences like $(a_0 + c, a_1 + c, a_2 + c, \ldots)$ or $(1 \cdot a_1, 2 \cdot a_2, 3 \cdot a_3, \ldots)$ in terms of $G(x)$. For example $$xG^{'}(x) = \sum\limits_{k=0}^{\infty} k a_k x^k$$

So I'm interested,

could something be done to express the function $$F(x) = \sum\limits_{k=0}^{\infty} \sqrt{k} a_k x^k$$ in terms of $G(x)$ and/or its derivatives?

  • $\begingroup$ I doubt there's an easy (or any) formula for that. $\endgroup$
    – Berci
    Apr 12, 2018 at 12:18
  • 1
    $\begingroup$ Just taking $a_k=2^{-k}$ and $x=1$, wolfram already spits out a value with the $\text{Li}$ function, so I'd put by bets on "no, that's can't be done" or at least "no, that can not be done in a nice enough manner for people to want to use it" $\endgroup$ Apr 12, 2018 at 12:19

1 Answer 1


Consider the function $$ G(x) = \sum_{n=1}^\infty x^n = \frac{x}{1-x} $$ This, and all its derivatives, are rational functions. But (as vrungtehagel noted) $$ F(x) = \sum_{n=1}^\infty \sqrt{n}\;x^n = \mathrm{Li}_{-1/2}(x) , $$ is a polylogarithm (of non-integer type), so certainly not a rational function.

You may think of the problem as trying to do the differential operator $$ x\;\frac{d}{dx} $$ a fractional numer of times: $$ F(x) = \left(x\frac{d}{dx}\right)^{1/2} G(x) \tag{?} $$

  • $\begingroup$ Since applying the same $\sqrt{k}$ to the $\sqrt{k}a_k$ gives us something simple, is it possible that even some non-degenerate integer sequences could end up in something simple? $\endgroup$
    – dEmigOd
    Apr 12, 2018 at 12:52
  • $\begingroup$ Good question. Is there some sequence $a_k$ (not eventually zero) so that both $F(x)$ and $G(x)$ have simple form? $\endgroup$
    – GEdgar
    Apr 12, 2018 at 12:58

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