6
$\begingroup$

Let $Li_s(z)$ denote the usual polylogarithm. The elementary functional equation $$Li_{-n}(z)=(-1)^{n-1}Li_{-n}(1/z)$$ holds for $n\geq 1$. I remember only that the proof used some reproducing property of the Stirling numbers of the second kind.

This functional equation is rather useful because, taking linear combinations, it amounts to saying that a meromorphic function having a power series on the open unit disc whose coefficients are the values of a polynomial $f$ of degree $d$ evaluated at integer arguments:

$$\sum_{k=0}^{\infty}f(k)z^k=\frac{1}{(1-z)^{d+1}}\sum_{k=0}^{d}\left(\sum_{j=0}^k{d+1\choose j}(-1)^jf(k-j)\right)z^k$$

is defined for $|z|>1$ by the series

$$-\sum_{k=1}^{\infty}f(-k)z^{-k}.$$

I would like to know how this functional equation (or the continuation by sum over negative powers) can be proved?

Please note that the use of statements like

$$\sum_{k=-\infty}^{\infty}k^nz^k=\left(z\frac{d}{dz}\right)^n\sum_{k=-\infty}^{\infty}z^k=\left(z\frac{d}{dz}\right)^n 0=0$$ is acceptable only if you can also prove that the sums of positive and negative powers converge on a common arc of the unit circle.

In his answer below, Fedor Petrov has pointed out that the use of divergent series as formal generating functions is justified in some cases, so the last statement is incorrect.

$\endgroup$
13
$\begingroup$

Let's try to understand in which sense this equality $$\sum_{k=-\infty}^{\infty}k^nz^k=\left(z\frac{d}{dz}\right)^n\sum_{k=-\infty}^{\infty}z^k=\left(z\frac{d}{dz}\right)^n 0=0$$ may be understood. Of course, there is no convergence in usual sense on the unit circle (because terms do not tend to 0). Actually, we do not use analysis at all, as it is purely combinatorial-algebraic statement. For two-sided Laurent series like $\sum_{k=-\infty}^{\infty} a_k z^k$ there are following well-defined operations: taking derivative and multiplying by polynomials, and we have $(fg)'=f'g+fg'$ for polynomial $f$ and two-sided series $g$. For $H(z)=\sum_{k=-\infty}^{\infty} z^k$ we have $(z-1)H(z)=0$. Hence $(z-1)H'(z)+H(z)=0$, multiplying by $(z-1)$ we get $(z-1)^2H'(z)=0$. Take a derivative again to get $(z-1)^3H''(z)=0$ and so on. This implies that $$0=(z-1)^{n+1}\left(\left(z\frac{d}{dz}\right)^n H(z)\right)=(z-1)^{n+1}(Li_{-n}(z)+(-1)^nLi_{-n}(1/z))$$ But we know that both expressions $(z-1)^{n+1}Li_{-n}(z)$ and $(z-1)^{n+1}Li_{-n}(1/z)$ are actually polynomials in $z$. What we have proved is that the sum of those two polynomials does vanish, as desired.

$\endgroup$
  • $\begingroup$ The proof I've found is an explicit version of what you've said. Yet, despite that it comes down to simple properties of factorials, it is necessary to pass through the partial fraction formula else I can't prove equality on a common open set. Hats off to generating functions. $\endgroup$ – Kevin Smith Oct 7 '14 at 19:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.