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I know that the terms are $0$ for odd $n > 1$, but I haven't had any luck proving this. Computing them directly verifies this for small $n$; the function is also analytic, so I've tried taking the integrals $$f^{(n)}(0) = \frac{n!}{2\pi i}\oint_C \frac{f(t)\,\mathrm dt}{t^{n+1}},$$ but I haven't found a way to show that the answer is $0$ for odd $n$.

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up vote 11 down vote accepted

Every function $f(z)$ defined on, say, a centrally symmetric open subset of $\mathbb{C}$ has a unique decomposition into even and odd parts

$$f(z) = \frac{f(z) + f(-z)}{2} + \frac{f(z) - f(-z)}{2}.$$

If $f$ has a Taylor series, then the even part is the sum of the even terms and the odd part is the sum of the odd terms. The odd part of $\frac{z}{e^z - 1}$ is given by

$$\frac{1}{2} \left( \frac{z}{e^z - 1} - \frac{-z}{e^{-z} - 1} \right) = \frac{1}{2} \left( \frac{z}{e^z - 1} - \frac{z e^z}{e^z - 1} \right) = - \frac{z}{2}$$

The point here is that $\frac{z}{e^z - 1}$ is "almost even," and the computation of the odd part is precisely a computation of how far the function is from being even.

The above computation generalizes to a decomposition of a Taylor series into the terms with exponents congruent to $a \bmod n$ for all $a$ and some $n$: it's essentially the discrete Fourier transform.

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+1, that is much nicer! – Eric Naslund May 8 '11 at 22:52
That's fantastic, thank you. – Ben Lerner May 8 '11 at 22:52

This follow by series bisection, e.g. bisecting into even and odd parts the power series for $\,e^{ix}$

$$\begin{align} f(x) \ &= \ \frac{f(x)+f(-x)}{2} \;+\; \frac{f(x)-f(-x)}{2} \\[6pt] \Rightarrow\quad e^{ix} \ &=\ \cos(x) \ +\ i \sin(x) \end{align}\qquad$$

Similarly one can perform multisections into $\,n\,$ parts using $\,n$'th roots of unity - see this answer for some examples and see Riordan's classic textbook Combinatorial Identities for many applications. Briefly, with $ \,\zeta\ $ a primitive $ \,n$'th root of unity, the $ \,m$'th $ \,n$-section $ $ selects the $ \, m+k\,n\,$ indexed terms from a series $ \ f(x)\ =\ a_0 + a_1 x + a_2 x^2 +\,\cdots\ $ as follows

$$\begin{align} &\ \ a_m x^m + a_{m+n}\ x^{m+n} + a_{m+2n} x^{m+2\,n}\ +\:\cdots\\[3pt] =\ & \frac{1}{n} \big(f(x) + f(x\zeta)\ \zeta^{-m} + f(x\zeta^{\,2})\ \zeta^{-2m} +\,\cdots +f(x\zeta^{\ n-1})\ \zeta^{\ (1-n)\,m}\big) \end{align}$$

For further discussion see this answer.

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$$\frac{z}{e^z-1} + \frac{z}{2} = \frac{z}{2}\, \coth{\frac{z}{2}}$$

product of two odd functions

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