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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 9 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
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This follow by series bisection, e.g. bisecting into even and odd parts the power series for $\:\rm e^{\:i\:x} \:,\;$

$$\begin{align} \rm f(x) \ \ &=\ \ \rm\frac{f(x)+f(-x)}{2} \;+\; \frac{f(x)-f(-x)}{2} \\\\ \\\\ \Rightarrow\quad\quad \rm e^{\:i\:x} \ \ &=\ \ \rm\cos(x) \ +\ i \ \sin(x) \end{align}$$

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

$\rm\quad\quad\quad\quad a_m\ x^m\ +\ a_{m+n}\ x^{m+n}\ +\ a_{m+2\:n}\ x^{m+2\:n}\ +\:\cdots $

$\rm\quad\quad =\ \frac{1}{n} \big(f(x)\ +\ f(x\zeta)\ \zeta^{-m}\ +\ f(x\zeta^{\:2})\ \zeta^{-2m}\ +\:\cdots\: +\ f(x\zeta^{\ n-1})\ \zeta^{\ (1-n)\:m}\big)$

For much further discussion see my post here.

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