6
$\begingroup$

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 $n$.

$\endgroup$
11
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ That's fantastic, thank you. $\endgroup$ – Ben Lerner May 8 '11 at 22:52
4
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

HINT:

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

product of two odd functions

$\endgroup$

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.