Fall 2018 - Problem 9

Solution.

1. Since $\cot\p{\pi z} = \frac{\cos\p{\pi z}}{\sin\p{\pi z}}$, it follows that $\cot{\pi z}$ has simple poles at every integer. Thus, $g$ has a simple pole at every non-zero integer, and these have residue given by

   $$\begin{aligned} \Res{g}{k} &= \lim_{z \to k} \,\p{z - k}g\p{z} \\ &= \p{\lim_{z \to k} -\pi z\cos\p{\pi z}} \p{\lim_{x \to k} \frac{x - k}{\sin\p{\pi x}}} \\ &= -\pi k\cos\p{\pi k} \p{\lim_{x \to k} \frac{1}{\pi\cos\p{\pi x}}} \\ &= -k. \end{aligned}$$

2. Recall that $\pi\cot\p{\pi z}$ has the expansion

   $$\pi \cot\p{\pi z} = \frac{1}{z} + \sum_{n=1}^\infty \frac{2z}{z^2 - n^2} \implies g\p{z} = -1 - \sum_{n=1}^\infty \frac{2z^2}{z^2 - n^2}.$$

   Let $f\p{z} = -1 - \sum_{n=1}^\infty \frac{2z}{z - n^2}$ so that $f\p{z^2} = g\p{z}$, i.e., the $k$-th coefficient in the Maclaurin series expansion of $f$ corresponds to the $2k$-th coefficient of $g$. Observe that if $k \geq 1$,

   $$\begin{aligned} \p{\deriv{^k}{x^k} f\p{z}}\p{0} &= -\sum_{n=1}^\infty \p{\sum_{m=0}^k \binom{k}{m} \p{\deriv{^m}{x^m} z}\p{0} \p{\deriv{^{k-m}}{x^{k-m}} \frac{2}{z - n^2}}\p{0}} \\ &= -\sum_{n=1}^\infty \binom{k}{1} \p{\deriv{^{k-1}}{x^{k-1}} \frac{2}{z - n^2}}\p{0} \\ &= -\sum_{n=1}^\infty \frac{2k\p{-1}^{k-1}\p{k-1}!}{\p{-n^2}^k} \\ &= \sum_{n=1}^\infty \frac{2 \cdot k!}{n^{2k}}. \end{aligned}$$

   Thus, if $b_k$ is the $k$-th coefficient of the Maclaurin series of $f$, then

   $$a_{2k} = b_k = \frac{1}{k!} \sum_{n=1}^\infty \frac{2 \cdot k!}{n^{2k}} = \sum_{n=1}^\infty \frac{2}{n^{2k}}.$$