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Week 11 Worksheet - Solutions
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Summing Series
- We take $f : \C \setminus \{0\} \to \C$ to be $f(z) = \dfrac{1}{z^4}$. Recall that $g(z) = \dfrac{\sin(\pi z)}{\cos(\pi z)}$ is uniformly bounded on $\Gamma_N$. We can estimate that
\[
\left| \int\limits_{\Gamma_N} fg \right| \le \ell(\Gamma_N) \dfrac{M}{N^4} \le \dfrac{16M}{N^3}
\]
because $|z| \ge N$ for all $z \in \Gamma_N$. Thus the contour integral converges to zero as $N \to \infty$. Cauchy's residue theorem gives
\[
\int\limits_{\Gamma_N} fg = 2 \pi i \sum_{n=-N}^N \Res(fg,n) \wind(\Gamma_N,n)
\]
because the only poles of $fg$ are at $n \in \{-N,\dots,N\}$. As usual
\[
\Res(fg,n) = \dfrac{1}{\pi n^4}
\]
for all non-zero $n$ and it reamins to calculate the residue at zero. The order of the pole at zero is five. From our lemma
\[
\Res(fg,0) = \lim_{z \to 0} \dfrac{1}{4!} \left( \dfrac{z^5 \cos(\pi z)}{z^4 \sin(\pi z)} \right)^{(4)} = \lim_{z \to 0} \dfrac{1}{4!} \left( \dfrac{z}{\sin(\pi z)} \cos(\pi z) \right)^{(4)}
\]
and the power series
\[
\dfrac{w}{\sin w} = 1 + \dfrac{w^2}{6} + \dfrac{w^4}{240} + \cdots
\]
we get
\begin{align*}
\Res(fg,0) & = \lim_{z \to 0} \dfrac{1}{4!} \left( \left( \dfrac{1}{\pi} + \dfrac{\pi}{6} z^2 + \dfrac{7\pi^3}{360} z^4 - \cdots \right) \left( 1 - \dfrac{\pi^2}{2} z^2 + \dfrac{\pi^4}{24} z^4 - \cdots \right) \right)^{(4)} \\
& = \lim_{z \to 0} \dfrac{1}{4!} \left( \dfrac{1}{\pi} - \dfrac{\pi}{3} z^2 - \dfrac{\pi^3}{45} z^4 + \cdots \right)^{(4)} \\
& = -\dfrac{\pi^3}{45}
\end{align*}
and we can put everything together. Cauchy's residue theorem gives
\[
2 \pi i \left( 2 \sum_{n=1}^N \dfrac{1}{\pi n^4} - \dfrac{\pi^3}{45} \right) = \int\limits_{\Gamma_N} fg
\]
and the limit as $N \to \infty$ gives $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^4} = \dfrac{\pi^4}{90}$
- We would want to use $f(z) = 1/z^3$ but with that choice
\[
\Res(fg,-n) = -\dfrac{1}{\pi n^3} = - \Res(fg,n)
\]
for all $n \in \N$ and Cauchy's residue theorem only gives
\[
\int\limits_{\Gamma_N} fg = 2 \pi i \Res(fg,0)
\]
which does not involve $\displaystyle\sum_{n=1}^N \dfrac{1}{n^3}$