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Spring 2013 - Problem 11

calculation, residue theorem

Let f(z)=−πzcot⁡(πz)f\p{z} = -\pi z\cot\p{\pi z} as a meromorphic function on the whole of C\C.

  1. Locate all poles of ff and determine their residues.

  2. Show that for each n≥1n \geq 1 the coefficient of z2nz^{2n} in the Taylor expansion of f(z)f\p{z} about z=0z = 0 coincides with

    an=∑k=1∞2k2n.a_n = \sum_{k=1}^\infty \frac{2}{k^{2n}}.
Solution.

  1. The poles of ff are precisely the poles of zsin⁡πz\frac{z}{\sin{\pi z}} since cos⁡πz\cos{\pi z} is entire. Since sin⁡πz\sin{\pi z} has a simple pole at 00, it follows that zsin⁡πz\frac{z}{\sin{\pi z}} is analytic near the origin, so the poles of ff are at Z∖{0}\Z \setminus \set{0}. Given k∈Z∖{0}k \in \Z \setminus \set{0}, sin⁡πz\sin{\pi z} has a simple zero at kk, so we may take the following limit from the real direction and apply L'Hôpital's:

    Res⁡(f;k)=lim⁡z→k(z−k)f(z)=−πlim⁡x→kx(x−k)sin⁡πx=−πlim⁡x→kx+x−kπcos⁡πx=−kcos⁡πk=(−1)k+1k.\begin{aligned} \Res{f}{k} &= \lim_{z \to k} \p{z - k}f\p{z} \\ &= -\pi \lim_{x \to k} \frac{x\p{x - k}}{\sin{\pi x}} \\ &= -\pi \lim_{x \to k} \frac{x + x - k}{\pi\cos{\pi x}} \\ &= -\frac{k}{\cos{\pi k}} \\ &= \p{-1}^{k+1} k. \end{aligned}

  2. We have the representation

    πcot⁡πz=1z+∑k=1∞2zz2−k2  ⟹  f(z)=−1−∑k=1∞2z2z2−k2.\pi \cot{\pi z} = \frac{1}{z} + \sum_{k=1}^\infty \frac{2z}{z^2 - k^2} \implies f\p{z} = -1 - \sum_{k=1}^\infty \frac{2z^2}{z^2 - k^2}.

    Set g(z)=∑k=1∞2z2−k2g\p{z} = \sum_{k=1}^\infty \frac{2}{z^2 - k^2} so that f(z)=−1−z2g(z)f\p{z} = -1 - z^2g\p{z}. Then

    f(2n)(0)=−∑j=02n(2nj)(djdzjz2)(0)g(2n−j)(0)=−2(2n)!2!(2n−2)!g(2n−2)(0)=−(2n)(2n−1)g(2n−2)(0)\begin{aligned} f^{\p{2n}}\p{0} &= -\sum_{j=0}^{2n} \binom{2n}{j} \p{\frac{\diff^j}{\diff{z}^j} z^2}\p{0} g^{\p{2n-j}}\p{0} \\ &= -2\frac{\p{2n}!}{2!\p{2n - 2}!} g^{\p{2n-2}}\p{0} \\ &= -\p{2n}\p{2n - 1} g^{\p{2n-2}}\p{0} \\ \end{aligned}

    Since gg is a Laurent series, it converges uniformly on compact sets, so we may differentiate term-by-term. Notice that

    2z2z2−k2=2+2k2z2−k2=2+k(1z−k−1z+k)\frac{2z^2}{z^2 - k^2} = 2 + \frac{2k^2}{z^2 - k^2} = 2 + k\p{\frac{1}{z - k} - \frac{1}{z + k}}

    and so

    g(2n−2)(z)=∑k=1∞k((−1)2n−2(2n−2)!(z−k)2n−1−(−1)2n−2(2n−2)!(z+k)2n−1)=∑k=1∞k((2n−2)!(z−k)2n−1−(2n−2)!(z+k)2n−1)g(2n−2)(0)=∑k=1∞k((2n−2)!(−k)2n−1−(2n−2)!k2n−1)=−2∑k=1∞(2n−2)!k2n.\begin{aligned} g^{\p{2n-2}}\p{z} &= \sum_{k=1}^\infty k\p{\p{-1}^{2n-2}\frac{\p{2n - 2}!}{\p{z - k}^{2n-1}} - \p{-1}^{2n-2}\frac{\p{2n - 2}!}{\p{z + k}^{2n-1}}} \\ &= \sum_{k=1}^\infty k\p{\frac{\p{2n - 2}!}{\p{z - k}^{2n-1}} - \frac{\p{2n - 2}!}{\p{z + k}^{2n-1}}} \\ g^{\p{2n-2}}\p{0} &= \sum_{k=1}^\infty k\p{\frac{\p{2n - 2}!}{\p{-k}^{2n-1}} - \frac{\p{2n - 2}!}{k^{2n-1}}} \\ &= -2 \sum_{k=1}^\infty \frac{\p{2n - 2}!}{k^{2n}}. \end{aligned}

    Hence,

    f(2n)(0)=2∑k=1∞n!k2n  ⟹  f(2n)(0)(2n)!=∑k=1∞2k2n.f^{\p{2n}}\p{0} = 2\sum_{k=1}^\infty \frac{n!}{k^{2n}} \implies \frac{f^{\p{2n}}\p{0}}{\p{2n}!} = \sum_{k=1}^\infty \frac{2}{k^{2n}}.