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Find the Laurent Series for $f(z) = \cot(\pi z)$ at $z= 0$.

I just need hints; I am not sure how to do this at all. I am very new to this.

Cookie
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Amad27
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2 Answers2

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\begin{align} \pi \cot(\pi z) &=\frac{d}{dz}\ln(\sin(\pi z))\\ &=\frac{d}{dz}\left[\ln\pi z\prod^\infty_{n=1}\left(1-\frac{z^2}{n^2}\right)\right]\\ &=\frac{1}{z}-2\sum^\infty_{n=1}\frac{1}{n^2}\frac{z}{1-\frac{z^2}{n^2}}\\ &=\frac{1}{z}-2\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{z^{2k-1}}{n^{2k}}\\ &=\frac{1}{z}-2\sum^\infty_{k=1}\zeta(2k)z^{2k-1} \end{align}

Guest543212345
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Summon the Maclaurin series (which is the Taylor series centered at $z=0$) of both $\cos(\pi z)$ and $\sin(\pi z)$, to obtain $$\cot(\pi z) = \frac{\cos(\pi z)}{\sin(\pi z)}=\frac{1-\frac 1{2!}(\pi z)^2+\frac 1{4!}(\pi z)^4-\frac 1{6!}(\pi z)^6+\cdots}{\pi z-\frac 1{3!}(\pi z)^3+\frac 1{5!}(\pi z)^5-\frac 1{7!}(\pi z)^7+\cdots}.$$

Can you do long division of the following? $$\pi z-\frac 1{3!}(\pi z)^3+\frac 1{5!}(\pi z)^5-\frac 1{7!}(\pi z)^7+\cdots\overline{\left){1-\frac 1{2!}(\pi z)^2+\frac 1{4!}(\pi z)^4-\frac 1{6!}(\pi z)^6+\cdots} \right. }$$ Your quotient will be the first few terms of your Laurent series of $\cot(\pi z)$ centered at $z=0$.

Cookie
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  • how to do this division? – Amad27 Jan 12 '15 at 10:02
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    Here is a hint to get you started: $$\qquad \qquad , \frac 1{\pi z} - \cdots \ \pi z-\frac 1{3!}(\pi z)^3+\frac 1{5!}(\pi z)^5-\frac 1{7!}(\pi z)^7+\cdots\overline{\left){1-\frac 1{2!}(\pi z)^2+\frac 1{4!}(\pi z)^4-\frac 1{6!}(\pi z)^6+\cdots} \right. \ -1 \ \quad \overline{0-\frac 1{2!}(\pi z)^2} \ \qquad \qquad \vdots }$$ – Cookie Jan 13 '15 at 07:47
  • I dont get it. In the fist step you subtracted $1$ right? But you should have subtracted $(\frac{1}{\pi z})(\pi z - \frac{(\pi z)^3}{3!} ....)$ All of that? But you only subtracted $1$? – Amad27 Jan 13 '15 at 12:16
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    Becuase $\frac 1{\pi z} \cdot (\pi z)=1$. I didn't finish everything, but yes, you should also subtract $\frac 1{\pi z}\cdot[-\frac 1{3!}(\pi z)^3] + \frac 1{\pi z}\cdot[\frac 1{5!}(\pi z)^5] - \frac 1{\pi z}\cdot[-\frac 1{7!}(\pi z)^7]$ in the same line, which I forgot to do in my last comment. – Cookie Jan 13 '15 at 18:47