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In a past exam question we prove that the following function is well-defined and holomorphic on $\mathbb{C}$ \ $\mathbb{Z}$, and then we are asked to find the closed form. Let $$ f(z)=\sum_{n=-\infty}^{\infty}\frac{1}{(z+n)^2}. $$

The mark scheme says:

We have that $f$ is periodic as $f(z) = f(z+2) \forall z$. See that $f(z)$ has double poles at every integer with residue $(-1)^k$.

$(*)$ Note that $f(z) = −g'(z)$, where $g(z)$ has single poles with residue $(−1)^k$ at each integer. Then by periodicity it follows that $g(z) = \frac{\pi}{sin(\pi z)}$ and we obtain $f$ by differentiating.

I honestly have no idea why this argument is right. I can see why $f$ is periodic and its residues are as given but everything from $(*)$ is not resonating.

Any help in understanding this would be great! Thanks

jcneek
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  • The argument is definitely incomplete. Did you mean to write $g’(z)$ rather than $-g(z)$? Also all of $f$’s residues are zero. The term residue only applies to the degree $-1$ coefficient in the Laurent expansion. – Vik78 Jan 13 '22 at 01:13
  • Right. Should be: $f(z) =- g'(z)$ as @Vik78 suggests. – GEdgar Jan 13 '22 at 01:31
  • Periodicity, plus the condition on the residues of $g$, does not by itself imply that $g(z) = \pi \textrm{csc} \pi z$. After all one could add any polynomial function of $\textrm{sin} \pi z$ and preserve those two conditions. I think your exam may have assumed that you had prior knowledge of the identity of the closed form of the series. Otherwise it is quite difficult to find. – Vik78 Jan 13 '22 at 02:24

2 Answers2

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What you wrote is very odd, it suggests a lot of confusion. First of all this is not a Laurent series, the terminology is "Mittag Leffler expansion".

Looking only at the poles and the periodicity is not enough, compare $\frac{\pi^2}{\sin^2(\pi z)}$ with $\frac{\pi^2}{\sin^2(\pi z)}+e^{2i\pi z}-e^{4i\pi z} $

The obvious solution is to say that $$\frac{\pi^2}{\sin^2(\pi z)}-\sum_n \frac1{(z+n)^2}$$ is a $1$-periodic entire function vanishing as $\Im(z)\to \pm\infty$ on $\Re(z)\in [0,1]$. This implies that it is bounded, constant, and identically zero.

reuns
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  • The solution works, but it I would call it a "solution by knowing the answer" (if you don't know about this representation of $1/\sin$, you are out of luck in the exam). – Maximilian Janisch Jan 13 '22 at 19:16
  • @MaximilianJanisch Not sure of what you mean. The OP's question is a mess, differentiating $\pi/\sin \pi z$ doesn't give what we want, we need to differentiate $2i\pi /(e^{2i\pi z}-1)$ which is harder because its Mittag Leffler series doesn't converges absolutely. So (in complex analysis) my answer is the simplest route to this problem and the related ones, obtained by integrating or differentiating the series for $\pi^2/\sin^2(\pi z)$. – reuns Jan 13 '22 at 23:15
  • I completely agree with you; I am saying that, while your proof is totally fine, it is hard to come up with if you do not know in before what the closed form for $$\sum_{n\in\mathbb Z} \frac1{(z+n)^2}$$ is. – Maximilian Janisch Jan 13 '22 at 23:41
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    You mean if you don't see that $\sum_n 1/(z+n)^2-h(z) $ will be entire for some trigonometric function $h$. Well that was Euler's genious to spot it. In complex analysis given a function it is natural and common to search for other functions having the same poles, branch points, zeros, growth, and so on. – reuns Jan 13 '22 at 23:44
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I guess some of the information is not properly copied.

This looks to me like the following well known calculation.

You start with Euler's infinite product formula for $\sin$: $$\frac{\sin(\pi z)}{\pi z} = \prod_{n = 1}^\infty (1 - \frac{z^2}{n^2})$$ and take $\log$ then $\frac d{dz}$ on both sides to get $$\pi \cot(\pi z) = \frac 1z + \sum_{n = 1}^\infty\left(\frac1{z - n} + \frac1{z + n}\right).$$ Taking $\frac d{dz}$ again gives $$\frac{\pi^2}{\sin^2(\pi z)} = \sum_{n = -\infty}^\infty \frac 1{(z + n)^2}.$$

This procedure appears e.g. in GTM 97, Introduction to Elliptic Curves and Modular Forms, by Neal Koblitz, in Chapter III, page 110.

WhatsUp
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