Series and product development (Ahlfors)

1. Jun 15, 2015

gianeshwar

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2. Jun 15, 2015

RUber

Hello. It seems like these series are related to p-series, with n^2 it is more like the Basel problem (https://en.wikipedia.org/wiki/Basel_problem). I do not immediately see the proper conversion, but I will continue to think about it.

3. Jun 15, 2015

gianeshwar

Thank you very much RUber.I was struggling with these problems after studying Jensons formulae and Mittag Lefflers theorems.

4. Jun 15, 2015

RUber

I had to ask wolframalpha.com, but was provided with the following solutions. These types of problems usually require a great deal of manipulation to show, even when the answer is known.

http://www4f.wolframalpha.com/Calculate/MSP/MSP652520diec13gdedd2dh00003i3bedh373981c66?MSPStoreType=image/gif&s=20&w=168.&h=45. [Broken]
My image didn't copy properly, so here it is in Tex.
$\sum_{-\infty}^{\infty} \frac{1}{z^2 - n^2 } = \frac{\pi \cot(\pi z)}{z}$

$\sum_{-\infty}^{\infty} \frac{1}{a^2 +(z + n)^2 } = \frac{\pi \sinh(2\pi a)}{a( \cosh(2 \pi a ) - \cos(2\pi z))}$

Last edited by a moderator: May 7, 2017
5. Jun 15, 2015

gianeshwar

Thanks RUber! I have progressed a little further.Will share soon.

6. Jun 16, 2015

Svein

A bit further up on the page, marked as (11) it says: $\pi\cot(\pi z)=\frac{1}{z}+\sum_{n=1}^{\infty}\frac{2z}{z^{2}-n^{2}}$. Keep that in mind. Now $\sum_{-\infty}^{\infty}\frac{1}{z^{2}-n^{2}}=\sum_{n=-\infty}^{-1}\frac{1}{z^{2}-n^{2}}+\frac{1}{z^{2}}+\sum_{n=1}^{\infty}\frac{1}{z^{2}-n^{2}}$. Since (-n)2=n2, we have $\sum_{-\infty}^{\infty}\frac{1}{z^{2}-n^{2}}=\frac{1}{z^{2}}+2 \cdot\sum_{n=1}^{\infty}\frac{1}{z^{2}-n^{2}}$. Now we are almost there - divide the expression in (11) by z (remembering to keep away from 0), you have $\frac{\pi\cot(\pi z)}{z}=\frac{1}{z^{2}}+2\cdot\sum_{n=1}^{\infty}\frac{1}{z^{2}-n^{2}}$. Conclusion: $\sum_{-\infty}^{\infty}\frac{1}{z^{2}-n^{2}}=\frac{\pi\cot(\pi z)}{z}$.

7. Jun 16, 2015

gianeshwar

Thanks svein!

8. Jun 16, 2015

gianeshwar

9. Jun 17, 2015

Svein

I shall give you some hints:
1. $\frac{1}{(z+n)^{2}+a^{2}}=\frac{1}{2ia}(\frac{1}{z+n-ia}-\frac{1}{z+n-ia})$
2. $\sum_{n=-\infty}^{\infty}\frac{1}{z+n+ia}=\sum_{n=-\infty}^{\infty}\frac{1}{z-n+ia}$
3. Let w = z+ia. We already know that $\sum_{n=-\infty}^{\infty}\frac{1}{w-n}=\pi\cot\pi w$.
4. Do the same thing with w = z-ia.
5. Substituting back, you get $\sum_{-\infty}^{\infty}\frac{1}{(z+n)^{2}+a^{2}}=\frac{1}{2ia}(\pi\cot\pi (z+ia)-\pi\cot\pi (z-ia))$
The rest is left to the student...

10. Jun 17, 2015

gianeshwar

Thank You Svein!

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