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Series and product development (Ahlfors)

  1. Jun 15, 2015 #1
    Dear friends !Please help me to solve these two problems.Thanks!
     

    Attached Files:

  2. jcsd
  3. Jun 15, 2015 #2

    RUber

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    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.
     
  4. Jun 15, 2015 #3
    Thank you very much RUber.I was struggling with these problems after studying Jensons formulae and Mittag Lefflers theorems.
     
  5. Jun 15, 2015 #4

    RUber

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    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}##

    http://www4f.wolframalpha.com/Calculate/MSP/MSP39321aggihfid3ad8g3700003g26gfg4f3h9dd14?MSPStoreType=image/gif&s=2&w=310.&h=45. [Broken]
    ##\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
  6. Jun 15, 2015 #5
    Thanks RUber! I have progressed a little further.Will share soon.
     
  7. Jun 16, 2015 #6

    Svein

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    A bit further up on the page, marked as (11) it says: [itex]\pi\cot(\pi z)=\frac{1}{z}+\sum_{n=1}^{\infty}\frac{2z}{z^{2}-n^{2}} [/itex]. Keep that in mind. Now [itex]\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}} [/itex]. Since (-n)2=n2, we have [itex] \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}}[/itex]. Now we are almost there - divide the expression in (11) by z (remembering to keep away from 0), you have [itex] \frac{\pi\cot(\pi z)}{z}=\frac{1}{z^{2}}+2\cdot\sum_{n=1}^{\infty}\frac{1}{z^{2}-n^{2}}[/itex]. Conclusion: [itex]\sum_{-\infty}^{\infty}\frac{1}{z^{2}-n^{2}}=\frac{\pi\cot(\pi z)}{z}[/itex].
     
  8. Jun 16, 2015 #7
    Thanks svein!
     
  9. Jun 16, 2015 #8
  10. Jun 17, 2015 #9

    Svein

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    I shall give you some hints:
    1. [itex] \frac{1}{(z+n)^{2}+a^{2}}=\frac{1}{2ia}(\frac{1}{z+n-ia}-\frac{1}{z+n-ia})[/itex]
    2. [itex]\sum_{n=-\infty}^{\infty}\frac{1}{z+n+ia}=\sum_{n=-\infty}^{\infty}\frac{1}{z-n+ia} [/itex]
    3. Let w = z+ia. We already know that [itex] \sum_{n=-\infty}^{\infty}\frac{1}{w-n}=\pi\cot\pi w[/itex].
    4. Do the same thing with w = z-ia.
    5. Substituting back, you get [itex] \sum_{-\infty}^{\infty}\frac{1}{(z+n)^{2}+a^{2}}=\frac{1}{2ia}(\pi\cot\pi (z+ia)-\pi\cot\pi (z-ia))[/itex]
    The rest is left to the student...
     
  11. Jun 17, 2015 #10
    Thank You Svein!
     
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