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Summation of exponential terms

  1. Mar 23, 2015 #1
    I found the following identity in a paper:
    ##
    \sum_{l=1}^{\infty}exp(-\pi\alpha l^2)=(\frac{1}{2\sqrt{\alpha}}-\frac{1}{2})+\frac{1}{\sqrt{\alpha}}\sum_{l=1}^{\infty}exp(\frac{-\pi l^2}{\alpha}) ##
    Someone please let me give some hints on how to prove this.
     
  2. jcsd
  3. Mar 23, 2015 #2

    jfizzix

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    It looks like it's the same sum on both sides, but scaled.

    I would see what happens to the exponential sum when you change [itex]\alpha[/itex] to [itex]\alpha\beta[/itex]. Can you make the result identical with the right scaling factor in front (twice as wide, but half as tall)?
     
  4. Mar 27, 2015 #3
    Yes, it's the same sum but scaled. One thing I have found that if you approximate the summation as integral, it can be proved easily as both are usual Gaussian Integral. But I was worrying about the factor that involves to transform the summation into integral. Any thoughts?
     
  5. Mar 27, 2015 #4

    jfizzix

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    So, you can write it alternatively as:
    [itex]\frac{1}{2} +\sum_{l=1}^{\infty}exp(-\pi\alpha l^2)=\frac{1}{\sqrt{\alpha}}\big(\frac{1}{2}+\sum_{l=1}^{\infty}exp(\frac{-\pi l^2}{\alpha})\big) [/itex]
    or better yet
    [itex]\sum_{l=-\infty}^{\infty}exp(-\pi\alpha l^2)=\frac{1}{\sqrt{\alpha}}\sum_{l=-\infty}^{\infty}exp(\frac{-\pi l^2}{\alpha}) [/itex]
    this sum is exactly the integral of a discrete gaussian. With the normalization constant in front, they should have the same area..

    At least, that is how I think it should work out.

    Does [itex]\alpha[/itex] have to be a positive integer?
     
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