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Gaussian Summation

  1. Jun 1, 2010 #1
    Hi,
    We know that the Gaussian integral is
    [tex]\int_{-\infty}^{+\infty}e^{-\frac{x^2}{a^2}}dx=a\sqrt{\pi}[/tex]
    However, if the gaussian function is discrete in x, what is the result of
    [tex]\sum_{n=0}^{+\infty}e^{-\frac{n^2}{a}} = \\?[/tex]
    where n is natural number, that is n=0,1,2,3.............
     
    Last edited: Jun 1, 2010
  2. jcsd
  3. Jun 1, 2010 #2

    Mute

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    Homework Helper

    I checked Wolfram alpha; it gives the result as

    [tex]\frac{1 + \vartheta_3(0,e^{-1/a})}{2},[/tex]

    where [itex]\vartheta_b(x,q)[/itex] is a theta function. Looking up the definition of the theta function on mathworld reveals that the result is pretty much by definition:

    [tex]\vartheta_3(z,q) = \sum_{n=-\infty}^\infty q^{n^2} e^{2\pi i z}[/tex]

    Rearrangement and plugging in z = 0, q = e^(-1/a) gives the result, although it's not very enlightening. The theta function is a special function that should be available in mathematica and perhaps matlab.

    I'm not sure if, setting a = 1 for example, there is a way to compute the resulting value without reference to the theta function.
     
  4. Jun 1, 2010 #3
    Thanks a lot! It's really a great help to me. :smile::smile::smile:
     
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