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Show complex summation property

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex] f(z) = \sum_{n =-\infty}^{\infty} e^{2 \pi i n z} e^{- \pi n^2}.[/itex] Show that [itex] f(z+i) = e^{\pi} e^{-2\pi i z}f(z). [/itex]


    2. Relevant equations
    Nothing specific I can think of; general complex analysis/summation techniques.


    3. The attempt at a solution
    [itex] f(z+i) = \sum_{n =-\infty}^{\infty} e^{-2 \pi n z} e^{2 \pi i n z} e^{- \pi n^2};[/itex] I can't factor the new term out of the sum because it contains an n. I feel like I might be missing some sort of summation identity that can accomplish this. I also tried completing the square, but it didn't really get me far.
     
  2. jcsd
  3. Mar 24, 2013 #2

    mfb

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    Staff: Mentor

    Messing around with the exponent,
    $$\exp(-2\pi n z + 2 \pi i n z - \pi n^2) = \exp(2 \pi i (n+1) z - \pi (n+1)^2 + 2\pi n + \pi - 2\pi i z - 2\pi n z) = e^\pi e^{- 2\pi i z} \exp(2 \pi i (n+1) z - \pi (n+1)^2 + 2\pi n(1-z))$$
    An index shift leads to
    $$e^\pi e^{- 2\pi i z} \exp(2 \pi i n z - \pi n^2 + 2\pi (n-1)(1-z))$$
    which looks quite nice. But how to get rid of the last part?
     
  4. Mar 25, 2013 #3
    That's really cool; it does simplify the problem a lot. But I'm having a bit of trouble getting rid of that last term. I tried shifting the index once more, but that doesn't seem to help. It can't be factored out, and it's definitely nontrivial for a given term. Could you give me some tips as to how to approach it?
     
  5. Mar 25, 2013 #4

    mfb

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    I don't know. The approach gives the required prefactors, so it might be something into it, but I don't know how to get rid of that additional term.
    It is trivial for z=1, and the function is periodic with a period of 1, therefore the equation is true for integer z.
     
  6. Mar 25, 2013 #5
    It seems z = 1 + ik, where k is an integer. It is also interesting to notice that f(z) is periodic with 1, as mfb has said. Then the relation between f(z) and f(z+i) must hold for z = r + ik, where both r and k are integers.
     
    Last edited: Mar 25, 2013
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