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I Sum of exponentials

  1. Sep 15, 2016 #1
    I am looking for a way to simplify the following expression:

    ##\sum\limits_{n=1}^{N}\ \sum\limits_{k=0}^{N-1}\ \sum\limits_{k'=0}^{N-1}\ \tilde{p}_{k}\ \tilde{p}_{k'}\ e^{2\pi in(k+k')/N}##.

    I presume that the sum of the exponentials over ##n## somehow reduce to a Kronecker delta.

    Am I wrong?
     
  2. jcsd
  3. Sep 15, 2016 #2

    Simon Bridge

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  4. Sep 16, 2016 #3
    This formula is handy:

    ##\sum_{k=1}^{N}e^{2\pi ikn/N}=N\delta_{\text{n mod N}, 0}##

    so that

    ##\sum\limits_{n=1}^{N}\ \sum\limits_{k=0}^{N-1}\ \sum\limits_{k'=0}^{N-1}\ \tilde{p}_{k}\ \tilde{p}_{k'}\ e^{2\pi in(k+k')/N}##

    ##=\frac{1}{2mN}\ \sum\limits_{k=0}^{N-1}\ \sum\limits_{k'=0}^{N-1}\ \tilde{p}_{k}\ \tilde{p}_{k'}\ N \delta_{(k+k') \text{mod N},0}##

    ##=\frac{1}{2m}\ \sum\limits_{k=0}^{N-1}\ \tilde{p}_{k}\ \tilde{p}_{N-k}##.

    What do you think?
     
  5. Sep 16, 2016 #4

    Simon Bridge

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    That first step requires that you can treat k+k' as a single integer, as required in the provided formula.
    ie. requires that: $$\sum_{m=1}^N e^{2\pi i mn/N} = \sum_{k=1}^N\sum_{k'=1}^N e^{2\pi i (k+k')n/N}$$
     
  6. Sep 16, 2016 #5
    thanks!
     
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