# I Sum of exponentials

1. Sep 15, 2016

### rock_pepper_scissors

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. Sep 15, 2016

### Simon Bridge

3. Sep 16, 2016

### rock_pepper_scissors

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?

4. Sep 16, 2016

### Simon Bridge

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

5. Sep 16, 2016

thanks!