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I'm trying to solve the following homework problem:

Show that, \begin{equation} 1 + 2\sum_{n=1}^{\infty} \cos(2\pi nx) = \sum_{k= - \infty}^{\infty}\delta(x-k), \end{equation} in the sense of distribution.

I can show that the Dirac comb is given by the expression on the left by a Fourier series argument. However, I am not sure how to show this by a distribution argument. Any suggestions? Just give hints as this is a HW problem.

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  • $\begingroup$ Perhaps use the Jacobi identities from Jaobi Theta function? $\endgroup$ – Somos Apr 9 '19 at 3:44
  • $\begingroup$ @Somos Presumably you had this section in mind. $\endgroup$ – J.G. Apr 9 '19 at 7:29
  • $\begingroup$ @J.G. Ah yes, an even better section. Thanks! $\endgroup$ – Somos Apr 9 '19 at 11:03
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$\ds{\sum_{k = -\infty}^{\infty}\delta\pars{x - k}}$ is even and periodic ( of period $\ds{1}$ ). Then,

\begin{align} &\sum_{k = -\infty}^{\infty}\delta\pars{x - k} = \sum_{n = 0}^{\infty}a_{n}\cos\pars{2\pi nx} \\[1cm] &\ \int_{-1/2}^{1/2}\cos\pars{2\pi nx}\sum_{k = -\infty}^{\infty}\delta\pars{x - k}\dd x \\[2mm] = &\ \sum_{m = 0}^{\infty} a_{m}\underbrace{\int_{-1/2}^{1/2}\cos\pars{2\pi nx}\cos\pars{2\pi mx}\dd x} _{\ds{=\ {1 + \delta_{n0} \over 2}\,\delta_{nm}}} \\[5mm] &\ \underbrace{\int_{-1/2}^{1/2}\cos\pars{2\pi nx}\delta\pars{x}\dd x}_{\ds{=\ 1}} = {1 + \delta_{n0} \over 2}\,a_{n} \\[5mm] &\ \implies a_{n} = 2 - \delta_{n0} \end{align}


$$ \implies \bbx{\sum_{k = -\infty}^{\infty}\delta\pars{x - k} = 1 + 2\sum_{n = 1}^{\infty}\cos\pars{2\pi nx}} $$

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  • $\begingroup$ My initial guess that to check that the effect of both the distributions (considered the 1 + infinite sum of cosines to be a generalize function) is the same on test functions. You're using a Fourier series argument, and I believe the way the question is phrased, I may have to use the other approach. Or maybe I'm looking too much into it. I suppose I can make a Fourier series argument as well. $\endgroup$ – user82261 Apr 10 '19 at 0:49
  • $\begingroup$ @user82261 I still find your question is an interesting one. $\endgroup$ – Felix Marin Apr 11 '19 at 17:38
  • $\begingroup$ Hmm, how so?... $\endgroup$ – user82261 Apr 12 '19 at 5:18

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