# Dirac Distribution

I'm trying to solve the following homework problem:

Show that, $$$$1 + 2\sum_{n=1}^{\infty} \cos(2\pi nx) = \sum_{k= - \infty}^{\infty}\delta(x-k),$$$$ 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.

• Perhaps use the Jacobi identities from Jaobi Theta function? – Somos Apr 9 '19 at 3:44
• @Somos Presumably you had this section in mind. – J.G. Apr 9 '19 at 7:29
• @J.G. Ah yes, an even better section. Thanks! – Somos Apr 9 '19 at 11:03

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