Define a Function$\mathit{h}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathit{f}\left(x+2k\pi \right)$, assuming that the series converges to a function satisfying Dirichlet conditions (Section 6). Verify that h(x)does have Period$\mathbf{2}\mathit{\pi }$.

(a) Expand h(x)in an exponential Fourier series$\mathit{h}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}{\mathit{c}}_{\mathbf{n}}{\mathit{e}}^{\mathbf{i}\mathbf{n}\mathbf{x}}$;show that ${\mathit{c}}_{\mathbf{n}}\mathbf{=}\mathit{g}\left(n\right)$where $\mathit{g}\left(\alpha \right)$is the Fourier transform of f(x). Hint: Write ${\mathit{c}}_{\mathbf{n}}$as anIntegral from 0to $\mathbf{2}\mathit{\pi }$and make the change of variable $\mathit{u}\mathbf{=}\mathit{x}\mathbf{+}\mathbf{2}\mathit{k}\mathit{\pi }$. Note that${\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{i}\mathbf{n}\mathbf{k}\mathbf{\pi }}\mathbf{=}\mathbf{1}$, and the sum onkgives a single integral from $\mathbf{-}\mathbf{\infty }\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathbf{\infty }$

(b) Let x=0in (a) to get Poisson’s summation formula $\mathbf{\sum }_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathbf{f}\left(2k\pi \right)\mathbf{=}\mathbf{\sum }_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathit{g}\left(n\right)$

This result has many applications; for example: statistical mechanics, communicationtheory, theory of optical instruments, scattering of light in a liquid and so on (See Problem 22).

The given function is $h\left(x\right)=\sum _{n=-\infty }^{\infty }f\left(x+2n\pi \right)$

For the given function$\mathit{u}\left(x\right)$define its Fourier transformation as

$\mathit{F}\left\{u\left(x\right)\right\}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{\pi }}}\underset{\mathbf{-}\mathbf{\infty }}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{k}\mathbf{x}}\mathit{u}\left(x\right)\mathit{d}\mathit{x}$

A function is said to be periodic functions if it repeats value after specific interval of time.

$$\begin{gathered} h\left(x\right)=\sum _{n=-\infty }^{\infty }f\left(x+2n\pi \right) \\ h\left(x+2\pi \right)=\sum _{n=-\infty }^{\infty }f\left(x+2\left(n+1\right)\pi \right) \\ =\sum _{n=-\infty }^{\infty }f\left(x+2m\pi \right)m=n+1 \end{gathered}$$

Thus the function is indeed periodic.

$$\begin{gathered} h\left(x\right)=\sum _{n=-\infty }^{\infty }{c}_n{e}^{inx} \\ {c}_n=\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}\left(\sum _{k=-\infty }^{\infty }f\left(x+2k\pi \right)\right)e^{-inx}dx \\ =\frac{1}{2\pi }\sum _{k=-\infty }^{\infty }\underset{0}{\overset{2\pi }{\int }}f\left(x+2k\pi \right)e^{-inx}dx \end{gathered}$$

As$u=x+2\pi kdu=dx$

$$\begin{gathered} c_n=\frac{1}{2\pi }\sum _{k=-\infty }^{\infty }\underset{2\pi k}{\overset{2\pi \left(k+1\right)}{\int }}f\left(u\right)e^{-in\left(u-2\pi k\right)}du \\ =\frac{1}{2\pi }\sum _{k=-\infty }^{\infty }{e}^{ink2\pi }\underset{2\pi k}{\overset{2\pi \left(k+1\right)}{\int }}f\left(u\right)e^{-inu}du \\ =\frac{1}{2\pi }\sum _{k=-\infty }^{\infty }\underset{2\pi k}{\overset{2\pi \left(k+1\right)}{\int }}f\left(u\right)e^{-inu}du \\ =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}f\left(u\right)e^{-inu}du=g\left(n\right) \end{gathered}$$

Where we used $e^{ink2\pi }=1$since is an integer

Simply put x=0

$$\begin{gathered} h\left(x\right)=\sum _{n=-\infty }^{\infty }{c}_n{e}^{inx} \\ =\sum _{k=-\infty }^{\infty }f\left(x+2\pi k\right) \end{gathered}$$

This implies that.$\sum _{-\infty }^{\infty }g\left(n\right)=\sum _{-\infty }^{\infty }f\left(2\pi k\right)$

Therefore,for the function, h(x)

Part (a)$c_n=g\left(n\right)$where,$g\left(\alpha \right)$is the Fourier transform of f(x)

Part (b)$\sum _{-\infty }^{\infty }g\left(n\right)=\sum _{-\infty }^{\infty }f\left(2\pi k\right)$