This problem is designed to guide you through a “proof” of Plancherel’s theorem, by starting with the theory of ordinary Fourier series on a finite interval, and allowing that interval to expand to infinity.

(a) Dirichlet’s theorem says that “any” function f(x) on the interval $\left[-a,+a\right]$can be expanded as a Fourier series:

$\mathbf{f}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\left[a_n\sin \left(\frac{\mathrm{n\pi x}}{a}\right)+b_n\cos \left(\frac{\mathrm{n\pi x}}{a}\right)\right]$

Show that this can be written equivalently as

$\mathbf{f}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}{\mathbf{c}}_{\mathbf{n}}{\mathbf{e}}^{\mathbf{in\pi x}\mathbf{/}\mathbf{a}}$.

What is ${\mathbf{c}}_{\mathbf{n}}$, in terms of ${\mathbf{a}}_{\mathbf{n}}$and ${\mathbf{b}}_{\mathbf{n}}$?

(b) Show (by appropriate modification of Fourier’s trick) that

${\mathbf{c}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{a}}{\mathbf{\int }}_{\mathbf{-}\mathbf{a}}^{\mathbf{+}\mathbf{a}}\mathbf{f}\left(x\right){\mathbf{e}}^{\mathbf{-}\mathbf{in\pi x}\mathbf{/}\mathbf{a}}\mathbf{dx}$

(c) Eliminate n and ${\mathbf{c}}_{\mathbf{n}}$in favor of the new variables $\mathbf{k}\mathbf{=}\left(\mathrm{n\tau \tau }/a\right)\mathbf{}\mathbf{andF}\mathbf{}\left(k\right)\mathbf{=}\sqrt{\mathbf{2}\mathbf{/}\mathbf{\pi }}{\mathbf{ac}}_{\mathbf{n}}$. Show that (a) and (b) now become

$\mathbf{f}\left(x\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{\pi }}}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathbf{F}\left(k\right){\mathbf{e}}^{\mathbf{ikx}}\mathbf{∆}\mathbf{k}\mathbf{;}\mathbf{}\mathbf{F}\left(k\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{\pi }}}{\mathbf{\int }}_{\mathbf{-}\mathbf{a}}^{\mathbf{+}\mathbf{a}}\mathbf{f}\left(x\right)\mathbf{.}{\mathbf{e}}^{\mathbf{ikx}}\mathbf{dx}$.

where $\mathbf{∆}\mathbf{k}$is the increment in k from one n to the next.

(d) Take the limit $\mathbf{a}\mathbf{\to }\mathbf{\infty }$to obtain Plancherel’s theorem. Comment: In view of their quite different origins, it is surprising (and delightful) that the two formulas—one for F(k) in terms of f(x), the other for f(x) terms of F(k) —have such a similar structure in the limit $\mathbf{a}\mathbf{\to }\mathbf{\infty }$.

Step by step solution

01

Given information:

The Fourier series is given by:

$\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }\left[{\mathrm{a}}_{\mathrm{n}}\sin \left(\frac{\mathrm{n\pi x}}{\mathrm{a}}\right)+{\mathrm{b}}_{\mathrm{n}}\cos \left(\frac{\mathrm{n\pi x}}{\mathrm{a}}\right)\right]$

02

 Step 2 : (a) Showing that the Fourier series can be equivalently written as:

Dirichlet’s theorem is written as:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{b}}_0+\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{a}}_{\mathrm{n}}}{2\mathrm{i}}\left({\mathrm{e}}^{\mathrm{in\pi x}/\mathrm{a}}-{\mathrm{e}}^{-\mathrm{in\pi x}/\mathrm{a}}\right)+\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{b}}_{\mathrm{n}}}{2}\left({\mathrm{e}}^{\mathrm{in\pi x}/\mathrm{a}}-{\mathrm{e}}^{-\mathrm{in\pi x}/\mathrm{a}}\right) \\ ={\mathrm{b}}_0+\sum _{\mathrm{n}=1}^{\infty }\left(\frac{{\mathrm{a}}_{\mathrm{n}}}{2\mathrm{i}}+\frac{{\mathrm{b}}_{\mathrm{n}}}{2}\right){\mathrm{e}}^{\mathrm{in\pi x}/\mathrm{a}}+\sum _{\mathrm{n}=1}^{\infty }\left(-\frac{{\mathrm{a}}_{\mathrm{n}}}{2\mathrm{i}}+\frac{{\mathrm{b}}_{\mathrm{n}}}{2}\right){\mathrm{e}}^{-\mathrm{in\pi x}/\mathrm{a}} \end{gathered}$$

So there are two values for , one for positive n’s, where the other is for negative values of n’s.

$$\begin{gathered} {\mathrm{c}}_0\equiv {\mathrm{b}}_0; \\ {\mathrm{c}}_{\mathrm{n}}=\frac{1}{2}\left(-{\mathrm{ia}}_{\mathrm{n}}+{\mathrm{b}}_{\mathrm{n}}\right),\mathrm{for}\mathrm{n}=1,2,3,...; \\ {\mathrm{c}}_{\mathrm{n}}\equiv \left({\mathrm{ia}}_{-\mathrm{n}}+{\mathrm{b}}_{-\mathrm{n}}\right),\mathrm{for}\mathrm{n}=-1,-2,-3,. \end{gathered}$$

$$\begin{gathered} {\mathrm{c}}_0\equiv {\mathrm{b}}_0; \\ {\mathrm{c}}_{\mathrm{n}}=\frac{1}{2}\left(-{\mathrm{ia}}_{\mathrm{n}}+{\mathrm{b}}_{\mathrm{n}}\right),\mathrm{for}\mathrm{n}=1,2,3,...; \\ {\mathrm{c}}_{\mathrm{n}}\equiv \left({\mathrm{ia}}_{-\mathrm{n}}+{\mathrm{b}}_{-\mathrm{n}}\right),\mathrm{for}\mathrm{n}=-1,-2,-3,. \end{gathered}$$

Thus,

$\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=-\infty }^{\infty }{\mathrm{c}}_{\mathrm{n}}{\mathrm{e}}^{\mathrm{in\pi x}/\mathrm{a}}$

03

(b) Showing by approximate modification of Fourier series

$$\begin{gathered} \mathrm{To}\mathrm{find}{\mathrm{C}}_{\mathrm{n}},\mathrm{multiply}\mathrm{both}\mathrm{sides}\mathrm{by}{\mathrm{\psi }}_{\mathrm{m}}^{*}\left(\mathrm{x}\right)\mathrm{then}\mathrm{integrate}\mathrm{from}-\mathrm{a}\mathrm{to}\mathrm{a}. \\ {\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{m\pi x}/\mathrm{a}}\mathrm{dx}=\sum _{\mathrm{n}=-\infty }^{\infty }{\mathrm{c}}_{\mathrm{n}}{\int }_{-\mathrm{a}}^{\mathrm{a}}{\mathrm{e}}^{\mathrm{i}\left(\mathrm{n}-\mathrm{m}\right)\mathrm{\pi x}/\mathrm{a}}\mathrm{dx}, \\ {\int }_{-\mathrm{a}}^{\mathrm{a}}{\mathrm{e}}^{\mathrm{i}\left(\mathrm{n}-\mathrm{m}\right)\mathrm{\pi x}/\mathrm{a}}\mathrm{dx}={\overline{)\frac{{\mathrm{e}}^{\mathrm{i}\left(\mathrm{n}-\mathrm{m}\right)\mathrm{\pi x}/\mathrm{a}}}{\mathrm{i}\left(\mathrm{n}-\mathrm{m}\right)\mathrm{\pi }/\mathrm{a}}}}_{-\mathrm{a}}^{\mathrm{a}} \\ =\frac{{\mathrm{e}}^{\mathrm{i}\left(\mathrm{n}-\mathrm{m}\right)\mathrm{\pi }}-{\mathrm{e}}^{{\mathrm{e}}^{-\mathrm{i}\left(\mathrm{n}-\mathrm{m}\right)\mathrm{\pi }}}}{\mathrm{i}\left(\mathrm{n}-\mathrm{m}\right)\mathrm{\pi }/\mathrm{a}} \\ =\frac{{\left(-1\right)}^{\mathrm{n}-\mathrm{m}}-{\left(-1\right)}^{\mathrm{n}-\mathrm{m}}}{\mathrm{i}\left(\mathrm{n}-\mathrm{m}\right)\mathrm{\pi }/\mathrm{a}} \\ =0 \end{gathered}$$

$$\begin{gathered} \mathrm{Whereas}\mathrm{for}\mathrm{n}=\mathrm{m}, \\ {\int }_{-\mathrm{a}}^{\mathrm{a}}{\mathrm{e}}^{\mathrm{i}\left(\mathrm{n}-\mathrm{m}\right)\mathrm{\pi x}/\mathrm{a}}\mathrm{dx}={\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{dx}=2\mathrm{a}. \\ \mathrm{So}\mathrm{all}\mathrm{terms}\mathrm{except}\mathrm{n}=\mathrm{m}\mathrm{are}\mathrm{zero},\mathrm{and} \\ {\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{im\pi x}/\mathrm{a}}=2{\mathrm{ac}}_{\mathrm{m}},\mathrm{so} \\ {\mathrm{c}}_{\mathrm{n}}=\frac{1}{2\mathrm{a}}{\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{im\pi x}/\mathrm{a}}\mathrm{dx}. \end{gathered}$$

04

 Step 3: (c) Showing that (a) and (b) are

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\sqrt{2\mathrm{\pi }}}\sum _{\mathrm{n}=-\infty }^{\infty }\mathrm{F}\left(\mathrm{k}\right){\mathrm{e}}^{\mathrm{ikx}}∆\mathrm{k};\mathrm{F}\left(\mathrm{k}\right)=\frac{1}{\sqrt{2\mathrm{\pi }}}{\int }_{-\mathrm{a}}^{+\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right).{\mathrm{e}}^{-\mathrm{ikx}}\mathrm{dx}. \\ \mathrm{using}\mathrm{k}=\frac{\mathrm{n\tau \tau }}{\mathrm{a}},\mathrm{and}\mathrm{F}\left(\mathrm{k}\right)=\sqrt{\frac{2}{\mathrm{\tau \tau }}}{\mathrm{ac}}_{\mathrm{n}} \\ \mathrm{We}\mathrm{can}\mathrm{write}: \\ \mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=-\infty }^{\infty }\sqrt{\frac{\mathrm{\pi }}{2}}\frac{1}{\mathrm{a}}\mathrm{F}\left(\mathrm{k}\right){\mathrm{e}}^{\mathrm{ikx}} \\ =\sqrt{\frac{1}{2\mathrm{\pi }}}\sum _{}\mathrm{F}\left(\mathrm{k}\right){\mathrm{e}}^{\mathrm{ikx}}∆\mathrm{k} \\ \mathrm{Where}∆\mathrm{k}=\frac{\mathrm{\tau \tau }}{\mathrm{a}}\mathrm{is}\mathrm{the}\mathrm{increment}\mathrm{in}\mathrm{k}\mathrm{from}\mathrm{n}\mathrm{to}\left(\mathrm{n}+1\right) \\ \mathrm{F}\left(\mathrm{k}\right)\sqrt{\frac{2}{\mathrm{\pi }}}\mathrm{a}\frac{1}{2\mathrm{a}}{\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right){\mathrm{e}}^{\mathrm{ikx}}\mathrm{dx}=\frac{1}{\sqrt{2\mathrm{\pi }}}{\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{ikx}}\mathrm{dx} \end{gathered}$$

05

(d) Obtaining Plancherel’s theorem 

As a → ∞, k becomes a continuous variable, and the sum becomes an integration, therefore,

$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\sqrt{2\mathrm{\pi }}}{\int }_{-\infty }^{\infty }\mathrm{F}\left(\mathrm{k}\right){\mathrm{e}}^{\mathrm{ikx}}\mathrm{dk};$

For F (k) the limits of the integration will change,

$\mathrm{f}\left(\mathrm{k}\right)=\frac{1}{\sqrt{2\mathrm{\pi }}}{\int }_{-\infty }^{\infty }f\left(x\right){\mathrm{e}}^{-\mathrm{ikx}}dx;$

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