Question

A general form of Parseval’s theorem says that if two functions are expanded in Fourier series

$$\begin{gathered} \mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{a}}_{\mathbf{0}}\mathbf{+}\mathbf{\sum }_{\mathbf{1}}^{\mathbf{\infty }}{\mathbf{a}}_{\mathbf{n}}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{n}\mathbf{x}\mathbf{+}\mathbf{\sum }_{\mathbf{1}}^{\mathbf{\infty }}{\mathbf{b}}_{\mathbf{n}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{n}\mathbf{x} \\ \mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{a}\mathbf{\text{'}}}_{\mathbf{0}}\mathbf{+}\mathbf{\sum }_{\mathbf{1}}^{\mathbf{\infty }}{\mathbf{a}\mathbf{\text{'}}}_{\mathbf{n}}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{n}\mathbf{x}\mathbf{+}\mathbf{\sum }_{\mathbf{1}}^{\mathbf{\infty }}{\mathbf{b}\mathbf{\text{'}}}_{\mathbf{n}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{n}\mathbf{x} \end{gathered}$$

then the average value of$\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}{\mathbf{a}}_{\mathbf{0}}\mathbf{a}{\mathbf{\text{'}}}_{\mathbf{0}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\sum }_{\mathbf{1}}^{\mathbf{\infty }}{\mathbf{a}}_{\mathbf{n}}{\mathbf{a}\mathbf{\text{'}}}_{\mathbf{n}}\mathbf{+}\mathbf{\sum }_{\mathbf{1}}^{\mathbf{\infty }}{\mathbf{b}}_{\mathbf{n}}{\mathbf{b}\mathbf{\text{'}}}_{\mathbf{n}}$.Prove this.

Short answer

The average value of$f\left(x\right)g\left(x\right)$ is proved to be$\frac{1}{4}{a}_{0}a{\text{'}}_0+\frac{1}{2}\sum _1^{\infty }{a}_n{a\text{'}}_n+\sum _1^{\infty }{b}_n{b\text{'}}_n$

Step-by-step solution

Definition of Parseval’s Theorem

In mathematics, Parseval's theorem generally refers to the result that the Fourier transfigure is unitary; approximately, that the sum of the forecourt of a function is equal to the sum of the forecourt of it is transfigure.

Given Parameters

Given two functions

$$\begin{gathered} f\left(x\right)=\frac{1}{2}{a}_0+\sum _1^{\infty }{a}_n\cos nx+\sum _1^{\infty }{b}_n\sin nx \\ g\left(x\right)=\frac{1}{2}{a\text{'}}_0+\sum _1^{\infty }{a\text{'}}_n\cos nx+\sum _1^{\infty }{b\text{'}}_n\sin nx \end{gathered}$$.

It is to be proven that average if these two functions is

$f\left(x\right)g\left(x\right)=\frac{1}{4}{a}_{0}a{\text{'}}_0+\frac{1}{2}\sum _1^{\infty }{a}_n{a\text{'}}_n+\sum _1^{\infty }{b}_n{b\text{'}}_n$

Use Parseval theorem to prove the required

Calculate the mean value of the product of the two functions as

Further solve and observe that here only sum of square of sine and cosine terms will survive and integrate of 1/2.

Further solve and get mean value of two functions as

Therefore, the average value of$f\left(x\right)g\left(x\right)$ is proved to be$\frac{1}{4}{a}_{0}a{\text{'}}_0+\frac{1}{2}\sum _1^{\infty }{a}_n{a\text{'}}_n+\sum _1^{\infty }{b}_n{b\text{'}}_n$