91Ó°ÊÓ

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 two functions

$$\begin{gathered} f\left(x\right)=\frac{1}{2}{a}_0+\underset{1}{\overset{∞}{∑}}{a}_n\cos nx+\underset{1}{\overset{∞}{∑}}{b}_n\sin nx \\ g\left(x\right)=\frac{1}{2}{a\text{'}}_0+\underset{1}{\overset{∞}{∑}}{a\text{'}}_n\cos nx+\underset{1}{\overset{∞}{∑}}{b\text{'}}_n\sin nx \end{gathered}$$.

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

$f\left(x\right)g\left(x\right)=\frac{1}{4}{a}_{0}a{\text{'}}_0+\frac{1}{2}\underset{1}{\overset{∞}{∑}}{a}_n{a\text{'}}_n+\underset{1}{\overset{∞}{∑}}{b}_n{b\text{'}}_n$

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$\frac{1}{4}{a}_{0}a{\text{'}}_0+\frac{1}{2}\underset{1}{\overset{∞}{∑}}{a}_n{a\text{'}}_n+\underset{1}{\overset{∞}{∑}}{b}_n{b\text{'}}_n$