Question

Given $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\left|}\mathbf{x}\mathbf{\right|}$on $\mathbf{(}\mathbf{-}\mathbf{\pi }\mathbf{,}\mathbf{\pi }\mathbf{)}$, expand $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$in an appropriate Fourier series of period$\mathbf{2}\mathit{\pi }$.

Short answer

With given function $f\left(x\right)=\left|x\right|$on interval $(-\pi ,\pi )$, an appropriate Fourier series of period $2\text{π}$is:

$f\left(x\right)=\frac{\pi }{2}-\frac{4}{\pi }\sum _{nodd}^{\infty }\frac{\cos \left(nx\right)}{n^2}$

Step-by-step solution

Meaning of the Fourier series

A Fourier series is defined as an infinite series used in the analysis of periodic functions, in which the terms are constants multiplied by sine or cosine functions of integer multiples of the variable.

Given parameter

Given the function $f\left(x\right)=\left|x\right|$and interval$(-\pi ,\pi )$.

 Expanding the function

The function $f\left(x\right)=\left|x\right|$is even about Zero.

This means$b_n=0$.

As a result,

$\begin{array}{c}{a}_0=\frac{2}{\pi }{\int }_0^{\pi }xdx\\ =\frac{2}{\pi }\frac{1}{2}{\pi }^2\\ =\pi \end{array}$

$\begin{array}{c}{a}_n=\frac{2}{\pi }{\int }_0^{\pi }x\cos \left(nx\right)dx\\ ={\frac{2}{\pi }\left[\frac{x}{n}\sin \left(nx\right)+\frac{1}{n^2}\cos \left(nx\right)\right]|}_0^{\pi }\\ =\frac{2}{n^2\pi }({(-1)}^n-1)\\ =-\frac{1}{n{\pi }^2}\end{array}$

Note: n is odd

Then the function will be:

$f\left(x\right)=\frac{\pi }{2}-\frac{4}{\pi }\sum _{nodd}^{\infty }\frac{\cos \left(nx\right)}{n^2}$

So, the Fourier series will be $f\left(x\right)=\frac{\pi }{2}-\frac{4}{\pi }\sum _{nodd}^{\infty }\frac{\cos \left(nx\right)}{n^2}$.