Answer to Question #282523 in Real Analysis for Khaja

Question

1. Find the fourier cosine Series for the function f(x)=x^2 for 0<x<π
2. Obtain a fourier expression for f(x)=x^2 for -π<x<π where f(x) is a even function

Answer 1

1.

$a_0=\dfrac{1}{\pi}\displaystyle\int_{0}^{\pi}x^2dx=\dfrac{1}{\pi}[\dfrac{x^3}{3}]\begin{matrix} \pi \\ 0 \end{matrix}=\dfrac{\pi^2}{3}$

$a_n=\dfrac{2}{\pi}\displaystyle\int_{0}^{\pi}x^2\cos(nx)dx$

$\int x^2\cos(nx)dx=\dfrac{1}{n}x^2\sin(nx)-\dfrac{2}{n}\int x\sin(nx)dx$

$=\dfrac{1}{n}x^2\sin(nx)+\dfrac{2}{n^2}x\cos(nx)-\dfrac{2}{n^2}\int \cos(nx)dx$

$=\dfrac{1}{n}x^2\sin(nx)+\dfrac{2}{n^2}x\cos(nx)-\dfrac{2}{n^3}\sin(nx)+C$

$a_n=\dfrac{2}{\pi}[\dfrac{x^2}{n}\sin(nx)+\dfrac{2x}{n^2}\cos(nx)-\dfrac{2}{n^3}\sin(nx)]\begin{matrix} \pi \\ 0 \end{matrix}$

$=\dfrac{4}{n^2}\cos(n\pi)=\dfrac{4(-1)^n}{n^2}$

Therefore, we have

$f(x)\sim \dfrac{\pi^2}{3}-4\cos(x)+\cos(2x)-\dfrac{4}{9}\cos(3x)+...$

2.

$a_0=\dfrac{1}{2\pi}\displaystyle\int_{0}^{2\pi}x^2dx=\dfrac{1}{2\pi}[\dfrac{x^3}{3}]\begin{matrix} 2\pi \\ 0 \end{matrix}=\dfrac{4\pi^2}{3}$

$a_n=\dfrac{1}{\pi}\displaystyle\int_{0}^{2\pi}x^2\cos(nx)dx$

$\int x^2\cos(nx)dx=\dfrac{1}{n}x^2\sin(nx)-\dfrac{2}{n}\int x\sin(nx)dx$

$=\dfrac{1}{n}x^2\sin(nx)+\dfrac{2}{n^2}x\cos(nx)-\dfrac{2}{n^2}\int \cos(nx)dx$

$=\dfrac{1}{n}x^2\sin(nx)+\dfrac{2}{n^2}x\cos(nx)-\dfrac{2}{n^3}\sin(nx)+C$

$a_n=\dfrac{1}{\pi}[\dfrac{x^2}{n}\sin(nx)+\dfrac{2x}{n^2}\cos(nx)-\dfrac{2}{n^3}\sin(nx)]\begin{matrix} 2\pi \\ 0 \end{matrix}$

$=\dfrac{4}{n^2}$

Therefore, we have

$f(x)\sim \dfrac{4\pi^2}{3}+4\cos(x)+\cos(2x)+\dfrac{4}{9}\cos(3x)+...$