MA 1505 Tutorial 5: Fourier Series

In this tutorial, we will learn how to calculate the Fourier series of periodic functions.

Assume $f(x)$ is a periodic function with period $2\pi$ , i.e. $f(x)=f(x+2\pi)$ for all $x \in \mathbb{R}$ . The Fourier Series of $f(x)$ is defined as $a_{0}+\sum_{n=1}^{\infty} (a_{n} \cos(nx) +b_{n} \sin(nx)),$ where

$a_{0}= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx,$

$a_{n}= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx$ for all $n\geq 1,$

$b_{n}= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx$ for all $n\geq 1,$

Theorem 1. If $f(x)$ satisfies Lipchitz condition on $(-\pi, \pi)$ , then

$f(x) =a_{0}+\sum_{n=1}^{\infty} (a_{n} \cos(nx) +b_{n} \sin(nx)).$

Theorem 2. Parseval’s Identity.

$\frac{1}{\pi} \int_{-\pi}^{\pi} |f(x)|^{2} dx= 2a_{0}^{2}+ \sum_{n=1}^{\infty} (a_{n}^{2}+b_{n}^{2}).$

Question 1. Assume $f(x)=f(x+2\pi)$ for all $x\in \mathbb{R}$ and $f(x)=1505+1506x+1507x^{2}+1508x^{3}$ on $[-\pi, \pi).$

What is the value of $a_{0}+\sum_{n=1}^{\infty}a_{n} ?$

Solution. From Theorem 1, $f(x)=a_{0}+\sum_{n=1}^{\infty} (a_{n} \cos(nx) +b_{n} \sin(nx))$ on $(-\pi, \pi)$ . Therefore, $f(0)=a_{0}+\sum_{n=1}^{\infty}a_{n}$ and $f(0)=1505$ . Hence, $a_{0}+\sum_{n=1}^{\infty}a_{n}=1505.$

Question 2. Prove these identities:

$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{2}}=\frac{\pi^{2}}{8}$

$\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$

$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{4}}=\frac{\pi^{4}}{96}$

$\sum_{n=1}^{\infty} \frac{1}{n^{4}}=\frac{\pi^{4}}{90}$

Solution.

Choose the function $f(x)=|x|$ on $(-\pi, \pi)$ and f(x) is a periodic function with period $2\pi$ .

Use the formulas of $a_{n}$ and $b_{n}$ , we can prove that the Fourier series of $f(x)=|x|$ is

$\frac{\pi}{2} + \sum_{n=1}^{\infty} \frac{2((-1)^{n}-1)}{\pi} \cdot \frac{cos(nx)}{n^{2}}$

From Theorem 1, take $x=0$ , then

$0= \frac{\pi}{2} + \sum_{n=1}^{\infty} \frac{2((-1)^{n}-1)}{n^{2} \pi} = \frac{\pi}{2} + \sum_{m=1}^{\infty} \frac{-4}{(2m-1)^{2}\pi} = \frac{\pi}{2} - \frac{4}{\pi} \sum_{m=1}^{\infty} \frac{1}{(2m-1)^{2}}$

Therefore, $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{2}}=\frac{\pi^{2}}{8}$ .

Assume $S=\sum_{n=1}^{\infty} \frac{1}{n^{2}}$ , we get

$S=\sum_{odd} \frac{1}{n^{2}} + \sum_{even} \frac{1}{n^{2}} = \frac{\pi^{2}}{8} + \frac{1}{4} S$ .

Therefore $S=\frac{\pi^{2}}{6}$ .

From Parserval’s identity, we know

$\frac{2\pi^{2}}{3}= \frac{1}{\pi} \int_{-\pi}^{\pi} x^{2}dx = 2\cdot (\frac{\pi}{2})^{2} + \sum_{n=1}^{\infty} \frac{4((-1)^{n}-1)^{2}}{\pi^{2}\cdot n^{4}} = \frac{\pi^{2}}{2} + \sum_{m=1}^{\infty} \frac{16}{\pi^{2} (2m-1)^{4}}$