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}}

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

Assume S=\sum_{n=1}^{\infty} \frac{1}{n^{4}} , we get

S=\sum_{odd} \frac{1}{n^{4}} + \sum_{even} \frac{1}{n^{4}}  = \frac{\pi^{4}}{96} + \frac{1}{16} S

Therefore, S=\frac{\pi^{4}}{90} .

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