Sine and Cosine Series

Theory

Fourier series can be simplified if the function exhibits symmetry.

Odd Functions: $f(-x) = -f(x)$ .
The Fourier series of an odd function is a Sine Series, where $a_0 = a_n = 0$ .
Even Functions: $f(-x) = f(x)$ .
The Fourier series of an even function is a Cosine Series, where $b_n = 0$ .

Exercise 1

Determine if the following are even or odd:

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Solution

$f(-x) = (-x)^2 = x^2 = f(x)$ , so $f(x)$ is even.
$f(-x) = (-x)^3 = -x^3 = -f(x)$ , so $f(x)$ is odd.
$f(-x) = e^{-x} \neq f(x)$ and $\neq -f(x)$ , so $f(x)$ is neither even nor odd.

Exercise 2

Write the Fourier series of $f(x) = |x|$ , $x \in (-\pi, \pi)$ .

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Solution

The function $f(x)=|x|$ is even.
Therefore, its Fourier series is a cosine series:

f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos(nx).

a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}|x|dx =\frac{2}{\pi}\int_{0}^{\pi}x\,dx=\pi.

a_n=\frac{2}{\pi}\int_{0}^{\pi}x\cos(nx)\,dx =\frac{2}{\pi}\frac{(-1)^n-1}{n^2}.

Thus,

|x|=\frac{\pi}{2} -\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\cos((2n-1)x)}{(2n-1)^2}, \quad -\pi&lt;x&lt;\pi.

Exercise 3

Write the Fourier series of

f(x)= \begin{cases} x+1, & -1&lt;x&lt;0,\\ x-1, & 0&lt;x&lt;1. \end{cases}

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Solution

Check symmetry:

f(-x)= \begin{cases} -x+1, & 0&lt;x&lt;1,\\ -x-1, & -1&lt;x&lt;0, \end{cases} = -f(x).

Thus, $f(x)$ is an odd function, and its Fourier series is a sine series.

f(x)=\sum_{n=1}^{\infty}b_n\sin(n\pi x), \quad (-1&lt;x&lt;1).

b_n=\int_{-1}^{1}f(x)\sin(n\pi x)\,dx =2\int_{0}^{1}(x-1)\sin(n\pi x)\,dx.

Evaluating,

b_n=\frac{2(-1)^n}{n\pi}.

Hence,

f(x)=\sum_{n=1}^{\infty}\frac{2(-1)^n}{n\pi}\sin(n\pi x).

Exercise 4

Write the Fourier series of $f(x)=\sin(2x)$ , $-\pi<x<\pi$ .

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Solution

The function $f(x)=\sin(2x)$ is odd.
Therefore, its Fourier series contains only sine terms.

Since $\sin(2x)$ is already a sine function,

b_2=1,\quad b_n=0 \text{ for } n\neq2.

Thus, the Fourier series is simply

f(x)=\sin(2x).