Fourier Series of Even and Odd Functions [With Examples]

The Fourier Series of even and odd functions have a nice description. Recall that the Fourier series expansion of a function f(x) with period 2π is given by

f(x) = a₀ + ∑_n=1[ a_ncos(nx) + b_n sin(nx) ]

where a₀, a_n, b_n are the Fourier coefficients of f(x) (provided that the sum exists). In this article, we learn how to find the Fourier series of an even and of an odd Function.

Even Function Definition

A function f(x) is called an even function if f(-x) = f(x) for all x.

For example, f(x) = x² is an even function.

Fourier Series of Even Functions

If f(x) is an even function with period 2L over the interval [-L, L], then the formula of the Fourier series of f(x) is given by

f(x) = a₀ + $\displaystyle \sum_{n=1}^\infty$ a_n cos$\big( \dfrac{n\pi x}{L}\big)$.

with Fourier coefficients

a0 = $\dfrac{1}{L} \int_0^L$ f(x) dx an = $\dfrac{2}{L} \displaystyle \int_0^L$ f(x) cos$\big( \dfrac{n\pi x}{L}\big)$ dx, for n = 1, 2, 3, … bn = 0, for n = 1, 2, 3, …

This series expansion is known as Fourier cosine series of f(x) as only cosine functions are involved.

Main Article: Fourier Series: Definition, Formula, Solved Examples

Odd Function Definition

A function f(x) is called an odd function if f(-x) = -f(x) for all x.

For example, f(x) = x is an even function.

Fourier Series of Odd Functions

If f(x) is an odd function with period 2L over the integral [-L, L], then the formula of the Fourier series of f(x) is given by

f(x) = $\displaystyle \sum_{n=1}^\infty$ b_n sin$\big( \dfrac{n\pi x}{L}\big)$.

with Fourier coefficients

a0 = 0 an = 0, for n = 1, 2, 3, … bn = $\dfrac{2}{L} \displaystyle \int_0^L$ f(x) sin$\big( \dfrac{n\pi x}{L}\big)$ dx, for n = 1, 2, 3, …

This series expansion is known as Fourier sine series as only sine functions are involved.

Note:

The above two results can only be used when the domain of the function is [-L, L] or [-π, π]. Do not use these, if the domain is [0, L] or so.

Also Study:

• Even and Odd Functions
• Orthogonal and Orthonormal Functions
• Periodic Functions
• Fourier Series of a Periodic Function

Solved Problems

Question 1: Find the Fourier series expansion of

$f(x) = \begin{cases} -1, & \text{ if } -\pi < x < 0 \\ 1, & \text{ if } 0 < x < \pi. \end{cases}$

Solution:

Observe that f(x) is an odd function.

Also, f(x+2π) = f(x) for all x ∈ [-π, π].

So f(x) is a periodic function with period 2π.

Thus, noting L=π in the above, the Fourier series expansion of f(x) is given by

f(x) = $\displaystyle \sum_{n=1}^\infty$ b_n sin$\big( \dfrac{n\pi x}{L}\big)$

where b_n‘s are given by

b_n = $\dfrac{2}{\pi} \displaystyle \int_0^\pi$ f(x) sin$\big( \dfrac{n\pi x}{\pi}\big)$ dx

= $\dfrac{2}{\pi} \displaystyle \int_0^\pi$ sin(nx) dx as we have f(x) = 1 in 0 < x < π.

= $\dfrac{2}{\pi} \Big[ \dfrac{-\cos(nx)}{n}\Big]_0^\pi$

= $\dfrac{2}{n\pi} \Big[ -\cos(n\pi) + \cos 0\Big]$

= $\begin{cases} \dfrac{4}{n\pi}, & \text{ if } n \text{ is odd} \\ 0, & \text{ if } n \text{ is even}. \end{cases}$

Thus, we can write b_n = $\dfrac{4}{(2n-1)\pi}$ where n = 1, 2, 3, ….

Thus the Fourier series expansion of the given odd function f(x) is given by

f(x) = $\displaystyle \sum_{n=1}^\infty \dfrac{4}{(2n-1)\pi}$ sin$\big( \dfrac{n\pi x}{L}\big)$.

FAQs