Fourier Series of a Periodic Function [With Examples]

Q: Q1: Define a periodic function.

Answer: If f(x+P) = f(x) for all x, then f(x) is called a periodic function with period P.

The Fourier series of a periodic function with period P is determined based on the following properties of integrals:

$\int_a^{a+P} f(x) ~ dx = \int_0^p f(x) ~dx$.

This means the area under the curve y=f(x) over any interval of length P is always the same. Using this property, one can express a function of period 2π as a Fourier series in [0, 2π].

Table of Contents

Fourier Series of Periodic Functions Formula

The formula of the Fourier series expansion of a periodic function f(x) of period 2π in the interval [0, 2π] is given as follows:

$f(x) =a_0+\sum_{n=1}^\infty [ a_n \cos(nx)$ $+b_n \sin(nx) ]$

where

$a_0= \dfrac{1}{2\pi} \int_0^{2\pi} f(x)~dx$

$a_n= \dfrac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx)~dx$

$b_n= \dfrac{1}{\pi} \int_0^{2\pi} f(x) \sin(nx)~dx$.

Solved Problems

Question: Find the Fourier series representation of the following function

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

Answer:

See that

f(x+2π) = f(x).

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

By the above formula, the Fourier series representation of the periodic function f(x) with period 2π is given by

f(x) = a₀ + $\sum_{n=1}^\infty$ [a_n cos(nx) + b_n sin(nx)]

whose Fourier coefficients are given by

a₀ = $\dfrac{1}{2\pi} \int_{-\pi}^{\pi} f(x)~dx$

= $\dfrac{1}{2\pi} \big[ \int_{-\pi}^0 -k~dx$ $+\int_{0}^\pi k~dx \big]$

= $\dfrac{1}{2\pi} \left\{ [-kx]_{-\pi}^0 + [kx]_{0}^\pi \right\}$

= $\dfrac{k}{2\pi} \left( -0-\pi+\pi-0 \right)$

= 0.

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

Also,

a_n = $\dfrac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx)~dx$

= $\dfrac{1}{\pi} \big[ \int_{-\pi}^0 -k\cos(nx)~dx$ $+\int_{0}^\pi k \cos(nx)~dx \big]$

= $\dfrac{k}{\pi} \left\{ -\Big[\dfrac{\sin(nx)}{n} \Big]_{-\pi}^0 + \Big[\dfrac{\sin(nx)}{n} \Big]_{0}^\pi \right\}$

= 0, since sin(nπ)=0.

And

b_n = $\dfrac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx)~dx$

= $\dfrac{1}{\pi} \big[ \int_{-\pi}^0 -k\sin(nx)~dx$ $+\int_{0}^\pi k \sin(nx)~dx \big]$

= $\dfrac{k}{\pi} \left\{ \Big[\dfrac{\cos(nx)}{n} \Big]_{-\pi}^0 – \Big[\dfrac{\cos(nx)}{n} \Big]_{0}^\pi \right\}$

= $\dfrac{k}{n\pi} \left\{ \cos 0 – \cos(-n\pi) -\cos(n\pi) +\cos 0\right\}$

= $\dfrac{2k}{n\pi} \left\{ 1 -\cos(n\pi) \right\}$, because cos(-θ) = cosθ.

= $\dfrac{2k}{n\pi} \left\{ 1 -(-1)^n \right\}$, since cos(nπ)=(-1)ⁿ.

Therefore, the Fourier series representation of f(x) is given by

f(x) = $\sum_{n=1}^\infty \frac{2k}{n \pi}$ [1-(-1)ⁿ] sin(nx)

= $\sum_{m=0}^\infty \frac{4k}{(2m+1) \pi}$ sin{(2m+1)x}, since 1-(-1)ⁿ = 2 if n odd and 0 otherwise.

= 4k/π [sinx + 1/3 sin3x + 1/5 sin5x + ….].

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