A Fourier series is an expansion of a periodic function

in terms of an infinite sum of sines and cosines. It decomposes any periodic function or periodic signal into the sum of a set of simple oscillating functions, namely sines and cosines.

$$f(x)=\frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}{a}_{n}cosnx+\sum _{n=1}^{\mathrm{\infty }}{b}_{n}sinnx$$

Where
a_o =

a_n = 

b_n = 

n = 1, 2, 3…..

Solved Example

Question: Expand the function f(x) = e^kx in the interval [ –

, ] using fourier series.

Solution:

Let the Fourier series for f(x) be

f(x) =

a_o + (a_ncos(nx) + b_nsin(nx))

Here

a₀ =

f(x) dx

=

e^kxdx

=

[e^kx

=

(e^kπ – e^-kπ )

=

sinh( k )

Now,

a_n =

e^kx cos (nx) dx

=

=

=

 

= 2k (-1)ⁿ

and

b_n =

e^kx sin(nx) dx

=

$\begin{array}{l}[ksin(nx)–ncos(nx){]}_{-\pi }^{\pi }\end{array}$

=

=

=

Substituting these values of a_o ,a_n ,b_n we get

f(x) = e^kx =

sinh(k ) +

$\begin{array}{l}\frac{2(-1{)}^{n}sinh(k\pi )}{\pi (k^2+n^2)}\end{array}$

 (acos(nx) – n sin(nx))

$\begin{array}{l}\frac{2sinh(k\pi )}{\pi }\end{array}$

$\begin{array}{l}\frac{1}{2k}\end{array}$

+

$\begin{array}{l}\frac{(-1{)}^n}{k^2+n^2}\end{array}$

(k cos(nx) – n sin(nx))]

or

e^kx =

– k [

$\begin{array}{l}\frac{cosx}{k^2+1^2}\end{array}$

–

$\begin{array}{l}\frac{cos2x}{k^2+2^2}\end{array}$

+

$\begin{array}{l}\frac{cos3x}{k^2+3^2}\end{array}$

– ….] + [

$\begin{array}{l}\frac{sinx}{k^2+1^2}\end{array}$

–

$\begin{array}{l}\frac{2sin2x}{k^2+2^2}\end{array}$

+

$\begin{array}{l}\frac{3sin3x}{k^2+3^2}\end{array}$

-…..) ]