Subject: Calculus

# Fourier Series Fourier series were introduced by Joseph Fourier (1768�1830) for the purpose of solving problems in physics.

So far we have already discussed Taylor series and Maclaurin series which are special types of function expansion. With Taylor series, we give an approximate value of a function by expanding it as an infinite series of powers of the form,

\sum_{n=0}^{\infty}\frac{f^{n}(c)}{n!}\left ( x-c \right )^n=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}\left (x-c \right )^2+\frac{f''(c)}{3!}\left (x-c \right )^3+...

Maclaurin series is a special case of Taylor series when c is equal to zero.
With Fourier series, we are interested in expanding a function as an infinite series of sines and cosines. Other functions can also be used instead of sines and cosines like the Legendre polynomials for more general expansions but we will be using sines and cosines for the examples. A Fourier series expansion is written as,

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

where a_0, a_n, and b_n for n =1, 2, 3,..., are called the Fourier coefficients.

## Deriving the Coefficients for 2π-periodic function

Suppose that the 2π-periodic function f has the Fourier series representation

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

The coefficients can then be solved using Euler formulas and are illustrated below.

### Computing a_0

From Equation (2), we integrate both sides over the interval [-\pi, \pi].

\int_{-\pi}^{\pi}f(x)dx=a_0 \int_{-\pi}^{\pi}dx+\sum_{n=1}^{\infty}\left [a_n \int_{-\pi}^{\pi}\cos{nx}dx+b_n \int_{-\pi}^{\pi}\sin{nx}dx\right ]

Since \int_{-\pi}^{\pi}\cos{nx}dx=\int_{-\pi}^{\pi}\sin{nx}dx=0 for n=1,2,3,..., we are then left with

\int_{-\pi}^{\pi}f(x)dx=a_0\int_{-\pi}^{\pi}dx
=2\pi a_0

Therefore,

a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx

### Computing a_n

From Equation (2), we multiply each side by \cos{mx} for a fixed integer m \geq 1 and integrate both sides over the interval [-\pi, \pi ].

\int_{-\pi}^{\pi}\cos{mx}dx+\sum_{n=1}^{\infty}\left [a_n \int_{-\pi}^{\pi}\cos{mx}\cos{nx}dx+b_n \int_{-\pi}^{\pi}\cos{mx}\sin{nx}dx\right ]

We know that

\int_{-\pi}^{\pi} \cos{mx}dx=0

and

\int_{-\pi}^{\pi}\cos{mx}\sin{nx}dx

is 0 when m \neq n, and \pi when m=n. We are then left with

\int_{-\pi}^{\pi}f(x)\cos{mx}dx = \pi a_{n}

Therefore

a_n=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)\cos{mx}dx

### Computing b_{n}

From Equation (2), we now multiply each side by \sin{mx} for a fixed integer m \geq 1 and integrate both sides over the interval [-\pi, \pi]. Following the same steps we have,

b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin{mx}dx

for m \geq 1.

NEXT TOPIC: Euler Maclaurin Formula