 # Fourier Series Formula

Many of the phenomena studied in the domain of Engineering and Science are periodic in nature. For example current and voltage existing in an alternating current circuit. We can analyze these periodic functions into their constituent components by using a process called Fourier analysis. In this article, we will discuss the Fourier series and Fourier Series Formula. Let us begin learning!

## Fourier Series Formula

### What is the Fourier Series?

Periodic functions occur frequently in the problems studied during engineering problem-solving. Their representation in terms of simple periodic functions like sine and cosine functions, which are leading towards the Fourier series. Fourier series is a very powerful and versatile tool in connection with the partial differential equations.

A Fourier series is nothing but the expansion of a periodic function f(x) with the terms of an infinite sum of sins and cosine values. Fourier series is making use of the orthogonal relationships of the sine and cosine functions. A difficult thing to understand here is to motivate the fact that arbitrary periodic functions have Fourier series representations.

### Fourier Analysis for Periodic Functions

The Fourier series representation of analytic functions has been derived from the Laurent expansions. We use the elementary complex analysis to derive additional fundamental results in the harmonic analysis, which includes representation of the periodic functions by the Fourier series.

The representation of rapidly decreasing functions by Fourier integrals, and Shannon’s sampling theorem. The ideas are classical and of transcendent beauty.

The trigonometric functions sin x and cos x are examples of periodic functions with fundamental period 2π and tan x is periodic with fundamental period \pi.

Therefore a Fourier series is a method to represent a periodic function as a sum of sine and cosine functions possibly till infinity. It is analogous to the famous Taylor series, which represents functions as possibly infinite sums of monomial terms. ### Fourier Series Formula

$$\large f(x)=\frac{1}{2}a_{0}+\sum_{n=1}^{\infty}a_{n}cos\;nx+\sum_{n=1}^{\infty}b_{n}sin\;nx$$

Where,

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

$$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi}f(x)sin\;nx\;dx$$

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

For the functions that are not periodic, the Fourier series is replaced by the Fourier transform. For the functions of two variables that are periodic in both variables, the trigonometric basis in the Fourier series is replaced by the spherical harmonics.

## Solved Examples

Q.1: Expand the function f(x) = e^{kx} in the interval [ – \pi , \pi ] using Fourier series?

Solution: Applying the formula:

$$\large f(x)=\frac{1}{2}a_{0}+\sum_{n=1}^{\infty}a_{n}cos\;nx+\sum_{n=1}^{\infty}b_{n}sin\;nx$$

Here,

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

$$=\frac{1}{\pi}\int_{-\pi}^{\pi} e^{kx}dx$$

$$a_0 = \frac{2}{k \pi} sinh (k\pi)$$

Also,

$$a_n = \frac{1}{\pi}\int_{- \pi}^{\pi}e^{kx}cos nx dx$$

$$a_n= \frac{e^{(kx)}}{\pi(k^2+n^2))} [(k cos(nx) + n sin(nx))]_{-\pi}^{\pi}$$

$$a_n = \frac{k cos(n \pi)}{\pi (k^2 + n^2)} [e^{k \pi} – e^{-k \pi}]$$

$$a_n = 2k(-1)^n \frac{sinh(k \pi)}{\pi (k^2 + n^2)}$$

Also,

$$b_n = \frac{1}{\pi}\int_{- \pi}^{\pi}e^{kx}sin nx dx$$

$$b_n = \frac{e^{(kx)}}{\pi(k^2 + n^2)} [k sin(nx) – n cos(nx) ]_{-\pi}^{\pi}$$

$$= \frac{-n cos(n \pi)}{\pi(k^2 + n^2)} \frac{-n cos(n \pi)}{\pi(k^2 + n^2)}|$$

$$b_n = \frac{-2n(-1)^n sinh(k \pi)}{\pi(k^2 + n^2)}$$

Putting all above values in the equation, we get the expansion of the above function:

$$f(x) = e^{kx}= \frac{2sinh(k \pi)}{\pi} \frac{1}{2k}- k[\frac{cos x}{k^{2} + 1^{2}}- \frac{cos 2x}{k^{2} + 2^{2}}+ \frac{cos 3x}{k^{2} + 3^{2}}-…]+ [\frac{sin x}{k^{2} + 1^{2}} – \frac{2\;sin\;2x}{k^{2} + 2^{2}} + \frac{3\; sin\; 3x}{k^{2} + 3^{2}}-…]$$

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##### Maths Formulas 4 Followers

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KUCKOO B

I get a different answer for first example.
I got Q1 as 20.5
median 23 and
Q3 26 Guest
Yashitha

Hi
Same Guest
virat

yes