Fourier Series Formula

A Fourier series is an enlargement of a periodic function f(x) with respect to an infinite sum of sines and cosines. The Fourier series considers the orthogonality links of the sine and cosine functions. The study and measure of Fourier series is referred to harmonic analysis and is tremendously useful to break up an arbitrary periodic function into a set of simple terms, which can be plugged in, solved separately, and then recombined to gain the solution to the actual problem or estimation to it to whatever appropriateness is desired or practical.

Fourier Series Formula

Following is the Fourier series formula:-

f(x) = 1/2a0 + ∑∞n = 1a_n cosnx + ∑∞n=1 b_n sinnx

Where,

a0 = 1/π∫^π_−pi f(x)dx

a_n = 1/π∫^π_−π f(x)sin nx dx

b_n = 1/π∫^π_−π f(x)sin nx dx

To know more about the Fourier series formula derivation as well as Fourier series notes, refer to Vedantu— a top educational portal.

Finding the Coefficients Using Fourier Series Coefficients Formula

How did we know to implement sin (5x)/5, sin (7x)/7, etc?

There are formulas! Formulas like Fourier coefficients formula.

Below is a series of sines and cosines having a name for all coefficients:

f(x) = a0 + ∞n=1an cos(nxπ/L) + ∞n=1 bn sin(nxπ/L)

Where:

f(x) =  the function we desire (such as a square wave)

L = half of the period of the function

a0, an and bn = coefficients that we require to calculate!

To calculate the coefficients a0, an and bn we use the given below Fourier series formula list:

a0 = 1/2L ∫L−L f(x) dx

an = 1/L ∫L−L f(x) cos(nxπ/L) dx

bn = 1L ∫L−L f(x) sin(nxπ/L) dx

Solved Examples

Example:

Identify the Fourier series for f(x)=x f(x)=x on −L≤x≤L−L≤x≤L.

Solution:

Let’s begin with the integrals for An

A0=1/2L∫L−Lf(x)dx=1/2L∫L−Lxdx=0

An=1/L∫L−Lf(x)cos(nπx/L)dx=1/L∫L−Lxcos(nπx/L)dx=0

In both cases remember that we are integrating an odd function (x is odd and cosine is even, thus the product is odd) over the interval [−L, L] and thus we know that both of these integrals will be zero (0).

Next, is the integral for Bn

Bn=1/L∫L−Lf(x)sin(nπx/L)dx=1/L∫L−Lxsin(nπx/L)dx=2/L∫L0xsin(nπx/L)dx

In this case we are integrating an even function (x and sine are both odd thus the product is even) on the interval [−L,L] and thus we can also “simplify” the integral as above. Now, Using the previous result we get,

Bn=(−1)n+12L/nπ   n=1,2,3…

In the case the Fourier series is,

f(x) = ∞∑n =0A_ncos(nπx/L) + ∞∑n=1Bnsin(nπx/L) = ∞∑n=1(−1)n+12L/nπ sin(nπx/L)