Fourier Series : How to determine Fourier Coefficients ?

Fourier Series :

A series of sines & cosines of an angle and its multiples of the form :

$ \frac{a_0}{2} + a_1 cosx + a_2 cos2x + a_3 cos3x + …+ a_n cosnx + …+ b_1 sinx + b_2 sin2x + ..+ b_n sin nx + ..$ is called Fourier Series .

How to determine Fourier Coefficients ?

$ f(x) =\frac{a_0}{2} + a_1 cosx + a_2 cos2x + a_3 cos3x + …+ a_n cosnx + …+ b_1 sinx + b_2 sin2x + ..+ b_n sin nx + ..$

(i) To find a0 : Integrate f(x) from x = 0 to x = 2 π

(ii) To find an : Multiply f(x) by cos(nx) & Integrate from x = 0 to x = 2 π

(ii) To find bn : Multiply f(x) by sin(nx) & Integrate from x = 0 to x = 2 π

Example : Given that f(x) = x + x2 for -π < x < π , find the Fourier Expression of f(x) .

Solution : Suppose , $ x + x^2 = \frac{a_0}{2} + a_1 cosx + a_2 cos2x + a_3 cos3x + …+ b_1 sinx + b_2 sin2x + ..$ …(i)

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

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

$\displaystyle a_0 = \frac{1}{\pi} [\frac{x^2}{2} + \frac{x^3}{3}]_{-\pi}^{\pi} $

$\displaystyle a_0 = \frac{2 \pi^2}{3} $

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

$\displaystyle a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (x + x^2) cosnx \; dx $

$\displaystyle a_n = \frac{1}{\pi} [4 \pi \frac{cos n\pi}{n^2}] $

$\displaystyle a_n = \frac{4 (-1)^n}{n^2} $

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

$\displaystyle b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (x + x^2) sin nx \; dx $

$\displaystyle b_n = \frac{1}{\pi} [-\frac{2 \pi}{n} cos n\pi] $

$\displaystyle b_n = -\frac{2}{n} (-1)^n $

Now , on putting the value of a0 , an & bn

$ x + x^2 = \frac{\pi^2}{3} + 4 [- cosx + \frac{1}{2^2} cos2x – \frac{1}{3^2} cos3x + …] – 2[- sinx + \frac{1}{2} sin2x – \frac{1}{3} sin3x + ..] $

Even Function

A function f(x) is said to be Even if f(-x) = f(x)

$\displaystyle \int_{-\pi}^{\pi} f(x) dx = 2 \int_{0}^{\pi} f(x) dx$

Expansion of a Even function :

$\displaystyle a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx = \frac{2}{\pi} \int_{0}^{\pi} f(x) dx $

$\displaystyle a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) cos nx dx = \frac{2}{\pi} \int_{0}^{\pi} f(x) cos nx dx $

$\displaystyle b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) sin nx \; dx = 0 $

Odd Function

A function f(x) is said to be Odd if f(-x) = – f(x)

$\displaystyle \int_{-\pi}^{\pi} f(x) dx = 0 $

Expansion of an Odd function :

$\displaystyle a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \; dx = 0 $

$\displaystyle a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) cos nx \; dx = 0 $

$\displaystyle b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) sin nx \; dx = \frac{2}{\pi} \int_{0}^{\pi} f(x) sin nx \; dx $

Example: Obtain a Fourier expression for f(x) = x3 for −π < x < π

Solution : As f(x) = x3 is an Odd function

$\displaystyle a_0 = 0 \; a_n = 0 $

$\displaystyle b_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) sin nx \; dx $

$\displaystyle b_n = \frac{2}{\pi} \int_{0}^{\pi} x^3 sin nx \; dx $

$\displaystyle b_n = \frac{2}{\pi} [x^3 (\frac{cos nx}{n}) – 3 x^2 (- \frac{sin nx}{n^2}) + 6 x (\frac{cos nx}{n^3}) – 6 (\frac{sin nx}{n^4})]_{0}^{\pi} $

$\displaystyle b_n = \frac{2}{\pi}[\frac{- \pi^3 cos n \pi}{n} + \frac{6 \pi cos n \pi}{n^3}] $

$\displaystyle b_n = 2 (-1)^n [- \frac{\pi^2}{n} + \frac{6}{n^3}] $

$\displaystyle x^3 = 2 [- ( – \frac{\pi^2}{1} + \frac{6}{1^3})sin x + ( -\frac{\pi^2}{2} + \frac{6}{2^3}) sin 2x – ( – \frac{\pi^2}{3} + \frac{6}{3^3})sin 3x ….] $

Example : Find the Fourier series expansion of the periodic function of period 2π ; f(x) = x2 , −π ≤ x ≤ π

Solution : As f(x) = x is an even function

$\displaystyle a_0 = \frac{2}{\pi} \int_{0}^{\pi} f(x) dx $

$\displaystyle a_0 = \frac{2}{\pi} \int_{0}^{\pi} x^2 dx $

$\displaystyle a_0 = \frac{2}{\pi} [\frac{x^3}{3}]_{0}{\pi} = \frac{2 \pi^2}{3}$

$\displaystyle a_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) cos nx dx $

$\displaystyle a_n = \frac{2}{\pi} \int_{0}^{\pi} x^2 cos nx dx $

$\displaystyle a_n = \frac{2}{\pi} [x^2 (\frac{sin nx}{n}) – 2 x (- \frac{cos nx}{n^2}) + 2 (- \frac{sin nx}{n^3})]_{0}^{\pi} $

$\displaystyle a_n = \frac{2}{\pi}[\frac{\pi^2 sin n\pi}{n} + \frac{2 \pi cos n \pi}{n^2} – \frac{2 sin n\pi}{n^3}] $

$\displaystyle a_n = \frac{4 (-1)^n}{n^2} $

and , $\displaystyle b_n = 0 $

The fourier series will be ,

$ f(x) = \frac{a_0}{2} + a_1 cosx + a_2 cos2x + a_3 cos3x + …+ a_n cos nx + ..$ ; (since bn = 0)

$ x^2 = \frac{\pi^2}{3} – 4 [\frac{cos x}{1^2} – \frac{cos 2x}{2^2} + \frac{cos 3x}{3^2} – \frac{cos 4x}{4^2} + …]$

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