Fourier series example solution

Fourier series example solution

ByAbdul Wahab Junaid August 17, 2024August 17, 2024

Let’s work through an example of finding the Fourier series for a specific function. We’ll use a simple function to illustrate the process: $( f(x) = x )$ over the interval $([- \pi, \pi])$ .

Step-by-Step Solution

1. Function and Interval:

We want to find the Fourier series for $( f(x) = x )$ in the interval $([- \pi, \pi])$ .

2. Fourier Series Formula:

The Fourier series of a function $( f(x) )$ with period $( 2\pi )$ is given by:

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

where the Fourier coefficients are calculated as follows:

Average (DC Component) $( a_0 )$ : $[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx ]$
Cosine Coefficients ( a_n ): $[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx ]$
Sine Coefficients $( b_n )$ : $[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx ]$

3. Compute $( a_0 )$ :

$[a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x \, dx]$

Since ( x ) is an odd function and the interval is symmetric about the origin, the integral of an odd function over a symmetric interval is zero:

$[a_0 = \frac{1}{\pi} \left[ \frac{x^2}{2} \right]_{-\pi}^{\pi} = \frac{1}{\pi} \left( \frac{\pi^2}{2} - \frac{\pi^2}{2} \right) = 0]$

4. Compute $( a_n$ ):

$[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cos(nx) \, dx]$

To solve this integral, use integration by parts. Let ( u = x ) and $( dv = \cos(nx) \, dx )$ . Then $( du = dx )$ and $( v = \frac{\sin(nx)}{n} )$ :

$[\int x \cos(nx) \, dx = x \cdot \frac{\sin(nx)}{n} - \int \frac{\sin(nx)}{n} \, dx]$

$[= \frac{x \sin(nx)}{n} + \frac{1}{n^2} \cos(nx)]$

Evaluating from $(-\pi) to (\pi)$ :

$[\left[ \frac{x \sin(nx)}{n} \right]<em>{-\pi}^{\pi} + \frac{1}{n^2} \left[ \sin(nx) \right]</em>{-\pi}^{\pi}]$

$[= \frac{\pi \sin(n\pi)}{n} - \frac{(-\pi) \sin(-n\pi)}{n} + \frac{1}{n^2} \left[ \sin(n\pi) - \sin(-n\pi) \right]]$

Since $(\sin(n\pi) = 0)$ , this simplifies to:

$[a_n = \frac{2(-1)^{n+1}}{n}]$

5. Compute $( b_n )$ :

$[b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx]$

Again, using integration by parts. Let ( u = x ) and $( dv = \sin(nx) \, dx )$ . Then $( du = dx )$ and $( v = -\frac{\cos(nx)}{n} )$ :

$[\int x \sin(nx) \, dx = -\frac{x \cos(nx)}{n} - \int -\frac{\cos(nx)}{n} \, dx]$

$[= -\frac{x \cos(nx)}{n} + \frac{1}{n^2} \sin(nx)]$

Evaluating from $(-\pi) to (\pi)$ :

$[\left[ -\frac{x \cos(nx)}{n} \right]<em>{-\pi}^{\pi} + \frac{1}{n^2} \left[ \sin(nx) \right]</em>{-\pi}^{\pi}]$

Since $(\sin(n\pi) = 0)$ and $(\cos(n\pi) = (-1)^n)$ , this simplifies to:

$[b_n = \frac{2}{n} \left(-1\right)^{n+1}]$

6. Construct the Fourier Series:

Putting it all together, the Fourier series for $( f(x) = x )$ is:

$[f(x) = \sum_{n=1}^\infty \frac{2 (-1)^{n+1}}{n} \sin(nx)]$

This series represents the function $( f(x) = x )$ in terms of its sine components.

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