ONE MORE STEP TOWARD BETTER TOMORROW : Fourier Sine and Cosine Series

Monday, 25 March 2013

Fourier Sine and Cosine Series

Recall that the Fourier series of f(x) is defined by

$\begin{displaymath}a_0 + \sum \Big(a_n\cos(nx) + b_n\sin(nx)\Big)\end{displaymath}$

where

$\begin{displaymath}\left\{\begin{array}{lclr} a_0 &=& \displaystyle \frac{1}{2\p... ..._{-\pi}^{\pi} f(x) \sin(nx)dx,& 1 \leq n.\\ \end{array}\right.\end{displaymath}$

We have the following result:

Theorem. Let f(x) be a function defined and integrable on interval $[-\pi, \pi]$ .

(1): If f(x) is even, then we have

$\begin{displaymath}b_n = 0 ,\;\;\; \mbox{for all } n \geq 1;\end{displaymath}$

and

$\begin{displaymath}a_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \cos(nx)dx.\end{displaymath}$
(2): If f(x) is odd, then we have

$\begin{displaymath}a_n = 0 ,\;\;\; \mbox{for all } n \geq 0;\end{displaymath}$

and

$\begin{displaymath}b_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin(nx)dx.\end{displaymath}$

This Theorem helps define the Fourier series for functions defined only on the interval $[0,\pi]$ . The main idea is to extend these functions to the interval $[-\pi, \pi]$ and then use the Fourier series definition.

Let f(x) be a function defined and integrable on $[0,\pi]$ . Set

$\begin{displaymath}f_1(x) = \left\{ \begin{array}{lrl} -f(-x),& -\pi \leq x < 0\\ f(x), & 0 \leq x \leq \pi \end{array} \right.\end{displaymath}$

and

$\begin{displaymath}f_2(x) = \left\{ \begin{array}{lrl} f(-x),& -\pi \leq x < 0\\ f(x), & 0 \leq x \leq \pi \end{array} \right.\end{displaymath}$

Then f₁ is odd and f₂ is even. It is easy to check that these two functions are defined and integrable on $[-\pi, \pi]$ and are equal to f(x) on $[0,\pi]$ . The function f₁ is called the odd extension of f(x),
while f₂ is called its even extension.

Definition. Let f(x), f₁(x), and f₂(x) be as defined above.

(1): The Fourier series of f₁(x) is called the Fourier Sine series of the function f(x), and is given by

$\begin{displaymath}f(x) \sim \sum_{n=1}^{\infty} b_n\sin(nx).\end{displaymath}$

where

$\begin{displaymath}b_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin(nx)dx.\end{displaymath}$
(2): The Fourier series of f₂(x) is called the Fourier Cosine series of the function f(x), and is given by

$\begin{displaymath}f(x) \sim a_0 + \sum_{n=1}^{\infty} a_n\cos(nx).\end{displaymath}$

where

$\begin{displaymath}a_0 = \frac{1}{\pi} \int_{0}^{\pi} f(x) \cos(nx)dx,\;\;\mbox{... ...}{\pi} \int_{0}^{\pi} f(x) \cos(nx)dx\;\;\mbox{for $n \geq 1$}.\end{displaymath}$

Example. Find the Fourier Cosine series of f(x) = x for $x \in [0,\pi]$ .
Answer. We have

$\begin{displaymath}a_0 = \frac{1}{\pi} \int_{0}^{\pi} xdx = \frac{\pi}{2},\end{displaymath}$

and

$\begin{displaymath}a_n = \frac{2}{\pi} \int_{0}^{\pi} x\cos(nx)dx = \frac{2}{n^2\pi} (\cos(n\pi) - 1) = \frac{2}{n^2\pi} ((-1)^n - 1).\end{displaymath}$

Therefore, we have

$\begin{displaymath}f(x) \sim \frac{\pi}{2} - \frac{4}{\pi}\left(\cos(x) + \frac{1}{9}\cos(3x) + \frac{1}{25}\cos(5x)+\ldots\right).\end{displaymath}$

Example. Find the Fourier Sine series of the function f(x) = 1 for $x \in [0,\pi]$ .
Answer. We have

$\begin{displaymath}b_n = \frac{2}{\pi} \int_{0}^{\pi} \sin(nx)dx = \frac{2}{n\pi}(-\cos(n\pi) + 1) = \frac{2}{n\pi}(1-(-1)^n).\end{displaymath}$

Hence

$\begin{displaymath}f(x) \sim \frac{4}{\pi}\left(\sin(x) + \frac{1}{3}\sin(3x) + \frac{1}{5}\sin(5x)...\right).\end{displaymath}$

Example. Find the Fourier Sine series of the function $f(x) = \cos(x)$ for $x \in [0,\pi]$ .
Answer. We have

$\begin{displaymath}b_n = \frac{2}{\pi} \int_{0}^{\pi}\cos(x)\sin(nx)dx\end{displaymath}$

which gives b₁ = 0 and for n > 1, we obtain

$\begin{displaymath}b_n = \frac{2n}{\pi}\left(\frac{1 + (-1)^n}{n^2 -1}\right).\end{displaymath}$

Hence

$\begin{displaymath}\cos(x) \sim \frac{8}{\pi} \sum_{n=1}^{\infty} \frac{n\sin(2nx)}{4n^2-1}.\end{displaymath}$

Special Case of 2L-periodic functions.
As we did for $2\pi$ -periodic functions, we can define the Fourier Sine and Cosine series for functions defined on the interval [-L,L]. First, recall the Fourier series of f(x)

$\begin{displaymath}f(t) \sim a_0 + \sum_{n=1}^{\infty} \left(a_n\cos\left(n\frac{\pi t} {L}\right) + b_n\sin\left(n\frac{\pi t}{L}\right)\right)\end{displaymath}$

where

$\begin{displaymath}\left\{\begin{array}{lclr} a_0 &=& \displaystyle \displaystyl... ...f(x) \sin\left(n\frac{\pi x}{L}\right)dx,\\ \end{array}\right.\end{displaymath}$

for $1\leq n$ .

1.: If f(x) is even, then b_n = 0, for $n \geq 1$ . Moreover, we have

$\begin{displaymath}a_0 = \frac{1}{L} \int_{0}^{L} f(x) \cos\left(n\frac{\pi x}{L}\right)dx,\end{displaymath}$

and

$\begin{displaymath}a_n = \frac{2}{L} \int_{0}^{L} f(x) \cos\left(n\frac{\pi x}{L}\right)dx.\end{displaymath}$

Finally, we have

$\begin{displaymath}f(x) \sim a_0 + \sum_{n=1}^{\infty}a_n\cos\left(n\frac{\pi x}{L} \right).\end{displaymath}$
2.: If f(x) is odd, then a_n = 0, for all $n \geq 0$ , and

$\begin{displaymath}b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(n\frac{\pi x}{L}\right)dx.\end{displaymath}$

Finally, we have

$\begin{displaymath}f(x) \sim \sum_{n=1}^{\infty} b_n\sin\left(n\frac{\pi x}{L}\right).\end{displaymath}$