ONE MORE STEP TOWARD BETTER TOMORROW : Fourier Series: Basic Results

Monday, 25 March 2013

Fourier Series: Basic Results

Recall that the mathematical expression

$\begin{displaymath}A_0 + \sum_{n = 1}^{\infty} (A_n\cos(nx) + B_n\sin(nx)).\end{displaymath}$

is called a Fourier series.
Since this expression deals with convergence, we start by defining a similar expression when the sum is finite.
Definition. A Fourier polynomial is an expression of the form

$\begin{displaymath}F_n(x) = a_0 + \Big(a_1\cos(x) +b_1\sin(x)\Big)+ \cdots + \Big(a_n\cos(nx) + b_n\sin(nx)\Big)\end{displaymath}$

which may rewritten as

$\begin{displaymath}F_n(x) = a_0 + \sum_{k= 1}^{k=n} \Big(a_k\cos(kx) + b_k\sin(kx)\Big).\end{displaymath}$

The constants a₀, a_i and b_i, $i=1,\cdots,n$ , are called the coefficients of F_n(x).

The Fourier polynomials are $2\pi$ -periodic functions. Using the trigonometric identities

$\begin{displaymath}\begin{array}{lcr} \sin(mx)\cos(nx) &=&\displaystyle \frac{1}... ...e \frac{1}{2}\Big[\cos((m-n)x) - \cos((m+n)x) \Big] \end{array}\end{displaymath}$

we can easily prove the integral formulas

(1): for $n \geq 0$ , we have

$\begin{displaymath}\int_{-\pi}^{\pi} \cos(nx)dx = 0,\;\;\;\mbox{and}\;\; \int_{-\pi}^{\pi}\sin(nx) dx = 0,\end{displaymath}$

for n>0 we have

$\begin{displaymath}\int_{-\pi}^{\pi} \cos(nx)dx = 0,\;\;\;\mbox{and}\;\; \int_{-\pi}^{\pi}\sin(nx) dx = 0,\end{displaymath}$
(2): for m and n, we have

$\begin{displaymath}\int_{-\pi}^{\pi} \sin(mx) \cos(nx)dx = 0,\end{displaymath}$
(3): for $n \neq m$ , we have

$\begin{displaymath}\int_{-\pi}^{\pi} \cos(mx)\cos(nx)dx = 0,\;\;\mbox{and}\;\; \int_{-\pi}^{\pi}\sin(mx) \sin(nx)dx=0,\end{displaymath}$
(4): for $n \geq 1$ , we have

$\begin{displaymath}\int_{-\pi}^{\pi}\cos^2(nx)dx = \pi,\;\;\mbox{and}\;\; \int_{-\pi}^{\pi} \sin^2(nx)dx = \pi.\end{displaymath}$

Using the above formulas, we can easily deduce the following result:

Theorem. Let

$\begin{displaymath}F_n(x) = a_0 + \sum_{k= 1}^{k=n} \Big(a_k\cos(kx) + b_k\sin(kx)\Big).\end{displaymath}$

We have

$\begin{displaymath}\left\{\begin{array}{lclr} a_0 &=&\displaystyle \frac{1}{2\pi... ...\pi} F_n(x) \sin(kx)dx,& 1 \leq k \leq n.\\ \end{array}\right.\end{displaymath}$

This theorem helps associate a Fourier series to any $2\pi$ -periodic function.

Definition. Let f(x) be a $2\pi$ -periodic function which is integrable on $[-\pi, \pi]$ . Set

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

The trigonometric series

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

is called the Fourier series associated to the function f(x). We will use the notation

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

Example. Find the Fourier series of the function

$\begin{displaymath}f(x) = x, \;\;\; -\pi \leq x \leq \pi.\end{displaymath}$

Answer. Since f(x) is odd, then a_n = 0, for $n \geq 0$ . We turn our attention to the coefficients b_n. For any $n \geq 1$ , we have

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

We deduce

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

Hence

$\begin{displaymath}f(x) \sim 2\left(\sin(x) - \frac{\sin(2x)}{2} + \frac{\sin(3x)}{3} ....\right).\end{displaymath}$

Example. Find the Fourier series of the function

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

Answer. We have

$\begin{displaymath}a_0 = \frac{1}{2\pi}\left(\int_{-\pi}^{0} 0dx + \int_{0}^{\pi... ...\;\;\; a_n = \int_{0}^{\pi} \pi\cos(nx)dx = 0, \;\;\; n \geq 1,\end{displaymath}$

and

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

We obtain b_2n = 0 and

$\begin{displaymath}b_{2n+1} =\frac{2}{2n+1}.\end{displaymath}$

Therefore, the Fourier series of f(x) is

$\begin{displaymath}f(x) \sim \frac{\pi}{2} + 2 \left(\sin(x) + \frac{\sin(3x)}{3} + \frac{\sin(5x)}{5}+\ldots\right).\end{displaymath}$

Example. Find the Fourier series of the function function

$\begin{displaymath}f(x) = \left\{ \begin{array}{rrr} -{\displaystyle \frac{\pi}... ...ystyle \frac{\pi}{2}}, & 0 \leq x \leq \pi \end{array} \right.\end{displaymath}$

Answer. Since this function is the function of the example above minus the constant $\displaystyle \frac{\pi}{2}$ . So Therefore, the Fourier series of f(x) is

$\begin{displaymath}f(x) \sim 2 \left(\sin(x) + \frac{\sin(3x)}{3} + \frac{\sin(5x)}{5}+\ldots \right).\end{displaymath}$

Remark. We defined the Fourier series for functions which are $2\pi$ -periodic, one would wonder how to define a similar notion for functions which are L-periodic.

Assume that f(x) is defined and integrable on the interval [-L,L]. Set

$\begin{displaymath}F(x) = f\left(\frac{Lx}{\pi}\right).\end{displaymath}$

The function F(x) is defined and integrable on $[-\pi, \pi]$ . Consider the Fourier series of F(x)

$\begin{displaymath}F(x) = f\left(\frac{Lx}{\pi}\right) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} \Big(a_n\cos(nx) + b_n\sin(nx)\Big).\end{displaymath}$

Using the substitution $t =\displaystyle \frac{Lx}{\pi}$ , we obtain the following definition:

Definition. Let f(x) be a function defined and integrable on [-L,L]. The Fourier series of f(x) is

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

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 $1\leq n$ .
Example. Find the Fourier series of

$\begin{displaymath}f(x) = \left\{ \begin{array}{lll} 0,& -2 \leq x < 0\\ x, & 0 \leq x \leq 2 \end{array} \right.\end{displaymath}$

Answer. Since L = 2, we obtain

$\begin{displaymath}\begin{array}{lll} a_0 &=& \displaystyle \frac{1}{4} \int_{0... ...{n\pi} = \displaystyle \frac{2}{n\pi}(-1)^{n+1}\\ \end{array}\end{displaymath}$

for $n \geq 1$ . Therefore, we have

$\begin{displaymath}f(x) \sim \frac{1}{2} + \sum_{n=1}^{\infty} \left[\frac{2}{n... ...rac{2}{n\pi}(-1)^{n+1}\sin\left(n\frac{\pi x}{2}\right)\right].\end{displaymath}$

Monday, 25 March 2013

Fourier Series: Basic Results

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