Fourier Example: Difference between revisions

Revision as of 23:55, 18 January 2010

Find the Fourier Series of the function:

f(x)={\begin{cases}0,&-\pi \leq x<0\\\pi ,&0\leq x\leq \pi \\\end{cases}}

Here we have

b_{n}=\int _{0}^{\pi }\pi \sin(nx)\,dx={\frac {1}{n}}(1-cos(x\pi ))={\frac {1}{n}}(1-(-1)^{n})

We obtain b_2n = 0 and

$b_{2n+1}={\frac {2}{2n+1}}$

Therefore, the Fourier series of f(x) is

$f(x)={\frac {\pi }{2}}+2(sin(x)+{\frac {sin(3x)}{3}}+{\frac {sin(5x)}{5}}+...)$

@@ Line 2: / Line 2: @@
-:<math>f(x) = \begin{cases}0,& -\pi<x<0\\
+:<math>f(x) = \begin{cases}0,& -\pi\le x<0\\
-\pi,& 0<x<\pi\\
+\pi,& 0 \le x \le \pi\\
 \end{cases}</math>
+'''
-:<math>b_n = \frac{1}{2\pi}\int_{0}^\pi  \pi\sin(nx)\, dx, = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math>
+== Solution ==
+'''
+Here we have
+:<math>b_n = \int_{0}^\pi  \pi\sin(nx)\, dx = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math>
@@ Line 18: / Line 24: @@
 <math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math>
+==References:==
+[[Fourier Series: Basic Results]]