Fourier Example: Difference between revisions

Revision as of 01:28, 19 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

$a_{0}={\frac {1}{2\pi }}(\int _{-\pi }^{0}0\ dx+\int _{0}^{\pi }\pi \ dx)={\frac {\pi }{2}}$

$a_{n}=\int _{0}^{\pi }\pi cos(nx)\ dx=0,n\geq 1,$

and

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 5: / Line 5: @@
 \pi,& 0 \le x \le \pi\\
 \end{cases}</math>
 '''
@@ Line 13: / Line 15: @@
 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>
+<math>a_0=\frac{1}{2\pi}(\int_{-\pi}^00\ dx+\int_{0}^\pi\pi\ dx)=\frac{\pi}{2}</math>
-We obtain   b_2n = 0 and
+<math>a_n=\int_{0}^\pi\pi cos(nx)\ dx=0,   n\ge1,</math>
+and
+:<math>b_n = \int_{0}^\pi  \pi\sin(nx)\, dx = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math>
+We obtain   <math>b_{2n}</math> = 0 and
 <math>b_{2n+1}=\frac{2}{2n+1}</math>
 Therefore, the Fourier series of f(x) is