Fourier Example: Difference between revisions

Revision as of 00:41, 19 January 2010

Find the Fourier Series of the function:

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

b_{n} = \frac{1}{2 π} \int_{0}^{π} π \sin (n x) d x, = \frac{1}{n} (1 - c o s (x π)) = \frac{1}{n} (1 - (- 1)^{n})

We obtain b_2n = 0 and

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

Therefore, the Fourier series of f(x) is

$f (x) = \frac{π}{2} + 2 (s i n (x) + \frac{s i n (3 x)}{3} + \frac{s i n (5 x)}{5} + . . .)$

@@ Line 1: / Line 1: @@
 Find the Fourier Series of the function:
 :<math>f(x) = \begin{cases}0,& -\pi<x<0\\
@@ Line 5: / Line 6: @@
 \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>
+:<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>
+We obtain   b_2n = 0 and
+<math>b_{2n+1}=\frac{2}{2n+1}</math>
+Therefore, the Fourier series of f(x) is
+<math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math>