Fourier Example: Difference between revisions

Revision as of 00:55, 19 January 2010

Find the Fourier Series of the function:

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

Here we have

b_{n} = \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 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]]