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, & - π \leq x < 0 \\ π, & 0 \leq x \leq π \end{cases}

Here we have

$a_{0} = \frac{1}{2 π} (\int_{- π}^{0} 0 d x + \int_{0}^{π} π d x) = \frac{π}{2}$

$a_{n} = \int_{0}^{π} π c o s (n x) d x = 0, n \geq 1,$

and

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 $b_{2 n}$ = 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 5: / Line 5: @@
 \pi,& 0 \le x \le \pi\\
 \end{cases}</math>
 '''
@@ Line 12: / Line 14: @@
 Here we have
+<math>a_0=\frac{1}{2\pi}(\int_{-\pi}^00\ dx+\int_{0}^\pi\pi\ dx)=\frac{\pi}{2}</math>
+<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
-We obtain   b_2n = 0 and
+<math>b_{2n+1}=\frac{2}{2n+1}</math>
-<math>b_{2n+1}=\frac{2}{2n+1}</math>
 Therefore, the Fourier series of f(x) is