# User:Egm4313.s12.team11.Suh/R6.1

## Report 6, Problem 6.1

### Problem Statement

• Find the fundamental period of $\cos (n\omega x) \!$ and $\sin (n\omega x) \!$
• Show that these functions also have period $p\!$.
• Show that the constants $a_{0}\!$ is also a periodic function with period $p\!$.

### Solution

• Find the fundamental period, and show that these functions have a period of $p\!$.

$p=2L=2\frac{\pi}{\omega}\!$

$L = \frac{\pi}{\omega}\!$

$f(x+np) = f(x)\!$

$f(x) = cos (n\omega x)\!$

$f(x+n\frac{2\pi}{n\omega }) = cos (n\omega (x+\frac{2\pi}{n\omega }))\!$

$f(x+\frac{2\pi}{\omega}) = cos (n\omega x+2\pi)\!$

$f(x+2L) = cos (n\omega x+2\pi)\!$

              $p=2L\!$ is the fundamental period for $cos (n\omega x)\!$.


$f(x+np) = f(x)\!$

$f(x) = sin (n\omega x)\!$

$f(x+n\frac{2\pi}{n\omega }) = sin (n\omega (x+\frac{2\pi}{n\omega }))\!$

$f(x+\frac{2\pi}{\omega }) = sin (n\omega x+2\pi)\!$

$f(x+2L) = sin (n\omega x+2\pi)\!$

              $p=2L\!$ is the fundamental period for $sin (n\omega x)\!$.


• Show that the constants $a_{0}\!$ is also a periodic function with period $p\!$.

From Fourier Series,
$a_{0}=\frac{1}{2L}\int_{-L}^{L}f(x)dx\!$

$f(x) = \cos (n\omega x)\!$

$a_{0}=\frac{\omega}{2\pi}\int_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}cos (n\omega x)dx\!$

$a_{0}=\frac{\omega}{2\pi}[\frac{sin (n\omega x)}{n\omega}|_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}]\!$

$a_{0}=\frac{\omega}{2\pi}[\frac{2sin (n\pi)}{n\omega}]\!$

          $a_{0}=0\!$


$f(x) = \sin (n\omega x)\!$

$a_{0}=\frac{\omega}{2\pi}\int_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}sin (n\omega x)dx\!$

$a_{0}=\frac{\omega}{2\pi}[\frac{cos (n\omega x)}{n\omega}|_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}}]\!$

$a_{0}=\frac{\omega}{2\pi}[\frac{-cos{n\pi}+cos{n\pi}}{n\omega}]\!$

          $a_{0}=0\!$