1. ## Fourier Serie

How do I determine the fourier series of interval function: f(x)= 8+2(x/9) , 0<x<4
, -8+2(x/9) , -4<x<0
and in addition we know: f(x+8)=f(x)

2. ## Nice

Originally Posted by DanielM99
How do I determine the fourier series of interval function: f(x)= 8+2(x/9) , 0<x<4
, -8+2(x/9) , -4<x<0
and in addition we know: f(x+8)=f(x)

$\displaystyle f(x)=\sum_{n=0}^{\infty}b_n\sin\left( \frac{n\pi}{p}x\right)$

where $\displaystyle b_n=\frac{2}{p}\int_{0}^{p}f(x) \sin\left( \frac{n\pi}{2}x\right)dx$

Our integral is

$\displaystyle b_n=\frac{1}{2}\int_{0}^{4}\left( 8+\frac{2x}{9}\right) \sin\left( \frac{n\pi}{p}x\right)dx =$

$\displaystyle =-\frac{4}{n\pi}\left(8+\frac{2x}{9} \right)\cos\left( \frac{n\pi}{2}x\right)+\frac{2}{9}\cdot \frac{4^2}{n^2\pi^2}\sin\left( \frac{n\pi}{2}x\right) \big]_{0}^{4}$

$\displaystyle =-\frac{4}{n\pi}\left( \frac{80}{9}\right)\cos(2\pi n)+\frac{32}{n\pi}= -\frac{32}{9 \pi n}$

so finally we get

$\displaystyle f(x)=\sum_{n=0}^{\infty} = -\frac{32}{9 \pi n} \sin\left( n \frac{\pi }{4}x\right)$