# Homework Help: Foorier represtaion of this function

1. Dec 23, 2009

### nhrock3

$$f(x)=\sin(\frac{px}{2})\\$$

$$a_0=0$$

$$a_n=0$$

$$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}\(\sin(\frac{px}{2}))\sin(nx)dx=\frac{1}{2\pi}[\sin(\frac{p\pi}{2}-n\pi)-\sin(\frac{-p\pi}{2}+n\pi)]+\frac{1}{2\pi}[\sin(\frac{p\pi}{2}+n\pi)-\sin(\frac{-p\pi}{2}-n\pi)]$$

i used trig identetied to splt into two cosines

andi solved

but i got sines

i need an expression of cosines to do cos nx=(-1)^n

i need to have a simple linear fracture without cosines or sines

i cant transform it here in the needed form

?

and if thinking thurely then i see that i have a trig function on a simetric period
so its zero

so the foorier representation of the given function is zero
?

where is the mistake
it cant be zero

2. Dec 23, 2009

### tiny-tim

Hi nhrock3!

These trigonometric identities need a factor of 1/2 inside the RHS brackets.

3. Dec 23, 2009

### nhrock3

i took the 0.5 out side of the integral

4. Dec 23, 2009

### tiny-tim

i'm getting confused

let's start again … shouldn't there be factors of 1/(p/2 ± n) after the integration?

5. Dec 23, 2009

### whs

foorier... Really? Did you really just do that? I'd attempt a response but I can't read whatever the hell you typed.

6. Dec 24, 2009

### LCKurtz

You need to state your problem more clearly. Are you trying to find a Fourier series that represents your function for all x? If not that, on what interval? Why are you choosing $(-\pi,\pi)$? Depending on the value of p, the periodic extension of your function from that interval may have discontinuities. Do your care about that? Do you want a half range expansion? A more careful statement of the problem please.

And what is "a simple linear fracture without sines or cosines?"