Foorier represtaion of this function

[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 can't 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 can't be zero

 

[tiny-tim]

Hi nhrock3!

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

 

[nhrock3]

i took the 0.5 out side of the integral

 

[tiny-tim]

i'm getting confused …

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

 

[whs]

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

 

[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?"