# Fourier series how can it be solved?

1. Nov 2, 2008

### KAS90

1. The problem statement, all variables and given/known data

find fourier series for:
sinnπ=0
(n=0,+1,-1,+2,-2...)

I really can't understand how fourier series works.. Like I tried solving problems..but till now, it didn't sink in my brain..
I just want 2 know the basic way of solving a problem regarding fourier series..then it will be much easier to understand I guess..

2. Nov 2, 2008

### Staff: Mentor

What's the function that you want to find the Fourier series for? sin (n*pi) = 0 for all integers, as you point out, but it's otherwise not very interesting. Is the function f(x) = 0?
If so, all of the coefficients in the Fourier series would be 0.

Is there more to this problem that you haven't shown us?

3. Nov 6, 2008

### KAS90

hi mark
sorry 4 being late..plus giving a completely wrong question..
ok..
the I realize the problem is for example :
f(x)= sinx
0<x<pi
it's a problem I want to understand like how is it really solved?

4. Nov 6, 2008

### HallsofIvy

A Fourier series for a function f(x) is an infinite sum such that
$$f(x)= \sum_{m=0}^\infty A_n cos(nx)+ B_n sin(nx)$$ (or other (x/L) or whatever inside the trig functions).

Of course, the right side of that is periodic with period $2\pi$ so if f is not itself periodic, that can only be true on some interval (which is one reason why you might need that "/L" to alter the interval).

But if f(x)= sin(x) is not only periodic with period $2\pi$, it is already of that form and it is obvious that $A_n= 0$ for all n, while $B_1= 1$ and $B_n= 0$ for all n greater than 1.

More generally, if f(x) is integrable on the interval [0, L], then on that interval $f(x)= A_n cos(2n x\pi/L)+ B_n sin(2n x\pi/L)$ where
$A_0= \frac{1}{L}\int_0^L f(x) dx$
$A_n= \frac{1}{2L}\int_0^L f(x)cos(2n x/L)$
for n> 0 and
$B_n= \frac{1}{2L}\int_0^L f(x)sin(2n x/L)$.

5. Nov 7, 2008

### KAS90

THANX a lot Hallsofivy..
u were a lot of help..as usual lol..