# Math Help - Fourier series

1. ## Fourier series

Expand the function $f(x)=10-x$ on the interval <5,15>.

How do i expand it when the interval isn't $<-l,l>$ ?

2. Originally Posted by Pinsky
Expand the function $f(x)=10-x$ on the interval <5,15>.

How do i expand it when the interval isn't $<-l,l>$ ?
Hmmmmm...

Hint: Can you find the F transform of $f(x + 10) = 10 - (x + 10) = -x$ on (-5, 5)?

-Dan

3. Could you not add ten to your function? then it would be the same as taking the integral from -5 to 5?

I'm not sure about that but that seems reasonable to me.

4. Originally Posted by Pinsky
Expand the function $f(x)=10-x$ on the interval <5,15>.

How do i expand it when the interval isn't $<-l,l>$ ?
Do what topsquark said: $f(x+10) = g(x)$ where $g(x) = x$ on $(-5,5)$. This means, $f(x) = g(x-10)$. Now we expand $g = x$ into a Fourier series on $(-5,5)$. This function is odd so we will just have sine terms. Meaning $\sum_{n=1}^{\infty} a_n \sin \frac{\pi n x}{5}$ where $a_n = \frac{1}{5} \int_{-5}^5 x\sin \frac{\pi n x}{5} dx = \frac{2}{5} \int_0^5 x\sin \frac{\pi n x}{5} dx$. Once you find that you will have Fourier for $g(x)$ this means $g(x-10) = \sum_{n=1}^{\infty} a_n \sin \frac{\pi n (x-10)}{5}$.

5. g(x) would be -x not x.

6. Originally Posted by TrevorP
g(x) would be -x not x.
That would be my mistake, not TPH's, since he was mostly copying from my post, I think. Thanks for the catch and I have fixed it in my original post.

-Dan

7. Isn't the formula fo $a_n=\frac {1}{l} \int_{-l}^{l} f(x)\cos{\frac{n\pi x}{l}} dx$?

U used sinus

8. Originally Posted by Pinsky
Isn't the formula fo $a_n=\frac {1}{l} \int_{-l}^{l} f(x)\cos{\frac{n\pi x}{l}} dx$?

U used sinus
$\frac {1}{l} \int_{-l}^{l} f(x)\cos{\frac{n\pi x}{l}} dx$ gives the coefficent for the cos terms in a Fourier Series ......

9. Originally Posted by ThePerfectHacker
Do what topsquark said: $f(x+10) = g(x)$ where $g(x) = x$ on $(-5,5)$. This means, $f(x) = g(x-10)$. Now we expand $g = x$ into a Fourier series on $(-5,5)$. This function is odd so we will just have sine terms. Meaning $\sum_{n=1}^{\infty} a_n \sin \frac{\pi n x}{5}$ where $a_n = \frac{1}{5} \int_{-5}^5 x\sin \frac{\pi n x}{5} dx = \frac{2}{5} \int_0^5 x\sin \frac{\pi n x}{5} dx$ *. Once you find that you will have Fourier for $g(x)$ this means $g(x-10) = \sum_{n=1}^{\infty} a_n \sin \frac{\pi n (x-10)}{5}$.
*Why sinus?

10. Originally Posted by Pinsky
*Why sinus?
The function g(x) = -x is an odd function. Therefore the F series expansion for it has to be odd as well. So we can have no cosine terms in the expansion since cosine is an even function.

-Dan

11. Oh, ok! I was a bit to radical on what $a_n$ and $b_n$ represent.