# Fourier series

• Feb 1st 2008, 07:04 AM
Pinsky
Fourier series
Expand the function $\displaystyle f(x)=10-x$ on the interval <5,15>.

How do i expand it when the interval isn't $\displaystyle <-l,l>$ ?
• Feb 1st 2008, 07:23 AM
topsquark
Quote:

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

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

Hmmmmm...

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

-Dan
• Feb 1st 2008, 07:26 AM
TrevorP
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.
• Feb 1st 2008, 07:44 AM
ThePerfectHacker
Quote:

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

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

Do what topsquark said: $\displaystyle f(x+10) = g(x)$ where $\displaystyle g(x) = x$ on $\displaystyle (-5,5)$. This means, $\displaystyle f(x) = g(x-10)$. Now we expand $\displaystyle g = x$ into a Fourier series on $\displaystyle (-5,5)$. This function is odd so we will just have sine terms. Meaning $\displaystyle \sum_{n=1}^{\infty} a_n \sin \frac{\pi n x}{5}$ where $\displaystyle a_n = \frac{1}{5} \int_{-5}^5 x\sin \frac{\pi n x}{5} dx = \frac{2}{5} \int_0^5 x\sin \frac{\pi n x}{5} dx$. Once you find that you will have Fourier for $\displaystyle g(x)$ this means $\displaystyle g(x-10) = \sum_{n=1}^{\infty} a_n \sin \frac{\pi n (x-10)}{5}$.
• Feb 1st 2008, 07:56 AM
TrevorP
g(x) would be -x not x.
• Feb 1st 2008, 09:07 AM
topsquark
Quote:

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
• Feb 2nd 2008, 02:09 AM
Pinsky
Isn't the formula fo $\displaystyle a_n=\frac {1}{l} \int_{-l}^{l} f(x)\cos{\frac{n\pi x}{l}} dx$?

U used sinus
• Feb 2nd 2008, 02:28 AM
mr fantastic
Quote:

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

U used sinus

$\displaystyle \frac {1}{l} \int_{-l}^{l} f(x)\cos{\frac{n\pi x}{l}} dx$ gives the coefficent for the cos terms in a Fourier Series ......
• Feb 2nd 2008, 02:33 AM
Pinsky
Quote:

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

*Why sinus?
• Feb 2nd 2008, 03:15 AM
topsquark
Quote:

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
• Feb 2nd 2008, 03:42 AM
Pinsky
Oh, ok! I was a bit to radical on what $\displaystyle a_n$ and $\displaystyle b_n$ represent.