Fourier series

**Pinsky** · February 1st 2008, 07:04 AM

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

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

**topsquark** · February 1st 2008, 07:23 AM

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

**TrevorP** · February 1st 2008, 07:26 AM

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.

**ThePerfectHacker** · February 1st 2008, 07:44 AM

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 $g(x-10) = \sum_{n=1}^{\infty} a_n \sin \frac{\pi n (x-10)}{5}$ .

**TrevorP** · February 1st 2008, 07:56 AM

g(x) would be -x not x.

**topsquark** · February 1st 2008, 09:07 AM

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

**Pinsky** · February 2nd 2008, 02:09 AM

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

**mr fantastic** · February 2nd 2008, 02:28 AM

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 ......

**Pinsky** · February 2nd 2008, 02:33 AM

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?

**topsquark** · February 2nd 2008, 03:15 AM

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

**Pinsky** · February 2nd 2008, 03:42 AM

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

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