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

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

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- Feb 1st 2008, 07:04 AMPinskyFourier 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 AMtopsquark
- Feb 1st 2008, 07:26 AMTrevorP
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 AMThePerfectHacker
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 AMTrevorP
g(x) would be -x not x.

- Feb 1st 2008, 09:07 AMtopsquark
- Feb 2nd 2008, 02:09 AMPinsky
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 AMmr fantastic
- Feb 2nd 2008, 02:33 AMPinsky
- Feb 2nd 2008, 03:15 AMtopsquark
- Feb 2nd 2008, 03:42 AMPinsky
Oh, ok! I was a bit to radical on what $\displaystyle a_n$ and $\displaystyle b_n$ represent.