Fourier Transformation l^2 - x^2

Fourier Transformation l^2 - x^2

Postby Guest » Sun Sep 09, 2018 2:17 pm

Hello, I need much help with this task. I have spent almost 3h working on it and still I do not know how to calculate it properly.

The task is: transform a function into Fourier

f(x) = [tex]l^{2}[/tex] - [tex]x^{2}[/tex] for < -l, l > and l > 0

Re: Fourier Transformation l^2 - x^2

Postby HallsofIvy » Tue Mar 12, 2019 9:53 am

You have "spent almost 3h working on it" and have no work at all to show what you have tried? You say "transform a function into Fourier". Fourier what? Fourier series or Fourier transform? Is "l" the length of the interval over which you wish to find the Fourier series?

If you are trying to find the Fourier series over interval l, you need to find the integrals
[tex]\frac{1}{\sqrt{l}}\int_0^{l} (l^2- x^2) sin\left(\frac{2\pi}{l}x\right)dx[/tex]
and [tex]\frac{1}{\sqrt{l}}\int_0^{l} (l^2- x^2) cos\left(\frac{2\pi}{l}x\right)dx[/tex]

To do the sin integral, write the integral as [tex]l^2\int sin\left(\frac{2\pi}{l}\right) dx+ \int x^2sin\left(\frac{2\pi}{l}x\right)dx[/tex]. The first is easy. For the second use "integration by parts" twice

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