Using inversion formula for the fourier transform

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To deduce that \hat{\hat{f}}(x) = 2\pi f(-x) using the Fourier transform inversion formula, start by writing the Fourier transform and its inverse. The Fourier transform is defined as \hat{f}(\xi) = ∫ f(x) e^{-i x \xi} dx, and the inverse as f(x) = (1/2π) ∫ \hat{f}(\xi) e^{i x \xi} dξ. By substituting \hat{f} into the expression for \hat{\hat{f}}(x) and manipulating the terms, you can show that it leads to f(-x). The key step is recognizing that substituting x with -x in the inversion formula simplifies the expression. This approach effectively demonstrates the relationship between \hat{\hat{f}}(x) and f(-x).
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I need to deduce that \hat{\hat{f}}(x)=2\pif(-x) using the inversion formula for the Fourier transform, I was wondering if someone could explain why there's f(-x) because i just can't get started on this problem!
 
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that is supposed to be 2*Pi not 2 to the power of Pi
 
Well, the first thing to do is to write down the Fourier transform formulas for \hat{f} and f, and try to manipulate them into the right form.
 
I'd written them out already and I've tried substituting them into the inversion formula and vice versa but getting nowhere! My lecturer has said stare at the inversion formula but I'm just going around in circles!
 
OK, so I assume based on how the question is posed that you have defined the the Fourier transform like this:

\hat{f}(\xi) = \intop_{-\infty}^{\infty} f(x) e^{-i x \xi} dx
and the inverse like this:
f(x) = \frac{1}{2\pi}\intop_{-\infty}^{\infty} \hat{f}(\xi) e^{i x \xi} d\xi

Now we can write an expression for:

\hat{\hat{f}}(x) = \intop_{-\infty}^{\infty} \hat{f}(\xi) e^{-i x \xi} d\xi

Now compare this to the inverse formula. Can you make them match up?
 
yes that's what i had, is it actually possible to just write if x=-x then...? thankyou so much
 
Yes, just write f(-x) in the inversion formula and things just fall out.
 

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