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Math Help - Fourier Inversion

  1. #1
    Super Member Deadstar's Avatar
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    Fourier Inversion

    Given that f is a function of moderate decrease, \hat{f}(\xi) is continuous and also satisfies,

    |\hat{f}(\xi)| \leq \frac{C}{|\xi|^{1+\alpha}}.

    Use the inversion theorem to derive an expression for f(x+h) - f(x). It should be in the form of an integral.

    How would I do this? Would I use a change of variable?
    Just typed out long possible solution but then realized inversion theorem is d \xi not dx...
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    Given that f is a function of moderate decrease, \hat{f}(\xi) is continuous and also satisfies,

    |\hat{f}(\xi)| \leq \frac{C}{|\xi|^{1+\alpha}}.

    Use the inversion theorem to derive an expression for f(x+h) - f(x). It should be in the form of an integral.

    How would I do this? Would I use a change of variable?
    Just typed out long possible solution but then realized inversion theorem is d \xi not dx...
    Well, by the inversion formula, f(x+h)-f(x)=\frac{1}{2\pi}\int (e^{-i\xi(x+h)}-e^{-i\xi x})\hat{f}(\xi)d\xi . You can factor by e^{-i\xi x}, but that's almost all you can do.
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  3. #3
    Super Member Deadstar's Avatar
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    Yeah that's what I had. Thought there might be more to it, oh well!
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  4. #4
    Super Member Deadstar's Avatar
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    Actually that's not what I got I never read it right!

    I think I'm getting majorly confused here. We have the Fourier Inversion defined as;

    f(x) = \int_{-\infty}^{\infty} \hat{f}(\xi) e^{2 \pi i x \xi} d \xi...

    So wouldn't it be.

    f(x+h) - f(x) = \int_{-\infty}^{\infty} (e^{2 \pi i (x+h) \xi} - e^{2 \pi i x \xi}) \hat{f}(\xi)  d \xi.

    Where does the \frac{1}{2 \pi} come from?
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  5. #5
    MHF Contributor

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    Quote Originally Posted by Deadstar View Post
    Where does the \frac{1}{2 \pi} come from?
    There are at least three commonly used definitions of the Fourier transform, and I simply used another one... (namely \mathcal{F}(f)(\xi)=\int e^{it\xi}f(t)dt, without 2\pi in the exponent) You switch from one to another by a change of variable. You formula is correct with you definition.
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