(New page: Definition of Fourier Transform: <math>\widehat{f} = \int_{-\infty}^{\infty} f(t) e^{-\imath xt} dt</math> Definition of Convolution: <math>f\ast g(x) = \int_{-\infty}^{\infty} f(x-y) g...) |
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Definition of Fourier Transform: | Definition of Fourier Transform: | ||
Revision as of 13:41, 29 July 2009
Back to The_Pirate's_Booty
Definition of Fourier Transform:
$ \widehat{f} = \int_{-\infty}^{\infty} f(t) e^{-\imath xt} dt $
Definition of Convolution:
$ f\ast g(x) = \int_{-\infty}^{\infty} f(x-y) g(y) dy $
Now, onto the problem:
$ \widehat{f}(x)\widehat{g}(x) = \int_{-\infty}^{\infty} f(t) e^{-\imath xt} dt \int_{-\infty}^{\infty} g(t') e^{-\imath xt'} dt' $
Now, since f and g are both $ L^{1} $, this integral exists, so by Fubini's Theorem, we may rewrite it as:
$ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) g(t') e^{-\imath xt - \imath xt'} dt' dt $
Now, use a change of variables (namely let $ t'-t = t' $)
$ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) g(t'-t) e^{-\imath xt'} dt' dt $
Now, apply Fubini's Theorem again (since all of these are equalities, we don't need to check that the integral exists, since it's automatic), to get:
$ = \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} f(t) g(t'-t) dt \right) e^{-\imath xt'} dt' = \widehat{f\ast g} $
"And that's all I have to say about that" -Forrest Gump
(Written by Nicholas Stull)
