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Fourier transform

Defines: 
first Parseval theorem
Type of Math Object: 
Definition
Major Section: 
Reference

Mathematics Subject Classification

42A38 no label found

Comments

I think it might be better to define the transform as for f, let

F(s) = int( f(x)*e^(-2*pi*i*s*x) dx, x=-oo,oo).

and then correspondingly,

f(x) = int( F(s)*e^(2*pi*s*x) ds, s=-oo,oo).

I have never seen it in that form, does it work?
"Do not meddle in the affairs of wizards for they are subtle and quick to anger."

I hope so, since that's the form we used my entire class last term. It seems a bit more symmetrical to me.

The form without the one-on-root(2*pi) is usually used for beginners (as it is easier) or statisticians as their characteristic functions of distributions come out nicely.

The form with the one-on-root(2*pi) is a bit nicer mathematically as it makes the fourier transform an isometry.

Putting the 2*pi in the exponent also makes it an isometry, though it does make some of the properties (eg, translation) seem a little strange. However, the payoff comes with any sort of representation thoery. Also see Poisson's summation formula.

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