Re: FourierTransform of Sign^2 ?
- To: mathgroup at smc.vnet.net
- Subject: [mg80140] Re: FourierTransform of Sign^2 ?
- From: "Dana DeLouis" <dana.del at gmail.com>
- Date: Mon, 13 Aug 2007 04:40:33 -0400 (EDT)
> Mathematica (v. 5.0, Mac) gives
> -Sqrt[2 Pi]*DiracDelta[q]
> with a minus sign.
There was a discussion in another group recently about 5.2 with something
similar. I'm not sure what the correct answer is, but I thought you might
find this interesting.
equ = Sech[x*a];
5.2 takes about 1 minute to do this...
(* 5.2 *)
Timing[FourierTransform[equ, x, s]]
{55.594,
(Sqrt[Pi/2]*Sech[(Pi*s)/ (2*a)])/a}
In Ver 6.0, we get a Minus sign. I thought it was ver 6.0 that has the bug,
but I'm not sure of the correct answer.
We do note that it is much faster in 6.0 though.
(* 6.0.1 *)
Timing[FourierTransform[equ, x, s]]
{0.703,
-((Sqrt[Pi/2]*Sech[(Pi*s)/ (2*a)])/a)}
The real discussion focused on the fact that if you change 'a to a number
like Log[2], then FourierTransform does not work.
Ie
FourierTransform[Sech[x*Log[2]], x, s]
Unevaluated after a long long time.
--
Dana
Windows XP & Mathematica 5.2
"Jung-Tsung Shen" <jushen at gmail.com> wrote in message
news:f9gv9e$b82$1 at smc.vnet.net...
>I tried to evaluate the following Fourier Transform:
>
> FourierTransform[Sign[y] Sign[y], y, q]
>
> where Sign is the sign function. Instead of Sqrt[2 Pi]*DiracDelta[q],
> Mathematica (v. 5.0, Mac) gives
>
> -Sqrt[2 Pi]*DiracDelta[q]
>
> with a minus sign. Is this a bug or I am doing something wrong here?
>
> Thanks.
>
> JT
>