Re: Fourier Transform
- To: mathgroup at smc.vnet.net
- Subject: [mg93491] Re: Fourier Transform
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Tue, 11 Nov 2008 07:46:25 -0500 (EST)
- Organization: Uni Leipzig
- References: <gf8rgp$poq$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
Hi,
ClearSystemCache[]
expr = (c^2 Sqrt[2 \[Pi]] DiracDelta[ky + sy] DiracDelta[
sz])/(-c^2 sx^2 - c^2 sy^2 - c^2 sz^2 + \[Omega]^2);
$Assumptions = {{x, y, z} \[Element] Reals};
InverseFourierTransform[expr, {sx, sy, sz}, {x, y, z}] // Timing
Integrate[
1/Sqrt[2 \[Pi]] Exp[-I sx x] Exp[-I sy y] Exp[-I sz z] expr, {sx, -\
\[Infinity], \[Infinity]}, {sy, -\[Infinity], \[Infinity]}, {sz, -\
\[Infinity], \[Infinity]}, GenerateConditions -> False] // Timing
is not so different.
Regards
Jens
Nikolaus Rath wrote:
> Hello,
>
> Consider the following expression:
>
> expr = (c^2 Sqrt[2 \[Pi]]
> DiracDelta[ky + sy] DiracDelta[sz])/(-c^2 sx^2 - c^2 sy^2 -
> c^2 sz^2 + \[Omega]^2);
> $Assumptions = {{x, y, z} \[Element] Reals};
> InverseFourierTransform[expr, {sx, sy, sz}, {x, y, z}] // Timing
> Integrate[
> 1/Sqrt[2 \[Pi]] Exp[-I sx x] Exp[-I sy y] Exp[-I sz z]
> expr, {sx, -\[Infinity], \[Infinity]}, {sy, -\[Infinity], \
> \[Infinity]}, {sz, -\[Infinity], \[Infinity]}] // Timing
>
> On my system with Mathematica 6, the explicit integration takes 3
> times as long as the InverseFourierTransform and also gives several
> additional required assumptions for the same result (e.g. Im[-ky^2 +
> \[Omega]^2/c^2] != 0 || Re[-ky^2 + \[Omega]^2/c^2] <= 0).
>
> How is this possible? Is Mathematica using some special tricks when
> evaluating the InverseFourierTransform?
>
>
> Best,
>
> -Nikolaus
>
> --
> =C2=BBIt is not worth an intelligent man's time to be in the majority.
> By definition, there are already enough people to do that.=C2=AB
> -J.H. Hardy
>
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>