Re: Fourier Transform

*To*: mathgroup at smc.vnet.net*Subject*: [mg93516] Re: Fourier Transform*From*: Nikolaus Rath <Nikolaus at rath.org>*Date*: Thu, 13 Nov 2008 04:03:47 -0500 (EST)*References*: <gf8rgp$poq$1@smc.vnet.net> <gfbur0$43e$1@smc.vnet.net>

Jens-Peer Kuska <kuska at informatik.uni-leipzig.de> writes: >> 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? > > 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. It's not? This is what I get: {9.98862, -(1/2) E^(-(Abs[x]/Sqrt[(c^2/(c^2 ky^2 - \[Omega]^2))])) Sqrt[c^2/(c^2 ky^2 - \[Omega]^2)] (Cos[ky y] + I Sin[ky y])} for the Fourier Transform and {13.5928, -E^( I ky y - Abs[x]/Sqrt[c^2/(c^2 ky^2 - \[Omega]^2)]) \[Pi]^(3/2) Sqrt[c^2/(c^2 ky^2 \[Pi] - \[Pi] \[Omega]^2)]} for the explicit integral. it's no longer a factor 3, but it's still quite a difference. Moreover, now even the results are different (look at the factor of 1/2 and pi). Any ideas? -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 PGP fingerprint: 5B93 61F8 4EA2 E279 ABF6 02CF A9AD B7F8 AE4E 425C