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

```

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