Re: question
- To: mathgroup at smc.vnet.net
- Subject: [mg80076] Re: [mg80065] question
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sun, 12 Aug 2007 07:11:19 -0400 (EDT)
- Reply-to: hanlonr at cox.net
$Version
6.0 for Mac OS X x86 (32-bit) (June 19, 2007)
expr = FourierTransform[Sign[y]*Sign[a*y], y, q]
(-Sqrt[2*Pi])*DiracDelta[q]*
UnitStep[-a] -
(I*Sqrt[2/Pi]*UnitStep[a]*
UnitStep[-a])/q +
Sqrt[2*Pi]*DiracDelta[q]*
UnitStep[a]
Simplify[expr, a > 0]
Sqrt[2*Pi]*DiracDelta[q]
Assuming[{a > 0}, FourierTransform[Sign[y]*Sign[a*y], y, q]]
Sqrt[2*Pi]*DiracDelta[q]
Simplify[expr, a < 0]
(-Sqrt[2*Pi])*DiracDelta[q]
Assuming[{a < 0}, FourierTransform[Sign[y]*Sign[a*y], y, q]]
(-Sqrt[2*Pi])*DiracDelta[q]
x1 = FullSimplify[expr, a != 0]
Sqrt[2*Pi]*DiracDelta[q]*
UnitStep[a] - Sqrt[2*Pi]*
DiracDelta[q]*UnitStep[-a]
x2 = Assuming[{a != 0}, FourierTransform[Sign[y]*Sign[a*y], y, q]]
2*Sqrt[2*Pi]*DiracDelta[q]*
UnitStep[a] - Sqrt[2*Pi]*
DiracDelta[q]
Simplify[x2 - x1, #] & /@ {a > 0, a < 0}
{0,0}
Bob Hanlon
---- dimitris <dimmechan at yahoo.com> wrote:
> Let's see if Mathematica 6 has become better.
>
> In Mathematica 5.2 (and as well 6; as it
> was mentioned in a recent thread).
>
> In[11]:=
> FourierTransform[Sign[y]*Sign[y], y, q]
>
> Out[11]=
> Sqrt[2*Pi]*DiracDelta[q]
>
> which is correct.
>
> In Mathematica 5.2
>
> In[5]:=
> FourierTransform[Sign[y]*Sign[a*y], y, q]
>
> Out[5]=
> (1/q)*((I*(1 - E^(I*q*Sqrt[1/Integrate`NLtheoremDump`myMax[0,
> 0]^2]*Integrate`NLtheoremDump`myMax[0, 0]^2) +
> E^(2*I*q*Sqrt[1/Integrate`NLtheoremDump`myMax[0,
> 0]^2]*Integrate`NLtheoremDump`myMax[0, 0]^2))*Sqrt[2/Pi]*
> (2 + DiscreteDelta[a] - UnitStep[-a] - UnitStep[a]))/E^(I*q*Sqrt[1/
> Integrate`NLtheoremDump`myMax[0, 0]^2]*
> Integrate`NLtheoremDump`myMax[0, 0]^2))
>
> I think outputs like this is a mini nightmare for the developers.
>
> What does version 6 returns?
>
> Thanks
> Dimitris
>
>