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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
> 
> 



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