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Re: FoourierTransform of a function defined in sections
*To*: mathgroup at smc.vnet.net
*Subject*: [mg70202] Re: [mg70176] FoourierTransform of a function defined in sections
*From*: Daniel Lichtblau <danl at wolfram.com>
*Date*: Sat, 7 Oct 2006 07:07:41 -0400 (EDT)
*References*: <200610060559.BAA15565@smc.vnet.net>
Eckhard Schlemm wrote:
> Hello,
>
> I want Mathematica to calculate the FourierTransform of a function which is
> defined by Sin[x]^2 for Abs[x]<PI and zero else. I tried and defined the
> function g as follows:
>
> g[x_]:=If[Abs[x]>PI,0,Sin[x]^2];
>
> That works fine. But if I have mathematica try to determine the
> FourierTransform by
>
> FourierTransform[g[x],x,p]
>
> I always get the error that the recursion limit and the iteration limit were
> exceeded...
>
> what am I'm doing wrong?
>
> Any help is appreciated
>
> thanks
>
> Eckhard
>
> --
> _________________________
> Ludwig Schlemm
> Tel: +49 (0) 160 91617114
> LudwigSchlemm at hotmail.com
>
Infinite recursion suggests a bug in the implementation of
FourierTransform. That said, it may not handle very well a programming
construct such as If. Instead one could use Piecewise or products of
UnitStep cutoffs.
In[3]:= ff = UnitStep[Pi-x]*UnitStep[Pi+x]*Sin[x]^2;
In[4]:= InputForm[FourierTransform[ff, x, w]]
Out[4]//InputForm= (-2*Sqrt[2/Pi]*Sin[Pi*w])/(-4*w + w^3)
In[7]:= hh[x_] = Piecewise[{{Sin[x]^2, Abs[x]<Pi}}];
In[11]:= InputForm[FourierTransform[hh[x], x, w]]
Out[11]//InputForm= (-2*Sqrt[2/Pi]*Sin[Pi*w])/(-4*w + w^3)
Version 5.2 of Mathematica does in fact handle your If formulation. My
guess is some preprocessing reformulates an integrand as a Piecewise
construct.
In[12]:= g[x_]:=If[Abs[x]>Pi,0,Sin[x]^2]
In[14]:= InputForm[FourierTransform[g[x],x,p]]
Out[14]//InputForm= (-2*Sqrt[2/Pi]*Sin[p*Pi])/(-4*p + p^3)
Daniel Lichtblau
Wolfram Research
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