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Re: FourierTransform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg96019] Re: [mg96001] FourierTransform
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Sun, 1 Feb 2009 04:42:22 -0500 (EST)
  • References: <200901311145.GAA03846@smc.vnet.net>

Fabian wrote:
> I just got the new version (7.0) of Mathematica and was full of hope
> that the function FourierTransform is finally improved. (If you have no
> idea what I'm talking about try FourierTransform[Cosh[x],x,t] on
> Mathematica 6) It turns out that the developers have added support
> for the Human Genome project but did not manage to improve the basic
> mathematical functions considerably. For example try:
> 
>  FourierTransform[1/Cosh[a x],x,t]

In version 7 I see:

In[1]:= FourierTransform[Cosh[x],x,t]
Out[1]= FourierTransform[Cosh[x], x, t]

In[2]:= InputForm[FourierTransform[1/Cosh[a*x],x,t]]
Out[2]//InputForm= -((Sqrt[Pi/2]*Sech[(Pi*t)/(2*a)])/a)

The first is definitely correct (the version 6 result of 0 being of 
course not correct). The second is correct at least up to scaling 
factors. What result were you expecting? Maybe you had meant to do 
FourierTransform[Cosh[a*x],x,t]?

In[3]:= InputForm[FourierTransform[Cosh[a*x],x,t]]
Out[3]//InputForm= Sqrt[Pi/2]*DiracDelta[(-I)*a + t] +
   Sqrt[Pi/2]*DiracDelta[I*a + t]

That one is more problematic, though the result is certainly viable if a 
is assumed pure imaginary. In any case it is not an outright bug in the 
sense that FourierTransform[Cosh[x],x,t] giving zero was a bug.


> Any ideas why this is the case? I always thought that an average user
> is more into Fourier transformation than into image processing?
> 
> Cheers,
>   Fabian

I do not claim expertise but, perhaps naively, I would expect the 
opposite. Many people like to play with images. I would think relatively 
fewer have need to use exact Fourier analysis.


Daniel Lichtblau
Wolfrram Research


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