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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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