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