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Re: FourierTransform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg96048] Re: FourierTransform
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Tue, 3 Feb 2009 06:31:58 -0500 (EST)
  • Organization: Uni Leipzig
  • References: <gm1dks$3nk$1@smc.vnet.net> <gm3r8h$mev$1@smc.vnet.net> <gm6kvu$a39$1@smc.vnet.net>
  • Reply-to: kuska at informatik.uni-leipzig.de

Hi,

this is called a distribution or generalized function
and not a function and it is only defined
inside of an integral as my Vladimirov

http://www.amazon.de/Methods-Generalized-Functions-Analytical-Special/dp/0415273560/ref=sr_1_33?ie=UTF8&s=books-intl-de&qid=1233576829&sr=8-33

say.

Regards
   Jens

John Doty wrote:
> Jens-Peer Kuska wrote:
>> Hi,
>>
>> the Fourier transform over the interval x in (-Infinity,Infinity)
>> converges only for quadratic integrable functions, i.e., functions
>> where Integrate[Conjugate[f[x]]*f[x],{x,-Infinity,Infinity}]< Infinity
>>
>> This is not the case for Cosh[x], and so no Fourier transform exist.
> 
> Depends on what you mean by "function". Mathematica tries in its 
> pragmatic way to do what you might want here:
> 
> In[1]:= FourierTransform[t^2,t,w]
> 
> Out[1]= -(Sqrt[2 Pi] DiracDelta''[w])
> 
> t^2 is certainly not square integrable, but this is the kind of useful 
> result scientists and engineers want.
> 
> Mathematica's support for "generalized functions" still has room for 
> improvement, but it has come a long way. The bizarre problems I saw in 
> the past trying Fourier methods to perform fractional differentiation 
> and integration 
> (http://forums.wolfram.com/mathgroup/archive/2000/Apr/msg00043.html) 
> seem no longer to be with us in Mathematica 7.
> 


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