Re: FourierTransform and removable singularities
- To: mathgroup at smc.vnet.net
- Subject: [mg75393] Re: FourierTransform and removable singularities
- From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
- Date: Sat, 28 Apr 2007 05:54:51 -0400 (EDT)
- References: <f0kbmk$qvt$1@smc.vnet.net> <f0pk9v$1t0$1@smc.vnet.net> <f0sfan$n4v$1@smc.vnet.net>
Peter Pein <petsie at dordos.net> wrote:
> David W.Cantrell schrieb:
> ...
>
> > Perhaps the best way to handle you problem would be to have the sine
> > cardinal function
> >
> > | { 1 if x = 0,
> > | sinc(x) = {
> > | { sin(x)/x otherwise
> >
> > implemented in Mathematica. But defining that function yourself, it
> > does not work as desired with FourierTransform.
>
> Hi David,
>
> sorry, I did not believe this.
Thanks, Peter!
Although rather embarrassed, I'm _glad_ you didn't believe it!
> And indeed:
>
> In[1]:= Off[General::spell];
> Sinc[t_] := Piecewise[{{1, t == 0}}, Sin[t]/t]
>
> In[3]:= FourierTransform[DiracDelta[t]*Sinc[t], t, w]
> Out[3]= 1/Sqrt[2*Pi]
Yes, that's the desired result. <blush>
Despite my having defined sinc above in a piecewise manner, when I defined
it _in Mathematica_, I did not do it in that manner. Rather, I had used
Sinc[t_] := Limit[Sin[u]/u, u -> t]
and then FourierTransform[DiracDelta[t]*Sinc[t], t, w] had given 0.
Regards,
David