Re: Re: Re: Folding Deltas
- To: mathgroup at smc.vnet.net
- Subject: [mg57516] Re: [mg56967] Re: [mg56906] Re: [mg56876] Folding Deltas
- From: Chris Chiasson <chris.chiasson at gmail.com>
- Date: Tue, 31 May 2005 04:59:07 -0400 (EDT)
- References: <200505090545.BAA13779@smc.vnet.net> <200505100742.DAA08259@smc.vnet.net> <200505110924.FAA24073@smc.vnet.net>
- Reply-to: Chris Chiasson <chris.chiasson at gmail.com>
- Sender: owner-wri-mathgroup at wolfram.com
It's strange how some pages disappear quickly ( http://xoom.virgilio.it/Home.php ), but at least archive.org can come to the rescue: http://web.archive.org/web/20040615001739/http://xoomer.virgilio.it/maurocer/Text07.htm On 5/11/05, Daniel Lichtblau <danl at wolfram.com> wrote: > Zhengji Li wrote: > > Integrate[DiracDelta[t], {t, -a, a}] = 1, where a > 0. > > > > Maybe Mathematica think Integrate[DiracDelta[t] DiracDelta[t - 2] , > > {t, -3, 3}] is a little bit complicated. (As far as I know, the result > > should be 0) > > > > But, Integrate[Anything, {t, a, b}] + Integrate[Anything, {t, b, a}] = > > 0, so you will get the result. > > > > On 5/9/05, baermic at yahoo.com <baermic at yahoo.com> wrote: > > > >>Can anyone help to verify in Mathematica the expression given by Rota > >>(http://xoomer.virgilio.it/maurocer/Text07.htm): > >> > >>Convolution ( Sum of DiracDeltaFct ** Sum of DiracDeltaFct) == Sum > >>(DiracDeltaFct + Values). > >> > >>I tied > >> > >>Integrate[DiracDelta[t] DiracDelta[t - 2] , {t, -3, 3} ] > >>which does not evaluate; > >>but > >> > >>Integrate[DiracDelta[t] DiracDelta[t - 2] , {t, -3, 1} ] + > >>Integrate[DiracDelta[t] DiracDelta[t - 2] , {t, 1, 3} ] == 0 > >>True > >> > >>( I use ver 5.1 with W2k) > >> > > > Actually the product of delta functions is not defined. This is because > it is not possible to make it well defined. For example I could give > limiting approximations to the integrand that make that integral give > any result, not just zero. There are also other ways to show it is not > well defined. > > The original note was a bit unclear but I the correct form of the > identity in question uses convolution of deltas, not product. Also there > is a minor typo at the web page for Rota's talk. The idenity boils down > simply to > > delta(a_i) * delta(b_j) = delta(a_i+b_j) > > where '*' denotes convolution, not ordinary product. This follows from > basic rules of convolution, translation, and the fact that delta(0) is > the identity element for convolution. > > > Daniel Lichtblau > Wolfram Research > > > > -- Chris Chiasson http://chrischiasson.com/ 1 (810) 265-3161
- References:
- Folding Deltas
- From: baermic@yahoo.com
- Re: Folding Deltas
- From: Zhengji Li <zhengji.li@gmail.com>
- Re: Re: Folding Deltas
- From: Daniel Lichtblau <danl@wolfram.com>
- Folding Deltas