Re: Re: (x-y) DiracDelta[x-y] does not simplify to 0

*To*: mathgroup at smc.vnet.net*Subject*: [mg56338] Re: [mg56297] Re: (x-y) DiracDelta[x-y] does not simplify to 0*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Fri, 22 Apr 2005 06:23:24 -0400 (EDT)*References*: <d42kg5$39t$1@smc.vnet.net> <d45agf$ieu$1@smc.vnet.net> <200504210936.FAA05048@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 21 Apr 2005, at 18:36, Alain Cochard wrote: > *This message was transferred with a trial version of CommuniGate(tm) > Pro* > yehuda ben-shimol writes: > >> As I remember, DiracDelta is singular and has a meaning only under >> integration. Anyway the properties of the DiracDelta are kept by >> Mathematica i.e., Integrate[(x - y)DiracDelta[x - y], {x, -1, 1}, >> {y, -1, 1}] returns 0 as expected > > Andrzej Kozlowski writes pretty much the same: > >> On the one hand I think the Mathematica implementation of DiracDelta >> (and KroneckerDelta) leaves a lot to be desired... and that is putting >> it mildly. (That means I have plenty of much worse examples...). >> >> On the other hand, I am not convinced that Mathematica ought to >> perform >> this sort of simplification at all. DiracDelta is a generalised >> function. The statement x DiracDelta[x] == 0 needs a lot of >> interpreting to make sense of (I prefer to think of it as nonsense). >> However >> >> >> Integrate[(x-y) DiracDelta[x-y], {x,-Infinity,Infinity}] >> >> 0 >> >> is correct. > > I don't understand these reservations. I learned the statement x delta > = 0 in my lectures on distributions at university, and I checked today > that it also appears in one of Laurent Schwartz's ("father" of > distribution theory) books. Plus I find it very intuitive and it's > straightforward to demonstrate. > As I wrote: it takes a lot of interpreting. You not only have to interpret both x and 0 as distributions, but you also have to interpret multiplication in a special way, quite different form the normal Mathematica interpretation. The statement x DiracDelta[x] ==0 is merely an informal short hand for the staatement: <x DiracDelta[x], f[x]>= <0, f[x]> for any "test function" f[x], where <f,g> stands for Integrate[f[x]*g[x],{-Infinity,Infinity}] (One can also approach distributions from the point of view of non-standard analysis but this is not implemented in Mathematica and the same remarks as below also apply). One can perform certain further informal manipulations on this 'identity". But such informal manipulations should only be performed carefully by people who know what they are doing and Mathematica certainly does not. Because the algebra of distributions is not like usual "algebra" (for a start multiplication is not defined for arbitrary distributions) so it is very easy to obtain nonsensical answers if you manipulate informal expressions containing DiracDelta blindly, e.g. the following is complete nonsense: Solve[x*DiracDelta[x] == 0, x] {{x -> 0}, {x -> InverseFunction[DiracDelta, 1, 1][0]}} One can manufacture plenty of such examples but I think they point I am making should be clear. Since very easy to obtain nonsensical conclusions by manipulating expressions involving the Dirac function blindly and since on the other hand there is actually nothing useful that can be done with the above "relationship" , I don't think Mathematica should perform any such "simplifications" except in the proper formal context, which means within Integrate. Andrzej Kozlowski Chiba, Japan http://www.akikoz.net/andrzej/index.html http://www.mimuw.edu.pl/~akoz/

**References**:**Re: (x-y) DiracDelta[x-y] does not simplify to 0***From:*Alain Cochard <alain@geophysik.uni-muenchen.de>

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**Re: Re: Exact Symbolic Notation**

**Re: Re: (x-y) DiracDelta[x-y] does not simplify to 0**

**Re: Re: (x-y) DiracDelta[x-y] does not simplify to 0**