Re: Re: Re: (x-y) DiracDelta[x-y] does not simplify to 0
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- Subject: [mg56380] Re: [mg56361] Re: [mg56297] Re: (x-y) DiracDelta[x-y] does not simplify to 0
- From: Pratik Desai <pdesai1 at umbc.edu>
- Date: Sat, 23 Apr 2005 01:16:12 -0400 (EDT)
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Andrzej Kozlowski wrote:
>On 22 Apr 2005, at 05:53, Alain Cochard wrote:
>
>
>
>>*This message was transferred with a trial version of CommuniGate(tm)
>>Pro*
>>Andrzej Kozlowski writes:
>>
>>
>>
>>>Since [use inside Integrate] is the only context in which [x
>>>DiracDelta[x] == 0] is useful and makes sense, I can't see any
>>>justification for your application of Simplify.
>>>
>>>
>>OK, I think I now understand what you say and I disagree with your
>>conclusion. I indeed do find practical justifications for the use of
>>that identity and had so far to perform the simplifications manually.
>>
>>
>>
>>
>
>Of course I did not mean to doubt that this identity or this particular
>way of treating the DiracDelta can't be useful for you or others. There
>is no end to unexpected and ingenious uses that people make of various
>built in functions in Mathematica. However, unless you actually
>integrate your expressions containing the DiracDelta I do not think you
>are really making use of the mathematical notion of a distribution. For
>example, I could define a function MyDiscreteDelta by
>
>MyDiscreteDelta[x_] := DiracDelta[x]/DiracDelta[0]
>
>
>and then I would have
>
>
>FullSimplify[x*MyDiscreteDelta[x]]
>
>0
>
>(Note that, as i pointed out earlier, the
>FullSimplify[x*DiscreteDelta[x]] does not return 0). So one could argue
>that I "used" the DiracDelta to get a "superior" version of the the
>DiscreteDelta, and this may be further useful for something else. Even
>if this were the case it would still be true that the mathematical
>notion of a distribution is not being used here at all because that
>notion is only meaningful when used in th context of integration.
>
>Andrzej
>
>
>
>
I agree with Andrzej Kozlowski, another way to look at the Dirac Delta
is that it represents a charge density for a unit charge placed at x=y,
where as the less eggregious (at least to the mathematician) Heaviside
Function Integrate[DiracDelta[x-y],{x,0,1}]
represents the corresponding cumulative charge distribution. So the
upshot is that the DiracDelta is a symbolic function, whereas the
Heaviside function is a respectable piecewise continuous function.
--
Pratik Desai
Graduate Student
UMBC
Department of Mechanical Engineering
Phone: 410 455 8134
- References:
- Re: (x-y) DiracDelta[x-y] does not simplify to 0
- From: Alain Cochard <alain@geophysik.uni-muenchen.de>
- Re: Re: (x-y) DiracDelta[x-y] does not simplify to 0
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Re: (x-y) DiracDelta[x-y] does not simplify to 0