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Apr 19, 2023 at 20:54 answer added Alessandro Della Corte timeline score: 2
Apr 19, 2023 at 15:37 history edited Davide Dal Bosco CC BY-SA 4.0
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Nov 3, 2022 at 2:06 comment added user44143 Can anyone explain the square root of the delta function as defined in Moerdijk & Reyes, Models for Smooth Infinitesimal Analysis, VII.3.11, p. 330? The key suggestion there, using invertible infinitesimals, is $$\delta_\epsilon^{1/2}(x)=\delta_\epsilon(x)/\sqrt{\delta(0)}$$
May 19, 2020 at 11:24 history edited Martin Sleziak CC BY-SA 4.0
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May 19, 2020 at 10:53 history edited Martin Sleziak CC BY-SA 4.0
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May 19, 2020 at 10:32 history edited User CC BY-SA 4.0
Fixed grammar and readded "Hello", that I removed accidetally...
May 17, 2020 at 17:06 history edited User CC BY-SA 4.0
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Aug 18, 2013 at 9:04 answer added Andreas timeline score: 1
Apr 12, 2012 at 21:37 comment added Vectornaut @RBega, @Bruce: I'm reading through Colombeau's book "New Generalized Functions and Multiplication of Distributions" right now. Colombeau's stated motivation is to make sense of the distributional products that show up in quantum field theory; in fact, the first chapter of his book is a quick introduction to quantum field theory.
Apr 12, 2012 at 21:08 answer added Bazin timeline score: 3
Sep 22, 2011 at 12:44 history made wiki Post Made Community Wiki by Anand
Jul 21, 2011 at 2:46 answer added Phil Isett timeline score: 4
Dec 6, 2010 at 11:54 history edited Anand CC BY-SA 2.5
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Dec 5, 2010 at 9:59 comment added Anand @Dirk, Thanks for your comments. I will consider it more carefully. :-)
Dec 3, 2010 at 14:10 vote accept Anand
Dec 3, 2010 at 14:10 vote accept Anand
Dec 3, 2010 at 14:10
Dec 3, 2010 at 14:09 vote accept Anand
Dec 3, 2010 at 14:10
Dec 3, 2010 at 13:02 answer added Tim van Beek timeline score: 6
Dec 3, 2010 at 10:30 answer added Denis Serre timeline score: 58
Dec 2, 2010 at 21:39 vote accept Anand
Dec 2, 2010 at 21:39
Dec 2, 2010 at 21:17 answer added Chris Elion timeline score: 1
Dec 2, 2010 at 19:45 answer added Charles Matthews timeline score: 9
Dec 2, 2010 at 18:58 comment added Dirk @Anad: I think you have to clarify what kind of object you want your square of $\delta$ to be. E.g. restricting the test funtions does not give you something like $\delta^2$ as long as you are looking for a linear functional (which you probably don't).
Dec 2, 2010 at 17:56 answer added Michael Hardy timeline score: 5
Dec 2, 2010 at 17:54 answer added user1504 timeline score: 15
Dec 2, 2010 at 17:54 comment added Anand @Otis. Thank you for your reference. I will have a look. Yes. As you said, I intend to restrict the space of the test functions, in order to give a rigorous definition of $\delta_0^2$.
Dec 2, 2010 at 17:47 comment added Otis Chodosh @Bruce the multiplication of distributions is sometimes defined. If you are interested you should look up wavefront sets, one good reference is Chapter 5 of math.mit.edu/~rbm/18.157-F09/18.157-F09.html, see e.g. Prop 5.12. For example, this would make rigorous the notion that the product of a delta function at $0$ and a delta function at $x\neq 0$ should be $0$. However, you are right that the product of two delta functions is not well defined as a distribution. However, I think that @Anand is asking about restricting the domain of distributions to allow it to exist.
Dec 2, 2010 at 17:41 history edited Anand CC BY-SA 2.5
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Dec 2, 2010 at 17:35 comment added Bruce Westbury I am no expert but my understanding is that the Dirac delta function is not a function but is a distribution. Furthermore multiplication of distributions is not defined and this is the cause of much frustration in quantum field theory.
Dec 2, 2010 at 17:29 comment added Rbega There is the <a href="en.wikipedia.org/wiki/…"> Colombeau algebra </a>. Though I have no idea what applications it has.
Dec 2, 2010 at 17:18 history edited Anand CC BY-SA 2.5
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Dec 2, 2010 at 17:10 history asked Anand CC BY-SA 2.5