Timeline for Simplifying products of DiracDelta
Current License: CC BY-SA 4.0
28 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 18, 2024 at 20:46 | answer | added | Michael E2 | timeline score: 0 | |
| Jul 18, 2024 at 15:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 21, 2024 at 16:23 | comment | added | Stephen Luttrell | I do not automatically use the Wolfram documentation as my primary reference source. If you are unsure about how to handle Dirac delta functions, then simply replace them with a convenient alternative representation such as finite width Gaussian bump functions, do whatever manipulations are needed, then take the limit of zero width. If you are solving physics problems, then this approach makes sense. | |
| Mar 20, 2024 at 16:03 | comment | added | user64494 | @StephenLuttrell: You are not right. Up to the documentation "Function domain of DiracDelta ... is restricted to real arguments:" FunctionDomain[DiracDelta[z], z, Complexes] results in z < 0 || z > 0. | |
| Mar 20, 2024 at 14:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 21, 2023 at 13:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 24, 2023 at 19:24 | comment | added | Stephen Luttrell | Yes, that is correct, because it makes k + z real, as is required. You have NOT yet integrated w.r.t. z and k so those variables are still free, but the t integral which you HAVE done requires Im[k] + Im[z] == 0. | |
| Oct 23, 2023 at 18:09 | comment | added | user64494 | @StephenLuttrell: Are you serious? Im[k] + Im[z] == 0allows k and z to be complex numbers. | |
| Oct 23, 2023 at 16:32 | comment | added | Stephen Luttrell | The ConditionalExpression is correct. You have not yet integrated w.r.t. z and k, so not all of the DiracDelta are inside Integrate. | |
| Oct 23, 2023 at 5:52 | comment | added | user64494 | whereas Integrate[ f[z, k, t] DiracDelta[z - 2] DiracDelta[k - 5] DiracDelta[ t - 16], {t, -Infinity, Infinity}] performs DiracDelta[-5 + k] DiracDelta[-2 + z] f[z, k, 16]. | |
| Oct 23, 2023 at 5:51 | comment | added | user64494 | @StephenLuttrell : Can you elaborate your "You can get Mathematica to simplify expressions involving DiracDelta by using it inside Integrate"? Yor reference to Wiki and your example are not clear to me. For example, Integrate[ f[z, k, t] DiracDelta[z - 2] DiracDelta[k - 5] DiracDelta[ t - z - k - 9], {t, -Infinity, Infinity}] considered as the action of the distribution on f[z,k,t], results in ConditionalExpression[ DiracDelta[-5 + k] DiracDelta[-2 + z] f[z, k, 9 + k + z], Im[k] + Im[z] == 0], | |
| Oct 22, 2023 at 11:44 | answer | added | Roland F | timeline score: 0 | |
| Oct 21, 2023 at 17:22 | history | edited | Mohamed Mostafa | edited tags | |
| Oct 21, 2023 at 15:28 | comment | added | Mohamed Mostafa | @StephenLuttrell Tried it, but unfortunatly it didn't work. | |
| Oct 21, 2023 at 15:20 | comment | added | Stephen Luttrell | To define transformation rules that address your problem of simplifying expr = DiracDelta[z-2] DiracDelta[k-5] DiracDelta[t-z-k-9] I would start by defining the rule transf = Times[DiracDelta[Plus[u_,z]], DiracDelta[Plus[v__,z]]] :> Times[DiracDelta[Plus[u,z]], DiracDelta[Plus[v,u]]], and then evaluate expr /. transf to check that it does what you want. The trick to constructing rules is to look at FullForm[expr], and to then use whichever part of expr you want to transform as the lefthand side of the rule, inserting appropriate patterns to make the rule as general as you require. | |
| Oct 21, 2023 at 14:32 | comment | added | Mohamed Mostafa | @StephenLuttrell how to solve this problem then? | |
| Oct 21, 2023 at 14:27 | comment | added | Stephen Luttrell | I am writing from the point of view of someone who used to do LOTS of theoretical particle physics products-of-DiracDelta manipulations. Simplify[DiracDelta[t-5]/t] simplifies to 1/5 DiracDelta[-5 + t] because Mathematica knows that these two (simple) expressions are equivalent when using DiracDelta as a distribution inside an integral. However, Mathematica does NOT know about all equivalences between distributions involving DiracDelta. | |
| Oct 21, 2023 at 11:54 | answer | added | Ulrich Neumann | timeline score: 1 | |
| Oct 21, 2023 at 11:19 | comment | added | Mohamed Mostafa | @StephenLuttrell If you run Simplify[DiracDelta[t-5]/t] in Mathematica you will get DiracDelta[t-5]/5. You can actually do this by hand following the properties of DiracDelta. | |
| Oct 21, 2023 at 8:54 | comment | added | Stephen Luttrell | DiracDelta is a distribution; it is not actually a function (see en.wikipedia.org/wiki/Dirac_delta_function). You can get Mathematica to simplify expressions involving DiracDelta by using it inside Integrate. Here is an example of this: Integrate[f[z, k, t] DiracDelta[z - 2] DiracDelta[k - 5] DiracDelta[t - z - k - 9], {z, z1, z2}, {k, k1, k2}, {t, t1, t2}, Assumptions -> {z2 > z1, k2 > k1, t2 > t1}] which evaluates to give f[2, 5, 16] UnitStep[2 - z1] UnitStep[-2 + z2] UnitStep[5 - k1, -5 + k2, 16 - t1, -16 + t2]. | |
| Oct 21, 2023 at 5:16 | review | Close votes | |||
| Oct 28, 2023 at 3:04 | |||||
| Oct 21, 2023 at 5:10 | comment | added | Mohamed Mostafa | We did before :). And the only reference you prefer is documentation of Mathematica and not textbooks ^^ | |
| Oct 21, 2023 at 5:08 | comment | added | user64494 | Sorry, I prefer arguments and references over empty words. | |
| Oct 21, 2023 at 5:05 | comment | added | Mohamed Mostafa | Yes it's true. We have discussed DiracDelta before in another question and I know your opinion. | |
| Oct 21, 2023 at 5:03 | comment | added | user64494 | Is that answer true? | |
| Oct 21, 2023 at 5:02 | comment | added | Mohamed Mostafa | Let's say I want to force Mathematica to get that answer, how could I do that? | |
| Oct 21, 2023 at 4:58 | comment | added | user64494 | For example if z==k-3, then this product is not defined at all (see the documentation to DiracDelta). Also the question arises: where do such products appear? | |
| Oct 21, 2023 at 4:49 | history | asked | Mohamed Mostafa | CC BY-SA 4.0 |