Timeline for Is delta function symmetric against real axis? [closed]
Current License: CC BY-SA 4.0
28 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 9, 2021 at 19:09 | history | closed |
R W Pedro Lauridsen Ribeiro Yemon Choi David Roberts♦ David Handelman |
Not suitable for this site | |
| Mar 9, 2021 at 8:20 | answer | added | bathalf15320 | timeline score: 2 | |
| Mar 9, 2021 at 5:42 | comment | added | Anixx | @YemonChoi what did I misunderstand? | |
| Mar 8, 2021 at 23:01 | comment | added | Yemon Choi | I’m voting to close this question because it was attracting feedback on MSE, it is just that the OP seems to have misunderstood things (and the question is founded on a misreading of an answer to another of the user's MSE questions) | |
| Mar 8, 2021 at 1:47 | review | Close votes | |||
| Mar 9, 2021 at 19:09 | |||||
| Mar 8, 2021 at 0:43 | comment | added | Michael Engelhardt | Of course, for complex $z$, the sensible definition is $\delta (z)=\delta (\Re z)\delta (\Im z)$, you can define both of those factors separately and straightforwardly by Fourier transforms, it's symmetric in your $b$, and there is no contradiction anywhere. | |
| Mar 8, 2021 at 0:33 | comment | added | Michael Engelhardt | If you want to use $z$ and $t$ as Fourier conjugate variables, but take $z$ to be complex, then you would have to also allow $t$ to be complex - and then who tells you that the integral over $t$ should be a one-dimensional integral along the real axis? You're entirely overstretching the concept of a "Fourier transform definition" of $\delta $ here. | |
| Mar 7, 2021 at 23:15 | comment | added | Anixx | @NickS okay, okay. Anyway, it is not symmetric against real axis. | |
| Mar 7, 2021 at 22:59 | comment | added | Nick S | @Anixx No they are not. The answer clearly states that this is NOT a distribution, but it can be interpreted in terms of "analytic functionals". And the answer is NOT using the FT, it is saying that for a very particular class of functions you can actually make sense of the RHS in a very specialized context. | |
| Mar 7, 2021 at 22:51 | comment | added | Anixx | @NickS "This means that the "multiple places" where is used are, from a formal mathematical point of view, wrong." Here they use Fourier transform to derive the same formula: math.stackexchange.com/a/4045521/2513 | |
| Mar 7, 2021 at 22:48 | comment | added | Nick S | Note that if you interpret everything as measures on the locally compact Abelian group $\mathbb C$, it is true that $2 \pi \delta_{a+ib}$ is the Fourier transform of the character $(c+di) \to e^{i (ac+bd)}$. Formally this means that for any "nice" function $f : \mathbb C \to \mathbb C$ you have $$ \widehat{f}(a+bi)=\frac{1}{2 \pi} \int_{\mathbb C} f(c+di) e^{i (ac+bd)} d (c+di) $$ which is very different than your formula. But this is the correct mathematical generalisation of $\delta_{a}$ being the Fourier transform of $e^{i ax }$. | |
| Mar 7, 2021 at 22:42 | comment | added | Nick S | That definition does contradict the definition of the Fourier transform, both in terms of measures and distributions. This means that the "multiple places" where is used are, from a formal mathematical point of view, wrong. | |
| Mar 7, 2021 at 21:48 | comment | added | Anixx | @bathalf15320 simply: this identity, which can be found in multiple places, $\int_{-\infty}^\infty \delta(t+bi)f(t)dt=f(-bi)$ appears to contradict the Fourier transform definition. Or I am missing something (I think so) | |
| Mar 7, 2021 at 21:46 | comment | added | Anixx | @bathalf15320 I am not sure about the points on the imaginary axis though. There the Fourier transform becomes divergent as well. | |
| Mar 7, 2021 at 21:45 | comment | added | bathalf15320 | I am, of course, not claiming hereby that the OP makes any sense--it certainly doesn't to me. | |
| Mar 7, 2021 at 21:43 | comment | added | bathalf15320 | @Nate Eldredge "You may certainly define $\delta$ as a distribution but then it doesn't make sense to plug a real or complex number into it". This is a non sequitur--the fact that not every distribution has a value at each point does not mean that no distribution has a value at any point. In fact most distributions which arise in practice have values at most points. And, trivially, the delta function (which can be regarded as a distribtion on the real line or complex plane--and in higher dimensions, of course) has a value at every point bar one. | |
| Mar 7, 2021 at 21:34 | comment | added | Anixx | @NateEldredge I am interested in whatever sense. And also, here they treat it as a function: researchgate.net/publication/… | |
| Mar 7, 2021 at 21:31 | comment | added | Nate Eldredge | But the definition you've given of "symmetric" is something that makes no sense for distributions! | |
| Mar 7, 2021 at 21:29 | comment | added | Anixx | @NateEldredge function or distribution, or functional, whatever. I wonder, if it is symmetric or not. If it depends on framework, okay, let the answer tell it. | |
| Mar 7, 2021 at 21:27 | comment | added | Nate Eldredge | In that case $\delta(i+x)$ is just convenient notation for thinking of the distribution $C^\infty_c \ni \phi \mapsto \phi(-i)$. It doesn't mean you can plug numbers in. Your question reads as if you don't have a clear understanding of what distributions are, and if that's the case, I don't think it's going to be productive to discuss it at an MO level until you do. | |
| Mar 7, 2021 at 21:27 | comment | added | Anixx | @NateEldredge also: researchgate.net/publication/… | |
| Mar 7, 2021 at 21:23 | comment | added | Anixx | @NateEldredge well, somehow people do it, considering dirac delta as a functional, one such link is in the question. | |
| Mar 7, 2021 at 20:57 | comment | added | Carlo Beenakker | see mathoverflow.net/questions/118101/… | |
| Mar 7, 2021 at 20:54 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
|
| Mar 7, 2021 at 20:49 | comment | added | Nate Eldredge | That doesn't answer my question. The definition you wrote down is an integral that doesn't exist in the usual sense. You may certainly define $\delta$ as a distribution, which is usually what "the Fourier transform definition" does, but then it doesn't make sense to plug a real or complex number into it. | |
| Mar 7, 2021 at 20:47 | comment | added | Anixx | @NateEldredge I am interested in an answer about Dirac Delta defined via Fourier transform. | |
| Mar 7, 2021 at 20:46 | comment | added | Nate Eldredge | What do you even mean by "$\delta(a+bi)$" since it's not a function? What kind of mathematical object would $\delta(a+bi)$ represent? Not a real or complex number, that's for sure. | |
| Mar 7, 2021 at 20:37 | history | asked | Anixx | CC BY-SA 4.0 |