Timeline for Anti-delta function?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 1, 2023 at 1:40 | comment | added | Anixx | So, using this variant of delta function math.stackexchange.com/questions/4627326/… it would be $\overline{\delta}(1/x)$. | |
| Jan 31, 2022 at 16:37 | comment | added | Willie Wong | I think this is basically what Gro-Tsen wrote, except instead of Stone-Čech you take the one-point compactification and require your functions to have a limit at $\infty$. | |
| Jan 30, 2022 at 23:12 | history | edited | LSpice | CC BY-SA 4.0 |
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| Jan 30, 2022 at 22:41 | comment | added | Anixx | As others state that $F(x)$ cannot be a distribution, what $1/x^2 \delta(1/x)$ from your answer supposed to be? A hyperfunction? Or some formal expression? | |
| Jan 30, 2022 at 22:38 | comment | added | Anixx | Very interesting, but still I do not understand :-) You said, $F(x)$ is dimensionless. Is it similar to an empty matrix math.stackexchange.com/questions/1913141/… ? | |
| Jan 30, 2022 at 22:32 | comment | added | Carlo Beenakker | just dimensional analysis, $F(x)dx$ should be dimensionless, so if you wish to map 0 onto $\infty$ by taking $F(x)\propto\delta(1/x)$ you need a factor $1/x^2$; $\delta_\epsilon(x)$ is a mollified delta function [$\lim_{\epsilon\rightarrow 0}\delta_\epsilon(x)=\delta(x)$]. | |
| Jan 30, 2022 at 22:28 | comment | added | Anixx | Well, I was wondering what thinking path led you to this development? Also, why it is assumed that the integral $\int_{-\infty}^\infty F_\epsilon(x)=1$ is continuous at $\epsilon=0$? Also, does $\delta_{\epsilon}(x)$ mean delta function with support at $\epsilon$? | |
| Jan 30, 2022 at 22:24 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 30, 2022 at 22:23 | comment | added | Anixx | Did you use any known properties when deriving this? For instance, identities for $\delta(1/x)$ or Laplace transform? | |
| Jan 30, 2022 at 22:19 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |