Timeline for Properties of convolutions
Current License: CC BY-SA 4.0
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| S Jan 22, 2020 at 12:44 | history | bounty ended | Landauer | ||
| S Jan 22, 2020 at 12:44 | history | notice removed | Landauer | ||
| Jan 22, 2020 at 12:43 | vote | accept | Landauer | ||
| Jan 22, 2020 at 12:43 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 22, 2020 at 1:46 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 21, 2020 at 3:01 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 20, 2020 at 21:15 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 20, 2020 at 1:22 | answer | added | fedja | timeline score: 7 | |
| Jan 20, 2020 at 0:37 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 19, 2020 at 17:15 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 19, 2020 at 2:30 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 18, 2020 at 15:21 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 18, 2020 at 7:56 | history | edited | user44143 | CC BY-SA 4.0 |
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| Jan 18, 2020 at 3:24 | comment | added | Landauer | @MattF. it is to verify the numerical findings. But I emphasized the question now more clearly. | |
| Jan 18, 2020 at 3:24 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 18, 2020 at 3:17 | comment | added | user44143 | What is the question? I do not see a question mark in the post. | |
| S Jan 18, 2020 at 2:00 | history | bounty started | Landauer | ||
| S Jan 18, 2020 at 2:00 | history | notice added | Landauer | Authoritative reference needed | |
| Jan 17, 2020 at 18:53 | comment | added | Landauer | Yes, $p=$ means I plotted the function $F_p$ with the respective label of $p.$ I agree it looks like a plot label but it is an axis label in mathematica for the $y$ axis. The derivative is still numerically, but I was more careful this time. The problem is if I differentiate it analytically and then do it numerically, it is more difficult for the numerics to compute, as one has many more integrals. | |
| Jan 17, 2020 at 18:48 | comment | added | Iosif Pinelis | @Martinique : In a couple of instances, you labeled the vertical axis by something like $p=...$. I think the vertical axis carries the values of your resulting function, whereas something like $p=...$ is a plot label (rather than an axis label). Also, it is unclear to me how you approximated the derivative at $p=2$: (i) numerically or (ii) analytically and then evaluated the involved integrals numerically. | |
| Jan 17, 2020 at 16:53 | comment | added | Landauer | @IosifPinelis okay, now the derivative is much more stable | |
| Jan 17, 2020 at 16:52 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 17, 2020 at 16:15 | comment | added | Iosif Pinelis | @Martinique : Your graph for the derivative in $p$ at $p=2$ (showing a lack of a phase transition for all $y$ at once) seems to contradict the graphs for $p=2\pm10^{-5}$. Have you tried to take the derivative analytically and then evaluate the involved integrals numerically? You can also try to use asymptotics to find the sign at the tails of the derivative, for large $|y|$. Also, can you label the axes (especially the vertical one)? | |
| Jan 17, 2020 at 15:18 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 17, 2020 at 15:13 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 17, 2020 at 12:59 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 17, 2020 at 11:00 | comment | added | Landauer | @IosifPinelis Yes, it is $p=2+10^{-5}$, I clarified it now. Let me check for the derivative. | |
| Jan 17, 2020 at 10:54 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 17, 2020 at 3:40 | comment | added | Iosif Pinelis | @Martinique : I don't understand what you got for $2+10^{-5}$. Did you mean for $p=2+10^{-5}$? Anyhow, what were the "desired results" you got? Did you try something like $p=2\pm10^{-3}$ or $p=2\pm10^{-2}$? (I mean both $+$ and $-$.) Did you try the derivative in $p$ at $p=2$? | |
| Jan 17, 2020 at 1:41 | comment | added | Landauer | @IosifPinelis if you want me to test some other conjectures numerically, please let me know. Your help is greatly appreciated. | |
| Jan 16, 2020 at 23:12 | comment | added | Landauer | @IosifPinelis Sorry, $2+10^{-5}$ was all I could do numerically but I gives the desired results. I also checked whether the functions are log-convex/log-concave in general. It seems they are not. So trying to verify some log-convexity/log-concavity here seems to be a misleading approach, at least as far as the numerics is concerned. The negativity was because I plotted logs. I clarified this now and plotted the actual function and one log. The actual functions are all positive by Cauchy-Schwarz (sorry for the confusion about what is plotted I hope it is much clearer now) | |
| Jan 16, 2020 at 23:10 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 16, 2020 at 22:39 | comment | added | Iosif Pinelis | Have you tried exponents $p=2\pm\epsilon$ for small (maybe infinitesimally small) real $\epsilon$, instead of exponents $p\in\{1,4\}$ in $e^{-|x|^p}$? I'd think that a "phase transition" at $p=2$ would not be very surprising, since the resulting function is constant for $p=2$ and thus is on the border between the (log-)convexity and (log-)concavity. Also, what I see as a lack of a "global" "phase transition" is that the resulting function is $>0$ for $p=2$ but seems to be $<0$ for $p\in\{1,4\}$ -- whereas $1<2<4$. | |
| Jan 16, 2020 at 19:32 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 16, 2020 at 15:18 | comment | added | LSpice | $x \mapsto e^{-x^2}$ isn't just some random medium-decaying function, though; it's an eigenfunction for the Fourier transform, and that's got to be significant. | |
| Jan 16, 2020 at 15:08 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 16, 2020 at 9:42 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 16, 2020 at 2:40 | history | edited | Landauer | CC BY-SA 4.0 |
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| Jan 16, 2020 at 1:54 | history | asked | Landauer | CC BY-SA 4.0 |