Timeline for Asymptotic estimation of an integral
Current License: CC BY-SA 4.0
14 events
when toggle format | what | by | license | comment | |
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Jan 23 at 10:59 | answer | added | Claude Leibovici | timeline score: 1 | |
Aug 20, 2022 at 2:06 | vote | accept | minxin jia | ||
Aug 17, 2022 at 0:10 | history | edited | minxin jia | CC BY-SA 4.0 |
edited title
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Aug 16, 2022 at 17:40 | review | Close votes | |||
Aug 28, 2022 at 3:01 | |||||
Aug 16, 2022 at 17:22 | comment | added | LSpice | Your title is misspelled, but the typo cannot easily be fixed since there is another question with the (correctly spelled) title. Please pick a new, correctly spelled title. | |
Aug 16, 2022 at 13:11 | answer | added | Brendan McKay | timeline score: 3 | |
Aug 16, 2022 at 12:38 | comment | added | Brendan McKay | @GiorgioMetafune Quite right, thanks. Upper and lower are interchanged. Also, for large $t$ the lower bound is sharper and that suggest the asymptotic value is $\sqrt{\pi/8t}$ which matches experiment. | |
Aug 16, 2022 at 10:27 | comment | added | Giorgio Metafune | @BrendanMcKay Upper and lower bounds look interchanged (without affecting the resut). | |
Aug 16, 2022 at 10:05 | comment | added | minxin jia | @BrendanMcKay: Thank you very much, Professor McKay. I think you are right because my estimate is the same as yours. | |
Aug 16, 2022 at 9:51 | comment | added | Brendan McKay | A lower bound is $\exp(-\frac45 t v^2)$. This completes the proof that the value is $\Theta(t^{-1/2})$. | |
Aug 16, 2022 at 9:34 | comment | added | Brendan McKay | The bound $\exp(-2tv^2)$ gives $O(t^{-1/2})$ for the integral, not $O(t^{-1})$. And indeed plotting the value of the integral strongly suggests an asymptotic value about $0.63 t^{-1/2}$. So I believe the claim is false. | |
Aug 16, 2022 at 6:34 | comment | added | Brendan McKay | The integrand is less than $\exp(-2tv^2)$. | |
S Aug 16, 2022 at 6:07 | review | First questions | |||
Aug 16, 2022 at 10:05 | |||||
S Aug 16, 2022 at 6:07 | history | asked | minxin jia | CC BY-SA 4.0 |