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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
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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