Timeline for Estimate of incomplete binomial integral
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 20, 2016 at 6:06 | vote | accept | Sergei | ||
| Apr 19, 2016 at 22:14 | comment | added | Iosif Pinelis | I have replaced the answer by a much simpler one. | |
| Apr 19, 2016 at 5:10 | comment | added | Iosif Pinelis | I have added details for large $n$. | |
| Apr 18, 2016 at 22:49 | comment | added | Iosif Pinelis | I have simplified the proof of the inequality $I_1\le I_2$. | |
| Apr 18, 2016 at 22:37 | comment | added | Iosif Pinelis | Instead of the deleted answer, I have posted a new one, hoping it's correct. | |
| Apr 18, 2016 at 22:36 | answer | added | Iosif Pinelis | timeline score: 2 | |
| Apr 18, 2016 at 13:58 | comment | added | Sergei | Comments are deleting, deleting, deleting... One by one. So no one will stay here... | |
| Apr 18, 2016 at 9:22 | comment | added | Iosif Pinelis | I made a miscalculation (switched the signs at one point). So, I am deleting my answer. | |
| Apr 18, 2016 at 5:31 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
added 4 characters in body
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| Apr 18, 2016 at 3:39 | comment | added | Brendan McKay | For fixed $\alpha\in(0,1)$ the value is $\frac12+O(n^{-1/2})$ as $n\to\infty$. The method is to expand the logarithm of the integrand about $t=k/n$. Expanding more carefully with explicit error terms will prove the inequality for large enough $n$. | |
| Apr 17, 2016 at 20:30 | comment | added | Sergei | I knew of this problem from Prof. Igor Novikov, he also said that proved the first part. I do not know his proof. | |
| Apr 17, 2016 at 17:55 | comment | added | Iosif Pinelis | Where can one find the proof for $k\le(n+1)/2$? | |
| Apr 17, 2016 at 14:54 | comment | added | Kevin O'Bryant | No, they don't seem relevant to me, either. But my first step was to track it down and see what was there. I posted the link for the convenience of the next reader. | |
| Apr 17, 2016 at 14:42 | comment | added | Sergei | @KevinO'Bryant - thank you for the reference. But it is not obvious how to use such transformations. | |
| Apr 17, 2016 at 14:31 | comment | added | Kevin O'Bryant | The integral is $B_{k/(n+1)}(k+1,n-k+1)$, in the notation of the Wolfram site. Various transformations and identities (though nothing obviously relevant) are cataloged here: functions.wolfram.com/GammaBetaErf/Beta3 | |
| Apr 17, 2016 at 14:08 | history | asked | Sergei | CC BY-SA 3.0 |