Timeline for Estimating a sum involving binomial coefficients [refined]
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Feb 27, 2014 at 16:56 | comment | added | The Masked Avenger | I suspect $\tau=1/2$. | |
| Feb 27, 2014 at 16:47 | comment | added | W-t-P | Resolved - but this was a formal problem only, for the values of $S(m,q)$ for $q=0$ and also for $q>m/2$ are small, anyway. | |
| Feb 27, 2014 at 16:45 | history | edited | W-t-P | CC BY-SA 3.0 |
added 4 characters in body
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| Feb 27, 2014 at 16:39 | comment | added | The Masked Avenger | Your refined version has different ranges for q in S and sigma. Please resolve. | |
| Feb 27, 2014 at 16:21 | history | edited | W-t-P | CC BY-SA 3.0 |
added 676 characters in body; edited title
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| Feb 27, 2014 at 5:29 | answer | added | The Masked Avenger | timeline score: 2 | |
| Feb 26, 2014 at 20:15 | answer | added | Roland Bacher | timeline score: 3 | |
| Feb 26, 2014 at 5:11 | comment | added | Douglas Zare | I think you should be able to use Laplace's method. Find the maximum term as something like $3/4 m - 1/2 q - 1/4 \sqrt{m^2+4mq-4q^2}$. Use Stirling's formula and estimate the second derivative of the logarithm of the terms there. Approximate the sum as a Gaussian whose logarithm has the same max and second derivative. | |
| Feb 25, 2014 at 21:34 | answer | added | Dima Pasechnik | timeline score: 1 | |
| Feb 25, 2014 at 21:15 | comment | added | The Masked Avenger | It feels like I've seen the title "Estimating a sum involving binomial coefficients" on six other questions on this forum alone. In any case, break the trend! You are allowed to add things like ${\binom{n-k}{k}}^2$ to the title. | |
| Feb 25, 2014 at 20:56 | comment | added | Tim de Laat | Perhaps you can use Stirling's Formula: $n! \sim {\sqrt {2 \pi n}}\left({\frac ne}\right)^{n}$ for large $n$. | |
| Feb 25, 2014 at 19:43 | review | First posts | |||
| Feb 25, 2014 at 19:44 | |||||
| Feb 25, 2014 at 19:27 | history | asked | W-t-P | CC BY-SA 3.0 |