Timeline for Sum of square roots of binomial coefficients
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2017 at 4:06 | comment | added | Noam D. Elkies | The alternating sum $\sum_k (-1)^{n+k} \sqrt{2n \choose k}$ that T. Amdeberhan refers to is much subtler to estimate. As it happens this sum (and likewise for other fractional powers) appeared here five years ago: mathoverflow.net/questions/85013/… | |
| Jan 5, 2017 at 3:25 | history | edited | Martin Zhang | CC BY-SA 3.0 |
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| Jan 4, 2017 at 14:18 | comment | added | Martin Sleziak | You might find something about asymptotic of this sum also in this math.SE post: How to evaluate $\sum\limits_{k=0}^{n} \sqrt{\binom{n}{k}} $. There is also a generalization: An asymptotic expression of sum of powers of binomial coefficients.. Found using Approach0. | |
| Jan 4, 2017 at 13:35 | comment | added | Pietro Majer | A quick bound from Cauchy-Schwarz is $\sum_{k=0}^n\sqrt{n\choose k}\le \sqrt{n+1}\sqrt{\sum_{i=0}^n {n\choose i}}=\sqrt{n+1} \ 2^{n/2} $. Moreover, since the mass of the binomial distribution is concentrated at the indices $|n/2 - i|\le \sqrt{n}$, the same inequality for $\sum_{|n/2 - i|\le \sqrt{n}}$ would lower the bound to $Cn^{1/4}2^{n/2}$. | |
| Jan 4, 2017 at 3:06 | comment | added | T. Amdeberhan | Please see page 109 and the estimates on page 119. Yes, in the earlier chapters the powers were integers. | |
| Jan 4, 2017 at 3:01 | comment | added | Brendan McKay | @T.Amdeberhan dB says that the power is an integer. I'm not sure the method works for $s=1/2$. | |
| Jan 4, 2017 at 2:59 | comment | added | T. Amdeberhan | I just looked at de Bruijn: the closest example is $\sum(-1)^{n+k}\sqrt{\binom{2n}k}$, and the proof runs through pp.109-119. It's complicated. Of course, it does consider other real powers of the binomial, not just $\frac12$ (always alternating sums). | |
| Jan 4, 2017 at 2:46 | comment | added | Brendan McKay | @RichardStanley I don't see it in de Bruijn's book, though I could miss it. There are some similar alternating sums. | |
| Jan 4, 2017 at 2:43 | answer | added | Brendan McKay | timeline score: 9 | |
| Jan 4, 2017 at 2:33 | comment | added | Richard Stanley | Note that rather than using Stirling's formula to prove $\sum_k \binom nk=2^n(1+o(1))$, one can instead use $\sum_k \binom nk=2^n$ to prove Stirling's formula. | |
| Jan 4, 2017 at 2:24 | comment | added | David E Speyer | The last example in <i>Concrete Mathematics</i> is to derive asymtotics for $\binom{n}{k}$ from Stirlings formula and verify $\sum_k \binom{n}{k} = 2^n(1+o(1))$. This somewhat ridiculous finale should be easy to adapt to this problem. | |
| Jan 4, 2017 at 2:00 | comment | added | Richard Stanley | If memory serves, this problem (for any positive exponent) is considered in de Bruijn, Asymptotic Methods in Analysis. I don't have access to this book at the moment. | |
| Jan 4, 2017 at 1:13 | comment | added | Noam D. Elkies | I wouldn't even expect a log term. The OP's heuristic (comparison with a square root of the matching Gaussian) should be easy to convert to a proof. | |
| Jan 4, 2017 at 0:40 | answer | added | T. Amdeberhan | timeline score: 4 | |
| Jan 4, 2017 at 0:24 | comment | added | Gerhard Paseman | You should be able to compare the size of the central binomial coefficient with nearby coefficients to see that a small number are close in size to contribute to your sum , and most of the rest do not. I might expect something like a log n term though. Gerhard "It's Not Quite Hump Day" Paseman, 2017.01.03 | |
| Jan 4, 2017 at 0:13 | history | asked | Martin Zhang | CC BY-SA 3.0 |