I'm looking for asymptotic estimate for the binomial coefficient: $$ \ln{\binom{n}{[\sqrt{n}]}} $$ I assume Stirling's approximation can help, but I'm not sure I will get any good estimation with this approach. Is there any good way to make an estimation for this coefficient? Thanks in advance.
Asked
Active
Viewed 2,342 times
2
-
1Use $\ln(n!)=(n+1/2)\ln(n)-n+O(1)$ for all the three terms. – Shitikanth Oct 05 '12 at 20:14
2 Answers
2
I shall derive a simple asymptotic formula for $$\binom{n}{\sqrt{n}+\xi}$$ where $\xi$ is bounded and $n\to+\infty$. It follows from the main result of this paper that $$ \log ((n + a)!) = \left( {n + a + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + \frac{{6a^2 + 6a + 1}}{{12n}} + \mathcal{O}\!\left( {\frac{{\max (\left| a \right|^3 ,1)}}{{n^2 }}} \right) $$ provided $n + a+1 \ge 0$ and $\left| a+1 \right| < \frac{3}{5}n$, where the implied constant does not depend on $n$ or $a$. Then $$ \log (n!) = \left( {n + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + \mathcal{O}\!\left( {\frac{1}{n}} \right), $$ $$ \log ((\sqrt n + \xi )!) = \left( {\sqrt n + \xi + \frac{1}{2}} \right)\log \sqrt n - \sqrt n + \frac{1}{2}\log (2\pi ) + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right) $$ and $$ \log ((n - \sqrt n - \xi )!) = \left( {n - \sqrt n - \xi + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + \frac{1}{2} + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right), $$ as $n\to+\infty$, uniformly with respect to bounded values of $\xi$. Accordingly, $$ \log \binom{n}{\sqrt{n}+\xi} = \left( {\sqrt n + \xi - \frac{1}{2}} \right)\log \sqrt n + \sqrt n - \frac{1}{2}\log (2\pi {\rm e}) + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right) $$ or $$ \binom{n}{\sqrt{n}+\xi}= \frac{1}{{\sqrt {2\pi {\rm e}} }}\sqrt n ^{\sqrt n + \xi - \frac{1}{2}} {\rm e}^{\sqrt n } \left( {1 + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right), $$ as $n\to+\infty$, uniformly with respect to bounded values of $\xi$. In particular, $$ \binom{n}{[\sqrt{n}]}= \frac{1}{{\sqrt {2\pi {\rm e}} }}\sqrt n ^{[\sqrt{n}]- \frac{1}{2}} {\rm e}^{\sqrt n } \left( {1 + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right), $$ as $n\to+\infty$.
Gary
- 31,845
1
Using Shitikanth's hint I think you're going to be coming up with $$\text{ln}{n \choose [\sqrt{n}]}\approx\text{ln}\left(\frac{n^{n+\sqrt{n}/2+3/4}}{\left(n-\sqrt{n}\right)^{1/2-\sqrt{n}+n}}\right).$$
Alexander Gruber
- 26,963
-
Thank you. But I'm thinking if there is any other hints for this? I'm confused with sqrt(n), because if there is no special hint, why didn't they use any other parameter or number (like a, or whatever) instead of sqrt? – Kos Oct 06 '12 at 04:27
-
@Kos Well, I started with $\text{ln}{n \choose [\sqrt{n}]}=\text{ln}{(n)!}-\text{ln}(n - [\sqrt{n}])! + \text{ln}[\sqrt{n}]!$ and then use Shitikanth's expansion for all three terms, then I used the fact that $\sqrt{n}=n^{1/2}$ to help simplify at the end. – Alexander Gruber Oct 06 '12 at 04:39
-
It may just be that ${n \choose [\sqrt{n}]}$ is an important term in a lot of algorithms and that's why they put it in there. For what it's worth my answer looks right based on the plot of it: http://i.imgur.com/JOPFU.gif – Alexander Gruber Oct 06 '12 at 04:41