the asymptotic growth of n choose floor(n/2)

Question

How can I find the asymptotic growth of n choose floor(n/2) ? I tried to use the expansion and got that it is equal to

[n*(n-1)*........*(floor(n/2)+1)] / (n-floor(n/2))!

Any idea how can i go from there? Any help is appreciated, prefer hints over answers

Try using [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation). — Gassa, Sep 12 '14 at 17:19
This question appears to be off-topic because it is about math, not programming, and, as such, belongs on math.SE — sds, Sep 12 '14 at 17:26
@sds: I'm tempted to say that this question is general enough and the method and result behind your answer are important enough in programming that it belongs just fine on SO. I'd keep it open. (I wouldn't say the same about many other similar "math questions"---this one gets special treatment from me because the asymptotics of binomial coefficients lie at the heart of a *lot* of important stuff.) — tmyklebu, Sep 12 '14 at 19:23
@tmyklebu: I agree with the facts you state but not with your conclusions ;-) — sds, Sep 12 '14 at 19:26

score 11 · Answer 1 · edited Feb 28 '20 at 18:47

I agree with the answer above but would like to provide more depth. Assuming n is even, we have:

$\binom{n}{n/2} \leq \frac{n!}{\left ( \frac{n}{2}! \right )^2}$

To upper bound this, we use the upper bound of Stirling's Approximation in the numerator and the lower bound in the denominator (e.g. we want the largest numerator and smallest denominator). This will give us the upper bound:

$\binom{n}{n/2} \leq \frac{e\cdot n^{n+\frac12}\cdot e^{-n}}{\left (\sqrt{2\pi}\cdot \left (\frac{n}{2} \right )^{\frac{n+1}{2}}\cdot e^{-\frac{n}{2}}\right )^2}$

We then distribute the exponent in the denominator to get:

$\binom{n}{n/2} \leq \frac{e\cdot n^{n+\frac12}}{2\pi\cdot \left (\frac{n}{2} \right )^{n+1}}$

Cancel out $e^{-n}$ , move $2^{n+1}$ from denominator to numerator and simplify; we get:

$\binom{n}{n/2} \leq \frac{e}{\pi} \cdot \frac{2^n}{\sqrt{n}}$

Follow the same process with the lower bound, put Stirling's approx upper bound in the denominator, and lower bound in the numerator. This will yield:

$\binom{n}{n/2} \geq \frac{2 \sqrt{2\pi}}{e^2} \cdot \frac{2^n}{\sqrt{n}}$

We then know it's lower bounded by some constant times $\frac{2^n}{\sqrt{n}}$ and it's upper bounded by another constant times $\frac{2^n}{\sqrt{n}}$ .

Thus, we conclude its asymptotic growth is $\binom{n}{n/2} \in \Theta \left (\frac{2^n}{\sqrt{n}} \right )$ .

Personally I can hardly read the equations due to stackoverflow dark mode, and the suggested edit queue is full. Can someone do something about this? — J. Schmidt, Oct 19 '22 at 12:01

sds · Accepted Answer · 2014-09-12T17:33:25.427

3

Using Stirling's approximation, you get

n! = \sqrt{2n\pi}(n/e)^n

If you substitute it into $\choose{n}{n/2}$, you should eventually end up with

2^{n+1/2}/\sqrt{n\pi}

PS. you might want to check my math before you actually use the answer :-)

edited Sep 12 '14 at 17:33

answered Sep 12 '14 at 17:26

sds

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thanks :) you gave me a way and an answer to look up too ! More than i can wish for :-) Definitely checking the math tho :) – hussein hammoud Sep 12 '14 at 17:37
for your answer u assumed that n is even right? and substituted floor(n/2) by n/2 ?? thx – hussein hammoud Sep 12 '14 at 17:45
1

we are talking about asymptotics anyway! – sds Sep 12 '14 at 17:46

the asymptotic growth of n choose floor(n/2)

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