问题
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
回答1:
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 :-)
回答2:
I agree with the answer above but would like to provide more depth. Assuming n is even, we have:
n choose n/2 = n! / ((n/2)!)^2
To upper bound this, we use the upper bound of Using 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:
n choose n/2 <= [e * n^(n + 1/2) * e^-n] / [sqrt(2*pi) * (n/2)^(n/2 + 1/2) * e^-n/2]^2
We then distribute the exponent in the denominator to get:
n choose n/2 <= [e * n^(n + 1/2) * e^-n] / [2*pi * (n/2)^(n+1) * e^-n]
Cancel out e^-n, move 2^(n+1) from denominator to numerator and simplify; we get:
n choose n/2 <= (e / pi) * [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:
n choose n/2 >= (2 * sqrt(2*pi) / e^2) * [2^n / sqrt(n)]
We then know it's lower bounded by some constant times 2^n / sqrt(n) and it's upper bounded by another constant times 2^n / sqrt(n).
Thus, we conclude its asymptotic growth is n choose n/2 = Theta( 2^n / sqrt(n) ).
来源：https://stackoverflow.com/questions/25813440/the-asymptotic-growth-of-n-choose-floorn-2