I am currently analyzing the complexity of an algorithm and I need to know if this binomial coefficient can be bounded by a polynomial, by an exponential function or even worse by something involving a factorial.
Whats the smallest function $f$ such that $\binom{m}{m/2} = \frac{m!}{(m/2)!^2} < f(m)$ holds, where $m \in \mathbb{N}$ and $m$ is an even number?
There is no smallest function satisfying the condition: If ${m \choose {m / 2}} < f(m)$, then $g(m) := \frac{1}{2}\left({m \choose {m / 2}} + f(m)\right)$ satisfies ${m \choose {m / 2}} < g(m) < f(m)$, i.e., is a better bound.
We can still say something meaningful, however: Stirling's approximation gives that $$n! = \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \left(1 + \frac{1}{12 n} + O\left(\frac{1}{n^2}\right)\right),$$ and substituting gives that for large $m$ we have $${m \choose {m / 2}} = \frac{m!}{\left(\frac{m}{2}\right)!^2} = \sqrt{\frac{2}{\pi m}} 2^m \left(1 - \frac{4}{m} + O\left(\frac{1}{m^2}\right)\right) .$$ So, for large $m$, $${m \choose {m / 2}} < \sqrt{\frac{2}{\pi m}} 2^m ,$$ and more careful analysis shows that this inequality holds for all positive integers $m$.
$$ \binom{2n}{n} < \frac{4^n}{\sqrt{\pi n}} $$
which is asymptotically tight
– HallaSurvivor Jul 02 '22 at 19:28