I have a recursive function f(n) with time complexity
$$O(f(n)) = O\left(\binom{n}{n/2} \cdot f\left(\frac{n}{2}\right)^2\right)$$
I tried to simplify it, but don't have enough mathematical skills. The only thing that I found out is that
$$ \binom{n}{n/2} = 2^n \cdot \frac{\Gamma\left(\frac{n}{2} + \frac{1}{2}\right)}{\sqrt{\pi} \cdot \Gamma(\frac{n}{2} + 1)}$$
I have allready solved it.
$\binom{n}{\frac{n}{2}} = 2^n \cdot \frac{\Gamma\left(\frac{n}{2} + \frac{1}{2}\right)}{\sqrt{\pi} \cdot \Gamma\left(\frac{n}{2} + 1\right)}$
As you can see here: https://www.wolframalpha.com/input/?i=binom(n,+n%2F2)
Now we define function
$a(n) := \frac{\Gamma\left(\frac{n}{2} + \frac{1}{2}\right)}{\sqrt{\pi} \cdot \Gamma\left(\frac{n}{2} + 1\right)}$
Because gamma function is growing function for every real number bigger than one, than $0 < a(n) < 1$. If we derive this function, as you can see here
https://www.wolframalpha.com/input/?i=derive(gamma(n%2F2+%2B+1%2F2)%2F(gamma(n%2F2+%2B+1)*sqrt(pi)))
we can see that derivation is negative, but very fast is very close to zero, so the function $a(n)$ is decreasing, but very slovly and in compare to function $2^n$, we can say, that it is a constant. So we have simplified the problem to
$O(f(n)) = O\left(2^n \cdot \left(f(\frac{n}{2})\right)^2\right) = O\left(2^n \cdot \left(\prod\limits_{i=1}^{Log_2(n)} 2^{n\cdot2^{-i}}\right)^2\right)$
And beacause
$ \prod\limits_{i=1}^{Log_2(n)} 2^{n\cdot2^{-i}} = 2^{n-1}$
As you can see here: https://www.wolframalpha.com/input/?i=product(2%5E(n*2%5E(-i))),+i+%3D+1+to+log_2(n)
Than we can make result
$O(f(n)) = O\left(2^n \cdot \left(2^{n-1}\right)^2\right) = O\left(2^n \cdot 2^{2n-2}\right) = O\left(2^{4n-2}\right) = O\left(2^n\right)$
So we can see that final complexity is exponential