1
$\begingroup$

I'm working on a problem and began suspecting that the following inequality holds.

Let $k\in\mathbb{N}$ be fixed, and define $f(n)={n\choose k}$. Then $f(n)$ is log-concave in $n$, in particular if $N$ is fixed then for any $n\in[N]$

$$f(n)f(N-n)\leqslant f(\left[\frac{N}{2}\right])^2$$

For example, taking $N=24,n=10,k=2$,

$$LHS={10\choose 2}{14\choose 2}=4095\leqslant 4356={12\choose 2}^2=RHS.$$

I tried doing the second derivative test on $\log{f}$, but there were many ugly terms. I was wondering if there is a neater way of showing it, or if the inequality is perhaps false in general.

Thanks!

$\endgroup$
1
$\begingroup$

Well, $$ \binom{N}{k}=\frac{\Gamma(N+1)}{\Gamma(N-k+1)\Gamma(k+1)} $$ hence:

$$\frac{d^2}{dN^2}\log\binom{N}{k} = \psi'(N+1)-\psi'(N-k+1) = \sum_{n\geq 0}\left(\frac{1}{(n+N+1)^2}-\frac{1}{(n+N-k+1)^2}\right)$$ and it is not difficult to discuss convexity.

$\endgroup$
  • $\begingroup$ Thanks, yes I eventually did something similar (assuming $n\geqslant k$, of course). $\endgroup$ – youngtableaux Aug 28 '15 at 22:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.