# Interesting behaviour of binomial coefficients

Let $$\binom{n}{k}:=\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(1-k+n)}$$ be the generalized binomial coefficient then I noticed by playing around with Mathematica that the function $$f:[0,n/2] \rightarrow \mathbb R$$

$$f(x) = \log\left(\binom{n}{n/2+x} \right)-n \alpha(1-(2x/n)^2)$$

has very interesting properties.

For $$\alpha\le \tfrac{1}{2}$$ the global maximum of this function is attained at $$x_n=0$$ for sufficiently large $$n.$$

Yet, for $$\alpha>\tfrac{1}{2}$$ the global maximum seems to be attained at some $$x_n>0.$$

My question is: Is it possible to analytically verify this property $$(x_n>0)$$ and in particular can anybody shed some light on what is so special about $$\alpha =1/2$$? (The first comment below this thread seems to indicate already that $$\alpha=1/2$$ is indeed the right threshold)

In particular, Mathematica gave me the following equation for the derivative of $$f$$

$$f'(x)= \frac{8 \alpha x}{n} + \operatorname{HarmonicNumber}(n/2 - x) - \operatorname{HarmonicNumber}(n/2 + x).$$

• Expanding $f'(x)$ to $O(n^{-3})$ gives $f'(x)=8 x (\alpha-\frac{1}{2})\ n^{-1}+ 4\ x\ n^{-2} -\frac{8}{3}\ x \ (1+2 x^2)\ n^{-3}+O(n^{-4})$. We get three solutions of $f'(x)=0$: $x=0$ and two other solutions, which are real for $\alpha\ge \alpha_0$. The critical alpha is $\alpha_0= \frac{1}{2}-\frac{1}{2n}+\frac{1}{3 n^2}+O(n^{-3})$. (all calculations done with Mathematica 11) Nov 11, 2018 at 9:10
• I think that you observed the de Moivre's local limit theorem that states $$2^{-n/2-x}\binom{n}{n/2+x}\sim\frac{1}{\sqrt{n\pi/2}} e^{-2x^2/n}$$ as $n\to\infty$ Nov 11, 2018 at 9:52

We have $$\begin{equation} f'(x)=\frac{8 \alpha x}{n}+\psi\left(\frac{n}{2}-x+1\right)-\psi\left(\frac{n}{2}+x+1\right), \end{equation}$$ where $$\psi:=(\ln\Gamma)'$$. By the Gauss formula, Theorem 1.6.1, page 26 $$\begin{equation} \psi(x)=\int_0^\infty\Big(\frac{e^{-z}}{z}-\frac{e^{-xz}}{1-e^{-z}}\Big)dz, \end{equation}$$ we have $$\begin{equation} f'(x)=\frac{8 \alpha x}{n}-2\int_0^\infty\frac{dz}{1-e^{-z}}\,e^{-(1+n/2)z}\sinh xz. \end{equation}$$ Since $$\sinh$$ is strictly convex on $$(0,\infty)$$, $$f'$$ is strictly concave on $$(0,n/2)$$. Also, $$f'(0)=0$$ and $$f'(\frac n2)\to-\infty$$ as $$n\to\infty$$. So, eventually (i.e., for all large enough $$n$$), either $$f'<0$$ on $$(0,n/2)$$ (which will be the case if $$f''(0)\le0$$) or $$f'$$ changes sign only once on $$(0,n/2)$$, from $$+$$ to $$-$$, at some $$x_n\in(0,n/2)$$. So, either $$f$$ is decreasing on $$(0,n/2)$$ (which will be the case if $$f''(0)\le0$$) or increasing-decreasing, attaining its only maximum at some $$x_n\in(0,n/2)$$.
As $$x\downarrow0$$, $$\begin{equation} f'(x)=\Big(\frac{8\alpha}{n}-2\psi'(n/2+1)\Big)x+O(x^3), \end{equation}$$ and $$\begin{equation} \frac{8\alpha}{n}-2\psi'(n/2+1)=\frac{4\alpha-2}{n}+\frac2{n^2}+O(n^{-3}) \end{equation}$$ as $$n\to\infty$$.
So, if $$\alpha<1/2$$, then $$f''(0)<0$$ eventually, and so, $$f$$ is decreasing on $$(0,n/2)$$, with the maximum at $$0$$.
If $$\alpha=1/2$$, then $$f'(c\sqrt n)=\left(4 c-\frac{16 c^3}{3}\right) \left(\frac{1}{n}\right)^{3/2}+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$$ for each real $$c>0$$ as $$n\to\infty$$, and so, eventually $$f$$ attains its only maximum at some $$x_n\sim c\sqrt n$$, where $$c=\sqrt3/2$$.
Finally, if $$\alpha>1/2$$ and $$c\in(0,1/2)$$, then $$f'(cn)=g(c)+O(1/n)$$ as $$n\to\infty$$, where $$g(c):=8 \alpha c+\ln \frac{1-2 c}{2 c+1}$$ for each real $$c>0$$. We have $$g(c)=0$$ iff $$\alpha=a(c):=-\frac1{8c}\,\ln\frac{1-2 c}{2 c+1}$$. Note that $$a(c)$$ increases from $$1/2$$ to $$\infty$$ as $$c$$ increases from $$0$$ to $$1/2$$. So, eventually $$f$$ attains its only maximum at some $$x_n\sim cn$$, where $$c=a^{-1}(\alpha)$$.