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 $f$

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

  • 1
    $\begingroup$ 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) $\endgroup$ Nov 11, 2018 at 9:10
  • 6
    $\begingroup$ 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 $n\to\infty$ $\endgroup$ Nov 11, 2018 at 9:52

1 Answer 1


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 $c=a^{-1}(\alpha)$.


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