Which changes faster $ f(m) = {n \choose m}$ or $ g(m) = n^{-2c \frac{mn - m^2}{n-1}}$ wrt $m$ I have two functions of $m$. $m$ and $n$ are both integers with $m <n$.
The functions are :-
$$ f(m) = {n \choose m}$$ and
$$ g(m) = n^{-2c \frac{mn - m^2}{n-1}}$$
$c$ is a constant which you may take to be 1.
I wish to show that as $m$ varies from $1$ to $n/2$ , $g(m)$ decreases at a faster rate than $f(m)$ grows. (My goal is to show that f(1)g(1)) is a maxima. )
To do so I am trying to find the derivative of the two . The derivative which is larger in magnitude will indicate a greater rate of change. Is my approach correct ? I am unable to find a derivative for $f(m)$. I was thinking of bounding the binomial coefficient too.
Also $n$ is very large.
 A: Use logarithms. Considering that $n\gg1$, we have (using Stirling's approximation) $$\ln f(m)=n\ln n-m\ln m-(n-m)\ln(n-m)+O(\ln m)$$
and
$$\ln g(m)=-2c\ln n\frac{mn-m^2}{n-1}.$$
The derivatives with respect to $m$ are
$$\frac{\mathrm d\ln f}{\mathrm dm}=\ln\frac{n-m}m+O(1),\qquad
\frac{\mathrm d^2\ln f}{\mathrm dm^2}=-\frac{n}{m(n-m)}+O(1)$$
$$\frac{\mathrm d\ln g}{\mathrm dm}=-2c\ln n\;\frac{n-2m}{n-1},\qquad
\frac{\mathrm d^2\ln g}{\mathrm dm^2}=4c\frac{\ln n}{n-1}.$$
Then you see that
$$\frac{\mathrm d^2\ln(fg)}{\mathrm dm^2}=\frac{\mathrm d^2\ln g}{\mathrm dm^2}+\frac{\mathrm d^2\ln g}{\mathrm dm^2}$$
cancels for $m=m_0$ of the order of $-1/(4c\ln n)$. ( EDIT I made a mistake in my original answer. The value of $m_0$ is the one that is here now.)
It follows that $\frac{\mathrm d^2\ln(fg)}{\mathrm dm^2}$ is positive for $m>0$. Since $\frac{\mathrm d\ln(fg)}{\mathrm dm}=0$ for $m=n/2$, it means that $\frac{\mathrm d\ln(fg)}{\mathrm dm}\leq0$ for $0\leq m\leq n/2$ and then that your claim is correct : $fg$ is maximum for $m=1$ (actually for $m=0$, but you seem to omit this value.)
