 Joel French

2022-07-08

An approximation by Stirling's formula
Let $0<\alpha <1$ be a real number and αn be $\alpha n$ integer. I want to prove the following fact
$\left(\genfrac{}{}{0}{}{n}{\alpha n}\right)=\frac{1+o\left(1\right)}{\sqrt{2\pi \alpha \left(1-\alpha \right)n}}{2}^{nH\left(\alpha \right)},$
where $H\left(\alpha \right)=-\alpha {\mathrm{log}}_{2}\alpha -\left(1-\alpha \right){\mathrm{log}}_{2}\left(1-\alpha \right)$. To do this, I've used the Stirling's formula for the factorial:
$n!=\left(\frac{n}{e}{\right)}^{n}\sqrt{2\pi n}{e}^{r\left(n\right)},$,
where $1/\left(12n+1\right).
What I've tried:
$\left(\genfrac{}{}{0}{}{n}{\alpha n}\right)=\frac{n!}{\left(\alpha n\right)!\left(\left(1-\alpha \right)n\right)!}=\frac{\left(\frac{n}{e}{\right)}^{n}\sqrt{2\pi n}{e}^{r\left(n\right)}}{\left(\frac{\alpha n}{e}{\right)}^{\alpha n}\sqrt{2\pi \alpha n}{e}^{r\left(\alpha n\right)}\left(\frac{\left(1-\alpha \right)n}{e}{\right)}^{\left(1-\alpha \right)n}\sqrt{2\pi \left(1-\alpha \right)n}{e}^{r\left(\left(1-\alpha \right)n\right)}}\phantom{\rule{0ex}{0ex}}\frac{{e}^{r\left(n\right)-r\left(\alpha n\right)-r\left(\left(1-\alpha \right)n\right)}}{\left({\alpha }^{\alpha }\left(1-\alpha {\right)}^{\left(1-\alpha \right)}{\right)}^{n}\sqrt{2\pi \alpha \left(1-\alpha \right)n}}=\frac{{e}^{r\left(n\right)-r\left(\alpha n\right)-r\left(\left(1-\alpha \right)n\right)}}{\sqrt{2\pi \alpha \left(1-\alpha \right)n}}{2}^{nH\left(\alpha \right)}.$
Now how can I show that ${e}^{r\left(n\right)-r\left(\alpha n\right)-r\left(\left(1-\alpha \right)n\right)}=1+o\left(1\right)$? poquetahr

Expert

Step 1
You are only having difficulty because the version of Stirling you're using is too precise. Instead, use $n!=\left(1+o\left(1\right)\right)\sqrt{2\pi n}\left(\frac{n}{e}{\right)}^{n}$ - replacing all of your ${e}^{r\left(n\right)}$ factors by $1+o\left(1\right)$.
Step 2
The step you're stuck on, where you're trying to show that ${e}^{r\left(n\right)-r\left(\alpha n\right)-r\left(\left(1-\alpha \right)n\right)}=1+o\left(1\right)$, becomes $\frac{1+o\left(1\right)}{\left(1+o\left(1\right)\right)\left(1+o\left(1\right)\right)}=1+o\left(1\right)$, which is straightforward.
(Technically, the $1+o\left(1\right)$ factor we get from Stirling's approximation is $1+{o}_{n\to \mathrm{\infty }}\left(1\right)$, so we also have to observe that $1+{o}_{\alpha n\to \mathrm{\infty }}\left(1\right)$ and $1+{o}_{\left(1-\alpha \right)n\to \mathrm{\infty }}\left(1\right)$ are the same asymptotically. This is fine as long as $\alpha$ is a constant, or even more generally than that.) Waldronjw

Expert

Step 1
Since $f\left(n\right)=o\left(1\right)$ just means $\frac{f\left(n\right)}{1}\to 0$ as $n\to \mathrm{\infty }$ (i.e. $f\left(n\right)=1+o\left(1\right)$ means $f\left(n\right)\to 1$), our goal is to show that ${e}^{r\left(n\right)-r\left(\alpha n\right)-r\left(\left(1-\alpha \right)n\right)}\to 1$, which can be done by showing $r\left(n\right)-r\left(\alpha n\right)-r\left(\left(1-\alpha \right)n\right)\to 0$. We can use the bounds on r(n) to show that $r\left(n\right)-r\left(\alpha n\right)-r\left(\left(1-\alpha \right)n\right)<\frac{1}{12n}-\frac{1}{12\alpha n+1}-\frac{1}{12\left(1-\alpha \right)n+1}\underset{n\to \mathrm{\infty }}{\to }0$ and $r\left(n\right)-r\left(\alpha n\right)-r\left(\left(1-\alpha \right)n\right)>\frac{1}{12n+1}-\frac{1}{12\alpha n}-\frac{1}{12\left(1-\alpha \right)n}\underset{n\to \mathrm{\infty }}{\to }0$
Step 2
So since $r\left(n\right)-r\left(\alpha n\right)-r\left(\left(1-\alpha \right)n\right)$ is squeezed between 0 on both sides asymptotically, indeed we have $r\left(n\right)-r\left(\alpha n\right)-r\left(\left(1-\alpha \right)n\right)\to 0$, and we are done.

Do you have a similar question?