0

I'm reading the following document. On page 2 we have the following formula:

$$\frac{N!}{(pN)!((1-p)N)!}\sim \frac{N^N}{(pN)^{pN}((1-p)N)^{((1-p)N)}}$$

It seems that $n!$ is being approximated with $n^n$, which is obviously false. However, I guess the following approximation works $\frac{m!}{n!}=\frac{m^m}{n^n}$. Any ideas why?

  • 2
    Most likely Stirling's approximation: https://en.wikipedia.org/wiki/Stirling%27s_approximation – DKS Jun 01 '18 at 12:13
  • @DKS- Using Stirling's approximation gives us a $\sqrt{N}$ factor in the denominator, which is not there in the formula above –  Jun 01 '18 at 12:23
  • 1
    This is true, but is a common abuse in physics. The $\sqrt{N}$ factor does not contribute too much as $N$ gets very large, so it is often dropped. I won't condone that practice, but my experience in physics during undergrad tells me that what is likely being done. – DKS Jun 01 '18 at 12:25

2 Answers2

0

Your approximation does not quite work, but the approximation in question should follow from Stirling's formula.

Kusma
  • 3,979
0

Recall Stirling's approximation $$ n!\approx n^ne^{-n}\sqrt{2\pi n}$$ so that we can approximate $\frac{n!}{m!k!}$ where $m+n=k$ as $$ \frac{n!}{m!k!}\approx\frac{n^ne^{-n}\sqrt{2\pi n}}{m^me^{-m}\sqrt{2\pi m}\cdot k^ke^{-k}\sqrt{2\pi k}}=\frac{n^n}{m^mk^k}\cdot \sqrt{\frac{n}{\pi mk}}$$ The error factor $\sqrt{\frac{n}{\pi mk}}$ can vary from $\approx \frac1{\sqrt\pi}$ when $m=1$ or $k=1$ to $\approx\sqrt{\frac{4}{\pi n}}$ when $m=k=\frac n2$.

Hagen von Eitzen
  • 374,180