Stirling's approximation

James Stirling (1692-1770, Scotland)

$\left.\ln N!\right.=\ln 1+\ln 2+\ln 3+...+\ln N=\sum _{k=1}^{N}\ln k.$

Because of Euler-MacLaurin formula

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{k=1}^{N}\ln k=\int _{1}^{N}\ln x\,dx+\sum _{k=1}^{p}{\frac {B_{2k}}{2k(2k-1)}}\left({\frac {1}{n^{2k-1}}}-1\right)+R,}

where B₁ = −1/2, B₂ = 1/6, B₃ = 0, B₄ = −1/30, B₅ = 0, B₆ = 1/42, B₇ = 0, B₈ = −1/30, ... are the Bernoulli numbers, and R is an error term which is normally small for suitable values of p.

Then, for large N,

$\ln N!\sim \int _{1}^{N}\ln x\,dx\sim N\ln N-N.$

after some further manipulation one arrives at

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle N!={\sqrt {2\pi N}}\;N^{N}e^{-N}e^{\lambda _{N}}}

where

${\frac {1}{12N+1}}<\lambda _{N}<{\frac {1}{12N}}.$

For example:

N	N! (exact)	N! (Stirling)	Error (%)
5	120	118.019168	1.016
6	720	710.078185	1.014
7	5040	4980.39583	1.012
8	40320	39902.3955	1.010
9	362880	359536.873	1.009
10	3628800	3598695.62	1.008

As one usually deals with number of the order of the Avogadro constant ($10^{23}$) this formula is essentially exact.

Applications in statistical mechanics

• Ideal gas Helmholtz energy function