Stirling's approximation

Stirling's approximation is named after the Scottish mathematician James Stirling (1692-1770).

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

using Euler-MacLaurin formula one has

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,

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \ln N!\sim \int _{1}^{N}\ln x\,dx\sim N\ln N-N.}

after some further manipulation one arrives at (apparently Stirling's contribution was the prefactor of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\sqrt {2\pi }}} )

$N!={\sqrt {2\pi N}}\;N^{N}e^{-N}e^{\lambda _{N}}$

where

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

For example:

N	N! (exact)	N! (Stirling)	Error (%)
2	2	1.91900435	4.22
3	6	5.83620959	2.81
4	24	23.5061751	2.10
5	120	118.019168	1.67
6	720	710.078185	1.40
7	5040	4980.39583	1.20
8	40320	39902.3955	1.05
9	362880	359536.873	0.93
10	3628800	3598695.62	0.84

When one is dealing with numbers of the order of the Avogadro constant ($10^{23}$) this formula is essentially exact. In computer simulations the number of atoms or molecules (N) is invariably greater than 100; for N=100 the percentage error is approximately 0.083%.

Applications in statistical mechanics

• Ideal gas Helmholtz energy function