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 (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln N! \sim \int_1^N \ln x\,dx \sim N \ln N -N .}

after some further manipulation one arrives at

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N! = \sqrt{2 \pi N} \; N^{N} e^{-N} e^{\lambda_N}}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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