Stirling's approximation

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James Stirling (1692-1770, Scotland)

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 (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 10^{23}} ) this formula is essentially exact.

Applications in statistical mechanics