Stirling's approximation
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.