Stirling's approximation: Difference between revisions

From SklogWiki
Jump to navigation Jump to search
m (Added applications section.)
(Added a table)
Line 12: Line 12:


:<math>\ln N! \sim \int_1^N \ln x\,dx \sim N \ln N -N .</math>
:<math>\ln N! \sim \int_1^N \ln x\,dx \sim N \ln N -N .</math>
after some further manipulation one arrives at
:<math>N! = \sqrt{2 \pi N} \; N^{N} e^{-N} e^{\lambda_N}</math>
where
:<math>\frac{1}{12N+1} < \lambda_N < \frac{1}{12N}.</math>
For example:
{| border="1"
|-
| 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 ]](<math>10^{23}</math>) this formula is essentially  exact.
==Applications in statistical mechanics==
==Applications in statistical mechanics==
*[[Ideal gas Helmholtz energy function]]
*[[Ideal gas Helmholtz energy function]]
[[Category: Mathematics]]
[[Category: Mathematics]]

Revision as of 20:14, 4 November 2008

James Stirling (1692-1770, Scotland)

Because of Euler-MacLaurin formula

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,

after some further manipulation one arrives at

where

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 () this formula is essentially exact.

Applications in statistical mechanics