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Difference between revisions of "Stirling's approximation"

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| N || N! (exact) || N! (Stirling)  || Error (%)
 
| N || N! (exact) || N! (Stirling)  || Error (%)
 
|-  
 
|-  
|5 ||  120   || 118.019168 || 1.016
+
|3 ||  6   || 5.83620959 || 2.81
 
|-  
 
|-  
||| 720 || 710.078185 || 1.014
+
|4 ||  24  || 23.5061751 || 2.10
 
|-  
 
|-  
||| 5040 || 4980.39583  || 1.012
+
|5 ||  120  || 118.019168 || 1.67
 
|-  
 
|-  
|8 ||  40320 ||   39902.3955 || 1.010
+
|6 || 720 || 710.078185 || 1.40
 
|-  
 
|-  
|9 ||   362880|| 359536.873  || 1.009
+
|7 || 5040  || 4980.39583  || 1.20
 
|-  
 
|-  
|10  || 3628800  ||  3598695.62  || 1.008
+
|8  ||  40320 ||  39902.3955 || 1.05
 +
|-
 +
|9  ||  362880||  359536.873  || 0.93
 +
|-
 +
|10  || 3628800  ||  3598695.62  || 0.84
 
|}
 
|}
  
As one usually deals with number of the order of the [[Avogadro constant ]](<math>10^{23}</math>) this formula is essentially  exact.
+
As one usually deals with number of the order of the [[Avogadro constant ]](<math>10^{23}</math>) this formula is essentially  exact.
 +
In [[Computer simulation techniques | computer simulations]] the number of atoms or molecules (N) is invariably greater than 100, where the
 +
percentage error is less than .  
 
==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 14:08, 5 November 2008

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

\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,

\ln N! \sim \int_1^N \ln x\,dx \sim N \ln N -N .

after some further manipulation one arrives at

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 (%)
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

As one usually deals with number 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, where the percentage error is less than .

Applications in statistical mechanics