Stirling's approximation: Difference between revisions
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'''Stirling's approximation''' | '''Stirling's approximation''' is named after the Scottish mathematician James Stirling (1692-1770). | ||
:<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k .</math> | :<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k .</math> | ||
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:<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 | after some further manipulation one arrives at (apparently Stirling's contribution was the prefactor of <math>\sqrt{2 \pi}</math>) | ||
:<math>N! = \sqrt{2 \pi N} \; N^{N} e^{-N} e^{\lambda_N}</math> | :<math>N! = \sqrt{2 \pi N} \; N^{N} e^{-N} e^{\lambda_N}</math> |
Revision as of 17:28, 5 November 2008
Stirling's approximation is named after the Scottish mathematician James Stirling (1692-1770).
using Euler-MacLaurin formula one has
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 (apparently Stirling's contribution was the prefactor of )
where
For example:
N | N! (exact) | N! (Stirling) | Error (%) |
2 | 2 | 1.91900435 | 4.22 |
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 |
When one is dealing with numbers of the order of the Avogadro constant () this formula is essentially exact. In computer simulations the number of atoms or molecules (N) is invariably greater than 100; for N=100 the percentage error is approximately 0.083%.