Stirling's approximation: Difference between revisions
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James Stirling (1692-1770, Scotland) | James Stirling (1692-1770, Scotland) | ||
:<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> | ||
Because of [http://en.wikipedia.org/wiki/Euler-Maclaurin_formula Euler-MacLaurin formula] | Because of [http://en.wikipedia.org/wiki/Euler-Maclaurin_formula Euler-MacLaurin formula] | ||
:<math>\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</math> | :<math>\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 ,</math> | ||
where ''B''<sub>1</sub> = −1/2, ''B''<sub>2</sub> = 1/6, ''B''<sub>3</sub> = 0, ''B''<sub>4</sub> = −1/30, ''B''<sub>5</sub> = 0, ''B''<sub>6</sub> = 1/42, ''B''<sub>7</sub> = 0, ''B''<sub>8</sub> = −1/30, ... are the [http://en.wikipedia.org/wiki/Bernoulli_numbers Bernoulli numbers], and ''R'' is an error term which is normally small for suitable values of ''p''. | where ''B''<sub>1</sub> = −1/2, ''B''<sub>2</sub> = 1/6, ''B''<sub>3</sub> = 0, ''B''<sub>4</sub> = −1/30, ''B''<sub>5</sub> = 0, ''B''<sub>6</sub> = 1/42, ''B''<sub>7</sub> = 0, ''B''<sub>8</sub> = −1/30, ... are the [http://en.wikipedia.org/wiki/Bernoulli_numbers Bernoulli numbers], and ''R'' is an error term which is normally small for suitable values of ''p''. | ||
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Then, for large ''N'', | Then, for large ''N'', | ||
:<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> | ||
[[Category: Mathematics]] | [[Category: Mathematics]] | ||
Revision as of 14:50, 28 March 2007
James Stirling (1692-1770, Scotland)
- 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 \left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k .}
Because of Euler-MacLaurin formula
- Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 .}