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# Difference between revisions of "Stirling's approximation"

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:<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 [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> | ||

+ | |||

+ | 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 [[Bernoulli numbers]], and ''R'' is an error term which is normally small for suitable values of ''p''. | ||

:<math>~\approx \int_1^N \ln x dx</math> | :<math>~\approx \int_1^N \ln x dx</math> |

## Revision as of 15:36, 28 March 2007

James Stirling (1692-1770, Scotland)

Because of [Euler-MacLaurin formula]

where *B*_{1} = −1/2, *B*_{2} = 1/6, *B*_{3} = 0, *B*_{4} = −1/30, *B*_{5} = 0, *B*_{6} = 1/42, *B*_{7} = 0, *B*_{8} = −1/30, ... are the Bernoulli numbers, and *R* is an error term which is normally small for suitable values of *p*.

Thus, for large *N*