Stirling's approximation: Difference between revisions

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where ''B''<sub>1</sub> = &minus;1/2, ''B''<sub>2</sub> = 1/6, ''B''<sub>3</sub> = 0, ''B''<sub>4</sub> = &minus;1/30, ''B''<sub>5</sub> = 0, ''B''<sub>6</sub> = 1/42, ''B''<sub>7</sub> = 0, ''B''<sub>8</sub> = &minus;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> = &minus;1/2, ''B''<sub>2</sub> = 1/6, ''B''<sub>3</sub> = 0, ''B''<sub>4</sub> = &minus;1/30, ''B''<sub>5</sub> = 0, ''B''<sub>6</sub> = 1/42, ''B''<sub>7</sub> = 0, ''B''<sub>8</sub> = &minus;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''.


Then
Then, for large ''N'',


:<math>\ln N! \approx \int_1^N \ln x dx = N \ln N -N +1</math>
:<math>\ln N! \approx \int_1^N \ln x dx = N \ln N -N</math>
 
Thus, for large ''N''
 
:<math>\ln N! \approx  N \ln N -N</math>
[[Category: Mathematics]]
[[Category: Mathematics]]

Revision as of 14:48, 28 March 2007

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,