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Difference between revisions of "Stirling's approximation"

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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 15:48, 28 March 2007

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! \approx \int_1^N \ln x dx = N \ln N -N