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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>
  
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Because of [Euler-MacLaurin formula]
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:<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>
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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 [[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)

\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.

~\approx \int_1^N \ln x dx


~= \left[ x \ln x - x \right]_1^N


~= N \ln N -N +1

Thus, for large N

\ln N! \approx  N \ln N -N