Stirling's approximation - Wikipedia, the free encyclopedia - 0 views
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Matthew Leingang on 20 Jul 09Far More Than You Ever Wanted To Know about where this formula comes from.
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Matthew Leingang on 20 Jul 09Stirling's approximation is for n! in terms of continuous functions.
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Marc Tourangeau on 21 Jul 09The derivation didn't require any complex analysis after all; thank goodness.