Stirling’s Formula: Ahlfors’ Derivation

If you’re reading this blog, you probably know Stirling’s formula:

n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n

It’s not hard to estimate n! to within a factor of √2; I wrote up a note on this and even easier derivations. It’s quite a bit harder to show that the ratio of the two sides approaches a definite limit as n→∞ and that this limit is 1. You can find a variety of proofs in a number of places, one being Ahfors’ Complex Analysis.  I wrote up a note about this too, expanding on some of the details.

Incidentally, the two sides are asymptotic not just for positive integers n. Replace n! with Γ(z+1) on the left, and both n‘s with z‘s on the right. Allow z to go to infinity in the complex plane, while staying at least a fixed finite distance to the right of the imaginary axis. Then the two sides remain asymptotic. Ahfors proves this stronger result, and uses it to derive the integral form for the Γ function.

Note that if you replace the n‘s with z‘s, you have zz on the right. So you’ve got to worry about branches of the complex logarithm (since zz is defined as ez log z). The note deals with this (and other things).

John Baez has a post outlining another derivation of the full Stirling formula, using Laplace’s method. It looks a lot easier than Ahlfors’!


Filed under Analysis

2 responses to “Stirling’s Formula: Ahlfors’ Derivation

  1. Nice! I was just thinking about Stirling’s formula. Mike Stay and I were trying to come up with a quick “by hand” proof that the partition function p(n) grows faster than any polynomial, and he did it using Stirling’s formula. So I wondered how hard that was to prove. Most of these simpler approximations would do the job.

  2. Pingback: Stirling’s Formula | Azimuth

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