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Stirling's Formula Is a Logarithmic Asymptotic Expansion by Vicfred

Last updated: Wed 08 November 2017

tags: mathematics, special functions, asymptotic analysis, distributions

Factorials grow like square root 2 pi n times n over e to n. I will separate the object being defined from the consequence being claimed.

Set-up

Special functions extend \(n!\), integrals, and differential equations beyond integer parameters. Asymptotic notation \(f(x)\sim g(x)\) means their ratio tends to one.

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

The typography mirrors the proof: first declare \(\mathsf D\), then state \(\mathsf C\). The symbol \(\Longrightarrow\) below is a logical dependency, not extra mathematical structure.

$$ \log n!=n\log n-n+\frac12\log(2\pi n)+O(1/n) $$

The calculation

This is the algebraic core of the note. Once this line is correct, the surrounding interpretation has something solid to refer to.

$$ \log n!=\left(n+\frac12\right)\log n-n+\frac12\log(2\pi)+\frac1{12n}-\frac1{360n^3}+\cdots $$

What survives abstraction

The aligned summary deliberately puts the datum and conclusion on different rows. Mathematically, this is the distinction between specifying an object and proving a property of it.

$$ \begin{aligned} \mathsf{D}\;&:\quad n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n,\\[5pt] \mathsf{C}\;&:\quad \log n!=n\log n-n+\frac12\log(2\pi n)+O(1/n). \end{aligned} $$

The boundary

An asymptotic expansion need not converge. Truncating near the smallest term can be useful even when the infinite series diverges.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \log n!=n\log n-n+\frac12\log(2\pi n)+O(1/n) \end{gathered}} $$

This is enough machinery for one note: an exact object, a worked case, a structural law, and a clearly marked boundary. Each layer can now be tested independently.

This article was posted on Tue 24 December 2013. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.