# Stirling’s approximation

Stirling’s formula gives an approximation for $n!$, the factorial . It is

 $n!\approx\sqrt{2n\pi}n^{n}e^{-n}$

We can derive this from the gamma function. Note that for large $x$,

 $\Gamma(x)=\sqrt{2\pi}x^{x-\frac{1}{2}}e^{-x+\mu(x)}$ (1)

where

 $\mu(x)=\sum_{n=0}^{\infty}\left(x+n+\frac{1}{2}\right)\ln\left(1+\frac{1}{x+n}% \right)-1=\frac{\theta}{12x}$

with $0<\theta<1$. Taking $x=n$ and multiplying by $n$, we have

 $n!=\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n+\frac{\theta}{12n}}$ (2)

Taking the approximation for large $n$ gives us Stirling’s formula.

There is also a big-O notation version of Stirling’s approximation:

 $n!=\left(\sqrt{2\pi n}\right)\left(\frac{n}{e}\right)^{n}\left(1+\mathcal{O}% \left(\frac{1}{n}\right)\right)$ (3)

We can prove this equality starting from (2). It is clear that the big-O portion of (3) must come from $e^{\frac{\theta}{12n}}$, so we must consider the asymptotic behavior of $e$.

First we observe that the Taylor series for $e^{x}$ is

 $e^{x}=1+\frac{x}{1}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots$

But in our case we have $e$ to a vanishing exponent. Note that if we vary $x$ as $\frac{1}{n}$, we have as $n\longrightarrow\infty$

 $e^{x}=1+\mathcal{O}\left(\frac{1}{n}\right)$

We can then (almost) directly plug this in to (2) to get (3) (note that the factor of 12 gets absorbed by the big-O notation.)

 Title Stirling’s approximation Canonical name StirlingsApproximation Date of creation 2013-03-22 12:00:36 Last modified on 2013-03-22 12:00:36 Owner drini (3) Last modified by drini (3) Numerical id 22 Author drini (3) Entry type Theorem Classification msc 68Q25 Classification msc 30E15 Classification msc 41A60 Synonym Stirling’s formula Synonym Stirling’s approximation formula Related topic MinkowskisConstant Related topic AsymptoticBoundsForFactorial