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 $x$,

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

where

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

with . 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:

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

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

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

$$e^x=1+𝒪\left(\frac{1}{n}\right)$$

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