Stirling's Approximation

Stirling's approximation gives an approximate value for the Factorial function $n!$ or the Gamma Function $\Gamma(n)$ for $n\gg 1$. The approximation can most simply be derived for $n$ an Integer by approximating the sum over the terms of the Factorial with an Integral, so that

$$\ln n! = \ln 1+\ln 2+\ldots+\ln n= \sum_{k=1}^n \ln k \approx \int_1^n \ln x\,dx$$

$$= [x\ln x-x]^n_1 = n\ln n-n+1 \approx n\ln n-n.$$ (1)

The equation can also be derived using the integral definition of the Factorial,

$$n!=\int_0^\infty e^{-x} x^n\,dx.$$ (2)

Note that the derivative of the Logarithm of the integrand can be written

$${d\over dx} \ln(e^{-x}x^n) = {d\over dx} (n\ln x-x) = {n\over x}-1.$$ (3)

The integrand is sharply peaked with the contribution important only near $x=n$. Therefore, let $x\equiv n+\xi$ where $\xi\ll n$, and write

$$\ln(x^n e^{-x})=n\ln x-x=n\ln(n+\xi)-(n+\xi).$$ (4)

Now,

$$\ln(n+\xi) = \ln\left[{n\left({1+{\xi\over n}}\right)}\right]= \ln n+\ln\left({1+{\xi\over n}}\right)$$

$$= \ln n+{\xi\over n}-{1\over 2} {\xi^2\over n^2}+\ldots,$$ (5)

so

$$\ln(x^ne^{-x}) = n\ln (n+\xi)-(n+\xi)$$

$$= n\ln n+\xi-{1\over 2} {\xi^2\over n}-n-\xi+\ldots$$

$$= n\ln n-n-{\xi^2\over 2n}+\ldots.$$ (6)

Taking the Exponential of each side then gives

$$x^n e^{-x} \approx e^{n\ln n} e^{-n}e^{-\xi^2/2n} = n^n e^{-n}e^{-\xi^2/2n}.$$ (7)

Plugging into the integral expression for $n!$ then gives

$$n! \approx \int_{-n}^\infty n^n e^{-n} e^{-\xi^2/2n}\, d\xi \approx n^ne^{-n} \int_{-\infty}^\infty e^{-\xi^2/2n} \, d\xi$$

$$= n^n.$$ (8)

Evaluating the integral gives

$$n! \approx n^ne^{-n}\sqrt{2\pi n},$$ (9)

$$\approx \sqrt{2\pi}\,n^{n+1/2} e^{-n}.$$ (10)

Taking the Logarithm of both sides then gives

$$\ln n!\approx n\ln n-n+{\textstyle{1\over 2}}\ln(2\pi n) = (n+{\textstyle{1\over 2}})\ln n-n+{\textstyle{1\over 2}}\ln(2\pi).$$ (11)

This is Stirling's Series with only the first term retained and, for large $n$, it reduces to Stirling's approximation

$$\ln n!\approx n\ln n-n.$$ (12)

Gosper notes that a better approximation to $n!$ (i.e., one which approximates the terms in Stirling's Series instead of truncating them) is given by

$$n!\approx \sqrt{(2n+{\textstyle{1\over 3}})\pi}\,n^n e^{-n}.$$ (13)

This also gives a much closer approximation to the Factorial of 0, $0!=1$, yielding $\sqrt{\pi/3}\approx 1.02333$ instead of 0 obtained with the conventional Stirling approximation.

See also Stirling's Series