Stirling approximation for factorial

Abstract

Stirling’s approximation for the factorial function.

Keywords

Thermodynamics, Math, factorial

\(\require{cancel}\)

\(\require{cancel}\) \(\def\oot{\frac{1}{2}}\)

Consider the following integral:

\[\begin{eqnarray} \int_0^\infty dx x^{n}e^{-x}= \left[(-1)^n\frac{d^n}{d\alpha^n}\int_0^\infty dx e^{-\alpha x}\right]_{\alpha=1} =\left[(-1)^n\frac{d^n}{d\alpha^n} \frac{1}{\alpha}\right]_{\alpha=1}=n!. \label{eq:gamma} \end{eqnarray}\]

Taking this definition, we can do the following: \[\begin{eqnarray} n!=\int_0^\infty dx x^{n}e^{-x}=\int_0^\infty dx e^{n \ln(x)-x}. \label{eq:gamma2} \end{eqnarray}\]

Let’s take a close look at the function in the exponent: \[\begin{eqnarray} u(x) &=&n \ln(x)-x, \label{eq:expo} \end{eqnarray}\] as shown in Figure 1. This function has its peak value at \(x=n\). Note that this function appears in the exponent, under the integral. The dominant contribution to the integral will come from the domain around \(x=n\). We can expand \(u(x)\) around \(x=n\):

\[\begin{eqnarray} u(x) &=&n \ln(x)-x =n \ln(x-n +n)-x=n \ln(n[1 +\frac{x-n}{n}])-x\nonumber\\ & \simeq& n \left( \ln(n)+\frac{x-n}{n} -\frac{1}{2}\left[\frac{x-n}{n}\right]^2 \right)-x =n \ln(n) -n -\frac{1}{2}\frac{(x-n)^2}{n}\equiv \tilde{u}(x). \label{eq:expand} \end{eqnarray}\]

The original function and the approximated functions are plotted in Figure 1.

\(n\)

Figure 1: Interactive plot showing \(u(x)\) (left), \(e^{u(x)}\), and \(e^{\tilde{u}(x)}\) (right).

From Figure 1, we also notice that if we extend the \(x\) range to include negative values, the integral would not change much since \(e^{\frac{1}{2}\frac{(x-n)^2}{n}}\) is rapidly decaying. Therefore we can change the lower limit of the integral from \(0\) to \(-\infty\) to get:

\[\begin{eqnarray} n!&=&\int_0^\infty dx x^{n}e^{-x}=\int_0^\infty dx e^{u(x)}\simeq \int_0^\infty dx e^{\tilde{u}(x)}=n^n e^{-n} \int_0^\infty dx e^{-\frac{1}{2}\frac{(x-n)^2}{n}}\nonumber\\ &\simeq&n^n e^{-n} \int_{-\infty }^\infty dx e^{-\frac{1}{2}\frac{(x-n)^2}{n}}=n^n e^{-n} \sqrt{2\pi n}=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n. \label{eq:gammai} \end{eqnarray}\]

Figure 2 shows the comparison of \(n!\) with the Stirling’s approximation given in Eq. \(\ref{eq:gammai}\).

Figure 2: The plot of \(n!\) and its Stirling’s approximation (left). The relative error(right) gets smaller as \(n\) increases.