Stirling's Factorial Approximation Formula
In the study of combinatorics, we will see that The Factorial Function, $n!$, comes up rather frequently. As a brief reminder, recall that for $n \in \{0, 1, 2, ... \}$ we have that $n!$ is defined to be the finite product:
(1)It would be nice if there existed a more compact way to compute $n!$ directly for large $n$ - or at the very least, approximate $n!$ accurately. Fortunately, such a formula exists known as Stirling's Formula which is given below to be:
(2)The graph of Stirling's formula is given below alongside the ordered pairs $(n, n!)$ for $n = 1, 2, 3$. We claim that Stirling's formula approximates $n!$ increasingly well for large $n$, that is:
(3)We will now prove this fact as it involves some rather involved calculus. Instead, we will look at an example of applying Stirling's formula.
Suppose that we want to approximate $11!$. By using Stirling's formula we see that:
(4)Meanwhile we have that:
(5)As we can see $11! \approx \sqrt{22\pi} \left ( \frac{11}{e} \right )^{11}$.