Stirling's Formula
Contents |
Theorem
The factorial function can be approximated by the formula:
- $\displaystyle n! \sim \sqrt {2 \pi} n^n n^{1/2} e^{-n} = \sqrt {2 \pi n} \left({\frac {n} {e}}\right)^n$
where $\sim$ denotes asymptotically equal.
Proof
Consider the sequence $\left \langle {d_n} \right \rangle$ defined as $\displaystyle d_n = \ln \left({n!}\right) - \left({n + \frac 1 2}\right) \ln n + n$.
What we want to do is show that $\left \langle {d_n} \right \rangle$ is decreasing.
So we examine the sign of $d_n - d_{n+1}$.
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle d_n - d_{n+1}\) | \(=\) | \(\displaystyle - \ln \left({n+1}\right) - \left({n + \frac 1 2}\right) \ln n + \left({n + \frac 3 2}\right) \ln \left({n+1}\right) - 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | (as $\ln \left({\left({n+1}\right)!}\right) = \ln \left({n+1}\right) + \ln \left({n!}\right)$) | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({n + \frac 1 2}\right) \ln \left({\frac {n+1} n}\right) - 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {2n + 1} 2 \ln \left({\frac {1 + \left({2n + 1}\right)^{-1} } {1 - \left({2n + 1}\right)^{-1} } }\right) - 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Work in progress. I'll get back to this.
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Source of Name
This entry was named for James Stirling.
It is otherwise known as Stirling's approximation.
