How much of Stirling is in Stirling's formula?

Question

This is a naive question about history.

My understanding is that Stirling's formula or something trivially equivalent to it first appeared in an early edition of Abraham de Moivre's book The Doctrine of Chances (and maybe in a journal article he wrote before that?), provided that we understand "Stirling's formula" to mean that $$ \lim_{n\to\infty} \frac{\sqrt{n}\cdot n^n e^{-n}}{n!} = \text{some finite non-zero number}. $$ If I'm not mistaken, de Moivre computed this number numerically and it was James Stirling who later showed just which number it is, and then de Moivre included it in a later edition of his book.

It's been a while since I read about all this and I don't remember enough detail to be 100% sure I've got all of the above right.

But today the term "Stirling's formula" seems to be used to refer to an asymptotic expansion.

How much of that series called "Stirling's formula" can be attributed to Stirling? And who should get credit for what he shouldn't get credit for?

Answer 1

As you note (Wikipedia)

The formula was first discovered by Abraham de Moivre in the form

$$n!\sim \text{constant}\cdot n^{n+1/2} e^{-n}.$$

De Moivre gave an expression for the constant in terms of its natural logarithm. Stirling's contribution consisted of showing that the constant is $\sqrt{2\pi}$. The more precise versions are due to Jacques Binet.

When it comes to getting credit, you should check that lemma that says that when a finding is attributted to someone, it isn't usually the real discoverer (ironically enough this finding was proved to be discovered by yet anoher person before, so the finding itself is an example of this controversy).

I guess the credit is given to him for his asymptotic expansion in series, which gives the "little" approximation:

$$n! \sim {n^n}{e^{ - n}}\sqrt {2\pi n} $$

As to how the approximation is found, I answered some days ago the following:

If you're familiar with [Wallis infinite product]

$$\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}\frac{1}{{\sqrt n }} = \sqrt \pi $$

Then you can use

$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {2n} \right)!{!^2}}}{{\left( {2n} \right)!}}\frac{1}{{\sqrt n }} = \sqrt \pi \cr & \mathop {\lim }\limits_{n \to \infty } \frac{{{2^{2n}}{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}}\frac{1}{{\sqrt n }} = \sqrt \pi \cr} $$

Now you can check that

$$\alpha = \mathop {\lim }\limits_{n \to \infty } \frac{{n!{e^n}}}{{{n^n}\sqrt n }} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {2n} \right)!{e^{2n}}}}{{{{\left( {2n} \right)}^{2n}}\sqrt {2n} }}$$

exists. Then square the first expression and divide by the latter to get

$$\alpha = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {n!} \right)}^2}{e^{2n}}}}{{{n^{2n}}n}}\frac{{{{\left( {2n} \right)}^{2n}}\sqrt {2n} }}{{\left( {2n} \right)!{e^{2n}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {n!} \right)}^2}{2^{2n}}\sqrt 2 }}{{\left( {2n} \right)!\sqrt n }} = \sqrt {2\pi } $$

Thus you have that

$$\mathop {\lim }\limits_{n \to \infty } \frac{{n!{e^n}}}{{{n^n}\sqrt {2n} }} = \sqrt \pi $$

or

$$n! \sim {n^n}{e^{ - n}}\sqrt {2\pi n} $$

Answer 2

Stirling's original result is the expansion $$ \log n! \sim \left( {n + \frac{1}{2}} \right)\log \left( {n + \frac{1}{2}} \right) - \left( {n + \frac{1}{2}} \right) + \frac{1}{2}\log \left( {2\pi } \right) - \frac{1}{{24\left( {n + \frac{1}{2}} \right)}} + \cdots , $$ as $n\to \infty$. De Moivre's version is that $$ \log n! \sim \left( {n + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log \left( {2\pi } \right) + \frac{1}{{12n}} - \cdots , $$ for large $n$. A nice reference for the topic is

A. Hald, A History of Probability and Statistics and Their Applications Before 1750. John Wiley $\&$ Sons, New York, 1990.