Is there an inverse to Stirling's approximation?

Question

The factorial function cannot have an inverse, $0!$ and $1!$ having the same value. However, Stirling's approximation of the factorial $x! \sim x^xe^{-x}\sqrt{2\pi x}$ does not have this problem, and could provide a ballpark inverse to the factorial function. But can this actually be derived, and if so how? Here is my work:

$$ \begin{align} y &= x^xe^{-x}\sqrt{2\pi x}\\ y^2 &= 2\pi x^{2x + 1}e^{-2x}\\ \frac{y^2}{2\pi} &= x^{2x + 1}e^{-2x}\\ \ln \frac{y^2}{2\pi} &= (2x + 1)\ln x - 2x\\ \ln \frac{y^2}{2\pi} &= 2x\ln x + \ln x - 2x\\ \ln \frac{y^2}{2\pi} &= 2x(\ln x - 1) + \ln x \end{align} $$

That is as far as I can go. I suspect the solution may require the Lambert W function.

Edit: I have just realized that after step 3 above, one can divide both sides by e to get

$$\left(\frac{x}{e}\right)^{2x + 1} = \frac{y^2}{2e\pi}$$

Can this be solved?

Answer 1

It is better to start with $$ x! \approx \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right)^{x + \frac{1}{2}} \sqrt {2\pi } . $$ Setting $$ y = \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right)^{x + \frac{1}{2}} \sqrt {2\pi } , $$ we find $$ \frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right) = \frac{{x + \frac{1}{2}}}{\mathrm{e}}\log \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right) = \exp \left( {\log \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right)} \right)\log \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right). $$ We use the Lambert $W$-function defined by $W(z)\mathrm{e}^{W(z)} = z$ for $z>0$. From the above we see that $$ \log \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right) = W\!\left( {\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right). $$ Whence $$ \frac{{x + \frac{1}{2}}}{\mathrm{e}} = \exp \left( {W\!\left( {\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)} \right) = \frac{{\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)}}{{W\!\left( {\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)}}, $$ i.e., $$ x = \frac{{\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)}}{{W\!\left( {\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)}} - \frac{1}{2}. $$ For example, if $y=720$ then $$ x = \frac{{\log \left( {\frac{{720}}{{\sqrt {2\pi } }}} \right)}}{{W\!\left( {\frac{1}{\mathrm{e}}\log \left( {\frac{{720}}{{\sqrt {2\pi } }}} \right)} \right)}} - \frac{1}{2} \approx 5.99658, $$ which is a very good approximation.

Answer 2

As $n$ increases to infinity we want to know roughly the size of the $x$ that satisfies the equation $x! = n$. By Stirling $$ x^x e^{-x} \sqrt{2\pi x} \sim n $$ Just focusing on $x^x$ a first approximation is $\log n / \log\log n$. Now writing $x = \log n / \log\log n + x_1$ and solving approximately for $x_1$, this time using $x^x e^{-x}$ we get $$ x = \frac{\log n}{\log\log n} + \frac{\log n \cdot ( \log\log\log n + 1)}{(\log\log n)^2} + x_2 $$ with a yet smaller $x_2$ which can be also determined by plugging this into $x^x e^{-x}$. You'll notice eventually that the $\sqrt{2\pi x}$ is too small to contribute. You can continue in this way, and this will give you an asymptotic (non-convergent) serie (in powers of $\log\log n$). For more I recommend looking at De Brujin's book "Asymptotic methods in analysis". He specifically focuses on the case of $n!$ case in one of the Chapters (don't have the book with me to check).

Answer 3

If you wish to make an inverse of the factorial for a fractional value, try starting with one of the aproximations here and solving for $x$. The two formulas are $$x!\approx\sqrt{2\pi}x^xe^{-x}\sqrt{x+\frac{1}{6}+\frac{1}{72x}-\frac{31}{6480x^2}-\frac{139}{155520x^3}+\frac{9871}{6531840x^4}}\\ x!\approx\sqrt{2\pi}x^xe^{-x}\root{\LARGE{4}}\of{x^2+\frac{x}{3}+\frac{1}{18}-\frac{2}{405x}-\frac{31}{9720x^2}}$$ They may not be as simple to solve as your Stirling approximation, but they're much more accurate for smaller values of $x$. Read the article while you're at it.

Update: apparently Ramanujan's approximation is even more accurate. $$x!\approx\sqrt{\pi}\left(\frac{x}{e}\right)^x\root{\LARGE{6}}\of{8x^3+4x^2+x+\frac{1}{30}}$$