11
$\begingroup$

Being familiar with Stirling's formula for factorial: $$n!\sim\sqrt{2\pi n}\left(\frac n e\right)^n,\quad\color{gray}{n\to\infty}$$ I naïvely assumed that for twice iterated factorial we can simply substitute the right-hand side into itself and write $$(n!)!\,\stackrel{\color{red}{\small\text{wrong}}}\sim\,\sqrt{2\pi \sqrt{2\pi n}\left(\frac n e\right)^n}\left(\frac{\sqrt{2\pi n}\left(\frac n e\right)^n}e\right)^{\sqrt{2\pi n}\left(\frac n e\right)^n},$$ but, as it turns out, I was wrong. Actually, it grows faster than that: $$(n!)!\,\succ\,\sqrt{2\pi \sqrt{2\pi n}\left(\frac n e\right)^n}\left(\frac{\sqrt{2\pi n}\left(\frac n e\right)^n}e\right)^{\sqrt{2\pi n}\left(\frac n e\right)^n}.$$

Can we express the correct asymptotic growth rate of twice iterated factorial $(n!)!$ using only elementary functions and, possibly, also their inverses, such as the Lambert W-function?

Update: Apparently, the same issue arises for simpler functions like $2^{n!}$. If we simply substitute Stirling's formula for the exponent $n!$, we will get an incorrect asymptotic.

$\endgroup$
6
  • 9
    $\begingroup$ You can use a bound on each side, but they won't agree with one another in the limit. The problem is that if the exponent behaves as $n^k$ then you need to resolve the relative error of the inside of the parentheses down to $n^{-k}$ in order to have bounded relative error overall...and in this case the exponent grows faster than any polynomial. So no finite number of Stirling series terms will get you a correct asymptotic expansion. $\endgroup$
    – Ian
    Commented Apr 2, 2022 at 3:10
  • $\begingroup$ This answer about inverse factorial might be related: math.stackexchange.com/a/461207/19661 $\endgroup$
    – Vladimir Reshetnikov
    Commented Apr 2, 2022 at 7:34
  • $\begingroup$ math.stackexchange.com/a/1587792/815585 $\endgroup$
    – FShrike
    Commented Apr 2, 2022 at 9:22
  • $\begingroup$ @Ian Is your argument a heuristic, or it can be made rigorous? Does it apply only to a truncated Stirling series, or any combinations of elementary functions? $\endgroup$
    – Vladimir Reshetnikov
    Commented Apr 2, 2022 at 23:04
  • 1
    $\begingroup$ What I said is only for trying to get an asymptotic by taking a Stirling series expression and plug it into itself. No idea about other possibilities. $\endgroup$
    – Ian
    Commented Apr 3, 2022 at 2:39

1 Answer 1

Reset to default
0
$\begingroup$

Let $$q=\frac{(12 n+1)\sqrt{2\pi}}{12 \sqrt{n}}\,\,\exp\big(n (\log (n)-1)\big)$$ $$\color{blue}{\log\big((n!)!\big) \sim \left(q+\frac{1}{2}\right) \log (q+1)-(q+1)+\frac{1}{2} \log (2 \pi )+\frac{1}{12 (q+1)}-\frac{1}{360 (q+1)^3}}$$ A few numbers for the logarithmic expression $$\left( \begin{array}{ccc} n & \text{asymptotic} & \log\big((n!)!\big) \\ 2 & 0.6921873 & 0.6931472 \\ 3 & 6.5761173 & 6.5792512 \\ 4 & 54.771574 & 54.784729 \\ 5 & 457.74604 & 457.81239 \\ 6 & 4020.8773 & 4021.2696 \\ 7 & 37929.320 & 37931.995 \\ 8 & 387243.08 & 387263.81 \\ \end{array} \right)$$

I made another version which is a bit more accurate. For $n=8$, it gives $387266.04$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .