Find a Stirling-formula like for $(x!)!$

Question

Hi working again on the gamma function I find that :

let $0<x\leq 1$ and $1\leq k\leq \infty$ then define :

$$f(x)=(x!)!,g(x)=\left(\left(x!\right)!\right)^{\frac{1}{e^{x^{k}}}}$$

Then a conjecture :

It seems $\exists k\in(1,\infty)$ such that $g(x)$ admits an asymptote as $x\to \infty$.

Then see here Trying to prove the Stirling approximation using concavity for $x\geq 1$ we can squeeze the function $f(x)$ and remains to determine some constant .

If my conjecture is true how to achieve this ?

Answer 1

Let $a$ to be the solution of $\psi^{(0)} (x+1)=0$ (I think that you know it).

Expand $f(x)=\log[(x!)!]$ around $x=a$ and write $$f(x)=f(a)+\Gamma (a+1)\sum_{n=2}^\infty \frac {c_n} {n!} \, (x-a)^n$$ The first terms are $$\color{red}{c_2}=\psi ^{(1)}(a+1) \psi ^{(0)}(\Gamma (a+1)+1)$$ $$\color{red}{c_3}=\psi ^{(2)}(a+1) \psi ^{(0)}(\Gamma (a+1)+1)$$ $$\color{red}{c_4}=3 \Gamma (a+1) \psi ^{(1)}(a+1)^2 \psi ^{(1)}(\Gamma (a+1)+1)+$$ $$\left(3 \psi ^{(1)}(a+1)^2+\psi ^{(3)}(a+1)\right) \psi ^{(0)}(\Gamma (a+1)+1)$$ $$\color{red}{c_5}=10 \psi ^{(1)}(a+1) \psi ^{(2)}(a+1) (\psi ^{(0)}(\Gamma (a+1)+1)+$$ $$\Gamma (a+1) \psi ^{(1)}(\Gamma (a+1)+1))+\psi ^{(4)}(a+1) \psi ^{(0)}(\Gamma (a+1)+1)$$ and so on for as many terms as required.

Using the expansion to $O\left((x-a)^{n+1}\right)$, computing the norm $$\Phi_n=\int_0^1 \Big[\log[(x!)!]-\text{approximation}_{(n)}\Big]^2\,dx$$

$$\left( \begin{array}{cc} n & \log_{10}\big[\Phi_n\big] \\ 2 & -5.0858 \\ 3 & -5.1547 \\ 4 & -6.3509 \\ 5 & -6.9713 \\ 6 & -7.9027 \\ 7 & -8.7240 \\ 8 & -9.5957 \\ 9 & -10.456 \end{array} \right)$$

Answer 2

As I wrote in comment, the same work can be done for $$g(x)=(x!)!$$ $$g(x)=g(a)+\Gamma (a+1) \Gamma (a!+1)\sum_{n=2}^\infty \frac {d_n} {n!} \, (x-a)^n$$ The first coefficients being $$\color{red}{d_2}=\psi ^{(1)}(a+1) \psi ^{(0)}(a!+1)$$ $$\color{red}{d_3}=\psi ^{(2)}(a+1) \psi ^{(0)}(a!+1)$$ $$\color{red}{d_4}=\left(3 \psi ^{(1)}(a+1)^2+\psi ^{(3)}(a+1)\right) \psi ^{(0)}(a!+1)+$$ $$3 \Gamma (a+1) \psi ^{(1)}(a+1)^2 \psi ^{(0)}(a!+1)^2+3 \Gamma (a+1) \psi ^{(1)}(a+1)^2 \psi ^{(1)}(a!+1)$$

Computing again

$$\Psi_n=\int_0^1 \Big[[(x!)!]-\text{approximation}_{(n)}\Big]^2\,dx$$

$$\left( \begin{array}{cc} n & \log_{10}\big[\Psi_n\big] \\ 2 & -5.0613 \\ 3 & -5.0973 \\ 4 & -6.3083 \\ 5 & -6.7855 \\ 6 & -7.7200 \\ 7 & -8.3849 \\ 8 & -9.2041 \\ 9 & -9.9228 \\ 10 & -10.691 \\ 11 & -11.423 \end{array} \right)$$ that is to say $$\log_{10}\big[\Psi_n\big] \sim -\frac {3n+13}4$$