A new infinite series for the minimum of the Gamma function?

Question

I continue my work on the minimum of the Gamma's function and recently I found a slightly different way via an infinite series as I have the beginning :

Let $0<x$ then define :

$$f(x)=x!,h(x)=f'(x)$$

Then we have :

$$h(k)=0$$

And :

$$k\simeq \frac{1}{2}\left(\pi-e+\frac{1}{2}\right)-\frac{1}{28}\left(\pi-e\right)^{\pi e}-\frac{10}{37}\left(\pi-e\right)^{2\pi e}-\frac{10000}{3675}\left(\pi-e\right)^{3\pi e}-\frac{1000}{499}\left(\pi-e\right)^{4\pi e}$$

I would like to find a infinite series like :

$$k=\frac{1}{2}\left(\pi-e+\frac{1}{2}\right)-\sum_{k=1}^{\infty}a_k\left(\pi-e\right)^{k\pi e}$$

Where $a_k>0$

I can progress numericaly not theoreticaly .

Edit following Tyma Gaidash's comment :

We have (if there is no mistake) 36 decimals right for the minimum value of the gamma function taking for $k$ :

$$k\simeq \frac{1}{2}\left(\pi-e+\frac{1}{2}\right)-\frac{1}{28}\left(\pi-e\right)^{\pi e}-\frac{10}{37}\left(\pi-e\right)^{2\pi e}-\frac{10000}{3675}\left(\pi-e\right)^{3\pi e}-\frac{1000}{499}\left(\pi-e\right)^{4\pi e}-\frac{736}{100}\left(\pi-e\right)^{5\pi e}-\frac{5}{10}\left(\pi-e\right)^{6\pi e}$$

How to find the sequence $a_k$ ?

Answer 1

Assume arbitrary constants $x_0,\xi\in\textbf{R}$. Then exists constants $a_0,a_1,a_2,\ldots$ and analytic function $f(x)$ such that (take for example $a_k=\left(\frac{x_0-1}{\xi x_0}\right)^k$): $$ x_0=\sum^{\infty}_{k=0}a_k\xi^k\tag 1 $$ and $$ \frac{f^{(k)}(\xi)}{k!}=a_k\tag 2 $$ and $$ f(2\xi)=x_0.\tag 3 $$

We expand $f$ into power series $$ f(x)=\sum^{\infty}_{k=0}\frac{f^{(k)}(\xi)}{k!}(x-\xi)^k\Rightarrow f(x+\xi)=\sum^{\infty}_{k=0}\frac{f^{(k)}(\xi)}{k!}x^k\Rightarrow f(2\xi)=\sum^{\infty}_{k=0}\frac{f^{(k)}(\xi)}{k!}\xi^k\Rightarrow $$ $$ x_0=\sum^{\infty}_{k=0}a_k\xi^k $$ Set now $\xi=\frac{1}{2}(\pi-e)^{\pi e}$. Then $f\left((\pi-e)^{\pi e}\right)=x_0$.

However, I want not make fun of the problem. To find such $a_k$ is a very great achievement. For example in [B] page 194 Entry 15, Ramanujan give us a way to construct such series:

Set

$$ \log a=\psi(x+1), $$ then $$ \left(\frac{x+\frac{1}{2}}{a}\right)^{4n}\approx 1-\frac{n}{6a^2}+\frac{10n^2+11n}{720a^4}-\frac{70n^3+231 n^2+891 n}{90720a^6}+\ldots\tag 4 $$ When $x=x_0$ i.e. as in your problem, then equivalently $x_0$ is root of $\psi(x+1)=0$. Hence $a=1$ and we get approximations of $x_0$ using (4).

References

[B]: Bruce C. Berndt. "Ramanujan's Notebooks I". Springer-Verlag. New York, Berlin, Heidelberg, Tokyo, 1985.