Minimum of the Gamma/factorial function using dichotomy and Lagrange inversion theorem .

Question

I want to evaluate the minimum of the Gamma's function via a kind of dichotomy .

We have to start the value :

$$\left(\frac{1}{\sqrt{5}}\right)!\simeq 0.8856$$

Solving $$x!=\left(\frac{1}{\sqrt{5}}\right)!$$

We have for $x>0$ two solutions : $x\simeq 0.4472,x\simeq 0.4761$

Taking the arithmetic mean of these values we have :

$$y\simeq 0.46165$$

And so on ...

... The bad things is , we need more and more accuracy which is a disavantage but obvious as we want a solution .

On the other hand it reminds me a Potential well see https://en.wikipedia.org/wiki/Potential_well.

All of this is very classic so I ask this question :

Can we build a solution using a double power series inversion (Lagrange inversion theorem) one for the Gamma's function and the other one for the two values of dichotomy starting from a tricky value (I mean for which the value is well-know ) or someting else ?

Any hint is very appreciated .

Answer 1

Are known the polygamma functions derivatives in the form of $$\left(\ln\Gamma(x)\right)^{(n+1)}=\psi_n(x),\quad \psi(x)=\psi^{(0)}(x)=\dfrac{\Gamma'(x)}{\Gamma(x)}\;\text{etc.}$$

Stationary points of $\Gamma(x)$ correspond to zeros of $\psi(x)$, which can be easily calculated numerically by the Newton iterative algorithm $$x_0=\dfrac32,\quad x_{n+1}= - \dfrac{\psi(x_n)}{\psi^{(1)}(x_n)}.$$ which provides $28$ correct decimal digits after 3 iterations, with $\Gamma(x_3)\approx 0.88560319441088870027881590058259.$

This possibility makes OP algorithms insufficiently effective.

Answer 2

I think I have found something :

Let define :

$$f\left(x\right)=\left(x^{2}-1\right)\left(x!-1\right)$$

Then define :

$$m(x)=f\left(f\left(f\left(f\left(f\left(f\left(f\left(f\left(f\left(f\left(f\left(f\left(f\left(f\left(f\left(f\left(x\cdots\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)+x!-\left(d\right)!$$

Where :

$$d=\frac{1}{v}\int_{0}^{\infty}\frac{1}{\left(x-1\right)!}dx$$

Then if $x_{min}$ is the minimum of the factorial for $x>0$ then $v>0$ is defined such that :

$$m((x_{min}))=0$$

Then define :

$$g\left(x\right)=\left(x^{2}-\frac{1}{\sqrt{5}}\right)\left(x!-\left(\frac{1}{\sqrt{5}}\right)!\right)$$

Then we have :

$$m(\left(2\cdot x_{min}-d\right))\simeq g(\left(2\cdot x_{min}-d\right))$$

Then we can use Newton's method and Faa di Bruno formula .