I would like to study the irrationality of ${{{{x}^{x}}^{x}}^{x}}^{\cdots } $
for $x=\frac{1}{2} $ using the irrationality of $\zeta(2)$ .
Some computations in wolfram alpha show to me that :
$${{{{x}^{x}}^{x}}^{x}}^{\cdots } $$ converge to $0.64...$ for $x=\frac{1}{2} $.
My Question here is: Is $x=\frac{1}{2}$ the solution of this equation $\zeta(2)= 1+{{{{x}^{x}}^{x}}^{x}}^{\cdots } $ and does ${{{{x}^{x}}^{x}}^{x}}^{\cdots } $ irrational for $x=\frac{1}{2}$ ?.
A solution of $a = 1/2^a$ is transcendental. First note that $a$ is not an integer. It can't be a non-integer rational, because $2^a$ for positive rational $a$ is an algebraic integer, and the only algebraic integers that are rational are the ordinary integers. It can't be an irrational algebraic number by the Gelfond-Schneider theorem.
Expanding on some of this...
I suppose it makes no sense unless $x$ is between 0 and 1. And in this case...
Define $y =x^{x^{x^{\cdots}}}$. Then we have $y= x^y$. For any fixed $0<x<1$, this defines a unique positive value for $y$ (intermediate value theorem will imply $y$ is between 0 and 1).
So that's one way to define it in the range we want. It seems to me that solving the above equation gives us:
$$x = e^{\log(y)/y} = y^{1/y}$$
or
$$ 1/x = (1/y) ^{1/y}.$$
This can be written in terms of the Lambert W function to give:
$$y = e^{-W(-\log(x))} = \dfrac{W(-\log(x))}{-\log(x)}.$$
We won't get more explicit than that.
But your situation is much easier since you already know $y$. So your question is the same as "is $y= \zeta(2)-1$ a solution to $y^{1/y} = 1/2$." (No.)
