tetration as an iterative solution for the transcendental equation $\sqrt[x]{x}=y$

Question

for positive integer $n$ use the notation $y^{[n]}$ to represent the $n$-th tetration of $y$, so $y^{[1]}=y$, $\, y^{[2]}= y^y$, $\,y^{[3]}=y^{y^y}$, and so on.

a few simulations suggest that on $(0,1]$ the sequence $y^{[n]}$ converges to a value $x$ which is the solution of the equation: $$\sqrt[x]{x} = y $$

the rationale for this solution procedure follows from the definition: $$ y^{[n+1]} = y^{y^{[n]}} $$ so for a fixed point: $$ x = y^x $$ i.e. $$ x^{\frac1x} = y $$ for an example, set $y=\frac12$. this gives, as an approximate solution $x = y^{[40]}=0.6411857445049887$ and: $$ x^{\frac1x} = 0.5000000000000047 $$

my question is really a request for ideas on how one might approach the task of formulating a rigorous treatment of a suspected convergence that is strongly indicated by computer simulation and by the existence of a fixed point. all helpful comments much appreciated

Answer 1

If $y^{y^{y^{y^{.^{.^.}}}}}$ is equal to $x$, then $y^x$ is equal to $x$, and therefore $y=\sqrt[x]x$. For a fixed point, simply solve for this. Obviously $1$ works. If you graph these, you can see it's the only solution.

Answer 2

here's the output for the first few tetrations of $0.2$ $$ \begin{align} 1\ &0.2\\ 2\ &0.7247796636776955\\ 3\ &0.31145890709837815\\ 4\ &0.6057585690219652\\ 5\ &0.37721845362498657\\ 6\ &0.5449235984185974\\ 7\ &0.4160205176450843\\ 8\ &0.5119341920025698\\ 9\ &0.4387057792655525\\ 10\ &0.4935803022174034\\ 11\ &0.45185820630979723\\ 12\ &0.4832419938771518\\ 13\ &0.45943951143174566\\ 14\ &0.4773814757637366\\ 15\ &0.46379351072870595\\ 16\ &0.47404792267501356\\ 17\ &0.46628851801869226\\ 18\ &0.4721481719525574\\ 19\ &0.46771639131236575\\ 20\ &0.4710643865777475\\ \end{align} $$ so the sequence $y^{[n]}$ is not monotonic, but consists of two interleaved monotonic subsequences, one increasing and one decreasing.