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

for positive integer $n$ use the notation $y^{[n]}$ to represent the $n$-th tetration of $y$, so $y^{}=y$, $\, y^{}= y^y$, $\,y^{}=y^{y^y}$, 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 $x = y^{}=0.6411857445049887$ 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

• It’s just a monotonic sequence that is bounded, hence it converges. As you pointed out, it converges to the value $x$ that satisfies the equation $x^{1/x}=y$. – Clayton Mar 7 '18 at 18:17
• i hadn't noticed that rather important fact ! thanks ;-) – David Holden Mar 7 '18 at 19:22
• but see also the table i posted below (as an answer for ease of visibilty) – David Holden Mar 7 '18 at 19:54
• You’re right; I jumped on something too quickly. The odd terms in the sequence are increasing while the even terms decrease. By the argument above, thee both converge. Now you just need to show their limit is the same. – Clayton Mar 7 '18 at 21:40
• yes. i framed my intuition of the non-uniformity of convergence wrongly. what actually seems to be the case is that for any $n \gt 0, \epsilon \gt 0$ there is an interval $(0,\delta_{n,\epsilon})$ in which $x^{[2n]} \gt 1-\epsilon$ and $x^{[2n+1]} \lt \epsilon$ – David Holden Mar 8 '18 at 0:07

## 2 Answers

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.

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.