About the solution of the infinite recurrence $f(x,f(x,f(x,f(x,f(...))))=a$

Solution 1:

For your example $x^{x^\ldots}$, for $x \in (0,1)$ the function $f(y) = x^y$ is decreasing on $(0,\infty)$ with limits $1$ and $0$ as $y \to 0+$ and $y \to \infty$, so there is a unique fixed point $L(x)$. This is not $2$, it is $-{\frac {{\it LambertW} \left( -\ln \left( x \right) \right) }{\ln \left( x \right) }}$.

Now we have to investigate stability. We have $f'(L(x)) = \exp(-LambertW(-\ln(x))) \ln(x)$. This is an increasing function on $(0,1)$, and is $-1$ at $x = x_0 \approx 0.06598803585$. So for $1 > x > x_0$ the fixed point is stable: from any starting point close enough to $L(x)$ the sequence will converge to $L(x)$. In fact, it appears that this will be the case for any starting point in $(0,\infty)$. For $x < x_0$ the fixed point is unstable, and the sequence will never converge to $L(x)$ unless it happens to hit exactly $L(x)$.

EDIT: That $x_0 = e^{-e}$. For this value, the fixed point is $1/e$.

Solution 2:

Assume $0<x<1$. As seen above, if $x^{x^{x^\cdots}}$ converges at all, then it converges to a number $a$ such that $x^a=a$, that is a fixpoint of the map $f\colon (0,1)\to(0,1)$, $y\mapsto x^y = e^{y\ln x}$. Since $f'(y) = x^y\ln x$, we have $|f'(y)|<1$ at least when $\frac1e<x<1$; then we find a unique fixpoint $a$. However, if $0< x<\frac 1 e$ I am no longer sure; maybe one can ensure contraction after a few steps?