Infinite tetration, convergence radius

First, I would recommend reading "Exponentials Reiterated" by R.A.Knoebel, http://mathdl.maa.org/mathDL/22/?pa=content&sa=viewDocument&nodeId=3087&pf=1. It is by far the most comprehensive resource on infinite tetration. Second, your question might have been answered sooner if you had posted on the Tetration Forum, http://math.eretrandre.org/tetrationforum/index.php. Lastly, I think it is very encouraging to see others interested in this subject, since I've been interested in it for quite a long time now.

Define

$$f(t)=x^t$$

In order to have this converge, we want

$$|f(t)-a|<|t-a|$$

where $a=f(a)$ and $t$ is sufficiently close to $a$. If there are multiple solutions, then $a=\min\{t=f(t)\}$. Dividing both sides by $|t-a|$ and rewriting, we see we need to have

$$\left|\frac{f(t)-f(a)}{t-a}\right|<1$$

By the mean value theorem, we need

$$|f'(\xi)|<1$$

for some $\xi$ between $t$ and $a$. Since $f'$ is continuous, this means we must have

$$|f'(a)|\le1$$

We proceed to find that

$$f'(a)=x^a\ln(x)=a\ln(x)=\ln(x^a)=\ln(a)$$

and thus, we want

$$|\ln(a)|\le1$$

leading to $a\in[1/e,e]$. Since $a=x^a$, we have $x=a^{1/a}$, and hence $x\in[1/e^e,e^{1/e}]$.

Convergence on the boundaries is not guaranteed. For $x=1/e^e$ and $a=1/e$, note that $f(f(t))$ is increasing and bounded above by $a$ for $t<a$, and decreasing and bounded below by $a$ for $t>a$. For $x=e^{1/e}$ and $a=e$, note that $f(t)$ is increasing and bounded above by $a$ for $t<a$. Hence it will converge on the boundaries.

Likewise note that it cannot converge outside of this range unless the initial value is equal to $a$, with $x=-a$, for example.

And so we have

$$x\in[1/e^e,e^{1/e}]$$

as was claimed.