Does the infinite power tower converge for all $0I know that if
$$y=x^{x^{x^{x^{x^{\dots}}}}}$$
then
$$x=y^\frac1y$$
for values of $x$ where the infinite power tower converges, so when $x\le e^\frac1e$. However, when I put the power tower into Desmos, it seems to stop being accurate at around $x\approx0.1$, no matter how many $x$'s I add.
If it is truly infinite, is it accurate all the way to $0$, or does it stop? How can it be proven either way?
 A: The power tower $x^{x^{x^{x^{x^{\dots}}}}}$ converges if $e^{-e}\le x\le e^{\frac1e}$. To prove this, we need to learn about derivatives. The derivative of a function $f(x)$ is essentially how much numbers close to $x$ are "spread out" from $f(x)$ with $x\leftarrow f(x)$. In this case, the variable is $y$, and $f(y)=x^y$. The derivative of $x^y$ with respect to $y$ is $x^y\ln(x)$. If the absolute value of the derivative is less than $1$, points near $y$ will be "pulled towards" $y$. Otherwise, points will be "pushed away" from $y$. All we have to do is solve for $y$:
$$\newcommand{W}{\operatorname W}$$
$$y=x^y$$
$$\ln(y)=y\ln(x)$$
$$\frac1y\ln(y)=\ln(x)$$
$$-\frac1y\ln\left(\frac1y\right)=\ln(x)$$
$$\frac1y\ln\left(\frac1y\right)=-\ln(x)$$
$$\ln\left(\frac1y\right)e^{\ln\left(\frac1y\right)}=-\ln(x)$$
$$\ln\left(\frac1y\right)=\W(-\ln(x))$$
$$\frac1y=e^{\W(-\ln(x))}$$
$$y=e^{-\W(-\ln(x))}$$
$$y=-\frac{\W(-\ln(x))}{\ln(x)}$$
Here $\W(x)$ is the Lambert W function. The equation just above works since $e^{\W(x)}=\frac x{\W(x)}$.
The derivative of $x^y$ with respect to $y$ is now $-\W(-\ln(x))$. To have the "pulled towards" effect, we must have $|\W(-\ln(x))|<1$. To find the bounds to our domain, we need to find two values such that $\W(-\ln(x))=\pm1$. Applying $x\leftarrow xe^x$ to both sides, this gives:
$$-\ln(x)=e\ \cup\ -\ln(x)=-\frac1e$$
$$\ln(x)=-e\ \cup\ \ln(x)=\frac1e$$
$$x=e^{-e}\ \cup\ x=e^{\frac1e}$$
However, we have to do more work at the endpoints. For case 2, where $x=e^{\frac1e}$, this is easy. $e^{\frac xe}$ is always greater than or equal to $x$. If $x$ is given a starting value $x_0$, like $1$, it will converge (slowly) towards $e$, because $1$ is below $e$ and $e^{\frac xe}$ is always greater than $x$. For case 1, where $x=e^{-e}$, the proof is much more complicated. However, it is known that $x\leftarrow e^{-ex}$ will converge towards $\frac1e$, but the proofs are not included here. Anyway:

$$\operatorname{Dom}\left(x^{x^{x^{x^{x^{\dots}}}}}\right)=\left[e^{-e}, e^{\frac1e}\right]$$

A: For $0<x<e^{-e}$, it diverges. Note that for such small values of $x$, we get $t>x^{x^t}$ for when $t>y$ and $t<x^{x^t}$ when $t<y$, where $y=x^y$. In other words, adding more powers of $x$ ends up pushing it farther and farther away from $y$ instead of closer. Particularly, it approaches $0$ and $1$, for even and odd powers of $x$.
