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I 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?

Anthony Lahmann
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Polygon
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    You need $x\ge \dfrac{1}{e^e}$ as well as the restriction you pointed out. Perhaps "$\approx 0.1$" is close enough to this limit that the software bugs out. – David P Aug 22 '16 at 02:59

2 Answers2

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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:

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

Anthony Lahmann
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    Just for the record. Write $$y_{n+2}=e^{-e^{1-ey_n}}$$ and check that for $y_n\geq e^{-1}$ i.e. $e^{1-ey_n}\leq 1$ it follows $$y_{n+2} \geq e^{-1}$$ i.e. $y_n$ is bounded below by $e^{-1}$. It is also monotonically decreasing since $$y_{n+2}=e^{-e^{1-ey_n}} \leq y_n$$ which can be checked with analysis methods if $y_n>0$. – Diger Oct 05 '20 at 15:21
  • Intersting this video: https://www.youtube.com/watch?v=oy_kPyTstqk – Sebastiano Jan 31 '24 at 19:44
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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 $x$.

Simply Beautiful Art
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