In the power tower $x^{x^{x^{x\cdots}}}$ where there is an infinite stack of $x$'s, what is the maximum convergent number? I know the answer by playing with the form $x^y=y$ and using Mathematica, but I don't know how to solve this by hand.

  • $\begingroup$ Isn't the solution just equal to the solution of $x^y = y$? So what part are you stuck on? $\endgroup$ – TMM Oct 1 '16 at 1:48
  • $\begingroup$ The maximum value for x would be 1. $\endgroup$ – kamoroso94 Oct 1 '16 at 1:48
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    $\begingroup$ No @kamoroso94. I can prove it concerges for $x=1.1$. Given the sequence $x, x^x, x^{x^x}, ..., x=1.1$. Let any member such as the first be less than $2$. Then the next member is less than $1.1^2=1.21$, thus also less than $2$. By induction the increasing sequence is bounded thus convergent. $\endgroup$ – Oscar Lanzi Oct 1 '16 at 2:07

Fix an $x > 0$. Since the map $a \mapsto x^a$ is continuous, we know that if the infinite tower $x^{x^{x^\cdots}}$ converges to some limit $y$ then $x^y = y$, which implies $x = y^{1/y}$. Elementary calculus shows that the maximum possible value of $y^{1/y}$ occurs at $y = e$, so it is impossible for the infinite tower to converge unless $x \le e^{1/e}$.

It remains to show that the sequence actually converges for $x = e^{1/e}$. To prove this we need only establish the inequality $1 < y \le x^y \le e$ for any $y$ in the range $1 < y \le e$. Then a simple induction will show that the sequence $x, x^x, x^{x^x}, \ldots$ is increasing and bounded, hence convergent. Finally, the desired inequality is easy to prove using the aforementioned calculation that shows $e^{1/e}$ is the unique maximum of the function $y^{1/y}$.

  • $\begingroup$ ...and as we all know the limit of the infinite power tower at this upper bound is $e$. $\endgroup$ – Parcly Taxel Oct 1 '16 at 2:47
  • $\begingroup$ How do you know it converges for $x < e^{1/e}$? The same argument works for $x \ge 1$, but the sequence is not increasing for $x < 1$. $\endgroup$ – arkeet Oct 1 '16 at 3:01
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    $\begingroup$ I conjecture that the sequence oscillates without converging when $x < e^{-e}$. I admit that you did answer the question, which only asks for the largest $x$. $\endgroup$ – arkeet Oct 1 '16 at 3:13
  • $\begingroup$ @ParclyTaxel No, not all of us know - the pair "you and I" does not constitute everyone. ;-)) ... But, yes certainly $(e^{1/e})^e=e$ and therefore the maximum value of the infinite tower is indeed $e$. $\endgroup$ – Mark Viola Oct 1 '16 at 3:50

Example 4 of the Wikipedia page on the Lambert W function tells how to solve $x^y=y=x^{x^{x^{x\cdots}}}$ for $y$ using the function: $$x=y^{\frac1y}$$ $$\frac1x=y^{-\frac1y}=\left(\frac1y\right)^{\frac1y}$$ $$-\ln x=\frac1y\cdot\ln\frac1y=\ln\frac1y\cdot e^{\ln\frac1y}$$ $$W(-\ln x)=\ln\frac1y$$ $$\frac1y=e^{W(-\ln x)}=\frac{-\ln x}{W(-\ln x)}$$ $$y=\frac{W(-\ln x)}{-\ln x}$$ The largest possible value of $x$ that will make this expression for $y$ defined over the reals satisfies $-\ln x=-\frac1e$, where the RHS is the lower limit of the domain of the real-valued Lambert W. Therefore the maximum convergent value of $x$ is $$e^{\frac1e}=1.444667861\dots$$

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    $\begingroup$ Can we explain it without reference to Lambert $W$? Not the value of the limit, just whether it converges. $\endgroup$ – arkeet Oct 1 '16 at 2:19
  • $\begingroup$ @arkeet I don't think there's such a way. Although Euler showed this convergence result, he lived at the same time as Lambert, and indeed the convergence was shown in Euler's paper on the Lambert W. The connection between $e$ and the infinite power tower is not obvious. $\endgroup$ – Parcly Taxel Oct 1 '16 at 2:27
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    $\begingroup$ @arkeet I will add an answer that addresses your comment. $\endgroup$ – Erick Wong Oct 1 '16 at 2:30

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