The title is fairly self explanatory: I have been trying to rigorously prove that $y(x)=x^{x^{x^{\ldots}}}$ is a strictly increasing function over the interval $[1,e^{\frac{1}{e}})$ for a while now, primarily by exploring various manipulations using logarithms and polylogarithms but have gotten nowhere. Although it is simple enough to show that $y(\sqrt{2})>y(1)$ and if $y'(x)>0$ for some $x \in [1,e^{\frac{1}{e}})$ then $y'(x)>0$ for all $x \in [1,e^{\frac{1}{e}})$ (since either $y$ must be strictly increasing or strictly decreasing), I am not satisfied by the rigor of this argument, although perhaps this is me being too finicky. This lack of progress has led me to explore the possibility that it is only strictly non-decreasing but this loosening of constraints has not helped at all. When it comes to proving that it is a function I've been at a loss as to where I might even begin. Any and all insights are welcome.
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2$\begingroup$ So, let me see if I understand: you define $$a_n(x):=\begin{cases} x&\text{if }n=0\\ x^{a_{n-1}(x)}&\text{if }n>0\end{cases}$$ and then define $$y(x):=\lim_{n\to\infty} a_n(x)$$ and you want to know whether $y$ is actually a function $[1,e^{1/e})\to\mathbb R$ and, in that case, if it's increasing? $\endgroup$– user228113Commented Jun 22, 2015 at 16:07
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$\begingroup$ You mean $x^{a_{n-1}}$. $\endgroup$– user217285Commented Jun 22, 2015 at 16:08
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$\begingroup$ Yes, @Nitin, I just noticed. $\endgroup$– user228113Commented Jun 22, 2015 at 16:09
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$\begingroup$ @G.Sassatelli I think that the OP should realise what is going on with your definition $\endgroup$– MartiganCommented Jun 22, 2015 at 16:22
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1$\begingroup$ Indeed, I find this definition to be very clear. $\endgroup$– ArchaickCommented Jun 22, 2015 at 17:09
2 Answers
It is easier to prove that the inverse function is strictly increasing. Since the inverse function is just: $$ g(x) = \left(\frac{1}{x}\right)^{-\frac{1}{x}}$$ with a change of variable everything boils down to proving that $h(x)=x^x$ is increasing over $\left[\frac{1}{e},1\right]$. That is trivial since: $$ h'(x) = h(x)\cdot\frac{d}{dx}\log h(x) = (1+\log x)\,h(x) \geq 0.$$
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2$\begingroup$ This is just a beautifully elegant solution. $\endgroup$– mweissCommented Jun 22, 2015 at 17:41
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2$\begingroup$ This is indeed a great solution. $\endgroup$ Commented Jun 22, 2015 at 17:46
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2$\begingroup$ You will receive a +100 bounty from me for this, but it may take quite a while before I can add it. (And if I forget it, feel free to leave a comment) $\endgroup$ Commented Jun 22, 2015 at 17:51
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$\begingroup$ inverse is ok, but I didnt get the idea how we reduce to $x^x$ over $[1/e,1]$ can some one justify more? $\endgroup$ Commented Dec 29, 2017 at 0:40
We write $$x^{x^{x^{...}}} = \frac{W(-\ln(x))}{-\ln(x)}, x \neq 1$$
We differentiate that. This becomes $$\frac{\ln(x)W'(-\ln(x))+W(-\ln(x))}{x\ln^2(x)} , x \neq 1$$
Using the quotient rule. (W|A verification)
Since $e^{\frac{1}{e}}>x>1$, the numerator is positive.
Therefore we want to show that $$\ln(x)W'(-\ln(x))+W(-\ln(x))>0$$
Let $-\frac{1}{e}<y<0$.
Since $y$ is negative, thus $W(y)$ is negative, we have
$$0>W(y)$$
$$1>1+W(y)$$
$W(y)>-1$, thus the RHS is postive. Therefore taking the reciprocal will change sign.
$$1<\frac{1}{1+W(y)}$$
$$-\frac{1}{1+W(y)}+1<0$$
$y$ is negative thus $W(y)$ is negative. Therefore multiplying by it changes sign.
$$W(y)\left(-\frac{1}{1+W(y)}+1\right)>0$$
$$-y\frac{W(y)}{y(1+W(y))}+W(y)>0$$
$$-yW'(y)+W(y)>0$$
Substitute $y=-\ln(x)$, $-\frac{1}{e}<y<0$, thus $e^{\frac{1}{e}}>x>1$. This gives $\ln(x)W'(-\ln(x))+W(-\ln(x))>0$.
Therefore the derivative is positive so the function is strictly increasing.
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$\begingroup$ I wouldn't count pulling out $W$ as an answer, especially in an elementary setting. $\endgroup$ Commented Jun 22, 2015 at 17:44
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1$\begingroup$ @Alizter Can you clarify? It is quite straightforward to obtain that the power tower equals the $W$ expression. $\endgroup$ Commented Jun 22, 2015 at 17:48