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Question: Find the maximum of $x^{x^{x^{⋰}}}.$

Let $y = x^{x^{x^{⋰}}}.$ Then \begin{align} y & = x^y \\ \Rightarrow \ln y & = y\ln x \\ \Rightarrow \frac{1}{y} \frac{dy}{dx} & = y\left(\frac{1}{x}\right) + \ln x \cdot \frac{dy}{dx}. \end{align} Since we are looking for maximum, we set $\frac{dy}{dx} = 0.$ So, $$\frac{y}{x} = 0$$ $$\Rightarrow y = 0.$$ I am not sure what's wrong here.

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    $\begingroup$ First, the graph of $y=x^y$ should suggest to you that it is not a function of $x$. Second, why would you imagine that the value is bounded? $\endgroup$ – Andrew Chin Dec 17 '19 at 14:43
  • $\begingroup$ No idea. This is an interview question. The above was my thought process during interview. $\endgroup$ – Idonknow Dec 17 '19 at 14:48
  • $\begingroup$ Does maximum imply the $y$ value furthest up, or can it suggest the $x$ value furthest right? If the second, you can find $dx/dy$. $\endgroup$ – Andrew Chin Dec 17 '19 at 15:03
  • $\begingroup$ There is no maximum unless you consider $\infty$ a number or restrict the range of x. Is this an interview for a math teaching job? $\endgroup$ – William Elliot Dec 17 '19 at 15:12
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    $\begingroup$ @AndrewChin Note that $x^{x^{x^{.^{.^.}}}}$ converges to one of the $y$ values satisfying $y=x^y$, if it converges at all. That is to say, it is a function of $x$. It's the like arguing $y=\sqrt x$ is not a function because $y^2=x$ is not a function. $\endgroup$ – Simply Beautiful Art Dec 17 '19 at 15:31
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If we are allowed to consider values of $x$ s.t. this tends to $\infty$, then the answer is trivially $\infty$. Assuming the question is concerned with the interval over which this converges to real numbers though:

Note that when $y=0$, you get $0=x^0$, which is a contradiction. Instead, the maxima in this case occurs when $y'=\infty$. Dividing everything by $y'$ and letting it go to infinity gives us

$$\frac1y=\frac y{xy'}+\ln(x)$$

$$\frac1y=\ln(x)\tag{as $y'\to\infty$}$$

$$1=y\ln(x)$$

Since we also know that $\ln(y)=y\ln(x)$, we end up with $\ln(y)=1$, or $y=e$, which occurs at $x=\sqrt[e]e$.

For more information on convergence, see here.

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    $\begingroup$ I've added a link concerning convergence. It does in fact converge for $y>1$. Consider $x=\sqrt[e]e$ and $y=e$. It is easy to prove by induction that the sequence given by $y_1=x$ and $y_{n+1}=x^{y_n}$ is increasing and bounded above by $y$, and hence must converge. $\endgroup$ – Simply Beautiful Art Dec 17 '19 at 15:21
  • $\begingroup$ (+1). I'd also like to add that the argument that $y'=0$ at suprema does not hold when the function increases to an end-point, as we see here. $\endgroup$ – Jam Dec 17 '19 at 15:22

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