# Find the maximum of $x^{x^{x^{⋰}}}.$

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.

• 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? – Andrew Chin Dec 17 '19 at 14:43
• No idea. This is an interview question. The above was my thought process during interview. – Idonknow Dec 17 '19 at 14:48
• Does maximum imply the $y$ value furthest up, or can it suggest the $x$ value furthest right? If the second, you can find $dx/dy$. – Andrew Chin Dec 17 '19 at 15:03
• 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? – William Elliot Dec 17 '19 at 15:12
• @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. – Simply Beautiful Art Dec 17 '19 at 15:31

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 $$x=\sqrt[e]e$$.

• 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. – Simply Beautiful Art Dec 17 '19 at 15:21
• (+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. – Jam Dec 17 '19 at 15:22