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I was looking at the graph of the equation $x^{x^y}=y$ (Desmos link). This graph has two components that cross at the point $(1/e^e,1/e)=(e^{-e},e^{-1})$.

Component 1 (as I'll call it) is the component $x^y=y$ which has the simple parametrization $$(x,y)=\left(t^{1/t},t\right),\qquad0<t<\infty.$$ Component 2 is a path between the points $(0,0)$ and $(0,1)$.

Does component 2 also admit a parameterization?

To clarify: Component 2 is a path so of course it abstractly admits a parameterization, but I'm asking if there is a parametrization that we can actually write down algebraically in terms of elementary functions.

My motivation for this question is from the limiting behavior of the sequence $0,1,x,x^x,x^{x^x},x^{x^{x^x}},\ldots$, whose behavior is closely related to the solutions to $x^{x^y}=y$. In particular, if $x$ is less than $e^{-e}$ then this sequence alternates between the upper and lower parts of component 2.

Thomas Browning
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    The second solution is $$x = \exp \left( \frac {W_k(y \ln y)} y \right), \quad k = \cases { 0 & $0 < y \leq e^{-1}$ \ -1 & $e^{-1} < y < 1$}.$$ – Maxim Sep 23 '20 at 10:23
  • I plugged in some values and that seems to work, but it would be nice if there was a parametrization without the Lambert W (it doesn't need to come from x=f(y)). – Thomas Browning Sep 23 '20 at 18:09

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Edit: In the comments, Maxim found a better way to get this solution. Here's Maxim's way in detail: Start with the equation $$x^{(x^y)}=y.$$ Raising both sides to the $y$th power gives the equation $$(x^y)^{(x^y)}=y^y.$$ In other words, if we set $z=x^y$ then we are looking for solutions to the equation $z^z=y^y$. There is the trivial solution $y=z$ which corresponds to component 1. We are interested in parametrizing the nontrivial solutions. We can rearrange the equation $z^z=y^y$ to get $$(1/z)^{1/y}=(1/y)^{1/z}.$$ This has the well-known parametrization $1/y=t^{1/(t-1)}$ and $1/z=t^{t/(t-1)}$ (see this answer). Then $y=t^{1/(1-t)}$ and $z=t^{t/(1-t)}$. Finally, solving for $x$ gives $$x=z^{1/y}=t^{(t^{-t/(1-t)})/(1-t)}.$$ We obtain the parametrization $$\boxed{(x,y)=(t^{(t^{-t/(1-t)})/(1-t)},t^{t/(1-t)}),\qquad0<t<\infty}.$$ Interestingly, the substitution $t\leftrightarrow\frac{1}{t}$ gives the similar parametrization $$\boxed{(x,y)=(t^{(t^{-t/(1-t)})/(1-t)},t^{1/(1-t)}),\qquad0<t<\infty}.$$

Thomas Browning
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    Neat. If we rewrite the equation as $x^y \ln(x^y) = y \ln y$, then, same as here, we can take $z = t y$ or $z = y^t$ to find a parametrization of $z^z = y^y$. That will give either the same parametrization as in your answer or $y = t^{t/(1 - t)}$ instead of $y = t^{1/(1 - t)}$. Incidentally, since the solutions of $x^x = y^y$ can be written as $y \ln(y)/W_k(y \ln y)$, that means we have an elementary parametrization for $W_k(y \ln y)$. – Maxim Sep 27 '20 at 15:35