# Ask Uncle Colin: Where does the asymptote go?

Dear Uncle Colin,

I’ve figured out that $x^{x^{x^{\dots}}} = 2$ when $x = \sqrt{2}$, but I’m struggling to make sense of the function - it seems to have a vertical gradient when $x = e^{\frac{1}{e}}$, but it doesn’t seem to have what I think of as an asymptote there. What gives?

- Puzzling Observation With Exponential Recursion - Thrash Out Whether Evaluation’s Right?

Hi, POWERTOWER, and thanks for your message!

The first thing is to verify your assertion that $x^{x^{x^{\dots}}} = 2$ when $x = \sqrt{2}$ – I’m not 100% convinced that’s a true statement!

Suppose $y = x^{x^{x^{\dots}}}$, which means that $y = x^y$, or $x = y^{\frac{1}{y}}$.

When $x = \sqrt{2}$, this equation is satisfied by both $y = 2$ and $y = 4$ – I’m not sure the power tower (at least, considered naively) is well-defined.

### So how should we consider it?

When I was trying to make sense of it, I found it easiest to consider its ‘inverse’ ((this is in quotes; I’m not sure you can take the inverse of something that isn’t a function)) function, $f(y) = y^{\frac{1}{y}}$, which *is* a well-defined function. For reasons of clarity, I’ll also consider an equivalent form, $f(y) = e^{ \frac{1}{y} \ln(y) }$.

We can work out several things about the graph of this. First, when $y$ approaches zero, the exponent is an enormous negative number - so $f(y)$ approaches 0 as $y$ does.

It’s also fairly clear that $f(1) = 1$, and because $\frac{\ln(y)}{y}$ approaches zero (from above) for large values of $y$, that $f(y)$ approaches 1 (from above) as $y$ gets large.

That presumably means we have a turning point somewhere! Let’s differentiate. In fact, let’s differentiate implicitly: let’s let $F = f(y)$:

- Take logs: $\ln(F) = \frac{1}{y}\ln(y)$
- Multiply by $y$: $y \ln(F) = \ln(y)$
- Differentiate with respect to $y$: $\ln(F) + \frac{y}{F} \diff{F}{y} = \frac{1}{y}$
- We’re looking for a turning point, so let $\diff{F}{y} = 0$: $\ln(F) = \frac{1}{y}$
- Replacing $\ln(F)$ and multiplying by $y$: $\ln(y) = 1$

So the turning point is at $\left(e, e^{\frac{1}{e}}\right)$.

Here’s what Desmos says it looks like.

### So what does that tell us?

Remember, this was the ‘inverse’ of the power tower. The curve $y = x^{x^{x^{\dots}}}$ is a reflection (in the line $y=x$) of the curve I’ve asked Desmos for.

For $1 \lt x \lt e^{\frac{1}{e}}$, there are two $y$ values for every $x$ - so it’s not a function, and $x^{x^{x^{\dots}}}$ is not properly defined.

In terms of what happens to the curve at the point where the tangent is vertical, it doubles back on itself. However, I think it’s reasonable - as long as you’re explicit about it - to define a power-tower function that returns the smallest value that satisfies the equation.

Hope that helps!

- Uncle Colin

* Updated 2020-09-16 to add graph links.