## The Power Tower

Some time ago, Foxmaths! had a nice post about a power tower

$f(x) = b^{b^{b^{...}}}$

To make this more precise, we say

$S_1 = b$

$S_n = b^{S_{n-1}}$.

And ask for the limit of $S_n$ as $x \to \infty$. A student of mine at physics camp asked me about this problem with the specific value $b= \sqrt{2}$. He was stuck. He assumed there was some limit $L$, and then wrote

$\left(\sqrt{2}\right)^L = L$

substituting an infinite tower of $\sqrt{2}$‘s for itself less one. This is equivalent to setting

$S_n = S_{n-1}$

in our recursion relation, meaning that we’re finding what the limit must be if it exists.

The problem is that by inspection there are two solutions: $2$ and $4$. But the real answer is two. Plugging it into the calculator and going a few iterations, we see that this is correct.

I didn’t know why it was two rather than four. Neither did anyone else around. But there were a lot of people around. Pretty soon we had six or eight people all trying to find why the thing went to two rather than four.

We tried using different values, like

$\sqrt(3)^{\sqrt{3}^{\sqrt{3}^{...}}}$,

which didn’t work at all – it blew up. But use $3^{1/3}$ rather than $3^{1/2}$ for the tower and it works fine – going to about $2.48$, although by straightforward algebra there still appear to be two solutions, the other being $3$. Use $4^{1/4}$ and it goes to $2$ again, rather than $4$. Use $5^{1/5}$ and it goes to $1.74$.

Then we tried fractional numbers to their own root. Take a power tower with a bunch of $1.6^{1/1.6}$‘s stacked and you converge to $1.6$. Same $2.3^{1/2.3}$ (converges to $2.3$). But $3.5^{1/3.5}$ converges towards $2.19$. Trial and error showed that for $x$ from zero to $e$, a power tower of $m^{1/m}$ converges to $m$. Then, above $e$, it converges to something smaller and smaller, until for very large $m$ it converges towards one. This makes some sense, because $m^{1/m}$ has a maximum at $m=e$ and converges to $1$ as $m\to \infty$.

That was as far as we got before 11PM, when I had to send them all to bed without any dessert. Until I realized: duh. Draw a picture.

The recurrence relation was

$S_n = b^{S_{n-1}}$

with $b$ some base we’re considering. So let’s draw $S_n$ as a function of $S_{n-1}$ That way, by going from a value on the x-axis to the value of the plotted function, we do one iteration. The relevant function is an exponential.

To do another iteration, we have to move to the correct x-value, so we can drop to the new y-value. That is, we want to go from the y-value we’re at, and make that the x-value. So plot the line $y=x$, then trace over horizontally to that line. Now you can drop down to the new value of the function, conducting another iteration. Doing this over and over, we get a series of steps, and each step represents a new iteration.

A heuristic representation of the series we're talking about here. Each blue step marks a new iteration. You start at the red line, drop to the green line to iterate, then cut over horizontally to the red line to get ready to do it again.

Repeating this process, we see that where the lines intersect there are solutions to $S_n = S_{n-1}$. But only one of those solutions is “stable”. If we’re close to the lower one, we’ll get closer on the next try. But if we’re close to the higher one, we’ll get further away on the next try. So the power tower converges towards the lower intersection.

As we change the base $b$, the roots move. As $b$ increases from a small value, the roots come in towards each other because the exponential is getting steeper. When $b$ gets to $e^{1/e}$, the two lines are tangent, which is an easy calculus problem. Then the roots coincide. If $b$ increases beyond that, the lines never cross and the series diverges.

We can also see why using $b = m^{1/m}$ converges to $m$ for $m. It's because $m$ is clearly a solution for $L$ in

$\left(m^{1/m}\right)^L = L$.

The two roots have one bigger than $e$ and one less. And for $m, we converge to $m$, because we converge to the smaller root.