The Experbola – The Prisoner's Dilemma

D13. The Experbola

Contributed by William Keehn, PMP

In Dilemma 8, we investigated a curve similar to an ellipse, but changing addition to multiplication in its definition. This time, PMP participant William Keehn proposes a similar exploration related to the hyperbola, another conic section, which can be thought of as all of the points $(x, y)$ in the plane where $x y = 1.$ Since that description already involves multiplication, this time we will move on to exponentiation.

Figure 1: A hyperbola (in blue) and the experbola (in red). Courtesy of desmos.com.

So, now consider all of the points where $x^{y} = y^{x}$ such that $x$ and $y$ are distinct positive real numbers. You get a curve somewhat resembling a hyperbola, that William has dubbed the “experbola.” Figure 1 shows the hyperbola and experbola side-by-side. And William is brimming with interesting questions about the experbola, so Dilemma 13 has lots of parts:

Find all of the points on the experbola where both coordinates are whole numbers. (And explain why there aren’t any others.)
An asymptote of a curve is a straight line that the curve becomes arbitrarily close to but never touches. For example, the hyperbola in Figure 1 has two asymptotes: the lines $x = 0$ and $y = 0.$ Does the experbola have any asymptotes? If so, find them.
The condition that $x$ and $y$ are distinct is needed because whenever they are the same, the equation $x^{y} = y^{x}$ is trivially satisfied. So without that condition, the experbola would include the entire diagonal line $x = y .$ But as you can see in Figure 1, that condition also cuts the experbola into two disconnected pieces. However, there is a unique single point on the line $x = y$ that you could add back to the experbola to “complete” it and make it a single continuous curve. (That’s the point indicated by the small red open circle in Figure 1.) What are the coordinates of that point?
Find (continuous, non-constant) functions $f (t)$ and $g (t)$ such that the point $(f (t), g (t))$ is on the experbola for all positive real numbers $t$ . For this one, William offers a hint, by way of an example: $(\sqrt{3})^{3 \sqrt{3}} = (3 \sqrt{3})^{\sqrt{3}} .$ Such a pair of functions is called a parametrization of the experbola, or at least of the part of the experbola it lands on.

This problem originally appeared in the Prisoner’s Dilemma in the 2023 Spring issue of the PMP Newsletter. Solutions are no longer being accepted for this Dilemma.

Show solution?

Solution.

The Prisoner’s Dilemma received submissions from three PMP participants: a full one from Christopher West, a nearly complete solution from Blazej Kot, and one from Andrew Trentacoste II. The Eagle Problem Solvers of Georgia Southern University also submitted a very thorough solution.

Every complete solution received relied on methods of calculus, so we follow such an approach here. Raising both sides of the equation defining the experbola to the $1 / x y$ power, we see that it is also the curve satisfying $x^{1 / x} = y^{1 / y}$ . So determining points on the experbola (and properties of the curve as asked about here) becomes entirely about finding distinct values of $t$ for which the function $E (t) = t^{1 / t}$ (shown in Figure 6) takes on the same value. In other words, we want to find horizontal lines which intersect the graph of $E$ twice. Now, if $t > 1$ , then $1 / t > 0$ , and so $E (t) = t^{1 / t} > 1$ . Therefore, any such horizontal line must satisfy $E > 1$ . Relating this observation back to the experbola, all points on it must satisfy $x > 1$ and $y > 1$ . Moreover, it appears from the graph in Figure 6 that $E$ is strictly increasing up to some value of $x$ , and strictly decreasing thereafter. Presuming so, it means that any horizontal line can intersect the graph of $E$ at most twice, once to the left of that maximum value, and once to its right.

Figure 6: The function $E (t) = t^{1 / t}$ .

We verify this property by considering the first derivative

$E^{'} (t) = t^{- 2 + 1 / t} (1 - \ln t) .$

Since $t^{- 2 + 1 / t}$ is always positive, this derivative is positive (i.e., $E$ is increasing) when $1 - \ln t > 0$ , or in other words, for $t < e$ . Similarly, $E$ is strictly decreasing for $t > e$ , and has a unique maximum at $t = e$ . Translating these facts back to the experbola, if $(x, y)$ is any point on the experbola, one of $x$ or $y$ is less than $e$ and the other is larger.

So for question (a), we already know that in any integer solution, both $x$ and $y$ must be larger than one, and one or the other must be less than $e$ , or in other words, equal to 2. Thus the only integer solutions are $x = 2$ , $y = 4$ and $x = 4$ , $y = 2$ .

As the original proposer William Keehn points out, it is easiest to next consider part (d): one can verify by direct computation that for all $t > 1$ ,

$(t^{1 / (t - 1)}, t^{t / (t - 1)})$

lies on the experbola. So the desired functions are $f (t) = t^{1 / (t - 1)}$ and $g (t) = t^{t / (t - 1)}$ .

Turning back to question (b), we want to find asymptotes of the experbola, so we let $t$ tend to infinity in $f (t)$ and $g (t) .$ It is easier to work with $F (t) = \ln f (t) = (\ln t) / (t - 1)$ and $G (t) = \ln g (t) = t (\ln t) / (t - 1) .$ Using L’Hôpital’s rule, the limits of $F$ and $G$ as $t$ goes to infinity are the limits of $1 / t$ and $\ln t + 1$ , respectively. So $F$ tends to 0 and $G$ to infinity, or in other words, $x = f (x) = e^{F (t)}$ tends to 1 and $y = g (x) = e^{G (t)}$ tends to infinity. Thus, the experbola has a vertical asymptote at $x = 1$ , and by symmetry, a horizontal asymptote at $y = 1.$

Finally, we can find the point that should be added to restore the continuity of the experbola by taking the limit in the parametrization of the experbola as $t$ goes to 1. Using L’Hôpital’s rule again in exactly the same way, we see that both and $F$ and $G$ approach 1 as $t$ tends to 1, which in turn means that $f$ and $g$ approach $e$ , so the “missing” point is $(e, e)$ .

May 1, 2023

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