Understanding the graph for $x^y = y^x$

Question

I graphed $x^y=y^x$ and it is a union of the line y=x, with some other curve. So my first question is, how do I derive that other curve? My next question is, why don't I get the same graph when I plot $x^{1/x}=y^{1/y}$?

Answer 1

Let $y=\mathcal{R}x$, $\mathcal{R}$ a variable. Then, $$x^{\mathcal{R}x}=(\mathcal{R}x)^x\\ \implies \mathcal{R}x\ln x=x\ln (\mathcal{R}x)=x\ln \mathcal{R}+x\ln x\\ \implies (\mathcal{R}-1)\ln x=\ln \mathcal{R}$$ We can choose $\mathcal{R}=1$, or, if $\mathcal{R}>1$, $$x=\exp\left(\dfrac{\ln \mathcal{R}}{r-1}\right)=\mathcal{R}^{1/(\mathcal{R}-1)}$$ I.e., $$y=\mathcal{R}^{\mathcal{R}/(\mathcal{R}-1)}$$ The non-straight graph is the graph for $y$ for which $\mathcal{R}\neq 1$.

Since $\mathcal{R}x\neq \dfrac{1}{x}$, you don't get the same graph.

Answer 2

Here is the way I first saw it, many years ago:

If $x^y = y^x$ with $x > 0$ and $y > 0$, let $r = y/x$, so $y = rx$. We will derive a parameterization for $x$ and $y$ in terms of $r$.

Then $x^{rx} = (rx)^x$ or, taking the $x^{th}$ root, $x^r = rx$ or $x^{r-1} = r$ or, if $r \ne 1$, $x = r^{1/(r-1)}$ and $y = rx = r^{1+1/(r-1)} = r^{r/(r-1)}$.

Letting $r$ go through the reals $> 1$ gives point $(x, y)$ with $x^y = y^x$.

Another way to look at this is to consider the curve $v = u^{1/u}$. For each $v$ such that $1 < v < e$, there are two values of $u$, $u_1 < e < u_2$, such that $u_1^{1/u_1} = v = u_2^{1/u_2}$.

Note that if $1/(r-1)$ is an integer, say $n$, then $r = 1+1/n$ and $x = (1+1/n)^n$ and $y = (1+1/n)^{n+1}$ are two rational numbers such that $x^y = y^x$.

$n=1$ gives $x=2$ and $y=4$ (the well known solution).

$n=2$ goves $x = (3/2)^2 = 9/4$ and $y = (3/2)^3 = 27/8$ (it is definitely less well known that $(9/4)^{27/8} = (27/8)^{9/4}$.

I believe that these are the only rational solutions to $x^y = y^x$, but I do not have a proof.

What the heck, I'll make it a problem and, I hope, harvest some points.

Answer 3

An alternative parameterization, which is a little more symmetric and found in the SE link provided in the comments by @SteveKass, can be derived as follows: $$x^y=y^x \\ x=a^b\quad y=a^c \\ a^cb\ln a=a^bc\ln a\implies a^cb=a^bc \\ a^{c-b}=\frac{c}{b} $$ Now we can let $c=b+1$ so that the equation becomes $$a=1+\frac{1}{b}$$ Therefore a parameterization for the second curve is $$\mathbf{x}(t)=\left(\left(1+\frac{1}{t}\right)^t\,,\,\left(1+\frac{1}{t}\right)^{t+1}\right)$$

As to why you're getting different graphs, it almost certainly is due to numerical errors in the CAS you're using. I don't know much about that area though, so I couldn't tell you what was causing the various problems.

Answer 4

Here is an analysis of what the graph looks like. I think it shows that the Maple graph (see comments on the OP) is correct. I haven't seen the Wolfram graph so can't comment on that.

Lemma. Let $c$ be a real number and consider the equation $$\frac{\ln x}{x}=c$$ for $x>0$. The equation has

• one solution if $c\le0$, and this solution satisfies $x\le1$;
• two solutions if $0<c<e^{-1}$, and the solutions satisfy $1<x<e$ and $x>e$ respectively;
• one solution $x=e$ if $c=e^{-1}$;
• no solution if $c>e^{-1}$.

Proof. Look at the graph of $(\ln x)/x$.

Corollary. The equation $x^y=y^x$ has no solutions with $x\le1$ or $y\le1$, except for $x=y$.

Proof. We have $$\frac{\ln x}{x}=\frac{\ln y}{y}\ ;$$ if $x\le1$ then $LHS\le0$; by the lemma there is a unique value for $y$, and clearly it is $y=x$.

Theorem. For any $x>1$ there is a unique $y\ne x$ such that $x^y=y^x$. Moreover, if $1<x_1<x_2$ then the corresponding values $y_1,y_2$ satisfy $y_1>y_2$. That is, the curve $$x^y=y^x\ ,\quad x>1\ ,\quad y>1\ ,\quad x\ne y$$ is decreasing. Finally, if $x\to1^+$ then $y\to\infty$, and if $x\to\infty$ then $y\to 1$; so the curve has asymptotes at $x=1$ and at $y=1$.

Proof. If $x>1$ then once again we have $$\frac{\ln x}{x}=\frac{\ln y}{y}\ ;$$ in this case $0<LHS<e^{-1}$ and so the lemma shows that there are two solutions for $y$. But one of them is $y=x$, so only the other one satisfies $y\ne x$. And if we look again at the graph of $(\ln x)/x$ we see that if $1<x_1<x_2\le e$ then $e\le y_2<y_1$; if $1<x_1<e<x_2$ then $y_2<e<y_1$; if $e\le x_1<x_2$ then $y_2<y_1\le e$; if $x\to1$ then $y\to\infty$; and if $x\to\infty$ then $y\to1$. This completes the proof.

Answer 5

The "top" part of the graph (the left side of the curve, and the right side of the straight line) is the limit of:

$$ y=x \log_x\left(x \log_x\left(\log_x\left(x \log_x\left(\cdots\right) \right)\right) \right) $$

The "bottom" part of the graph (the left side of the straight line, and the right side of the curve) is:

$$ y=\sqrt[x]{x^{\sqrt[x]{x^{\sqrt[x]{x^{\sqrt[x]{x^{\cdots}}}}}}}} $$

This could be of some use in numerically calculating it I suppose. But they converge very very slowly around x=e. Here's an approximation:

Approximation of y^x=x^y as a pair of functions