Parametric equation of $x^y=y^x$ curve

What is the easiest/most natural way to parametrize the following curve?

$$\{(x,y)\in\Bbb R^{+2}\mid x^y=y^x, x\neq y\}\cup\{(e,e)\}$$

The best I could do was taking it apart, and for $x>y$ use $y(x)=\sqrt[x]{x}^{\sqrt[x]{x}^{\sqrt[x]{x}^{.....}}}$, but I see no theoretical reason why I should break it into parts, since the curve looks so nice.

Maybe the niceness of the curve is just an illusion, and in fact the $x<y$ and $x>y$ cases must be considered separately?

EDIT:$$(x,y)=(e^{t/(e^t-1)},e^{t/(1-e^{-t})})\quad t \in\Bbb R$$ This is a very nice parametrization, but still doesn't include $(e,e)$.

Suppose you take $x=e^{r\cos\theta}$ and $y=e^{r\sin\theta}$ for $\theta\in(0,\pi/2)$. Then you want to solve $$e^{r\cos\theta e^{r\sin\theta}}=e^{r\sin\theta e^{r\cos\theta}}$$ for $r$. Taking the logarithm of both sides and dividing through by $r$ (losing the $r=0$ solution in the process) gives $$\cos\theta e^{r\sin\theta} = \sin\theta e^{r\cos\theta},$$ hence $$\tan\theta = e^{r(\sin\theta - \cos\theta)},$$ or $$r(\sin\theta - \cos\theta) = \ln\tan\theta.$$ This is satisfied for $\theta=\pi/4$, for any $r$, which gives the line $y=x$. The piece you're interested in is given by $$r=\frac{\ln\tan\theta}{\sin\theta - \cos\theta}.$$ Plugging this back in, you have $$(x,y)=\left(\exp\left(\frac{\ln\tan\theta}{\tan\theta - 1}\right), \exp\left(\frac{\ln\tan\theta}{1-\cot\theta}\right) \right) \\ =\left(\left(\tan\theta\right)^{\frac{1}{\tan\theta-1}}, \left(\tan\theta\right)^{\frac{1}{1-\cot\theta}}\right),$$ or simply $$(x,y)=\left(t^{1/(t-1)}, t^{t/(t-1)}\right)$$ for $t \in (0,\infty)$. (Yes, there is a hole at $t=1$.)

• Can you use a connected $\theta$ domain without hitting $y=x$? Oct 26, 2013 at 0:44
• @dfeuer: I don't see how. But there's not really a problem with this parameterization; it's analytic at $t=1$. Just looking at $\ln x(t)$, you have $\ln t / (t-1)$, which is $\ln(1+(t-1)) / (t-1) = \sum_{k=0}^{\infty} (t-1)^k / (k+1)$. Oct 26, 2013 at 1:28

I've done this here before, but it's quicker to write it again than to try to find it.

Let $y = rx$. $x^y = y^x$ becomes $x^{rx} = (rx)^x$ or $x^r = rx$ or $x^{r-1} = r$ or $x = r^{1/(r-1)}$ and $y = rx = r^{1+1/(r-1)} =r^{r/(r-1)}$.

If $r=1$, all values of $x$ work since the condition is $x^{r-1} = r$.

• Maybe I'm dense, but I can't make head or tail of this. Oct 26, 2013 at 1:59
• @dfeuer: after substitution, take the x-th root, then divide by x, then take the (r-1)-th root Oct 29, 2013 at 0:36

Edit: I just see that you already mention this one in your question... But unlike you state there, it does include $(e,e)$ namely for $t=0$.

Basically a reparameterisation of other answers but $$x(t) = \exp\left(\dfrac{t}{e^t-1}\right) \quad y(t)=\exp\left(\dfrac{t}{1-e^{-t}}\right)$$ for $t\in \mathbb{R}$ seems to be a nice and symmetric one. Note that these expressions extend nicely over $t=0$ and $$\left(x(0), y(0)\right) = (e,e)$$ and $$\left(x(-t), y(-t)\right) = \left(y(t), x(t)\right).$$

• Its limit at $t=0$ is $(e,e)$, but it is not itself defined there. Oct 29, 2013 at 21:17
• @dfeuer It does not show in this notation perhaps but it extends even holomorphically over $t=0$ in much the same way as $\sin(t)/t$ does. It is not uncommon to use this notation to mean the extended function, as I do in this case.
– WimC
Oct 29, 2013 at 21:23
• Is there a way to reformulate this expression so that we don't have this anomaly at $t=0$?
– Dave
Oct 30, 2013 at 10:31
• @Dave I don't think so. But again, in (complex) function theory it is very common to extend implicitly over all removable singularities. And a tip: it is perfectly fine to post your own answer and accept it! (Instead of stating this result in the question.)
– WimC
Oct 30, 2013 at 11:59
• Write it in the form $$x(t)=\exp\left(1/\int_0^1 e^{t\tau}\ d\tau\right),\quad y(t)=\exp\left(1/\int_0^1 e^{-t\tau}\ d\tau\right)\ .$$ Oct 30, 2013 at 20:58

I can't really answer your question, but I think I can see why it might split like that:

$x^y=y^x$ generally means that $e^{y\log x}=e^{x\log y}$, which occurs when $y\log x = x\log y$, so $\frac {\log y}y=\frac{\log x}x$.

If you graph $f(x)=\frac{\log x}x$, you will see that the only region where $f$ is not injective is $(1,\infty)$, in which $f$ first rises rapidly from $0$ to a maximum of $1/e$ at $x=e$ and then declines gradually forever, approaching $0$.

So giving $y$ in terms of $x$ or $x$ in terms of $y$ means giving one point in terms of the other, and it's not surprising that odd things would happen when they cross.

As stated in the edit of my question, and then poined out by WimC, the best result is $$x(t) = \exp\left(\dfrac{t}{e^t-1}\right) \quad y(t)=\exp\left(\dfrac{t}{1-e^{-t}}\right)$$For the simplest way to obtain this, take marty cohen's answer and substitute $r=e^t$. Or, alternatively, see mjqxxxx's post.

For $t=0$ we can holomorphically extend this function by defining $x(0)=y(0)=e$.