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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 $(e,e)$.

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5 Answers 5

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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 $t=1$.)

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  • $\begingroup$ Can you use a connected $\theta$ domain without hitting $y=x$? $\endgroup$
    – dfeuer
    Oct 26, 2013 at 0:44
  • $\begingroup$ @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 $\ln(1+(t-1)) / (t-1) = \sum_{k=0}^{\infty} (t-1)^k / (k+1)$. $\endgroup$
    – mjqxxxx
    Oct 26, 2013 at 1:28
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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 $x^{r-1} = r$.

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

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  • $\begingroup$ Its limit at $t=0$ is $(e,e)$, but it is not itself defined there. $\endgroup$
    – dfeuer
    Oct 29, 2013 at 21:17
  • $\begingroup$ @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. $\endgroup$
    – WimC
    Oct 29, 2013 at 21:23
  • $\begingroup$ Is there a way to reformulate this expression so that we don't have this anomaly at $t=0$? $\endgroup$
    – Dave
    Oct 30, 2013 at 10:31
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    $\begingroup$ @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.) $\endgroup$
    – WimC
    Oct 30, 2013 at 11:59
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    $\begingroup$ Write it in the form $$x(t)=\exp\left(1/\int_0^1 e^{t\tau}\ d\tau\right),\quad y(t)=\exp\left(1/\int_0^1 e^{-t\tau}\ d\tau\right)\ .$$ $\endgroup$ Oct 30, 2013 at 20:58
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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.

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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 $x(0)=y(0)=e$.

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