Graph of the function $x^y = y^x$, and $e$ (Euler's number).

Question

Earlier, I was using the Desmos Graphing Calculator, and I wanted to remind myself of what the graph of the function $x^y = y^x$ looked like.

If you have never seen what it looks like before, it is similar to the shape of the Greek letter, psi (ψ); composed of two graphs:

• $y = x$
• I am unsure as to what the equation of this graph is equal to, but I would say that it is similar in shape to the graph of $y = \frac{1}{x}$, but much steeper.

However, what I do know is that these two graphs intercept each other at the point $(e, e)$, where $e$ is Euler's number.

Please, if you could, check it out for yourself, but my question is, why is Euler's number related to this graph?

Edit: I believe, upon further inspection, that it is related to Euler's number due to the fact that $x^y = e^{x \ln y}$, and, therefore, $y^x = e^{y \ln x}$.

Due to this fact, we can take the natural logarithm of both sides of the equation $e^{x \ln y} = e^{y \ln x}$, in order to get that $x \ln y = y \ln x$.

I do not know how to continue with this explanation, but my new question is, what is the equation of the graph that I could not work out?

Thank you, and good luck!

Answer 1

Any equation which can be written $$A+Bx+C\log(D+Ex)=0$$ has solutions expressed in terms of Lambert function.

In the case of $x^y=y^x$, this writes $$y = -\frac{x}{\log x}\,W\left(-\frac{\log x}{x}\right)$$ and what Lambert and Euler showed is that, in the real domain, the function $W(z)$ exists if $z\geq -\frac 1e$. So, for the argument $-\frac{\log x}{x}\geq -\frac 1e$, this implies $x \geq e$. This corresponds to the second branch of the curve (after $y=x$). At $x=e$, the slope of the curve is $-1$.

Answer 2

For most points on $y=x$, you have $dy/dx=1$. There is one point where $dy/dx$ could be $1$ or $-1$. Call that point $(x,y)=(a,a)$. $$x\ln y=y\ln x\\ \ln y+\frac xy\frac{dy}{dx}=\frac{dy}{dx}\ln x+\frac yx\\ \ln a+\frac{dy}{dx}=\frac{dy}{dx}\ln a+1$$ This has a unique solution unless $\ln a=1$

Answer 3

This is a derivation of Claude's answer.

The definition of the Lambert-$W$ function is the solution to:

$$x = y e^y$$

Our equation is:

$$x \ln y = y \ln x$$

$$ \implies - \frac {\ln x}{x} = -\frac {\ln y}y = - \ln y e^{- \ln y}$$

$$\implies W\left({-\frac {\ln x}{x}}\right) = - \ln y = - y \frac{\ln x}{x}$$

$$\implies y = - \frac{x}{\ln x}W\left({-\frac {\ln x}{x}}\right) $$

Answer 4

Here is a relevant paper. They find a parametric equation for the non-$“x=y”$ branch of the graph (Desmos link), and then prove that the branches intersect at $(e,e)$. The rest of this answer is my own input.

The fact that the two branches of the graph of $x^y=y^x$ intersect at $(e,e)$ is related to the fact that the graphs of $y=x^e$ and $y=e^x$ are tangent. (See, we have two solutions for things like $x^3=3^x$, which is why there are two points with $y=3$ on that graph. However, there is only one solution to $x^e=e^x$, since they're tangent.)

Theorem: $x^e$ and $e^x$ are tangent.

Proof: Remember that $$e^t\ge t+1\quad\text{for all }t$$ That's one of my favorite formulae. It lets us prove so many things about $e$ without calculus.

Substituting in $t=\dfrac xe-1$ gives us: $$e^{x/e-1}\ge\dfrac xe$$ Multiplying by $e$: $$e^{x/e}\ge x$$ Raising to the power of $e$: $$e^x\ge x^e$$ Since they clearly intersect at $x=e$, they must be tangent.$\tag*{$\blacksquare$}$

P.S. That last equation solves the puzzle of which is larger, $e^\pi$ or $\pi^e$.

P.P.S. Extra credit: Using the fact that $e^t\ge t+1$, and without calculus, prove that $1-\frac12+\frac13-\dotsb=\ln2$. Warning: This is hard. (I say "no calculus," but you kind of need the Squeeze Theorem at the very end.)

Answer 5

For a given $y>0$, the equation $x\ln(y) = y\ln(x)$ is equivalent to $\ln(x)/x=\ln(y)/y$.

Here is the graph of the function $f(x)=\displaystyle\frac{\ln(x)}{x}$.

The function has a global maximum at $x=e$.

If $y\leq 1$, then $f(y)\leq0$ and the equation has only one solution: $x=y$.

If $y>1$, then $f(y)>0$ and $y\neq e$ and the equation has two solutions (you can see that an horizontal line will cut the graph of the function at two points, unless $y=e$, in which case the "two" solutions become equal.)

If $y=e$, then $x=e$ (This is the point where your two graphs intersect.)

I don't think there is a closed form for your second graph (that is, using only elementary functions.) But you can use the Lambert W function to get something close enough.

The Lambert W functions $(W_k)_{k \in \mathbb Z}$ are the solutions of the equation: $$\forall k \in \mathbb Z:\forall x\in \mathbb R: W_k(x)e^{W_k(x)}=x$$ If we want real valued solutions (which we do, here), then we need only $W_0$ and $W_{-1}$.

We have:

$$\frac{1}{x}\ln(x)=\frac{1}{y}\ln(y)$$ $$\frac{1}{x}\ln(\frac{1}{x})=\frac{1}{y}\ln(\frac{1}{y})$$ $$\ln(\frac{1}{x})e^{\ln(\frac{1}{x})}=\ln(\frac{1}{y})e^{\ln(\frac{1}{y})}$$ Therefore:

$$\ln(\frac{1}{y})=W_0(\frac{1}{x}\ln(\frac{1}{x}))\text{ or } \ln(\frac{1}{y})=W_{-1}(\frac{1}{x}\ln(\frac{1}{x}))$$

$$y=e^{-W_0(\frac{1}{x}\ln(\frac{1}{x}))}\text{ or } y=e^{-W_{-1}(\frac{1}{x}\ln(\frac{1}{x}))}$$

We simplify it a little to get the final result:

$$y=-\frac{xW_{0}(\frac{-\ln(x)}{x})}{\ln(x)}\text{ or } y=-\frac{xW_{-1}(\frac{-\ln(x)}{x})}{\ln(x)}$$

This gives us two graphs.

A superposition of the two graphs gives us the one you saw initially.

Answer 6

Note that $y^x=x^y$ is symmetrical across $y=x$. From here we can conclude that the curved part of your graph has a slope of $-1$ at the point in interest. If it didn't, then it wouldn't be symmetrical.

We can prove it symmetrical across $y=x$ by taking the inverse function and noting the inverse function is indeed still equal to the original equation.

Upon solving for $y$ we get $$y=e^{-W(-\frac{\ln(x)}{x})}$$

Differentiate:$$y'=[\frac{\ln(x)-1}{x^2}][\frac1{-\frac{\ln(x)}{x}+e^{W(-\frac{\ln(x)}{x})}}]e^{-W(-\frac{\ln(x)}{x})}$$

Where setting it equal to $-1$, we get $x=e$.

I have no idea how you would solve for that, but that is a solution.