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As the figure below shows, the graph of the implicit function $$x^x=y^y,(x,y >0)$$ composes of a straight line and an arc, which of the two have an intersection point $P$.

enter image description here

How to find the coordinates $(x_p,y_p)$ of $P$? Does there exist a closed-form solution?

mengdie1982
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  • The line in the figure is $x = y$, so $x_p = y_p$ should be true. (If there is any intersection : the answer below hints there may not be one). – Sarvesh Ravichandran Iyer Jun 06 '18 at 05:16
  • @астонвіллаолофмэллбэрг That's right! And then? – mengdie1982 Jun 06 '18 at 05:22
  • I have also gotten this issue which an answer below has recorded : that of there being no intersection. But I think this is an issue of Desmos rather than of the function. I am going to say more : I think the intersection point has $x$ (and $y$) coordinate $\frac 1e$, which is the minimum of the function $x^x$ on the interval $[0,\infty)$. This is what the coordinates seem to indicate. I can't come up with a proof, simply because I cannot even see the explicit relation between $y$ and $x$. I may not need the explicit relation, though. – Sarvesh Ravichandran Iyer Jun 06 '18 at 05:27
  • The only solution of equation $x^x=y^y$ is $x=y=1$. so the coordinates of p is $(1, 1)$. – sirous Jun 06 '18 at 05:30

2 Answers2

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Let's use the following theorem from multivariable calculus:

If the function $f$ is differentiable, the gradient of $f$ at a point is either zero, or perpendicular to the level set of $f$ at that point.

Taking the function $f=x^x-y^y$ we see that the gradient at the intersection point of yours must be zero (it is impossible for a nonzero planar vector to be perpendicular to two linearly independent vectors simultaneously).

Solving the system $$\nabla f=\left(x^x(1+\log(x)),y^y(1+\log(y)) \right)= \mathbf{0} $$ gives $$x=y=\mathrm{e}^{-1}. $$

user1337
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Another Solution

Since $x^x=y^y$, hence ${\rm d}x^x={\rm d}y^y,$ i.e. $x^x(\ln x+1){\rm d}x=y^y(\ln y+1){\rm d}y.$ We obtain the derivative $$\frac{{\rm d}y}{{\rm d}x}=\frac{x^x(\ln x+1)}{y^y(\ln y+1)}=\frac{\ln x+1}{\ln y+1}.$$ Notice that $$x_p=y_p,$$ and $$\dfrac{{\rm d}y}{{\rm d}x}\bigg|_{x=x_p}=-1.$$ Thus, we can set up an equation as follows $$\frac{\ln x_p+1}{\ln x_p+1 }=-1.$$ Therefore,$$\ln x_p=-1.$$ As a result, $$x_p=\frac{1}{e}.$$ It follows that $$(x_p,y_p)=\left(\frac{1}{e},\frac{1}{e}\right).$$

Am I right？ Indeed，there exists a flaw here， because the expression of $\dfrac{{\rm d}y}{{\rm d}x}=\dfrac{\ln x+1}{\ln y+1}$ has no definition at $x=y=\dfrac{1}{e}$. How to remedy the problem?

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mengdie1982
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