# Can someone help explain the behaviour of the implicit equation $x^n+y^n-n^x-n^y=0$, specifically the significance of the number 1.9667894071088...?

So I was looking at the graph of the implicit equation $$x^n+y^n-n^x-n^y=0$$ for different values of n in Desmos, and some weird stuff started to pop out. (I'm only really concerned with the graph in the first quadrant, since a negative raised to a fractional power goes into complex territory, so $$x^n$$ wouldn't make sense for most n.)

Firstly, for $$0\le n \le 1$$, there's just a single "branch" in the graph, and it's confined to the unit square $$[0,1]\times[0,1]$$. Fair enough. After that, though, for $$n>1$$, there's another branch that seems to "come in from infinity." It gets closer and closer to the curve that used to be in the unit square, but has now extended out of it for some reason.

Then (and this is what I found most strange), at a value of $$n$$ slightly less than $$2$$, the two curves merge at two different points. Through trial and error (again in Desmos), I have determined that the value is something like $$n\approx 1.9667894071088$$, and that the points of intersection are $$\sim(3.274,0.451)$$ and $$(0.451,3.274)$$.

My main question is where on earth the number $$1.966789...$$ came from! Is it a well-known number that I am just unaware of? If so, what are its properties? Is it rational, is it algebraic? Can it be written as a continued fraction, or an infinite sum/product?

If it is not a well-known number... is it useful? Can this be applied to any mathematics other than me just idly playing around on Desmos?

• Are you sure about the $1$ in $0.451$? I'm getting something closer to $0.456$. Dec 10, 2019 at 11:36
• @BarryCipra Ahhh yes, that is more accurate, I was just sort of eyeballing it I have to admit haha Dec 10, 2019 at 12:04
• Implicitly differentiating the function defined by the equation $x^n+y^n-n^x-n^y=0$, we get $$(ny^{n-1}-n^y\ln n)y'=n^x\ln n-nx^{n-1}$$ The derivative fails to be well defined when $$ny^{n-1}-n^y\ln n=n^x\ln n-nx^{n-1}=x^n+y^n-n^x-n^y=0$$ This is three equations in three unknowns. Dec 10, 2019 at 12:09

Let $$f_n(x,y)=x^n+y^n-n^x-n^y\qquad(n>0, \ x>0, \ y>0)\ .$$ The equation $$f_n(x,y)=0\tag{1}$$ defines a set $$S_n$$ in the first quadrant of the $$(x,y)$$-plane. Consider a point $$(x_0,y_0)$$ satisfying $$(1)$$. The implicit function theorem says the following: If $$\nabla f_n(x_0,y_0)\ne(0,0)$$ then there is a small window $$W$$ centered at $$(x_0,y_0)$$ such that $$S_n\cap W$$, i.e., the part of $$S_n$$ lying in $$W$$, is a smooth arc passing through $$(x_0,y_0)$$. When $$\nabla f_n(x_0,y_0)=(0,0)$$ then various things may happen, and the situation has to be studied in detail. E.g., we could have a selfintersection of $$S_n$$ at $$(x_0,y_0)$$.

We therefore compute $$\nabla f_n(x,y)=\bigl(n x^{n-1}-n^x\log n,n y^{n-1}-n^y\log n\bigl)=\bigl(g_n(x),g_n(y)\bigr)$$ with $$g_n(t):=nt^{n-1}-n^t\log n\ .$$ Plotting $$t\mapsto g_n(t)$$ for various $$n$$ shows that $$g_n$$ has two positive zeros $$t_1$$, $$t_2$$ (depending on $$n$$). This implies that the first component of $$\nabla f_n(x,y)$$ is zero along the two verticals $$x=t_i$$ and the second component is zero along the two horizontals $$y=t_i$$. We therefore obtain four points in the first quadrant where $$\nabla f_n(x,y)=(0,0)$$.

If it now happens that, by coincidence, the set $$S_n$$ contains one of these points then $$S_n$$ will probably be "special" there. In such a special situation the following three equations (in the variables $$n$$, $$x$$, $$y$$) will be fulfilled: $$f_n(x,y)=0,\quad g_n(x)=0,\quad g_n(y)=0\ .$$ You have found experimentally a triple $$(n_0,x_0,y_0)=(1.9667892,3.2725,0.45562)$$ where something special happens. Let's see at the numerics: • Ahhh, I see. I did actually use this technique to justify another strange value for n, namely n=e. At this point there seemed to be just a single point satisfying the equation, that is (e,e). I didn't realise it would apply to this, though. Thanks! Dec 21, 2019 at 22:27

Define $$\,h(n,x) := x^n - n^x.\,$$ For $$\,n>1\,$$ the exponential term $$\,n^x\,$$ gets eventually bigger than the first power term as $$\,x\to\infty.\,$$ Thus, $$\,h(n,0) = -1\,$$ and decreases to a relative minimum at $$\,x_-(n)\,$$ and then increases past $$\,0\,$$ at $$\,h(n,n) = 0\,$$ reaching a relative maximum at $$\,x_+(n)\,$$ and then decreases monotonically to $$\,-\infty.\,$$ We need to find the unique value of $$\,n\,$$ for which the relative minimum and maximum are negatives of each other. This unique value is $$\,n_0 \approx 1.9667894071088097346\,$$ for which $$\,x_-(n_0) \approx 0.456175623\,$$ and $$\,x_+(n_0) \approx 3.272562090\,$$ with minimum and maximum values $$\, -h(n_0, x_-(n_0)) = h(n_0,x_+(n_0)) \approx 1.1478691065689943883.\,$$

I have not yet been able to identify these approximate numbers.

Please. Consider this as a comment. Follows the plot showing in black the $$f(x,y,n) = 0$$ locus for $$n = (0,0.05, 0.10,\cdots , 3)$$ In red some interesting points. 