# How can we find some radius of circle which fully contains $x\arctan(x)-ax+y\arctan(y)-by=0$?

How can we find some radius of circle with center at origin which contains $$x\arctan(x)-ax+y\arctan(y)-by=0$$,

where $$\pi/2>a>0$$ and $$\pi/2>b>0$$.

I'm not sure how can we prove that these inequalities should hold so we can have closed curve: $$\pi/2>a$$ and $$\pi/2>b$$ .

Also very interesting similar equation $$x\arctan(x)-ax+(-x+y)\arctan(-x+y)-b(-x+y)+y\arctan(y)-cy=0$$.

I need some approximate estimation so I can prove that for some radius this estimation will be correct.

For example I've found such a circle for $$a=1.5$$ and $$b=1.5$$:

And for $$x\arctan(x)-1.5x+(-x+y)\arctan(-x+y)-1.5(-x+y)+y\arctan(y)-1.5y=0$$:

Maybe Lagrange multipliers can help? I'm also interested in higher dimensions where we can add $$z$$ and find some radius of a sphere which fully contains $$x\arctan(x)-ax+y\arctan(y)-by+z\arctan(z)-cz=0$$. But I think it can be done in similar way as for two dimensions.

• Are you looking for the smallest circle, or the smallest one centred on $O$?
– J.G.
Dec 10, 2018 at 17:09
• I need a circle centred on $O$ (origin). And not necessarily the smallest (but the smallest will be good, of course).
– Tag
Dec 10, 2018 at 17:11

Definition. Given $$a\in\mathbb R$$, define the function $$f_a(x)=x(\arctan x-a)$$.

Definition. Given $$a,b\in\mathbb R$$, consider the closed set $$\begin{split} E_{a,b} &= \{(x,y)\in\mathbb R^2 : x(\arctan x-a)+y(\arctan y-b)=0\} \\ &= \{(x,y)\in\mathbb R^2 : f_a(x)+f_b(y)=0\} \end{split}$$

Lemma 1. For every $$a\in\mathbb R$$, the set $$[0,\infty)$$ is in the image of $$f_a$$.

Proof. Assume $$a\geq0$$. Then $$f_a(0)=0$$ and $$\lim_{x\to-\infty}x(\arctan x-a)=(-\pi/2-a) \lim_{x\to-\infty}x=\infty.$$ Since $$f_a$$ is continous, these limits imply the thesis. The case $$a\leq0$$ is analogous, with the signs changed. □

Lemma 2. If $$|a|>\pi/2$$, then the function $$f_a:\mathbb R\to\mathbb R$$ is surjective.

Proof. Assume $$a>\pi/2$$. We have $$\pm\pi/2-a < 0$$, so $$\lim_{x\to+\infty} x(\arctan x-a) = (\pi/2-a)\lim_{x\to\infty} x = -\infty$$ and $$\lim_{x\to-\infty} x(\arctan x-a) = (-\pi/2-a)\lim_{x\to-\infty} x = \infty.$$ Since $$f_a$$ is continuous, these limits imply that it is surjective. The case $$a<-\pi/2$$ is analogous. □

Lemma 3. If $$|a|<\pi/2$$, then $$\lim_{x\to\pm\infty} f_a(x) = \infty.$$

Proof. A direct computation shows $$\lim_{x\to\infty} f_a(x) = (\pi/2-a)\lim_{x\to\infty}x = \infty$$ and the same holds for $$x\to-\infty$$. □

Proposition. If $$|a|>\pi/2$$. Then $$E_{a,b}$$ is unbounded.

Proof. Since $$f_a$$ is surjective by Lemma 2, for every $$y\in\mathbb R$$ there exists $$x\in\mathbb R$$ such that $$f_a(x)=-f_b(y)$$. This means that $$(x,y)\in E_{a,b}$$. Therefore the set $$E_{a,b}$$ is unbounded because we can find points with arbitrary large $$y$$.

Proposition. If $$|a|=\pi/2$$. Then $$E_{a,b}$$ is unbounded.

Proof. Assume $$a=\pi/2$$. The other case is analogous. Recall the well known limit $$\lim_{x\to\infty} f_{\pi/2}(x) = \lim_{x\to\infty} x(\arctan x-\pi/2) = -1.$$ For every $$x$$ sufficiently large we have that $$f_{\pi/2}(x)\leq0$$, so by Lemma 1 there exists $$y\in\mathbb R$$ such that $$f_b(y)=-f_a(x)\in[0,\infty)$$. Therefore we can find points $$(x,y)\in E_{\pi/2,b}$$ with arbitrary large $$x$$. □

Proposition. If $$|a|<\pi/2$$ and $$|b|<\pi/2$$, then $$E_{a,b}$$ is bounded.

Proof. By Lemma 3, the function $$f_a(x)+f_b(y)$$ is coercive, meaning that $$f_a(x)+f_b(y)\to\infty$$ if $$|(x,y)|\to\infty$$, therefore its sublevel sets are bounded. In particular $$E_{a,b}$$ is bounded as a consequence. □

Corollary. $$E_{a,b}$$ is bounded if and only if $$|a|<\pi/2$$ and $$|b|<\pi/2$$.

Now, given $$a,b\in(-\pi/2,\pi/2)$$, how can we find an estimate on $$\max_{(x,y)\in E_{a,b}} x^2+y^2$$? We could try the Lagrange multipliers approach again, similarly to what we did here. The stationary points must satisfy $$\bigl(f'_a(x), f'_b(y)\bigr) = \lambda (x, y) \qquad \text{for some \lambda\in\mathbb R},$$ which is equivalent to $$\frac1{1+x^2} + \frac{\arctan x-a}x = \frac{f'_a(x)}{x} = \lambda = \frac{f'_b(y)}{y} =\frac1{1+y^2} + \frac{\arctan y-b}y.$$

Unfortunately, this time I'm not able to find a closed form solution to the system of equations $$\left\{\begin{array}{l} f_a(x)+f_b(y)=0 , \\ \frac{f'_a(x)}x=\frac{f'_b(y)}y . \end{array}\right.$$

One can of course fall back to numerical solutions. I'm using Mathematica for that. Here I fix the values $$a=3/2$$ and $$b=5/4$$, then I find extremal points numerically both with the built-in NMaximize function and by solving the Lagrange multiplier system with FindRoot. The two solutions are the same, up to machine precision. Then I plot the set $$E_{a,b}$$ in blue, the root locus of the Lagrange multiplier equation in orange, and the smallest fitting circle in gray.

• Well, that falls in the broad question of numerical optimization. I proved that the set is bounded. Therefore the problem $\max_{(x,y)\in E_{a,b}} x^2+y^2$ is well posed. Then there are plenty of algorithms to perform numerical maximization. Here everything is smooth, so there should be no problems with running any numerical algorithm. There cannot be a divergence, because the set is bounded! Dec 11, 2018 at 18:05
• Ok, so for the $n$-dimensional version, the same statement holds. Namely, if you define $E_{a_1,\dots,a_n}=\{f_{a_1}(x_1)+\dots+f_{a_n}(x_n)=0\}$, then $E_{a_1,\dots,a_n}$ is bounded if and only if $|a_i|<\pi/2$ for all $i=1,\dots,n$. The proof is also the same Dec 13, 2018 at 14:33
• Once you have this condition ($|a_i|<\pi/2\ \forall i$), you can apply again a numerical scheme to maximize $x_1^2+\dots+x_n^2$ over $E_{a_1,\dots,a_n}$ and it should be able to find the optimum Dec 13, 2018 at 14:35
• You mention the symmetry in the case $a_1=\dots=a_n$. While it is true that in this case $E_{a_1,\dots,a_n}$ is symmetric w.r.t. permutations of the variables, be aware that the maximum of $x_1^2+\dots+x_n^2$ is not achieved when all $x_i$'s except one are zero, or when $x_1=\dots=x_n$. The maximum is achieved at a point which is not particularly special from the point of view of symmetries. This is something different from your other problem Dec 13, 2018 at 14:38
• @Tag The set $\{(x,y):f_a(x)+f_b(y)+f_c(y-x)=0\}$ is equal to the projection on the $(x,y)$ plane of $E_{a,b,c}\cap\{z=y-x\}=\{(x,y,z):f_a(x)+f_b(y)+f_c(z)=0\}\cap\{z=y-x\}$. So, if $a,b,c\in(-\pi/2,\pi/2)$, then $E_{a,b,c}$ is bounded and therefore slicing and projecting it gives a bounded set. Notice that in this case it is no longer an "if and only if". I'm not sure we can say that if $\{(x,y):f_a(x)+f_b(y)+f_c(y-x)=0\}$ is bounded then $a,b,c\in(-\pi/2,\pi/2)$. I don't know the answer because I haven't analyzed this problem in detail. But at least the other implication is true. Dec 17, 2018 at 16:55