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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 $b=1.5$:

enter image description here

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

enter image description here

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.

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

1 Answer 1

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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.

enter image description here

enter image description here

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    $\begingroup$ 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! $\endgroup$
    – Federico
    Dec 11, 2018 at 18:05
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    $\begingroup$ 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 $\endgroup$
    – Federico
    Dec 13, 2018 at 14:33
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    $\begingroup$ 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 $\endgroup$
    – Federico
    Dec 13, 2018 at 14:35
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    $\begingroup$ 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 $\endgroup$
    – Federico
    Dec 13, 2018 at 14:38
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    $\begingroup$ @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. $\endgroup$
    – Federico
    Dec 17, 2018 at 16:55

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