How can we find minimum radius of circle which contains $arctan^2(x)+arctan^2(y)=a$?












2












$begingroup$


How can we find minimum radius of circle which contains $arctan^2(x)+arctan^2(y)=a$, where I think $a<pi/2$?



For example I have some plots from WolframAlpha and I see it depends on $a$.



But have no idea how to find maximum radius because this is not simple implicit function.



enter image description here






maxima-minima implicit-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    This might be amenable to using polar coordinates.
    $endgroup$
    – hardmath
    Dec 7 '18 at 16:23










  • $begingroup$
    @hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
    $endgroup$
    – Federico
    Dec 7 '18 at 16:29










  • $begingroup$
    If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
    $endgroup$
    – Tag
    Dec 7 '18 at 16:34










  • $begingroup$
    @Federico yes, but perhaps we can do it neatly with a change of variables.
    $endgroup$
    – hardmath
    Dec 7 '18 at 16:35










  • $begingroup$
    @user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
    $endgroup$
    – Ross Millikan
    Dec 7 '18 at 16:39
















2












$begingroup$


How can we find minimum radius of circle which contains $arctan^2(x)+arctan^2(y)=a$, where I think $a<pi/2$?



For example I have some plots from WolframAlpha and I see it depends on $a$.



But have no idea how to find maximum radius because this is not simple implicit function.



enter image description here






maxima-minima implicit-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    This might be amenable to using polar coordinates.
    $endgroup$
    – hardmath
    Dec 7 '18 at 16:23










  • $begingroup$
    @hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
    $endgroup$
    – Federico
    Dec 7 '18 at 16:29










  • $begingroup$
    If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
    $endgroup$
    – Tag
    Dec 7 '18 at 16:34










  • $begingroup$
    @Federico yes, but perhaps we can do it neatly with a change of variables.
    $endgroup$
    – hardmath
    Dec 7 '18 at 16:35










  • $begingroup$
    @user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
    $endgroup$
    – Ross Millikan
    Dec 7 '18 at 16:39














2












2








2


0



$begingroup$


How can we find minimum radius of circle which contains $arctan^2(x)+arctan^2(y)=a$, where I think $a<pi/2$?



For example I have some plots from WolframAlpha and I see it depends on $a$.



But have no idea how to find maximum radius because this is not simple implicit function.



enter image description here






maxima-minima implicit-function







share|cite|improve this question











$endgroup$




How can we find minimum radius of circle which contains $arctan^2(x)+arctan^2(y)=a$, where I think $a<pi/2$?



For example I have some plots from WolframAlpha and I see it depends on $a$.



But have no idea how to find maximum radius because this is not simple implicit function.



enter image description here






maxima-minima implicit-function




maxima-minima implicit-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 16:13







Tag

















asked Dec 7 '18 at 16:07









TagTag

696




696












  • $begingroup$
    This might be amenable to using polar coordinates.
    $endgroup$
    – hardmath
    Dec 7 '18 at 16:23










  • $begingroup$
    @hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
    $endgroup$
    – Federico
    Dec 7 '18 at 16:29










  • $begingroup$
    If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
    $endgroup$
    – Tag
    Dec 7 '18 at 16:34










  • $begingroup$
    @Federico yes, but perhaps we can do it neatly with a change of variables.
    $endgroup$
    – hardmath
    Dec 7 '18 at 16:35










  • $begingroup$
    @user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
    $endgroup$
    – Ross Millikan
    Dec 7 '18 at 16:39


















  • $begingroup$
    This might be amenable to using polar coordinates.
    $endgroup$
    – hardmath
    Dec 7 '18 at 16:23










  • $begingroup$
    @hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
    $endgroup$
    – Federico
    Dec 7 '18 at 16:29










  • $begingroup$
    If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
    $endgroup$
    – Tag
    Dec 7 '18 at 16:34










  • $begingroup$
    @Federico yes, but perhaps we can do it neatly with a change of variables.
    $endgroup$
    – hardmath
    Dec 7 '18 at 16:35










  • $begingroup$
    @user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
    $endgroup$
    – Ross Millikan
    Dec 7 '18 at 16:39
















$begingroup$
This might be amenable to using polar coordinates.
$endgroup$
– hardmath
Dec 7 '18 at 16:23




$begingroup$
This might be amenable to using polar coordinates.
$endgroup$
– hardmath
Dec 7 '18 at 16:23












$begingroup$
@hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
$endgroup$
– Federico
Dec 7 '18 at 16:29




$begingroup$
@hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
$endgroup$
– Federico
Dec 7 '18 at 16:29












$begingroup$
If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
$endgroup$
– Tag
Dec 7 '18 at 16:34




$begingroup$
If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
$endgroup$
– Tag
Dec 7 '18 at 16:34












$begingroup$
@Federico yes, but perhaps we can do it neatly with a change of variables.
$endgroup$
– hardmath
Dec 7 '18 at 16:35




$begingroup$
@Federico yes, but perhaps we can do it neatly with a change of variables.
$endgroup$
– hardmath
Dec 7 '18 at 16:35












$begingroup$
@user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
$endgroup$
– Ross Millikan
Dec 7 '18 at 16:39




$begingroup$
@user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
$endgroup$
– Ross Millikan
Dec 7 '18 at 16:39










1 Answer
1






active

oldest

votes


















2












$begingroup$

Seems like the smallest circle is always tangent at $x=0$ or $y=0$. So you can put $arctan(x)^2=a$ and solve $x=tansqrt a$. This would be the radius.



Edit: let's make it rigorous.



Let's maximize $x^2+y^2$ with the constraint $arctan(x)^2+arctan(y)^2=a$. By symmetry we can work in the positive quadrant. With Lagrange multipliers we find
$$
left(frac{arctan(x)}{1+x^2},frac{arctan(y)}{1+y^2}right)
= lambda (x,y),
$$
therefore (unless $x=0$ or $y=0$, which are easily seen to be stationary points) we have
$$
frac{arctan(x)}{(1+x^2)x} = frac{arctan(y)}{(1+y^2)y}.
$$
We want to show that the only possibility is $x=y$. The we need an argument to say that this is a minimum and not a maximum.



As a matter of fact, the function $frac{arctan(x)}{(1+x^2)x}$ is decreasing, so this shows that $x=y$ are the only solutions. Its derivative is in fact
$$
begin{split}
frac{d}{dx}frac{arctan(x)}{(1+x^2)x} &=
frac{2}{x
left(x^2+1right)^2}-frac{2 arctan(x)}{x^2
left(x^2+1right)}-frac{4 arctan(x)}{left(x^2+1right)^2} \
&< frac{2}{x
left(x^2+1right)^2}-frac{2 arctan(x)}{x^2
left(x^2+1right)} \
&= -2 frac{(1+x^2)arctan(x)-x}{(x+x^3)^2} < 0.
end{split}
$$



You may rightfully wonder why is $(1+x^2)arctan(x)-x>0$? Well, it vanishes at $0$ and its derivative is $2xarctan(x)>0$.



So, why is $x=y$ a minimum? Well, we just need to compare it with the point at $y=0$.



The values of $x^2+y^2$ at the two points are respectively $2(tansqrt{a/2})^2$ and $(tansqrt a)^2$. Since the function $f(t)=(tansqrt t)^2$ is convex and vanishes at $0$, it is superadditive, implying that $2f(a/2)leq f(a)$.




Bonus



I'd like to add one little piece, related to proving that $(1+x^2)arctan(x)-x$ is increasing. The insight is that it is of the form $frac{g(x)}{g'(x)}-x$ with $g$ concave. Then it's derivative is $-frac{g(x)g''(x)}{g'(x)^2} geq 0$. I don't know if this general detail might be useful to someone.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
    $endgroup$
    – Tag
    Dec 7 '18 at 16:29










  • $begingroup$
    @user8053696 not at the moment
    $endgroup$
    – Federico
    Dec 7 '18 at 16:30










  • $begingroup$
    Could you, please, clarify why we are sure that $x=y$ is a minimum?
    $endgroup$
    – Tag
    Dec 7 '18 at 17:11












  • $begingroup$
    @user8053696 I added the proof of this last bit
    $endgroup$
    – Federico
    Dec 7 '18 at 17:19










  • $begingroup$
    Wow, this is amazing! Many thanks.
    $endgroup$
    – Tag
    Dec 7 '18 at 17:23











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Seems like the smallest circle is always tangent at $x=0$ or $y=0$. So you can put $arctan(x)^2=a$ and solve $x=tansqrt a$. This would be the radius.



Edit: let's make it rigorous.



Let's maximize $x^2+y^2$ with the constraint $arctan(x)^2+arctan(y)^2=a$. By symmetry we can work in the positive quadrant. With Lagrange multipliers we find
$$
left(frac{arctan(x)}{1+x^2},frac{arctan(y)}{1+y^2}right)
= lambda (x,y),
$$
therefore (unless $x=0$ or $y=0$, which are easily seen to be stationary points) we have
$$
frac{arctan(x)}{(1+x^2)x} = frac{arctan(y)}{(1+y^2)y}.
$$
We want to show that the only possibility is $x=y$. The we need an argument to say that this is a minimum and not a maximum.



As a matter of fact, the function $frac{arctan(x)}{(1+x^2)x}$ is decreasing, so this shows that $x=y$ are the only solutions. Its derivative is in fact
$$
begin{split}
frac{d}{dx}frac{arctan(x)}{(1+x^2)x} &=
frac{2}{x
left(x^2+1right)^2}-frac{2 arctan(x)}{x^2
left(x^2+1right)}-frac{4 arctan(x)}{left(x^2+1right)^2} \
&< frac{2}{x
left(x^2+1right)^2}-frac{2 arctan(x)}{x^2
left(x^2+1right)} \
&= -2 frac{(1+x^2)arctan(x)-x}{(x+x^3)^2} < 0.
end{split}
$$



You may rightfully wonder why is $(1+x^2)arctan(x)-x>0$? Well, it vanishes at $0$ and its derivative is $2xarctan(x)>0$.



So, why is $x=y$ a minimum? Well, we just need to compare it with the point at $y=0$.



The values of $x^2+y^2$ at the two points are respectively $2(tansqrt{a/2})^2$ and $(tansqrt a)^2$. Since the function $f(t)=(tansqrt t)^2$ is convex and vanishes at $0$, it is superadditive, implying that $2f(a/2)leq f(a)$.




Bonus



I'd like to add one little piece, related to proving that $(1+x^2)arctan(x)-x$ is increasing. The insight is that it is of the form $frac{g(x)}{g'(x)}-x$ with $g$ concave. Then it's derivative is $-frac{g(x)g''(x)}{g'(x)^2} geq 0$. I don't know if this general detail might be useful to someone.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
    $endgroup$
    – Tag
    Dec 7 '18 at 16:29










  • $begingroup$
    @user8053696 not at the moment
    $endgroup$
    – Federico
    Dec 7 '18 at 16:30










  • $begingroup$
    Could you, please, clarify why we are sure that $x=y$ is a minimum?
    $endgroup$
    – Tag
    Dec 7 '18 at 17:11












  • $begingroup$
    @user8053696 I added the proof of this last bit
    $endgroup$
    – Federico
    Dec 7 '18 at 17:19










  • $begingroup$
    Wow, this is amazing! Many thanks.
    $endgroup$
    – Tag
    Dec 7 '18 at 17:23
















2












$begingroup$

Seems like the smallest circle is always tangent at $x=0$ or $y=0$. So you can put $arctan(x)^2=a$ and solve $x=tansqrt a$. This would be the radius.



Edit: let's make it rigorous.



Let's maximize $x^2+y^2$ with the constraint $arctan(x)^2+arctan(y)^2=a$. By symmetry we can work in the positive quadrant. With Lagrange multipliers we find
$$
left(frac{arctan(x)}{1+x^2},frac{arctan(y)}{1+y^2}right)
= lambda (x,y),
$$
therefore (unless $x=0$ or $y=0$, which are easily seen to be stationary points) we have
$$
frac{arctan(x)}{(1+x^2)x} = frac{arctan(y)}{(1+y^2)y}.
$$
We want to show that the only possibility is $x=y$. The we need an argument to say that this is a minimum and not a maximum.



As a matter of fact, the function $frac{arctan(x)}{(1+x^2)x}$ is decreasing, so this shows that $x=y$ are the only solutions. Its derivative is in fact
$$
begin{split}
frac{d}{dx}frac{arctan(x)}{(1+x^2)x} &=
frac{2}{x
left(x^2+1right)^2}-frac{2 arctan(x)}{x^2
left(x^2+1right)}-frac{4 arctan(x)}{left(x^2+1right)^2} \
&< frac{2}{x
left(x^2+1right)^2}-frac{2 arctan(x)}{x^2
left(x^2+1right)} \
&= -2 frac{(1+x^2)arctan(x)-x}{(x+x^3)^2} < 0.
end{split}
$$



You may rightfully wonder why is $(1+x^2)arctan(x)-x>0$? Well, it vanishes at $0$ and its derivative is $2xarctan(x)>0$.



So, why is $x=y$ a minimum? Well, we just need to compare it with the point at $y=0$.



The values of $x^2+y^2$ at the two points are respectively $2(tansqrt{a/2})^2$ and $(tansqrt a)^2$. Since the function $f(t)=(tansqrt t)^2$ is convex and vanishes at $0$, it is superadditive, implying that $2f(a/2)leq f(a)$.




Bonus



I'd like to add one little piece, related to proving that $(1+x^2)arctan(x)-x$ is increasing. The insight is that it is of the form $frac{g(x)}{g'(x)}-x$ with $g$ concave. Then it's derivative is $-frac{g(x)g''(x)}{g'(x)^2} geq 0$. I don't know if this general detail might be useful to someone.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
    $endgroup$
    – Tag
    Dec 7 '18 at 16:29










  • $begingroup$
    @user8053696 not at the moment
    $endgroup$
    – Federico
    Dec 7 '18 at 16:30










  • $begingroup$
    Could you, please, clarify why we are sure that $x=y$ is a minimum?
    $endgroup$
    – Tag
    Dec 7 '18 at 17:11












  • $begingroup$
    @user8053696 I added the proof of this last bit
    $endgroup$
    – Federico
    Dec 7 '18 at 17:19










  • $begingroup$
    Wow, this is amazing! Many thanks.
    $endgroup$
    – Tag
    Dec 7 '18 at 17:23














2












2








2





$begingroup$

Seems like the smallest circle is always tangent at $x=0$ or $y=0$. So you can put $arctan(x)^2=a$ and solve $x=tansqrt a$. This would be the radius.



Edit: let's make it rigorous.



Let's maximize $x^2+y^2$ with the constraint $arctan(x)^2+arctan(y)^2=a$. By symmetry we can work in the positive quadrant. With Lagrange multipliers we find
$$
left(frac{arctan(x)}{1+x^2},frac{arctan(y)}{1+y^2}right)
= lambda (x,y),
$$
therefore (unless $x=0$ or $y=0$, which are easily seen to be stationary points) we have
$$
frac{arctan(x)}{(1+x^2)x} = frac{arctan(y)}{(1+y^2)y}.
$$
We want to show that the only possibility is $x=y$. The we need an argument to say that this is a minimum and not a maximum.



As a matter of fact, the function $frac{arctan(x)}{(1+x^2)x}$ is decreasing, so this shows that $x=y$ are the only solutions. Its derivative is in fact
$$
begin{split}
frac{d}{dx}frac{arctan(x)}{(1+x^2)x} &=
frac{2}{x
left(x^2+1right)^2}-frac{2 arctan(x)}{x^2
left(x^2+1right)}-frac{4 arctan(x)}{left(x^2+1right)^2} \
&< frac{2}{x
left(x^2+1right)^2}-frac{2 arctan(x)}{x^2
left(x^2+1right)} \
&= -2 frac{(1+x^2)arctan(x)-x}{(x+x^3)^2} < 0.
end{split}
$$



You may rightfully wonder why is $(1+x^2)arctan(x)-x>0$? Well, it vanishes at $0$ and its derivative is $2xarctan(x)>0$.



So, why is $x=y$ a minimum? Well, we just need to compare it with the point at $y=0$.



The values of $x^2+y^2$ at the two points are respectively $2(tansqrt{a/2})^2$ and $(tansqrt a)^2$. Since the function $f(t)=(tansqrt t)^2$ is convex and vanishes at $0$, it is superadditive, implying that $2f(a/2)leq f(a)$.




Bonus



I'd like to add one little piece, related to proving that $(1+x^2)arctan(x)-x$ is increasing. The insight is that it is of the form $frac{g(x)}{g'(x)}-x$ with $g$ concave. Then it's derivative is $-frac{g(x)g''(x)}{g'(x)^2} geq 0$. I don't know if this general detail might be useful to someone.






share|cite|improve this answer











$endgroup$



Seems like the smallest circle is always tangent at $x=0$ or $y=0$. So you can put $arctan(x)^2=a$ and solve $x=tansqrt a$. This would be the radius.



Edit: let's make it rigorous.



Let's maximize $x^2+y^2$ with the constraint $arctan(x)^2+arctan(y)^2=a$. By symmetry we can work in the positive quadrant. With Lagrange multipliers we find
$$
left(frac{arctan(x)}{1+x^2},frac{arctan(y)}{1+y^2}right)
= lambda (x,y),
$$
therefore (unless $x=0$ or $y=0$, which are easily seen to be stationary points) we have
$$
frac{arctan(x)}{(1+x^2)x} = frac{arctan(y)}{(1+y^2)y}.
$$
We want to show that the only possibility is $x=y$. The we need an argument to say that this is a minimum and not a maximum.



As a matter of fact, the function $frac{arctan(x)}{(1+x^2)x}$ is decreasing, so this shows that $x=y$ are the only solutions. Its derivative is in fact
$$
begin{split}
frac{d}{dx}frac{arctan(x)}{(1+x^2)x} &=
frac{2}{x
left(x^2+1right)^2}-frac{2 arctan(x)}{x^2
left(x^2+1right)}-frac{4 arctan(x)}{left(x^2+1right)^2} \
&< frac{2}{x
left(x^2+1right)^2}-frac{2 arctan(x)}{x^2
left(x^2+1right)} \
&= -2 frac{(1+x^2)arctan(x)-x}{(x+x^3)^2} < 0.
end{split}
$$



You may rightfully wonder why is $(1+x^2)arctan(x)-x>0$? Well, it vanishes at $0$ and its derivative is $2xarctan(x)>0$.



So, why is $x=y$ a minimum? Well, we just need to compare it with the point at $y=0$.



The values of $x^2+y^2$ at the two points are respectively $2(tansqrt{a/2})^2$ and $(tansqrt a)^2$. Since the function $f(t)=(tansqrt t)^2$ is convex and vanishes at $0$, it is superadditive, implying that $2f(a/2)leq f(a)$.




Bonus



I'd like to add one little piece, related to proving that $(1+x^2)arctan(x)-x$ is increasing. The insight is that it is of the form $frac{g(x)}{g'(x)}-x$ with $g$ concave. Then it's derivative is $-frac{g(x)g''(x)}{g'(x)^2} geq 0$. I don't know if this general detail might be useful to someone.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 '18 at 17:30

























answered Dec 7 '18 at 16:17









FedericoFederico

5,014514




5,014514












  • $begingroup$
    Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
    $endgroup$
    – Tag
    Dec 7 '18 at 16:29










  • $begingroup$
    @user8053696 not at the moment
    $endgroup$
    – Federico
    Dec 7 '18 at 16:30










  • $begingroup$
    Could you, please, clarify why we are sure that $x=y$ is a minimum?
    $endgroup$
    – Tag
    Dec 7 '18 at 17:11












  • $begingroup$
    @user8053696 I added the proof of this last bit
    $endgroup$
    – Federico
    Dec 7 '18 at 17:19










  • $begingroup$
    Wow, this is amazing! Many thanks.
    $endgroup$
    – Tag
    Dec 7 '18 at 17:23


















  • $begingroup$
    Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
    $endgroup$
    – Tag
    Dec 7 '18 at 16:29










  • $begingroup$
    @user8053696 not at the moment
    $endgroup$
    – Federico
    Dec 7 '18 at 16:30










  • $begingroup$
    Could you, please, clarify why we are sure that $x=y$ is a minimum?
    $endgroup$
    – Tag
    Dec 7 '18 at 17:11












  • $begingroup$
    @user8053696 I added the proof of this last bit
    $endgroup$
    – Federico
    Dec 7 '18 at 17:19










  • $begingroup$
    Wow, this is amazing! Many thanks.
    $endgroup$
    – Tag
    Dec 7 '18 at 17:23
















$begingroup$
Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
$endgroup$
– Tag
Dec 7 '18 at 16:29




$begingroup$
Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
$endgroup$
– Tag
Dec 7 '18 at 16:29












$begingroup$
@user8053696 not at the moment
$endgroup$
– Federico
Dec 7 '18 at 16:30




$begingroup$
@user8053696 not at the moment
$endgroup$
– Federico
Dec 7 '18 at 16:30












$begingroup$
Could you, please, clarify why we are sure that $x=y$ is a minimum?
$endgroup$
– Tag
Dec 7 '18 at 17:11






$begingroup$
Could you, please, clarify why we are sure that $x=y$ is a minimum?
$endgroup$
– Tag
Dec 7 '18 at 17:11














$begingroup$
@user8053696 I added the proof of this last bit
$endgroup$
– Federico
Dec 7 '18 at 17:19




$begingroup$
@user8053696 I added the proof of this last bit
$endgroup$
– Federico
Dec 7 '18 at 17:19












$begingroup$
Wow, this is amazing! Many thanks.
$endgroup$
– Tag
Dec 7 '18 at 17:23




$begingroup$
Wow, this is amazing! Many thanks.
$endgroup$
– Tag
Dec 7 '18 at 17:23


















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