How can we find minimum radius of circle which contains $arctan^2(x)+arctan^2(y)=a$?
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.
maxima-minima implicit-function
asked Dec 7 at 16:07
Tag
696
|
show 3 more comments
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.
maxima-minima implicit-function
asked Dec 7 at 16:07
Tag
696
This might be amenable to using polar coordinates.
– hardmath
Dec 7 at 16:23
@hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
– Federico
Dec 7 at 16:29
If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
– Tag
Dec 7 at 16:34
@Federico yes, but perhaps we can do it neatly with a change of variables.
– hardmath
Dec 7 at 16:35
@user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
– Ross Millikan
Dec 7 at 16:39
|
show 3 more comments
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.
maxima-minima implicit-function
asked Dec 7 at 16:07
Tag
696
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.
maxima-minima implicit-function
maxima-minima implicit-function
asked Dec 7 at 16:07
Tag
696
asked Dec 7 at 16:07
Tag
696
edited Dec 7 at 16:13
asked Dec 7 at 16:07
Tag
696
asked Dec 7 at 16:07
Tag
696
asked Dec 7 at 16:07
Tag
696
696
This might be amenable to using polar coordinates.
– hardmath
Dec 7 at 16:23
@hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
– Federico
Dec 7 at 16:29
If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
– Tag
Dec 7 at 16:34
@Federico yes, but perhaps we can do it neatly with a change of variables.
– hardmath
Dec 7 at 16:35
@user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
– Ross Millikan
Dec 7 at 16:39
|
show 3 more comments
This might be amenable to using polar coordinates.
– hardmath
Dec 7 at 16:23
@hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
– Federico
Dec 7 at 16:29
If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
– Tag
Dec 7 at 16:34
@Federico yes, but perhaps we can do it neatly with a change of variables.
– hardmath
Dec 7 at 16:35
@user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
– Ross Millikan
Dec 7 at 16:39
This might be amenable to using polar coordinates.
– hardmath
Dec 7 at 16:23
This might be amenable to using polar coordinates.
– hardmath
Dec 7 at 16:23
@hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
– Federico
Dec 7 at 16:29
@hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
– Federico
Dec 7 at 16:29
If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
– Tag
Dec 7 at 16:34
If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
– Tag
Dec 7 at 16:34
@Federico yes, but perhaps we can do it neatly with a change of variables.
– hardmath
Dec 7 at 16:35
@Federico yes, but perhaps we can do it neatly with a change of variables.
– hardmath
Dec 7 at 16:35
@user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
– Ross Millikan
Dec 7 at 16:39
@user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
– Ross Millikan
Dec 7 at 16:39
|
show 3 more comments
1 Answer
1
active
oldest
votes
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.
answered Dec 7 at 16:17
Federico
4,459512
Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
– Tag
Dec 7 at 16:29
@user8053696 not at the moment
– Federico
Dec 7 at 16:30
Could you, please, clarify why we are sure that $x=y$ is a minimum?
– Tag
Dec 7 at 17:11
@user8053696 I added the proof of this last bit
– Federico
Dec 7 at 17:19
Wow, this is amazing! Many thanks.
– Tag
Dec 7 at 17:23
|
show 6 more comments
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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.
answered Dec 7 at 16:17
Federico
4,459512
Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
– Tag
Dec 7 at 16:29
@user8053696 not at the moment
– Federico
Dec 7 at 16:30
Could you, please, clarify why we are sure that $x=y$ is a minimum?
– Tag
Dec 7 at 17:11
@user8053696 I added the proof of this last bit
– Federico
Dec 7 at 17:19
Wow, this is amazing! Many thanks.
– Tag
Dec 7 at 17:23
|
show 6 more comments
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.
answered Dec 7 at 16:17
Federico
4,459512
Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
– Tag
Dec 7 at 16:29
@user8053696 not at the moment
– Federico
Dec 7 at 16:30
Could you, please, clarify why we are sure that $x=y$ is a minimum?
– Tag
Dec 7 at 17:11
@user8053696 I added the proof of this last bit
– Federico
Dec 7 at 17:19
Wow, this is amazing! Many thanks.
– Tag
Dec 7 at 17:23
|
show 6 more comments
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.
answered Dec 7 at 16:17
Federico
4,459512
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.
answered Dec 7 at 16:17
Federico
4,459512
edited Dec 7 at 17:30
answered Dec 7 at 16:17
Federico
4,459512
answered Dec 7 at 16:17
Federico
4,459512
answered Dec 7 at 16:17
Federico
4,459512
4,459512
Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
– Tag
Dec 7 at 16:29
@user8053696 not at the moment
– Federico
Dec 7 at 16:30
Could you, please, clarify why we are sure that $x=y$ is a minimum?
– Tag
Dec 7 at 17:11
@user8053696 I added the proof of this last bit
– Federico
Dec 7 at 17:19
Wow, this is amazing! Many thanks.
– Tag
Dec 7 at 17:23
|
show 6 more comments
Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
– Tag
Dec 7 at 16:29
@user8053696 not at the moment
– Federico
Dec 7 at 16:30
Could you, please, clarify why we are sure that $x=y$ is a minimum?
– Tag
Dec 7 at 17:11
@user8053696 I added the proof of this last bit
– Federico
Dec 7 at 17:19
Wow, this is amazing! Many thanks.
– Tag
Dec 7 at 17:23
Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
– Tag
Dec 7 at 16:29
Can we prove that smallest circle is always tangent at $x=0$ or $y=0$?
– Tag
Dec 7 at 16:29
@user8053696 not at the moment
– Federico
Dec 7 at 16:30
@user8053696 not at the moment
– Federico
Dec 7 at 16:30
Could you, please, clarify why we are sure that $x=y$ is a minimum?
– Tag
Dec 7 at 17:11
Could you, please, clarify why we are sure that $x=y$ is a minimum?
– Tag
Dec 7 at 17:11
@user8053696 I added the proof of this last bit
– Federico
Dec 7 at 17:19
@user8053696 I added the proof of this last bit
– Federico
Dec 7 at 17:19
Wow, this is amazing! Many thanks.
– Tag
Dec 7 at 17:23
Wow, this is amazing! Many thanks.
– Tag
Dec 7 at 17:23
|
show 6 more comments
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This might be amenable to using polar coordinates.
– hardmath
Dec 7 at 16:23
@hardmath I believe it becomes an atrocious mess. Lagrange multipliers might be better?
– Federico
Dec 7 at 16:29
If we take derivative with respect to $x$ we and set $y'=0$ we will find that $x=0$. Isn't that correct proof?
– Tag
Dec 7 at 16:34
@Federico yes, but perhaps we can do it neatly with a change of variables.
– hardmath
Dec 7 at 16:35
@user8053696: no, the maximum radius might come at $x=y$. In that case the derivative would be $pm 1$
– Ross Millikan
Dec 7 at 16:39