Mixing
Proving that a function derived from $arctan(x/y)$ is continuous on $yne0$
$begingroup$
$Omega_1 = {y > 0} $
$Omega_2 = {y < 0} $
$Omega_3 = mathbb R^2 backslash {x leq 0 ; y = 0 } $
begin{equation}
f(x,y) = left {
begin{aligned}
&- arctan frac x y + pi, && text{if} (x,y) in Omega_1 \
& frac {pi} 2, && text{if} y = 0 ; x > 0 \
&- arctan frac x y, && text{if} (x,y) in Omega_2
end{aligned} right.
end{equation}
Prove that :$$ f in C^1( Omega_3) $$
So the obviously only problem is on the boundary of $Omega_1$ and $Omega_2$. I don't know how to prove neither continuity nor differentiability. Can you help me ?
multivariable-calculus
edited Jan 6 at 12:50
Did
247k23222458
asked Jan 5 at 21:25
Niels Niels
186
$endgroup$
add a comment |
$begingroup$
$Omega_1 = {y > 0} $
$Omega_2 = {y < 0} $
$Omega_3 = mathbb R^2 backslash {x leq 0 ; y = 0 } $
begin{equation}
f(x,y) = left {
begin{aligned}
&- arctan frac x y + pi, && text{if} (x,y) in Omega_1 \
& frac {pi} 2, && text{if} y = 0 ; x > 0 \
&- arctan frac x y, && text{if} (x,y) in Omega_2
end{aligned} right.
end{equation}
Prove that :$$ f in C^1( Omega_3) $$
So the obviously only problem is on the boundary of $Omega_1$ and $Omega_2$. I don't know how to prove neither continuity nor differentiability. Can you help me ?
multivariable-calculus
edited Jan 6 at 12:50
Did
247k23222458
asked Jan 5 at 21:25
Niels Niels
186
$endgroup$
add a comment |
$begingroup$
$Omega_1 = {y > 0} $
$Omega_2 = {y < 0} $
$Omega_3 = mathbb R^2 backslash {x leq 0 ; y = 0 } $
begin{equation}
f(x,y) = left {
begin{aligned}
&- arctan frac x y + pi, && text{if} (x,y) in Omega_1 \
& frac {pi} 2, && text{if} y = 0 ; x > 0 \
&- arctan frac x y, && text{if} (x,y) in Omega_2
end{aligned} right.
end{equation}
Prove that :$$ f in C^1( Omega_3) $$
So the obviously only problem is on the boundary of $Omega_1$ and $Omega_2$. I don't know how to prove neither continuity nor differentiability. Can you help me ?
multivariable-calculus
edited Jan 6 at 12:50
Did
247k23222458
asked Jan 5 at 21:25
Niels Niels
186
$endgroup$
$Omega_1 = {y > 0} $
$Omega_2 = {y < 0} $
$Omega_3 = mathbb R^2 backslash {x leq 0 ; y = 0 } $
begin{equation}
f(x,y) = left {
begin{aligned}
&- arctan frac x y + pi, && text{if} (x,y) in Omega_1 \
& frac {pi} 2, && text{if} y = 0 ; x > 0 \
&- arctan frac x y, && text{if} (x,y) in Omega_2
end{aligned} right.
end{equation}
Prove that :$$ f in C^1( Omega_3) $$
So the obviously only problem is on the boundary of $Omega_1$ and $Omega_2$. I don't know how to prove neither continuity nor differentiability. Can you help me ?
multivariable-calculus
multivariable-calculus
edited Jan 6 at 12:50
Did
247k23222458
asked Jan 5 at 21:25
Niels Niels
186
edited Jan 6 at 12:50
Did
247k23222458
asked Jan 5 at 21:25
Niels Niels
186
edited Jan 6 at 12:50
Did
247k23222458
edited Jan 6 at 12:50
Did
247k23222458
edited Jan 6 at 12:50
Did
247k23222458
247k23222458
asked Jan 5 at 21:25
Niels Niels
186
asked Jan 5 at 21:25
Niels Niels
186
asked Jan 5 at 21:25
Niels Niels
186
186
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Ok well, I have something of which I'm not completly sure. Do not hesitate to comment my proof.
What I'm going to show is that the $f$ function is differentiable, meaning it has continuous partial derivatives. This will give continuity.
First : (I note $f_x$ the partial derivative according to the first variable.)
$$ g(x,y) := f_x(x,y) = frac {y}{x^2+y^2} $$
$$ h(x,y) := f_y(x,y) = -frac {x}{x^2+y^2} $$
So, let's take a look at those two functions.
We fixe $A(x,y)$ ; $B( x, overline{y}) $ ; $ C( overline{x} , overline{y})$
We want to prove that :
$$lim_{ (x,y) → (overline{x} , overline{y}) } g(x,y) = g(overline{x} , overline{y}) $$
Nevertheless :
$$|g(x,y) - g(overline{x} , overline{y})| leq |g(x,y) - g(x , overline{y})| + |g(x , overline{y}) -g(overline{x} , overline{y}) | $$
and :
$$|g(x,y) - g(x , overline{y})| = | frac {y}{x^2 + y^2}| →_{(x,y) → (x , overline{y})} 0$$
$$|g(x , overline{y}) -g(overline{x} , overline{y}) | = 0$$
Then $g$ is continuous on $Omega_3$.
We do the same thing for $h$ and this will conclude the proof.
We fixe another 3 points : $A(x,y)$ ; $B( overline{x}, y) $ ; $ C( overline{x} , overline{y})$
We want to prove that :
$$lim_{ (x,y) → (overline{x} , overline{y}) } h(x,y) = h(overline{x} , overline{y}) $$
Nevertheless :
$$|h(x,y) - h(overline{x} , overline{y})| leq |h(x,y) - h(overline{x} , y)| + |h(overline{x} , y) -h(overline{x} , overline{y}) | $$
and :
$$|h(x,y) - h(overline{x} , y)| →_{(x,y) → (overline{x} , y)} 0$$
$$|h(overline{x} , y) -h(overline{x} , overline{y}) | →_{(overline{x} , y) → (overline{x} , overline{y})} 0$$
Then $h$ is continuous on $Omega_3$.
answered Jan 6 at 12:39
Marine GalantinMarine Galantin
786216
$endgroup$
add a comment |
$begingroup$
To see what's going on, let's consider a circle centered at the origin, and trace what happens to $f$ as we go around. Parametrize the circle in the standard way as $(x,y)=(cos t,sin t)$ for $-pi < tle pi$ (That means it has radius $1$).
For $t=0$, we get $x=1,y=0$. That's the middle case of the definition of $f$, so $f(x(0),y(0))=frac{pi}{2}$.
For $0<t<pi$, we have $y=sin t>0$ and $f(x(t),y(t)) = pi -arctanfrac{cos t}{sin t} = pi - (frac{pi}{2}-t) = frac{pi}{2}+t$. We used $cot t =tan(frac{pi}{2}-t)$ there.
For $-pi < t < 0$, we have $y=sin t < 0$ and $f(x(t),y(t))=-arctanfrac{cos t}{sin t} = -(-frac{pi}{2}-t)=frac{pi}{2}+t$. It's still true that $cot t=tan(frac{pi}{2}-t)$, but since $arctan$ takes values in $(-frac{pi}{2},frac{pi}{2})$, we use a different branch instead.
For $t=pi$, we get $x=-1,y=0$. We can't fill in the value in a continuous way - it would have to be both $frac{3pi}{2}$ approaching from positive $t$ and $-frac{pi}{2}$ approaching from negative $t$ - but since the point isn't in the domain of $f$, we don't need to.
So then, on this nearly complete circle, $f(cos t,sin t) = frac{pi}{2}+t$. That's continuous, differentiable, and more. How do we extend it to the whole domain? Polar form; we can write an arbitrary point in the domain as $(x(r,theta),y(r,theta))=(rcostheta,rsintheta)$ for $r>0$ and $-pi<theta<pi$. Then $f(x(r,theta),y(r,theta)) = frac{pi}{2}+theta$. Verify that the coordinate functions for the transformation to polar form and its inverse are smooth, and we have it.
Added material:
Moreover, we can find more formulas for $f$ on various other subsets of the region it's defined on. Any half-plane in the domain has a shifted version of the arctangent formula; the easiest to find is the one for the right half-plane $Omega_0={(x,y): x>0}$. We get
$$f(x,y) = arctanfrac yx +frac{pi}{2}text{ if } x>0$$
That formula agrees with the ones we already have. Every point in the domain of $f$ is in at least one of the open regions $Omega_0,Omega_1,Omega_2$, and $f$ is clearly smooth (infinitely differentiable) on each of those regions.
answered Jan 5 at 21:59
jmerryjmerry
4,702514
$endgroup$
$begingroup$
I have a question, because you have check if it is C^1 on circles, but with this proof, did you check for every direction ? You know there are those classical counter example where the function is not C^1 but the partial derivatives exists.
$endgroup$
– Niels
Jan 5 at 23:13
$begingroup$
Moreover, I don't see the step where you show it is differentiable. You only wrote the continuity part, am I right?
$endgroup$
– Niels
Jan 5 at 23:14
$begingroup$
The circle stuff was exploratory work, not intended to be a complete proof - but once we see it, we know how it all fits together. What we have here is a shifted version of the function that finds the angle in the polar form of a point. Actually, I just had a thought...
$endgroup$
– jmerry
Jan 5 at 23:21
$begingroup$
yeah I mean it does look nice, but it's not sufficiently rigorous... I'm trying to understand how to write it down for an exam so I don't think it will be enough
$endgroup$
– Niels
Jan 6 at 11:08
add a comment |
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$begingroup$
Ok well, I have something of which I'm not completly sure. Do not hesitate to comment my proof.
What I'm going to show is that the $f$ function is differentiable, meaning it has continuous partial derivatives. This will give continuity.
First : (I note $f_x$ the partial derivative according to the first variable.)
$$ g(x,y) := f_x(x,y) = frac {y}{x^2+y^2} $$
$$ h(x,y) := f_y(x,y) = -frac {x}{x^2+y^2} $$
So, let's take a look at those two functions.
We fixe $A(x,y)$ ; $B( x, overline{y}) $ ; $ C( overline{x} , overline{y})$
We want to prove that :
$$lim_{ (x,y) → (overline{x} , overline{y}) } g(x,y) = g(overline{x} , overline{y}) $$
Nevertheless :
$$|g(x,y) - g(overline{x} , overline{y})| leq |g(x,y) - g(x , overline{y})| + |g(x , overline{y}) -g(overline{x} , overline{y}) | $$
and :
$$|g(x,y) - g(x , overline{y})| = | frac {y}{x^2 + y^2}| →_{(x,y) → (x , overline{y})} 0$$
$$|g(x , overline{y}) -g(overline{x} , overline{y}) | = 0$$
Then $g$ is continuous on $Omega_3$.
We do the same thing for $h$ and this will conclude the proof.
We fixe another 3 points : $A(x,y)$ ; $B( overline{x}, y) $ ; $ C( overline{x} , overline{y})$
We want to prove that :
$$lim_{ (x,y) → (overline{x} , overline{y}) } h(x,y) = h(overline{x} , overline{y}) $$
Nevertheless :
$$|h(x,y) - h(overline{x} , overline{y})| leq |h(x,y) - h(overline{x} , y)| + |h(overline{x} , y) -h(overline{x} , overline{y}) | $$
and :
$$|h(x,y) - h(overline{x} , y)| →_{(x,y) → (overline{x} , y)} 0$$
$$|h(overline{x} , y) -h(overline{x} , overline{y}) | →_{(overline{x} , y) → (overline{x} , overline{y})} 0$$
Then $h$ is continuous on $Omega_3$.
answered Jan 6 at 12:39
Marine GalantinMarine Galantin
786216
$endgroup$
add a comment |
$begingroup$
Ok well, I have something of which I'm not completly sure. Do not hesitate to comment my proof.
What I'm going to show is that the $f$ function is differentiable, meaning it has continuous partial derivatives. This will give continuity.
First : (I note $f_x$ the partial derivative according to the first variable.)
$$ g(x,y) := f_x(x,y) = frac {y}{x^2+y^2} $$
$$ h(x,y) := f_y(x,y) = -frac {x}{x^2+y^2} $$
So, let's take a look at those two functions.
We fixe $A(x,y)$ ; $B( x, overline{y}) $ ; $ C( overline{x} , overline{y})$
We want to prove that :
$$lim_{ (x,y) → (overline{x} , overline{y}) } g(x,y) = g(overline{x} , overline{y}) $$
Nevertheless :
$$|g(x,y) - g(overline{x} , overline{y})| leq |g(x,y) - g(x , overline{y})| + |g(x , overline{y}) -g(overline{x} , overline{y}) | $$
and :
$$|g(x,y) - g(x , overline{y})| = | frac {y}{x^2 + y^2}| →_{(x,y) → (x , overline{y})} 0$$
$$|g(x , overline{y}) -g(overline{x} , overline{y}) | = 0$$
Then $g$ is continuous on $Omega_3$.
We do the same thing for $h$ and this will conclude the proof.
We fixe another 3 points : $A(x,y)$ ; $B( overline{x}, y) $ ; $ C( overline{x} , overline{y})$
We want to prove that :
$$lim_{ (x,y) → (overline{x} , overline{y}) } h(x,y) = h(overline{x} , overline{y}) $$
Nevertheless :
$$|h(x,y) - h(overline{x} , overline{y})| leq |h(x,y) - h(overline{x} , y)| + |h(overline{x} , y) -h(overline{x} , overline{y}) | $$
and :
$$|h(x,y) - h(overline{x} , y)| →_{(x,y) → (overline{x} , y)} 0$$
$$|h(overline{x} , y) -h(overline{x} , overline{y}) | →_{(overline{x} , y) → (overline{x} , overline{y})} 0$$
Then $h$ is continuous on $Omega_3$.
answered Jan 6 at 12:39
Marine GalantinMarine Galantin
786216
$endgroup$
add a comment |
$begingroup$
Ok well, I have something of which I'm not completly sure. Do not hesitate to comment my proof.
What I'm going to show is that the $f$ function is differentiable, meaning it has continuous partial derivatives. This will give continuity.
First : (I note $f_x$ the partial derivative according to the first variable.)
$$ g(x,y) := f_x(x,y) = frac {y}{x^2+y^2} $$
$$ h(x,y) := f_y(x,y) = -frac {x}{x^2+y^2} $$
So, let's take a look at those two functions.
We fixe $A(x,y)$ ; $B( x, overline{y}) $ ; $ C( overline{x} , overline{y})$
We want to prove that :
$$lim_{ (x,y) → (overline{x} , overline{y}) } g(x,y) = g(overline{x} , overline{y}) $$
Nevertheless :
$$|g(x,y) - g(overline{x} , overline{y})| leq |g(x,y) - g(x , overline{y})| + |g(x , overline{y}) -g(overline{x} , overline{y}) | $$
and :
$$|g(x,y) - g(x , overline{y})| = | frac {y}{x^2 + y^2}| →_{(x,y) → (x , overline{y})} 0$$
$$|g(x , overline{y}) -g(overline{x} , overline{y}) | = 0$$
Then $g$ is continuous on $Omega_3$.
We do the same thing for $h$ and this will conclude the proof.
We fixe another 3 points : $A(x,y)$ ; $B( overline{x}, y) $ ; $ C( overline{x} , overline{y})$
We want to prove that :
$$lim_{ (x,y) → (overline{x} , overline{y}) } h(x,y) = h(overline{x} , overline{y}) $$
Nevertheless :
$$|h(x,y) - h(overline{x} , overline{y})| leq |h(x,y) - h(overline{x} , y)| + |h(overline{x} , y) -h(overline{x} , overline{y}) | $$
and :
$$|h(x,y) - h(overline{x} , y)| →_{(x,y) → (overline{x} , y)} 0$$
$$|h(overline{x} , y) -h(overline{x} , overline{y}) | →_{(overline{x} , y) → (overline{x} , overline{y})} 0$$
Then $h$ is continuous on $Omega_3$.
answered Jan 6 at 12:39
Marine GalantinMarine Galantin
786216
$endgroup$
Ok well, I have something of which I'm not completly sure. Do not hesitate to comment my proof.
What I'm going to show is that the $f$ function is differentiable, meaning it has continuous partial derivatives. This will give continuity.
First : (I note $f_x$ the partial derivative according to the first variable.)
$$ g(x,y) := f_x(x,y) = frac {y}{x^2+y^2} $$
$$ h(x,y) := f_y(x,y) = -frac {x}{x^2+y^2} $$
So, let's take a look at those two functions.
We fixe $A(x,y)$ ; $B( x, overline{y}) $ ; $ C( overline{x} , overline{y})$
We want to prove that :
$$lim_{ (x,y) → (overline{x} , overline{y}) } g(x,y) = g(overline{x} , overline{y}) $$
Nevertheless :
$$|g(x,y) - g(overline{x} , overline{y})| leq |g(x,y) - g(x , overline{y})| + |g(x , overline{y}) -g(overline{x} , overline{y}) | $$
and :
$$|g(x,y) - g(x , overline{y})| = | frac {y}{x^2 + y^2}| →_{(x,y) → (x , overline{y})} 0$$
$$|g(x , overline{y}) -g(overline{x} , overline{y}) | = 0$$
Then $g$ is continuous on $Omega_3$.
We do the same thing for $h$ and this will conclude the proof.
We fixe another 3 points : $A(x,y)$ ; $B( overline{x}, y) $ ; $ C( overline{x} , overline{y})$
We want to prove that :
$$lim_{ (x,y) → (overline{x} , overline{y}) } h(x,y) = h(overline{x} , overline{y}) $$
Nevertheless :
$$|h(x,y) - h(overline{x} , overline{y})| leq |h(x,y) - h(overline{x} , y)| + |h(overline{x} , y) -h(overline{x} , overline{y}) | $$
and :
$$|h(x,y) - h(overline{x} , y)| →_{(x,y) → (overline{x} , y)} 0$$
$$|h(overline{x} , y) -h(overline{x} , overline{y}) | →_{(overline{x} , y) → (overline{x} , overline{y})} 0$$
Then $h$ is continuous on $Omega_3$.
answered Jan 6 at 12:39
Marine GalantinMarine Galantin
786216
answered Jan 6 at 12:39
Marine GalantinMarine Galantin
786216
answered Jan 6 at 12:39
Marine GalantinMarine Galantin
786216
answered Jan 6 at 12:39
Marine GalantinMarine Galantin
786216
786216
add a comment |
add a comment |
$begingroup$
To see what's going on, let's consider a circle centered at the origin, and trace what happens to $f$ as we go around. Parametrize the circle in the standard way as $(x,y)=(cos t,sin t)$ for $-pi < tle pi$ (That means it has radius $1$).
For $t=0$, we get $x=1,y=0$. That's the middle case of the definition of $f$, so $f(x(0),y(0))=frac{pi}{2}$.
For $0<t<pi$, we have $y=sin t>0$ and $f(x(t),y(t)) = pi -arctanfrac{cos t}{sin t} = pi - (frac{pi}{2}-t) = frac{pi}{2}+t$. We used $cot t =tan(frac{pi}{2}-t)$ there.
For $-pi < t < 0$, we have $y=sin t < 0$ and $f(x(t),y(t))=-arctanfrac{cos t}{sin t} = -(-frac{pi}{2}-t)=frac{pi}{2}+t$. It's still true that $cot t=tan(frac{pi}{2}-t)$, but since $arctan$ takes values in $(-frac{pi}{2},frac{pi}{2})$, we use a different branch instead.
For $t=pi$, we get $x=-1,y=0$. We can't fill in the value in a continuous way - it would have to be both $frac{3pi}{2}$ approaching from positive $t$ and $-frac{pi}{2}$ approaching from negative $t$ - but since the point isn't in the domain of $f$, we don't need to.
So then, on this nearly complete circle, $f(cos t,sin t) = frac{pi}{2}+t$. That's continuous, differentiable, and more. How do we extend it to the whole domain? Polar form; we can write an arbitrary point in the domain as $(x(r,theta),y(r,theta))=(rcostheta,rsintheta)$ for $r>0$ and $-pi<theta<pi$. Then $f(x(r,theta),y(r,theta)) = frac{pi}{2}+theta$. Verify that the coordinate functions for the transformation to polar form and its inverse are smooth, and we have it.
Added material:
Moreover, we can find more formulas for $f$ on various other subsets of the region it's defined on. Any half-plane in the domain has a shifted version of the arctangent formula; the easiest to find is the one for the right half-plane $Omega_0={(x,y): x>0}$. We get
$$f(x,y) = arctanfrac yx +frac{pi}{2}text{ if } x>0$$
That formula agrees with the ones we already have. Every point in the domain of $f$ is in at least one of the open regions $Omega_0,Omega_1,Omega_2$, and $f$ is clearly smooth (infinitely differentiable) on each of those regions.
answered Jan 5 at 21:59
jmerryjmerry
4,702514
$endgroup$
$begingroup$
I have a question, because you have check if it is C^1 on circles, but with this proof, did you check for every direction ? You know there are those classical counter example where the function is not C^1 but the partial derivatives exists.
$endgroup$
– Niels
Jan 5 at 23:13
$begingroup$
Moreover, I don't see the step where you show it is differentiable. You only wrote the continuity part, am I right?
$endgroup$
– Niels
Jan 5 at 23:14
$begingroup$
The circle stuff was exploratory work, not intended to be a complete proof - but once we see it, we know how it all fits together. What we have here is a shifted version of the function that finds the angle in the polar form of a point. Actually, I just had a thought...
$endgroup$
– jmerry
Jan 5 at 23:21
$begingroup$
yeah I mean it does look nice, but it's not sufficiently rigorous... I'm trying to understand how to write it down for an exam so I don't think it will be enough
$endgroup$
– Niels
Jan 6 at 11:08
add a comment |
$begingroup$
To see what's going on, let's consider a circle centered at the origin, and trace what happens to $f$ as we go around. Parametrize the circle in the standard way as $(x,y)=(cos t,sin t)$ for $-pi < tle pi$ (That means it has radius $1$).
For $t=0$, we get $x=1,y=0$. That's the middle case of the definition of $f$, so $f(x(0),y(0))=frac{pi}{2}$.
For $0<t<pi$, we have $y=sin t>0$ and $f(x(t),y(t)) = pi -arctanfrac{cos t}{sin t} = pi - (frac{pi}{2}-t) = frac{pi}{2}+t$. We used $cot t =tan(frac{pi}{2}-t)$ there.
For $-pi < t < 0$, we have $y=sin t < 0$ and $f(x(t),y(t))=-arctanfrac{cos t}{sin t} = -(-frac{pi}{2}-t)=frac{pi}{2}+t$. It's still true that $cot t=tan(frac{pi}{2}-t)$, but since $arctan$ takes values in $(-frac{pi}{2},frac{pi}{2})$, we use a different branch instead.
For $t=pi$, we get $x=-1,y=0$. We can't fill in the value in a continuous way - it would have to be both $frac{3pi}{2}$ approaching from positive $t$ and $-frac{pi}{2}$ approaching from negative $t$ - but since the point isn't in the domain of $f$, we don't need to.
So then, on this nearly complete circle, $f(cos t,sin t) = frac{pi}{2}+t$. That's continuous, differentiable, and more. How do we extend it to the whole domain? Polar form; we can write an arbitrary point in the domain as $(x(r,theta),y(r,theta))=(rcostheta,rsintheta)$ for $r>0$ and $-pi<theta<pi$. Then $f(x(r,theta),y(r,theta)) = frac{pi}{2}+theta$. Verify that the coordinate functions for the transformation to polar form and its inverse are smooth, and we have it.
Added material:
Moreover, we can find more formulas for $f$ on various other subsets of the region it's defined on. Any half-plane in the domain has a shifted version of the arctangent formula; the easiest to find is the one for the right half-plane $Omega_0={(x,y): x>0}$. We get
$$f(x,y) = arctanfrac yx +frac{pi}{2}text{ if } x>0$$
That formula agrees with the ones we already have. Every point in the domain of $f$ is in at least one of the open regions $Omega_0,Omega_1,Omega_2$, and $f$ is clearly smooth (infinitely differentiable) on each of those regions.
answered Jan 5 at 21:59
jmerryjmerry
4,702514
$endgroup$
$begingroup$
I have a question, because you have check if it is C^1 on circles, but with this proof, did you check for every direction ? You know there are those classical counter example where the function is not C^1 but the partial derivatives exists.
$endgroup$
– Niels
Jan 5 at 23:13
$begingroup$
Moreover, I don't see the step where you show it is differentiable. You only wrote the continuity part, am I right?
$endgroup$
– Niels
Jan 5 at 23:14
$begingroup$
The circle stuff was exploratory work, not intended to be a complete proof - but once we see it, we know how it all fits together. What we have here is a shifted version of the function that finds the angle in the polar form of a point. Actually, I just had a thought...
$endgroup$
– jmerry
Jan 5 at 23:21
$begingroup$
yeah I mean it does look nice, but it's not sufficiently rigorous... I'm trying to understand how to write it down for an exam so I don't think it will be enough
$endgroup$
– Niels
Jan 6 at 11:08
add a comment |
$begingroup$
To see what's going on, let's consider a circle centered at the origin, and trace what happens to $f$ as we go around. Parametrize the circle in the standard way as $(x,y)=(cos t,sin t)$ for $-pi < tle pi$ (That means it has radius $1$).
For $t=0$, we get $x=1,y=0$. That's the middle case of the definition of $f$, so $f(x(0),y(0))=frac{pi}{2}$.
For $0<t<pi$, we have $y=sin t>0$ and $f(x(t),y(t)) = pi -arctanfrac{cos t}{sin t} = pi - (frac{pi}{2}-t) = frac{pi}{2}+t$. We used $cot t =tan(frac{pi}{2}-t)$ there.
For $-pi < t < 0$, we have $y=sin t < 0$ and $f(x(t),y(t))=-arctanfrac{cos t}{sin t} = -(-frac{pi}{2}-t)=frac{pi}{2}+t$. It's still true that $cot t=tan(frac{pi}{2}-t)$, but since $arctan$ takes values in $(-frac{pi}{2},frac{pi}{2})$, we use a different branch instead.
For $t=pi$, we get $x=-1,y=0$. We can't fill in the value in a continuous way - it would have to be both $frac{3pi}{2}$ approaching from positive $t$ and $-frac{pi}{2}$ approaching from negative $t$ - but since the point isn't in the domain of $f$, we don't need to.
So then, on this nearly complete circle, $f(cos t,sin t) = frac{pi}{2}+t$. That's continuous, differentiable, and more. How do we extend it to the whole domain? Polar form; we can write an arbitrary point in the domain as $(x(r,theta),y(r,theta))=(rcostheta,rsintheta)$ for $r>0$ and $-pi<theta<pi$. Then $f(x(r,theta),y(r,theta)) = frac{pi}{2}+theta$. Verify that the coordinate functions for the transformation to polar form and its inverse are smooth, and we have it.
Added material:
Moreover, we can find more formulas for $f$ on various other subsets of the region it's defined on. Any half-plane in the domain has a shifted version of the arctangent formula; the easiest to find is the one for the right half-plane $Omega_0={(x,y): x>0}$. We get
$$f(x,y) = arctanfrac yx +frac{pi}{2}text{ if } x>0$$
That formula agrees with the ones we already have. Every point in the domain of $f$ is in at least one of the open regions $Omega_0,Omega_1,Omega_2$, and $f$ is clearly smooth (infinitely differentiable) on each of those regions.
answered Jan 5 at 21:59
jmerryjmerry
4,702514
$endgroup$
To see what's going on, let's consider a circle centered at the origin, and trace what happens to $f$ as we go around. Parametrize the circle in the standard way as $(x,y)=(cos t,sin t)$ for $-pi < tle pi$ (That means it has radius $1$).
For $t=0$, we get $x=1,y=0$. That's the middle case of the definition of $f$, so $f(x(0),y(0))=frac{pi}{2}$.
For $0<t<pi$, we have $y=sin t>0$ and $f(x(t),y(t)) = pi -arctanfrac{cos t}{sin t} = pi - (frac{pi}{2}-t) = frac{pi}{2}+t$. We used $cot t =tan(frac{pi}{2}-t)$ there.
For $-pi < t < 0$, we have $y=sin t < 0$ and $f(x(t),y(t))=-arctanfrac{cos t}{sin t} = -(-frac{pi}{2}-t)=frac{pi}{2}+t$. It's still true that $cot t=tan(frac{pi}{2}-t)$, but since $arctan$ takes values in $(-frac{pi}{2},frac{pi}{2})$, we use a different branch instead.
For $t=pi$, we get $x=-1,y=0$. We can't fill in the value in a continuous way - it would have to be both $frac{3pi}{2}$ approaching from positive $t$ and $-frac{pi}{2}$ approaching from negative $t$ - but since the point isn't in the domain of $f$, we don't need to.
So then, on this nearly complete circle, $f(cos t,sin t) = frac{pi}{2}+t$. That's continuous, differentiable, and more. How do we extend it to the whole domain? Polar form; we can write an arbitrary point in the domain as $(x(r,theta),y(r,theta))=(rcostheta,rsintheta)$ for $r>0$ and $-pi<theta<pi$. Then $f(x(r,theta),y(r,theta)) = frac{pi}{2}+theta$. Verify that the coordinate functions for the transformation to polar form and its inverse are smooth, and we have it.
Added material:
Moreover, we can find more formulas for $f$ on various other subsets of the region it's defined on. Any half-plane in the domain has a shifted version of the arctangent formula; the easiest to find is the one for the right half-plane $Omega_0={(x,y): x>0}$. We get
$$f(x,y) = arctanfrac yx +frac{pi}{2}text{ if } x>0$$
That formula agrees with the ones we already have. Every point in the domain of $f$ is in at least one of the open regions $Omega_0,Omega_1,Omega_2$, and $f$ is clearly smooth (infinitely differentiable) on each of those regions.
answered Jan 5 at 21:59
jmerryjmerry
4,702514
edited Jan 5 at 23:30
answered Jan 5 at 21:59
jmerryjmerry
4,702514
answered Jan 5 at 21:59
jmerryjmerry
4,702514
answered Jan 5 at 21:59
jmerryjmerry
4,702514
4,702514
$begingroup$
I have a question, because you have check if it is C^1 on circles, but with this proof, did you check for every direction ? You know there are those classical counter example where the function is not C^1 but the partial derivatives exists.
$endgroup$
– Niels
Jan 5 at 23:13
$begingroup$
Moreover, I don't see the step where you show it is differentiable. You only wrote the continuity part, am I right?
$endgroup$
– Niels
Jan 5 at 23:14
$begingroup$
The circle stuff was exploratory work, not intended to be a complete proof - but once we see it, we know how it all fits together. What we have here is a shifted version of the function that finds the angle in the polar form of a point. Actually, I just had a thought...
$endgroup$
– jmerry
Jan 5 at 23:21
$begingroup$
yeah I mean it does look nice, but it's not sufficiently rigorous... I'm trying to understand how to write it down for an exam so I don't think it will be enough
$endgroup$
– Niels
Jan 6 at 11:08
add a comment |
$begingroup$
I have a question, because you have check if it is C^1 on circles, but with this proof, did you check for every direction ? You know there are those classical counter example where the function is not C^1 but the partial derivatives exists.
$endgroup$
– Niels
Jan 5 at 23:13
$begingroup$
Moreover, I don't see the step where you show it is differentiable. You only wrote the continuity part, am I right?
$endgroup$
– Niels
Jan 5 at 23:14
$begingroup$
The circle stuff was exploratory work, not intended to be a complete proof - but once we see it, we know how it all fits together. What we have here is a shifted version of the function that finds the angle in the polar form of a point. Actually, I just had a thought...
$endgroup$
– jmerry
Jan 5 at 23:21
$begingroup$
yeah I mean it does look nice, but it's not sufficiently rigorous... I'm trying to understand how to write it down for an exam so I don't think it will be enough
$endgroup$
– Niels
Jan 6 at 11:08
$begingroup$
I have a question, because you have check if it is C^1 on circles, but with this proof, did you check for every direction ? You know there are those classical counter example where the function is not C^1 but the partial derivatives exists.
$endgroup$
– Niels
Jan 5 at 23:13
$begingroup$
I have a question, because you have check if it is C^1 on circles, but with this proof, did you check for every direction ? You know there are those classical counter example where the function is not C^1 but the partial derivatives exists.
$endgroup$
– Niels
Jan 5 at 23:13
$begingroup$
Moreover, I don't see the step where you show it is differentiable. You only wrote the continuity part, am I right?
$endgroup$
– Niels
Jan 5 at 23:14
$begingroup$
Moreover, I don't see the step where you show it is differentiable. You only wrote the continuity part, am I right?
$endgroup$
– Niels
Jan 5 at 23:14
$begingroup$
The circle stuff was exploratory work, not intended to be a complete proof - but once we see it, we know how it all fits together. What we have here is a shifted version of the function that finds the angle in the polar form of a point. Actually, I just had a thought...
$endgroup$
– jmerry
Jan 5 at 23:21
$begingroup$
The circle stuff was exploratory work, not intended to be a complete proof - but once we see it, we know how it all fits together. What we have here is a shifted version of the function that finds the angle in the polar form of a point. Actually, I just had a thought...
$endgroup$
– jmerry
Jan 5 at 23:21
$begingroup$
yeah I mean it does look nice, but it's not sufficiently rigorous... I'm trying to understand how to write it down for an exam so I don't think it will be enough
$endgroup$
– Niels
Jan 6 at 11:08
$begingroup$
yeah I mean it does look nice, but it's not sufficiently rigorous... I'm trying to understand how to write it down for an exam so I don't think it will be enough
$endgroup$
– Niels
Jan 6 at 11:08
add a comment |
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