Mixing
How do I prove that two equations in Cartesian and Polar coordinates are equivalent?
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I'm asked to verify that the set of points described by the cartesian equation $$(x-1)^2 + y^2 = 1$$ is equal to the set of points described by the polar equation $$r = 2 cos{theta}, cos{theta} > 0$$
Clearly the point described by $(0,0)$ under the Cartesian coordinate system is an element of the first set, since $(0-1)^2 + 0^2 = 1$. However, this point corresponds to polar coordinates of the form $(0, theta)$, for arbitrary $theta$, but $0 = 2cos{theta} iff cos{theta} = 0 iff theta=frac{pi}{2} + npi$; so there are plenty of pairs $(r, theta)$ that do satisfy the second condition, but there are plenty of pairs that do not. Do I just need to prove that there exists at least one such pair in the second set? How do I think about problems like this formally, and what would a formal proof of this look like?
Let $(x, y)$ such that $(x-1)^2 + y^2 = 1$. The pair $(r, theta)$ which corresponds to $(x, y)$ satisfies $x = rcos{theta}$, $y = rsin{theta}$. We must show that such a pair also satisfies $r = 2cos{theta}$.
If $r neq 0$, we have that $1 = (x-1)^2 + y^2 iff 1 = (rcos{theta} - 1)^2 + (rsin{theta})^2 iff 0 = r^2 - 2rcos{theta} iff r = 2cos{theta}$.
My problem is with the case of $r=0$. Is the argument below formal enough?
If $r = 0$, then we must have had that $x = rcos{theta} = 0$, $y = rsin{theta} = 0$. The pair $(0, frac{pi}{2})$ in polar coordinates satisfies $r = 2cos{theta}$, and so the origin is also contained in the second set.
polar-coordinates
asked Jan 25 at 22:39
TheProofisTriviumTheProofisTrivium
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I'm asked to verify that the set of points described by the cartesian equation $$(x-1)^2 + y^2 = 1$$ is equal to the set of points described by the polar equation $$r = 2 cos{theta}, cos{theta} > 0$$
Clearly the point described by $(0,0)$ under the Cartesian coordinate system is an element of the first set, since $(0-1)^2 + 0^2 = 1$. However, this point corresponds to polar coordinates of the form $(0, theta)$, for arbitrary $theta$, but $0 = 2cos{theta} iff cos{theta} = 0 iff theta=frac{pi}{2} + npi$; so there are plenty of pairs $(r, theta)$ that do satisfy the second condition, but there are plenty of pairs that do not. Do I just need to prove that there exists at least one such pair in the second set? How do I think about problems like this formally, and what would a formal proof of this look like?
Let $(x, y)$ such that $(x-1)^2 + y^2 = 1$. The pair $(r, theta)$ which corresponds to $(x, y)$ satisfies $x = rcos{theta}$, $y = rsin{theta}$. We must show that such a pair also satisfies $r = 2cos{theta}$.
If $r neq 0$, we have that $1 = (x-1)^2 + y^2 iff 1 = (rcos{theta} - 1)^2 + (rsin{theta})^2 iff 0 = r^2 - 2rcos{theta} iff r = 2cos{theta}$.
My problem is with the case of $r=0$. Is the argument below formal enough?
If $r = 0$, then we must have had that $x = rcos{theta} = 0$, $y = rsin{theta} = 0$. The pair $(0, frac{pi}{2})$ in polar coordinates satisfies $r = 2cos{theta}$, and so the origin is also contained in the second set.
polar-coordinates
asked Jan 25 at 22:39
TheProofisTriviumTheProofisTrivium
827
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There are two sets $A$ and $B$, so that all points in $A$ fulfil the 1th equation and all points in $B$ fulfil the 2nd equation. Now you have to show two things: All points in $A$ are in $B$ and all points in $B$ are in $A$. As you noted, it is enough to prove the existence of the point in the respective other set. Have you tried drawing the sets? I would however think you are on a good track :)
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– Imago
Jan 25 at 22:49
add a comment |
$begingroup$
I'm asked to verify that the set of points described by the cartesian equation $$(x-1)^2 + y^2 = 1$$ is equal to the set of points described by the polar equation $$r = 2 cos{theta}, cos{theta} > 0$$
Clearly the point described by $(0,0)$ under the Cartesian coordinate system is an element of the first set, since $(0-1)^2 + 0^2 = 1$. However, this point corresponds to polar coordinates of the form $(0, theta)$, for arbitrary $theta$, but $0 = 2cos{theta} iff cos{theta} = 0 iff theta=frac{pi}{2} + npi$; so there are plenty of pairs $(r, theta)$ that do satisfy the second condition, but there are plenty of pairs that do not. Do I just need to prove that there exists at least one such pair in the second set? How do I think about problems like this formally, and what would a formal proof of this look like?
Let $(x, y)$ such that $(x-1)^2 + y^2 = 1$. The pair $(r, theta)$ which corresponds to $(x, y)$ satisfies $x = rcos{theta}$, $y = rsin{theta}$. We must show that such a pair also satisfies $r = 2cos{theta}$.
If $r neq 0$, we have that $1 = (x-1)^2 + y^2 iff 1 = (rcos{theta} - 1)^2 + (rsin{theta})^2 iff 0 = r^2 - 2rcos{theta} iff r = 2cos{theta}$.
My problem is with the case of $r=0$. Is the argument below formal enough?
If $r = 0$, then we must have had that $x = rcos{theta} = 0$, $y = rsin{theta} = 0$. The pair $(0, frac{pi}{2})$ in polar coordinates satisfies $r = 2cos{theta}$, and so the origin is also contained in the second set.
polar-coordinates
asked Jan 25 at 22:39
TheProofisTriviumTheProofisTrivium
827
$endgroup$
I'm asked to verify that the set of points described by the cartesian equation $$(x-1)^2 + y^2 = 1$$ is equal to the set of points described by the polar equation $$r = 2 cos{theta}, cos{theta} > 0$$
Clearly the point described by $(0,0)$ under the Cartesian coordinate system is an element of the first set, since $(0-1)^2 + 0^2 = 1$. However, this point corresponds to polar coordinates of the form $(0, theta)$, for arbitrary $theta$, but $0 = 2cos{theta} iff cos{theta} = 0 iff theta=frac{pi}{2} + npi$; so there are plenty of pairs $(r, theta)$ that do satisfy the second condition, but there are plenty of pairs that do not. Do I just need to prove that there exists at least one such pair in the second set? How do I think about problems like this formally, and what would a formal proof of this look like?
Let $(x, y)$ such that $(x-1)^2 + y^2 = 1$. The pair $(r, theta)$ which corresponds to $(x, y)$ satisfies $x = rcos{theta}$, $y = rsin{theta}$. We must show that such a pair also satisfies $r = 2cos{theta}$.
If $r neq 0$, we have that $1 = (x-1)^2 + y^2 iff 1 = (rcos{theta} - 1)^2 + (rsin{theta})^2 iff 0 = r^2 - 2rcos{theta} iff r = 2cos{theta}$.
My problem is with the case of $r=0$. Is the argument below formal enough?
If $r = 0$, then we must have had that $x = rcos{theta} = 0$, $y = rsin{theta} = 0$. The pair $(0, frac{pi}{2})$ in polar coordinates satisfies $r = 2cos{theta}$, and so the origin is also contained in the second set.
polar-coordinates
polar-coordinates
asked Jan 25 at 22:39
TheProofisTriviumTheProofisTrivium
827
asked Jan 25 at 22:39
TheProofisTriviumTheProofisTrivium
827
edited Jan 25 at 23:05
TheProofisTrivium
asked Jan 25 at 22:39
TheProofisTriviumTheProofisTrivium
827
asked Jan 25 at 22:39
TheProofisTriviumTheProofisTrivium
827
asked Jan 25 at 22:39
TheProofisTriviumTheProofisTrivium
827
827
$begingroup$
There are two sets $A$ and $B$, so that all points in $A$ fulfil the 1th equation and all points in $B$ fulfil the 2nd equation. Now you have to show two things: All points in $A$ are in $B$ and all points in $B$ are in $A$. As you noted, it is enough to prove the existence of the point in the respective other set. Have you tried drawing the sets? I would however think you are on a good track :)
$endgroup$
– Imago
Jan 25 at 22:49
add a comment |
$begingroup$
There are two sets $A$ and $B$, so that all points in $A$ fulfil the 1th equation and all points in $B$ fulfil the 2nd equation. Now you have to show two things: All points in $A$ are in $B$ and all points in $B$ are in $A$. As you noted, it is enough to prove the existence of the point in the respective other set. Have you tried drawing the sets? I would however think you are on a good track :)
$endgroup$
– Imago
Jan 25 at 22:49
$begingroup$
There are two sets $A$ and $B$, so that all points in $A$ fulfil the 1th equation and all points in $B$ fulfil the 2nd equation. Now you have to show two things: All points in $A$ are in $B$ and all points in $B$ are in $A$. As you noted, it is enough to prove the existence of the point in the respective other set. Have you tried drawing the sets? I would however think you are on a good track :)
$endgroup$
– Imago
Jan 25 at 22:49
$begingroup$
There are two sets $A$ and $B$, so that all points in $A$ fulfil the 1th equation and all points in $B$ fulfil the 2nd equation. Now you have to show two things: All points in $A$ are in $B$ and all points in $B$ are in $A$. As you noted, it is enough to prove the existence of the point in the respective other set. Have you tried drawing the sets? I would however think you are on a good track :)
$endgroup$
– Imago
Jan 25 at 22:49
add a comment |
3 Answers
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In $mathbb{R}^2$ you can formally pass from Cartesian to polar coordinates in this way:
$$begin{cases}
x = rho cos theta \
y = rho sin theta.
end{cases} $$
Of course I need to specify more about $rho$ and $theta$, for now $rho geq 0$ and $theta in [0,2pi)$. Note that
$$begin{cases}
mathrm{d}x = cos theta ,mathrm{d}rho -rhosin theta ,mathrm{d}theta \
mathrm{d}y = sin theta ,mathrm{d}rho + rho cos theta ,mathrm{d}theta,
end{cases}$$ hence
begin{align}
mathrm{d}x wedge mathrm{d}y & = (cos theta ,mathrm{d}rho -rhosin theta ,mathrm{d}theta) wedge (sin theta ,mathrm{d}rho + rho cos theta ,mathrm{d}theta) \
& = rho ,mathrm{d}rho wedge mathrm{d}theta.
end{align}
Therefore the transformation is singular when $rho = 0$. Also, reversing the equations above you get
$$
begin{cases}
rho^2 = x^2+y^2 \
tan theta = y/x,
end{cases}
$$
so when $theta = pi/2+kpi$, where $k$ is an integer, the inverse transformation is not defined. This is why one usually restricts to the conditions $rho > 0$ and $theta neq pi/2 +kpi$. This way the coordinate transformation is invertible.
Now using this change of coordinates your equation $(x-1)^2+y^2 = 1 $
becomes
begin{align}
(rho cos theta-1)^2+rho^2sin^2theta -1 & = 0 \
rho^2-2rhocos theta& = 0 \
rho(rho-2cos theta)&=0,
end{align}
which implies $rho = 2cos theta$, because $rho neq 0$. Actually $rho > 0$, which implies $cos theta > 0$. This is how you go from one set to the other when the transformation is invertible, but $rho = 0$ is also a solution of $rho(rho-2cos theta)=0$, thus just study this case on its own. It is trivial, because $rho = 0$ corresponds to the origin $(x,y) = (0,0)$.
answered Jan 25 at 23:16
GibbsGibbs
5,4183827
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Thank you! I'm having some trouble understanding the first argument; if you don't mind, what function are you differentiating there and what does $hat{}$ stand for? Should I just assume that there is some function $T:mathbb{R}^2tomathbb{R}^2$ which transforms $(x,y)$ to $(r, theta)$?
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– TheProofisTrivium
Jan 25 at 23:28
1
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Do you know what the Jacobian of a transformation is? The symbol $wedge$ is the exterior product.
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– Gibbs
Jan 25 at 23:29
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Ah okay, I haven't learned about those yet. Might you be able to recommend a reference you like which treats this material formally? I'm working through Apostol's Calculus Vol. 1 at the moment.
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– TheProofisTrivium
Jan 25 at 23:34
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Before recommending a reference let me ask a question: have you ever seen calculus in two or more variables?
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– Gibbs
Jan 25 at 23:38
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No, I haven't taken a multivariable calculus course specifically. I have taken courses in linear algebra (we used Axler's book) and mathematical analysis (with metric spaces), so I can follow elementary rigorous arguments, but my fundamentals are weak.
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– TheProofisTrivium
Jan 25 at 23:44
|
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Note that your original circle (in Cartesian coordinates) is a circle centered at $(1,0)$ of radius $1$. So the entire circle is in quadrants 1 and 4. So $theta$ must be in the interval $[-pi/2,pi/2]$. Otherwise you just go over the circle again and again. You could have also noticed that you were given $costhetagt 0$, so it will also restrain your solutions.
The way I tend to approach this problem is to explicitly use $x=rcostheta$ and $y=rsintheta$ in the first equation on the left hand side. If you get $1$, then the proof is done.
answered Jan 25 at 22:50
AndreiAndrei
13.2k21230
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Wouldn't your approach only prove containment in one direction? Why does $(x-1)^2 + y^2 = 1$ imply that $cos{theta} > 0$?
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– TheProofisTrivium
Jan 25 at 22:52
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Because $xge 0$. Do you want to use negative radius? That would also work. If you go with $theta$ from $pi/2$ to $3pi/2$ you would traverse the circle using a negative radius.
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– Andrei
Jan 25 at 22:55
add a comment |
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Those sets are not equal.
The point $(0, 0) in mathbb{R}^2$ is in your first set of points, but not on the second. Proof: if $x = y = 0$, then $r = 0$, and since the second equation says $r = 2 cos theta$, then we must have $cos theta = 0$, violating the restriction that $cos theta > 0$.
That said, you probably meant $cos theta geq 0$, so let's assume that.
As you said yourself, you have to prove that the set of points described by one equation is equal to the set of points described by another equation. So you have to prove equality of two sets. There is one super standard way to do this rigorously: to prove that two sets $A$ and $B$ are equal, you prove that $A subseteq B$ and $B subseteq A$.
So let's take your example and tackle it rigorously. Let's start defining the two sets.
$$A = { , (x,y) in mathbb{R}^2 ; : ; (x - 1)^2 + y^2 = 1 , }$$
$$B = { , (x,y) in mathbb{R}^2 ; : ; exists , r,theta in [0,infty) ; x = r cos theta , wedge y = r sin theta , wedge r = 2 cos theta , wedge cos theta geq 0 , }$$
Part 1: $A subseteq B$
Let $(x,y) in A$. We need to prove that $(x,y) in B$. By definition of $B$ this happens if and only if there exists $r,theta in [0,infty)$ satisfying the following four properties simultaneously:
$$x = r cos theta$$
$$y = r sin theta$$
$$r = 2 cos theta$$
$$cos theta geq 0$$
If we can show how to construct $r, theta$ based on $x, y$ we're done.
Our intuition says that we should probably try $r = sqrt{x^2 + y^2}$ and $theta = text{atan2}(y,x)$. Note, if you are not familiar with atan2, it is a better $arctan$. You might have thought of $theta = arctan(dfrac{y}{x})$, but that would have the hassle of dealing with $x = 0$ differently. Instead, atan2 is much better because is only undefined for $x = 0$ and $y = 0$, for which case we shall explicitly choose $theta = dfrac{pi}{2}$ (the reason will be clear later). Let's see if that works:
Checking property 1:
$$x = r cos theta$$
If $x = y = 0$, this holds because $0 = 0 cos dfrac{pi}{2}$. Otherwise, this holds because $x = sqrt{x^2 + y^2} cos text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).
Checking property 2:
$$y = r sin theta$$
If $x = y = 0$, this holds because $0 = 0 sin dfrac{pi}{2}$. Otherwise, this holds because $y = sqrt{x^2 + y^2} sin text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).
Checking property 3:
This one is not so immediate. What do we know about $x, y$ ? We know that $(x,y) in A$ therefore we know that $(x - 1)^2 + y^2 = 1$. Therefore, the following is also true:
$$(r cos theta - 1)^2 + (r sin theta)^2 = 1$$
$$iff r^2 - 2r cos theta + 1 = 1$$
$$iff r^2 = 2r cos theta$$
$$iff r = 0 vee r = 2 cos theta$$
We wanted to prove that $r = 2 cos theta$, but it looks like we didn't get there. It looks like we proved something slightly weaker: that $r = 2 cos theta$ or $r = 0$. We now have to prove that if it happens to be the case that $r = 0$, then $r = 2 cos theta$ holds as well, so we can conclude that property 3 holds. This is easy: the only way for $r$ to be zero is $x = y = 0$, for which case our construction explicitly chooses $theta = dfrac{pi}{2}$ which forces $r = 2 cos theta$ to hold.
Checking property 4:
$$cos theta geq 0$$
Since we know that $x = r cos theta$ and $r geq 0$ (because we defined it with a square root), we know that $cos theta geq 0$ if and only if $x > 0$. So we have to prove that $x > 0$. How to do that? We know that $(x - 1)^2 + y^2 = 1$, and since $y^2 geq 0$, this means that $(x - 1)^2 leq 1$, which in turn means that $-1 leq (x - 1) leq 1$ which means implies $x geq 0$.
Part 2: $B subseteq A$
Analogously, let $(x,y) in B$. We need to prove that $(x,y) in A$. By definition of $A$ this happens if and only if $(x - 1)^2 + y^2 = 1$ holds. We know that $(x,y) in B$, so we know that there exists $r, theta in [0, infty)$ satisfying those four properties. In particular, by the third property alone, we have:
$$r = 2 cos theta$$
$$iff r^2 = 2 r cos theta$$
$$iff r^2 - 2 r cos theta = 0$$
$$iff r^2 - 2 r cos theta + 1 = 1$$
$$iff r^2 ((sin theta)^2 + (cos theta)^2) - 2 r cos theta + 1 = 1$$
$$iff r^2 (cos theta)^2 - 2 r cos theta + 1 + r^2 (sin theta)^2 = 1$$
$$iff (r cos theta)^2 - 2 r cos theta + 1 + (r sin theta)^2 = 1$$
$$iff (r cos theta - 1)^2 + (r sin theta)^2 = 1$$
$$iff x^2 + y^2 = 1$$
Which completes the proof.
Final thoughts
The proof above is probably too rigorous. But it should be enlightening anyway. We should always remember that when we make a not-so-rigorous proof, it is because we believe the proof can be made rigorous straightforwardly and we want to save both our time and our reader's time. But often we get carried away and make non-rigorous proofs without being actually convinced that we could make it rigorous if necessary.
In fact, I only noticed the problem with the strict inequality $cos theta > 0$ because I was going through this super rigorous proof, and on step 3 of part 1 I got stuck because I couldn't get the inequality. Then I realized that maybe they weren't equal in the first place!
At least I like to follow super rigorous proofs every now and then...
answered Jan 26 at 0:44
Pedro APedro A
2,0661827
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Thank you for writing this out, I sincerely appreciate it. I actually double-checked and the strict inequality is there in the book exercise (Calculus, Apostol, Ex. 2.11.1) So I'm not really sure what to make of that.
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– TheProofisTrivium
Jan 28 at 3:35
add a comment |
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3 Answers
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3 Answers
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$begingroup$
In $mathbb{R}^2$ you can formally pass from Cartesian to polar coordinates in this way:
$$begin{cases}
x = rho cos theta \
y = rho sin theta.
end{cases} $$
Of course I need to specify more about $rho$ and $theta$, for now $rho geq 0$ and $theta in [0,2pi)$. Note that
$$begin{cases}
mathrm{d}x = cos theta ,mathrm{d}rho -rhosin theta ,mathrm{d}theta \
mathrm{d}y = sin theta ,mathrm{d}rho + rho cos theta ,mathrm{d}theta,
end{cases}$$ hence
begin{align}
mathrm{d}x wedge mathrm{d}y & = (cos theta ,mathrm{d}rho -rhosin theta ,mathrm{d}theta) wedge (sin theta ,mathrm{d}rho + rho cos theta ,mathrm{d}theta) \
& = rho ,mathrm{d}rho wedge mathrm{d}theta.
end{align}
Therefore the transformation is singular when $rho = 0$. Also, reversing the equations above you get
$$
begin{cases}
rho^2 = x^2+y^2 \
tan theta = y/x,
end{cases}
$$
so when $theta = pi/2+kpi$, where $k$ is an integer, the inverse transformation is not defined. This is why one usually restricts to the conditions $rho > 0$ and $theta neq pi/2 +kpi$. This way the coordinate transformation is invertible.
Now using this change of coordinates your equation $(x-1)^2+y^2 = 1 $
becomes
begin{align}
(rho cos theta-1)^2+rho^2sin^2theta -1 & = 0 \
rho^2-2rhocos theta& = 0 \
rho(rho-2cos theta)&=0,
end{align}
which implies $rho = 2cos theta$, because $rho neq 0$. Actually $rho > 0$, which implies $cos theta > 0$. This is how you go from one set to the other when the transformation is invertible, but $rho = 0$ is also a solution of $rho(rho-2cos theta)=0$, thus just study this case on its own. It is trivial, because $rho = 0$ corresponds to the origin $(x,y) = (0,0)$.
answered Jan 25 at 23:16
GibbsGibbs
5,4183827
$endgroup$
$begingroup$
Thank you! I'm having some trouble understanding the first argument; if you don't mind, what function are you differentiating there and what does $hat{}$ stand for? Should I just assume that there is some function $T:mathbb{R}^2tomathbb{R}^2$ which transforms $(x,y)$ to $(r, theta)$?
$endgroup$
– TheProofisTrivium
Jan 25 at 23:28
1
$begingroup$
Do you know what the Jacobian of a transformation is? The symbol $wedge$ is the exterior product.
$endgroup$
– Gibbs
Jan 25 at 23:29
$begingroup$
Ah okay, I haven't learned about those yet. Might you be able to recommend a reference you like which treats this material formally? I'm working through Apostol's Calculus Vol. 1 at the moment.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:34
$begingroup$
Before recommending a reference let me ask a question: have you ever seen calculus in two or more variables?
$endgroup$
– Gibbs
Jan 25 at 23:38
$begingroup$
No, I haven't taken a multivariable calculus course specifically. I have taken courses in linear algebra (we used Axler's book) and mathematical analysis (with metric spaces), so I can follow elementary rigorous arguments, but my fundamentals are weak.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:44
|
show 1 more comment
$begingroup$
In $mathbb{R}^2$ you can formally pass from Cartesian to polar coordinates in this way:
$$begin{cases}
x = rho cos theta \
y = rho sin theta.
end{cases} $$
Of course I need to specify more about $rho$ and $theta$, for now $rho geq 0$ and $theta in [0,2pi)$. Note that
$$begin{cases}
mathrm{d}x = cos theta ,mathrm{d}rho -rhosin theta ,mathrm{d}theta \
mathrm{d}y = sin theta ,mathrm{d}rho + rho cos theta ,mathrm{d}theta,
end{cases}$$ hence
begin{align}
mathrm{d}x wedge mathrm{d}y & = (cos theta ,mathrm{d}rho -rhosin theta ,mathrm{d}theta) wedge (sin theta ,mathrm{d}rho + rho cos theta ,mathrm{d}theta) \
& = rho ,mathrm{d}rho wedge mathrm{d}theta.
end{align}
Therefore the transformation is singular when $rho = 0$. Also, reversing the equations above you get
$$
begin{cases}
rho^2 = x^2+y^2 \
tan theta = y/x,
end{cases}
$$
so when $theta = pi/2+kpi$, where $k$ is an integer, the inverse transformation is not defined. This is why one usually restricts to the conditions $rho > 0$ and $theta neq pi/2 +kpi$. This way the coordinate transformation is invertible.
Now using this change of coordinates your equation $(x-1)^2+y^2 = 1 $
becomes
begin{align}
(rho cos theta-1)^2+rho^2sin^2theta -1 & = 0 \
rho^2-2rhocos theta& = 0 \
rho(rho-2cos theta)&=0,
end{align}
which implies $rho = 2cos theta$, because $rho neq 0$. Actually $rho > 0$, which implies $cos theta > 0$. This is how you go from one set to the other when the transformation is invertible, but $rho = 0$ is also a solution of $rho(rho-2cos theta)=0$, thus just study this case on its own. It is trivial, because $rho = 0$ corresponds to the origin $(x,y) = (0,0)$.
answered Jan 25 at 23:16
GibbsGibbs
5,4183827
$endgroup$
$begingroup$
Thank you! I'm having some trouble understanding the first argument; if you don't mind, what function are you differentiating there and what does $hat{}$ stand for? Should I just assume that there is some function $T:mathbb{R}^2tomathbb{R}^2$ which transforms $(x,y)$ to $(r, theta)$?
$endgroup$
– TheProofisTrivium
Jan 25 at 23:28
1
$begingroup$
Do you know what the Jacobian of a transformation is? The symbol $wedge$ is the exterior product.
$endgroup$
– Gibbs
Jan 25 at 23:29
$begingroup$
Ah okay, I haven't learned about those yet. Might you be able to recommend a reference you like which treats this material formally? I'm working through Apostol's Calculus Vol. 1 at the moment.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:34
$begingroup$
Before recommending a reference let me ask a question: have you ever seen calculus in two or more variables?
$endgroup$
– Gibbs
Jan 25 at 23:38
$begingroup$
No, I haven't taken a multivariable calculus course specifically. I have taken courses in linear algebra (we used Axler's book) and mathematical analysis (with metric spaces), so I can follow elementary rigorous arguments, but my fundamentals are weak.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:44
|
show 1 more comment
$begingroup$
In $mathbb{R}^2$ you can formally pass from Cartesian to polar coordinates in this way:
$$begin{cases}
x = rho cos theta \
y = rho sin theta.
end{cases} $$
Of course I need to specify more about $rho$ and $theta$, for now $rho geq 0$ and $theta in [0,2pi)$. Note that
$$begin{cases}
mathrm{d}x = cos theta ,mathrm{d}rho -rhosin theta ,mathrm{d}theta \
mathrm{d}y = sin theta ,mathrm{d}rho + rho cos theta ,mathrm{d}theta,
end{cases}$$ hence
begin{align}
mathrm{d}x wedge mathrm{d}y & = (cos theta ,mathrm{d}rho -rhosin theta ,mathrm{d}theta) wedge (sin theta ,mathrm{d}rho + rho cos theta ,mathrm{d}theta) \
& = rho ,mathrm{d}rho wedge mathrm{d}theta.
end{align}
Therefore the transformation is singular when $rho = 0$. Also, reversing the equations above you get
$$
begin{cases}
rho^2 = x^2+y^2 \
tan theta = y/x,
end{cases}
$$
so when $theta = pi/2+kpi$, where $k$ is an integer, the inverse transformation is not defined. This is why one usually restricts to the conditions $rho > 0$ and $theta neq pi/2 +kpi$. This way the coordinate transformation is invertible.
Now using this change of coordinates your equation $(x-1)^2+y^2 = 1 $
becomes
begin{align}
(rho cos theta-1)^2+rho^2sin^2theta -1 & = 0 \
rho^2-2rhocos theta& = 0 \
rho(rho-2cos theta)&=0,
end{align}
which implies $rho = 2cos theta$, because $rho neq 0$. Actually $rho > 0$, which implies $cos theta > 0$. This is how you go from one set to the other when the transformation is invertible, but $rho = 0$ is also a solution of $rho(rho-2cos theta)=0$, thus just study this case on its own. It is trivial, because $rho = 0$ corresponds to the origin $(x,y) = (0,0)$.
answered Jan 25 at 23:16
GibbsGibbs
5,4183827
$endgroup$
In $mathbb{R}^2$ you can formally pass from Cartesian to polar coordinates in this way:
$$begin{cases}
x = rho cos theta \
y = rho sin theta.
end{cases} $$
Of course I need to specify more about $rho$ and $theta$, for now $rho geq 0$ and $theta in [0,2pi)$. Note that
$$begin{cases}
mathrm{d}x = cos theta ,mathrm{d}rho -rhosin theta ,mathrm{d}theta \
mathrm{d}y = sin theta ,mathrm{d}rho + rho cos theta ,mathrm{d}theta,
end{cases}$$ hence
begin{align}
mathrm{d}x wedge mathrm{d}y & = (cos theta ,mathrm{d}rho -rhosin theta ,mathrm{d}theta) wedge (sin theta ,mathrm{d}rho + rho cos theta ,mathrm{d}theta) \
& = rho ,mathrm{d}rho wedge mathrm{d}theta.
end{align}
Therefore the transformation is singular when $rho = 0$. Also, reversing the equations above you get
$$
begin{cases}
rho^2 = x^2+y^2 \
tan theta = y/x,
end{cases}
$$
so when $theta = pi/2+kpi$, where $k$ is an integer, the inverse transformation is not defined. This is why one usually restricts to the conditions $rho > 0$ and $theta neq pi/2 +kpi$. This way the coordinate transformation is invertible.
Now using this change of coordinates your equation $(x-1)^2+y^2 = 1 $
becomes
begin{align}
(rho cos theta-1)^2+rho^2sin^2theta -1 & = 0 \
rho^2-2rhocos theta& = 0 \
rho(rho-2cos theta)&=0,
end{align}
which implies $rho = 2cos theta$, because $rho neq 0$. Actually $rho > 0$, which implies $cos theta > 0$. This is how you go from one set to the other when the transformation is invertible, but $rho = 0$ is also a solution of $rho(rho-2cos theta)=0$, thus just study this case on its own. It is trivial, because $rho = 0$ corresponds to the origin $(x,y) = (0,0)$.
answered Jan 25 at 23:16
GibbsGibbs
5,4183827
edited Jan 25 at 23:55
answered Jan 25 at 23:16
GibbsGibbs
5,4183827
answered Jan 25 at 23:16
GibbsGibbs
5,4183827
answered Jan 25 at 23:16
GibbsGibbs
5,4183827
5,4183827
$begingroup$
Thank you! I'm having some trouble understanding the first argument; if you don't mind, what function are you differentiating there and what does $hat{}$ stand for? Should I just assume that there is some function $T:mathbb{R}^2tomathbb{R}^2$ which transforms $(x,y)$ to $(r, theta)$?
$endgroup$
– TheProofisTrivium
Jan 25 at 23:28
1
$begingroup$
Do you know what the Jacobian of a transformation is? The symbol $wedge$ is the exterior product.
$endgroup$
– Gibbs
Jan 25 at 23:29
$begingroup$
Ah okay, I haven't learned about those yet. Might you be able to recommend a reference you like which treats this material formally? I'm working through Apostol's Calculus Vol. 1 at the moment.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:34
$begingroup$
Before recommending a reference let me ask a question: have you ever seen calculus in two or more variables?
$endgroup$
– Gibbs
Jan 25 at 23:38
$begingroup$
No, I haven't taken a multivariable calculus course specifically. I have taken courses in linear algebra (we used Axler's book) and mathematical analysis (with metric spaces), so I can follow elementary rigorous arguments, but my fundamentals are weak.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:44
|
show 1 more comment
$begingroup$
Thank you! I'm having some trouble understanding the first argument; if you don't mind, what function are you differentiating there and what does $hat{}$ stand for? Should I just assume that there is some function $T:mathbb{R}^2tomathbb{R}^2$ which transforms $(x,y)$ to $(r, theta)$?
$endgroup$
– TheProofisTrivium
Jan 25 at 23:28
1
$begingroup$
Do you know what the Jacobian of a transformation is? The symbol $wedge$ is the exterior product.
$endgroup$
– Gibbs
Jan 25 at 23:29
$begingroup$
Ah okay, I haven't learned about those yet. Might you be able to recommend a reference you like which treats this material formally? I'm working through Apostol's Calculus Vol. 1 at the moment.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:34
$begingroup$
Before recommending a reference let me ask a question: have you ever seen calculus in two or more variables?
$endgroup$
– Gibbs
Jan 25 at 23:38
$begingroup$
No, I haven't taken a multivariable calculus course specifically. I have taken courses in linear algebra (we used Axler's book) and mathematical analysis (with metric spaces), so I can follow elementary rigorous arguments, but my fundamentals are weak.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:44
$begingroup$
Thank you! I'm having some trouble understanding the first argument; if you don't mind, what function are you differentiating there and what does $hat{}$ stand for? Should I just assume that there is some function $T:mathbb{R}^2tomathbb{R}^2$ which transforms $(x,y)$ to $(r, theta)$?
$endgroup$
– TheProofisTrivium
Jan 25 at 23:28
$begingroup$
Thank you! I'm having some trouble understanding the first argument; if you don't mind, what function are you differentiating there and what does $hat{}$ stand for? Should I just assume that there is some function $T:mathbb{R}^2tomathbb{R}^2$ which transforms $(x,y)$ to $(r, theta)$?
$endgroup$
– TheProofisTrivium
Jan 25 at 23:28
1
1
$begingroup$
Do you know what the Jacobian of a transformation is? The symbol $wedge$ is the exterior product.
$endgroup$
– Gibbs
Jan 25 at 23:29
$begingroup$
Do you know what the Jacobian of a transformation is? The symbol $wedge$ is the exterior product.
$endgroup$
– Gibbs
Jan 25 at 23:29
$begingroup$
Ah okay, I haven't learned about those yet. Might you be able to recommend a reference you like which treats this material formally? I'm working through Apostol's Calculus Vol. 1 at the moment.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:34
$begingroup$
Ah okay, I haven't learned about those yet. Might you be able to recommend a reference you like which treats this material formally? I'm working through Apostol's Calculus Vol. 1 at the moment.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:34
$begingroup$
Before recommending a reference let me ask a question: have you ever seen calculus in two or more variables?
$endgroup$
– Gibbs
Jan 25 at 23:38
$begingroup$
Before recommending a reference let me ask a question: have you ever seen calculus in two or more variables?
$endgroup$
– Gibbs
Jan 25 at 23:38
$begingroup$
No, I haven't taken a multivariable calculus course specifically. I have taken courses in linear algebra (we used Axler's book) and mathematical analysis (with metric spaces), so I can follow elementary rigorous arguments, but my fundamentals are weak.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:44
$begingroup$
No, I haven't taken a multivariable calculus course specifically. I have taken courses in linear algebra (we used Axler's book) and mathematical analysis (with metric spaces), so I can follow elementary rigorous arguments, but my fundamentals are weak.
$endgroup$
– TheProofisTrivium
Jan 25 at 23:44
|
show 1 more comment
$begingroup$
Note that your original circle (in Cartesian coordinates) is a circle centered at $(1,0)$ of radius $1$. So the entire circle is in quadrants 1 and 4. So $theta$ must be in the interval $[-pi/2,pi/2]$. Otherwise you just go over the circle again and again. You could have also noticed that you were given $costhetagt 0$, so it will also restrain your solutions.
The way I tend to approach this problem is to explicitly use $x=rcostheta$ and $y=rsintheta$ in the first equation on the left hand side. If you get $1$, then the proof is done.
answered Jan 25 at 22:50
AndreiAndrei
13.2k21230
$endgroup$
$begingroup$
Wouldn't your approach only prove containment in one direction? Why does $(x-1)^2 + y^2 = 1$ imply that $cos{theta} > 0$?
$endgroup$
– TheProofisTrivium
Jan 25 at 22:52
$begingroup$
Because $xge 0$. Do you want to use negative radius? That would also work. If you go with $theta$ from $pi/2$ to $3pi/2$ you would traverse the circle using a negative radius.
$endgroup$
– Andrei
Jan 25 at 22:55
add a comment |
$begingroup$
Note that your original circle (in Cartesian coordinates) is a circle centered at $(1,0)$ of radius $1$. So the entire circle is in quadrants 1 and 4. So $theta$ must be in the interval $[-pi/2,pi/2]$. Otherwise you just go over the circle again and again. You could have also noticed that you were given $costhetagt 0$, so it will also restrain your solutions.
The way I tend to approach this problem is to explicitly use $x=rcostheta$ and $y=rsintheta$ in the first equation on the left hand side. If you get $1$, then the proof is done.
answered Jan 25 at 22:50
AndreiAndrei
13.2k21230
$endgroup$
$begingroup$
Wouldn't your approach only prove containment in one direction? Why does $(x-1)^2 + y^2 = 1$ imply that $cos{theta} > 0$?
$endgroup$
– TheProofisTrivium
Jan 25 at 22:52
$begingroup$
Because $xge 0$. Do you want to use negative radius? That would also work. If you go with $theta$ from $pi/2$ to $3pi/2$ you would traverse the circle using a negative radius.
$endgroup$
– Andrei
Jan 25 at 22:55
add a comment |
$begingroup$
Note that your original circle (in Cartesian coordinates) is a circle centered at $(1,0)$ of radius $1$. So the entire circle is in quadrants 1 and 4. So $theta$ must be in the interval $[-pi/2,pi/2]$. Otherwise you just go over the circle again and again. You could have also noticed that you were given $costhetagt 0$, so it will also restrain your solutions.
The way I tend to approach this problem is to explicitly use $x=rcostheta$ and $y=rsintheta$ in the first equation on the left hand side. If you get $1$, then the proof is done.
answered Jan 25 at 22:50
AndreiAndrei
13.2k21230
$endgroup$
Note that your original circle (in Cartesian coordinates) is a circle centered at $(1,0)$ of radius $1$. So the entire circle is in quadrants 1 and 4. So $theta$ must be in the interval $[-pi/2,pi/2]$. Otherwise you just go over the circle again and again. You could have also noticed that you were given $costhetagt 0$, so it will also restrain your solutions.
The way I tend to approach this problem is to explicitly use $x=rcostheta$ and $y=rsintheta$ in the first equation on the left hand side. If you get $1$, then the proof is done.
answered Jan 25 at 22:50
AndreiAndrei
13.2k21230
answered Jan 25 at 22:50
AndreiAndrei
13.2k21230
answered Jan 25 at 22:50
AndreiAndrei
13.2k21230
answered Jan 25 at 22:50
AndreiAndrei
13.2k21230
13.2k21230
$begingroup$
Wouldn't your approach only prove containment in one direction? Why does $(x-1)^2 + y^2 = 1$ imply that $cos{theta} > 0$?
$endgroup$
– TheProofisTrivium
Jan 25 at 22:52
$begingroup$
Because $xge 0$. Do you want to use negative radius? That would also work. If you go with $theta$ from $pi/2$ to $3pi/2$ you would traverse the circle using a negative radius.
$endgroup$
– Andrei
Jan 25 at 22:55
add a comment |
$begingroup$
Wouldn't your approach only prove containment in one direction? Why does $(x-1)^2 + y^2 = 1$ imply that $cos{theta} > 0$?
$endgroup$
– TheProofisTrivium
Jan 25 at 22:52
$begingroup$
Because $xge 0$. Do you want to use negative radius? That would also work. If you go with $theta$ from $pi/2$ to $3pi/2$ you would traverse the circle using a negative radius.
$endgroup$
– Andrei
Jan 25 at 22:55
$begingroup$
Wouldn't your approach only prove containment in one direction? Why does $(x-1)^2 + y^2 = 1$ imply that $cos{theta} > 0$?
$endgroup$
– TheProofisTrivium
Jan 25 at 22:52
$begingroup$
Wouldn't your approach only prove containment in one direction? Why does $(x-1)^2 + y^2 = 1$ imply that $cos{theta} > 0$?
$endgroup$
– TheProofisTrivium
Jan 25 at 22:52
$begingroup$
Because $xge 0$. Do you want to use negative radius? That would also work. If you go with $theta$ from $pi/2$ to $3pi/2$ you would traverse the circle using a negative radius.
$endgroup$
– Andrei
Jan 25 at 22:55
$begingroup$
Because $xge 0$. Do you want to use negative radius? That would also work. If you go with $theta$ from $pi/2$ to $3pi/2$ you would traverse the circle using a negative radius.
$endgroup$
– Andrei
Jan 25 at 22:55
add a comment |
$begingroup$
Those sets are not equal.
The point $(0, 0) in mathbb{R}^2$ is in your first set of points, but not on the second. Proof: if $x = y = 0$, then $r = 0$, and since the second equation says $r = 2 cos theta$, then we must have $cos theta = 0$, violating the restriction that $cos theta > 0$.
That said, you probably meant $cos theta geq 0$, so let's assume that.
As you said yourself, you have to prove that the set of points described by one equation is equal to the set of points described by another equation. So you have to prove equality of two sets. There is one super standard way to do this rigorously: to prove that two sets $A$ and $B$ are equal, you prove that $A subseteq B$ and $B subseteq A$.
So let's take your example and tackle it rigorously. Let's start defining the two sets.
$$A = { , (x,y) in mathbb{R}^2 ; : ; (x - 1)^2 + y^2 = 1 , }$$
$$B = { , (x,y) in mathbb{R}^2 ; : ; exists , r,theta in [0,infty) ; x = r cos theta , wedge y = r sin theta , wedge r = 2 cos theta , wedge cos theta geq 0 , }$$
Part 1: $A subseteq B$
Let $(x,y) in A$. We need to prove that $(x,y) in B$. By definition of $B$ this happens if and only if there exists $r,theta in [0,infty)$ satisfying the following four properties simultaneously:
$$x = r cos theta$$
$$y = r sin theta$$
$$r = 2 cos theta$$
$$cos theta geq 0$$
If we can show how to construct $r, theta$ based on $x, y$ we're done.
Our intuition says that we should probably try $r = sqrt{x^2 + y^2}$ and $theta = text{atan2}(y,x)$. Note, if you are not familiar with atan2, it is a better $arctan$. You might have thought of $theta = arctan(dfrac{y}{x})$, but that would have the hassle of dealing with $x = 0$ differently. Instead, atan2 is much better because is only undefined for $x = 0$ and $y = 0$, for which case we shall explicitly choose $theta = dfrac{pi}{2}$ (the reason will be clear later). Let's see if that works:
Checking property 1:
$$x = r cos theta$$
If $x = y = 0$, this holds because $0 = 0 cos dfrac{pi}{2}$. Otherwise, this holds because $x = sqrt{x^2 + y^2} cos text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).
Checking property 2:
$$y = r sin theta$$
If $x = y = 0$, this holds because $0 = 0 sin dfrac{pi}{2}$. Otherwise, this holds because $y = sqrt{x^2 + y^2} sin text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).
Checking property 3:
This one is not so immediate. What do we know about $x, y$ ? We know that $(x,y) in A$ therefore we know that $(x - 1)^2 + y^2 = 1$. Therefore, the following is also true:
$$(r cos theta - 1)^2 + (r sin theta)^2 = 1$$
$$iff r^2 - 2r cos theta + 1 = 1$$
$$iff r^2 = 2r cos theta$$
$$iff r = 0 vee r = 2 cos theta$$
We wanted to prove that $r = 2 cos theta$, but it looks like we didn't get there. It looks like we proved something slightly weaker: that $r = 2 cos theta$ or $r = 0$. We now have to prove that if it happens to be the case that $r = 0$, then $r = 2 cos theta$ holds as well, so we can conclude that property 3 holds. This is easy: the only way for $r$ to be zero is $x = y = 0$, for which case our construction explicitly chooses $theta = dfrac{pi}{2}$ which forces $r = 2 cos theta$ to hold.
Checking property 4:
$$cos theta geq 0$$
Since we know that $x = r cos theta$ and $r geq 0$ (because we defined it with a square root), we know that $cos theta geq 0$ if and only if $x > 0$. So we have to prove that $x > 0$. How to do that? We know that $(x - 1)^2 + y^2 = 1$, and since $y^2 geq 0$, this means that $(x - 1)^2 leq 1$, which in turn means that $-1 leq (x - 1) leq 1$ which means implies $x geq 0$.
Part 2: $B subseteq A$
Analogously, let $(x,y) in B$. We need to prove that $(x,y) in A$. By definition of $A$ this happens if and only if $(x - 1)^2 + y^2 = 1$ holds. We know that $(x,y) in B$, so we know that there exists $r, theta in [0, infty)$ satisfying those four properties. In particular, by the third property alone, we have:
$$r = 2 cos theta$$
$$iff r^2 = 2 r cos theta$$
$$iff r^2 - 2 r cos theta = 0$$
$$iff r^2 - 2 r cos theta + 1 = 1$$
$$iff r^2 ((sin theta)^2 + (cos theta)^2) - 2 r cos theta + 1 = 1$$
$$iff r^2 (cos theta)^2 - 2 r cos theta + 1 + r^2 (sin theta)^2 = 1$$
$$iff (r cos theta)^2 - 2 r cos theta + 1 + (r sin theta)^2 = 1$$
$$iff (r cos theta - 1)^2 + (r sin theta)^2 = 1$$
$$iff x^2 + y^2 = 1$$
Which completes the proof.
Final thoughts
The proof above is probably too rigorous. But it should be enlightening anyway. We should always remember that when we make a not-so-rigorous proof, it is because we believe the proof can be made rigorous straightforwardly and we want to save both our time and our reader's time. But often we get carried away and make non-rigorous proofs without being actually convinced that we could make it rigorous if necessary.
In fact, I only noticed the problem with the strict inequality $cos theta > 0$ because I was going through this super rigorous proof, and on step 3 of part 1 I got stuck because I couldn't get the inequality. Then I realized that maybe they weren't equal in the first place!
At least I like to follow super rigorous proofs every now and then...
answered Jan 26 at 0:44
Pedro APedro A
2,0661827
$endgroup$
$begingroup$
Thank you for writing this out, I sincerely appreciate it. I actually double-checked and the strict inequality is there in the book exercise (Calculus, Apostol, Ex. 2.11.1) So I'm not really sure what to make of that.
$endgroup$
– TheProofisTrivium
Jan 28 at 3:35
add a comment |
$begingroup$
Those sets are not equal.
The point $(0, 0) in mathbb{R}^2$ is in your first set of points, but not on the second. Proof: if $x = y = 0$, then $r = 0$, and since the second equation says $r = 2 cos theta$, then we must have $cos theta = 0$, violating the restriction that $cos theta > 0$.
That said, you probably meant $cos theta geq 0$, so let's assume that.
As you said yourself, you have to prove that the set of points described by one equation is equal to the set of points described by another equation. So you have to prove equality of two sets. There is one super standard way to do this rigorously: to prove that two sets $A$ and $B$ are equal, you prove that $A subseteq B$ and $B subseteq A$.
So let's take your example and tackle it rigorously. Let's start defining the two sets.
$$A = { , (x,y) in mathbb{R}^2 ; : ; (x - 1)^2 + y^2 = 1 , }$$
$$B = { , (x,y) in mathbb{R}^2 ; : ; exists , r,theta in [0,infty) ; x = r cos theta , wedge y = r sin theta , wedge r = 2 cos theta , wedge cos theta geq 0 , }$$
Part 1: $A subseteq B$
Let $(x,y) in A$. We need to prove that $(x,y) in B$. By definition of $B$ this happens if and only if there exists $r,theta in [0,infty)$ satisfying the following four properties simultaneously:
$$x = r cos theta$$
$$y = r sin theta$$
$$r = 2 cos theta$$
$$cos theta geq 0$$
If we can show how to construct $r, theta$ based on $x, y$ we're done.
Our intuition says that we should probably try $r = sqrt{x^2 + y^2}$ and $theta = text{atan2}(y,x)$. Note, if you are not familiar with atan2, it is a better $arctan$. You might have thought of $theta = arctan(dfrac{y}{x})$, but that would have the hassle of dealing with $x = 0$ differently. Instead, atan2 is much better because is only undefined for $x = 0$ and $y = 0$, for which case we shall explicitly choose $theta = dfrac{pi}{2}$ (the reason will be clear later). Let's see if that works:
Checking property 1:
$$x = r cos theta$$
If $x = y = 0$, this holds because $0 = 0 cos dfrac{pi}{2}$. Otherwise, this holds because $x = sqrt{x^2 + y^2} cos text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).
Checking property 2:
$$y = r sin theta$$
If $x = y = 0$, this holds because $0 = 0 sin dfrac{pi}{2}$. Otherwise, this holds because $y = sqrt{x^2 + y^2} sin text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).
Checking property 3:
This one is not so immediate. What do we know about $x, y$ ? We know that $(x,y) in A$ therefore we know that $(x - 1)^2 + y^2 = 1$. Therefore, the following is also true:
$$(r cos theta - 1)^2 + (r sin theta)^2 = 1$$
$$iff r^2 - 2r cos theta + 1 = 1$$
$$iff r^2 = 2r cos theta$$
$$iff r = 0 vee r = 2 cos theta$$
We wanted to prove that $r = 2 cos theta$, but it looks like we didn't get there. It looks like we proved something slightly weaker: that $r = 2 cos theta$ or $r = 0$. We now have to prove that if it happens to be the case that $r = 0$, then $r = 2 cos theta$ holds as well, so we can conclude that property 3 holds. This is easy: the only way for $r$ to be zero is $x = y = 0$, for which case our construction explicitly chooses $theta = dfrac{pi}{2}$ which forces $r = 2 cos theta$ to hold.
Checking property 4:
$$cos theta geq 0$$
Since we know that $x = r cos theta$ and $r geq 0$ (because we defined it with a square root), we know that $cos theta geq 0$ if and only if $x > 0$. So we have to prove that $x > 0$. How to do that? We know that $(x - 1)^2 + y^2 = 1$, and since $y^2 geq 0$, this means that $(x - 1)^2 leq 1$, which in turn means that $-1 leq (x - 1) leq 1$ which means implies $x geq 0$.
Part 2: $B subseteq A$
Analogously, let $(x,y) in B$. We need to prove that $(x,y) in A$. By definition of $A$ this happens if and only if $(x - 1)^2 + y^2 = 1$ holds. We know that $(x,y) in B$, so we know that there exists $r, theta in [0, infty)$ satisfying those four properties. In particular, by the third property alone, we have:
$$r = 2 cos theta$$
$$iff r^2 = 2 r cos theta$$
$$iff r^2 - 2 r cos theta = 0$$
$$iff r^2 - 2 r cos theta + 1 = 1$$
$$iff r^2 ((sin theta)^2 + (cos theta)^2) - 2 r cos theta + 1 = 1$$
$$iff r^2 (cos theta)^2 - 2 r cos theta + 1 + r^2 (sin theta)^2 = 1$$
$$iff (r cos theta)^2 - 2 r cos theta + 1 + (r sin theta)^2 = 1$$
$$iff (r cos theta - 1)^2 + (r sin theta)^2 = 1$$
$$iff x^2 + y^2 = 1$$
Which completes the proof.
Final thoughts
The proof above is probably too rigorous. But it should be enlightening anyway. We should always remember that when we make a not-so-rigorous proof, it is because we believe the proof can be made rigorous straightforwardly and we want to save both our time and our reader's time. But often we get carried away and make non-rigorous proofs without being actually convinced that we could make it rigorous if necessary.
In fact, I only noticed the problem with the strict inequality $cos theta > 0$ because I was going through this super rigorous proof, and on step 3 of part 1 I got stuck because I couldn't get the inequality. Then I realized that maybe they weren't equal in the first place!
At least I like to follow super rigorous proofs every now and then...
answered Jan 26 at 0:44
Pedro APedro A
2,0661827
$endgroup$
$begingroup$
Thank you for writing this out, I sincerely appreciate it. I actually double-checked and the strict inequality is there in the book exercise (Calculus, Apostol, Ex. 2.11.1) So I'm not really sure what to make of that.
$endgroup$
– TheProofisTrivium
Jan 28 at 3:35
add a comment |
$begingroup$
Those sets are not equal.
The point $(0, 0) in mathbb{R}^2$ is in your first set of points, but not on the second. Proof: if $x = y = 0$, then $r = 0$, and since the second equation says $r = 2 cos theta$, then we must have $cos theta = 0$, violating the restriction that $cos theta > 0$.
That said, you probably meant $cos theta geq 0$, so let's assume that.
As you said yourself, you have to prove that the set of points described by one equation is equal to the set of points described by another equation. So you have to prove equality of two sets. There is one super standard way to do this rigorously: to prove that two sets $A$ and $B$ are equal, you prove that $A subseteq B$ and $B subseteq A$.
So let's take your example and tackle it rigorously. Let's start defining the two sets.
$$A = { , (x,y) in mathbb{R}^2 ; : ; (x - 1)^2 + y^2 = 1 , }$$
$$B = { , (x,y) in mathbb{R}^2 ; : ; exists , r,theta in [0,infty) ; x = r cos theta , wedge y = r sin theta , wedge r = 2 cos theta , wedge cos theta geq 0 , }$$
Part 1: $A subseteq B$
Let $(x,y) in A$. We need to prove that $(x,y) in B$. By definition of $B$ this happens if and only if there exists $r,theta in [0,infty)$ satisfying the following four properties simultaneously:
$$x = r cos theta$$
$$y = r sin theta$$
$$r = 2 cos theta$$
$$cos theta geq 0$$
If we can show how to construct $r, theta$ based on $x, y$ we're done.
Our intuition says that we should probably try $r = sqrt{x^2 + y^2}$ and $theta = text{atan2}(y,x)$. Note, if you are not familiar with atan2, it is a better $arctan$. You might have thought of $theta = arctan(dfrac{y}{x})$, but that would have the hassle of dealing with $x = 0$ differently. Instead, atan2 is much better because is only undefined for $x = 0$ and $y = 0$, for which case we shall explicitly choose $theta = dfrac{pi}{2}$ (the reason will be clear later). Let's see if that works:
Checking property 1:
$$x = r cos theta$$
If $x = y = 0$, this holds because $0 = 0 cos dfrac{pi}{2}$. Otherwise, this holds because $x = sqrt{x^2 + y^2} cos text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).
Checking property 2:
$$y = r sin theta$$
If $x = y = 0$, this holds because $0 = 0 sin dfrac{pi}{2}$. Otherwise, this holds because $y = sqrt{x^2 + y^2} sin text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).
Checking property 3:
This one is not so immediate. What do we know about $x, y$ ? We know that $(x,y) in A$ therefore we know that $(x - 1)^2 + y^2 = 1$. Therefore, the following is also true:
$$(r cos theta - 1)^2 + (r sin theta)^2 = 1$$
$$iff r^2 - 2r cos theta + 1 = 1$$
$$iff r^2 = 2r cos theta$$
$$iff r = 0 vee r = 2 cos theta$$
We wanted to prove that $r = 2 cos theta$, but it looks like we didn't get there. It looks like we proved something slightly weaker: that $r = 2 cos theta$ or $r = 0$. We now have to prove that if it happens to be the case that $r = 0$, then $r = 2 cos theta$ holds as well, so we can conclude that property 3 holds. This is easy: the only way for $r$ to be zero is $x = y = 0$, for which case our construction explicitly chooses $theta = dfrac{pi}{2}$ which forces $r = 2 cos theta$ to hold.
Checking property 4:
$$cos theta geq 0$$
Since we know that $x = r cos theta$ and $r geq 0$ (because we defined it with a square root), we know that $cos theta geq 0$ if and only if $x > 0$. So we have to prove that $x > 0$. How to do that? We know that $(x - 1)^2 + y^2 = 1$, and since $y^2 geq 0$, this means that $(x - 1)^2 leq 1$, which in turn means that $-1 leq (x - 1) leq 1$ which means implies $x geq 0$.
Part 2: $B subseteq A$
Analogously, let $(x,y) in B$. We need to prove that $(x,y) in A$. By definition of $A$ this happens if and only if $(x - 1)^2 + y^2 = 1$ holds. We know that $(x,y) in B$, so we know that there exists $r, theta in [0, infty)$ satisfying those four properties. In particular, by the third property alone, we have:
$$r = 2 cos theta$$
$$iff r^2 = 2 r cos theta$$
$$iff r^2 - 2 r cos theta = 0$$
$$iff r^2 - 2 r cos theta + 1 = 1$$
$$iff r^2 ((sin theta)^2 + (cos theta)^2) - 2 r cos theta + 1 = 1$$
$$iff r^2 (cos theta)^2 - 2 r cos theta + 1 + r^2 (sin theta)^2 = 1$$
$$iff (r cos theta)^2 - 2 r cos theta + 1 + (r sin theta)^2 = 1$$
$$iff (r cos theta - 1)^2 + (r sin theta)^2 = 1$$
$$iff x^2 + y^2 = 1$$
Which completes the proof.
Final thoughts
The proof above is probably too rigorous. But it should be enlightening anyway. We should always remember that when we make a not-so-rigorous proof, it is because we believe the proof can be made rigorous straightforwardly and we want to save both our time and our reader's time. But often we get carried away and make non-rigorous proofs without being actually convinced that we could make it rigorous if necessary.
In fact, I only noticed the problem with the strict inequality $cos theta > 0$ because I was going through this super rigorous proof, and on step 3 of part 1 I got stuck because I couldn't get the inequality. Then I realized that maybe they weren't equal in the first place!
At least I like to follow super rigorous proofs every now and then...
answered Jan 26 at 0:44
Pedro APedro A
2,0661827
$endgroup$
Those sets are not equal.
The point $(0, 0) in mathbb{R}^2$ is in your first set of points, but not on the second. Proof: if $x = y = 0$, then $r = 0$, and since the second equation says $r = 2 cos theta$, then we must have $cos theta = 0$, violating the restriction that $cos theta > 0$.
That said, you probably meant $cos theta geq 0$, so let's assume that.
As you said yourself, you have to prove that the set of points described by one equation is equal to the set of points described by another equation. So you have to prove equality of two sets. There is one super standard way to do this rigorously: to prove that two sets $A$ and $B$ are equal, you prove that $A subseteq B$ and $B subseteq A$.
So let's take your example and tackle it rigorously. Let's start defining the two sets.
$$A = { , (x,y) in mathbb{R}^2 ; : ; (x - 1)^2 + y^2 = 1 , }$$
$$B = { , (x,y) in mathbb{R}^2 ; : ; exists , r,theta in [0,infty) ; x = r cos theta , wedge y = r sin theta , wedge r = 2 cos theta , wedge cos theta geq 0 , }$$
Part 1: $A subseteq B$
Let $(x,y) in A$. We need to prove that $(x,y) in B$. By definition of $B$ this happens if and only if there exists $r,theta in [0,infty)$ satisfying the following four properties simultaneously:
$$x = r cos theta$$
$$y = r sin theta$$
$$r = 2 cos theta$$
$$cos theta geq 0$$
If we can show how to construct $r, theta$ based on $x, y$ we're done.
Our intuition says that we should probably try $r = sqrt{x^2 + y^2}$ and $theta = text{atan2}(y,x)$. Note, if you are not familiar with atan2, it is a better $arctan$. You might have thought of $theta = arctan(dfrac{y}{x})$, but that would have the hassle of dealing with $x = 0$ differently. Instead, atan2 is much better because is only undefined for $x = 0$ and $y = 0$, for which case we shall explicitly choose $theta = dfrac{pi}{2}$ (the reason will be clear later). Let's see if that works:
Checking property 1:
$$x = r cos theta$$
If $x = y = 0$, this holds because $0 = 0 cos dfrac{pi}{2}$. Otherwise, this holds because $x = sqrt{x^2 + y^2} cos text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).
Checking property 2:
$$y = r sin theta$$
If $x = y = 0$, this holds because $0 = 0 sin dfrac{pi}{2}$. Otherwise, this holds because $y = sqrt{x^2 + y^2} sin text{atan2}(y,x)$ is a known transformation between these coordinates (you can prove it explicitly if you want, but that would be some extra work).
Checking property 3:
This one is not so immediate. What do we know about $x, y$ ? We know that $(x,y) in A$ therefore we know that $(x - 1)^2 + y^2 = 1$. Therefore, the following is also true:
$$(r cos theta - 1)^2 + (r sin theta)^2 = 1$$
$$iff r^2 - 2r cos theta + 1 = 1$$
$$iff r^2 = 2r cos theta$$
$$iff r = 0 vee r = 2 cos theta$$
We wanted to prove that $r = 2 cos theta$, but it looks like we didn't get there. It looks like we proved something slightly weaker: that $r = 2 cos theta$ or $r = 0$. We now have to prove that if it happens to be the case that $r = 0$, then $r = 2 cos theta$ holds as well, so we can conclude that property 3 holds. This is easy: the only way for $r$ to be zero is $x = y = 0$, for which case our construction explicitly chooses $theta = dfrac{pi}{2}$ which forces $r = 2 cos theta$ to hold.
Checking property 4:
$$cos theta geq 0$$
Since we know that $x = r cos theta$ and $r geq 0$ (because we defined it with a square root), we know that $cos theta geq 0$ if and only if $x > 0$. So we have to prove that $x > 0$. How to do that? We know that $(x - 1)^2 + y^2 = 1$, and since $y^2 geq 0$, this means that $(x - 1)^2 leq 1$, which in turn means that $-1 leq (x - 1) leq 1$ which means implies $x geq 0$.
Part 2: $B subseteq A$
Analogously, let $(x,y) in B$. We need to prove that $(x,y) in A$. By definition of $A$ this happens if and only if $(x - 1)^2 + y^2 = 1$ holds. We know that $(x,y) in B$, so we know that there exists $r, theta in [0, infty)$ satisfying those four properties. In particular, by the third property alone, we have:
$$r = 2 cos theta$$
$$iff r^2 = 2 r cos theta$$
$$iff r^2 - 2 r cos theta = 0$$
$$iff r^2 - 2 r cos theta + 1 = 1$$
$$iff r^2 ((sin theta)^2 + (cos theta)^2) - 2 r cos theta + 1 = 1$$
$$iff r^2 (cos theta)^2 - 2 r cos theta + 1 + r^2 (sin theta)^2 = 1$$
$$iff (r cos theta)^2 - 2 r cos theta + 1 + (r sin theta)^2 = 1$$
$$iff (r cos theta - 1)^2 + (r sin theta)^2 = 1$$
$$iff x^2 + y^2 = 1$$
Which completes the proof.
Final thoughts
The proof above is probably too rigorous. But it should be enlightening anyway. We should always remember that when we make a not-so-rigorous proof, it is because we believe the proof can be made rigorous straightforwardly and we want to save both our time and our reader's time. But often we get carried away and make non-rigorous proofs without being actually convinced that we could make it rigorous if necessary.
In fact, I only noticed the problem with the strict inequality $cos theta > 0$ because I was going through this super rigorous proof, and on step 3 of part 1 I got stuck because I couldn't get the inequality. Then I realized that maybe they weren't equal in the first place!
At least I like to follow super rigorous proofs every now and then...
answered Jan 26 at 0:44
Pedro APedro A
2,0661827
answered Jan 26 at 0:44
Pedro APedro A
2,0661827
answered Jan 26 at 0:44
Pedro APedro A
2,0661827
answered Jan 26 at 0:44
Pedro APedro A
2,0661827
2,0661827
$begingroup$
Thank you for writing this out, I sincerely appreciate it. I actually double-checked and the strict inequality is there in the book exercise (Calculus, Apostol, Ex. 2.11.1) So I'm not really sure what to make of that.
$endgroup$
– TheProofisTrivium
Jan 28 at 3:35
add a comment |
$begingroup$
Thank you for writing this out, I sincerely appreciate it. I actually double-checked and the strict inequality is there in the book exercise (Calculus, Apostol, Ex. 2.11.1) So I'm not really sure what to make of that.
$endgroup$
– TheProofisTrivium
Jan 28 at 3:35
$begingroup$
Thank you for writing this out, I sincerely appreciate it. I actually double-checked and the strict inequality is there in the book exercise (Calculus, Apostol, Ex. 2.11.1) So I'm not really sure what to make of that.
$endgroup$
– TheProofisTrivium
Jan 28 at 3:35
$begingroup$
Thank you for writing this out, I sincerely appreciate it. I actually double-checked and the strict inequality is there in the book exercise (Calculus, Apostol, Ex. 2.11.1) So I'm not really sure what to make of that.
$endgroup$
– TheProofisTrivium
Jan 28 at 3:35
add a comment |
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There are two sets $A$ and $B$, so that all points in $A$ fulfil the 1th equation and all points in $B$ fulfil the 2nd equation. Now you have to show two things: All points in $A$ are in $B$ and all points in $B$ are in $A$. As you noted, it is enough to prove the existence of the point in the respective other set. Have you tried drawing the sets? I would however think you are on a good track :)
$endgroup$
– Imago
Jan 25 at 22:49