Balls in R^k - Rudin PMA C1 16

[TPAINE]

Homework Statement

Let x,y be in R^k, |x-y|=d. Prove there are an infinite number of solutions to |z-x|=|z-y|=r when 2r>d.

Homework Equations

I have done this problem for 2r=d and 2r<d. For 2r<d, the triangle inequality works nicely. For 2r=d, the equality gives me additional information to solve it. However, just setting |z-x|=|z-y|and expanding just get me a mess.

The Attempt at a Solution

The solution is obvious geometrically (the two "balls" intersect in a "circle") but I'd like to avoid undefined topological notions and just do it with the information in Rudin.

 

[micromass]

First, note that this is not true in the rational numbers, so we will have to use some kind of property of R here. (Specifically, we'll use that R is connected, but you can forget that)

First, fix some notations. Let S_x be all the points z such that |x-z|=r. Let S_y be the points z such that |y-z|=r. Let B_x, B_y be the closed balls that correspond to S_x, S_y. Define a function

$$f:S_x\rightarrow \mathbb{R}:~z\rightarrow d(z,B_y)-d(z,\overline{\mathbb{R}^k\setminus B_y})$$

You will only need to show that this function has a zero.

 

[jbunniii]

First, note that this is not true in the rational numbers

Are you sure about this? I don't know if it's true or not but if not, I find it surprising.

I think the equivalent statement is that there are spheres in R^k, centered at rational coordinates and with rational radius, such that there are not infinitely many rational points on the sphere.

I know the unit circle in R^2 has infinitely many rational points, and so if I translate it so the center is at any rational coordinate, the same will be true. Same thing if I scale the radius by a rational multiple.

Doesn't this carry over to R^k?

P.S. Sorry TPAINE for the digression :-)

 

[TPAINE]

Could you explain a bit more? I'm having difficulty parsing that notation. Also, wouldn't you need to show that function has infinitely many zeroes?

 

[micromass]

Are you sure about this? I don't know if it's true or not but if so, I find it surprising.

We can easily find two spheres in R² such that their circle of intersection lies in the plane x=pi. When viewed over the rationals, their will thus be no intersection...

 

[TPAINE]

We can easily find two spheres in R² such that their circle of intersection lies in the plane x=pi. When viewed over the rationals, their will thus be no intersection...

I think he means for any dimension higher than this.

 

[micromass]

Also, wouldn't you need to show that function has infinitely many zeroes?

Ah yes, I'm sorry, I misread the question. But, let's first prove that there is at least one zero. We'll see if we can fix this later on...

As for why I came with this function. I just wanted a continuous function that is negative on the inside of S_y and positive on the outside of S_y...

 

[micromass]

I think he means for any dimension higher than this.

Sorry, typo, I meant R³ in that post...

 

[jbunniii]

We can easily find two spheres in R² such that their circle of intersection lies in the plane x=pi. When viewed over the rationals, their will thus be no intersection...

How can the plane be x = pi if the spheres are centered at rational coordinates? That means the midpoint between the two centers also has rational coordinates. Maybe I'm just being dense, but I am having trouble seeing it.

 

[TPAINE]

Micromass, I don't think that's the way to go. There should be some way to directly solve the equations for the equation of a k-1 sphere, but I'm not seeing it.

 

[jbunniii]

TPAINE - Can you translate the problem so that one of the points, say x, is at the origin? It should be less messy that way, and if you can prove it in that case it's clearly true for general x.

 

[micromass]

Well, the only problem would be the irrational centers. But we can make the radius irrational however. So this could provide an intersecting circle at z=pi. I should find an example of this...

When the centers and the radius are rational, however, then there probably are solutions...

 

[jbunniii]

Well, the only problem would be the irrational centers. But we can make the radius irrational however. So this could provide an intersecting circle at z=pi. I should find an example of this...

When the centers and the radius are rational, however, then there probably are solutions...

Hmm, OK, I'll buy that. Varying r doesn't move the plane of intersection, but it does change the radius of the circle within the plane. If you make r irrational, then the circle has irrational radius and you're in business.

 

[TPAINE]

Yeah, if you make one center 0 and the other lie on an axis, then k-1 sphere they intersect in has a nice analytic form because it's orthogonal to the axis, but I get in trouble translating it back to the original coordinate frame.

EDIT: Although I guess youe don't really need the exact expression because it's only an existence statement...

 

[micromass]

Micromass, I don't think that's the way to go. There should be some way to directly solve the equations for the equation of a k-1 sphere, but I'm not seeing it.

Hmm, after actually taking a look at Rudin, I agree that this is not what Rudin meant. My idea was simply to apply theorem 4.23 of Rudin, which would give the answer almost immediately.
But if you only want to use chapter 1, then I see no other way to explicitely solve these things.

If you want to make it easy on yourself, first reduce the situation to x is zero by translating the space. Then rotate the space such that y is on the x-axis. Then shrink/expand the space such that y=(1,0,0,0,...,0). This gives a system that is very simple.

 

[TPAINE]

What's he best way to solve it in the original coordinate system? I'm doing this independently, but I'm sure for a class with this level of attention to foundation and details I would have to go through the tedious process of justifying why affine transformation preserve the properties I want.

 

[micromass]

What's he best way to solve it in the original coordinate system? I'm doing this independently, but I'm sure for a class with this level of attention to foundation and details I would have to go through the tedious process of justifying why affine transformation preserve the properties I want.

I actually think that this tedious process will go faster then actually solving the system of equations.

While the tedious process is a mighty fine exercise, you don't need to go through it. Just try to find the solutions in case of x=0 and y on one axis. And then try to use your affine transformations to lift these solutions to the original case. The only thing you need to show is that these numbers are actually solutions, and you can this by simply plugging them in your equation.

So, in any case, I think your best bet lies in solving the system in the simpler case, and then actually using these solutions to find the solutions of the original case...

 

[jbunniii]

I don't think it is actually that tedious to justify why the answer is independent of coordinate frame.

I would start as follows:

Let u = z - x, w = y - x.

Then |z - x| = |z - y|

if and only if

|u| = |u - w|

if and only if

|u|^2 = |u - w|^2

if and only if

2<u,w> = |w|^2

Then note that |w| = |x - y| = d and check that there are infinitely many u satisfying this equation if the condition on d is satisfied. (This is where the work is.)

Then because every link in the chain was an "if and only if", there must be infinitely many solutions in the original coordinates.

 

[micromass]

I don't think it is actually that tedious to justify why the answer is independent of coordinate frame.

I would start as follows:

Let u = z - x, w = y - x.

Then |z - x| = |z - y|

if and only if

|u| = |u - w|

if and only if

|u|^2 = |u - w|^2

if and only if

2<u,w> = |w|^2

Then note that |w| = |x - y| = d and check that there are infinitely many u satisfying this equation if the condition on d is satisfied. (This is where the work is.)

Then because every link in the chain was an "if and only if", there must be infinitely many solutions in the original coordinates.

Ah yes! This is indeed easy. But I think he meant that the tedious part is in why you can assume that one point lies on one of the axis. (Thus why you can rotate your coordinate frame). This is a little harder to justify...

 

[TPAINE]

That looks really slick. However, I lose you at the key point "Then note that |w| = |x - y| = d and check that there are infinitely many u satisfying this equation if the condition on d is satisfied." Could you explain further?
How do we know there are infinitely many solutions?
It looks like he's getting away without rotations.

 

[jbunniii]

That looks really slick. However, I lose you at the key point "Then note that |w| = |x - y| = d and check that there are infinitely many u satisfying this equation if the condition on d is satisfied." Could you explain further?
How do we know there are infinitely many solutions?
It looks like he's getting away without rotations.

OK, you want to know how many vectors u satisfy

2<u,w> = |w|^2

given that |w| = d.

Write

<u,w> = |u| |w| cos(theta)

where theta is the angle between u and w.

Then the above simplifies to

2|u| cos(theta) = |w| = d

or equivalently

|u| cos(theta) = d/2

Now cos(theta) <= 1 so if there is a solution, it must satisfy

d/2 = |u| cos(theta) <= |u|

or equivalently

2|u| >= d

But what is |u|? We defined it as |z - x|, which has to be r for any solution (see original problem statement).

Thus in order for there to be solutions, we must have 2r >= d. Luckily, that's also given.

So we've shown that there are NO solutions if d > 2r, and that there CAN be solutions if d <= 2r.

Now, look at two cases:

if 2r = d, can you show that there is exactly one solution (and what is that solution?)

and if 2r < d, that there are infinitely many?

 

[TPAINE]

I've already solved the case 2r=d. For 2r>d, how do we show |u| cos(theta) = d/2 has infinitely many solutions? Is it just because we can take a ball with radius r around x, and that surface has infinitely many points? That seems wrong...

EDIT: Yeah I don't think |u|=d/(wcos(theta)) is a sufficient condition.

To be clear we want to show there are infinitely many solutions to |u| cos(theta) = d/2 or |u|=d/(wcos(theta))=r, and the ball with radius d/(wcos(theta))=r around x will produce such u, but not every u is going to be a solution.

EDIT2: I have an idea for R^3. Ok, so if we rearrange the above 2<u,w> = |w|^2, we get w(2u-w). This means that w and the vector from the tip of w to the tip of 2u must be orthogonal. So consider the plane passing through the tip of w perpendicular to it. Because 2r>d the points 2r away from the origin will form a circle on this plane, and everything on that circle will work. How I can bump this up in dimension and make it more rigorous?

 

[jbunniii]

Sorry, I've been having to multitask so my solution sketch is a bit scatterbrained.

OK, the cosine was useful for understanding why d has to be <= 2r, but I think it's just obscuring the final solution.

We seek the set of all vectors u that satisfy the following two conditions:

<u,w> = d^2 / 2
and
|u| = r

The first equation is that of a plane, and the second is that of a sphere. These objects intersect in exactly one of three ways: as a circle, as a point, or not at all. It remains to show that these correspond precisely to the cases d < 2r, d = 2r, and d > 2r.

Furthermore, let D be the distance from the origin to the plane defined by the first equation. It's clear geometrically that the type of intersection depends on D as follows: if D > r, then there is no intersection; if D = r, then the intersection is a single point, and if D < r, then the intersection is a circle (infinitely many points).

So all that remains is to relate D to d. In particular, if you can show that D = d/2, you're done.

 

[TPAINE]

Sorry, I've been having to multitask so my solution sketch is a bit scatterbrained.

OK, the cosine was useful for understanding why d has to be <= 2r, but I think it's just obscuring the final solution.

We seek the set of all vectors u that satisfy the following two conditions:

<u,w> = d^2 / 2
and
|u| = r

The first equation is that of a plane, and the second is that of a sphere. These objects intersect in exactly one of three ways: as a circle, as a point, or not at all. It remains to show that these correspond precisely to the cases d < 2r, d = 2r, and d > 2r.

I am interested in the case 2r>d. The other two I have figured out.

It's so obvious but I can't do it rigorously.

 

[jbunniii]

P.S. Sorry, I forgot we're dealing with k dimensions, not necessarily 3. For "circle" above you should read "lower dimensional sphere of the appropriate dimension." All that really matters is that this is the case where there are infinitely many solutions.

 

[jbunniii]

Sorry, I edited post #23, bad habit I know. Please refresh to see the added paragraphs. That's the key thing you need to prove.

 

[TPAINE]

I see why it would get me the result, but I have no clue how to prove it. How do we calculate the distance from the origin to that plane?

 

[TPAINE]

Actually, just take an orthogonal coordinate system u,v,w with u perpendicular to x-y. Then the circle defined by (x+y)/2+R(cos(t)*v+sin(t)*w) works, where R=(r^2-(d/2)^2) by the Pythagorean theorem.

 

[jbunniii]

The distance from a point to a plane is the length of a line segment normal to the plane and passing through the point.

If the plane is defined by

<u,w> = c,

where w is fixed, then w is normal to the plane.

If you constrain u to point in the direction of w, then u = aw for some a.

Furthermore, we can solve for a:

<u, w> = c

means

<aw, w> = c

or equivalently

a<w, w> = c

which is the same as

a = c / <w,w> = c / |w|^2

Now in our case, c = d^2 / 2, and |w| = d, so substituting, we get

a = d^2 / (2 d^2) = 1/2

Thus u = w/2.

And the length of u is the distance from the origin to the plane:

D = |u| = |w|/2 = d/2. Voila.

 

[jbunniii]

Just as a clarifying remark, the reason I am setting u = aw above is because I want to find the unique vector u that points from the origin to the plane and which is normal to the plane. The distance of this unique vector is by definition the distance from the origin to the plane.

I know that u must point in the direction of w since w is normal to the plane, so the only degree of freedom I have is the scale factor a, which I choose so that the endpoint of u is in the plane.

How do I ensure the endpoint of u is in the plane? By forcing it to satisfy the equation of the plane: <u,w> = c.

 

[jbunniii]

I just thought of a much easier way!

We had defined u = z - x, w = y - x

and we have

|z - x|^2 = |z - y|^2

iff

|u|^2 = |u - w|^2

iff

<u,u> = <u,u> - 2<u,w> + <w,w>

iff

2<u,w> = <w,w>

iff

<2u - w, w> = 0

We seek all vectors u which satisfy the above equation, and we also have the constraints

|w| = d
|u| = r

This is really easy to see geometrically. Draw a vector v1 from 0 to w, and another vector v2 from 0 to 2u. Then 2u - w is the vector from the end of v1 to the end of v2.

The condition <2u - w, w> = 0 says precisely that 2u - w is orthogonal to w.

Under what circumstances is this possible?

There are two possibilities: case 1 (sort of a degenerate case) is when 2u = w [in which case 2r = d]

The other possibility is when you can draw a right triangle with a base of length |w| = d and a hypotenuse of length 2|u| = 2r. This is of course possible if and only if 2r > d, and it's equally clear that in that case there are infinitely many choices for u.