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Outer content and polars

  1. Nov 24, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Let S be the part of the spiral ( given in polars ) r = eθ which is inside the square with vertices [±1, ±1]. Prove that C(S) = 0.

    2. Relevant equations

    I know that [itex]C(S) = inf \left\{{\sum A_k}\right\}[/itex] where Ak is the area of the rectangle Rk

    3. The attempt at a solution

    I'm not really sure how to start this one. I drew my square with center (0,0) on my plane and observed that I wanted to show anything outside the square is zero.

    Any help would be appreciated.
     
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  3. Nov 24, 2012 #2

    haruspex

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    Where the Rk are?
    Does this C(S) function have a name I can Google?
    Edit: Just figured it out from the title - it's Outer Measure, right?
     
  4. Nov 24, 2012 #3

    Zondrina

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    I was at work, sorry for the late reply.

    Yes indeed. I believe outer content is really outer measure ( I did some Googling myself earlier ).

    EDIT : Also, the Rk are the rectangles created by partitioning the region into a finite amount of rectangles.
     
    Last edited: Nov 24, 2012
  5. Nov 25, 2012 #4

    haruspex

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    Perhaps you mean they are a cover for S?
     
  6. Nov 25, 2012 #5

    Zondrina

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    Yeah, I drew my axes and I plotted my square with its vertices. Then I drew my spiral inside of my square.

    I know that a curve of finite length must have outer content zero I believe?
     
  7. Nov 25, 2012 #6

    haruspex

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    Certainly, in fact the question seems more awkward than interesting. But you agree the Rk are just a cover, i.e. a set whose union includes S? So the trick is to define some parametrised set, Rk, δ, k = 1...Nδ, whose total area tends to 0 as δ→0, right?
    Do the rectangles need to be oriented parallel to X and Y axes? If not, could try rectangles with vertices (r,θ), (reδθcos(δθ),θ), (reδθ,θ+δθ) (and whatever the fourth vertex would be).
     
  8. Nov 25, 2012 #7

    Zondrina

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    Okay here's what I did.

    I plotted my region first ( which is the square ). Then I drew my spiral inside of my square which extends only to the boundary of my region.

    I then enclose a finite amount of points on my spiral inside of smaller squares and then extend the sides of these squares out to the boundary of my region thus partitioning my region into a mesh.

    So [itex]\sum A_k[/itex] should be less than or equal to the total area of all of my squares in my partition.

    So if I can calculate the area of my squares to be some N, then if the total area of my squares tends to zero as N→∞ I would get that C(S) = 0 if my drawing isn't lying to me?
     
    Last edited: Nov 25, 2012
  9. Nov 25, 2012 #8

    haruspex

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    But you don't want all squares in the mesh, or you'll end up with the area of the entire 2x2 square. You just want the subset that intersects S. (And every part of S needs to lie in some square.)
     
  10. Nov 25, 2012 #9

    Zondrina

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    Yes ^ this is why I said that nothing goes beyond the boundary of S. So as soon as I figure out the length of the side of a square, I can figure out its area and take the limit as the total area of the squares goes to infinity.

    So hmm... Lets say there are n squares. What if I took the side of a square to be [itex]\frac{L}{n}[/itex]?

    Then : [itex]\sum A_k ≤ n(\frac{L}{n})^2 = \frac{L}{n}[/itex]

    So we have : [itex]C(S) ≤ \frac{L}{n} → 0 \space as \space n → ∞[/itex]

    Thus C(S) = 0. Would this be good? ( Since we can't have negative outer content ).
     
    Last edited: Nov 25, 2012
  11. Nov 25, 2012 #10

    haruspex

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    No, you said nothing goes beyond the boundary of the region. S is the curve, and does not constitute any obvious boundary.
    Exactly where are these n little squares? Their placement is critical, or you won't be able to show they cover S.
    I have the feeling we're still not understanding the problem in the same way. Let me take a simpler example. Suppose S is the straight line from (0,0) to (1,1). You could cover that with little squares (nδ,(n+1)δ) x (nδ,(n+1)δ). There would be (about) 1/δ of these. Every point of S would lie in at least one of these, and their total area would be (about) δ.
     
  12. Nov 25, 2012 #11

    Zondrina

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    Ohhh I see what you're saying now ( Sorry for my technicality about S ).

    Just a few questions about what you said though for your example. You claim the squares are (nδ,(n+1)δ) x (nδ,(n+1)δ), but one side can't possibly be longer than the other if you're talking about a square right?

    Also, why do you say 1/δ of them? Wouldn't there be n squares?
     
  13. Nov 25, 2012 #12

    haruspex

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    Square n is the product of the intervals (nδ,(n+1)δ) in the x direction and (nδ,(n+1)δ) in the y direction, so measures δxδ.
     
  14. Nov 25, 2012 #13

    Zondrina

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    Ahh yes I see what you're saying now. What about : "Also, why do you say 1/δ of them? Wouldn't there be n squares?"

    After you clarify that I think I understand.
     
    Last edited: Nov 25, 2012
  15. Nov 25, 2012 #14

    haruspex

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    I'm just using n as the index for a square. Each is width δ, and they're arranged on a diagonal. So after r of them they run from (0,0) to (rδ,rδ). When r > 1/δ you've reached (1,1).
     
  16. Nov 25, 2012 #15

    Zondrina

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    Wait, do you mean when r ≥ 1/δ then we've reached (1,1)? Since (1,1) is still on the line right? So it exists at both boundaries, (0,0) and (1,1)? So when rδ has gone far enough, we've reached the boundary of the curve. So if it exists at both boundaries... then the outer content would have to be zero?

    Since every neighborhood of a point at that point would contain points not in S otherwise? The curve should be defined to be zero everywhere outside the boundary otherwise it wouldn't be a curve of finite length.

    So these co-ordinates you said earlier about this curve... [itex](r,θ), (re^{δθ}cos(δθ),θ), (re^{δθ},θ+δθ) \space and \space (re^{δθ}sin(δθ),θ+δθ)[/itex]

    I think I understand what you mean now if I'm not mistaken.
     
    Last edited: Nov 25, 2012
  17. Nov 26, 2012 #16

    haruspex

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    Yes.
    Now you've lost me. The rectangles cover the curve, i.e. the points of the curve are a subset of the union of the rectangles. The sum of the areas of the rectangles < (1/δ+1)*δ2 < δ + δ2. So tends to 0 as δ→0.
    Yes, just trying to find a set of rectangles that cover the curve but with combined area O(δθ). If you try to it with rectangles aligned to the axes you get into strife where the gradient approaches horizontal or vertical.
     
  18. Nov 26, 2012 #17

    Zondrina

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    Okay so let me sort of try to clean up all of my thoughts here into one post.

    Suppose S is a part of the spiral r = eθ inside the square R with vertices [±1, ±1].

    Then S is a curve of finite length and thus has finitely many points along its path, say [itex]Q_1, ..., Q_n[/itex]. Suppose we enclose each point Qi, 1 ≤ i ≤ n on the path of S inside of a square Rk with vertices at : [itex][r,θ], [re^{δθ},(1+δ)θ], [re^{δθ}cos(δθ),θ] \space and \space [re^{δθ}sin(δθ),(1+δ)θ][/itex] so that we have δ little squares covering our curve.

    Now, we extend the sides of the δ squares out to the boundary of R thus obtaining a partition P of R.

    Suppose that Ak is the area of a square Rk where all of the Rk are contained within the boundary of our square R. The area of a square is at most : ( Having a bit of trouble here ).

    Then we have that :

    [itex]\sum A_k [/itex] ≤ δ*(the area of a square)

    Just having a bit of trouble finding the area of a square now. These vertices are tricky. After I find the area though the rest should be obvious I'd assume?
     
    Last edited: Nov 26, 2012
  19. Nov 26, 2012 #18

    haruspex

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    No, but you may have some way of choosing a finite set of points (which you would need to state).
    The first three of those are vertices of a rectangle (not a square), but that fourth point will not complete it. The r value is quite wrong, and the angle should be a little larger. Might not need to specify it exactly.
    Why would you want to do that? You don't need to partition R.
    You only need a set of rectangles the union of which contains all the points of S. Nothing outside that set is of interest. I keep saying this, but you don't seem to be taking it in.
     
  20. Nov 26, 2012 #19

    Zondrina

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    Then I would just say : "Choose a finite set of points which are a subset of S, say [itex]Q_1, ..., Q_n[/itex]"

    I'm having trouble visualizing the rectangle then. Where does each vertices go when I'm drawing this? I wouldn't have a clue where the fourth vertices would go.

    I understand this.
     
  21. Nov 26, 2012 #20

    haruspex

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    The construction I gave for the rectangles selects points for some chosen δθ: in r, θ co-ordinates they are (enδθ, nδθ). These are the vertices of the rectangles nearest the origin. I.e. for each rectangle , its vertex nearest the origin is one of those points.
    Draw a line out from the origin to the point Qn = (rn, θn) = (enδθ, nδθ) for some n. That's the first vertex. Continue that line to point (e(n+1)δθcos(δθ), nδθ). That's about rnδθ further. Turn left one right angle and connect to the point (e(n+1)δθ, (n+1)δθ). That's a distance of e(n+1)δθsin(δθ), also about rnδθ. Notice that this point is Qn+1. The fourth point is harder to express algebraically, but the details are uninteresting. You can figure out the distance hn from the origin by Pythagoras, the shorter two sides being rn and e(n+1)δθsin(δθ). The angle from the x-axis is nδθ+atan(e(n+1)δθsin(δθ)/rn) = nδθ+atan(eδθsin(δθ)).
    You'll need to show all parts of S lie in one or more of these rectangles.
    I'm not saying this is the best way - all looks far more painful than the object of the exercise warrants.
     
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