1. Cape Town
    Joined
    14 Apr '05
    Moves
    52945
    22 Aug '11 14:41
    Originally posted by Palynka
    If all the points are at the exact same location, what properties are different?
    The points can never be in the exact same location as in the circle, one point can never be vertically above another (in the arc above the x-axis) whereas in the jagged shape for any given point on the circle there is always another point not on the circle vertically above it. There are always point not on the circle.
  2. Standard memberPalynka
    Upward Spiral
    Halfway
    Joined
    02 Aug '04
    Moves
    8702
    22 Aug '11 14:513 edits
    Originally posted by twhitehead
    The points can never be in the exact same location as in the circle, one point can never be vertically above another (in the arc above the x-axis) whereas in the jagged shape for any given point on the circle there is always another point not on the circle vertically above it. There are always point not on the circle.
    Which point of the circle does it not converge to?
    Which points outside of the circle does it converge to?

    Those are again properties of the approximating configurations that are not properties of the limiting configuration (to use the language of the link I provided).
  3. Joined
    07 Sep '05
    Moves
    35068
    22 Aug '11 15:531 edit
    Originally posted by Palynka
    If all the points are at the exact same location, what properties are different?
    Well, "the circumference" is the obvious answer, given the circumstances. 🙂

    But yes, I agree with the general idea. If the limit exists, the circle is the only possible shape it can be.
  4. Joined
    29 Dec '08
    Moves
    6788
    22 Aug '11 17:12
    Originally posted by mtthw
    Well, "the circumference" is the obvious answer, given the circumstances. 🙂

    But yes, I agree with the general idea. If the limit exists, the circle is the only possible shape it can be.
    The limit never exists, as the demon doing the nibbling lacks a curved bite. 🙂
  5. Germany
    Joined
    27 Oct '08
    Moves
    3118
    22 Aug '11 17:39
    Originally posted by Palynka
    If all the points are at the exact same location, what properties are different?
    For example curvature. Assume you have such a circle with d = 1 with the origin at the middle, and consider the point (1/2) (sqrt(2)/2, sqrt(2)/2). The slope on a circle will be -1, but the slope on the "jagged" circle is undefined.
  6. Standard memberPalynka
    Upward Spiral
    Halfway
    Joined
    02 Aug '04
    Moves
    8702
    22 Aug '11 18:091 edit
    Originally posted by KazetNagorra
    For example curvature. Assume you have such a circle with d = 1 with the origin at the middle, and consider the point (1/2) (sqrt(2)/2, sqrt(2)/2). The slope on a circle will be -1, but the slope on the "jagged" circle is undefined.
    How is there "jaggedness" if the points are all in the same location? You keep confusing the approximating configurations with the limiting one.
  7. Standard memberPalynka
    Upward Spiral
    Halfway
    Joined
    02 Aug '04
    Moves
    8702
    22 Aug '11 18:103 edits
    Originally posted by JS357
    The limit never exists, as the demon doing the nibbling lacks a curved bite. 🙂
    And 0.9 + 0.09 + 0.009 +... never reaches 1 because it always lacks something, right?

    Every point of the circumference will be part of the limit, it's a simple bisection-type argument.
  8. Standard memberPalynka
    Upward Spiral
    Halfway
    Joined
    02 Aug '04
    Moves
    8702
    22 Aug '11 18:191 edit
    More explanations for the nay-sayers:

    This site explains it with a similar case where the arithmetic is more pallatable.

    http://qntm.org/trollpi

    Quotes:

    "In case there is any confusion or ambiguity, let me make it explicit: the limit curve of this sequence of jagged sawtoothed curves is not a sawtoothed curve, not an "infinitely jagged" sawtoothed curve, not a fractal, and not undefined. It is simply a straight horizontal line. It has a gradient of 0. We have just proved this."



    So the limit of the lengths is sqrt(2) and the length of the limit is 1.
    Therefore we have one good counterexample and the original statement is false.
  9. Standard memberSoothfast
    0,1,1,2,3,5,8,13,21,
    Planet Rain
    Joined
    04 Mar '04
    Moves
    2701
    22 Aug '11 19:52
    Originally posted by Palynka
    Wrong again, not all properties need to converge, there can be (and there is) discontinuity in some. In this case, the perimeter of the shape remains 4 but at the limit there is a discontinuity.

    Which point of the circle does it not converge to?
    Which points outside of the circle does it converge to?

    If your answers to these two questions are both "none", then clearly it converges to the circle.
    Since the figure being attained by the limit process is a planar curve with length 4, clearly it is not a circle. The curve will be everywhere continuous but nowhere differentiable.
  10. Standard memberPalynka
    Upward Spiral
    Halfway
    Joined
    02 Aug '04
    Moves
    8702
    22 Aug '11 20:03
    Originally posted by Soothfast
    Since the figure being attained by the limit process is a planar curve with length 4, clearly it is not a circle. The curve will be everywhere continuous but nowhere differentiable.
    Sigh.
  11. Joined
    29 Dec '08
    Moves
    6788
    22 Aug '11 20:03
    Originally posted by Palynka
    And 0.9 + 0.09 + 0.009 +... never reaches 1 because it always lacks something, right?

    Every point of the circumference will be part of the limit, it's a simple bisection-type argument.
    Is the limit a member of the set of polygons derived by the procedure?
  12. Standard memberPalynka
    Upward Spiral
    Halfway
    Joined
    02 Aug '04
    Moves
    8702
    22 Aug '11 20:06
    Originally posted by JS357
    Is the limit a member of the set of polygons derived by the procedure?
    Sorry, I don't understand the question. From what I understand, the answer is "no". See the link and quotes I posted above for an example.
  13. Standard memberSoothfast
    0,1,1,2,3,5,8,13,21,
    Planet Rain
    Joined
    04 Mar '04
    Moves
    2701
    22 Aug '11 20:152 edits
    Originally posted by Palynka
    Sigh.
    Again, the length of the figure being constructed is 4, while the circumference of the circle is pi. How can you contend that they are one and the same?

    The usual way to approximate a curved path of any kind, including a circle, is to connect points on the path with straight line segments in an iterative fashion that introduces more points on the path and more line segments connecting them. The process being employed in the site linked to above does not do this exactly, because new points are introduced that do not lie on the path.

    There is a definition for the length of a curve given in basic calculus texts that will tell us that the length of a circle with diameter 1 is pi, but it obtains from a definite integral which is a limit of Riemann sums that are constructed in the manner that I've just described.
  14. Standard memberPalynka
    Upward Spiral
    Halfway
    Joined
    02 Aug '04
    Moves
    8702
    22 Aug '11 20:203 edits
    Originally posted by Soothfast
    Again, the length of the figure being constructed is 4, while the circumference of the circle is pi. How can you contend that they are one and the same?

    The usual way to construct a curved path of any kind, including a circle, is to connect points on the circle with straight line segments in an iterative fashion that introduces more points and more l ...[text shortened]... l which is a limit of Riemann sums that are constructed in the manner that I've just described.
    The area between the two figures converges to zero, for any point in the circumference, one can find a sub-sequence that will converges it, one cannot find any sub-sequence that converges to a point outside the circumference. The limit is the circumference.

    That there are other sequences that converge to it is irrelevant.
  15. Standard memberSoothfast
    0,1,1,2,3,5,8,13,21,
    Planet Rain
    Joined
    04 Mar '04
    Moves
    2701
    22 Aug '11 20:21
    Originally posted by JS357
    Is the limit a member of the set of polygons derived by the procedure?
    I'd say no, not a polygon in the usual sense, since a polygon in the usual sense must have a finite number of sides. There are more esoteric definitions of the term "polygon" in mathematics that may call it a polygon, but it wouldn't be your grandfather's polygon.
Back to Top

Cookies help us deliver our Services. By using our Services or clicking I agree, you agree to our use of cookies. Learn More.I Agree