Originally posted by PalynkaThe 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.
If all the points are at the exact same location, what properties are different?
3 edits
Originally posted by twhiteheadWhich point of the circle does it not converge to?
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 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).
1 edit
Originally posted by PalynkaWell, "the circumference" is the obvious answer, given the circumstances. 🙂
If all the points are at the exact same location, what properties are different?
But yes, I agree with the general idea. If the limit exists, the circle is the only possible shape it can be.
Originally posted by PalynkaFor 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.
If all the points are at the exact same location, what properties are different?
1 edit
Originally posted by KazetNagorraHow is there "jaggedness" if the points are all in the same location? You keep confusing the approximating configurations with the limiting one.
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.
1 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.
Originally posted by PalynkaSince 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.
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.
2 edits
Originally posted by PalynkaAgain, 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?
Sigh.
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.
3 edits
Originally posted by SoothfastThe 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.
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.
That there are other sequences that converge to it is irrelevant.
Originally posted by JS357I'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.
Is the limit a member of the set of polygons derived by the procedure?