- 21 Aug '11 19:09

Yes I also guess that that is the case even though I don't have a rigorous proof of that.*Originally posted by Palynka***The length of the perimeter at the limit is not necessarily the limit of the approaching perimeters.**

People hated on me when I argued something like this during the "balls in a bag" thread.

Another mistake is that this shows that Pi=2 and not Pi=4 as claimed. - 21 Aug '11 19:41

It should be trivial to give a proof, if a similar attempt is made to show that the length of the line y=x from the origin to (1,1) is 2 units.*Originally posted by adam warlock***Yes I also guess that that is the case even though I don't have a rigorous proof of that.**

The line is divided into two and two right angled triangles are drawn, and the process is iterated. It can be shown that for any given iteration, the line y=x remains sqrt(2) units and the length of the horizontal and vertical line segments remain 2. - 22 Aug '11 12:43

The length of the perimiter in the limit is 4, but the curvature at some point on the circle is not the same as it would be for a circle (at a given point the line is either horizontal or vertical), i.e. the limiting shape is not a circle, it just looks like one.*Originally posted by Palynka***The length of the perimeter at the limit is not necessarily the limit of the approaching perimeters.**

People hated on me when I argued something like this during the "balls in a bag" thread. - 22 Aug '11 13:14 / 2 edits

What do you mean by "in the limit"? And "it just looks like a circle"? Seems to me you're talking out of your backside.*Originally posted by KazetNagorra***The length of the perimiter in the limit is 4, but the curvature at some point on the circle is not the same as it would be for a circle (at a given point the line is either horizontal or vertical), i.e. the limiting shape is not a circle, it just looks like one.**

The sequence converges to a circle indeed. Take any point of the circle, the sequence will approximate to it in the limit (some points it converges to in finite iterations, the rest in infinite ones) and it will converge to no point outside of the circle. - 22 Aug '11 13:29

Like your link in your next post says - not all properties converge to it "being a circle". Yes, every point on the "jagged" circle corresponds to a point on a "real" circle, but the curvature and circumference are different, so it's not a "real" circle. The infinitely-sided "jagged" circle will have a circumference 4, not pi.*Originally posted by Palynka***What do you mean by "in the limit"? And "it just looks like a circle"? Seems to me you're talking out of your backside.**

The sequence converges to a circle indeed. Take any point of the circle, the sequence will approximate to it in the limit (some points it converges to in finite iterations, the rest in infinite ones) and it will converge to no point outside of the circle. - 22 Aug '11 13:35

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.*Originally posted by KazetNagorra***Like your link in your next post says - not all properties converge to it "being a circle". Yes, every point on the "jagged" circle corresponds to a point on a "real" circle, but the curvature and circumference are different, so it's not a "real" circle. The infinitely-sided "jagged" circle will have a circumference 4, not pi.**

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. - 22 Aug '11 13:58

Yes, like I said all points converge to points on a circle with d = 1, but the final shape is not a circle because the definition of a circle entails more than the location of its points.*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. - 22 Aug '11 14:41

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.*Originally posted by Palynka***If all the points are at the exact same location, what properties are different?**