Originally posted by PalynkaYes I also guess that that is the case even though I don't have a rigorous proof of that.
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.
Originally posted by adam warlockIt 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.
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.
Originally posted by PalynkaThe 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 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.
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Originally posted by KazetNagorraWhat 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 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.
Originally posted by PalynkaLike 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.
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.
Originally posted by KazetNagorraWrong 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.
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.
Originally posted by PalynkaYes, 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.
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.