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You make a good point intuitively, but I still don't quite buy your argument. The circle is continuously differentiable along its circumference. The fractal is
by kwantam 16y ago
You make a good point intuitively, but I still don't quite buy your argument.
The circle is continuously differentiable along its circumference. The fractal is not---it's still got a large number of discontinuities (offhand, it looks like the number of discontinuities is equal to 2^(n-1) after n steps).
The area enclosed by this curve approaches that of a circle, but the curve is not a circle: it cannot be described by x^2+y^2=1 because its derivatives are not equal to those of same.
- martincmartin 16y agoEvery point on the limit of the fractal is within epsilon of some point on x^2 + y^2 = 1, for every epsilon > 0. So, if some point on the limit of the fractal were not on the circle, it would be some finite distance d from the circle. By choosing epsilon = d/2 you have a contradiction. The circumference of the limit doesn't equal the limit of the circumferences.