31 May
2014
31 May
'14
3:04 a.m.
Argh, I was correct the first time - in the rotattoinal case (quaternion) as I said the surface around the real; axis is not fractal so the points that would make it 3D from the 2D cross-section *are not surface points*. This also answer's James' point I think/ On 31 May 2014, at 00:41, Warren D Smith wrote:
Just rotate the 2D Mandelbrot set to get a body of revolution with a 3D boundary given that the 2D set has 2D boundary.
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