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Geometric convergence is tricky. Specifically, the issue here is that uniform convergence does not imply convergence in length or area or any other such measure
by cscheid 16y ago
Geometric convergence is tricky. Specifically, the issue here is that uniform convergence does not imply convergence in length or area or any other such measures. (see figure 6 here: http://www.sci.utah.edu/~etiene/publications/verifiable-vis.pdf http://www.sci.utah.edu/~etiene/publications/verifiable-vis.... . disclaimer - I'm a co-author)
Imagine a circle and its diameter. Now imagine two circles with half of the diameter, lined up so that the diameter lines align. Now split those two in four, etc. The circle becomes a snaking line whose total length doesn't change, and the snaking line converges uniformly to the line. Clearly, however, pi is not 1.
What you need is convergence in position _and_ angle. A curve that converges in position and angle _does_ converge in length: the reference I know which shows this is reference [9] on the above-mentioned paper. (edit: in case you don't want to download the gigantic file --- yay for publishing in graphicsy places --- the reference is: K. Hildebrandt, K. Polthier, and M. Wardetzky. On the convergence of metric and geometric properties of polyhedral surfaces. Geometriae Dediacata, (123):89–112, 2006.)
- pixcavator 16y ago>>...uniform convergence does not imply convergence in length or area... Under uniform convergence, the limit of the integrals is equal to the integral of the limit. So, this works fine for areas, or are you talking about something else?
- cscheid 16y agoSorry, I should have been clearer. I was trying to say that convergence in area for two-dimensional surfaces in R^3 requires convergence in normals in the same way that convergence in length for one-dimensional curves in R^2 requires convergence in normals. For area in R^2, volume in R^3, and so on, you're definitely right.