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Sep 16, 2011 at 1:22 comment added t3suji Out of curiosity, we may also look at the following: Suppose the curvature changes sign, but if it does, it stays above -1/r, so the corresponding radius of curvature is greater than r. In this case, we do not need to doodle backwards'. It seems likely to me that the formulas still hold (but the proof by polynomial approximation would not work as polynomial approximation has negative infinite curvature now). It would seem that the direct write the line integral' approach applies.
Sep 16, 2011 at 1:08 comment added Paul Siegel Yeah, that sort of example is why I added the restriction that the curvature doesn't change sign. The problem isn't with the winding number - we only care about the winding number of the original curve, not the doodle. The problem is that the path made by the marker is too complicated when you hit a corner where you have to turn by more than 180 degrees. I'm not sure what the resolution is (or, come to think of it, if I even believe the formula in such cases).
Sep 16, 2011 at 0:24 comment added Jiji Cool, thanks. But I am still confused with something like the figure 8. (see picture of the doodle around 8 in the link). What if the doodle goes "inside" the curve as in the picture? I am having a hard time seeing how your answer would apply in that case. (because whenever the doodle is "inside" the curve, the winding number goes down).
Sep 15, 2011 at 23:37 comment added Paul Siegel I suppose we need to assume that the curvature of the curve never changes sign, as was guaranteed by the convexity assumption in the case where the winding number was 1.
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Sep 15, 2011 at 23:10 history answered Paul Siegel CC BY-SA 3.0