Olga
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 Dec16 awarded Caucus Mar25 comment How to prove Lyapounov stability of a circle orbit? @ABC, the definition is correct. Although I know it. I won't ask a question without knowing what I actually want to obtain. That's also why you shouldn't ask me to marry you. Mar25 comment How to prove Lyapounov stability of a circle orbit? @Artem still, they can be close to a circle, even if the period depends on initial conditions? $\alpha=2$ is a harmonic oscillator, orbits are ellipses Mar24 awarded Yearling Mar24 awarded Self-Learner Mar24 asked How to prove Lyapounov stability of a circle orbit? Dec17 awarded Revival Dec17 awarded Scholar Dec17 accepted Why do Zoll metrics exist only on $S^2$ and $RP^2$? Dec17 answered Why do Zoll metrics exist only on $S^2$ and $RP^2$? Oct25 comment Why do Zoll metrics exist only on $S^2$ and $RP^2$? Rasmus, I updated a post after your comment. Yes, of course, I want to prove a fact which really holds. It's mentioned in lots of places in the litterature, in particular, in the book that I mentioned above which is considered a main book on this topic. Oct25 revised Why do Zoll metrics exist only on $S^2$ and $RP^2$? added 229 characters in body Oct25 asked Why do Zoll metrics exist only on $S^2$ and $RP^2$? Jul24 awarded Autobiographer Oct29 awarded Teacher Oct21 answered Is there a map from a segment to a triangle? Oct14 comment Is there a map from a segment to a triangle? Thank you very much for help! By the way, doest it follow, that diameters of the circle go to straight lines in hexagon from your argument? Oct13 awarded Supporter Oct13 comment Is there a map from a segment to a triangle? I would like to precise the argument via symmetry principle: so, I do have a map from the unit circle to a hexagon. Let me now choose a point that maps to a vertice, and a corresponding diameter of a unit circle. Why does this diameter map to a line connecting edges of hexagon? Because only in this case we can apply symmetry principle. Oct13 comment Is there a map from a segment to a triangle? Do I understand correctly that I actually can choose $Arg f'(0)$ to be any angle I want (by Riemann theorem construction) and that determines my map? And as far as I understand, for any choice of $Arg f'(0)$ the map will still map vertices to vertices, isn't it?