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Dec
16
awarded  Caucus
Mar
25
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.
Mar
25
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
Mar
24
awarded  Yearling
Mar
24
awarded  Self-Learner
Mar
24
asked How to prove Lyapounov stability of a circle orbit?
Dec
17
awarded  Revival
Dec
17
awarded  Scholar
Dec
17
accepted Why do Zoll metrics exist only on $S^2$ and $RP^2$?
Dec
17
answered Why do Zoll metrics exist only on $S^2$ and $RP^2$?
Oct
25
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.
Oct
25
revised Why do Zoll metrics exist only on $S^2$ and $RP^2$?
added 229 characters in body
Oct
25
asked Why do Zoll metrics exist only on $S^2$ and $RP^2$?
Jul
24
awarded  Autobiographer
Oct
29
awarded  Teacher
Oct
21
answered Is there a map from a segment to a triangle?
Oct
14
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?
Oct
13
awarded  Supporter
Oct
13
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.
Oct
13
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?