Timeline for Continuous-time extension of a discrete dynamical system
Current License: CC BY-SA 4.0
8 events
when toggle format | what | by | license | comment | |
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Jan 14, 2020 at 9:08 | vote | accept | ABIM | ||
Jan 9, 2020 at 14:37 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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Jan 9, 2020 at 10:57 | answer | added | user37691 | timeline score: 3 | |
Jan 9, 2020 at 10:40 | comment | added | Mateusz Kwaśnicki | Then this is essentially a question whether there is a smooth simple curve passing through given $N$ points $x_1, \ldots, x_N$ in a given order. Or, more precisely, we require this curve to pass through $M$ points $x_1, \ldots, x_M$ in a given order and have no self-intersections except possibly at $x_M$, in which case it must be tangent to itself; here $M$ is the least number such that $x_M \in \{x_1, \ldots, x_{M-1}\}$, or $M = N$ if no such number exists. This seems to be fairly straightforward, no? | |
Jan 9, 2020 at 10:22 | comment | added | ABIM | @MateuszKwaśnicki No, $x$ should be specified (so the "anti-discritization" is dependent on the choice of x); It would only need to hold for an (arbitrarily large but fixed) number of iterations. | |
Jan 9, 2020 at 10:20 | history | edited | ABIM | CC BY-SA 4.0 |
added 50 characters in body
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Jan 9, 2020 at 10:10 | comment | added | Mateusz Kwaśnicki | If you mean "for all $x$", then no, it is not possible. For every $F$, the Jacobian determinant of the corresponding flow is non-zero, and hence it remains positive. Thus, $f(x) = (-x_1,x_2,x_3,\ldots,x_d)$ provides a simple counter-example. On the other hand, if $x$ is fixed, then I do not see the answer right away. | |
Jan 9, 2020 at 9:28 | history | asked | ABIM | CC BY-SA 4.0 |