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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
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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
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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