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Aug 5, 2014 at 8:23 vote accept Andrej Bauer
Aug 5, 2014 at 8:23 vote accept Andrej Bauer
Aug 5, 2014 at 8:23
Aug 5, 2014 at 0:55 answer added Vladimir Dotsenko timeline score: 6
Aug 4, 2014 at 23:00 answer added John Mount timeline score: 2
Aug 4, 2014 at 20:26 history rollback Andrej Bauer
Rollback to Revision 2
S Aug 4, 2014 at 20:21 history suggested John Mount CC BY-SA 3.0
make want of an interior solution more prominent
Aug 4, 2014 at 20:03 review Suggested edits
S Aug 4, 2014 at 20:21
Aug 2, 2014 at 4:11 comment added Brendan McKay Always $x_k=1\pm x_0$, so for solutions in $[0,1]^{k+1}$ it must be that $x_k=1-x_0$. This can continue to $x_{k-1}$ and so on, but I don't know how to see that it remains real, let alone in $[0,1]$.
Aug 1, 2014 at 20:54 history edited Andrej Bauer CC BY-SA 3.0
added 65 characters in body
Aug 1, 2014 at 16:03 answer added John Mount timeline score: 3
Aug 1, 2014 at 13:40 comment added Vít Tuček Did you try to express $S(p,x)$ in terms of Bernstein basis? Can Mathematica handle that at least for some small $k$?
Aug 1, 2014 at 13:32 comment added Vít Tuček The polynomial $S(p,x)$ is actually divisible by $p$, which means that we have a well defined mapping $\mathbb{R}^{k+1} \to \mathbb{R}^{k+1}$. This allows us to use topological methods such as this one: en.wikipedia.org/wiki/…
Aug 1, 2014 at 13:09 comment added Vít Tuček In other words, we have a family of one-sheeted hyperboloids $$ \sum_{i=1}^k {k \choose i} p^{i-1} (1 - p)^{k - i-1}(x_i-p)^2 - \frac{1}{1-p}(x_0-(p-1))^2 = 1 $$ in $\mathbb{R}^{k+1}$ parametrised by $p \in [0,1]$ and we want to know whether there is a point lying in all of them at once.
Aug 1, 2014 at 9:41 history asked Andrej Bauer CC BY-SA 3.0