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Sep 30, 2016 at 17:12 comment added Ilya Bogdanov Well, if $x$ has coordinates of both signs (or a zero coordinate), then you may choose $w$ orthogonal to $x$. Othewise we may assume that $x>0$. Then $w$ is clearly a vector in the direction of the axis corresponding to smallest coordinate of $x$.
Sep 30, 2016 at 17:06 history edited JohnA CC BY-SA 3.0
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Sep 30, 2016 at 17:02 comment added JohnA I love good counterexamples like this! The idea of thinking about the special case $\Vert x_j\Vert=1$ did not occur to me. Any thoughts on the case $w_i\ge 0$ and $j=1$?
Sep 30, 2016 at 15:57 comment added Ilya Bogdanov I doubt there is an analytic solution. Assume that $\|x_j\|=1$ for all $j$, so the $x_j$ are the points on the sphere. If there is a large ball on this sphere which is free of the points, and if the rest ot the sphere is covered more or less densely, then the answer is the center of this ball, regardless of the specific ppositions of the points.
Sep 30, 2016 at 15:49 review First posts
Sep 30, 2016 at 16:17
Sep 30, 2016 at 15:48 history asked JohnA CC BY-SA 3.0