How to find radius of covering of sphere?

Suppose we have unit sphere in space $R^n$ which is inscribed sphere of a hypercube. Let we have epsilon-net on the facets of hypercube. For examle, in 3-D this epsilon-net is given as the set of points with coordinates:

$(-1,\hspace{2mm} -1+i*h,\hspace{2mm} -1+j*h)$

$(1,\hspace{2mm} -1+i*h,\hspace{2mm} -1+j*h)$

$(-1+i*h, \hspace{2mm} -1,\hspace{2mm} -1+j*h)$

$(-1+i*h, \hspace{2mm} 1,\hspace{2mm} -1+j*h)$

$(-1+i*h, \hspace{2mm} -1+j*h,\hspace{2mm} -1)$

$(-1+i*h, \hspace{2mm} -1+j*h,\hspace{2mm} 1)$

In R^n we form the set of points analogically.

Then we join every point from this set with the origin ( in 3-D - $(0,0,0)$) by straight line. This lines intersect with our unit sphere in some kind of points. So we have a set of points at sphere which ic called 'base of covering'.

It's need to find by analitycal way the radius of covering (or try to get up and low appreciations).

2022-07-25 20:43:44
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Answers: 1

The sort of point you'll have most trouble covering will be at half distance between the projections of $(1,0,0,\dotsc,h,h)$ and $(1,h,h,\dotsc,0,0)$, where $\lfloor(n-1)/2\rfloor$ entries are $0$ in one vector and $h$ in the other and $\lceil(n-1)/2\rceil$ the other way around. The geodesic distance between those two points on the sphere is

$$\arccos\frac1{\sqrt{1+\lfloor(n-1)/2\rfloor h^2}\sqrt{1+\lceil(n-1)/2\rceil h^2}}\;,$$

and the covering radius is half of that. For odd $n$, this simplifies to


and for small $h$ this is approximately


(Again the covering radius is half of that.)

2022-07-25 22:23:40