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Consider the $n-1$ dimensional unit sphere embedded in $\mathbb{R}^n$. For example, when $n=3$, the sphere is characterized by $x_1^2+x_2^2+x_3^2=1$. Define a special point as a point whose coordinates are all zero except one coordinate that is either $1$ or $-1$. For example, if $n=3$, there are six special points: $(1,0,0)$, $(-1,0,0)$, $(0,1,0)$, $(0,-1,0)$, $(0,0,1)$, $(0,0,-1)$. In general, there are $2n$ special points in $\mathbb{R}^n$.

Consider a spherical cap centered at each special point. So there are $2n$ caps as well. Suppose all caps have the same height $h$. I am interested in finding the smallest $h$ such that allows these caps to entirely cover the sphere.

Thank you!

Golabi

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2 Answers 2

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Assuming you are measuring height along the relevant coordinate axis, $1-\frac{1}{\sqrt{n}}$. Because you have to be at least $1/\sqrt{n}$ in some direction from the center to make it onto the sphere.

I should add that if you are measuring "height" along a great circle, it's the arccosine of $1/\sqrt{n}$.

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    $\begingroup$ Thank you very much for your response. Unfortunately I did not fully see how you got $1/\sqrt{n}$ and why subtracting that from one. Could you please provide a bit more detail about the derivation? $\endgroup$
    – Golabi
    Commented Apr 28, 2018 at 6:18
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    $\begingroup$ I think the idea is that any point on the sphere must have at least one co-ordinate at least $1/\sqrt{n}$, otherwise the sum of the squares of the co-ordinates will be less than 1. Moreover, putting a cap on all the special points of height $h$ is sufficient to cover all points which have a co-ordinate $\ge 1 - h$. Since all points have a co-ordinate $\ge 1/\sqrt{n}$, setting $h = 1 - 1/\sqrt{n}$ will ensure you get all the points. $\endgroup$ Commented Apr 28, 2018 at 10:44
  • $\begingroup$ Yes, that's right. Another way to think about it is that the hardest case will be the points $(\pm1/\sqrt{n},...,\pm1/\sqrt{n})$, so you have to cover those. $\endgroup$
    – C Monsour
    Commented Apr 28, 2018 at 12:54
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Consider the cube with vertices $(\pm 1, \ldots ,\pm 1)$ in $\mathbb{R}^n$. It is inscribed in the sphere of radius $\sqrt{n}$ centered at the origin. Note that if we consider hyperplanes parallel to the faces of the cube then the corresponding caps on the sphere cover the whole sphere (if you slice the sphere along those planes you are left with the cube). However, if we took the caps higher than the faces then the vertices of the cube are left out. The ratio we want $$\frac{\text{ height of cap}}{\text{ radius of sphere} }= \frac{\sqrt{n}-1}{\sqrt{n}}=1- \frac{1}{\sqrt{n}}$$ like the previous answer

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    $\begingroup$ Thank you! It was very clear argument and easy to understand. $\endgroup$
    – Golabi
    Commented Apr 29, 2018 at 16:09

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