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Let the surface $S_n$ of the unit ball in $\mathbb{R}^n$ centered at the origin $O$ be defined as the set of points $P(x_1,x_2,…,x_n )$ such that $x_1^2+x_2^2+⋯+x_n^2=1$. Let the spherical cap $C(α)$ of angular radius $\alpha≤π$ centered at $T$ on $S_n$ be defined as the set of all points $Q$ in $S$ such that $∠QOT≤α$. Let $N(α)$ be the maximum number of non-overlapping spherical caps $C(α)$ that can be placed on $S_n$ for $n≥2$.

Is there some function $\Upsilon (n)$ such that $N(α)\sim\Upsilon(n)\alpha^{1-n}$ as $\alpha→0$?

EDIT: An estimate for the first few values of $\Upsilon$ is $\Upsilon(n)\approx1.83,1.46,1.3$ when $n=3,4,5$. The only exact value so far is $\Upsilon(2)=\pi$.

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up vote 2 down vote accepted

Presumably $\Upsilon (n)$ is (up to a constant) the optimal packing density for balls in ${\mathbb R}^{n-1}$.

EDIT: As $\alpha \to 0$, the $n-1$-dimensional Hausdorff measure of the spherical cap $C(\alpha)$ is asymptotic to the Lebesgue measure in ${\mathbb R}^{n-1}$ of a ball of radius $\alpha$, which is $\pi^{(n-1)/2}/\Gamma((n+1)/2) \alpha^{n-1}$. The $n-1$-dimensional Hausdorff measure of the sphere is $n \pi^{n/2}/\Gamma(n/2+1)$. So if we could pack them perfectly, we'd have asymptotically $$ \dfrac{n \pi^{n/2}}{\Gamma(n/2+1)} \dfrac{\Gamma((n+1)/2)}{\pi^{(n-1)/2}} \alpha^{1-n} = \dfrac{2 \sqrt{\pi}\; \Gamma((n+1)/2)}{\Gamma(n/2)} \alpha^{1-n}$$ If the packing density for balls in ${\mathbb R}^{n-1}$ is $\rho(n-1)$, this is reduced by a factor of $\rho(n-1)$. So what I meant was

$$ \Upsilon(n) = \dfrac{2 \sqrt{\pi}\; \Gamma((n+1)/2)}{\Gamma(n/2)} \rho(n-1)$$

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I don't think so; with the data here estimates the first few values $n=2,3,4,5$ at $\Upsilon(n)\approx\pi,1.83,1.46,1.3$. Sorry; I should have mentioned that in the question. –  hombre Nov 27 '12 at 20:13
    
Thank you! Well said. –  hombre Nov 27 '12 at 21:12
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