## Doubling dimension of a Euclidean space

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The doubling dimension of a metric space $X$ is the smallest positive integer $k$ such that every ball of $X$ can be covered by $2^k$ balls of half the radius.

It is well known that the doubling dimension $d(n)$ of the Euclidean space $\mathbb R^n$ is $O(n)$, which means that there is a constant $C$ such that for large $n$ one has $d(n)\leq Cn$. A posteriori, I can find a new constant $D$ that works for all $n$. I would like to have an explicit description of this new constant. In other words,

Question: What explicit and possibly nice and small constant $D>0$ would guarantee that $d(n)\leq Dn$, for all $n$?

Edit. As observed by Igor Rivin, $D=\log 2$ should be good for $n\geq7$, by a theorem of Verger-Gaugry. Any idea for all $n$? I have to clarify that at the moment I am not interested in the best possible constant, but in some good-looking constant, something to make aesthetically pleasant a certain formula that I found out.

Thank you in advance,

Valerio

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## 1 Answer

As shown in this paper,Theorem 1.2, $D \leq \log 2.$ I remark that this paper came up in my answer to this question, and there is a bug for small $n$ ($n < 7$), but the author's interest was apparently similar to yours, so the large $n$ results should be correct. (the paper is: "Covering a Ball with Smaller Equal Balls in \$\mathbb{R}^n," by Jean-Louis Verger-Gaugry)

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Thank you very much. What can we say for small $n$? I am really interested in all values of $n$. Maybe also to know that say $D=2$ is good enough would be OK. Indeed, for the moment I want to put this constant in a as nice as possible formula for all $n$ and then maybe discuss the fact that can be sharpened for $n\geq7$... – Valerio Capraro Jun 27 at 14:23
I am having a look at the paper and maybe I am missing something. He fixes a radius $T>\frac12$ and answers the question of how many balls of radius $\frac12$ are needed to cover a ball of radius $T$. My case is little different, because the covering balls have radius $T/2$. Well, it is possible that one can go down inductively and apply that formula, but I am little in trouble with that terrible formula. Moreover, that formula holds only for $T\leq\frac{n}{2\log(n)}$... In two words: I'm confused! – Valerio Capraro Jun 27 at 14:40
For your problem $T=1,$ and the inequality is vacuously satisfied... – Igor Rivin Jun 27 at 15:15
Yes, indeed! thanks again – Valerio Capraro Jun 27 at 15:33
I am sorry, but I cannot understand how you get $\log 2$. Indeed, let $f(n)$ be Verger-Gaugry's estimation in Theorem 1.2 but without $2^n$. It seems to me that $d$ should verify the property that $\log_2(f(n))\leq(d-1)n$, for all $n$. One can easily see that every $d>1$ is eventually good, but the function $f(n)$ diverges and then it seems to me impossible to find a constant $d\leq1$. Am I missing something? – Valerio Capraro Jun 30 at 16:48
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