Covering Spheres with Spheres
Discrete Comput Geom
Covering Spheres with Spheres
Ilya Dumer
Given a sphere of any radius r in an ndimensional Euclidean space, we study the coverings of this sphere with solid spheres of radius one. Our goal is to design a covering of the lowest covering density, which defines the average number of solid spheres covering a point in a bigger sphere. For growing dimension n, we design a covering that gives the covering density of order (n ln n)/2 for a sphere of any radius r > 1 and a complete Euclidean space. This new upper bound reduces two times the order n ln n established in the classic Rogers bound. Spherical coverings. Let Brn(x) be a ball (solid sphere) of radius r centered at some point x = (x1, . . . , xn) of an ndimensional Euclidean space Rn:

1 Introduction
n
i=1
We also use a simpler notation Brn if a ball is centered at the origin x = 0. For any
subset A ⊆ Rn, we say that a subset Cov(A, ε) ⊆ Rn forms an εcovering (an εnet) of A
if A is contained in the union of the balls of radius ε centered at points x ∈Cov(A, ε).
In this case, we use notation
Cov(A, ε) : A ⊆
Bεn(x).
x∈Cov(A,ε)
By changing the scale in Rn, we can always consider the rescaled set A/ε and the
new covering Cov(A/ε, 1) with unit balls B1n(x). Without loss of generality, below
we consider these (unit) coverings. One of the classical problems is to obtain tight
bounds on the covering size Cov(Brn, 1) for any ball Brn of radius r and dimension n.
Another related covering problem arises for a sphere
Srn d=ef z ∈ Rn+1
zi2 = r2 .
n+1
Then a unit ball B1n+1(x) intersects this sphere with a spherical cap
Crn(ρ , y) = Srn ∩ B1n+1(x),
which has some center y ∈ Srn, halfchord ρ ≤ 1, and the corresponding halfangle
α = arcsin ρ/r . The biggest possible cap Cn(1, y) is obtained if the center x of the
r
corresponding ball B1n+1(x) is centered at the distance
(
1
)
(
2
)
(
3
)
x =
from the origin. To obtain a minimal covering, we shall consider the biggest caps
Crn(1, y) assuming that all the centers x satisfy (
1
).
Covering density. Given a set A ⊆ Rn, let A denote ndimensional volume
(Lebesque measure) of A. We then consider any unit covering Cov(A, 1) and define
minimum covering density
min
ϑ (A) = Cov(A,1) x∈Cov(A,1)
B1n(x) ∩ A
A
.
Minimal coverings have been long studied for the spheres Srn and the balls Brn. The
celebrated Coxeter–Few–Rogers lower bound [1] shows that for a sufficiently large
ball Brn,
ϑ (Srn) ≥ c0 n.
Here and below ci denote some universal constants. A similar result also holds for
any sphere Srn of radius r ≥ n (see Example 6.3 in [4]).
Various upper bounds on the minimum covering density are obtained for Brn and
Srn by Rogers in the classic papers [2] and [3]. In particular, it follows from these
papers that for a sufficiently large radius r , a ball Brn and a sphere Srn can be covered
with density
ϑ ≤
Despite recent improvements obtained in [4] and [5], respectively, for spheres Srn and
balls Brn of a relatively small radius r , the Rogers bound (
3
) is still the best asymptotic
bound known for sufficiently large spheres, balls, and complete spaces Rn of growing
dimension n.
For a sphere Srn of any dimension n ≥ 3 and an arbitrary radius r > 1, the best
universal upper bound known to date is obtained in [4] (see Corollary 1.2 and
Remark 5.1):
ϑ (Srn) ≤
2
1 + ln n
Our main result is presented in Theorem 1, which reduces about two times the present
upper bounds (
3
) and (
4
) for n → ∞.
Theorem 1 Unit balls can cover a sphere Srn of any radius r > 1 and any dimension
n ≥ 3 with density
For n → ∞, there exists o(
1
) → 0 such that
ϑ (Srn) ≤
ϑ (Srn) ≤
The following corollary to Theorem 1 (see also [8]) shows that the Rogers
bound can also be reduced about two times for the coverings of complete
Euclidean spaces Rn.
Corollary 2 For n → ∞, unit balls can cover the entire Euclidean space Rn with
density
ϑ (Rn) ≤
1
2 + o(
1
) n ln n.
2 Preliminaries: Embedded Coverings
In this section, we obtain the estimates on ϑ (Srn) that are similar to (
3
) and (
4
).
However, we will introduce here a slightly different technique of embedded coverings that
will be substantially extended in Section 3 to improve the former bounds (
3
) and (
4
).
We will also employ most of our calculations performed in this section.
Consider a sphere Srn of some dimension n ≥ 3 and radius r > 1. We use notation
C(ρ , y) for a cap Crn(ρ , y) whenever parameters n and r are fixed; we also use a
shorter notation C(ρ) when a specific center y is of no importance. In this case,
Cov(ρ) will denote any covering of Srn with spherical caps C(ρ). By definition, a
covering Cov(ρ) has covering density
where Ωρ is the fraction of the surface of the sphere Srn covered by a cap C(ρ),
ϑρ = Ωρ Cov(ρ)
Ωρ =
C(ρ)
Srn
.
(
4
)
(
5
)
(
6
)
(
7
)
For any τ < ρ ≤ 1, we extensively use inequality (see Corollary 3.2 (ii) in [4]):
and its particular version Ωτ ≥ Ω1τ n obtained for ρ = 1. We begin with two
preliminary lemmas, which will simplify our calculations. Let f1(x) and f2(x) be two
positive differentiable functions. We say that f1(x) moderates f2(x) for x ≥ a if for
all x ≥ a,
Lemma 3 Consider m functions fi (x) such that f1(x) moderates each function
fi (x), i ≥ 2, for x ≥ a. Then inequality
Ωτ ≥ Ωρ
τ n
ρ
f1(x) f2(x)
f1(x) ≥ f2(x)
.
f1(x) ≥
fi (x)
si (x) d=ef
x fi (t )
a fi (t )
dt.
m
i=2
m
i=2
(
8
)
(9)
(10)
Proof For n = 4, . . . , 7, the above inequality is verified numerically. Let t = ln1n . We
take the logarithm of the left part of (9) and use power series. This gives the following
inequalities
t
−n ln 1 − n
=
∞
i=1
t i
ni−1i
< t + t 2
∞
i=1
Here we observe that t < 12 for n ≥ 8, and replace the lefthand side of the first
inequality with geometric series in its righthand side, which in turn is bounded by
holds for any x ≥ a if it is valid for x = a.
Proof Note that fi (x) = fi (a) exp{si (x)}, where
Also, si (x) ≤ s1(x) for all i ≥ 2. Therefore,
which completes the proof.
Lemma 4 For any n ≥ 4,
1
1 − n ln n
−n
< 1 + 1/ln n + 1/ln2 n.
m
i=2
f1(x) ≥
fi (a) exp{s1(x)} ≥
fi (a) exp{si (x)} =
fi (x),
m
i=2
taking n = 8. We also note that for any n ≥ 8,
For any z < 12 we can also use inequality
To design a covering Cov(
1
), we also use another covering
1
ε = n ln n ,
λ = 1 +
ρ = 1 − ε,
with smaller caps C(ε, u). In our first step, we randomly choose N points y ∈Srn,
Consider the set {C(ρ , y)} of N caps. In the second step, we take all centers u ∈
Cov(ε) that are left uncovered by the set {C(ρ , y)} and form the extended set
This set {x} covers the entire set Cov(ε) and therefore forms a unit covering, if the
caps C(ρ , x) are expanded to the caps C(1, x),
Lemma 5 For any n ≥ 8, covering {x} has density
n
Cov(
1
) : Sr ⊆
C(1, x).
x∈{x}
ϑ∗ ≤
Proof Any point u is covered by some cap C(ρ , y) with probability Ωρ . Our goal is
to estimate the expected number N of centers u left uncovered after N trials. Here
N = (1 − Ωρ )N · Cov(ε).
Note that 1 − Ωρ ≤ e−Ωρ (1+Ωρ/2). Also, according to estimate (15), N Ωρ > λn ln n −
Ωρ . Thus,
(1 − Ωρ )N ≤ e−(λn ln n−Ωρ )(1+Ωρ/2) ≤ e−λn ln n.
Here the second inequality holds since Ωρ < Ω1 < 1/2, whereas λn ln n > 16.
Next, we estimate the covering size Cov(ε). Let this covering have some
density ϑε. Then inequality (
8
) gives the estimates
(17)
(18)
(19)
(20)
(21)
Cov(ε) = ϑε/Ωε ≤ (n ln n)nϑε/Ω1,
N ≤ e−λn ln n(n ln n)nϑε/Ω1 ≤ ϑε/(n2Ω1).
According to our design, there exists a covering {x} with caps C(1, x) that has size at
most N + N . Next, we combine last inequality with (15), and estimate the covering
density of {x}:
ϑ1 = Ω1(N + N ) ≤ (λn ln n)Ω1/Ωρ + ϑε/n2.
Finally, we estimate Ω1/Ωρ using inequalities (
8
) and (9):
Ω1/Ωρ ≤ (1 − ε)−n ≤ 1 + 1/ln n + 1/ln2 n.
Thus, we can rewrite our estimate ϑ1 as
using notation
ϑ1 ≤ ϑ∗(1 − 1/n2) + ϑε/n2
ϑ∗ = λn ln n
1 + 1/ln n + 1/ln2 n
1 − 1/n2
.
For any given n, bound (20) only depends on the density ϑε. Now let us assume that
ϑ1 meets some known upper bound ϑ˜1 valid for all (or sufficiently large) radii r . For
example, we can use (
4
) or any weaker estimate. Next, we can change the scale in
Rn+1 and replace a covering Cov(
1
) of a sphere Srn/ε with the covering Cov(ε) of the
sphere Srn. This rescaling shows that we can replace ϑε in (20) with any known upper
bound ϑ˜1. In turn, inequality (20) shows that the bound ϑ˜1 will be further reduced
by this replacement as long as ϑ˜1 > ϑ∗. Thus, this iteration process yields the upper
bound ϑ1 ≤ ϑ∗.
Our final step is to verify that the upper bound ϑ∗ of (21) also satisfies
inequality (16). Here we use Lemma 3. Then straightforward calculations show that the
righthand side of (16) exceeds the righthand side of (21) for n = 8 and moderates it
for n ≥ 8, due to the bigger remaining term 3/ ln n in (16).
Remark More detailed arguments show that (16) holds for n ≥ 3, whereas for n ≥ 8,
ϑ∗ ≤
Covering design. In this section, we obtain a covering of the sphere Srn with
asymptotic density (n ln n)/2. The new design will use both the former covering Cov(ε)
(with slightly different parameters) and another covering Cov(μ) with a larger radius
μ that has asymptotic order of n−1/2. Namely, we use parameters
and proceed as follows.
A. Let a sphere Srn be covered with two different coverings Cov(μ) and Cov(ε):
,
Cov(μ) : Srn ⊆
Cov(ε) : Srn ⊆
z∈Cov(μ)
We assume that both coverings have the former density ϑ∗ of (16) or less.
B. Randomly choose N points y ∈ Srn and consider the corresponding spherical caps
C(ρ , y), where
C. Let C(μ, z¯) be any cap in Cov(μ) that contains at least one center u ∈ Cov(ε)
not covered by the ρcaps. We consider all such centers z¯ and form the joint
set {x} = {y} ∪ {z¯}. This set covers Cov(ε) with ρcaps and therefore forms the
required covering
Cov(
1
) : Srn ⊆
C(1, x).
x∈{x}
We now proceed with preliminary discussion, which outlines the main steps of
the proof.
(22)
(23)
Outline of the proof. Let us first assume that we keep the design of Section 2
but apply it to the new covering Cov(μ) instead of Cov(ε). This will require taking
ρ = 1 − μ to cover the centers of the caps C(μ, z) and then expanding ρ to 1 to
cover the whole μcaps. Contrary to our former choice of ρ = 1 − ε in (19), this
expansion will exponentially increase the covering density. Namely, straightforward
calculations show that
Ω1/Ω1−ε → 1,
Ω1/Ω1−μ = exp{n1/2},
To circumvent this problem, we keep ρ = 1 − ε in (22) but change our design as
follows.
1. Given any cap C(μ, z), we say that a cap C(ρ , y) is d close if y falls within
distance d < ρ to z. In our proof, we refine the selection of the caps C(ρ , y) and
count only d close caps, instead of the ρclose caps considered in Section 2. It is
easy to verify that distance d of (22) is so close to ρ that
(24)
(25)
(26)
Ωρ /Ωd → 1,
For this reason, counting only d close caps instead of the former ρclose caps will
carry no overhead to the covering size (23).
2. On the other hand, we will show in Lemma 6 that the μcap becomes almost
completely covered by a cap C(ρ , y) when the latter becomes d close instead of
ρclose. Namely, only a vanishing fraction ω ≈ exp(− 32 ln2 n) of a μcap is left
uncovered in this case.
3. We shall also use the fact that a typical μcap is covered by multiple d close caps.
According to (23), the average number Ωd N of these caps has the exact order of
λn ln n:
λn ln n − Ωd < Ωd N ≤ λn ln n.
In our proof, we first define insufficiently covered μcaps. Namely, we call a cap
C(μ, z ) nonsaturated if it has only s or fewer d close caps, where
s = n/q ,
q = 3 ln ln n.
This choice of s will achieve two goals.
4. We prove in Lemma 7 that nonsaturated caps typically form a very small fraction
of order exp[−λn ln n] among all μcaps. On the other hand, it is easy to see that
the quantity
Cov(μ) ≤ ϑ∗/Ωμ
exceeds N by the exponential factor Ωd /Ωμ ∼ exp[βn ln n] or less. Then our
choice of λ and β in (22) gives the expected number N = o(N ) of nonsaturated
caps. Thus, nonsaturated caps typically form a vanishing fraction of not only
μcaps but also ρcaps.
5. Next, we proceed with saturated μcaps and count all those centers u ∈ Cov(ε)
that are left uncovered by random ρcaps. All caps C(μ, z ) that contain
uncovered centers u are called porous. For a given s, we show in Lemma 8 that
the set {u } forms a very small portion of Cov(ε), with expected fraction of
ωs+1 ∼ exp[− 23 n ln n]. Note that the quantity
Cov(ε) ≤ ϑ∗/Ωε
exceeds N by the exponential factor Ωd /Ωε ∼ exp[n ln n] or less. Therefore, the
expected size of {u } is N = o(N ).
6. Finally, the centers of all nonsaturated and porous caps are combined into the set
z¯ ={z , z }. Then the set {x} = {y, z¯} completely covers the set Cov(ε) with the
caps C(ρ , x). Therefore, {x} also covers Srn with unit caps.
Main proofs. To prove Theorem 1, we first observe (by numerical comparison) that
the existing bound (16) is tighter for n ≤ 100 than bound (
5
) of Theorem 1 or even its
refined version (39) considered below. Thus, Theorem 1 holds for n ≤ 100. For this
reason, we shall only consider dimensions n ≥ 100. In the end of the proof, we also
address the asymptotic case n → ∞, wherein the calculations are more
straightforward. The proof is based on three lemmas.
Consider two caps C(μ, Z) and C(ρ , Y) with centers Y and Z, which are d close.
These caps are represented in Fig. 1. Here the origin O is the center of Srn.
Lemma 6 For any cap C(μ, Z), a randomly chosen d close cap C(ρ , Y) fails to
cover any given point x of C(μ, Z) with probability p(x) ≤ ω, where
1
ω = 4 ln n
1 3
≤ 4 ln n exp − 2 ln2 n .
(27)
Proof The caps C(μ, Z) and C(ρ , Y) have bases PQRSA and PMRTB, which form
the balls Bμn (A) and Bρn(B). Below we consider the boundary PQRS of C(μ, Z),
which forms the sphere Sn−1(A). The bigger cap C(ρ , Y) covers this boundary, with
μ
the exception of the cap PQR centered at Q. We first consider the case, when x is a
boundary point and belongs to PQRS. Then the probability p(x) is the fraction Ω of
the entire boundary contained in uncovered cap PQR. We first estimate the halfangle
α = PAQ formed by the cap PQR.
Let d(H, G) denote the distance between any two points H and G. Also, let σ (H)
be the distance from a point H to the line OBY that connects the origin O with the
center B of the bigger base Bρn(B) and with the center Y of the cap C(ρ , Y). We use
inequalities
σ (A) ≤ σ (Z) ≤ d(Z, Y) ≤d.
On the other hand, consider the base PNR of the uncovered cap PQR. Here N denotes
the center of this base. Then both lines AN and BN are orthogonal to this base. Also,
d(B, P) = ρ, and d(N, P) ≤ d(A, P) = μ. Thus,
d(B, N) =
d2(B, P) − d2(N, P) ≥
ρ2 − μ2 ≥ ρ − μ2.
The latter inequality follows from the trivial inequality 2ρ − 1 ≥ μ2. Finally, the
center line OBY is orthogonal to the entire base PMRTB and its line BN. Then
d(B, N) = σ (N) and
d(A, N) ≥ σ (N) − σ (A) ≥ ρ − μ2 − d ≥ ε.
Now we consider the right triangle ANP and deduce that
(28)
(29)
(30)
cos α = d(A, N)/d(A, P) ≥ ε/μ =
3
n
ln n.
(Note that the latter expression exceeds 1 for n < 42, which simply implies that a
dclose cap entirely covers μcap for small n.) Given α, we can estimate the
fraction Ω of the boundary sphere Sn−1(A) contained in the uncovered cap PQR. This
μ
fraction can be defined as (see [4])
Ω < {2π(n − 1)}−1/2 sinn−1 α < 6π
cos α
n
n − 1 −1/2 sinn−1 α
ln n
<
sinn−1 α
4 ln n
.
approximation 1 − x ≤ e−x−x2/2. This gives inequalities
1 ) > 16. Now we use (30) and
The last inequality follows from the fact that 6π(1 − n
sinn−1 α ≤
3
1 − n ln2 n
≤ exp
− 32 ln2 n − 49n ln4 n
Finally, consider the second case, when a point x does not belong to the boundary
PQRS. Therefore, x is taken from a smaller cap C(μ , Z) ⊂ C(μ, Z) with the same
center Z and a halfchord μ < μ. Similarly to the first case, we define the boundary
Sn−1(A ) of C(μ , Z), where A is some center on the line AZ. Again, p(x) is the
μ
fraction Ω of this boundary left uncovered by the bigger cap C(ρ , Y). To obtain the
upper bound on Ω , we only need to replace μ with a smaller μ in (30). This gives
the smaller angle α ≤ arccos(ε/μ ), which is reduced to 0 if ε ≥ μ . In particular,
the center Z of the cap is always covered by any d close cap. Thus, we see that any
internal layer Sn−1(A ) of the cap C(μ, Z) has a smaller uncovered fraction Ω ≤ Ω .
μ
This gives the required condition
p(x) ≤ Ω ≤ Ω < ω
for any point x and proves our lemma.
Remark Our choice of d in (22) is central to the above proof, and even a marginal
increase in d will completely change our setting. Namely, it can be proven that about
half the base of the μcap is uncovered if a ρcap is (d + ε)close.
Our next goal is to estimate the expected number N of nonsaturated caps
C(μ, z ) left after N trials.
Lemma 7 For n ≥ 100, the expected number N of nonsaturated caps C(μ, z ) is
N < 2−n/4N .
(31)
Proof Given any center z, a randomly chosen center y is d close to z with the
probability Ωd . Then the probability to obtain s or fewer such caps is
P =
s
i=0
N
i
Ωdi (1 − Ωd )N−i .
Note that for any i ≤ s,
(1 − Ωd )N−i ≤ exp{−(Ωd + Ωd2/2)(N − i)}.
Now we use inequality Ωd ≤ 1/2 and the lower bound N Ωd > λn ln n − Ωd of (25).
Then
(Ωd + Ωd2/2)(N − i)
Ωd
= N Ωd 1 + 2
Ωd
− iΩd 1 + 2
≥ (λn ln n − Ωd ) + Ω2d [λn ln n − Ωd − s(2 + Ωd )] ≥ λn ln n.
Here last inequality follows from the fact that s(2 + Ωd ) ≤ 5n/(6 ln ln n) and
therefore
Then
λn ln n − Ωd − s(2 + Ωd ) ≥ 2.
P ≤ e−λn ln n
s (Ωd N )i
i=0
i!
≤ e−λn ln n
.
i=0
i!
(32)
Note that consecutive summands in (32) differ at least (λn ln n)/s ≥ λq ln n times.
Therefore
≤ (eλq ln n)n/q .
Here the sum of the geometric series (λq ln n)−i was first bounded from above by
cn < c100 < 2. Then we used the Sterling formula in the form s! > (2π s)1/2(s/e)s and
removed (for simplicity) the vanishing term 2(2π s)−1/2. Finally, the last inequality
follows from the fact that its lefthand side increases in s for any s < λn ln n.
Summarizing, these substitutions give
where
P ≤ exp{nhn − λn ln n},
hn =
Next, we recalculate ϑ∗ of (16) using parameter λ of (22) and the lower bound of (25).
It is easy to verify that for any n,
Now we estimate the size of Cov(μ) using (
8
) and (22) as follows
Cov(μ) ≤ ϑ∗/Ωμ ≤ 2N Ωd /Ωμ ≤ 2N μn ≤ 2N exp{n[β ln n + ln(12)/2]}. (33)
Thus, the expected number of nonsaturated caps is
N ≤ Cov(μ)P ≤ 2N exp{n[hn − (λ − β) ln n + ln(12)/2]}
≤ 2N exp{n[hn − (5 − ln 12)/2]}.
(34)
Now we see that the quantity Ψn in the brackets of (34) consists of the declining
positive function hn and the negative constant. Thus, Ψn is a declining function of n.
Direct calculation shows that Ψ100 < −0.257. Therefore estimate (31) holds.
Consider now the saturated caps C(μ, z) and the centers u ∈ Cov(ε) inside them.
Lemma 8 For any n ≥ 100, the number of centers u ∈ Cov(ε) left uncovered in all
saturated caps C(μ, z) has expectation
N
< 2−n/2N .
Proof We first estimate the total number Cov(ε) of centers u. Similarly to (33),
Cov(ε) ≤ ϑ∗/Ωε ≤ 2N Ωd /Ωε ≤ 2N (2n ln n)n.
Any cap C(μ, z) intersects with at least s + 1 randomly chosen caps C(ρ , y).
According to Lemma 6, any single ρcap fails to cover any given point x ∈ C(μ, z) with
probability ω or less. Therefore any point u ∈ C(μ, z) is not covered with
probability ωs+1 or less. Note that ωs+1 < ωn/q where q = 3 ln ln n. Then we use the upper
bound (27) for ω and deduce that
where
N
≤ Cov(ε) · ωn/q ≤ 2N exp{−nΦn},
Φn =
ln2 n ln 4
2 ln ln n + 3 ln ln n
1
− ln n − ln ln n − ln 2 + 3 .
(35)
(36)
Direct calculation shows that Φ100 > 0.71. It is also easy to see that the first term of
Φn (in parentheses) moderates both terms ln n and ln ln n. Thus, Φn > 0.71 for all
n ≥ 100, and the lemma is proved.
Proof of Theorem 1 Consider any cap C(μ, z¯) that contains at least one
uncovered center u ∈Cov(ε). Such a cap is either nonsaturated or porous and therefore
{¯} = {z } ∪ {z }. (Equivalently, we can directly consider the set {z } ∪ {u }.) Then,
z
according to Lemmas 7 and 8, {z¯} has expected size
N¯ ≤ N + N
< 21−n/4N .
Thus, there exist N randomly chosen centers y that leave at most 21−n/4N centers z¯.
The extended set {x} = {y} ∪ {z¯} forms a unit covering of Srn. This covering has
density
ϑ ≤ Ω1(N + N¯ ) ≤ Ω1N (1 + 21−n/4) ≤ λn ln n(1 + 21−n/4)Ω1/Ωd .
Similarly to inequality (9), we now verify that for n ≥ 100,
Ω1/Ωd ≤
1
1 − n ln n − n−2β
−n
Indeed, replace parameter t = ln1n used in the proof of Lemma 4 with the new value
1 1 1
t = ln n + n1−2β = ln n + ln4 n .
This new t also satisfies inequality (11) for n ≥ 100, which in turn proves
inequality (37). As a secondary remark, note that (37) also proves asymptotic equality (24).
Finally, we take λ of (9) and combine the last inequalities for ϑ and Ω1/Ωd as
follows
5
+ 2 ln n
1 1
1 + ln n + ln2 n
(1 + 21−n/4)
5
+ ln n .
Here we again used Lemma 3. Namely, we verify numerically that the last expression
exceeds the previous one at n = 100 and moderates it for larger n, due to its bigger
remaining term 5/ ln n. This completes the nonasymptotic case n ≥ 3.
To complete the proof of Theorem 1, we now present similar estimates for
n → ∞. We take any constant b > 3/2 and redefine the parameters in (22) and (26)
as follows:
1 ln ln n
β = 2 + b ln n
3 1
λ = β + 4 ln n ,
,
q = ln2 ln n.
First, bounds (30) and (27) can be replaced with
Second, bound (34) can be rewritten as
Note that the first term ln(eλq ln n)/q vanishes for n → ∞, and N declines
exponentially in n. Thus, Lemma 7 holds. Finally, bound (35) is replaced with
N ≤ 2N exp n
.
,
which vanishes (faster than exponent in n) for any given b > 3/2. Thus, Lemma 8
also holds. Then we proceed similarly to (37). In this case, for sufficiently large n,
we obtain the density
which proves (
6
). This completes the proof of Theorem 1.
For finite n, we can slightly refine bounds (28–29), and also use exact expression
for ω in (27). However, these refinements only marginally improve the main bound
(
5
), which is replaced with
ϑ 1
n ln n ≤ 2 +
+
4 ln ln n
ln n
.
Finally, note that Theorem 1 directly leads to Corollary 2. Indeed, here we can use
the well known fact
ϑ (Rn−1) = rl→im∞ ϑ (Srn)
(see a sketch of the proof in [6] or Theorem
detailed for packings of Rn).
.1 in [7], where a similar proof is
Concluding remarks In summary, we prove in this paper that the classic Rogers
bound (
3
) on covering density of a sphere Srn or the Euclidean space Rn can be
reduced about two times for large dimensions n. The main open problem is to
further reduce the gap between this bound and its lower counterpart (
2
), which is linear
in n. In this regard, note that our design holds if we employ any constant
parameter β > 1/2 introduced in (22). However, it can be verified that choosing a smaller
constant β < 1/2 will first increase parameter μ = n−β and then reduce parameter d
of (22) to the order of 1 − μ2/2. The latter reduction will exponentially increase the
covering size, by a factor of
Ωρ /Ωd → exp{1 − 2β}.
Therefore, our conjecture is that a completely new design is needed for further
asymptotic reductions.
Another important problem is to extend the above results to the balls Brn of an
arbitrary radius r . Our conjecture is that ϑ (Brn) ≤ ( 12 + o(
1
))n ln n for any r and
Acknowledgements The author is grateful to an anonymous referee for many helpful comments and
remarks. This research was supported in part by NSF grants CCF0622242 and CCF0635339.
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