Asymptotic series and inequalities associated to some expressions involving the volume of the unit ball

0.11 MB·English

by  Cristinel Mortici

Asymptotic series and inequalities associated to
some expressions involving the volume of the
unit ball
Cristinel Mortici1,2
5
1 1Department of Mathematics, Valahia University of Tˆargovi¸ste,
0
Bd. Unirii 18, 130082Tˆargovi¸ste, Romania
2
2Academy of Romanian Scientists, Splaiul Independen¸tei 54, 050094

n
Bucharest, Romania
a
E-Mail: [email protected]
J

7

] Abstract
A
C The aim of this work is to expose some asymptotic series associated
. to some expressions involving thevolumeof then-dimensional unit ball.
h
All proofs and the methods used for improving the classical
t
a inequalities announced in the final part of the first section
m
are presented in an extended form in a paper submitted by
[ the author to a journal for publication.

1
v 2010 Mathematics Subject Classification. 34E05; 41A60; 33B15;26D15.
2
6
Key Words and Phrases. Gammafunction;Volumeoftheunitball;Asymp-
4
totic series; Approximations and estimates; Two-sided inequalities.
1
0
.
1 1 Introduction and Motivation
0
5
In the recent past, inequalities about the volume of the unit ball in Rn :
1
:
v πn/2
(1) Ω = (n N)
Xi n Γ n +1 ∈
2
r haveattracted(cid:0)theat(cid:1)tentionofmanyauthors. See,e.g., [2]-[17]. HereΓdenotes
a
theEuler’sgammafunctiondefinedforeveryrealnumberx>0,bytheformula:
Γ(x+1)= ∞txe tdt,
−
Z0
while N denotes the set of all positive integers and N =N 0 .
0
∪{ }
Our products improve the following classical results:
Chen and Lin [9] (a= e 1, b= 1):
• 2 − 3
n n
1 2πe 2 1 2πe 2
Ω < (n N);
n
π(n+a) n ≤ π(n+b) n ∈
(cid:18) (cid:19) (cid:18) (cid:19)
p p
1
Borgwardt[7] (a=0, b=1), Alzer [3] and Qiu and Vuorinen [17] (a= 1,
• 2
b= π 1):
2 −
n+a Ω n+b
n−1 (n N);
2π ≤ Ω ≤ 2π ∈
r n r
• Alzer [3] (α∗ = 34ππ√+62, β∗ =√2π):
α∗ Ωn < β∗ (n N);
√n ≤ Ω +Ω √n ∈
n 1 n+1
−
Chen and Lin [9] (a= π(1+π)2 1, b= 1 +4π):
• 2 − 2
2π Ω 2π
n < (n N);
n+a ≤ Ω +Ω n+b ∈
r n 1 n+1 r
−
Chen and Lin in [9] (λ=1, µ= 2ln2 lnπ ):
• 2ln3−3ln2
−
1 3 λ Ω2 1 3 µ
1+ < n 1+ (n N);
2n − 8n2 Ω Ω ≤ 2n − 8n2 ∈
(cid:18) (cid:19) n−1 n+1 (cid:18) (cid:19)
Anderson et al. [5] and Klain and Rota [12]:
•
Ω2 1
1< n <1+ (n N);
Ω Ω n ∈
n 1 n+1
−
Alzer [3] (α=2 log π, β = 1):
• − 2 2
1 α Ω2 1 β
1+ n < 1+ (n N);
n ≤ Ω Ω n ∈
(cid:18) (cid:19) n−1 n+1 (cid:18) (cid:19)
Merkle [13]:
•
1 21 Ω2
1+ < n (n N);
n+1 Ω Ω ∈
(cid:18) (cid:19) n−1 n+1
Chen and Lin [9] (α= 1, β = 2ln2 lnπ):
• 2 ln3−ln2
−
1 α Ω2 1 β
1+ < n 1+ (n N).
n+1 Ω Ω ≤ n+1 ∈
(cid:18) (cid:19) n−1 n+1 (cid:18) (cid:19)
2
2 Classical results and new achievements
2.1 Asymptotic series and estimates for Ω and Ω1/n
n n
Mortici [15, Rel. 17] established the following asymptotic series as n :
→∞
1 n+1 n 1 ln2π
lnΩ ln + ln(πe)
n
n ∼ − 2n 2 2 − 2n
1 1 8 8 1
+ + + .
− 6n2 − 45n4 315n6 − 105n8 128n10 ···
(cid:18) (cid:19)
The entire series is given below:
Theorem 1 The following asymptotic series holds true, as n :
→∞
1 n+1 n 1 ln2π ∞ 22j−2B2j
(2) lnΩ ln + ln(πe) .
n n ∼− 2n 2 2 − 2n − j(2j 1)n2j
j=1 −
X
(B are the Bernoulli numbers).
j
The following double inequality [15, Theorem 2] was presented:
1
(3) α(n)< lnΩ <β(n) (n N),
n
n ∈
where
n+1 n 1 ln2π
α(n) = ln + ln(πe) λ(n)
− 2n 2 2 − 2n −
n+1 n 1 ln2π
β(n) = ln + ln(πe) µ(n),
− 2n 2 2 − 2n −
with
1 1 8 8
µ(n) = +
6n2 − 45n4 315n6 − 105n8
1
λ(n) = µ(n)+ .
128n10
Theorem 2 The following inequality holds true:
α(n) β(n+1)>0 (n N).
− ∈
In consequence, the sequence Ω1/n decreases monotonically (to 0).
n
n 1
n o ≥
Theorem 3 The following double inequality holds true, for every integer n 3
≥
in the left-hand side and n 1 in the right-hand side:
≥
n n
1 2πe 2 1 2πe 2
(4) <Ω < ,
n
π(n+θ(n)) n π(n+ν(n)) n
(cid:18) (cid:19) (cid:18) (cid:19)
p p
3
where
1 1 31
θ(n) = +
3 18n − 810n2
139
ν(n) = θ(n) .
− 9720n3
NextweconstructanasymptoticseriesfortheratioΩ1/n/Ω1/(n+1),thenwe
n n+1
give some lower and upper bounds.
Theorem 4 The following asymptotic series holds true, as n :
→∞
1 1 ∞ ψj
(5) lnΩ lnΩ Ψ(n)
n n− n+1 n+1 ∼ − nj
j=1
X
where
n+1 n n+2 n+1 ln2π
Ψ(n)= ln + ln
− 2n 2 2n+2 2 − 2n(n+1)
and the coefficients ψ are given by:
j
k+2s 1
ψ = ( 1)k+1 − (t,k,s N)
2t+1 − k ∈
k+2s=2t+1 (cid:18) (cid:19)
X
ψ = 22t−2B2t + ( 1)k+1 k+2s−1 (t,k,s N),
2t t(2t 1) − k ∈
− k+2s=2t (cid:18) (cid:19)
X
with
v v(v 1) (v k+1)
= − ··· − (v R, k N ).
k k! ∈ ∈ 0
(cid:18) (cid:19)
We have:
1 1 1 1 26 11
lnΩ lnΩ Ψ(n) + + + .
n n− n+1 n+1 ∼ − 3n3 2n4 − 45n5 18n6 ···
Theorem 5 The following double inequality holds true:
1 1 1 1 1
Ψ(n) < lnΩ lnΩ <Ψ(n) + (n N).
− 3n3 n n− n+1 n+1 − 3n3 2n4 ∈
Ω
2.2 Asymptotic series and estimates for n−1
Ω
n
Theorem 6 The following asymptotic series holds true, as n :
→∞
(6) lnΩn−1 = 1ln n + ∞ µj,
Ω 2 2π nj
n
j=1
X
where
1 2j
µ =( 1)j B B (1) (j N)
j − j+1 2 − j+1 j(j+1) ∈
(cid:20) (cid:18) (cid:19) (cid:21)
(B are the Bernoulli polynomials).
j
4
In a concrete form, (6) can be written as:
Ω 1 n 1 1 1 17 31
n 1
ln − = ln + + + + .
Ω 2 2π 4n − 24n3 20n5 − 112n7 36n9 ···
n (cid:18) (cid:19)
Remarkthatonlyoddpowersofn 1 appearinthisseries. Thiscanbe justified
−
usingthefollowingrepresentationformulasoftheBernoullipolynomialsinterms
of the Bernoulli numbers:
1
B (1)=( 1)tB , B = 21 t 1 B (t N).
t t t − t
− 2 − ∈
(cid:18) (cid:19)
(cid:0) (cid:1)
See, e.g., [18]. As the Bernoulli numbers of odd order vanish, it results that
1
B =B (1)=0 (j 2N)
j+1 j+1
2 ∈
(cid:18) (cid:19)
and consequently, µ =0, whenever j is a positive even integer.
j
We present the following estimates:
Theorem 7 The following double inequality holds true:
Ω
(7) a(n)<ln n−1 <b(n) (n N),
Ω ∈
n
where
1 n 1 1 1 17
a(n) = ln + +
2 2π 4n − 24n3 20n5 − 112n7
31
b(n) = a(n)+ .
36n9
Further we deduce a new asymptotic series and one of the resulting double
inequality:
Theorem 8 The following asymptotic series holds true, as n :
→∞
Ωn 1 n ∞ cj
(8) − = ,
Ω 2π  nj
n r Xj=0 
where c0 =1 and  
1 j 1 2k
c = ( 1)k B B (1) c (j N).
j k+1 k+1 j k
j − 2 − k+1 − ∈
k=1 (cid:20) (cid:18) (cid:19) (cid:21)
X
By listing the first terms, we get:
Ω n 1 1 5 21 399 869
n 1
− = 1+ + + + + .
Ω 2π 4n 32n2 − 128n3 − 2048n4 8192n5 65536n6 ···
n r (cid:18) (cid:19)
Related to this asymptotic expansion, we prove the following estimates:
5
Theorem 9 Thefollowingdoubleinequalityholdstrue,foreveryintegern 12
≥
in the left-hand side and n 1 in the right-hand side:
≥
n Ω n
n 1
(9) c(n)< − < d(n),
2π Ω 2π
r n r
where
1 1 5 21 399
c(n) = 1+ + +
4n 32n2 − 128n3 − 2048n4 8192n5
869
d(n) = c(n)+ .
65536n6
Theorem 10 The following asymptotic formula holds true as n :
→∞
(10) Ωn−1 = n+ 12 + ∞ sj,
Ω v 2π nj
n u j=1
u X
t
where
j+1
1
s = c c (j N).
j k j+1 k
2π − ∈
k=0
X
The first terms are indicated below:
Ω n+ 1 1 1 5
n−1 = 2 +
Ω 2π 16πn − 32πn2 − 256πn3
n (cid:18)
1
23 53 593 5165 2
+ + + .
512πn4 2048πn5 − 4096πn6 − 65536πn7 ···
(cid:19)
By truncation of this series, increasingly accurate under- and upper- approxi-
mationsfortheratio Ωn−1 areobtained. Asanexample,weshowthefollowing:
Ωn
Theorem 11 Thefollowingdoubleinequalityholdstrue,foreveryintegern 1
≥
in the left-hand side and n 2 in the right-hand side:
≥
n+ 1 1 1 5 Ω n+ 1 1 1
2 + < n−1 < 2 + .
s 2π 16πn − 32πn2 − 256πn3 Ω s 2π 16πn − 32πn2
n
Theorem 12 The following double inequality holds true:
(11)
2π Ω 2π
+ε (n)< n < +ε (n) (n N),
sn+4π+ 21 1 Ωn−1+Ωn+1 sn+4π+ 21 2 ∈
where
1π 4π2+8π3
ε (n) = 4 −
1 − n3
3π 7π2 12π3+64π4
ε (n) = ε (n)+ 8 − − .
2 1 n4
6
Ω2
2.3 Asymptotic series and estimates for n
Ω Ω
n−1 n+1
We start this section by establishing new asymptotic expansions for the ratio
Ω2n and some associated inequalities.
Ωn−1Ωn+1
Theorem 13 The following asymptotic series holds true, as n :
→∞
(12) ln Ω2n = ∞ λj,
Ω Ω nj
n 1 n+1
− Xj=1
where
1 3 2j
λ =( 1)j 2B (1) B B (j N).
j j+1 j+1 j+1
− − 2 − 2 j(j+1) ∈
(cid:26) (cid:18) (cid:19) (cid:18) (cid:19)(cid:27)
As the first terms in this series are
Ω2 1 1 5 1 1 1
ln n = + + + ,
Ω Ω 2n − 2n2 12n3 − 4n4 10n5 − 6n6 ···
n 1 n+1
−
we are entitled to present the following estimates:
Theorem 14 The following double inequality holds true:
Ω2
(13) p(n)<ln n <q(n) (n N),
Ω Ω ∈
n 1 n+1
−
where
1 1 5 1 1 1
p(n) = + +
2n − 2n2 12n3 − 4n4 10n5 − 6n6
1
q(n) = p(n)+ .
6n6
Theorem 15 The following asymptotic series holds true, as n :
→∞
Ω2n = ∞ dj,
Ω Ω nj
n 1 n+1
− Xj=0
where d =1 and d s, j N, are defined by the recursive relation:
0 ′j ∈
1 j 1 3 2k
d = ( 1)k 2B (1) B B d .
j k+1 k+1 k+1 j k
j − − 2 − 2 k+1 −
k=1 (cid:20) (cid:18) (cid:19) (cid:18) (cid:19)(cid:21)
X
More exactly, we have:
Ω2 1 3 3 3
(14) n =1+ + + +
Ω Ω 2n − 8n2 16n3 128n4 ···
n 1 n+1
−
We propose the following estimates associated to the series (14):
7
Theorem 16 Thefollowingdoubleinequalityholdstrue,foreveryintegern 6
≥
in the left-hand side and n 1 in the right-hand side:
≥
Ω2
(15) r(n)< n <s(n),
Ω Ω
n 1 n+1
−
where
1 3 3
r(n) = 1+ +
2n − 8n2 16n3
3
s(n) = r(n)+ .
128n4
Theorem 17 The following asymptotic formula holds true as n :
→∞
(16) Ω2n 1+ 1 P∞j=0 ntjj ,
Ω Ω ∼ n
n−1 n+1 (cid:18) (cid:19)
where t = 1 and t s, j N, are the solution of the infinite system:
0 2 ′j ∈
m
t
( 1)j+1 m−j =λm (m N).
− j ∈
j=1
X
Theorem 18 Thefollowingdoubleinequalityholdstrue,foreveryintegern 5
≥
in the left-hand side and n 1 in the right-hand side:
≥
(17) 1+ 1 12−41n+8n12 < Ω2n < 1+ 1 12−41n+8n12+481n3 .
n Ω Ω n
(cid:18) (cid:19) n−1 n+1 (cid:18) (cid:19)
Here we only list the following results:
Ω2n 1+ 1 21+41n−8n32+482n33−321n54+···
Ω Ω ∼ n+1
n−1 n+1 (cid:18) (cid:19)
and
Ω2n 1+ 1 21+4(n1+1)−8(n+11)2−48(n1+1)3+32(n3+1)4+···.
Ω Ω ∼ n+1
n−1 n+1 (cid:18) (cid:19)
Acknowledgements. 1. This work was supported by a Grant of the Ro-
manian National Authority for Scientific Research CNCS-UEFISCDI, no. PN-
II-ID-PCE-2011-3-0087. The computations made in this paper were performed
using the Maple software for symbolic computation.
2. All proofs and the methods used for improving the classical
inequalities announced in the final part of the first section are pre-
sented in an extended form in a paper submitted by the author to a
journal for publication.
8
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