Two monotonic functions involving gamma function and volume of unit ball

0.1 MB·English

by  Feng Qi

TWO MONOTONIC FUNCTIONS INVOLVING GAMMA
FUNCTION AND VOLUME OF UNIT BALL
0
1 FENGQIANDBAI-NIGUO
0
2
Abstract. In present paper, weprove the monotonicity of two functions in-

n volvingthegammafunctionΓ(x)andrelatingtothen-dimensionalvolumeof
a theunitballBn inRn.
J

0
1

] 1. Introduction
A
It is well-known that the classical Euler’s gamma function may be defined by
C
∞
.
h Γ(x)= tx−1e−tdt (1)
t Z
a 0
m for x > 0 and that the n-dimensional volume of the unit ball Bn in Rn is denoted
[ by

πn/2
1 Ω = , n∈N. (2)
n
v Γ(1+n/2)
6
For x≥0, define the function
9
4 lnΓ(x+1)
1 , x6=0,1,
ln(x2+1)−ln(x+1)
. F(x)= (3)
1 γ, x=0,
0
2(1−γ), x=1.
0
1 
Recently, the function F(x) was proved in [7] to be strictly increasing on [0,1].
:
v Moreover,as a remark in [7], the function F(x) was also conjectured to be strictly
i
X increasing on (1,∞).
The firstaimofthis paperis to verifyabove-mentionedconjecturewhichcanbe
r
a recited as the following theorem.
Theorem 1. The function F(x) defined by (3) is strictly increasing on [0,∞).
The second aim of this paper is to derive the monotonicity of the sequence
Ω1/[ln(n2/4+1)−ln(n/2+1)] (4)
n
for n∈N by establishing the following general conclusion.
2000 Mathematics Subject Classification. Primary33B15;Secondary 26A48.
Key words and phrases. monotonic function, gamma function, volume, unit ball, Descartes’
SignRule.
TheauthorsweresupportedinpartbytheScienceFoundationofTianjinPolytechnicUniver-
sity.
ThispaperwastypesetusingAMS-LATEX.
1
2 F.QIANDB.-N.GUO
Theorem 2. The function
πx 1/[ln(x2+1)−ln(x+1)]
G(x)= (5)
(cid:20)Γ(x+1)(cid:21)
isstrictlydecreasingon(1,∞). Consequently,thesequence (4)isstrictlydecreasing
for n≥3.
2. Two lemmas
In order to prove Theorem 1, we need the following lemma which can be found
in [13, pp. 9–10, Lemma 2.9], [14, p. 71, Lemma 1] or closely-related references
therein.
Lemma 1. Let f and g be continuouson [a,b] anddifferentiable on (a,b)suchthat
g′(x) 6= 0 on (a,b). If f′(x) is increasing (or decreasing) on (a,b), then so are the
g′(x)
functions f(x)−f(b) and f(x)−f(a) on (a,b).
g(x)−g(b) g(x)−g(a)
We also need the following elementary conclusions.
Lemma 2. The functions
p (x)=x3+3x2−x−1,
1
p (x)=x3+3x2−x−1,
2
p (x)=3x4+8x3+2x2−1,
3
p (x)=x5+3x4+2x3+2x2+x−1,
4
p (x)=x5+5x4+6x3−3x−1,
5
p (x)=120(15−4lnπ)x3+240(20−7lnπ)x2
6
+48(59−32lnπ)x+72(3−4lnπ)
are positive on (1,∞).
Proof. An easy calculation shows that
15−4lnπ =10.421··· , 20−7lnπ =11.986··· ,
59−32lnπ =22.368··· , 3−4lnπ =−1.578··· .
Then Descartes’ Sign Rule tells us that the function p (x) for 1 ≤ i ≤ 6 have just
i
one possible positive root. Since
p (0)=−1, p (0)=−1, p (0)=−1, p (0)=−1, p (0)=−1,
1 2 3 4 5
p (1)=2, p (1)=2, p (1)=12, p (1)=8, p (1)=8,
1 2 3 4 5
and
p (0)=−72(4lnπ−3)
6
=−113.68··· ,
p (1)=−[120(4lnπ−15)+72(4lnπ−3)
6
+240(7lnπ−20)+48(32lnπ−59)]
=5087.39··· ,
these functions are positive on [1,∞). (cid:3)
TWO MONOTONIC FUNCTIONS INVOLVING GAMMA FUNCTION 3
3. Proof of Theorem 1
The monotonicity of the function F(x) on [0,1] was proved in [7].
For x∈[1,∞), it is easy to see that
lnΓ(x+1) lnΓ(x+1)−lnΓ(1+1) f(x)−f(1)
= = , (6)
ln(x2+1)−ln(x+1) lnx2+1 −ln12+1 g(x)−g(1)
x+1 1+1
where
x2+1
f(x)=lnΓ(x+1) and g(x)=ln
x+1
on [1,∞). Easy computation and simplification yield
f′(x) (x+1) x2+1 ψ′(x+1)
=
g′(x) x(cid:0)2+2x(cid:1)−1
and
′
d f (x) q(x)
= ,
dx(cid:20)g′(x)(cid:21) (x2+2x−1)2
where
q(x)= x4+4x3−2x2−4x−3 ψ(x+1)
+(cid:0) (x+1) x2+1 x2+2x(cid:1)−1 ψ′(x+1)
(cid:0) (cid:1)(cid:0) (cid:1)
and
q′(x)=4 x3+3x2−x−1 ψ(x+1)
+(cid:0) x2+2x−1 2 3(cid:1)x2+2x+1 ψ′(x+1)
+((cid:0)x+1) x2+(cid:1)1(cid:2) ψ(cid:0)′′(x+1) . (cid:1)
(cid:0) (cid:1) (cid:3)
By virtue of
1 1
lnx− <ψ(x)<lnx− (7)
x 2x
and
(k−1)! k! (k−1)! k!
+ < ψ(k)(x) < + (8)
xk 2xk+1 xk xk+1
(cid:12) (cid:12)
for x > 0 and n ∈ N, see [3, p. 131],(cid:12) [4, Lem(cid:12)ma 3], [9, p. 79], [11, Lemma 3] or
related texts in [6, 10], and by using of the positivity of p (x) in Lemma 2 and
1
2t t(2+t)
≤ln(1+t)≤ (9)
2+t 2(1+t)
on (0,∞), see [5] or [12, p. 245, Remark 1], we obtain
1
q′(x)>4 x3+3x2−x−1 ln(x+1)−
(cid:20) x+1(cid:21)
(cid:0) (cid:1)
1 1
+ x2+2x−1 2 3x2+2x+1 +
(cid:26) (cid:20)x+1 2(x+1)2(cid:21)
(cid:0) (cid:1) (cid:0) (cid:1)
1 2
−(x+1) x2+1 +
(cid:20)(x+1)2 (x+1)3(cid:21)(cid:27)
(cid:0) (cid:1)
4+x−4x2+6x3+16x4+5x5
=4 x3+3x2−x−1 ln(x+1)+
(cid:20) 4(x+1)2(x3+3x2−x−1) (cid:21)
(cid:0) (cid:1)
4 F.QIANDB.-N.GUO
2x 4+x−4x2+6x3+16x4+5x5
≥4 x3+3x2−x−1 +
(cid:20)x+2 4(x+1)2(x3+3x2−x−1) (cid:21)
(cid:0) (cid:1)
13x6+66x5+86x4+8x3−31x2−2x+8
=
(x+1)2(x+2)
on [1,∞). Because
13x6+66x5+86x4+8x3−31x2−2x+8=13(x−1)6+144(x−1)5
+611(x−1)4+1272(x−1)3+1364(x−1)2+712(x−1)+148>0
on[1,∞),itfollowsthatq′(x)>0,andsothefunctionq(x)isincreasing,on[1,∞).
Due to
π2
q(1)=8 −1 −4(1−γ)=3.468··· ,
(cid:18) 6 (cid:19)
the function q(x) is positive on [1,∞). Therefore,
′
d f (x)
>0
dx(cid:20)g′(x)(cid:21)
on [1,∞), which means that the function f′(x) is strictly increasing on [1,∞).
g′(x)
Furthermore, from Lemma 1 and the equation (6), it follows that the function (3)
is strictly increasing on [1,∞). The proof of Theorem 1 is complete.
4. Proof of Theorem 2
Taking the logarithm of the function G(x) and differentiating yield
(lnπ)x−lnΓ(x+1)
lnG(x)=
ln(x2+1)−ln(x+1)
and
g(x)
′
[lnG(x)] = ,
[ln(x+1)−ln(x2+1)]2
where
x2+1 x2+2x−1
g(x)=[lnπ−ψ(x+1)]ln − [(lnπ)x−lnΓ(x+1)]
x+1 (x+1)(x2+1)
x2+2x−1 (x+1)(x2+1)[lnπ−ψ(x+1)] x2+1
= ln
(x+1)(x2+1)(cid:26) x2+2x−1 x+1
−(lnπ)x+lnΓ(x+1)
(cid:27)
x2+2x−1
, h(x),
(x+1)(x2+1)
with
ln(x+1)−ln x2+1
h′(x)= x4+4x3−2x2−4x−3 [ψ(x+1)−lnπ]
(x2+2x−(cid:0)1)2 (cid:1)
(cid:8)(cid:0) (cid:1)
+(x+1) x2+1 x2+2x−1 ψ′(x+1)
,h (x) (cid:0) (cid:1)(cid:0) (cid:1) (cid:9)
1
and
h′(x)=4 x3+3x2−x−1 ψ(x+1)
1
(cid:0) (cid:1)
TWO MONOTONIC FUNCTIONS INVOLVING GAMMA FUNCTION 5
+2 3x4+8x3+2x2−1 ψ′(x+1)
+ x(cid:0)5+3x4+2x3+2x2(cid:1)+x−1 ψ′′(x+1)
−4(cid:0) x3+3x2−x−1 lnπ. (cid:1)
(cid:0) (cid:1)
Utilizing Lemma 2 and employing (7), (8) for k =1 and (9) give
1
h′(x)>4 x3+3x2−x−1 ln(x+1)−
1 (cid:20) x+1(cid:21)
(cid:0) (cid:1)
1 1
+2 3x4+8x3+2x2−1 +
(cid:20)x+1 2(x+1)2(cid:21)
(cid:0) (cid:1)
1 2
− x5+3x4+2x3+2x2+x−1 +
(cid:20)(x+1)2 (x+1)3(cid:21)
(cid:0) (cid:1)
−4 x3+3x2−x−1 lnπ
(cid:0)1 (cid:1)
= (7−4lnπ)x5+(26−20lnπ)x4−6(4lnπ−3)x3
(x+1)2
(cid:2)
+(11+12lnπ)x−2+4ln(π)
+4 x5+5x4+6x3−3x−1 ln(x+1)
(cid:0)1 (cid:1) (cid:3)
> (7−4lnπ)x5+(26−20lnπ)x4−6(4lnπ−3)x3
(x+1)2(cid:20)
+(11+12lnπ)x−2+4ln(π)
2x
+4 x5+5x4+6x3−3x−1
x+2(cid:21)
(cid:0) (cid:1)
1
=− (4lnπ−15)x6+4(7lnπ−20)x5
(x+1)2(x+2)
(cid:2)
+2(32lnπ−59)x4+12(4lnπ−3)x3
+(13−12lnπ)x2−4(3+7lnπ)x+4−8lnπ
1 (cid:3)
,− h (x)
(x+1)2(x+2) 2
and
h′(x)=6(4lnπ−15)x5+20(7lnπ−20)x4+8(32lnπ−59)x3
2
+36(4lnπ−3)x2+2(13−12lnπ)x−4(3+7lnπ),
h′′(x)=30(4lnπ−15)x4+80(7lnπ−20)x3
2
+24(32lnπ−59)x2+72(4lnπ−3)x+2(13−12lnπ),
h′′′(x)=120(4lnπ−15)x3+240(7lnπ−20)x2
2
+48(32lnπ−59)x+72(4lnπ−3).
ByLemma 2,itfollowsthath′′(x) is negativeon[1,∞),so h′′(x) isdecreasingand
3 2
h′(x) is concave on [1,∞). Since
2
′′
h (1)=2(13−12lnπ)+30(4lnπ−15)+72(4lnπ−3)
2
+80(7lnπ−20)+24(32lnπ−59)
=−1696.22···,
6 F.QIANDB.-N.GUO
′′ ′
the derivativeh (x) is negative,andthus h (x) is concaveand h (x) is decreasing,
2 2 2
on [1,∞). From
′
h (1)=2(13−12lnπ)+6(4lnπ−15)+36(4lnπ−3)
2
+20(7lnπ−20)−4(3+7lnπ)+8(32lnπ−59)
=−469.89···,
′
itisimmediatelydeducedthath (x)isnegativeandthefunctionh (x)isdecreasing
2 2
on [1,∞). Due to
h (1)=2−16lnπ+12(4lnπ−3)+4(7lnπ−20)
2
−4(3+7lnπ)+2(32lnπ−59)
=−134.10···,
we derive that the function h (x) is negative on (1,∞), so h′(x) > 0 and h (x) is
2 1 1
increasing on (1,∞). From
π2
h (1)=8 −1 −4(1−γ)+4lnπ =8.04··· ,
1
(cid:18) 6 (cid:19)
itfollowsthatthefunctionh (x)ispositiveon(1,∞),andthusthederivativeh′(x)
1
is negative and h(x) is decreasing on (1,∞). Since h(1) = −lnπ = −1.14···, it
follows that the function h(x) is negative, that the function g(x) is negative, and
that the derivative [lnG(x)]′ is negative on (1,∞). As a result, the function G(x)
is strictly decreasing on (1,∞).
It is clear that the sequence (4) equals G n , so the sequence (4) decreases for
2
n>2. The proof of Theorem 2 is complete.(cid:0) (cid:1)
5. Remarks
Remark 1. In [2, Lemma 2.40], it was proved that the sequence Ω1/n decreases
n
strictly to 0 as n→∞, that the series ∞ Ω1/lnn is convergent,and that
n=2 n
lim Ω1/(nPlnn) =e−1/2. (10)
n→∞ n
1/(nlnn)
In [1, Corollary3.1], it was obtainedthat that the sequence Ω is strictly
n
decreasing for n≥2.
In [8], it was procured that the sequence Ω1/(nlnn) is strictly logarithmically
n
convex for n≥2.
By L’Hospital rule, we have
(lnπ)−ψ(x+1)
lim lnG(x)= lim =−∞,
x→∞ x→∞(x2+2x−1)/(x3+x2+x+1)
hence,
πx 1/ln xx2++11
lim G(x)= lim (cid:0) (cid:1) =0
x→∞ x→∞(cid:20)Γ(x+1)(cid:21)
and the sequence (4) converges to 0 as n→∞.
Remark 2. We conjecture that the sequence (4) and the function (5) are both
logarithmically convex on (1,∞).
TWO MONOTONIC FUNCTIONS INVOLVING GAMMA FUNCTION 7
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(F.Qi)DepartmentofMathematics,CollegeofScience,TianjinPolytechnicUniver-
sity,Tianjin City, 300160,China
E-mail address: [email protected], [email protected], [email protected]
URL:http://qifeng618.spaces.live.com
(B.-N.Guo) School of Mathematicsand Informatics,Henan Polytechnic University,
JiaozuoCity, HenanProvince,454010,China
E-mail address: [email protected], [email protected]