Volume 8 (2007), Issue 2, Article 50, 6 pp.
AN INEQUALITY AND ITS q-ANALOGUE
MINGJIN WANG
D EPARTMENT OF M ATHEMATICS ,
E AST C HINA N ORMAL U NIVERSITY,
S HANGHAI , 200062,
P EOPLE ’ S R EPUBLIC OF C HINA
wmj@jpu.edu.cn
Received 16 July, 2006; accepted 11 June, 2007
Communicated by J. Sándor
A BSTRACT. In this paper, we establish a new inequality and its q-analogue by means of the
Gould-Hsu inversions, the Carlitz inversions and the Grüss inequality.
Key words and phrases: Gould-Hsu inversions; Carlitz inversions; Grüss inequality; q-series.
2000 Mathematics Subject Classification. Primary 26D15; Secondary 33D15.
1. I NTRODUCTION AND S OME K NOWN R ESULTS
q-series, which are also called basic hypergeometric series, plays a very important role in
many fields, such as affine root systems, Lie algebras and groups, number theory, orthogonal
polynomials and physics, etc. In this paper, first we establish an inequality by means of the
Gould-Hsu inversions, and then we obtain a q-analogue of the inequality.
We first state some notations and known results which will be used in the next sections. It is
supposed in this paper that 0 < q < 1. The q-shifted factorial is defined by
(1.1)
(a; q)0 = 1,
(a; q)n =
n−1
Y
k
(1 − aq ),
(a; q)∞ =
k=0
∞
Y
(1 − aq k ).
k=0
The q-binomial coefficient is defined by
hni
(q; q)n
(1.2)
=
.
k
(q; q)k (q; q)n−k
The following inverse series relations are due to Gould-Hsu [4]:
Theorem 1.1. Let {ai } and {bj } be two real or complex sequences such that the polynomials
defined by

n−1
 ψ(x, n) = Q (a + xb ), (n = 1, 2, . . . ),
k
k=0

190-06
ψ(x, 0) = 1,
k
2
M INGJIN WANG
differ from zero for any non-negative integer x. Then we have the following inverse series
relations

n
P


f
(n)
=
(−1)k nk ψ(k, n)g(k),

k=0
(1.3)
n
a +kb
P


k
k
 g(n) =
(−1)k nk ψ(n,
f (k),
k+1)
k=0
where
n
k
=
n!
.
k!(n−k)!
Carlitz [2] gave the following q-analogue of the Gould-Hsu inverse series relations:
Theorem 1.2. Let {ai } and {bj } be two real or complex sequences such that the polynomials
defined by

n−1
 φ(x, n) = Q (a + q x b ), (n = 1, 2, . . . ),
k
k
k=0

φ(x, 0) = 1,
differ from zero for x = q n with n being non-negative integers. Then we have the following
inverse series relations

n
n−k
P

k n ( 2 )

f
(n)
=
(−1)
q
φ(k, n)g(k);

k
k=0
(1.4)
n
ak +qk bk
P


 g(n) =
(−1)k nk φ(n;
f (k).
k+1)
k=0
We also need the following inequality, which is well known in the literature as the Grüss
inequality [5]:
Theorem 1.3. We have
Z b
Z b
Z b
1
1
1
(1.5)
f
(x)g(x)dx
−
f
(x)dx
g(x)dx
b − a
b
−
a
b
−
a
a
a
a
(M − m)(N − n)
≤
,
4
provided that f, g : [a, b] → R are integrable on [a, b] and m ≤ f (x) ≤ M, n ≤ g(x) ≤ N for
all x ∈ [a, b], where m, M, n, N are given constants.
The discrete version of the Grüss inequality can be stated as:
Theorem 1.4. If a ≤ ai ≤ A and b ≤ bi ≤ B for i = 1, 2, . . . , n, then we have
n
n
n
1 X
(A − a)(B − b)
X
X
1
1
(1.6)
ai b i −
ai ·
bi ≤
,
n
n i=1
n i=1 4
i=1
where a, A, ai , b, B, bi are real numbers.
2. A N EW I NEQUALITY
In this section we obtain an inequality about series by using both the Gould-Hsu inversions
and the Grüss inequality.
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 50, 6 pp.
http://jipam.vu.edu.au/
A N I NEQUALITY AND ITS q-A NALOGUE
3
P
Theorem 2.1. Suppose 0 ≤ a ≤ f (k) ≤ A, g(k) = ki=0 ki f (i), k = 1, 2, . . . , n, then the
following inequality holds
2
n
X
n
n+k
f
(k)g(k)
−
f
(n)g(n)
(2.1)
(n
+
1)
(−1)
k
k=0
n
n
2 n−3
≤ 3(n + 1) 2 A
A
−a ,
k0
k0
where k0 = [ n−1
], [x] denotes the greatest integer less than or equal x .
2
Proof. Letting ai = −1, bi = 0 in (1.3), we have

n
P


f
(n)
=
(−1)n+k nk g(k),

k=0
(2.2)
n
P

n

 g(n) =
f (k).
k
k=0
Since 0 ≤ a ≤ f (k) ≤ A, we obtain
k k k X
X
X
k
k
k
a·
≤ g(k) =
f (i) ≤ A ·
.
i
i
i
i=0
i=0
i=0
Pk k
k
Substituting i=0 i = 2 into the above inequality we get
(2.3)
a · 2k ≤ g(k) ≤ A · 2k ,
On the other hand, we know that
n
k+1
n
k
=
k = 0, 1, . . . , n.
n!/(k + 1)!(n − k − 1)!
n−k
=
,
n!/(k)!(n − k)!
k+1
consequently
 n
(k+1)


 (n) ≥ 1 when k ≤ k0 ,
k
n

 (k+1) ≤ 1, when k ≥ k ,

0
(nk)
]. So, we get
where k0 = [ n−1
2
n
n
(2.4)
1≤
≤
, k = 0, 1, . . . , n.
k
k0
Let Ak = nk f (k) and Bk = (−1)n+k nk g(k), then
n
(2.5)
a ≤ Ak ≤ A
.
k0
From (2.3) and (2.4), we know that

 0 ≤ Bk ≤ 2n A kn0
if n − k is even,
 −2n−1 A n ≤ B ≤ 0, if n − k is odd.
k
k0
So, for k = 1, 2, . . . , n, we have
(2.6)
n−1
−2
n
n
n
A
≤ Bk ≤ 2 A
.
k0
k0
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 50, 6 pp.
http://jipam.vu.edu.au/
4
M INGJIN WANG
Combining (1.6), (2.5) and (2.6) we obtain
!
n
n
1 X
X
1
Ai Bi −
Ai ·
n + 1
n
+
1
i=0
i=0
!
n
1 X
Bi n + 1 i=0
A kn0 − a 2n A
≤
which can be written as
2
n
1 X
n+k n
(−1)
f (k)g(k)
n + 1
k
k=0
!
n 1 X n
−
f (k) ·
n + 1 k=0 k
n
k0
n−1
+2
A
n
k0
4
!
n
1 X
n
(−1)n+k
g(k) n + 1 k=0
k
A kn0 − a 2n A kn0 + 2n−1 A
≤
4
,
n
k0
.
Substituting (2.2) into the above inequality, we get (2.1).
3. A q-A NALOGUE OF THE I NEQUALITY
In this section we give a q-analogue of the inequality (2.1) by means of the Carlitz inversions.
First, we have the following lemma.
P Lemma 3.1. Suppose 0 ≤ f (k) ≤ A and g(k) = ki=0 ki f (i), then for any k = 1, 2, . . . , n,
we have
n h i
X
n
(3.1)
0 ≤ g(k) ≤ A
.
i
i=0
Proof. It is obvious that g(k) ≥ 0. If k ≤ n1 ≤ n2 , then we have
hn i
1 − q n1 +1
1 − q n1 +2
1 − q n2 h n1 i
2
=
·
···
.
k
1 − q n1 +1−k 1 − q n1 +2−k
1 − q n2 −k k
Since
1 − q n1 +1
1 − q n1 +2
1 − q n2
·
···
≥ 1,
1 − q n1 +1−k 1 − q n1 +2−k
1 − q n2 −k
we get
hn i
2
k
≥
hn i
1
k
.
Consequently,
k k h i
n h i
X
X
X
k
n
n
g(k) =
f (i) ≤
f (i) ≤
f (i).
i
i
i
i=0
i=0
i=0
The main result of this section is the following theorem.
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 50, 6 pp.
http://jipam.vu.edu.au/
A N I NEQUALITY AND ITS q-A NALOGUE
5
P Theorem 3.2. Suppose 0 ≤ a ≤ f (k) ≤ A, g(k) = ki=0 ki f (i), k = 1, 2, . . . , n, then the
following inequality holds
n
h n i2 n−k
X
(3.2) (n + 1)
(−1)n+k
q ( 2 ) f (k)g(k) − f (n)g(n)
i
k=0
X
!
n h i
n−1 X
n
n
n−1
A(n + 1)2 n
−a
+
,
A
≤
i
i
4
k0
k0
i=0
i=0
where k0 = [ n−1
], [x] denotes the greatest integer less than or equal x.
2
Proof. Letting ai = −1, bi = 0 in (1.4) we get

n
(n−k
P
2 )


(−1)n+k nk q
g(k),
 f (n) =
k=0
(3.3)
n P

n

 g(n) =
f (k).
k
k=0
Using the lemma, we have
(3.4)
a·
k X
k
i=0
i
k n h i
X
X
k
n
≤ g(k) =
f (i) ≤ A ·
.
i
i
i=0
i=0
On the other hand, we notice that
n (q; q)n /(q; q)k+1 (q; q)n−k−1
1 − q n−k
k+1
=
=
,
n
(q; q)n /(q; q)k (q; q)n−k
1 − q k+1
k
consequently
 n
[ k+1 ]


 [ n ] ≥ 1, when k ≤ k0 ,
k
n

]

 [ k+1
≤ 1, when k ≥ k0 ,
n
[k]
where k0 = [ n−1
]. So, we have
2
n
(3.5)
1≤
≤
, k = 0, 1, . . . , n.
k
k0
(n−k
2 )
Let Ak = nk f (k) and Bk = (−1)n+k nk q
g(k), then
n
(3.6)
a ≤ Ak ≤ A
.
k0
hni
From (3.4) and (3.5), we know that

h iP
n n
n

0
≤
B
≤
A
,

k

k0
i
if n − k is even,
i=0
h i n−1
P n−1 

 −A kn
≤ Bk ≤ 0, if n − k is odd.
i
0
i=0
So, for k = 1, 2, . . . , n, we get
X
X
n−1 n h i
n
n−1
n
n
(3.7)
−A
≤ Bk ≤ A
.
k0 i=0
i
k0 i=0 i
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 50, 6 pp.
http://jipam.vu.edu.au/
6
M INGJIN WANG
Combining (1.6), (3.6) and (3.7) we obtain
!
!
n
n
n
1 X
X
X
1
1
Ai Bi −
Ai ·
Bi n + 1
n
+
1
n
+
1
i=0
i=0
i=0
X
X
!
n−1 n h i
1
n
n
n
n
n−1
+A
,
≤
A
−a
A
k0 i=0
i
4
k0
k0 i=0 i
which can be written as
n
1 X
h i2 n−k
n+k n
q ( 2 ) f (k)g(k)
(−1)
n + 1
k
k=0
!
!
n h i
n
h
i
n−k
X
X
(
)
1
n
1
2
n+k n
−
f (k)
(−1)
q
g(k) n + 1 k=0 k
n + 1 k=0
k
X
!
n h i
n−1 X
A n
n
n
n−1
≤
A
−a
+
.
4 k0
k0
i
i
i=0
i=0
Substituting (3.3) into the above inequality, we get (3.2).
From [3], we know
h n i n lim
=
.
q→1 i
i
Let q → 1 in both sides of the inequality (3.2) to get
2
n
X
n+k n
(−1)
f (k)g(k) − f (n)g(n)
(n + 1)
k
k=0
"
X
#
n n−1 X
n
n
n−1
A(n + 1)2 n
A
−a
+
≤
4
k0
k0
2
2
i=0
i=0
n
n
A(n + 1)2 n
n
n
n−1
2 n−3
A
− a [2 + 2 ] = 3(n + 1) 2 A
A
−a ,
=
4
k0
k0
k0
k0
which is the inequality (2.1). So the inequality (3.2) is the q-analogue of the inequality (2.1).
R EFERENCES
[1] G.E. ANDREWS, R. ASKEY AND R. ROY, Special Functions, Cambridge University Press, 2000.
[2] L. CARLITZ, Some inverse relations, Duke Math. J., 40 (1973), 893–901
[3] G. GASPER AND M. RAHMAN, Basic Hypergeometric Series, Encyclopedia of Mathematics and
Its Applications, 35, Cambridge University Press, Cambridge and New York, 1990.
[4] H.W. GOULD
891
AND
L.C. HSU, Some new inverse series relations, Duke Math. J., 40 (1973), 885–
[5] G. GRÜSS, Über das maximum des absoluten Betrages von
Rb
Rb
1
1
( b−a
f
(x)dx)(
b−a a g(x)dx), Math.Z., 39 (1935), 215–226.
a
1
b−a
Rb
a
f (x)g(x)dx −
[6] MINGJIN WANG, An inequality about q-series, J. Ineq. Pure and Appl. Math., 7(4) (2006), Art.
136. [ONLINE: http://jipam.vu.edu.au/article.php?sid=756].
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 50, 6 pp.
http://jipam.vu.edu.au/