AN INEQUALITY AND ITS q-ANALOGUE
MINGJIN WANG
Department of Mathematics
East China Normal University,
Shanghai, 200062,
People’s Republic of China
EMail: wmj@jpu.edu.cn
An Inequality and its
q-Analogue
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
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Received:
16 July, 2006
Accepted:
11 June, 2007
Communicated by:
JJ
II
J. Sándor
J
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2000 AMS Sub. Class.:
Primary 26D15; Secondary 33D15.
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Key words:
Gould-Hsu inversions; Carlitz inversions; Grüss inequality; q-series.
Abstract:
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.
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1
Introduction and Some Known Results
3
2
A New Inequality
6
3
A q-Analogue of the Inequality
9
An Inequality and its
q-Analogue
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
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1.
Introduction and Some Known Results
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
(1 − aq k ),
(a; q)∞ =
k=0
∞
Y
(1 − aq k ).
An Inequality and its
q-Analogue
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
Title Page
k=0
The q-binomial coefficient is defined by
hni
(q; q)n
(1.2)
=
.
k
(q; q)k (q; q)n−k
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Page 3 of 14
The following inverse series relations are due to Gould-Hsu [4]:
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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
k=0

ψ(x, 0) = 1,
differ from zero for any non-negative integer x. Then we have the following inverse
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series relations

n
P


(−1)k nk ψ(k, n)g(k),
 f (n) =
(1.3)


 g(n) =
where
n
k
=
k=0
n
P
(−1)k
k=0
n ak +kbk
f (k),
k ψ(n, k+1)
An Inequality and its
q-Analogue
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]:
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
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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 a
b−a a
b−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:
An Inequality and its
q-Analogue
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
Theorem 1.4. If a ≤ ai ≤ A and b ≤ bi ≤ B for i = 1, 2, . . . , n, then we have
n
n
n
1 X
1X
1 X (A − a)(B − b)
ai b i −
ai ·
bi ≤
,
(1.6)
n
n i=1
n i=1 4
i=1
JJ
II
where a, A, ai , b, B, bi are real numbers.
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2.
A New Inequality
In this section we obtain an inequality about series by using both the Gould-Hsu
inversions and the Grüss inequality.
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
(2.1)
(−1)n+k
f (k)g(k) − f (n)g(n)
(n + 1)
k
k=0
n
n
2 n−3
≤ 3(n + 1) 2 A
A
−a ,
k0
k0
An Inequality and its
q-Analogue
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
Title Page
], [x] denotes the greatest integer less than or equal x .
where k0 = [ n−1
2
Proof. Letting ai = −1, bi = 0 in (1.3), we have

n
P
n+k n


f
(n)
=
(−1)
g(k),

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
P
Substituting ki=0 ki = 2k into the above inequality we get
(2.3)
a · 2k ≤ g(k) ≤ A · 2k ,
k = 0, 1, . . . , n.
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On the other hand, we know that
n
n!/(k + 1)!(n − k − 1)!
n−k
k+1
=
=
,
n
n!/(k)!(n − k)!
k+1
k
consequently
 n
(k+1)


 (n) ≥ 1 when k ≤ k0 ,
k
n

)

 (k+1
≤ 1, when k ≥ k0 ,
(nk)
where k0 = [ n−1
]. So, we get
2
n
n
≤
, k = 0, 1, . . . , n.
(2.4)
1≤
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
An Inequality and its
q-Analogue
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
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Combining (1.6), (2.5) and (2.6) we obtain
!
!
n
n
n
1 X
1 X
1 X
Ai Bi −
Ai ·
Bi n + 1
n + 1 i=0
n + 1 i=0
i=0
A kn0 − a 2n A kn0 + 2n−1 A
≤
4
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
,
An Inequality and its
q-Analogue
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
Title Page
!
1
n
n+k
(−1)
g(k) n + 1 k=0
k
A kn0 − a 2n A kn0 + 2n−1 A
n
X
Substituting (2.2) into the above inequality, we get (2.1).
4
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n
k0
.
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A q-Analogue of the Inequality
3.
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.
Pk k Lemma 3.1. Suppose 0 ≤ f (k) ≤ A and g(k) =
i=0 i f (i), then for any
k = 1, 2, . . . , n, we have
0 ≤ g(k) ≤ A
(3.1)
n h
X
i=0
ni
.
i
An Inequality and its
q-Analogue
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
Proof. It is obvious that g(k) ≥ 0. If k ≤ n1 ≤ n2 , then we have
hn i
2
k
Since
we get
1 − q n1 +1
1 − q n1 +2
1 − q n2 h n1 i
=
·
·
·
·
.
1 − q n1 +1−k 1 − q n1 +2−k
1 − q n2 −k k
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
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hn i
2
k
≥
hn i
1
k
.
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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.
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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
n+k
(
)
2
(3.2) (n + 1)
(−1)
q
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

n+k n ( 2 )

f
(n)
=
(−1)
q
g(k),

k
k=0
(3.3)
n P

n

 g(n) =
f (k).
k
k=0
(3.4)
a·
i=0
i
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
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Page 10 of 14
Using the lemma, we have
k X
k
An Inequality and its
q-Analogue
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
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consequently
 n
[ k+1 ]


 [ n ] ≥ 1, when k ≤ k0 ,
k
n

]

 [ k+1
≤ 1, when k ≥ k0 ,
[ nk ]
where k0 = [ n−1
]. So, we have
2
hni
n
≤
,
1≤
k
k0
An Inequality and its
q-Analogue
(3.5)
Let Ak =
(3.6)
k = 0, 1, . . . , n.
n
n−k
n+k n ( 2 )
f
(k)
and
B
=
(−1)
q
g(k), then
k
k
k
n
a ≤ Ak ≤ A
.
k0
From (3.4) and (3.5), we know that

h iP
n n
n

0
≤
B
≤
A
,
if n − k is even,

k

k0
i
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
n−1 X
n h i
n
n
n X n−1
−A
≤ Bk ≤ A
.
k0 i=0
i
k0 i=0 i
(3.7)
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
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Combining (1.6), (3.6) and (3.7) we obtain
!
!
n
n
n
1 X
1 X
1 X
Ai Bi −
Ai ·
Bi n + 1
n + 1 i=0
n + 1 i=0
i=0
!
X
X
n−1 n h i
n−1
1
n
n
n
n
,
+A
≤
−a
A
A
k0 i=0
i
k0 i=0 i
4
k0
which can be written as
n
1 X
h n i2 n−k
(−1)n+k
q ( 2 ) f (k)g(k)
n + 1
k
k=0
!
!
n
n
h n i (n−k
)
1 X
1 X hni
2
f (k)
(−1)n+k
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
n+k
(−1)
f (k)g(k) − f (n)g(n)
(n + 1)
k
k=0
An Inequality and its
q-Analogue
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
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"X
#
n n−1 X
n
n−1
A(n + 1)2 n
n
−a
+
≤
A
2
2
k0
k0
4
i=0
i=0
n
A(n + 1)2 n
n
n
n
n−1
2 n−3
A
−a ,
=
A
− a [2 + 2 ] = 3(n + 1) 2 A
k0
k0
k0
4
k0
which is the inequality (2.1). So the inequality (3.2) is the q-analogue of the inequality (2.1).
An Inequality and its
q-Analogue
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
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References
[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.
An Inequality and its
q-Analogue
Mingjin Wang
vol. 8, iss. 2, art. 50, 2007
[4] H.W. GOULD AND L.C. HSU, Some new inverse series relations, Duke Math.
J., 40 (1973), 885–891
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[5] G. R GRÜSS, Über dasR maximum Rdes absoluten Betrages von
b
b
b
1
1
1
f (x)g(x)dx − ( b−a
f (x)dx)( b−a
g(x)dx), Math.Z., 39 (1935),
b−a a
a
a
215–226.
[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].
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