AN INEQUALITY AND ITS q-ANALOGUE

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An Inequality and its q-Analogue Mingjin Wang vol. 8, iss. 2, art. 50, 2007

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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

Received: 16 July, 2006

Accepted: 11 June, 2007

Communicated by: J. Sándor

2000 AMS Sub. Class.: Primary 26D15; Secondary 33D15.

Key words: Gould-Hsu inversions; Carlitz inversions; Grüss inequality;q-series.

Abstract: In this paper, we establish a new inequality and itsq-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

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 aq-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=0

(1−aq^k), (a;q)∞ =

∞

Y

k=0

(1−aq^k).

Theq-binomial coefficient is defined by

(1.2) hn

k i

= (q;q)_n (q;q)_k(q;q)n−k

.

The following inverse series relations are due to Gould-Hsu [4]:

Theorem 1.1. Let {a_i} and {b_j} be two real or complex sequences such that the polynomials defined by







ψ(x, n) =

n−1

Q

k=0

(a_k+xb_k), (n = 1,2, . . .), ψ(x,0) = 1,

differ from zero for any non-negative integerx. Then we have the following inverse

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series relations

(1.3)











f(n) =

n

P

k=0

(−1)^{k n}_k

ψ(k, n)g(k), g(n) =

n

P

k=0

(−1)^{k n}_k _a_k_+kb_k

ψ(n, k+1)f(k), where ⁿ_k

= _k!(n−k)!^n! .

Carlitz [2] gave the followingq-analogue of the Gould-Hsu inverse series relations:

Theorem 1.2. Let {a_i} and {b_j} be two real or complex sequences such that the polynomials defined by







φ(x, n) =

n−1

Q

k=0

(a_k+q^xb_k), (n= 1,2, . . .), φ(x,0) = 1,

differ from zero forx = qⁿ with n being non-negative integers. Then we have the following inverse series relations

(1.4)











f(n) =

n

P

k=0

(−1)^k_n

k

q(^n−k2 )

φ(k, n)g(k);

g(n) =

n

P

k=0

(−1)^k_n

k

_a_k_+q^k_b_k

φ(n;k+1)f(k).

We also need the following inequality, which is well known in the literature as the Grüss inequality [5]:

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Theorem 1.3. We have (1.5)

1 b−a

Z b a

f(x)g(x)dx− 1

b−a Z b

a

f(x)dx 1

b−a Z b

a

g(x)dx

≤ (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 allx∈[a, b], wherem, M, n, N are given constants.

The discrete version of the Grüss inequality can be stated as:

Theorem 1.4. Ifa ≤a_i ≤Aandb≤b_i ≤Bfori= 1,2, . . . , n, then we have (1.6)

1 n

n

X

i=1

a_ib_i− 1 n

n

X

i=1

a_i· 1 n

n

X

i=1

b_i

≤ (A−a)(B−b)

4 ,

wherea,A,a_i,b,B,b_i 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.

Theorem 2.1. Suppose0≤a≤ f(k)≤A,g(k) =Pk i=0

k i

f(i), k = 1,2, . . . , n, then the following inequality holds

(2.1)

(n+ 1)

n

X

k=0

(−1)^n+k n

k 2

f(k)g(k)−f(n)g(n)

≤3(n+ 1)²2ⁿ⁻³A n

k₀ A n

k₀

−a

, wherek₀ = [ⁿ⁻¹₂ ],[x]denotes the greatest integer less than or equalx. Proof. Lettinga_i =−1, b_i = 0in (1.3), we have

(2.2)











f(n) =

n

P

k=0

(−1)^{n+k n}_k g(k), g(n) =

n

P

k=0 n k

f(k).

Since0≤a≤f(k)≤A, we obtain a·

k

X

i=0

k i

≤g(k) =

k

X

i=0

k i

f(i)≤A·

k

X

i=0

k i

. SubstitutingPk

i=0 k

i

= 2^kinto the above inequality we get (2.3) a·2^k ≤g(k)≤A·2^k, k= 0,1, . . . , n.

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On the other hand, we know that

n k+1

n k

= n!/(k+ 1)!(n−k−1)!

n!/(k)!(n−k)! = n−k k+ 1, consequently









 (k+1ⁿ )

(ⁿ_k) ≥1 whenk ≤k₀, (k+1ⁿ )

(ⁿk) ≤1, whenk ≥k₀, wherek₀ = [ⁿ⁻¹₂ ]. So, we get

(2.4) 1≤

n k

≤ n

k₀

, k = 0,1, . . . , n.

LetAk= ⁿ_k

f(k)andBk = (−1)^{n+k n}_k

g(k), then

(2.5) a≤A_k ≤A

n k₀

. From (2.3) and (2.4), we know that







0≤B_k ≤2ⁿA _kⁿ

0

ifn−kis even,

−2ⁿ⁻¹A _kⁿ

0

≤Bk≤0, ifn−kis odd.

So, fork = 1,2, . . . , n, we have

(2.6) −2ⁿ⁻¹A

n k₀

≤B_k≤2ⁿA n

k₀

.

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Combining (1.6), (2.5) and (2.6) we obtain

1 n+ 1

n

X

i=0

AiBi− 1 n+ 1

n

X

i=0

Ai

!

· 1 n+ 1

n

X

i=0

Bi

!

≤

A _kⁿ

0

−a 2ⁿA _kⁿ

0

+ 2ⁿ⁻¹A _kⁿ

0

4 ,

which can be written as

1 n+ 1

n

X

k=0

(−1)^n+k n

k 2

f(k)g(k)

− 1 n+ 1

n

X

k=0

n k

f(k)

!

· 1 n+ 1

n

X

k=0

(−1)^n+k n

k

g(k)

!

≤

A _kⁿ

0

−a 2ⁿA _kⁿ

0

+ 2ⁿ⁻¹A _kⁿ

0

4 .

Substituting (2.2) into the above inequality, we get (2.1).

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3. A q-Analogue of the Inequality

In this section we give aq-analogue of the inequality (2.1) by means of the Carlitz inversions. First, we have the following lemma.

Lemma 3.1. Suppose 0 ≤ f(k) ≤ A and g(k) = Pk i=0

_k

i

f(i), then for any k = 1,2, . . . , n, we have

(3.1) 0≤g(k)≤A

n

X

i=0

hn i i

.

Proof. It is obvious thatg(k)≥0. Ifk ≤n₁ ≤n₂, then we have hn₂

k i

= 1−qⁿ¹⁺¹

1−qⁿ¹^+1−k · 1−qⁿ¹⁺²

1−qⁿ¹^+2−k· · · 1−qⁿ² 1−qⁿ²^−k

hn₁ k

i . Since

1−qⁿ¹⁺¹

1−qⁿ¹^+1−k · 1−qⁿ¹⁺²

1−qⁿ¹^+2−k · · · 1−qⁿ² 1−qⁿ²^−k ≥1, we get

hn₂ k

i≥hn₁ k

i . Consequently,

g(k) =

k

X

i=0

k i

f(i)≤

k

X

i=0

hn i i

f(i)≤

n

X

i=0

hn i i

f(i).

The main result of this section is the following theorem.

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Theorem 3.2. Suppose0≤a≤f(k)≤A,g(k) =Pk i=0

_k

i

f(i),k = 1,2, . . . , n, then the following inequality holds

(3.2)

(n+ 1)

n

X

k=0

(−1)^n+khn i

i2

q(^n−k2 )f(k)g(k)−f(n)g(n)

≤ A(n+ 1)² 4

n k0

A n

k0

−a ⁿ

X

i=0

hn i i

+

n−1

X

i=0

n−1 i

! , wherek₀ = [ⁿ⁻¹₂ ],[x]denotes the greatest integer less than or equalx.

Proof. Lettinga_i =−1, b_i = 0in (1.4) we get

(3.3)











f(n) =

n

P

k=0

(−1)^n+k_n

k

q(^n−k2 ) g(k), g(n) =

n

P

k=0

_n

k

f(k).

Using the lemma, we have

(3.4) a·

k

X

i=0

k i

≤g(k) =

k

X

i=0

k i

f(i)≤A·

n

X

i=0

hn i i

. On the other hand, we notice that

_n

k+1

_n

k

= (q;q)_n/(q;q)_k+1(q;q)n−k−1

(q;q)_n/(q;q)_k(q;q)n−k

= 1−q^n−k 1−q^k+1,

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consequently









 [_k+1ⁿ ]

[ⁿ_k] ≥1, whenk ≤k₀, [_k+1ⁿ ]

[ⁿ_k] ≤1, whenk ≥k₀, wherek₀ = [ⁿ⁻¹₂ ]. So, we have

(3.5) 1≤hn

k i ≤

n k₀

, k = 0,1, . . . , n.

LetA_k=_n

k

f(k)andB_k = (−1)^n+k_n

k

q(^n−k2 )

g(k), then

(3.6) a ≤Ak≤A

n k₀

. From (3.4) and (3.5), we know that











0≤B_k ≤Ah

n k0

i ⁿ P

i=0

_n

i

, ifn−kis even,

−Ah

n k0

iⁿ⁻¹ P

i=0

n−1 i

≤Bk ≤0, ifn−kis odd.

So, fork = 1,2, . . . , n, we get

(3.7) −A

n k₀

ⁿ⁻¹ X

i=0

n−1 i

≤B_k ≤A n

k₀ n

X

i=0

hn i i

.

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Combining (1.6), (3.6) and (3.7) we obtain

1 n+ 1

n

X

i=0

AiBi− 1 n+ 1

n

X

i=0

Ai

!

· 1 n+ 1

n

X

i=0

Bi

!

≤ 1 4

A

n k₀

−a

A n

k₀ n

X

i=0

hn i i

+A n

k₀ ⁿ⁻¹

X

i=0

n−1 i

! , which can be written as

1 n+ 1

n

X

k=0

(−1)^n+khn k

i2

q(^n−k₂ )f(k)g(k)

− 1 n+ 1

n

X

k=0

hn k i

f(k)

! 1 n+ 1

n

X

k=0

(−1)^n+khn k i

q(^n−k2 ) g(k)

!

≤ A 4

n k₀ A

n k₀

−a n

X

i=0

hn i i

+

n−1

X

i=0

n−1 i

! . Substituting (3.3) into the above inequality, we get (3.2).

From [3], we know

limq→1

hn i i

= n

i

. Letq →1in both sides of the inequality (3.2) to get

(n+ 1)

n

X

k=0

(−1)^n+k n

k 2

f(k)g(k)−f(n)g(n)

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≤ A(n+ 1)² 4

n k₀ A

n k₀

−a " _n

X

i=0

n 2

+

n−1

X

i=0

n−1 2

#

= A(n+ 1)² 4

n k0

A n

k0

−a

[2ⁿ+ 2ⁿ⁻¹] = 3(n+ 1)²2ⁿ⁻³A n

k0

A n

k0

−a

, which is the inequality (2.1). So the inequality (3.2) is theq-analogue of the inequality (2.1).

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References

[1] G.E. ANDREWS, R. ASKEYANDR. ROY, Special Functions, Cambridge Uni- versity Press, 2000.

[2] L. CARLITZ, Some inverse relations, Duke Math. J., 40 (1973), 893–901 [3] G. GASPERANDM. RAHMAN, Basic Hypergeometric Series, Encyclopedia of

Mathematics and Its Applications, 35, Cambridge University Press, Cambridge and New York, 1990.

[4] H.W. GOULDAND L.C. HSU, Some new inverse series relations, Duke Math.

J., 40 (1973), 885–891

[5] G. GRÜSS, Über das maximum des absoluten Betrages von

1 b−a

Rb

a f(x)g(x)dx − (_b−a¹ Rb

af(x)dx)(_b−a¹ Rb

a g(x)dx), Math.Z., 39 (1935), 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].