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volume 7, issue 2, article 47, 2006.

Received 21 June, 2005;

accepted 16 March, 2006.

Communicated by:J.E. Peˇcari´c

Abstract Contents

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Journal of Inequalities in Pure and Applied Mathematics

SOME INEQUALITIES OF PERTURBED TRAPEZOID TYPE

ZHENG LIU

Institute of Applied Mathematics Faculty of Science

Anshan University of Science and Technology Anshan 114044, Liaoning, China.

EMail:lewzheng@163.net

c

2000Victoria University ISSN (electronic): 1443-5756 186-05

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Some Inequalities of Perturbed Trapezoid Type

Zheng Liu

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Abstract

A new generalized perturbed trapezoid type inequality is established by Peano kernel approach. Some related results are also given.

2000 Mathematics Subject Classification:26D15.

Key words: Perturbed trapezoid inequality;n-times continuously differentiable map- ping; Absolutely continuous.

Contents

1 Introduction. . . 3 2 For Differentiable Mappings With Bounded Derivatives . . . . 5 3 Bounds In Terms of Some Lebesgue Norms . . . 10 4 Non-Symmetric Bounds. . . 12

References

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Some Inequalities of Perturbed Trapezoid Type

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1. Introduction

In recent years, some authors have considered the perturbed trapezoid inequality

Z b a

f(x)dx−b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

≤C(Γ2−γ2)(b−a)3, where f : [a, b] → R is a twice differentiable mapping on (a, b) with γ2 = infx∈[a,b]f00(x)>−∞andΓ2 = supx∈[a,b]f00(x)<+∞whileC is a constant.

(e.g. see [1] – [8]) It seems that the best result C =

3

108 was separately and independently discovered by the authors of [5] and [8]. The perturbed trapezoid inequality has been established as

(1.1)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

√3

108(Γ2−γ2)(b−a)3. Moreover, we can also find in [5] the following two perturbed trapezoid inequal- ities as

(1.2)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

≤ 1

384(Γ3−γ3)(b−a)4,

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where f : [a, b] → R is a third-order differentiable mapping on (a, b) with γ3 = infx∈[a,b]f000(x)>−∞andΓ3 = supx∈[a,b]f000(x)<+∞, and

(1.3)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

≤ 1

720M4(b−a)5, where f : [a, b] → R is a fourth-order differentiable mapping on (a, b) with M4 = sup

x∈[a,b]

|f(4)(x)|<+∞.

The purpose of this paper is to extend these above results to a more general version by choosing appropriate harmonic polynomials such as the Peano ker- nel. A new generalized perturbed trapezoid type inequality is established and some related results are also given.

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Some Inequalities of Perturbed Trapezoid Type

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2. For Differentiable Mappings With Bounded Derivatives

Theorem 2.1. Let f : [a, b] → R be an n-times continuously differentiable mapping,n≥2and such thatMn := supx∈[a,b]|f(n)(x)|<∞. Then

(2.1)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

≤Mn×





3(b−a)3

54 ifn = 2;

n(n−2)(b−a)n+1

3(n+1)!2n ifn ≥3, where[n−12 ]denotes the integer part of n−12 .

Proof. It is not difficult to find the identity

(2.2) (−1)n Z b

a

Tn(x)f(n)(x)dx

= Z b

a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

,

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Some Inequalities of Perturbed Trapezoid Type

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whereTn(x)is the kernel given by

(2.3) Tn(x) =

(x−a)n

n!(b−a)(x−a)2(n−1)!n−1 + (b−a)12(n−2)!2(x−a)n−2 if x∈ a,a+b2

,

(x−b)n

n! +(b−a)(x−b)2(n−1)!n−1 + (b−a)12(n−2)!2(x−b)n−2 if x∈ a+b2 , b . Using the identity (2.2), we get

(2.4)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

=

Z b a

Tn(x)f(n)(x)dx

≤Mn Z b

a

|Tn(x)|dx.

For brevity, we put

Pn(x) := (x−a)n

n! − (b−a)(x−a)n−1

2(n−1)! +(b−a)2(x−a)n−2 12(n−2)!

= (x−a)n−2 n!

(x−a)2− n(b−a)(x−a)

2 +n(n−1)(b−a)2 12

, x∈

a,a+b 2

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Some Inequalities of Perturbed Trapezoid Type

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and

Qn(x) := (x−b)n

n! +(b−a)(x−b)n−1

2(n−1)! +(b−a)2(x−b)n−2 12(n−2)!

= (x−b)n−2 n!

(x−b)2 +n(b−a)(x−b)

2 +n(n−1)(b−a)2 12

, x∈

a+b 2 , b

.

It is clear that Pn(x) and Qn(x) are symmetric with respect to the line x =

a+b

2 for n even, and symmetric with respect to the point (a+b2 ,0) for n odd.

Therefore, Z b a

|Tn(x)|dx= 2 Z a+b2

a

|Pn(x)|dx

= (b−a)n+1 n!2n

Z 1 0

tn−2

t2−nt+n(n−1) 3

dt by substitutionx=a+b−a2 t, and it is easy to find that

rn(t) := tn−2

t2−nt+ n(n−1) 3

is always nonnegative on[0,1]forn ≥3. Thus we have Z 1

0

|rn(t)|dt= Z 1

0

tn−2

t2−nt+ n(n−1) 3

dt = n(n−2) 3(n+ 1)

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forn≥3, and Z 1

0

|r2(t)|dt = Z 1

0

t2−2t+2 3

dt

= Z t0

0

t2 −2t+2 3

dt−

Z 1 t0

t2−2t+2 3

dt, wheret0 = 1−

3

3 is the unique zero ofr2(t)in(0,1). Hence, (2.5)

Z b a

|Tn(x)|dx=

3(b−a)3

54 , n= 2,

n(n−2)(b−a)n+1

3(n+1)!2n , n≥3.

Consequently, the inequality (2.1) follows from (2.4) and (2.5).

Remark 1. If in the inequality (2.1) we choosen = 2,3,4, then we get

Z b a

f(x)dx−b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

√3

54M2(b−a)3,

Z b a

f(x)dx−b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

≤ 1

192M3(b−a)4 and the inequality (1.3), respectively.

For convenience in further discussions, we will now collect some technical results related to (2.3) which are not difficult to obtain by elementary calculus

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as:

(2.6)

Z b a

Tn(x)dx=

0, nodd,

n(n−2)(b−a)n+1

3(n+1)!2n , neven.

(2.7) max

x∈[a,b]|Tn(x)|=









(b−a)2

12 , n = 2,

3(b−a)3

216 , n = 3,

(n−1)(n−3)(b−a)n

3(n!)2n , n ≥4.

(2.8) max

x∈[a,b]

T2m(x)− 1 b−a

Z b a

T2m(x)dx

=

(b−a)4

720 , m = 2,

(8m3−16m2+2m+3)(b−a)2m

3(2m+1)!22m , m ≥3.

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3. Bounds In Terms of Some Lebesgue Norms

Theorem 3.1. Letf : [a, b] → Rbe a mapping such that the derivativef(n−1) (n ≥2)is absolutely continuous on[a, b]. Iff(n) ∈L[a, b], then we have (3.1)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

≤ kf(n)k×

3(b−a)3

54 , n = 2,

n(n−2)(b−a)n+1

3(n+1)!2n , n ≥3, where[n−12 ]denotes the integer part ofn−12 andkf(n)k :=esssupx∈[a,b]|f(n)(x)|

is the usual Lebesgue norm onL[a, b].

The proof of inequality (3.1) is similar to the proof of inequality (2.1) and so is omitted.

Theorem 3.2. Let f : [a, b] → R be a mapping such that the derivative f(n−1)(n ≥ 2) is absolutely continuous on [a, b]. If f(n) ∈ L1[a, b], then we have

(3.2)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

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Some Inequalities of Perturbed Trapezoid Type

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[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

≤ kf(n)k1×









(b−a)2

12 , n = 2,

3(b−a)3

216 , n = 3,

(n−1)(n−3)(b−a)n

3(n!)2n , n ≥4, wherekf(n)k1 :=Rb

a |f(x)|dxis the usual Lebesgue norm onL1[a, b].

Proof. By using the identity (2.2), we get

Z b a

f(x)dx−b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

=

Z b a

Tn(x)f(n)(x)dx

≤ max

x∈[a,b]|Tn(x)|

Z b a

|f(n)(x)|dx.

Then the inequality (3.2) follows from (2.7).

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4. Non-Symmetric Bounds

Theorem 4.1. Let f : [a, b] → R be a mapping such that the derivative f(n)(n ≥ 2) is integrable with γn = infx∈[a,b]f(n)(x) > −∞ and Γn = supx∈[a,b]f(n)(x)<+∞. Then we have

(4.1)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

≤ Γn−γn

2 ×





3(b−a)3

54 , n= 2,

n(n−2)(b−a)n+1

3(n+1)!2n , n≥3and odd,

(4.2)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

≤[f(n−1)(b)−f(n−1)(a)−γn(b−a)]

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×









(b−a)2

12 , n= 2,

3(b−a)3

216 , n= 3,

(n−1)(n−3)(b−a)n

3(n!)2n , n≥5and odd, (4.3)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

≤[Γn(b−a)−f(n−1)(b) +f(n−1)(a)]

×









(b−a)2

12 , n= 2,

3(b−a)3

216 , n= 3,

(n−1)(n−3)(b−a)n

3(n!)2n , n≥5and odd, (4.4)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

−m(m−1)(b−a)2m

3(2m+ 1)!22m−2 [f(2m−1)(b)−f(2m−1)(a)]

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≤[f(2m−1)(b)−f(2m−1)(a)−γ2m(b−a)]

×

(b−a)4

720 , m = 2,

(8m3−16m2+2m+3)(b−a)2m

3(2m+1)!22m , m ≥3,

(4.5)

Z b a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

−m(m−1)(b−a)2m

3(2m+ 1)!22m−2 [f(2m−1)(b)−f(2m−1)(a)]

≤[Γ2m(b−a)−f(2m−1)(b)+f(2m−1)(a)]

(b−a)4

720 , m= 2,

(8m3−16m2+2m+3)(b−a)2m

3(2m+1)!22m , m≥3.

Proof. Fornodd andn= 2, by (2.2) and (2.6) we get (−1)n

Z b a

Tn(x)[f(n)(x)−C]dx

= Z b

a

f(x)dx− b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

.

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whereC ∈Ris a constant.

If we chooseC = γn2 n, then we have

Z b a

f(x)dx−b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

≤ Γn−γn 2

Z b a

|Tn(x)|dx.

and hence the inequality (4.1) follows from (2.5).

If we chooseC =γn, then we have

Z b a

f(x)dx−b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

≤ max

x∈[a,b]|Tn(x)|

Z b a

|f(n)(x)−γn|dx, and hence the inequality (4.2) follows from (2.7).

Similarly we can prove that the inequality (4.3) holds.

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By (2.2) and (2.6) we can also get

Z b a

f(x)dx−b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

−m(m−1)(b−a)2m

3(2m+ 1)!22m−2 [f(2m−1)(b)−f(2m−1)(a)]

=

Z b a

T2m(x)− 1 b−a

Z b a

T2m(x)dx

[f2m(x)−C]dx , whereC ∈Ris a constant.

If we chooseC =γ2m, then we have

Z b a

f(x)dx−b−a

2 [f(a) +f(b)] + (b−a)2

12 [f0(b)−f0(a)]

[n−12 ]

X

k=2

k(k−1)(b−a)2k+1 3(2k+ 1)!22k−2 f(2k)

a+b 2

−m(m−1)(b−a)2m

3(2m+ 1)!22m−2 [f(2m−1)(b)−f(2m−1)(a)]

≤ max

x∈[a,b]

T2m(x)− 1 b−a

Z b a

T2m(x)dx

Z b a

|f(2m)(x)−γ2m|dx and hence the inequality (4.4) follows from (2.8).

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Similarly we can prove that the inequality (4.5) holds.

Remark 2. It is not difficult to find that the inequality (4.1) is sharp in the sense that we can choose f to attain the equality in (4.1). Indeed, for n = 2, we construct the functionf(x) =Rx

a

Ry

a j(z)dz

dy, where

j(x) =









Γ2, a≤x < (3+

3)a+(3− 3)b

6 ,

γ2, (3+

3)a+(3− 3)b

6 ≤x < (3−

3)a+(3+ 3)b

6 ,

Γ2, (3−

3)a+(3+ 3)b

6 ≤x≤b,

and forn ≥3and odd, we construct the function f(x) =

Z x a

Z yn

a

· · · Z y2

a

j(y1)dy1· · ·

dyn−1

dyn, where

j(x) =

Γn, a≤x < a+b2 , γn, a+b2 ≤x≤b.

Remark 3. If in the inequality (4.1) we choosen = 2,3, then we recapture the inequalities (1.1) and (1.2), respectively.

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References

[1] P. CERONE, On perturbed trapezoidal and midpoint rules, Korean J. Com- put. Appl. Math., 2 (2002), 423–435.

[2] P. CERONE AND S.S. DRAGOMIR, Trapezoidal type rules from an in- equalities point of view, Handbook of Analytic-Computational Methods in Applied Mathematics, CRC Press N.Y.(2000), 65–134.

[3] P. CERONE, S.S. DRAGOMIR ANDJ. ROUMELIOTIS, An inequality of Ostrowski-Grüss type for twice differentiable mappings and applications in numerical integration, Kyungpook Math. J., 39 (1999), 331–341.

[4] X.L. CHENG, Improvement of some Ostrowski-Grüss type inequalities, Comput. Math. Appl., 42 (2001), 109–114.

[5] X.L. CHENG ANDJ. SUN, A note on the perturbed trapezoid inequality, J.

Inequal. in Pure and Appl. Math., 3(2) (2002), Art. 29. [ONLINE:http:

//jipam.vu.edu.au/article.php?sid=181].

[6] S.S. DRAGOMIR, P. CERONEANDA. SOFO, Some remarks on the trape- zoid rule in numerical integration, Indian J. of Pure and Appl. Math., 31(5) (2000), 489–501.

[7] M. MATI ´C, J. PE ˇCARI ´CAND N. UJEVI ´C, Improvement and further gen- eralization of inequalities of Ostrowski-Grüss type, Computer Math. Appl., 39 (2000), 161–175.

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