Also we give a new Iyengar’s type inequality involv- ing a second order bounded derivative for the perturbed trapezoid inequality

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http://jipam.vu.edu.au/

Volume 3, Issue 2, Article 29, 2002

A NOTE ON THE PERTURBED TRAPEZOID INEQUALITY

XIAO-LIANG CHENG AND JIE SUN

DEPARTMENT OFMATHEMATICS

ZHEJIANGUNIVERSITY, XIXICAMPUS, ZHEJIANG310028, THEPEOPLE’SREPUBLIC OFCHINA.

xlcheng@mail.hz.zj.cn

Received 21 May, 2001; accepted 01 February, 2002.

Communicated by N.S. Barnett

ABSTRACT. In this paper, we utilize a variant of the Grüss inequality to obtain some new perturbed trapezoid inequalities. We improve the error bound of the trapezoid rule in numerical integration in some recent known results. Also we give a new Iyengar’s type inequality involving a second order bounded derivative for the perturbed trapezoid inequality.

Key words and phrases: Grüss inequality, Perturbed trapezoid inequality, Sharp bounds.

2000 Mathematics Subject Classification. 26D15, 26D10.

In the literature [2], [4] – [8], [11], [12] on numerical integration, the following estimation is well known as the trapezoid inequality:

(1.1)

Z b a

f(x)dx−1

2(b−a)(f(a) +f(b))

≤ 1

12M₂(b−a)³,

where the mappingf : [a, b]→ Ris supposed to be twice differentiable on the interval (a, b), with the second derivative bounded on (a, b)by M₂ = sup_x∈(a,b)|f⁰⁰(x)| < +∞. In [5], the authors derived the error bounds for the trapezoid inequality (1.1) by different norm of mapping f. In [2, 7, 11], the authors obtained the trapezoid inequality by the difference of sup and inf bound of the first derivative, that is,

Z b a

f(x)dx− 1

2(b−a)(f(a) +f(b))

≤ 1

8(Γ₁−γ₁)(b−a)², whereΓ₁ = sup_x∈(a,b)f⁰(x)<+∞andγ₁ = infx∈(a,b)f⁰(x)>−∞.

ISSN (electronic): 1443-5756

046-01

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For the perturbed trapezoid inequality, S. Dragomir et al. [5] obtained the following inequality by an application of the Grüss inequality:

(1.2)

Z b a

f(x)dx−1

2(b−a)(f(a) +f(b)) + 1

12(b−a)²(f⁰(b)−f⁰(a))

≤ 1

32(Γ₂−γ₂)(b−a)³, wheref is supposed to be twice differentiable on the interval(a, b), with the second derivative bounded on (a, b) by Γ2 = sup_x∈(a,b)f⁰⁰(x) < +∞ and γ2 = infx∈(a,b)f⁰⁰(x) > −∞. The constant ₃₂¹ is smaller than ¹

6√

5 given in [11] and ¹

18√

3 given in [2].

In this note we first improve the constant ₃₂¹ in the inequality (1.2) to the best possible one of ¹

36√

3. Then we give two new perturbed trapezoid inequalities for high-order differentiable mappings. We need the following variant of the Grüss inequality:

Theorem 1.1. Leth, g : [a, b] → Rbe two integrable functions such that φ ≤ g(x) ≤ Φfor some constantsφ, Φfor allx∈[a, b], then

(1.3)

1 b−a

Z b a

h(x)g(x)dx− 1 (b−a)²

Z b a

h(x)dx Z b

a

g(x)dx

≤ 1 2

Z b a

h(x)− 1 b−a

Z b a

h(y)dy

dx

(Φ−φ).

Proof. We write the left hand of inequality (1.3) as Z b

a

h(x)g(x)dx− 1 b−a

Z b a

h(x)dx Z b

a

g(x)dx= Z b

a

(h(x)− 1 b−a

Z b a

h(y)dy)g(x)dx.

Denote

I⁺ = Z b

a

max(h(x)− 1 b−a

Z b a

h(y)dy,0)dx and

I⁻= Z b

a

min(h(x)− 1 b−a

Z b a

h(y)dy,0)dx.

ObviouslyI⁺+I⁻= 0. Forφ≤g(x)≤Φ, then Z b

a

(h(x)− 1 b−a

Z b a

h(y)dy)g(x)dx≤I⁺Φ +I⁻φ and

− Z b

a

(h(x)− 1 b−a

Z b a

h(y)dy)g(x)dx≤ −I⁺φ−I⁻Φ

and hence the obtained result (1.3) follows.

Theorem 1.2. Let f : [a, b] → R be a twice differentibale mapping on (a, b) with Γ₂ = sup_x∈(a,b)f⁰⁰(x)<+∞andγ₂ = infx∈(a,b)f⁰⁰(x)>−∞, then we have the estimation

(1.4)

Z b a

f(x)dx−1

2(b−a)(f(a) +f(b)) + 1

12(b−a)²(f⁰(b)−f⁰(a))

≤ 1 36√

3(Γ2−γ2)(b−a)³, where the constant ₃₆¹^√₃ is the best one in the sense that it cannot be replaced by a smaller one.

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Proof. We choose in (1.3),h(x) =−¹₂(x−a)(b−x)andg(x) = f⁰⁰(x), we get 1

2 Z b

a

h(x)− 1 b−a

Z b a

h(y)dy

dx =

Z x2

x1

h(x) + 1

12(b−a)²

dx

= 1

36√

3(b−a)³, wherex1 =a+ ³⁻

√ 3

6 (b−a)andx2 =a+³⁺

√ 3

6 (b−a). Thus from (1.3), we derive

Z b a

f(x)dx− 1

2(b−a)(f(a) +f(b)) + 1

12(b−a)²(f⁰(b)−f⁰(a))

=

Z b a

−1

2(x−a)(b−x)f⁰⁰(x)dx− 1 b−a

Z b a

−1

2(x−a)(b−x)dx Z b

a

f⁰⁰(x)dx

≤ 1 36√

3(Γ₂−γ₂)(b−a)³. To explain the best constant ₃₆¹^√₃ in the inequality (1.4), we can construct the functionf(x) = Rx

a

Ry

a j(z)dz

dyto attain the inequality in (1.4),

j(x) =











γ₂, a ≤x < x₁ =a+ 3−√ 3

6 (b−a), Γ₂, x₁ ≤x < x₂ =a+3 +√

3

6 (b−a), γ₂, x₂ ≤x≤b.

The proof is complete.

Theorem 1.3. Let f : [a, b] → Rbe a third-order differentibale mapping on (a, b)withΓ₃ = sup_x∈(a,b)f⁰⁰⁰(x)<+∞andγ₃ = infx∈(a,b)f⁰⁰⁰(x)>−∞, then we have the estimation

(1.5)

Z b a

f(x)dx−1

2(b−a)(f(a) +f(b)) + 1

12(b−a)²(f⁰(b)−f⁰(a))

≤ 1

384(Γ₃−γ₃)(b−a)⁴, where the constant ₃₈₄¹ is the best one in the sense that it cannot be replaced by a smaller one.

Proof. We choose in (1.3),h(x) = ₁₂¹(x−a)(2x−a−b)(b−x),g(x) =f⁰⁰⁰(x), to get 1

2 Z b

a

|h(x)− 1 b−a

Z b a

h(y)dy|dx= Z ^a+b₂

a

h(x)dx= 1

384(b−a)⁴,

Thus from (1.3) in Theorem 1.1, we can derive the inequality (1.5) immediately. Finally, we construct the functionf(x) =Rx

a

Ry a

Rz

j(s)ds dz

dy, wherej(x) = Γ₃ fora ≤ x < ^a+b₂ andj(x) = γ₃ for ^a+b₂ ≤x≤b, then the equality holds in (1.5).

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Theorem 1.4. Let f : [a, b] → R be a fourth-order differentibale mapping on (a, b) with M₄ = sup_x∈(a,b)|f⁽⁴⁾(x)|<+∞, then

(1.6)

Z b a

f(x)dx−1

2(b−a)(f(a) +f(b)) + 1

12(b−a)²(f⁰(b)−f⁰(a))

≤ 1

720M₄(b−a)⁵, where ₇₂₀¹ is the best possible constant.

Proof. We may write the remainder of the perturbed trapezoid inequality in the kernel form (1.7)

Z b a

f(x)dx−1

2(b−a)(f(a) +f(b)) + 1

12(b−a)²(f⁰(b)−f⁰(a)) = Z b

a

f⁽⁴⁾(x)k4(x)dx, wherek₄(x) = ₂₄¹(x−a)²(b−x)². Then we get

(1.8)

Z b a

|k₄(x)|dx= 1 24

Z 1 0

x²(1−x)²dx= 1 720.

Then (1.7) – (1.8) imply (1.6). The equality holds forf(x) = x⁴,a ≤x≤bin inequality (1.6).

Remark 1.5. We also can prove Theorem 1.2 and 1.3 in the kernel form

(1.9) Z b

a

f(x)dx−1

2(b−a)(f(a) +f(b)) + 1

12(b−a)²(f⁰(b)−f⁰(a)) = Z b

a

f⁽ⁿ⁾(x)k_n(x)dx, wherek₂(x) = −¹₂(x−a)(b−x) + ₁₂¹ andk₃(x) = ₁₂¹ (x−a)(2x−a−b)(b−x). By the formula (1.7) and (1.9), and derive the perturbed trapezoid inequality for different norms as shown in [5].

Now we present the composite perturbed trapezoid quadrature for an equidistant partitioning of interval[a, b]intonsubintervals. Applying Theorems 1.2 – 1.4, we obtain

Z b a

f(x)dx=T_n(f) +R_n(f), where

T_n(f) = b−a 2n

n−1

X

i=0

f

a+ib−a n

+f

a+ (i+ 1)b−a n

− (b−a)²

12n² (f⁰(b)−f⁰(a)), and the remainderR_n(f)satisfies the error estimate

(1.10) |R_n(f)| ≤











(b−a)³ 36√

3n²(Γ2−γ2), ifγ2 ≤f⁰⁰(x)≤Γ2, ∀x∈(a, b), (b−a)⁴

384n³ (Γ3−γ3), ifγ3 ≤f⁰⁰⁰(x)≤Γ3, ∀x∈(a, b) (b−a)⁵

720n⁴ M₄, if|f⁽⁴⁾(x)| ≤M₄, ∀x∈(a, b).

Then we can use (1.10) to get different error estimates of the composite perturbed trapezoid quadrature.

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As in [5], we may also apply the Theorems 1.2, 1.3 and 1.4 to special means. In this case we may improve some of the bounds related to inequalities about special means as given in [5, p.

492-494].

Furthermore, we discuss the Iyengar’s type inequality for the perturbed trapezoidal quadrature rule for functions whose first and second order derivatives are bounded. In [1, 3, 9, 10] they proved the following interesting inequality involving bounded derivatives.

Iff is a differentiable function on(a, b)and|f⁰(x)| ≤M₁, then

Z b a

f(x)dx− 1

2(b−a)(f(a) +f(b))

≤ M₁(b−a)²

4 −(f(b)−f(a))² 4M₁ . If|f⁰⁰(x)| ≤M₂, x∈[a, b]for positive constantM₂ ∈R, then

Z b a

f(x)dx− 1

2(b−a)(f(a) +f(b)) + 1

8(b−a)²(f⁰(b)−f⁰(a))

≤ M₂

24 (b−a)³− |∆|

M₂ 3!

,

Z b a

f(x)dx− 1

2(b−a)(f(a) +f(b)) + 1

8(b−a)²(f⁰(b)−f⁰(a))

≤ M₂

24(b−a)³− ∆²₁(b−a) 16M2

, where

(1.11) ∆ =f⁰(a)−2f⁰

a+b 2

+f⁰(b), ∆₁ =f⁰(a)−2(f(b)−f(a))

b−a +f⁰(b).

We will prove the following inequality.

Theorem 1.6. Letf :I →R, whereI ⊆Ris an interval. Suppose thatfis twice differentiable in the interior

◦

I of I, and let a, b ∈ I^◦ with a < b. If |f⁰⁰(x)| ≤ M₂, x ∈ [a, b] for positive constantM₂ ∈R. Then

(1.12)

Z b a

f(x)dx−1

2(b−a)(f(a) +f(b)) + 1

8(b−a)²(f⁰(b)−f⁰(a))

≤ 1

24M₂(b−a)³− s

|∆₁|³(b−a)³ 72M₂ , where∆₁ is defined as (1.11).

Proof. Denote J_f =

Z b a

f(x)dx− 1

2(b−a)(f(a) +f(b)) + 1

8(b−a)²(f⁰(b)−f⁰(a)).

It is easy to see that

J_f = Z b

a

1 2

x− a+b 2

2

f⁰⁰(x)dx, and

(1.13) ∆₁ =f⁰(a)−2(f(b)−f(a))

b−a +f⁰(b) = 1 b−a

Z b a

2

x− a+b 2

f⁰⁰(x)dx.

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For any|ε| ≤ ¹₈, we get for|f⁰⁰(x)| ≤M₂, x∈[a, b], J_f +ε(b−a)²∆₁ =

Z b a

1 2

x− a+b 2

2

+ 2ε(b−a)

x− a+b 2

!

f⁰⁰(x)dx

≤ F(ε)M₂(b−a)³, where

F(ε) = 1

(b−a)³ Z b

a

1 2

x− a+b 2

2

+ 2ε(b−a)

x−a+b 2

dx

= Z 1

0

1 2

x− 1

2 2

+ 2ε

x−1 2

dx.

For the case0≤ε≤ ¹₈, we have F(ε) =

Z 1 0

1 2

x− 1

2 2

+ 2ε

x− 1 2

dx

=

(Z ¹₂−4ε

0

1 2

x−1

2 2

+ 2ε

x− 1 2

! dx

− Z ¹₂

1 2−4ε

1 2

x− 1

2 2

+ 2ε

x− 1 2

! dx

+ Z 1

1 2

x− 1

2 2

+ 2ε

x− 1 2

! dx

)

= 1 24 +32

3ε³.

For the case−¹₈ ≤ε≤0, we have similarly F(ε) = 1

24 −32 3 ε³.

We can prove|∆₁| ≤ ¹₂(b−a)M₂easily from (1.13). Thus we choose the parameter ε∗ =sign(∆₁)

s |∆₁| 32(b−a)M2

, |ε∗| ≤ 1 8. By the above inequalities, we obtain

J_f ≤F(ε∗)M₂−ε∗(b−a)²∆₁ ≤ 1

24(b−a)³M₂− s

(b−a)³|∆₁|³ 72M₂ . Replacingf with−f, we have

J−f =−J_f ≤ 1

24(b−a)³M₂− s

(b−a)³|∆₁|³ 72M2

.

Thus we obtain bounds for|J_f|and prove the inequality (1.12).

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Remark 1.7. As|∆₁| ≤ ¹₂M₂(b−a), we have s

|∆₁| M₂ ≥

r 2 b−a

|∆₁| M₂ ,

Z b a

f(x)dx− 1

2(b−a)(f(a) +f(b)) + 1

8(b−a)²(f⁰(b)−f⁰(a))

≤ M₂

24(b−a)³− ∆²₁(b−a) 6M2

. For the casef⁰(a) =f⁰(b) = 0, we have

(1.14)

Z b a

f(x)dx−1

2(b−a)(f(a) +f(b))

≤ M2

24(b−a)³ −2 3

|f(b)−f(a)|² M₂(b−a) . The inequality (1.14) is sharper than that stated in [9, p. 69].

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