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volume 3, issue 2, article 26, 2002.

Received 17 September, 2001;;

accepted 23 January, 2001.

Communicated by:G. Anastassiou

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Journal of Inequalities in Pure and Applied Mathematics

SOME INTEGRAL INEQUALITIES INVOLVING TAYLOR’S REMAINDER

HILLEL GAUCHMAN

Department of Mathematics, Eastern Illinois University, Charleston, IL 61920, USA EMail:cfhvg@ux1.cts.eiu.edu

c

2000Victoria University ISSN (electronic): 1443-5756 068-01

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Some Integral Inequalities Involving Taylor’s Remainder. I

Hillel Gauchman

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J. Ineq. Pure and Appl. Math. 3(2) Art. 26, 2002

http://jipam.vu.edu.au

Abstract

In this paper, using Steffensen’s inequality we prove several inequalities involving Taylor’s remainder. Among the simplest particular cases we obtain Iyengar’s inequality and one of Hermite-Hadamard’s inequalities for convex functions.

2000 Mathematics Subject Classification:26D15.

Key words: Taylor’s remainder, Steffensen’s inequality, Iyengar’s inequality, Hermite- Hadamard’s inequality.

1. Introduction and Statement of Main Results

In this paper, using Steffensen’s inequality we prove several inequalities (The- orems 1.1 and1.2) involving Taylor’s remainder. In Sections 3and4 we give several applications of Theorems 1.1 and 1.2. Among the simplest particular cases we obtain Iyengar’s inequality and one of Hermite-Hadamard’s inequalities for convex functions. We prove Theorems1.1and1.2in Section2.

In what followsndenotes a non-negative integer, I ⊆ R is a generic inter- val, and I^◦ is the interior of I. We will denote by R_n,f(c, x)the nth Taylor’s remainder of functionf(x)with centerc, i.e.

Rn,f(c, x) = f(x)−

n

X

k=0

f^(k)(c)

k! (x−c)^k.

The following two theorems are the main results of the present paper.

Theorem 1.1. Letf :I → Randg :I → Rbe two mappings,a, b ∈I^◦ with a < b, and letf ∈Cⁿ⁺¹([a, b]),g ∈C([a, b]). Assume thatm ≤f⁽ⁿ⁺¹⁾(x)≤ M,m 6=M, andg(x)≥0for allx∈[a, b]. Set

λ= 1 M −m

f⁽ⁿ⁾(b)−f⁽ⁿ⁾(a)−m(b−a) . Then

(i) 1

(n+ 1)!

Z b

b−λ

(x−b+λ)ⁿ⁺¹g(x)dx

≤ 1 M −m

Z b

a

Rn,f(a, x)−m(x−a)ⁿ⁺¹ (n+ 1)!

g(x)dx

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≤ 1 (n+ 1)!

Z b

a

(x−a)ⁿ⁺¹−(x−a−λ)ⁿ⁺¹

g(x)dx + (−1)ⁿ⁺¹

(n+ 1)!

Z a+λ

a

(a+λ−x)ⁿ⁺¹g(x)dx;

and

1 (n+ 1)!

Z a+λ

a

(a+λ−x)ⁿ⁺¹g(x)dx (ii)

≤ (−1)ⁿ⁺¹ M −m

Z b

a

R_n,f(b, x)−m(x−b)ⁿ⁺¹ (n+ 1)!

g(x)dx

≤ 1 (n+ 1)!

Z b

a

(b−x)ⁿ⁺¹−(b−λ−x)ⁿ⁺¹

g(x)dx +(−1)ⁿ⁺¹

(n+ 1)!

Z b

b−λ

(x−b+λ)ⁿ⁺¹g(x)dx.

Theorem 1.2. Let f : I → R and g : I → R be two mappings, a, b ∈ I^◦ with a < b, and let f ∈ Cⁿ⁺¹([a, b]), g ∈ C([a, b]). Assume thatfⁿ⁺¹(x)is increasing on[a, b]andm≤g(x)≤M,m 6=M, for allx∈[a, b]. Set

λ₁ = 1

(M −m)(b−a)ⁿ⁺¹ Z b

a

(x−a)ⁿ⁺¹g(x)dx− m

M −m · b−a n+ 2,

λ₂ = 1

(M −m)(b−a)ⁿ⁺¹ Z b

a

(b−x)ⁿ⁺¹g(x)dx− m

M −m · b−a n+ 2.

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Then

f⁽ⁿ⁾(a−λ₁)−f⁽ⁿ⁾(a) (i)

≤ (n+ 1)!

(M −m)(b−a)ⁿ⁺¹ Z b

a

R_n,f(a, x)(g(x)−m)dx

≤f⁽ⁿ⁾(b)−f⁽ⁿ⁾(b−λ₁);

and

f⁽ⁿ⁾(a+λ₂)−f⁽ⁿ⁾(a) (ii)

≤(−1)ⁿ⁺¹ (n+ 1)!

(M −m)(b−a)ⁿ⁺¹ Z b

a

R_n,f(b, x) (g(x)−m)dx

≤f⁽ⁿ⁾(b)−f⁽ⁿ⁾(b−λ₂).

Remark 1.1. It is easy to verify that the inequalities in Theorems 1.1 and1.2 become equalities iff(x)is a polynomial of degree≤n+ 1.

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2. Proofs of Theorems 1.1 and 1.2

The following is well-known Steffensen’s inequality:

Theorem 2.1. [4]. Suppose the f and g are integrable functions defined on (a, b), f is decreasing and for each x ∈ (a, b), 0 ≤ g(x) ≤ 1. Set λ = Rb

a g(x)dx. Then Z b

b−λ

f(x)dx≤ Z b

a

f(x)g(x)dx≤ Z a+λ

a

f(x)dx.

Proposition 2.2. Letf : I → Randg : I → Rbe two maps, a, b ∈ I^◦ with a < band letf ∈Cⁿ⁺¹([a, b]),g ∈C[a, b]. Assume that0≤f⁽ⁿ⁺¹⁾(x)≤1for all x ∈ [a, b]and Rb

x(t−x)ⁿg(t)dtis a decreasing function of xon [a, b]. Set λ =f⁽ⁿ⁾(b)−f⁽ⁿ⁾(a). Then

1 (n+ 1)!

Z b

b−λ

(x−b+λ)ⁿ⁺¹g(x)dx (2.1)

≤ Z b

a

R_n,f(a, x)g(x)dx

≤ 1 (n+ 1)!

Z b

a

(x−a)ⁿ⁺¹−(x−a−λ)ⁿ⁺¹

g(x)dx +(−1)ⁿ⁺¹

(n+ 1)!

Z a+λ

a

(a+λ−x)ⁿ⁺¹g(x)dx.

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Proof. Set

F_n(x) = 1 n!

Z b

x

(t−x)ⁿg(t)dt, G_n(x) = fⁿ⁺¹(x),

λ = Z b

a

G_n(x)dx=f⁽ⁿ⁾(b)−f⁽ⁿ⁾(a).

ThenF_n(x),G_n(x), andλsatisfy the conditions of Theorem2.1. Therefore (2.2)

Z b

b−λ

Fn(x)dx≤ Z b

a

Fn(x)Gn(x)dx≤ Z a+λ

a

Fn(x)dx.

It is easy to see thatF_n⁰(x) = −Fn−1(x). Hence Z b

a

F_n(x)G_n(x)dx= Z b

a

F_n(x)df⁽ⁿ⁾(x)

=f⁽ⁿ⁾(x)Fn(x)

b

a

+ Z b

a

f⁽ⁿ⁾(x)Fn−1(x)dx

=−f⁽ⁿ⁾(a) n!

Z b

a

(x−a)ⁿg(x)dx+ Z b

a

Fn−1(x)Gn−1(x)dx

=−f⁽ⁿ⁾(a) n!

Z b

a

(x−a)ⁿg(x)dx

− f⁽ⁿ⁻¹⁾(a) (n−1)!

Z b

a

(x−a)ⁿ⁻¹g(x)dx+ Z b

a

Fn−2(x)Gn−2(x)dx

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=. . .

=−f⁽ⁿ⁾(a) n!

Z b

a

(x−a)ⁿg(x)dx− f⁽ⁿ⁻¹⁾(a) (n−1)!

Z b

a

(x−a)ⁿ⁻¹g(x)dx

− · · · −f(a) Z b

a

g(x)dx+ Z b

a

f(x)g(x)dx.

Thus (2.3)

Z b

a

F_n(x)G_n(x)dx= Z b

a

R_n,f(a, x)g(x)dx.

In addition

Z a+λ

a

F_n(x)dx= 1 n!

Z a+λ

a

Z b

x

(t−x)ⁿg(t)dt

dx.

Changing the order of integration, we obtain Z a+λ

a

F_n(x)dx

= 1 n!

Z a+λ

a

Z t

a

(t−x)ⁿg(t)dx

dt+ 1 n!

Z b

a+λ

Z a+λ

a

(t−x)ⁿg(t)dx

dt

=−1 n!

Z a+λ

a

g(t)(t−x)ⁿ⁺¹ n+ 1

x=t

x=a

dt− 1 n!

Z b

a+λ

g(t)(t−x)ⁿ⁺¹ n+ 1

x=a+λ

x=a

dt

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= 1

(n+ 1)!

Z a+λ

a

(t−a)ⁿ⁺¹g(t)dt

− 1 (n+ 1)!

Z b

a+λ

(t−a−λ)ⁿ⁺¹−(t−a)ⁿ⁺¹ g(t)dt

= 1

(n+ 1)!

Z b

a

(t−a)ⁿ⁺¹g(t)dt− 1 (n+ 1)!

Z b

a

(t−a−λ)ⁿ⁺¹g(t)dt

+ 1

(n+ 1)!

Z a+λ

a

(t−a−λ)ⁿ⁺¹g(t)dt.

Thus, (2.4)

Z a+λ

a

Fn(x)dx= 1 (n+ 1)!

Z b

a

(x−a)ⁿ⁺¹−(x−a−λ)ⁿ⁺¹

g(x)dx + (−1)ⁿ⁺¹

(n+ 1)!

Z a+λ

a

Similarly we obtain (2.5)

Z b

b−λ

F_n(x)dx= 1 (n+ 1)!

Z b

b−λ

(x−b+λ)ⁿ⁺¹g(x)dx Substituting (2.3), (2.4), and (2.5) into (2.2), we obtain (2.1).

Proposition 2.3. Letf : I → Randg : I → Rbe two maps, a, b ∈ I^◦ with a < b and letf ∈ Cⁿ⁺¹([a, b]), g ∈ C([a, b]). Assume that m ≤ f⁽ⁿ⁺¹⁾(x)≤

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M for allx∈[a, b]andRb

x(t−x)ⁿg(t)dtis a decreasing function ofxon[a, b].

Setλ= _M_−m¹

f⁽ⁿ⁾(b)−f⁽ⁿ⁾(a)−m(b−a) . Then

1 (n+ 1)!

Z b

b−λ

(x−b+λ)ⁿ⁺¹g(x)dx (2.6)

≤ 1 M−m

Z b

a

R_n,f(a, x)−m(x−a)ⁿ⁺¹ (n+ 1)!

g(x)dx

≤ 1 (n+ 1)!

Z b

a

(x−a)ⁿ⁺¹−(x−a−λⁿ⁺¹)

g(x)dx +(−1)ⁿ⁺¹

(n+ 1)!

Z a+λ

a

Proof. Set

f˜(x) = 1 M−m

f(x)−m(x−a)ⁿ⁺¹ (n+ 1)!

. Then0≤f˜⁽ⁿ⁺¹⁾(x)≤1and

λ = 1 M −m

f⁽ⁿ⁾(b)−f⁽ⁿ⁾(a)−m(b−a)

= ˜f⁽ⁿ⁾(b)−f˜⁽ⁿ⁾(a).

Hencef˜(x), g(x), andλsatisfy the conditions of Proposition2.2. Substituting f˜(x)instead off(x)into (2.1), we obtain (2.6).

Proof of Theorem1.1(i). If g(x) ≥ 0for all x ∈ [a, b], then Rb

x(t−x)ⁿg(t)dt is a decreasing function ofxon[a, b]. Hence Proposition2.3implies Theorem 1.1(i).

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Proof of Theorems1.1(ii),1.2(i), and1.2(ii). Proofs of Theorems1.1(ii),1.2(i), and 1.2(ii) are similar to the above proof of Theorem 1.1(i). For the proof of Theorem1.1(ii) we take

F_n(x) = −1 n!

Z x

a

(x−t)ⁿg(t)dt, G_n(x) = fⁿ⁺¹(x).

For the proof of Theorem1.2(i) we take

F_n(x) =−f⁽ⁿ⁺¹⁾(x), G_n(x) = 1 n!

Z b

x

(t−x)ⁿg(t)dt.

For the proof of Theorem1.2(ii) we take Fn(x) = −f⁽ⁿ⁺¹⁾(x), Gn(x) = 1

n!

Z x

a

(x−t)ⁿg(t)dt.

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3. Applications of Theorem 1.1

Theorem 3.1. Let f : I → R be a mapping, a, b ∈ I^◦ with a < b, and let f ∈Cⁿ⁺¹([a, b]). Assume thatm≤f⁽ⁿ⁺¹⁾(x)≤M,m 6=M, for allx∈[a, b].

Set

λ= 1 M −m

1 (n+ 2)!

m(b−a)ⁿ⁺²+ (M −m)λⁿ⁺² (i)

≤ Z b

a

R_n,f(a, x)dx

≤ 1 (n+ 2)!

M(b−a)ⁿ⁺²−(M −m)(b−a−λ)ⁿ⁺²

;

and

1 (n+ 2)!

m(b−a)ⁿ⁺²+ (M −m)λⁿ⁺² (ii)

≤(−1)ⁿ⁺¹ Z b

a

R_n,f(b, x)dx

≤ 1 (n+ 2)!

M(b−a)ⁿ⁺²−(M −m)(b−a−λ)ⁿ⁺² . Proof. Takeg(x)≡1on[a, b]in Theorem1.1.

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Two inequalities of the form A ≤ X ≤ B and A ≤ Y ≤ B imply two new inequalities A ≤ ¹₂(X +Y) ≤ B and|X −Y| ≤ B −A. Applying this construction to inequalities (i) and (ii) of Theorem3.1, we obtain the following two more symmetric with respect toaandbinequalities:

Theorem 3.2. Let f : I → R be a mapping, a, b ∈ I^◦ with a < b, and let f ∈ Cⁿ⁺¹([a, b]). Assume thatm ≤fⁿ⁺¹(x)≤M,m 6=M, for allx∈ [a, b].

Set

λ= 1 M −m

1

(n+ 2)![m(b−a)ⁿ⁺²+ (M −m)λⁿ⁺²] (i)

≤ Z b

a

1 2

R_n,f(a, x) + (−1)ⁿ⁺¹R_n,f(b, x) dx

≤ 1 (n+ 2)!

M(b−a)ⁿ⁺²−(M −m)(b−a−λ)ⁿ⁺²

;

and (ii)

Z b

a

[Rn,f(a, x) + (−1)ⁿRn,f(b, x)]dx

≤ M −m (n+ 2)!

(b−a)ⁿ⁺²−λⁿ⁺²−(b−a−λ)ⁿ⁺² . We now consider the simplest cases of inequalities (i) and (ii) of Theorem 3.2, namely the cases whenn = 0or 1.

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Case 3.1. n = 0

Inequality (i) of Theorem3.2forn = 0gives us the following result.

Theorem 3.3. Let f : I → R be a mapping, a, b ∈ I^◦ with a < b and let f ∈C¹([a, b]). Assume thatm ≤f⁰(x)≤M,m6=M, for allx∈[a, b]. Set

λ= 1

M −m[f(b)−f(a)−m(b−a)]. Then

m+(M −m)λ²

(b−a)² ≤ f(b)−f(a)

b−a ≤M −(M −m)(b−a−λ)² (b−a)² . Remark 3.1. Theorem 3.3 is an improvement of a trivial inequality m ≤

f(b)−f(a) b−a ≤M.

Forn = 0, inequality (ii) of Theorem3.2gives the following result:

Theorem 3.4. Let f : I → R be a mapping, a, b ∈ I^◦ with a < b, and let f ∈C¹([a, b]). Assume thatm ≤f⁰(x)≤M,m6=M for allx∈[a, b]. Then

Z b

a

f(x)dx−f(a) +f(b)

2 (b−a)

≤ [f(b)−f(a)−m(b−a)] [M(b−a)−f(b) +f(a)]

2(M −m) .

Theorem 3.4 is a modification of Iyengar’s inequality due to Agarwal and Dragomir [1]. If |f⁰(x)| ≤ M, then taking m = −M in Theorem 3.4, we

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obtain

Z b

a

f(x)dx− f(a) +f(b)

2 (b−a)

≤ M(b−a)²

4 − 1

4M [f(b)−f(a)]². This is the original Iyengar’s inequality [2]. Thus, inequality (ii) of Theorem 3.2can be considered as a generalization of Iyengar’s inequality.

Case 3.2. n = 1

In the casen= 1, inequality (i) of Theorem3.2gives us the following result:

Theorem 3.5. Let f : I → R be a mapping, a, b ∈ I^◦ with a < b and let f ∈C²([a, b]). Assume thatm ≤f⁰⁰(x)≤M,m 6=M, for allx∈[a, b]. Set

λ = 1

M −m[f⁰(b)−f⁰(a)−m(b−a)]. Then

1 6

m(b−a)³+ (M −m)λ³

≤ Z b

a

f(x)dx− f(a) +f(b)

2 (b−a) + f⁰(b)−f⁰(a)

4 (b−a)²

≤ 1 6

M(b−a)³−(M −m)(b−a−λ)³ .

In the casen= 1, inequality (ii) of Theorem3.2implies that iff ∈C²([a, b])

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andm≤f⁰⁰(x)≤M, then

f(b)−f(a)

b−a − f⁰(a) +f⁰(b) 2

≤ [f⁰(b)−f⁰(a)−m(b−a)] [M(b−a)−f⁰(b) +f⁰(a)]

2(b−a)(M −m) .

This result follows readily from Iyengar’s inequality if we take f⁰(x)instead of f(x)in Theorem3.4.

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4. Applications of Theorem 1.2

Take g(x) = M on[a, b]in Theorem1.2. Thenλ₁ = λ₂ = _n+2^b−a and Theorem 1.2implies

Theorem 4.1. Let f : I → R be a mapping, a, b ∈ I^◦ with a < b, and let f ∈Cⁿ⁺¹([a, b]). Assume thatfⁿ⁺¹(x)is increasing on[a, b]. Then

(i) (b−a)ⁿ⁺¹ (n+ 1)!

f⁽ⁿ⁾

a+ b−a n+ 2

−f⁽ⁿ⁾(a)

≤ Z b

a

R_n,f(a, x)dx≤ (b−a)ⁿ⁺¹ (n+ 1)!

f⁽ⁿ⁾(b)−f⁽ⁿ⁾

b− b−a n+ 2

;

and

(b−a)ⁿ⁺¹ (n+ 1)!

f⁽ⁿ⁾

a+ b−a n+ 2

−f⁽ⁿ⁾(a) (ii)

≤(−1)ⁿ⁺¹ Z b

a

R_n,f(b, x)dx

≤ (b−a)ⁿ⁺¹ (n+ 1)!

b− b−a n+ 2

.

The next theorem follows from Theorem 4.1 in exactly the same way as Theorem3.2follows from Theorem3.1.

Theorem 4.2. Let f : I → R be a mapping, a, b ∈ I^◦ with a < b, and let

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f ∈Cⁿ⁺¹([a, b]). Assume thatfⁿ⁺¹(x)is increasing on[a, b]. Then (b−a)ⁿ⁺¹

(n+ 1)!

f⁽ⁿ⁾

a+ b−a n+ 2

−f⁽ⁿ⁾(a) (i)

≤ 1 2

Z b

a

R_n,f(a, x) + (−1)ⁿ⁺¹R_n,f(b, x) dx

≤ (b−a)ⁿ⁺¹ (n+ 1)!

b− b−a n+ 2

;

(ii)

Z b

a

[Rn,f(a, x) + (−1)ⁿRn,f(b, x)]dx

≤ (b−a)ⁿ⁺¹ (n+ 1)!

b− b−a n+ 2

−f⁽ⁿ⁾

a+ b−a n+ 2

+f⁽ⁿ⁾(a)

. We now consider inequalities (i) and (ii) of Theorem4.2in the simplest cases whenn= 0or1.

Case 4.1. n = 0.

Inequality (i) of Theorem4.2gives a trivial fact: Iff⁰(x)increases thenf ^a+b₂

≤

f(a)+f(b)

2 . Inequality (ii) of Theorem 4.2 gives the following result: Iff⁰(x)is increasing, then

(4.1) f

a+b 2

≤ 1 b−a

Z b

a

f(x)dx ≤f(a) +f(b)−f

a+b 2

.

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The left inequality (2.1) is a half of the famous Hermite-Hadamard’s inequality [3]: Iff(x)is convex, then

(4.2) f

a+b 2

≤ 1 b−a

Z b

a

f(x)dx≤ f(a) +f(b)

2 .

Note that the right inequality (4.1) is weaker than the right inequality (4.2).

Thus, inequality (ii) of Theorem 4.2 can be considered as a generalization of the Hermite-Hadamard’s inequality f ^a+b₂

≤ _b−a¹ Rb

a f(x)dx, where f(x)is convex.

Case 4.2. n = 1.

In this case Theorem4.2implies the following two results:

Theorem 4.3. Let f : I → R be a mapping, a, b ∈ I^◦ with a < b, and let f ∈C²([a, b]). Assume thatf⁰⁰(x)is increasing on[a, b]. Then

(b−a)² 2

f⁰

a+b−a 3

+f⁰(a) (i)

≤ Z b

a

f(x)dx−f(a) +f(b)

2 (b−a) + f⁰(b)−f⁰(a)

4 (b−a)²

≤ (b−a)² 2

f⁰(b)−f⁰

b−a 3

;

(ii)

f(b)−f(a)

b−a − f⁰(a) +f⁰(b) 2

≤ 1 2

f⁰(a)−f⁰

a+b−a 3

−f⁰

b− b−a 3

+f⁰(b)

.

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References

[1] R.P. AGARWAL AND S.S. DRAGOMIR, An application of Hayashi’s inequality for differentiable functions, Computers Math. Applic., 32(6) (1996), 95–99.

[2] K.S.K. IYENGAR, Note on an inequality, Math. Student, 6 (1938), 75–76.

[3] J.E. PE ˇCARI ´C, F. PROSCHANANDY.L. TONG, Convex functions, Partial Orderings, and Statistical Applications, Acad. Press, 1992.

[4] J.F. STEFFENSEN, On certain inequalities and methods of approximation, J. Inst. Actuaries, 51 (1919), 274–297.