View of Some estimations for the Taylor's remainder

Rev. Anal. Num´er. Th´eor. Approx., vol. 42 (2013) no. 2, pp. 161–169 ictp.acad.ro/jnaat

SOME ESTIMATIONS FOR THE TAYLOR’S REMAINDER^‡

HUI SUN,^∗ BO-YONG LONG^† and YU-MING CHU^†

Abstract. In this paper, we establish several integral inequalities for the Tay- lor’s remainder by Gr¨uss and Cheyshev inequalities.

MSC 2000. 26D15.

Keywords. Taylor remainder, Gr¨uss inequality, Cheyshev inequality.

1. INTRODUCTION

For two given integrable functions f and g on [a, b], the Chebychev func- tionalT(f, g) is defined by

T(f, g) = _b−a¹ Z b

a

f(x)g(x)dx−_b−a¹ Z b

a

f(x)dx·_b−a¹ Z b

a

g(x)dx.

In 1935, Gr¨uss [1] proved that

|T(f, g)| ≤ ¹₄(M−m)(L−l) (1.1)

if

m≤f(x)≤M, l≤g(x)≤L

for all x∈[a, b], whereM, m, L and lare constants. Inequality (1.1) is called Gr¨uss inequality.

The well-known Chebyshev inequality [2] can be stated as follows: if both f and g are increasing or decreasing, then

T(f, g)≥0.

If one of the functionsf and g is increasing and the other decreasing, then the above inequality is reversed.

∗ School of Mathematics and Computation Science, Hunan City University, Yiyang 413000, Hunan, China, e-mail: [email protected].

†Department of Mathematics, Huzhou Teachers College, Huzhou 313000, Zhejiang, China, e-mail: [email protected],[email protected].

‡ The work of the first author has been supported by the Natural Science Foundation of the Department of Education of Hunan Province (Grant No. 13C127) and the work of the last author has been supported by the Natural Science Foundation of China (Grant No.

61374086).

In what followsndenotes a non-negative integer. We denote byR_n,f(x0, x) thenth Taylor remainder of the function f(x) with center x₀, i.e.

R_n,f(x0, x) =f(x)−

n

X

k=0

(x−x₀)^k

k! f^(k)(x0).

The Taylor remainder has been the subject of intensive research [3]–[9].

In particular, many remarkable integral inequalities for the Taylor remainder can be found in the literature [5]–[7]. The following Theorems A and B were proved by Gauchman in [6].

Theorem A. Let f(x) be a function defined on [a, b] such that f(x) ∈ Cⁿ⁺¹[a, b] and m ≤f⁽ⁿ⁺¹⁾(x) ≤ M for each x ∈ [a, b], where m and M are constants. Then

Z b a

Rn,f(a, x)dx−^f(n)(b)−f_(n+2)!^(n)(a)(b−a)ⁿ⁺¹

≤ ^(b−a)_4(n+1)!ⁿ⁺²(M −m) and

(−1)ⁿ⁺¹ Z b

a

R_n,f(b, x)dx−^f(n)(b)−f_(n+2)!^(n)(a)(b−a)ⁿ⁺¹

≤ ^(b−a)_4(n+1)!ⁿ⁺²(M−m).

Theorem B. Let f(x) be a function defined on [a, b] such that f(x) ∈ Cⁿ⁺¹[a, b]. Iff⁽ⁿ⁺¹⁾(x) is increasing on [a, b], then

−^f(n+1)_4(n+1)!^{(b)−f(n+1)(a)}(b−a)ⁿ⁺² ≤

≤ Z b

a

R_n,f(a, x)dx− ^f(n)(b)−f_(n+2)!^(n)(a)(b−a)ⁿ⁺¹≤0 and

0≤(−1)ⁿ⁺¹ Z b

a

R_n,f(b, x)dx−^f(n)(b)−f_(n+2)!^(n)(a)(b−a)ⁿ⁺¹

≤ ^f(n+1)_4(n+1)!^{(b)−f(n+1)(a)}(b−a)ⁿ⁺². If f⁽ⁿ⁺¹⁾(x) is decreasing on [a, b], then

0≤ Z b

a

Rn,f(a, x)dx−^f(n)(b)−f_(n+2)!^(n)(a)(b−a)ⁿ⁺¹≤

≤ ^{f(n+1)(a)−f}_4(n+1)!^(n+1)(b)(b−a)ⁿ⁺² and

−^f(n+1)_4(n+1)!^{(a)−f(n+1)(b)}(b−a)ⁿ⁺² ≤

≤(−1)ⁿ⁺¹ Z b

a

Rn,f(b, x)dx−^f(n)(b)−f_(n+2)!^(n)(a)(b−a)ⁿ⁺¹ ≤0.

It is the aim of this paper to establish several new inequalities for the Taylor remainder by Gr¨uss and Cheyshev inequalities.

2. MAIN RESULTS

Lemma 2.1. Let f(x) be a function defined on [a, b] and x₀ ∈ (a, b). If f(x)∈Cⁿ⁺¹[a, b], then

Z b x0

Rn,f(x0, x)dx= Z b

x0

(b−x)ⁿ⁺¹

(n+1)! f⁽ⁿ⁺¹⁾(x)dx (2.1)

and

Z x0

a

R_n,f(x0, x)dx= Z x0

a

(a−x)ⁿ⁺¹

(n+1)! f⁽ⁿ⁺¹⁾(x)dx.

(2.2)

Proof. We only give the proof of (2.1) in detail, the similar argument leads to (2.2). It follows from the formula of integration by parts that

Z b x0

(b−x)ⁿ⁺¹

(n+1)! f⁽ⁿ⁺¹⁾(x)dx=

= ^(b−x)_(n+1)!ⁿ⁺¹f⁽ⁿ⁾(x)|^b_x0 + Z b

x0

(b−x)ⁿ

n! f⁽ⁿ⁾(x)dx

=−^(b−x_(n+1)!⁰⁾ⁿ⁺¹f⁽ⁿ⁾(x₀) +^(b−x)_n! ⁿf⁽ⁿ⁻¹⁾(x)|^b_x

0 + Z b

x0

(b−x)ⁿ⁻¹

(n−1)! f⁽ⁿ⁻¹⁾(x)dx

=−^(b−x_(n+1)!⁰⁾ⁿ⁺¹f⁽ⁿ⁾(x0)−^(b−x_n!⁰⁾ⁿf⁽ⁿ⁻¹⁾(x0)− · · · −(b−x0)f(x0) + Z b

x0

f(x)dx

= Z _b

x0

[f(x)−

n

X

k=0

f^(k)(x0)

k! (x−x0)^k]dx

= Z b

x0

R_n,f(x0, x)dx.

Theorem 2.2. Let f(x) ∈Cⁿ⁺¹[a, b], such that m₁ ≤f⁽ⁿ⁺¹⁾(x) ≤M₁ for x ∈[a, x0] and m2 ≤f⁽ⁿ⁺¹⁾(x) ≤M2 for x ∈ [x0, b], where m1, m2, M1 and M₂ are constants. Then

Z b a

Rn,f(x0, x)dx−^f(n)(b)−f_(n+2)!^(n)(x0)(b−x0)ⁿ⁺¹ (2.3)

−^f(n)(x_(n+2)!^0)−f(n)(a)(a−x0)ⁿ⁺¹

≤ ^(x_4(n+1)!^0−a)n+2(M1−m1) +^(b−x_4(n+1)!⁰⁾ⁿ⁺²(M2−m2).

Proof. Let

F(x) =f⁽ⁿ⁺¹⁾(x), G(x) = ^(b−x)_(n+1)!ⁿ⁺¹.

Then for any x∈[x0, b], we clearly see that

m2≤F(x)≤M2, 0≤G(x)≤ ^(b−x_(n+1)!⁰⁾ⁿ⁺¹. Making use of Gr¨uss inequality (1.1) one has

Z b x0

F(x)G(x)dx−_b−x¹

0

Z b x0

F(x)dx Z b

x0

G(x)dx = (2.4)

=

Z b x0

(b−x)ⁿ⁺¹

(n+1)! f⁽ⁿ⁺¹⁾(x)dx−_b−x¹

0

Z b x0

(b−x)ⁿ⁺¹ (n+1)! dx

Z b x0

f⁽ⁿ⁺¹⁾(x)dx

≤ ^(b−x_4(n+1)!⁰⁾ⁿ⁺²(M2−m2).

Equation (2.1) and inequality (2.4) lead to the conclusion that

Z b x0

R_n,f(x₀, x)dx−^f(n)(b)−f_(n+2)!^(n)(x0)(b−x₀)ⁿ⁺¹ ≤ (2.5)

≤ ^(b−x_4(n+1)!⁰⁾ⁿ⁺²(M₂−m₂).

Similarly, if x∈[a, x₀], then Gr¨uss inequality (1.1) leads to

Z x0

a

(a−x)ⁿ⁺¹

(n+1)! f⁽ⁿ⁺¹⁾(x)dx (2.6)

−_x¹

0−a

Z x0

a

(a−x)ⁿ⁺¹ (n+1)! dx

Z x0

a

f⁽ⁿ⁺¹⁾(x)dx ≤

≤ ^(x_4(n+1)!^0−a)n+2(M1−m1).

Equation (2.2) and inequality (2.6) imply that

Z _x0

a

R_n,f(x₀, x)dx−^f(n)(x_(n+2)!^0)−f(n)(a)(a−x₀)ⁿ⁺¹ ≤ (2.7)

≤ ^(x_4(n+1)!^0−a)n+2(M₁−m₁).

Therefore, inequality (2.3) follows form inequalities (2.5) and (2.7).

If take n= 1 in Theorem 2.2, then we have

Corollary 2.3. Let f(x) ∈ C²[a, b] and m1 ≤ f⁽²⁾(x) ≤ M1 for any x∈[a, x₀],m₂≤f⁽²⁾(x)≤M₂ for anyx∈[x₀, b], wherem₁, m₂,M₁ and M₂ are constants. Then

Z b a

f(x)dx−(b−a)f(x0)−^f0₆^(b)(b−x0)²+^f0(a)₆ (a−x0)²

−^f0(x₃⁰⁾(a+b−2x₀)(b−a) ≤

≤ ^(x0−a)₈ ³(M₁−m₁) +^(b−x₈⁰⁾³(M₂−m₂).

In particular, if x₀ = ^a+b₂ , then Corollary 2.3 becomes

Corollary 2.4. Let f(x) ∈ C²[a, b] and m1 ≤ f⁽²⁾(x) ≤ M1 for any x∈[a,^a+b₂ ], m₂ ≤f⁽²⁾(x)≤M₂ for any x ∈[^a+b₂ , b], wherem₁, m₂, M₁ and M2 are constants. Then

1 b−a

Z b a

f(x)dx−f(^a+b₂ )−₂₄¹(b−a)(f⁰(b)−f⁰(a)) ≤

≤ ^(b−a)₆₄ ²(M1−m1+M2−m2).

If take n= 0 in Theorem 2.2, then we have

Corollary 2.5. Let f(x) ∈ C¹[a, b] and m₁ ≤ f⁰(x) ≤ M₁ for any x ∈ [a, x0], m2 ≤f⁰(x)≤ M2 for any x ∈[x0, b], where m1, m2, M1 and M2 are constants. Then

Z b a

f(x)dx−(b−a)f(x₀)− ^f(b)−f₂ ^(x0)(b−x₀)− ^f(x0)−f₂ ^(a)(a−x₀) ≤

≤ ^(x0−a)₄ ²(M₁−m₁) + ^(b−x₄⁰⁾²(M₂−m₂).

In particular, if x₀ = ^a+b₂ , then Corollary 2.5 becomes

Corollary 2.6. Let f(x) ∈ C¹[a, b] and m1 ≤ f⁰(x) ≤ M1 for any x ∈ [a,^a+b₂ ], m2 ≤ f⁰(x) ≤ M2 for any x ∈ [^a+b₂ , b], where m1, m2, M1 and M2

are constants. Then

1 b−a

Z b a

f(x)dx−¹₂f(^a+b₂ )−^f(a)+f(b)₄ ≤

≤ ₁₆¹(b−a)(M1+M2−m1−m2).

Theorem 2.7. Let f(x) ∈ Cⁿ⁺¹[a, b] and x0 ∈ [a, b], then the following statements are true:

(1) If nis an odd number and f⁽ⁿ⁺¹⁾(x) is increasing in [a, b], then

(x0−a)ⁿ⁺²

4(n+1)! (f⁽ⁿ⁺¹⁾(x₀)−f⁽ⁿ⁺¹⁾(a))≥ (2.8)

≥ Z b

a

R_n,f(x₀, x)dx−^f(n)(x_(n+2)!^0)−f(n)(a)(a−x₀)ⁿ⁺¹

−^f(n)(b)−f_(n+2)!^(n)(x0)(b−x₀)ⁿ⁺¹

≥ −^(b−x_4(n+1)!⁰⁾ⁿ⁺²(f⁽ⁿ⁺¹⁾(b)−f⁽ⁿ⁺¹⁾(x0));

(2) If nis an odd number and f⁽ⁿ⁺¹⁾(x) is decreasing in[a, b], then

−^(x_4(n+1)!^0−a)n+2(f⁽ⁿ⁺¹⁾(a)−f⁽ⁿ⁺¹⁾(x₀))≤ (2.9)

≤ Z _b

a

R_n,f(x₀, x)dx−^f(n)(x_(n+2)!^0)−f(n)(a)(a−x₀)ⁿ⁺¹

−^f(n)(b)−f_(n+2)!^(n)(x0)(b−x₀)ⁿ⁺¹

≤ ^(b−x_4(n+1)!⁰⁾ⁿ⁺²(f⁽ⁿ⁺¹⁾(x0)−f⁽ⁿ⁺¹⁾(b));

(3) If nis an even number and f⁽ⁿ⁺¹⁾(x) is increasing in [a, b], then

−^(x_4(n+1)!^0−a)n+2(f⁽ⁿ⁺¹⁾(x0)−f⁽ⁿ⁺¹⁾(a)) (2.10)

−^(b−x_4(n+1)!⁰⁾ⁿ⁺²(f⁽ⁿ⁺¹⁾(b)−f⁽ⁿ⁺¹⁾(x0))≤

≤ Z _b

a

R_n,f(x₀, x)dx−^f(n)(x_(n+2)!^0)−f(n)(a)(a−x₀)ⁿ⁺¹

−^f(n)(b)−f_(n+2)!^(n)(x0)(b−x₀)ⁿ⁺¹

≤0;

(4) If nis an even number and f⁽ⁿ⁺¹⁾(x) is decreasing in [a, b], then

(x0−a)ⁿ⁺²

4(n+1)! (f⁽ⁿ⁺¹⁾(a)−f⁽ⁿ⁺¹⁾(x0)) (2.11)

+^(b−x_4(n+1)!⁰⁾ⁿ⁺²(f⁽ⁿ⁺¹⁾(x0)−f⁽ⁿ⁺¹⁾(b))

≥ Z b

a

R_n,f(x0, x)dx−^f(n)(x_(n+2)!^0)−f(n)(a)(a−x0)ⁿ⁺¹

−^f(n)(b)−f_(n+2)!^(n)(x0)(b−x₀)ⁿ⁺¹

≥0.

Proof. We divide the proof into two cases.

Case 1. x ∈ [x₀, b]. Let F(x) =f⁽ⁿ⁺¹⁾(x) and G(x) = ^(b−x)_(n+1)!ⁿ⁺¹, then we clearly see that G(x) is decreasing in [x0, b]. We divide this case into two subcases.

Subcase 1.1. F(x) = f⁽ⁿ⁺¹⁾(x) is increasing in [x0, b]. It follows from the Chebyshev inequality that

Z b x0

(b−x)ⁿ⁺¹

(n+1)! f⁽ⁿ⁺¹⁾(x)dx−_b−x¹

0

Z b x0

(b−x)ⁿ⁺¹ (n+1)! dx

Z b x0

f⁽ⁿ⁺¹⁾(x)dx≤0.

Making use of equation (2.1) we get Z b

x0

R_n,f(x0, x)dx−^f(n)(b)−f_(n+2)!^(n)(x0)(b−x0)ⁿ⁺¹ ≤0.

(2.12)

From the monotonicity of F and Gwe have

f⁽ⁿ⁺¹⁾(x₀)≤F(x)≤f⁽ⁿ⁺¹⁾(b) and

0≤G(x)≤ ^(b−x_(n+1)!⁰⁾ⁿ⁺¹

for x∈ [x0, b]. Therefore, inequalities (2.5) and (2.12) lead to the conclusion that

−^(b−x_4(n+1)!⁰⁾ⁿ⁺²(f⁽ⁿ⁺¹⁾(b)−f⁽ⁿ⁺¹⁾(x₀))≤ (2.13)

≤ Z b

x0

R_n,f(x₀, x)dx−^f(n)(b)−f_(n+2)!^(n)(x0)(b−x₀)ⁿ⁺¹≤0.

Subcase 1.2. F(x) = f⁽ⁿ⁺¹⁾(x) is decreasing in [x0, b]. The Chebyshev inequality implies that

Z b x0

(b−x)ⁿ⁺¹

(n+1)! f⁽ⁿ⁺¹⁾(x)dx−_b−x¹

0

Z b x0

(b−x)ⁿ⁺¹ (n+1)! dx

Z b x0

f⁽ⁿ⁺¹⁾(x)dx≥0.

Then equation (2.1) and inequality (2.5) lead to the conclusion that

(b−x0)ⁿ⁺²

4(n+1)! (f⁽ⁿ⁺¹⁾(x₀)−f⁽ⁿ⁺¹⁾(b))≥ (2.14)

≥ Z b

x0

R_n,f(x₀, x)dx−^f(n)(b)−f_(n+2)!^(n)(x0)(b−x₀)ⁿ⁺¹≥0.

Case 2. x ∈ [a, x0]. Let F(x) = f⁽ⁿ⁺¹⁾(x) and H(x) = ^(a−x)_(n+1)!ⁿ⁺¹. We divide the discussion into four subcases.

Subcase 2.1. n is an odd number and F(x) = f⁽ⁿ⁺¹⁾(x) is increasing in [a, x0]. ThenH(x) = ^(a−x)_(n+1)!ⁿ⁺¹ is increasing in [a, x0] and

f⁽ⁿ⁺¹⁾(a)≤F(x)≤f⁽ⁿ⁺¹⁾(x₀) for all x∈[a, x₀].

Making use of the Chebyshev inequality we get Z x0

a

(a−x)ⁿ⁺¹

(n+1)! f⁽ⁿ⁺¹⁾(x)dx−_x¹

0−a

Z x0

a

(a−x)ⁿ⁺¹ (n+1)! dx

Z x0

a

f⁽ⁿ⁺¹⁾(x)dx≥0.

Then equation (2.2) and inequality (2.7) imply that

(x0−a)ⁿ⁺²

4(n+1)! (f⁽ⁿ⁺¹⁾(x0)−f⁽ⁿ⁺¹⁾(a))≥ (2.15)

≥ Z x0

a

Rn,f(x0, x)dx− ^f(n)(x_(n+2)!^0)−f(n)(a)(a−x0)ⁿ⁺¹ ≥0.

Subcase 2.2. n is an odd number and F(x) = f⁽ⁿ⁺¹⁾(x) is decreasing in [a, x0]. Then H(x) is increasing in [a, x0]. It follows from equation (2.2) and inequality (2.7) together with the Chebyshev inequality that

−^(x_4(n+1)!^0−a)n+2(f⁽ⁿ⁺¹⁾(a)−f⁽ⁿ⁺¹⁾(x₀))≤ (2.16)

≤ Z x0

a

R_n,f(x₀, x)dx− ^f(n)(x_(n+2)!^0)−f(n)(a)(a−x₀)ⁿ⁺¹ ≤0.

Subcase 2.3. n is an even number and F(x) = f⁽ⁿ⁺¹⁾(x) is increasing in [a, x0]. Then H(x) = ^(a−x)_(n+1)!ⁿ⁺¹ is decreasing in [a, x0]. Therefore, equation

(2.2) and inequality (2.7) together with the Chebyshev inequality lead to the conclusion that

−^(x_4(n+1)!^0−a)n+2(f⁽ⁿ⁺¹⁾(x0)−f⁽ⁿ⁺¹⁾(a))≤ (2.17)

≤ Z x0

a

Rn,f(x0, x)dx− ^f(n)(x_(n+2)!^0)−f(n)(a)(a−x0)ⁿ⁺¹ ≤0.

Subcase 2.4. n is an even number and F(x) = f⁽ⁿ⁺¹⁾(x) is decreasing in [a, x0]. Then H(x) and F(x) have the same monotonicity in [a, x0]. It follows from equation (2.2) and inequality (2.7) together with the Chebyshev inequality that

(x0−a)ⁿ⁺²

4(n+1)! (f⁽ⁿ⁺¹⁾(a)−f⁽ⁿ⁺¹⁾(x₀))≥ (2.18)

≥ Z x0

a

R_n,f(x₀, x)dx− ^f(n)(x_(n+2)!^0)−f(n)(a)(a−x₀)ⁿ⁺¹ ≥0.

Therefore, inequality (2.8) follows from inequalities (2.13) and (2.15), in- equality (2.9) follows from inequalities (2.14) and (2.16), inequality (2.10) follows from inequalities (2.13) and (2.17), and inequality (2.11) follows from

inequalities (2.14) and (2.18).

If take n= 0 and x0 = ^a+b₂ in Theorem 2.7, then we have

Corollary2.8. Letf(x)∈C¹[a, b], then the following statements are true:

(1) If f⁰(x) is increasing in[a, b], then

−₁₆¹(b−a)(f⁰(b)−f⁰(a))≤ (2.19)

≤ _b−a¹ Z b

a

f(x)dx−¹₂f(^a+b₂ )−^f(a)+f₄ ^(b) ≤0.

(2) If f⁰(x) is decreasing in[a, b], then

1

16(b−a)(f⁰(a)−f⁰(b))≥ (2.20)

≥ _b−a¹ Z _b

a

f(x)dx−¹₂f(^a+b₂ )−^f(a)+f₄ ^(b) ≥0.

Let us recall the well known Hermite-Hadamard inequality:

f(^a+b₂ )≤(≥)_b−a¹ Z b

a

f(x)dx≤(≥)^f(a)+f₂ ^(b) (2.21)

iff(x) is convex (concave) in [a, b].

Inequality (2.19) can be rewritten as

1

2f(^a+b₂ ) +^f(a)+f(b)₄ −₁₆¹(b−a)(f⁰(b)−f⁰(a))≤ (2.22)

≤ _b−a¹ Z b

a

f(x)dx≤ ¹₂f(^a+b₂ ) +^f(a)+f(b)₄ .

We clearly see that ^f(a)+f(b)₂ ≥ ¹₂f(^a+b₂ ) +^f(a)+f(b)₄ iff(x) is convex in [a, b].

Therefore, inequality (2.22) is an improvement of inequality (2.21) if f⁰(x) is decreasing in [a, b].

Acknowledgement. The authors wish to thank the anonymous referees for their very careful reading of the manuscript and fruitful comments and suggestions.

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Received by the editors: April 30, 2013.