Journal of Inequalities in Pure and
Applied Mathematics
NOTE ON INEQUALITIES INVOLVING INTEGRAL TAYLOR’S
REMAINDER
volume 6, issue 3, article 72,
2005.
ZHENG LIU
Institute of Applied Mathematics, Faculty of Science
Anshan University of Science and Technology
Anshan 114044, Liaoning, China
Received 08 April, 2005;
accepted 02 June, 2005.
Communicated by: H. Gauchman
EMail: lewzheng@163.net
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Abstract
In this paper, some inequalities involving the integral Taylor’s remainder are
obtained by using various well-known methods.
2000 Mathematics Subject Classification: 26D15.
Key words: Taylor’s remainder, Leibniz formula, Variant of Grüss inequality, Taylor’s
formula, Steffensen inequality.
Contents
Note On Inequalities Involving
Integral Taylor’s Remainder
Zheng Liu
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Results Obtained via the Leibniz Formula . . . . . . . . . . . . . . . . 5
3
Results Obtained by a Variant of the Grüss Inequality . . . . . . 8
4
Results Obtained via the Steffensen Inequality . . . . . . . . . . . . 10
References
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1.
Introduction
In [4] – [5], H. Gauchman has derived some new types of inequalities involving
Taylor’s remainder.
In [1], L. Bougoffa continued to create several integral inequalities involving
Taylor’s remainder.
The purpose of this paper is to give some supplements and improvements for
the results obtained in [1] – [3].
In [1], two notations Rn,f (c, x) and rn,f (a, b) have been adopted to denote
the nth Taylor’s remainder of function f with center c and the integral Taylor’s
remainder respectively, i.e.,
Rn,f (c, x) = f (x) −
n
X
f (n) (c)
k=0
n!
(x − c)k ,
Note On Inequalities Involving
Integral Taylor’s Remainder
Zheng Liu
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Contents
and
Z
rn,f (a, b) =
a
b
(b − x)n (n+1)
f
(x)dx.
n!
However, it is evident that
Z b
(b − x)n (n+1)
Rn,f (a, b) =
f
(x) dx = rn,f (a, b),
n!
a
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and
n
Z
(−1) Rn,f (b, a) =
a
b
(x − a)n (n+1)
f
(x) dx = (−1)n rn,f (b, a).
n!
J. Ineq. Pure and Appl. Math. 6(3) Art. 72, 2005
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So, we would like only to keep the notation Rn,f (·, ·) in what follows.
We start by changing the order of integration to give a simple different proof
of Lemma 1.1 and Lemma 1.2 in [5] and [1]. i.e.,
Z b
Z b Z x
(x − t)n (n+1)
Rn,f (a, x) dx =
(t)dt dx
f
n!
a
a
a
Z b Z b
(x − t)n (n+1)
=
f
(t)dx dt
n!
a
t
Z b
(b − t)n+1 (n+1)
f
(t)dt.
=
(n + 1)!
a
Note On Inequalities Involving
Integral Taylor’s Remainder
Zheng Liu
and
n+1
Z
(−1)
a
b
Z b Z
b
(t − x)n (n+1)
Rn,f (b, x)dx =
f
(t)dt dx
n!
a
x
Z b Z t
(t − x)n (n+1)
=
f
(t)dx dt
n!
a
a
Z b
(t − a)n+1 (n+1)
=
f
(t)dt.
(n + 1)!
a
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2.
Results Obtained via the Leibniz Formula
We prove the following theorem by using the Leibniz formula.
Theorem 2.1. Let f be a function defined on [a, b]. Assume that f ∈ C n+1 ([a, b]).
Then
p
p−1
X
X
(n−k) (b − a)n−k
k k
k
f
,
(2.1)
Cp−1
(a)
(−1) Cp Rn−k,f (a, b) ≤
(n − k)!
k=0
k=0
(2.2)
p
p−1
X
X
(n−k) (b − a)n−k
n−k+1 k
k
f
Cp Rn−k,f (b, a) ≤
Cp−1
(b)
,
(−1)
(n − k)!
k=0
(2.3)
Note On Inequalities Involving
Integral Taylor’s Remainder
Zheng Liu
k=0
p
Z b
X
k k
(−1)
C
R
(a,
x)dx
n−k,f
p
a
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k=0
≤
p−1
X
(n−k) (b − a)n−k+1
k
f
,
Cp−1
(a)
(n − k + 1)!
k=0
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(2.4)
p
Z b
X
Rn−k,f (b, x)dx
(−1)n−k+1 Cpk
a
Close
k=0
p−1
≤
where Cpk =
p!
.
(p−k)!k!
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n−k+1
(n−k) (b − a)
k
f
Cp−1
(b)
,
(n
−
k
+
1)!
k=0
X
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Proof. We apply the following Leibniz formula
(F G)(p) = F (p) G + Cp1 F (p−1) G(1) + · · · + Cpp−1 F (1) G(p−1) + F G(P ) ,
provided the functions F, G ∈ C p ([a, b]).
n
Let F (x) = f (n−p+1) (x), G(x) = (b−x)
. Then
n!
p
(p)
X
(b − x)n
(b − x)n−k
(n−p+1)
(−1)k Cpk f (n−k+1) (x)
f
(x)
=
.
n!
(n − k)!
k=0
Integrating both sides of the preceding equation with respect to x from a to b
gives us
#x=b
"
n (p−1)
(b
−
x)
f (n−p+1) (x)
n!
x=a
Z b
p
X
(b − x)n−k
k k
=
(−1) Cp
f (n−k+1) (x)
dx.
(n
−
k)!
a
k=0
The integral on the right is Rn−k,f (a, x), and to evaluate the term on the left
hand side, we must again apply the Leibniz formula, obtaining
X
Zheng Liu
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p
p−1
−
Note On Inequalities Involving
Integral Taylor’s Remainder
k
(−1)k Cp−1
f (n−k) (a)
k=0
(b − a)n−k X
=
(−1)k Cpk Rn−k,f (a, b).
(n − k)!
k=0
Consequently,
p
p−1
X
X
(n−k) (b − a)n−k
k k
k
,
Cp−1 f
(a)
(−1) Cp Rn−k,f (a, b) ≤
(n − k)!
k=0
k=0
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which proves (2.1).
For the proof of (2.2), we take
F (x) = f (n−p+1) (x),
G(x) =
(x − a)n
.
n!
For the proof of (2.3), we take
F (x) = f (n−p+1) (x),
G(x) =
(b − x)n+1
.
(n + 1)!
Note On Inequalities Involving
Integral Taylor’s Remainder
For the proof of (2.4), we take
F (x) = f (n−p+1) (x),
Zheng Liu
G(x) =
(x − a)n+1
.
(n + 1)!
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Remark 1. It should be noticed that (2.3) and (2.4) have been mentioned and
proved in [1] with some misprints in the conclusion.
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3.
Results Obtained by a Variant of the Grüss Inequality
The following is a variant of the Grüss inequality which has been proved almost
at the same time by X.L. Cheng and J. Sun in [3] as well as M. Matić in [6]
respectively.
Let h, g : [a, b] → R be two integrable functions such that γ ≤ g(x) ≤ Γ for
some constants γ, Γ for all x ∈ [a, b]. Then
(3.1)
Z b
Z b
Z b
1
h(x)
dx
g(x)dx
h(x)g(x)
dx
−
b
−
a
a
a
a
Z b Z b
1
1
h(x) −
dx (Γ − γ).
h(y)dy
≤
2
b−a a
a
Theorem 3.1. Let f (x) be a function defined on [a, b] such that f ∈ C n+1 ([a, b])
and m ≤ f (n+1) (x) ≤ M for each x ∈ [a, b], where m and M are constants.
Then
f (n) (b) − f (n) (a)
n(b − a)n+1 (M − m)
n
√
(b − a) ≤
,
(3.2) Rn,f (a, b) −
(n + 1)!
(n + 1)!(n + 1) n n + 1
Note On Inequalities Involving
Integral Taylor’s Remainder
Zheng Liu
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(3.3)
(n)
(n)
(−1)n+1 Rn,f (b, a) − f (b) − f (a) (b − a)n (n + 1)!
≤
n(b − a)n+1 (M − m)
√
,
(n + 1)!(n + 1) n n + 1
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(3.4)
Z b
f (n) (b) − f (n) (a)
n+1 Rn,f (a, x) dx −
(b − a) (n + 2)!
a
≤
(n + 1)(b − a)n+2 (M − m)
√
(n + 2)!(n + 2) n+1 n + 2
and
(3.5)
Z b
(n)
(n)
f
(b)
−
f
(a)
n+1 (−1)n+1
R
(b,
x)
dx
−
(b
−
a)
n,f
(n + 2)!
a
n+2
≤
(n + 1)(b − a) (M − m)
√
.
(n + 2)!(n + 2) n+1 n + 2
Note On Inequalities Involving
Integral Taylor’s Remainder
Zheng Liu
n
Proof. To prove (3.2), setting g(x) = f (n+1) (x) and h(x) = (b−x)
in (3.1), we
n!
obtain
(n)
(n)
Rn,f (a, b) − f (b) − f (a) (b − a)n (n + 1)!
Z b
(b − x)n (b − a)n M −m
dx
≤
−
n!
2
(n + 1)! a
n(b − a)n+1 (M − m)
√
=
.
(n + 1)!(n + 1) n n + 1
The proofs of (3.3), (3.4) and (3.5) are similar and so are omitted.
Remark 2. It should be noticed that Theorem 3.1 improves Theorem 3.1 in [1]
and Theorem 2.1 in [5].
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4.
Results Obtained via the Steffensen Inequality
In [2] we can find a general version of the well-known Steffensen inequality
as follows: Let h : [a, b] → R be a nonincreasing mapping on [a, b] and g :
[a, b] → R be an integrable mapping on [a, b] with
φ ≤ g(x) ≤ Φ, for all x ∈ [a, b],
then
Z
(4.1) φ
b−λ
Z
b
h(x)dx ≤
h(x)dx + Φ
a
b
Z
b−λ
Note On Inequalities Involving
Integral Taylor’s Remainder
h(x)g(x)dx
a
Z
≤Φ
a+λ
Z
h(x)dx + φ
a
Zheng Liu
b
h(x)dx,
a+λ
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where
Z
(4.2)
λ=
b
G(x) dx,
a
g(x) − φ
G(x) =
,
Φ−φ
Contents
Φ 6= φ.
Theorem 4.1. Let f : [a, b] → R be a mapping such that f (x) ∈ C n+1 ([a, b])
and m ≤ f (n+1) (x) ≤ M for each x ∈ [a, b], where m and M are constants.
Then
(4.3)
m(b − a)n+1 + (M − m)λn+1
(n + 1)!
≤ Rn,f (a, b)
M (b − a)n+1 − (M − m)(b − a − λ)n+1
≤
,
(n + 1)!
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(4.4)
(4.5)
m(b − a)n+1 + (M − m)λn+1
(n + 1)!
≤ (−1)n+1 Rn,f (b, a)
M (b − a)n+1 − (M − m)(b − a − λ)n+1
≤
,
(n + 1)!
m(b − a)n+2 + (M − m)λn+2
(n + 2)!
Z b
≤
Rn,f (a, x)dx
Note On Inequalities Involving
Integral Taylor’s Remainder
Zheng Liu
a
≤
M (b − a)n+2 − (M − m)(b − a − λ)n+2
,
(n + 2)!
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and
(4.6)
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m(b − a)n+2 + (M − m)λn+2
(n + 2)!
Z b
n+1
≤ (−1)
Rn,f (b, x)dx
a
n+2
≤
where λ =
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M (b − a)
f (b)−f (a)−m(b−a)
.
M −m
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n+2
− (M − m)(b − a − λ)
(n + 2)!
,
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n
Proof. Observe that (b−x)
is a decreasing function of x on [a, b], then by (4.1)
n!
and (4.2) we have
Z b−λ
Z b
(b − x)n
(b − x)n
m
dx + M
dx
n!
n!
a
b−λ
Z b
(b − x)n (n+1)
≤
f
(x)dx
n!
a
Z a+λ
Z b
(b − x)n
(b − x)n
≤M
dx + m
dx
n!
n!
a
a+λ
Note On Inequalities Involving
Integral Taylor’s Remainder
with
Zheng Liu
Z
λ=
a
b
f (n+1) (x) − m
f (n) (b) − f (n) (a) − m(b − a)
dx =
,
M −m
M −m
and (4.3) follows.
n
Since (x−a)
is a increasing function of x on [a, b], then
n!
Z a+λ
Z b
(x − a)n
(x − a)n
M
dx + m
dx
n!
n!
a
a+λ
Z b
(x − a)n (n+1)
≤
f
(x)dx
n!
a
Z b−λ
Z b
(x − a)n
(x − a)n
≤m
dx + M
dx,
n!
n!
a
b−λ
and (4.4) follows.
The proofs of (4.5) and (4.6) are similar and so are omitted.
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Remark 3. It should be mentioned that (4.5) and (4.6) have also been proved
in [4]
Note On Inequalities Involving
Integral Taylor’s Remainder
Zheng Liu
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References
[1] L. BOUGOFFA, Some estimations for the integral Taylor’s remainder, J.
Inequal. Pure and Appl. Math., 4(5) (2003), Art. 86. [ONLINE: http:
//jipam.vu.edu.au/article.php?sid=327].
[2] P. CERONE, Generalised trapezoidal rules with error involving bounds of
the nth derivative, Math. Ineq. and Applic., 5(3) (2002), 451–462.
[3] X.L.CHENG AND J. SUN, A note on the perturbed trapezoid inequality,
J. Inequal. Pure and Appl. Math., 3(2) (2002), Art. 29. [ONLINE: http:
//jipam.vu.edu.au/article.php?sid=181].
[4] H. GAUCHMAN, Some integral inequalities involving Taylor’s remainder.
I, J. Inequal. Pure and Appl. Math., 3(2) (2002), Art. 26. [ONLINE: http:
//jipam.vu.edu.au/article.php?sid=178].
[5] H. GAUCHMAN, Some integral inequalities involving Taylor’s remainder.
II, J. Inequal. Pure and Appl. Math., 4(1) (2003), Art. 1. [ONLINE: http:
//jipam.vu.edu.au/article.php?sid=237].
[6] M. MATIĆ, Improvement of some estimations related to the remainder in
generalized Taylor’s formula, Math. Ineq. and Applic., 5(4) (2002), 617–
648.
Note On Inequalities Involving
Integral Taylor’s Remainder
Zheng Liu
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