Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 3, Article 72, 2005
NOTE ON INEQUALITIES INVOLVING INTEGRAL TAYLOR’S REMAINDER
ZHENG LIU
I NSTITUTE OF A PPLIED M ATHEMATICS , FACULTY OF S CIENCE
A NSHAN U NIVERSITY OF S CIENCE AND T ECHNOLOGY
A NSHAN 114044, L IAONING , C HINA
lewzheng@163.net
Received 08 April, 2005; accepted 02 June, 2005
Communicated by H. Gauchman
A BSTRACT. In this paper, some inequalities involving the integral Taylor’s remainder are obtained by using various well-known methods.
Key words and phrases: Taylor’s remainder, Leibniz formula, Variant of Grüss inequality, Taylor’s formula, Steffensen inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
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 ,
and
Z
rn,f (a, b) =
a
b
(b − x)n (n+1)
f
(x)dx.
n!
However, it is evident that
Z
Rn,f (a, b) =
a
ISSN (electronic): 1443-5756
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111-04
b
(b − x)n (n+1)
f
(x) dx = rn,f (a, b),
n!
2
Z HENG L IU
and
b
(x − a)n (n+1)
f
(x) dx = (−1)n rn,f (b, a).
n!
a
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 =
f
(t)dt dx
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
Z
n
(−1) Rn,f (b, a) =
and
n+1
Z
(−1)
b
Z b Z
b
Rn,f (b, x)dx =
a
a
x
Z b Z
t
=
a
a
b
Z
=
a
(t − x)n (n+1)
f
(t)dt dx
n!
(t − x)n (n+1)
f
(t)dx dt
n!
(t − a)n+1 (n+1)
f
(t)dt.
(n + 1)!
2. R ESULTS O BTAINED VIA
THE
L EIBNIZ F ORMULA
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
(2.1)
(−1)
C
R
(a,
b)
≤
,
C
(a)
p n−k,f
p−1 f
(n − k)!
k=0
k=0
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)!
(2.2)
k=0
(2.3)
p
p−1
Z b
X
X
(n−k) (b − a)n−k+1
k
f
,
Rn−k,f (a, x)dx ≤
Cp−1
(a)
(−1)k Cpk
(n − k + 1)!
a
k=0
(2.4)
k=0
k=0
p
p−1
Z b
X
X
(n−k) (b − a)n−k+1
n−k+1 k
k
(−1)
C
R
(b,
x)dx
C
(b)
,
≤
n−k,f
p
p−1 f
(n − k + 1)!
a
k=0
where Cpk =
k=0
p!
.
(p−k)!k!
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]).
J. Inequal. Pure and Appl. Math., 6(3) Art. 72, 2005
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N OTE O N I NEQUALITIES I NVOLVING I NTEGRAL TAYLOR ’ S R EMAINDER
3
n
. Then
Let F (x) = f (n−p+1) (x), G(x) = (b−x)
n!
(p) X
p
(b − x)n−k
(b − x)n
(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
"
Z b
p
n (p−1)
X
(b − x)n−k
(b
−
x)
k k
(n−k+1)
(n−p+1)
f
(x)
=
(−1) Cp
f
(x)
dx.
n!
(n − k)!
a
k=0
x=a
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
−
p−1
X
k
(−1)
p
− a)n−k X
=
(−1)k Cpk Rn−k,f (a, b).
(n − k)!
k=0
(b
k
Cp−1
f (n−k) (a)
k=0
Consequently,
p
p−1
X
X
(n−k) (b − a)n−k
k k
k
f
Cp−1
(a)
,
(−1) Cp Rn−k,f (a, b) ≤
(n − k)!
k=0
k=0
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)!
G(x) =
(x − a)n+1
.
(n + 1)!
For the proof of (2.4), we take
F (x) = f (n−p+1) (x),
Remark 2.2. It should be noticed that (2.3) and (2.4) have been mentioned and proved in [1]
with some misprints in the conclusion.
3. R ESULTS O BTAINED BY A VARIANT OF THE G RÜSS I NEQUALITY
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
Z b
Z b
Z b
1
(3.1) h(x)g(x) dx −
h(x) dx
g(x)dx
b−a a
a
a
Z b Z b
1
1
dx (Γ − γ).
≤
h(x)
−
h(y)dy
2
b
−
a
a
a
J. Inequal. Pure and Appl. Math., 6(3) Art. 72, 2005
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4
Z HENG L IU
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
n+1
(n)
(n)
f
(b)
−
f
(a)
n
Rn,f (a, b) −
≤ n(b − a) (M√− m) ,
(3.2)
(b
−
a)
(n + 1)!(n + 1) n n + 1
(n + 1)!
(3.3)
(3.4)
n+1
(n)
(n)
f
(b)
−
f
(a)
n+1
n
(−1) Rn,f (b, a) −
≤ n(b − a) (M√− m) ,
(b
−
a)
(n + 1)!(n + 1) n n + 1
(n + 1)!
Z b
(n)
(n)
(n + 1)(b − a)n+2 (M − m)
f
(b)
−
f
(a)
n+1
≤
√
R
(a,
x)
dx
−
(b
−
a)
n,f
(n + 2)!
(n + 2)!(n + 2) n+1 n + 2
a
and
Z b
(n)
(n)
f
(b)
−
f
(a)
n+1
n+1
(−1)
(b
−
a)
R
(b,
x)
dx
−
n,f
(n + 2)!
(3.5)
a
≤
(n + 1)(b − a)n+2 (M − m)
√
.
(n + 2)!(n + 2) n+1 n + 2
n
Proof. To prove (3.2), setting g(x) = f (n+1) (x) and h(x) = (b−x)
in (3.1), we obtain
n!
Z
b
(n)
(n)
n
n
Rn,f (a, b) − f (b) − f (a) (b − a)n ≤ M − m
(b − x) − (b − a) dx
n!
(n + 1)!
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 3.2. It should be noticed that Theorem 3.1 improves Theorem 3.1 in [1] and Theorem
2.1 in [5].
4. R ESULTS O BTAINED VIA
THE
S TEFFENSEN I NEQUALITY
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 b−λ
Z b
Z b
Z a+λ
Z b
(4.1) φ
h(x)dx+Φ
h(x)dx ≤
h(x)g(x)dx ≤ Φ
h(x)dx+φ
h(x)dx,
a
b−λ
a
a
a+λ
where
Z
(4.2)
λ=
b
G(x) dx,
a
G(x) =
g(x) − φ
,
Φ−φ
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
≤ Rn,f (a, b)
(n + 1)!
M (b − a)n+1 − (M − m)(b − a − λ)n+1
≤
,
(n + 1)!
J. Inequal. Pure and Appl. Math., 6(3) Art. 72, 2005
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N OTE O N I NEQUALITIES I NVOLVING I NTEGRAL TAYLOR ’ S R EMAINDER
(4.4)
(4.5)
5
m(b − a)n+1 + (M − m)λn+1
≤ (−1)n+1 Rn,f (b, a)
(n + 1)!
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
a
M (b − a)n+2 − (M − m)(b − a − λ)n+2
,
(n + 2)!
and
(4.6)
where λ =
Z b
m(b − a)n+2 + (M − m)λn+2
n+1
≤ (−1)
Rn,f (b, x)dx
(n + 2)!
a
M (b − a)n+2 − (M − m)(b − a − λ)n+2
≤
,
(n + 2)!
f (b)−f (a)−m(b−a)
.
M −m
n
Proof. Observe that (b−x)
is a decreasing function of x on [a, b], then by (4.1) and (4.2) we
n!
have
Z b
Z b
Z b−λ
(b − x)n
(b − x)n (n+1)
(b − x)n
m
dx + M
dx ≤
f
(x)dx
n!
n!
n!
a
b−λ
a
Z a+λ
Z b
(b − x)n
(b − x)n
≤M
dx + m
dx
n!
n!
a
a+λ
with
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
Z b
(x − a)n
(x − a)n
(x − a)n (n+1)
M
dx + m
dx ≤
f
(x)dx
n!
n!
n!
a
a+λ
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.
Remark 4.2. It should be mentioned that (4.5) and (4.6) have also been proved in [4]
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