Journal of Inequalities in Pure and
Applied Mathematics
INTEGRAL INEQUALITIES OF THE OSTROWSKI TYPE
A. SOFO
School of Communications and Informatics
Victoria University of Technology
PO Box 14428
Melbourne City MC
8001 Victoria, Australia
EMail: sofo@matilda.vu.edu.au
URL: http://rgmia.vu.edu.au/sofo/
volume 3, issue 2, article 21,
2002.
Received December, 2000;
accepted 16 November, 2001.
Communicated by: T. Mills
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
Integral inequalities of Ostrowski type are developed for n−times differentiable
mappings, with multiple branches, on the L∞ norm. Some particular inequalities are also investigated, which include explicit bounds for perturbed trapezoid,
midpoint, Simpson’s, Newton-Cotes and left and right rectangle rules. The results obtained provide sharper bounds than those obtained by Dragomir [5] and
Cerone, Dragomir and Roumeliotis [2].
Integral Inequalities of the
Ostrowski Type
2000 Mathematics Subject Classification: 26D15, 41A55.
Key words: Ostrowski Integral Inequality, Quadrature Formulae.
A. Sofo
I wish to acknowledge the useful comments of an anonymous referee which led to
an improvement of this work.
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Contents
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Integral Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Integral Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
The Convergence of a General Quadrature Formula . . . . . . . .
5
Grüss Type Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
Some Particular Integral Inequalities . . . . . . . . . . . . . . . . . . . .
7
Applications for Numerical Integration . . . . . . . . . . . . . . . . . .
8
Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
6
20
27
33
43
64
80
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1.
Introduction
In 1938 Ostrowski [18] obtained a bound for the absolute value of the difference
of a function to its average over a finite interval. The theorem is as follows.
Theorem 1.1. Let f : [a, b] → R be a differentiable mapping on [a, b] and let
|f 0 (t)| ≤ M for all t ∈ (a, b), then the following bound is valid
"
#
Z b
a+b 2
x
−
1
1
2
f (x) −
(1.1)
f (t) dt ≤ (b − a) M
+
b−a a
4
(b − a)2
Integral Inequalities of the
Ostrowski Type
A. Sofo
for all x ∈ [a, b].
The constant 14 is sharp in the sense that it cannot be replaced by a smaller one.
Dragomir and Wang [12, 13] extended the result (1.1) and applied the extended result to numerical quadrature rules and to the estimation of error bounds
for some special means. Dragomir [8, 9, 10] further extended the result (1.1) to
incorporate mappings of bounded variation, Lipschitzian mappings and monotonic mappings.
Cerone, Dragomir and Roumeliotis [3] as well as Dedić, Matić and Pečarić
[4] and Pearce, Pečarić, Ujević and Varošanec [19] further extended the result
(1.1) by considering n−times differentiable mappings on an interior point x ∈
[a, b].
In particular, Cerone and Dragomir [1] proved the following result.
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(n−1)
Theorem 1.2. Let f : [a, b] → R be a mapping such that f
is absolutely
continuous on [a, b], (AC[a, b] for short). Then for all x ∈ [a, b] the following
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
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bound is valid:
Z
!
n−1
b
X
(b − x)j+1 + (−1)j (x − a)j+1
f (j) (x)
(1.2) f (t) dt −
a
(j + 1)!
j=0
(n) f ∞
≤
(x − a)n+1 + (b − x)n+1 if f (n) ∈ L∞ [a, b] ,
(n + 1)!
where
(n) f := sup f (n) (t) < ∞.
∞
t∈[a,b]
Integral Inequalities of the
Ostrowski Type
A. Sofo
Dragomir also generalised the Ostrowski inequality for k points, x1 , . . . , xk
and obtained the following theorem.
Theorem 1.3. Let Ik := a = x0 < x1 < ... < xk−1 < xk = b be a division
of the interval [a, b], αi (i = 0, ..., k + 1) be “k + 2” points so that α0 = a,
αi ∈ [xi−1 , xi ] (i = 1, ..., k) and αk+1 = b. If f : [a, b] → R is AC[a, b], then we
have the inequality
Z
k
b
X
(1.3)
f (x) dx −
(αi+1 − αi ) f (xi )
a
i=0
" k−1
2 #
k−1 1X 2 X
xi + xi+1
≤
h +
αi+1 −
kf 0 k∞
4 i=0 i
2
i=0
≤
1 0
kf k∞
2
k−1
X
i=0
h2i ≤
1
(b − a) kf 0 k∞ ν (h) ,
2
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where hi := xi+1 −xi (i = 0, ..., k − 1) and ν (h) := max {hi |i = 0, ..., k − 1}.
The constant 14 in the first inequality and the constant 12 in the second and third
inequalities are the best possible.
The main aim of this paper is to develop the upcoming Theorem 3.1 (see
page 10) of integral inequalities for n−times differentiable mappings. The motivation for this work has been to improve the order of accuracy of the results
given by Dragomir [5], and Cerone and Dragomir [1]. In the case of Dragomir
[5], the bound of (1.3) is of order 1, whilst in this paper we improve the bound
of (1.3) to order n.
We begin the process by obtaining the following integral equalities.
Integral Inequalities of the
Ostrowski Type
A. Sofo
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2.
Integral Identities
Theorem 2.1. Let Ik : a = x0 < x1 < · · · < xk−1 < xk = b be a division of
the interval [a, b] and αi (i = 0, . . . , k + 1) be ‘k + 2’ points so that α0 = a,
αi ∈ [xi−1 , xi ] (i = 1, . . . , k) and αk+1 = b. If f : [a, b] → R is a mapping such
that f (n−1) is AC[a, b], then for all xi ∈ [a, b] we have the identity:
Z
(2.1)
a
b
k−1
n
X
(−1)j X n
f (t) dt +
(xi+1 − αi+1 )j f (j−1) (xi+1 )
j!
i=0
j=1
Z b
o
j (j−1)
n
− (xi − αi+1 ) f
(xi ) = (−1)
Kn,k (t) f (n) (t) dt,
Integral Inequalities of the
Ostrowski Type
A. Sofo
a
where the Peano kernel
(2.2)
Kn,k (t) :=
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
(t − α1 )n


,


n!




(t − α2 )n


,



n!

t ∈ [a, x1 )
t ∈ [x1 , x2 )
..
.


n


(t − αk−1 )


, t ∈ [xk−2 , xk−1 )


n!



n


 (t − αk ) ,
t ∈ [xk−1 , b] ,
n!
n and k are natural numbers, n ≥ 1, k ≥ 1 and f (0) (x) = f (x).
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Proof. The proof is by mathematical induction. For n = 1, from (2.1) we have
the equality
Z b
k−1 h
i
X
(2.3)
f (t) dt =
(xi+1 − αi+1 )j f (xi+1 ) − (xi − αi+1 )j f (xi )
a
i=0
Z
−
b
K1,k (t) f 0 (t) dt,
a
where


(t − α1 ) ,




(t
 − α2 ) ,
Integral Inequalities of the
Ostrowski Type
t ∈ [a, x1 )
t ∈ [x1 , x2 )
A. Sofo
..
.



(t − αk−1 ) , t ∈ [xk−2 , xk−1 )


 (t − α ) ,
t ∈ [xk−1 , b] .
k
To prove (2.3), we integrate by parts as follows
Z b
K1,k (t) f 0 (t) dt
K1,k (t) :=
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a
=
=
=
k−1 Z
X
xi+1
i=0 xi
k−1
X
i=0
k−1
X
i=0
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(t − αi+1 ) f 0 (t) dt
x
(t − αi+1 ) f (t) xi+1
−
i
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Z
xi+1
f (t) dt
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xi
[(xi+1 − αi+1 ) f (xi+1 ) − (xi − αi+1 ) f (xi )] −
Page 7 of 83
k−1 Z xi+1
X
i=0
xi
f (t) dt,
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
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Z
b
Z
f (t) dt +
a
b
K1,k (t) f 0 (t) dt
a
=
k−1
X
[(xi+1 − αi+1 ) f (xi+1 ) − (xi − αi+1 ) f (xi )] .
i=0
Hence (2.3) is proved.
Assume that (2.1) holds for ‘n’ and let us prove it for ‘n + 1’. We need to
prove the equality
Z
b
f (t) dt +
(2.4)
a
Integral Inequalities of the
Ostrowski Type
k−1 X
n+1
X
(−1)j
j!
i=0 j=1
A. Sofo
n
o
× (xi+1 − αi+1 )j f (j−1) (xi+1 ) − (xi − αi+1 )j f (j−1) (xi )
Z b
n+1
= (−1)
Kn+1,k (t) f (n+1) (t) dt,
a
where from (2.2)
Kn+1,k (t) :=



























(t−α1 )n+1
,
(n+1)!
t ∈ [a, x1 )
(t−α2 )n+1
,
(n+1)!
t ∈ [x1 , x2 )
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(t−αk−1 )n+1
,
(n+1)!
t ∈ [xk−2 , xk−1 )
(t−αk )n+1
,
(n+1)!
t ∈ [xk−1 , b] .
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Consider
Z b
Kn+1,k (t) f
(n+1)
(t) dt =
a
k−1 Z
X
i=0
xi+1
xi
(t − αi+1 )n+1 (n+1)
(t) dt
f
(n + 1)!
and upon integrating by parts we have
Z
b
Kn+1,k (t) f (n+1) (t) dt
xi+1 Z
"
#
k−1
xi+1
X
(t − αi+1 )n+1 (n) (t − αi+1 )n (n)
=
f (t)
−
f (t) dt
(n
+
1)!
n!
x
i
i=0
xi
(
)
k−1
n+1
X (xi+1 − αi+1 )
(xi − αi+1 )n+1 (n)
(n)
=
f (xi+1 ) −
f (xi )
(n + 1)!
(n + 1)!
i=0
Z b
−
Kn,k (t) f (n) (t) dt.
a
a
Upon rearrangement we may write
Z
a
Integral Inequalities of the
Ostrowski Type
A. Sofo
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b
(n)
Kn,k (t) f (t) dt
(
)
k−1
X
(xi+1 − αi+1 )n+1 (n)
(xi − αi+1 )n+1 (n)
f (xi+1 ) −
f (xi )
=
(n
+
1)!
(n
+
1)!
i=0
Z b
−
Kn+1,k (t) f (n+1) (t) dt.
a
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Now substitute
that
n
(−1)
Rb
a
Kn,k (t) f (n) (t) dt from the induction hypothesis (2.1) such
" n
k−1 X
X
(−1)j n
(xi+1 − αi+1 )j f (j−1) (xi+1 )
f (t) dt + (−1)
j!
a
i=0
j=1
oi
− (xi − αi+1 )j f (j−1) (xi )
(
)
k−1
X
(xi+1 − αi+1 )n+1 (n)
(xi − αi+1 )n+1 (n)
=
f (xi+1 ) −
f (xi )
(n + 1)!
(n + 1)!
i=0
Z b
−
Kn+1,k (t) f (n+1) (t) dt.
Z
b
n
Integral Inequalities of the
Ostrowski Type
A. Sofo
a
Collecting the second and third terms and rearranging, we can state
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Z
b
f (t) dt +
a
k−1 X
n+1
X
i=0 j=1
(−1)j
j!
JJ
J
n
o
j (j−1)
j (j−1)
× (xi+1 − αi+1 ) f
(xi+1 ) − (xi − αi+1 ) f
(xi )
Z b
= (−1)n+1
Kn+1,k (t) f (n+1) (t) dt,
a
which is identical to (2.4), hence Theorem 2.1 is proved.
The following corollary gives a slightly different representation of Theorem
2.1, which will be useful in the following work.
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Corollary 2.2. From Theorem 2.1, the equality (2.1) may be represented as
Z b
(2.5)
f (t) dt
a
" k
#
n
j X
n
o
X
(−1)
+
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi )
j!
i=0
j=1
Z b
n
= (−1)
Kn,k (t) f (n) (t) dt.
a
Integral Inequalities of the
Ostrowski Type
Proof. From (2.1) consider the second term and rewrite it as
k−1
X
(2.6) S1 + S2 :=
+
{− (xi+1 − αi+1 ) f (xi+1 ) + (xi − αi+1 ) f (xi )}
i=0
" n
k−1
X X
i=0
A. Sofo
j=2
Contents
(−1)j n
(xi+1 − αi+1 )j f (j−1) (xi+1 )
j!
j
− (xi − αi+1 ) f
Now
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(j−1)
o
(xi ) .
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S1 = (a − α1 ) f (a) +
k−1
X
(xi − αi+1 ) f (xi )
i=1
+
k−2
X
i=0
{− (xi+1 − αi+1 ) f (xi+1 )} − (b − αk ) f (b)
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= (a − α1 ) f (a) +
k−1
X
(xi − αi+1 ) f (xi )
i=1
+
k−1
X
{− (xi − αi ) f (xi )} − (b − αk ) f (b)
i=1
= − (α1 − a) f (a) −
k−1
X
(αi+1 − αi ) f (xi ) − (b − αk ) f (b) .
i=1
Integral Inequalities of the
Ostrowski Type
Also,
#
" n
k−2 X
o
X
(−1)j n
S2 =
(xi+1 − αi+1 )j f (j−1) (xi+1 )
j!
i=0
j=2
+
n
X
j=2
n
X
j
o
(−1) n
(xk − αk )j f (j−1) (xk )
j!
o
(−1) n
j (j−1)
−
(x0 − α1 ) f
(x0 )
j!
j=2
" n
#
k−1 X
o
X
(−1)j n
−
(xi − αi+1 )j f (j−1) (xi )
j!
i=1
j=2
=
j=2
j!
(b − αk )j f (j−1) (b) −
n
X
(−1)j
j=2
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j
n
X
(−1)j
A. Sofo
j!
(a − α1 )j f (j−1) (a)
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" n
#
k−1 X
o
X
(−1)j n
+
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi ) .
j!
i=1
j=2
From (2.6)
S 1 + S2
(
:= − (b − αk ) f (b) + (α1 − a) f (a) +
k−1
X
)
(αi+1 − αi ) f (xi )
Integral Inequalities of the
Ostrowski Type
i=1
n
X
o
(−1)j n
(b − αk )j f (j−1) (b) − (a − α1 )j f (j−1) (a)
j!
j=2
#
" n
k−1 X
o
X
(−1)j n
j
j
(j−1)
(xi )
+
(xi − αi ) − (xi − αi+1 ) f
j!
i=1
j=2
(
)
k−1
X
= − (α1 − a) f (a) +
(αi+1 − αi ) f (xi ) + (b − αk ) f (b)
+
A. Sofo
Title Page
Contents
JJ
J
i=1
+
n
X
(−1)j
j!
j=2
+
k−1 n
X
"
Go Back
− (a − α1 )j f (j−1) (a)
Close
o
#
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi ) + (b − αk )j f (j−1) (b) .
i=1
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Keeping in mind that x0 = a, α0 = 0, xk = b and αk+1 = b we may write
S 1 + S2 = −
k
X
(αi+1 − αi ) f (xi )
i=0
" k
#
n
o
X
(−1)j X n
+
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi )
j!
j=2
i=0
#
" k
n
j
n
o
X (−1) X
j
j
(j−1)
=
(xi − αi ) − (xi − αi+1 ) f
(xi ) .
j!
j=1
i=0
Integral Inequalities of the
Ostrowski Type
A. Sofo
And substituting S1 + S2 into the second term of (2.1) we obtain the identity
(2.5).
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If we now assume that the points of the division Ik are fixed, we obtain the
following corollary.
Corollary 2.3. Let Ik : a = x0 < x1 < · · · < xk−1 < xk = b be a division of
the interval [a, b]. If f : [a, b] → R is as defined in Theorem 2.1, then we have
the equality
" k
#
Z b
n
o
X
1 Xn j
j j
−hi + (−1) hi−1 f (j−1) (xi )
(2.7)
f (t) dt +
j j!
2
a
j=1
i=0
Z b
n
= (−1)
Kn,k (t) f (n) (t) dt,
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a
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
where hi := xi+1 − xi , h−1 := 0 and hk := 0.
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Proof. Choose
a + x1
x1 + x2
, α2 =
, ...,
2
2
xk−2 + xk−1
xk−1 + xk
=
, αk =
and αk+1 = b.
2
2
α0 = a, α1 =
αk−1
From Corollary 2.2, the term
(b − αk ) f (b) + (α1 − a) f (a) +
=
1
2
(
k−1
X
Integral Inequalities of the
Ostrowski Type
(αi+1 − αi ) f (xi )
i=1
k−1
X
h0 f (a) +
A. Sofo
)
(hi + hi−1 ) f (xi ) + hk−1 f (b) ,
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i=1
Contents
the term
JJ
J
n
o
X
(−1)j n
(b − αk )j f (j−1) (b) − (a − α1 )j f (j−1) (a)
j!
j=2
n
o
X
(−1)j n j
j j (j−1)
(j−1)
h
f
(b)
−
(−1)
h
f
(a)
=
0
k−1
j!2j
j=2
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and the term
k−1
X
i=1
Page 15 of 83
"
n
X
j=2
#
o
(−1)j n
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi )
j!
J. Ineq. Pure and Appl. Math. 3(2) Art. 21, 2002
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#
" n
k−1 X
o
X
(−1)j n j
=
hi−1 − (−1)j hji f (j−1) (xi ) .
j
j!2
i=1
j=2
Putting the last three terms in (2.5) we obtain
(
)
Z b
k−1
X
1
h0 f (a) +
(hi + hi−1 ) f (xi ) + hk−1 f (b)
f (t) dt −
2
a
i=1
n
o
X
(−1)j n j
j j (j−1)
(j−1)
h
f
(b)
−
(−1)
h
f
(a)
+
0
k−1
j!2j
j=2
" n
#
k−1 X
o
X
(−1)j n j
j j
(j−1)
(xi )
+
hi−1 − (−1) hi f
j!2j
i=1
j=2
Z b
n
= (−1)
Kn,k (t) f (n) (t) dt.
Integral Inequalities of the
Ostrowski Type
A. Sofo
Title Page
Contents
a
Collecting the inner three terms of the last expression, we have
Z
b
f (t) dt +
a
n
k
o
X
(−1)j X n j
j j
(j−1)
h
−
(−1)
h
(xi )
i−1
i f
j
j!2
j=1
i=0
Z b
n
= (−1)
Kn,k (t) f (n) (t) dt,
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a
which is equivalent to the identity (2.7).
The case of equidistant partitioning is important in practice, and with this in
mind we obtain the following corollary.
Page 16 of 83
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Corollary 2.4. Let
Ik : xi = a + i
(2.8)
b−a
k
, i = 0, . . . , k
be an equidistant partitioning of [a, b], and f : [a, b] → R be a mapping such
that f (n−1) is AC[a, b] , then we have the equality
Z
(2.9)
a
b
j "
n X
1
b−a
f (t) dt +
− f (j−1) (a)
2k
j!
j=1
Integral Inequalities of the
Ostrowski Type
A. Sofo
#
k−1 n
o
X
j
j (j−1)
(j−1)
+
(−1) − 1 f
(xi ) + (−1) f
(b)
i=1
Title Page
= (−1)n
Z
b
Kn,k (t) f (n) (t) dt.
a
It is of some interest to note that the second term of (2.9) involves only even
derivatives at all interior points xi , i = 1, . . . , k − 1.
Proof. Using (2.8) we note that
h0
hi
and hi−1
b−a
b−a
= x1 − x0 =
, hk−1 = (xk − xk−1 ) =
,
k
k
b−a
= xi+1 − xi =
k
b−a
, (i = 1, . . . , k − 1)
= xi − xi−1 =
k
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and substituting into (2.7) we have
" ( j
j
Z b
n
k−1
X
X
b−a
b−a
1
(j−1)
f (t) dt +
−
f
(a) +
−
j!2j
2k
k
a
j=1
i=0
)
#
j
j
b
−
a
b
−
a
+ (−1)j
f (j−1) (xi ) + (−1)j
f (j−1) (b)
k
k
Z b
n
= (−1)
Kn,k (t) f (n) (t) dt,
a
Integral Inequalities of the
Ostrowski Type
which simplifies to (2.9) after some minor manipulation.
A. Sofo
The following Taylor-like formula with integral remainder also holds.
Corollary 2.5. Let g : [a, y] → R be a mapping such that g (n) is AC[a, y]. Then
for all xi ∈ [a, y] we have the identity
" n
k−1 X
X
(−1)j n
(2.10) g (y) = g (a) −
(xi+1 − αi+1 )j g (j) (xi+1 )
j!
i=0
j=1
Z y
o
j (j)
n
− (xi − αi+1 ) g (xi ) + (−1)
Kn,k (y, t) g (n+1) (t) dt
a
or
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(2.11) g (y)
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"
= g (a) −
#
n
k
o
X
(−1)j X n
(xi − αi )j − (xi − αi+1 )j g (j) (xi )
j!
j=1
i=0
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n
Z
+ (−1)
y
Kn,k (y, t) g (n+1) (t) dt.
a
The proof of (2.10) and (2.11) follows directly from (2.1) and (2.5) respectively upon choosing b = y and f = g 0 .
Integral Inequalities of the
Ostrowski Type
A. Sofo
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3.
Integral Inequalities
In this section we utilise the equalities of Section 2 and develop inequalities for
the representation of the integral of a function with respect to its derivatives at
a multiple number of points within some interval.
Theorem 3.1. Let Ik : a = x0 < x1 < · · · < xk−1 < xk = b be a division of
the interval [a, b] and αi (i = 0, . . . , k + 1) be ‘k + 2’ points so that α0 = a,
αi ∈ [xi−1 , xi ] (i = 1, . . . , k) and αk+1 = b. If f : [a, b] → R is a mapping such
that f (n−1) is AC[a, b], then for all xi ∈ [a, b] we have the inequality:
Z
" k
#
n
b
o
X
(−1)j X n
(3.1)
f (t) dt +
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi ) a
j!
j=1
i=0
(n) k−1
f X ∞
(αi+1 − xi )n+1 + (xi+1 − αi+1 )n+1
≤
(n + 1)! i=0
(n) k−1
f X
∞
hn+1
≤
(n + 1)! i=0 i
(n) f ∞
≤
(b − a) ν n (h) if f (n) ∈ L∞ [a, b] ,
(n + 1)!
Integral Inequalities of the
Ostrowski Type
A. Sofo
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where
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(n) f := sup f (n) (t) < ∞,
∞
t∈[a,b]
hi := xi+1 − xi and
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ν (h) := max {hi |i = 0, . . . , k − 1} .
Proof. From Corollary 2.2 we may write
Z b
f (t) dt
(3.2)
a
#
" k
n
o
X
(−1)j X n
+
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi ) j!
j=1
i=0
Z b
n
(n)
= (−1)
Kn,k (t) f (t) dt ,
Integral Inequalities of the
Ostrowski Type
A. Sofo
a
and
Z b
Z b
(n) (n)
(−1)n
Kn,k (t) f (t) dt ≤ f
|Kn,k (t)| dt,
∞
a
Z
a
b
|Kn,k (t)| dt =
a
=
k−1 Z
X
xi
1
(n + 1)!
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n
|t − αi+1 |
dt
n!
i=0 xi
k−1 Z αi+1
X
i=0
=
xi+1
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(αi+1 − t)n
dt +
n!
k−1
X
Z
xi+1
αi+1
(t − αi+1 )n
dt
n!
(αi+1 − xi )n+1 + (xi+1 − αi+1 )n+1
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i=0
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and thus
Z b
n
(n)
(−1)
K
(t)
f
(t)
dt
n,k
a
(n) k−1
f X ∞
(αi+1 − xi )n+1 + (xi+1 − αi+1 )n+1 .
≤
(n + 1)! i=0
Hence, from (3.2), the first part of the inequality (3.1) is proved. The second
and third lines follow by noting that
(n) k−1
(n) k−1
f X f X
∞
∞
(αi+1 − xi )n+1 + (xi+1 − αi+1 )n+1 ≤
hn+1 ,
(n + 1)! i=0
(n + 1)! i=0 i
Integral Inequalities of the
Ostrowski Type
A. Sofo
Title Page
since for 0 < B < A < C it is well known that
(3.3)
n+1
(A − B)
n+1
+ (C − A)
Contents
n+1
≤ (C − B)
.
Also
(n) k−1
(n) (n) k−1
f X
f f X
∞
∞ n
∞
hn+1
≤
ν
(h)
h
=
(b − a) ν n (h) ,
i
(n + 1)! i=0 i
(n + 1)!
(n
+
1)!
i=0
where ν (h) := max {hi |i = 0, . . . , k − 1} and therefore the third line of the
inequality (3.1) follows, hence Theorem 3.1 is proved.
When the points of the division Ik are fixed, we obtain the following inequality.
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Corollary 3.2. Let f : [a, b] → R be a mapping such that f (n−1) is AC[a, b] ,
and Ik be defined as in Corollary 2.3, then
(3.4)
Z
" k
#
n
b
o
X
(−1)j X n j
j j
(j−1)
f (t) dt +
−hi + (−1) hi+1 f
(xi ) j
a
2 j! i=0
j=1
(n) k−1
f X
∞
≤
hn+1
(n + 1)!2n i=0 i
for f (n) ∈ L∞ [a, b] .
Integral Inequalities of the
Ostrowski Type
A. Sofo
Proof. From Corollary 2.3 we choose
a + x1
, ...,
2
xk−2 + xk−1
xk−1 + xk
=
, αk =
and αk+1 = b.
2
2
α0 = a, α1 =
αk−1
Now utilising the first line of the inequality (3.1), we may evaluate
k−1
X
n+1
(αi+1 − xi )
n+1 + (xi+1 − αi+1 )
i=0
n+1
k−1 X
hi
=
2
2
i=0
and therefore the inequality (3.4) follows.
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For the equidistant partitioning case we have the following inequality.
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Corollary 3.3. Let f : [a, b] → R be a mapping such that f (n−1) is AC[a, b] and
let Ik be defined by (2.8). Then
Z
j
n b
X
b−a
1
(3.5) f (t) dt +
a
2k
j!
j=1
"
#
k−1 n
o
X
× −f (j−1) (a) +
(−1)j − 1 f (j−1) (xi ) + (−1)j f (j−1) (b) i=1
(n) f n+1
∞
≤
n (b − a)
(n + 1)! (2k)
for f (n) ∈ L∞ [a, b] .
Proof. We may utilise (2.9) and from (3.1), note that
b−a
and
k
b−a
hi = xi+1 − xi =
, i = 1, . . . , k − 1
k
h0 = x1 − x0 =
in which case (3.5) follows.
The following inequalities for Taylor-like expansions also hold.
Integral Inequalities of the
Ostrowski Type
A. Sofo
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Corollary 3.4. Let g be defined as in Corollary 2.5. Then we have the inequality
n
X
(−1)j
(3.6) g (y) − g (a) +
j!
j=1
" k
#
o
Xn
×
(xi − αi )j − (xi − αi+1 )j g (j) (xi ) i=0
(n+1) k−1
g
X
∞
≤
(αi+1 − xi )n+1 + (xi+1 − αi+1 )n+1 if g (n+1) ∈ L∞ [a, b]
(n + 1)! i=0
Integral Inequalities of the
Ostrowski Type
A. Sofo
for all xi ∈ [a, y] where
(n+1) g
:= sup g (n+1) (t) < ∞.
∞
Title Page
Proof. Follows directly from (2.11) and using the norm as in (3.1).
Contents
t∈[a,y]
When the points of the division Ik are fixed we obtain the following.
Corollary 3.5. Let g be defined as in Corollary 2.5 and Ik : a = x0 < x1 <
· · · < xk−1 < xk = y be a division of the interval [a, y]. Then we have the
inequality
" k
#
n
o
X
1 Xn j
j j
(j)
−hi + (−1) hi+1 g (xi ) (3.7) g (y) − g (a) +
j
2 j! i=0
j=1
(n+1) k−1
g
X
∞
≤
hn+1 if g (n+1) ∈ L∞ [a, y] .
(n + 1)!2n i=0 i
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Proof. The proof follows directly from using (2.7).
For the equidistant partitioning case we have:
Corollary 3.6. Let g be defined as in Corollary 2.5 and
y−a
Ik : xi = a + i ·
, i = 0, . . . , k
k
be an equidistant partitioning of [a, y], then we have the inequality:
(3.8)
j
n X
y−a
1
g (y) − g (a) +
2k
j!
j=1
"
#
k−1 n
o
X
j
j (j)
(j)
(j)
× −g (a) +
(−1) − 1 g (xi ) + (−1) g (y) i=1
(n+1) g
∞
≤
(y − a)n+1 if g (n+1) ∈ L∞ [a, y] .
(n + 1)! (2k)n
Proof. The proof follows directly upon using (2.9) with f 0 = g and b = y.
Integral Inequalities of the
Ostrowski Type
A. Sofo
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4.
The Convergence of a General Quadrature Formula
Let
(m)
∆m : a = x0
(m)
< x1
(m)
< · · · < xm−1 < x(m)
m = b
be a sequence of division of [a, b] and consider the sequence of real numerical
integration formula
(4.1) Im f, f 0 , . . . , f (n) , ∆m , wm
" m (
!r
j−1
n
m
X
X
X
(−1)r X
(m)
(m)
(m)
−
:=
wj f xj
xj − a −
ws(m)
r!
r=2
s=0
j=0
j=0
!
)
#
r
j
X
(m)
(m)
(r−1)
(m)
− xj − a −
ws
f
xj
,
s=0
(m)
j=0 wj
Pm
where wj (j = 0, . . . , m) are the quadrature weights and assume that
=
b−a.
(m)
The following theorem contains a sufficient condition for the weights wj
so
that
Rb
Im f, f 0 , . . . , f (n) , ∆m , wm approximates
the
integral
f
(x)
dx
with
an
era
(n) ror expressed in terms of f
.
∞
Integral Inequalities of the
Ostrowski Type
A. Sofo
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Theorem 4.1. Let f : [a, b] → R be a continuous mapping on [a, b] such that
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(m)
f (n−1) is AC[a, b]. If the quadrature weights, wj
(4.2)
(m)
xi
−a≤
i
X
(m)
wj
satisfy the condition
(m)
≤ xi+1 − a for all i = 0, . . . , m − 1
j=0
then we have the estimation
Z b
0
(n)
Im f, f , . . . , f , ∆m , wm −
(4.3)
f
(t)
dt
a

!n+1
(n) m−1
i
f X
X
(m)
(m)
∞
 a+
≤
wj − xi
(n + 1)! i=0
j=0
!n+1 
i
X (m)
(m)

− xi+1 − a −
wj
j=0
(n) m−1
n+1
f X (m)
∞
≤
hi
(n + 1)!
(n) i=0
f ∞
≤
ν h(m) (b − a) , where f (n) ∈ L∞ [a, b] .
(n + 1)!
Integral Inequalities of the
Ostrowski Type
A. Sofo
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Also
(m)
ν h
(m)
hi
:=
max
i=0,...,m−1
(m)
n
o
(m)
hi
(m)
:= xi+1 − xi .
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In particular, if f (n) ∞ < ∞, then
0
lim Im f, f , . . . , f
ν (h(m) )→0
(n)
Z
b
, ∆m , wm =
f (t) dt
a
uniformly by the influence of the weights wm .
Proof. Define the sequence of real numbers
(m)
αi+1
:= a +
i
X
(m)
wj ,
Integral Inequalities of the
Ostrowski Type
i = 0, . . . , m.
A. Sofo
j=0
Note that
(m)
αi+1
=a+
m
X
(m)
wj
Title Page
= a + b − a = b.
j=0
By the assumption (4.2), we have
h
i
(m)
(m)
(m)
αi+1 ∈ xi , xi+1 for all i = 0, . . . , m − 1.
(m)
Define α0
(m)
α1
(m)
αi+1
(m)
−
(m)
αi
= a+
j=0
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(m)
= w0
i
X
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= a and compute
− α0
Contents
(m)
wj
−a−
i−1
X
j=0
Page 29 of 83
(m)
wj
=
(m)
wi
(i = 0, . . . , m − 1)
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and
(m)
αm+1
−
(m)
αm
=a+
m
X
(m)
wj
−a−
m−1
X
j=0
(m)
wj
(m)
= wm
.
j=0
Consequently
m X
(m)
(m)
αi+1 − αi
m
X
(m)
(m)
(m)
f xi
=
wi f xi
,
i=0
i=0
Integral Inequalities of the
Ostrowski Type
and let
m
X
(m)
(m)
wj f xj
−
" m (
n
X
(−1)r X
r=2
j=0
r!
(m)
−a−
−a−
X
!r )
ws(m)
A. Sofo
!r
ws(m)
Title Page
s=0
j=0
j
− xj
(m)
xj
j−1
X
(m)
f (r−1) xj
#
Contents
s=0
0
:= Im f, f , . . . , f
(n)
, ∆m , wm .
Applying the inequality (3.1) we obtain the estimate (4.1).
The case when the partitioning is equidistant is important in practice. Consider the equidistant partition
(m)
Em := xi
:= a + i
b−a
, (i = 0, . . . , m)
m
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and define the sequence of numerical quadrature formulae
Im f, f 0 , . . . , f (n) , ∆m , wm
" m (
!r
X
j−1
m
n
r X
X
X
j
(b
−
a)
(−1)
j
(b
−
a)
(m)
:=
wj f a +
−
−
ws(m)
n
r!
n
r=2
s=0
j=0
j=0
!
)
r
#
j
j
(b
−
a)
j (b − a) X (m)
.
−
−
ws
f (r−1) a +
n
n
s=0
The following corollary holds.
Integral Inequalities of the
Ostrowski Type
A. Sofo
(m)
Corollary 4.2. Let f : [a, b] → R be AC[a, b]. If the quadrature weights wj
satisfy the condition
Title Page
i
i
1 X (m) i + 1
≤
w
≤
, i = 0, 1, . . . , n − 1,
m
b − a j=0 j
m
then the following bound holds:
Z b
Im f, f 0 , . . . , f (n) , ∆m , wm −
f (t) dt
a
(n) m−1  i
!n+1
f X
X (m)
b
−
a
∞

wj − i
≤
(n + 1)! i=0
m
j=0
− (i + 1)
b−a
m
−
i
X
j=0
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!n+1 
(m)

wj
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(n) n+1
f b−a
∞
≤
,
(n + 1)!
m
In particular, if f (n) ∞ < ∞, then
0
lim Im f, f , . . . , f
m→∞
(n)
where f (n) ∈ L∞ [a, b] .
Z
, wm =
b
f (t) dt
a
uniformly by the influence of the weights wm .
Integral Inequalities of the
Ostrowski Type
The proof of Corollary 4.2 follows directly from Theorem 4.1.
A. Sofo
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5.
Grüss Type Inequalities
The Grüss inequality [15], is well known in the literature. It is an integral inequality which establishes a connection between the integral of a product of two
functions and the product of the integrals of the two functions.
Theorem 5.1. Let h, g : [a, b] → R be two integrable functions such that φ ≤
h (x) ≤ Φ and γ ≤ g (x) ≤ Γ for all x ∈ [a, b], φ, Φ, γ and Γ are constants.
Then we have
(5.1)
Integral Inequalities of the
Ostrowski Type
1
|T (h, g)| ≤ (Φ − φ) (Γ − γ) ,
4
A. Sofo
where
1
(5.2) T (h, g) :=
b−a
Z
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b
h (x) g (x) dx
Contents
a
1
−
b−a
Z
a
b
1
h (x) dx ·
b−a
b
Z
and the inequality (5.1) is sharp, in the sense that the constant
replaced by a smaller one.
g (x) dx
a
1
4
cannot be
For a simple proof of this fact as well as generalisations, discrete variants, extensions and associated material, see [17]. The Grüss inequality is also utilised
in the papers [6, 7, 14] and the references contained therein.
A premature Grüss inequality is the following.
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Theorem 5.2. Let f and g be integrable functions defined on [a, b] and let γ ≤
g (x) ≤ Γ for all x ∈ [a, b]. Then
(5.3)
|T (h, g)| ≤
1
Γ−γ
(T (f, f )) 2 ,
2
where T (f, f ) is as defined in (5.2).
Theorem 5.2 was proved in 1999 by Matić, Pečarić and Ujević [16] and it
provides a sharper bound than the Grüss inequality (5.1). The term premature is
used to highlight the fact that the result (5.3) is obtained by not fully completing
the proof of the Grüss inequality. The premature Grüss inequality is completed
if one of the functions, f or g, is explicitly known.
We now give the following theorem based on the premature Grüss inequality
(5.3).
Theorem 5.3. Let Ik : a = x0 < x1 < · · · < xk−1 < xk = b be a division of
the interval [a, b], αi (i = 0, . . . , k + 1) be ‘k + 1’ points such that α0 = a, αi ∈
[xi−1 , xi ] (i = 1, . . . , k) and αk = b. If f : [a, b] → R is AC[a, b] and n time
differentiable on [a, b], then assuming that the nth derivative f (n) : (a, b) → R
satisfies the condition
m ≤ f (n) ≤ M for all x ∈ (a, b) ,
Integral Inequalities of the
Ostrowski Type
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we have the inequality
Z b
n
X
(−1)j
n
(5.4) (−1)
f (t) dt + (−1)n
j!
a
j=1
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×
" k−1 (
X
hi
i=0
2
j
− δi
f (j−1) (xi+1 ) −
hi
+ δi
2
j
k−1 n+1
f (n−1) (b) − f (n−1) (a) X hi
−
(b − a) (n + 1)!
2
i=0
#
" n+1 X n + 1 2δi r
{1 + (−1)r } ×
r
h
i
r=0
"
2n+1
k−1 X
b−a
M −m
hi
≤
2
(2n + 1) (n!)2 i=0 2
#
"2n+1 X 2n + 1 2δi r
{1 + (−1)r }
×
hi
r
r=0
n+1
k−1
X
1
hi
−
(n + 1)! i=0 2
" n+1 #!2  21
r
X n+1
2δi

×
{1 + (−1)r }
r
h
i
r=0
)#
f (j−1) (xi )
where
Integral Inequalities of the
Ostrowski Type
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hi := xi+1 − xi and
xi+1 + xi
, i = 0, . . . , k − 1.
δi := αi+1 −
2
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Proof. We utilise (5.2) and (5.3), multiply through by (b − a) and choose h (t) :=
Kn,k (t) as defined by (2.2) and g (t) := f (n) (t), t ∈ [a, b] such that
(5.5)
Z b
Z b
Z b
1
(n)
(n)
K
(t)
f
(t)
dt
−
f
(t)
dt
·
K
(t)
dt
n,k
n,k
b−a a
a
a
"
Z b
2 # 12
Z b
Γ−γ
2
≤
(b − a)
Kn,k
(t) dt −
Kn,k (t) dt
.
2
a
a
Integral Inequalities of the
Ostrowski Type
Now we may evaluate
Z b
A. Sofo
f
(n)
(t) dt = f
(n−1)
(b) − f
(n−1)
(a)
a
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and
Z
G1 :=
=
Contents
b
Kn,k (t) dt
a
k−1 Z xi+1
X
i=0
xi
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1
(t − αi+1 )n dt
n!
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k−1
=
X
1
(xi+1 − αi+1 )n+1 + (αi+1 − xi )n+1 .
(n + 1)! i=0
hi
− δi
2
and
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Using the definitions of hi and δi we have
xi+1 − αi+1 =
II
I
αi+1 − xi =
hi
+ δi
2
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such that
n+1 n+1 )
hi
hi
− δi
+
+ δi
2
2
" n+1 n+1−r
k−1 X
X
1
n+1
hi
r
=
(−δi )
(n + 1)! i=0 r=0
r
2
n+1−r #
n+1 X
n + 1 r hi
+
δi
r
2
r=0
"
#
n+1 X
r
k−1 n+1 X
hi
n+1
2δi
1
{1 + (−1)r } .
=
(n + 1)! i=0 2
r
h
i
r=0
k−1
X
1
G1 =
(n + 1)! i=0
(
Integral Inequalities of the
Ostrowski Type
A. Sofo
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Also,
Z b
2
G2 :=
Kn,k
(t) dt
=
a
k−1
X Z xi+1
i=0
xi
(t − αi+1 )2n
dt
(n!)2
k−1
X
1
=
(xi+1 − αi+1 )2n+1 + (αi+1 − xi )2n+1
2
(2n + 1) (n!) i=0
(
2n+1 2n+1 )
k−1
X
1
hi
hi
− δi
+
+ δi
=
2
2
(2n + 1) (n!)2 i=0
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#
2n+1 "2n+1
k−1 X 2n + 1 2δi r
X
1
hi
{1 + (−1)r } .
=
r
h
(2n + 1) (n!)2 i=0 2
i
r=0
From identity (2.1), we may write
Z
b
Kn,k (t) f
a
(n)
n
Z
(t) dt = (−1)
b
n
f (t) dt + (−1)
n
X
(−1)j
j!
j=1
" k−1
#
o
Xn
×
(xi+1 − αi+1 )j f (j−1) (xi+1 ) − (αi+1 − xi )j f (j−1) (xi )
a
Integral Inequalities of the
Ostrowski Type
i=0
A. Sofo
and from the left hand side of (5.5) we obtain
n
Z
G3 := (−1)
b
n
f (t) dt + (−1)
a
×
j=1
" k−1 (
X
hi
i=0
2
j
− δi
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n
X
(−1)j
Contents
j!
f (j−1) (xi+1 ) −
hi
+ δi
2
k−1 n+1
f (n−1) (b) − f (n−1) (a) X hi
−
(b − a) (n + 1)!
2
i=0
" n+1 #
r
X n+1
2δi
×
{1 + (−1)r }
r
hi
r=0
after substituting for G1 .
j
)#
f (j−1) (xi )
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From the right hand side of (5.5) we substitute for G1 and G2 so that
2
:= (b − a)
(t) dt −
Kn,k (t) dt
a
a
#
2n+1 "2n+1
k−1 X
X 2n + 1 2δi r
b−a
hi
=
{1 + (−1)r }
2
2
r
h
(2n + 1) (n!) i=0
i
r=0
"
#!2
n+1 X
r
k−1 n+1 X
hi
n+1
2δi
1
{1 + (−1)r }
.
−
(n + 1)! i=0 2
r
h
i
r=0
Z
G4
b
Z
2
Kn,k
b
A. Sofo
Hence,
|G3 | ≤
1
M −m
(G4 ) 2 ,
2
and Theorem 5.3 has been proved.
δ = αi+1 −
xi+1 + xi
2
for all i = 0, . . . , k − 1 such that
(5.7)
1
|δ| ≤ min {hi |i = 1, . . . , k} .
2
The following inequality applies:
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Corollary 5.4. Let f , Ik and αk be defined as in Theorem 5.3 and further define
(5.6)
Integral Inequalities of the
Ostrowski Type
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Z b
n
X
1
n
n
f (t) dt + (−1)
(5.8) (−1)
j!
a
j=1
" k−1 (
j
X
hi
j
×
(−i)
− δi f (j−1) (xi+1 )
2
i=0
)# j
f (n−1) (b) − f (n−1) (a)
hi
(j−1)
−
+ δi f
(xi )
−
2
(b − a) (n + 1)!
n+1−r
k−1 X
n+1
X
hi
r
{1 + (−1)r }
δ
2
i=0 r=0
k−1 2n+1
X
X 2n + 1 b−a
M −m
≤
δr
2 ×
r
2
(2n + 1) (n!)
i=0 r=0
#
2n+1−r
k−1 X
n+1 X
hi
1
n+1 r
r
×
{1 + (−1) } −
δ
2
(n + 1)! i=0 r=0
r
!2  21
n+1−r
hi
×
{1 + (−1)r }  .
2
The proof follows directly from (5.4) upon the substitution of (5.6) and some
minor simplification.
Integral Inequalities of the
Ostrowski Type
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Remark 5.1. If for any division Ik : a = x0 < x1 < · · · < xk−1 < xk = b of
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the interval [a, b], we choose δ = 0 in (5.6), we have the inequality
Z b
n
X
1
n
(5.9)
f (t) dt + (−1)n
(−1)
j!
a
j=1
j
k−1 o
X
hi n
×
(−1)j f (j−1) (xi+1 ) − f (j−1) (xi )
2
i=0
(n−1)
k−1 n+1 f
(b) − f (n−1) (a) X hi
−2
(b − a) (n + 1)!
2
i=0
"
2n+1
k−1 M −m
2 (b − a) X hi
≤
2
(2n + 1) (n!)2 i=0 2
1
n+1 !2 2
k−1 X
hi
2
 .
−
(n + 1)! i=0 2
The proof follows directly from (5.8).
Remark 5.2. Let f (n) be defined as in Theorem 5.3 and consider an equidistant
partitioning Ek of the interval [a, b], where
b−a
, i = 0, . . . , k.
Ek := xi = a + i
k
Integral Inequalities of the
Ostrowski Type
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The following inequality applies
j
Z b
n
X
1 b−a
n
n
(5.10)
f (t) dt + (−1)
(−1)
j!
2k
a
j=1
a (k − i − 1) + b (i − 1)
j (j−1)
× (−1) f
k
a (k − i) + ib
−f (j−1)
k
(n−1)
n+1 b−a
f
(b) − f (n−1) (a)
−2
(b − a) (n + 1)!
2k
n+1
nk
b−a
√
≤ (M − m) ·
.
2k
(n + 1)! 2n + 1
Proof. The proof follows upon noting that hi = xi+1 −xi = b−a
, i = 0, . . . , k.
k
Integral Inequalities of the
Ostrowski Type
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6.
Some Particular Integral Inequalities
In this subsection we point out some special cases of the integral inequalities in
Section 3. In doing so, we shall recover, subsume and extend the results of a
number of previous published papers [2, 5].
We shall recover the left and right rectangle inequalities, the perturbed trapezoid inequality, the midpoint and Simpson’s inequalities and the Newton-Cotes
three eighths inequality, and a Boole type inequality.
In the case when n = 1, for the kernel K1,k (t) of (2.3), the inequality (3.1),
reduces to the results obtained by Dragomir [5] for the cases when f : [a, b] →
R is absolutely continuous and f 0 belongs to the L∞ [a, b] space.
Similarly, for n = 1, Dragomir [11] extended Theorem 3.1 for the case when
f [a, b] → R is a function of bounded variation on [a, b].
For the two branch Peano kernel,

1

n


 n! (t − a) , t ∈ [a, x)
(6.1)
Kn,2 (t) :=


1


(t − b)n , t ∈ (x, b]
n!
Integral Inequalities of the
Ostrowski Type
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the inequality (3.1) reduces to the result (1.2) obtained by Cerone and Dragomir
[1] and [3]. A number of other particular cases are now investigated.
Close
Theorem 6.1. Let f : [a, b] → R be an absolutely continuous mapping on [a, b]
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and a ≤ x1 ≤ b, a ≤ α1 ≤ x1 ≤ α2 ≤ b. Then we have
Z
n
b
n
X
(−1)j (6.2)
f (t) dt +
− (a − α1 )j f (j−1) (a) + (x1 − α1 )j
a
j!
j=1
o
− (x1 − α2 )j f (j−1) (x1 ) + (b − α2 )j f (j−1) (b) (n) f ∞
≤
(α1 − a)n+1 + (x1 − α1 )n+1
(n + 1)!
+ (α2 − x1 )n+1 + (b − α2 )n+1
(n) f ∞
≤
(x1 − a)n+1 + (b − x1 )n+1
(n + 1)!
(n) f ∞
≤
(b − a)n+1 ,
f (n) ∈ L∞ [a, b] .
(n + 1)!
Proof. Consider the division a = x0 ≤ x1 ≤ x2 = b and the numbers α0 = a,
α1 ∈ [a, x1 ), α2 ∈ (x1 , b] and α3 = b.
From the left hand side of (3.1) we obtain
n
2
o
X
(−1)j X n
j
j
(xi − αi ) − (xi − αi+1 ) f (j−1) (xi )
j!
j=1
i=0
n
X (−1)j − (a − α1 )j f (j−1) (a)
=
j!
j=1
n
o
j
j
j (j−1)
(j−1)
+ (x1 − α1 ) − (x1 − α2 ) f
(x1 ) + (b − α2 ) f
(b) .
Integral Inequalities of the
Ostrowski Type
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From the right hand side of (3.1) we obtain
(n) 1
f X
∞
(αi+1 − xi )n+1 + (xi+1 − αi+1 )n+1
(n + 1)! i=0
(n) f ∞
(α1 − a)n+1 + (x1 − α1 )n+1 + (α2 − x1 )n+1 + (b − α2 )n+1
=
(n + 1)!
and hence the first line of the inequality (6.2) follows.
Notice that if we choose α1 = a and α2 = b in Theorem 6.1 we obtain the
inequality (1.2).
Integral Inequalities of the
Ostrowski Type
A. Sofo
The following proposition embodies a number of results, including the Ostrowski inequality, the midpoint and Simpson’s inequalities and the three-eighths
Newton-Cotes inequality including its generalisation.
Proposition 6.2. Let f be defined as in Theorem 6.1 and let a ≤ x1 ≤ b, and
≤ x1 ≤ a+(m−1)b
≤ b for m a natural number, m ≥ 2, then we
a ≤ (m−1)a+b
m
m
have the inequality
Z
"
j
n
j
b
X
b − a (j−1)
(−1)
(6.3)
|Pm,n | := f (t) dt +
f
(b)
a
j!
m
j=1
(
j
o
b−a
j (j−1)
− (−1) f
(a) +
(x1 − a) −
m
#
j )
b−a
(j−1)
−
− (b − x1 )
f
(x1 ) m
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(n) n+1 n+1
f b−a
b−a
∞
≤
2
+ x1 − a −
(n + 1)!
m
m
!
n+1
b−a
+ b − x1 −
if f (n) ∈ L∞ [a, b] .
m
Proof. From Theorem 6.1 we note that
α1 =
(m − 1) a + b
a + (m − 1) b
and α2 =
m
m
Integral Inequalities of the
Ostrowski Type
so that
A. Sofo
b−a
b−a
, b − α2 =
,
m
m
b−a
b−a
= x1 − a −
and x1 − α2 =
− (b − x1 ) .
m
m
a − α1 = −
x1 − α1
From the left hand side of (6.2) we have
n
− (a − α1 )j f (j−1) (a) + (x1 − α1 )j
o
j
− (x1 − α2 ) f (j−1) (x1 ) + (b − α2 )j f (j−1) (b)
j
b − a (j−1)
=
f
(b) − f (j−1) (a)
m
(
j j )
b−a
b−a
+
x1 − a −
−
− (b − x1 )
f (j−1) (x1 ) .
m
m
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From the right hand side of (6.2),
(α1 − a)n+1 + (x1 − α1 )n+1 + (α2 − x1 )n+1 + (b − α2 )n+1
n+1 n+1
b−a
b−a
=2
+ x1 − a −
m
m
n+1
b−a
+ b − x1 −
m
and the inequality (6.3) follows, hence the proof is complete.
Integral Inequalities of the
Ostrowski Type
The following corollary points out that the optimum of Proposition 6.2 oc2
curs at x1 = α1 +α
= a+b
in which case we have:
2
2
A. Sofo
Corollary 6.3. Let f be defined as in Proposition 6.2 and let x1 = a+b
in which
2
case we have the inequality
a + b (6.4)
Pm,n
2
Z
"
j
n
b
X
(−1)j
b−a
:= f (t) dt +
a
j!
m
j=1
n
o
j (j−1)
(j−1)
× f
(b) − (−1) f
(a)
j
#
(m − 2) (b − a) a
+
b
j
+
1 − (−1) f (j−1)
2m
2
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2 f (n) ∞ b − a n+1
≤
(n + 1)!
m
1+
m−2
2
n+1 !
if f (n) ∈ L∞ [a, b] .
The proof follows directly from (6.3) upon substituting x1 = a+b
.
2
A number of other corollaries follow naturally from Proposition 6.2 and
Corollary 6.3 and will now be investigated.
The following two corollaries generalise the Simpson inequality and follow
directly from (6.3) and (6.4) for m = 6.
Integral Inequalities of the
Ostrowski Type
A. Sofo
Corollary 6.4. Let the conditions of Corollary 6.3 hold and put m = 6. Then
we have the inequality
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(6.5)
|P6,n |
Z b
:= f (t) dt
a
"
j
n
o
X
(−1)j
b − a n (j−1)
+
f
(b) − (−1)j f (j−1) (a)
j!
6
j=1
(
#
j j )
5a + b
a + 5b
− x1 −
f (j−1) (x1 ) +
x1 −
6
6
(n) n+1 n+1
f b−a
5a + b
∞
≤
2
+ x1 −
(n + 1)!
6
6
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a + 5b
+ −x1 +
6
n+1 !
,
f (n) ∈ L∞ [a, b] ,
which is the generalised Simpson inequality.
Corollary 6.5. Let the conditions of Corollary 6.3 hold and put m = 6,. Then
at the midpoint x1 = a+b
we have the inequality
2
a + b (6.6)
P6,n
2
Z b
:= f (t) dt
a
"
j
n
o
X
b − a n (j−1)
(−1)j
f
(b) − (−1)j f (j−1) (a)
+
j!
6
j=1
j
#
a
+
b
b−a j
+
1 − (−1) f (j−1)
3
2
(n) 2 f ∞ b − a n+1
≤
1 + 2n+1 , f (n) ∈ L∞ [a, b] .
(n + 1)!
6
Remark 6.1. Choosing n = 1 in (6.6) we have
a + b (6.7) P6,1
2
Z b
b
−
a
a
+
b
:= f (t) dt −
f (a) + 4f
+ f (b) 6
2
a
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≤
5
kf 0 k∞ (b − a)2 , f 0 ∈ L∞ [a, b] .
36
Remark 6.2. Choosing n = 2 in (6.6) we have a perturbed Simpson type inequality,
a
+
b
P6,2
(6.8)
2
Z b
b−a
a+b
:= f (t) dt −
f (a) + 4f
+ f (b)
6
2
a
2 0
f (b) − f 0 (a) b−a
+
6
2
(b − a)3 00
≤
kf k∞ , f 00 ∈ L∞ [a, b] .
72
Corollary 6.6. Let f be defined as in Corollary 6.3 and let m = 4, then we
have the inequality
a
+
b
P4,n
(6.9)
2
Z
j
n
b
X
(−1)j b − a
:= f (t) dt +
a
j!
4
j=1
h
× f (j−1) (b) − (−1)j f (j−1) (a)
a + b j
(j−1)
+ 1 − (−1) f
2
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(n) f ∞
≤
(b − a)n+1 , f (n) ∈ L∞ [a, b] .
n
(n + 1)!4
Remark 6.3. From (6.9) we choose n = 2 and we have the inequality
a + b (6.10) P4,2
2
Z b
b−a
a+b
:= f (t) dt −
f (b) + f (a) + 2f
4
2
a
2 0
f (b) − f 0 (a) b−a
+
4
2
≤
kf 00 k∞
(b − a)3 , f 00 ∈ L∞ [a, b] .
96
Integral Inequalities of the
Ostrowski Type
A. Sofo
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Contents
Theorem 6.7. Let f : [a, b] → R be an absolutely continuous mapping on [a, b]
and let a < x1 ≤ x2 ≤ b and α1 ∈ [a, x1 ), α2 ∈ [x1 , x2 ) and α3 ∈ [x2 , b]. Then
we have the inequality
Z
n
b
X
(−1)j h
(6.11) − (a − α1 )j f (j−1) (a)
f (t) dt +
a
j!
j=1
j
j
+ (x1 − α1 ) − (x1 − α2 ) f (j−1) (x1 )
+ (x2 − α2 )j − (x2 − α3 )j f (j−1) (x2 )
i
+ (b − α3 )j f (j−1) (b) JJ
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(n) f ∞
≤
(α1 − a)n+1 + (x1 − α1 )n+1 + (α2 − x1 )n+1
(n + 1)!
+ (x2 − α2 )n+1 + (α3 − x2 )n+1 + (b − α3 )n+1 ,
f (n) ∈ L∞ [a, b] .
Proof. Consider the division a = x0 < x1 < x2 = b, α1 ∈ [a, x1 ), α2 ∈
[x1 , x2 ), α3 ∈ [x2 , b], α0 = a, x0 = a, x3 = b and put α4 = b. From the left
hand side of (3.1) we obtain
3
n
o
X
(−1)j X n
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi )
j! i=0
j=1
=
n
X
(−1)j h
j=1
j!
− (a − α1 )j f (j−1) (a)
+ (x1 − α1 )j − (x1 − α2 )j f (j−1) (x1 )
i
+ (x2 − α2 )j − (x2 − α3 )j f (j−1) (x2 ) + (b − α3 )j f (j−1) (b) .
From the right hand side of (3.1) we obtain
(n) 2
f X
∞
(αi+1 − xi )n+1 + (xi+1 − αi+1 )n+1
(n + 1)! i=0
(n) f ∞
=
(α1 − a)n+1 + (x1 − α1 )n+1 + (α2 − x1 )n+1
(n + 1)!
+ (x2 − α2 )n+1 + (α3 − x2 )n+1 + (b − α3 )n+1
Integral Inequalities of the
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and hence the first line of the inequality (6.11) follows and Theorem 6.7 is
proved.
Corollary 6.8. Let f be defined as in Theorem 6.7 and consider the division
a ≤ α1 ≤
(m − 1) a + b
a + (m − 1) b
≤ α2 ≤
≤ α3 ≤ b
m
m
for m a natural number, m ≥ 2. Then we have the inequality
(6.12)
|Qm,n |
Z
n
b
X
(−1)j h
:= f (t) dt +
− (a − α1 )j f (j−1) (a)
a
j!
j=1
(
j
(m − 1) a + b
+
− α1
m
j )
(m − 1) a + b
−
− α2
f (j−1) (x1 )
m
(
j
a + (m − 1) b
+
− α2
m
j )
a + (m − 1) b
−
− α3
f (j−1) (x2 )
m
i
+ (b − α3 )j f (j−1) (b) Integral Inequalities of the
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(n) f ∞
≤
Rm,n
(n + 1)!
(n) n+1
f (m − 1) a + b
n+1
∞
:=
(α1 − a)
+
− α1
(n + 1)!
m
n+1 n+1
(m − 1) a + b
a + (m − 1) b
+ α2 −
+
− α2
m
m
!
n+1
a + (m − 1) b
+ α3 −
+ (b − α3 )n+1 ,
m
Integral Inequalities of the
Ostrowski Type
A. Sofo
f (n) ∈ L∞ [a, b] .
Proof. Choose in Theorem 6.7, x1 =
theorem is proved.
(m−1)a+b
,
m
and x2 =
a+(m−1)b
,
m
hence the
Contents
Remark 6.4. For particular choices of the parameters m and n, Corollary 6.8
contains a generalisation of the three-eighths rule of Newton and Cotes.
The following corollary is a consequence of Corollary 6.8.
Corollary 6.9. Let f be defined as in Theorem 6.7 and choose α2 =
x1 +x2
, then we have the inequality
2
Q̄m,n (6.13)
Z
n
b
X
(−1)j h
:= f (t) dt +
− (a − α1 )j f (j−1) (a)
a
j!
j=1
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a+b
2
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(
j )
(m − 2) (b − a)
+ (x1 − α1 ) − (−1)
f (j−1) (x1 )
2m
(
)
j
(m − 2) (b − a)
j
+
− (x2 − α3 ) f (j−1) (x2 )
2m
i
+ (b − α3 )j f (j−1) (b) (n) f ∞
≤
R̄m,n
(n + 1)!
(n) n+1
f (m − 1) a + b
n+1
∞
:=
(α1 − a)
+
− α1
(n + 1)!
m
n+1 n+1
a + (m − 1) b
(m − 2)
+ 2 (b − a)
+ α3 −
2m
m
n+1 + (b − α3 )
,
f (n) ∈ L∞ [a, b] .
j
Proof. If we put α2 = a+b
=
2
and the corollary is proved.
j
x1 +x2
2
into (6.12), we obtain the inequality (6.13)
The following corollary contains an optimum estimate for the inequality
(6.13).
Corollary 6.10. Let f be defined as in Theorem 6.7 and make the choices
4−m
3m − 4
a+
b and
α1 =
2m
2m
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α3 =
4−m
2m
a+
3m − 4
2m
b
then we have the best estimate
(6.14)
Q̂
m,n Z
n
b
X
(−1)j
:= f (t) dt +
a
j!
j=1
"
j
o
(b − a) (4 − m) n (j−1)
×
f
(b) − (−1)j f (j−1) (a)
2m
#
j
o
(b − a) (m − 2) n
j
(j−1)
(j−1)
+
1 − (−1)
f
(x1 ) + f
(x2 ) 2m
(n) 2 f ∞ b − a n+1
≤
(4 − m)n+1 + 2 (m − 2)n+1 ,
(n + 1)!
2m
Integral Inequalities of the
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A. Sofo
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f (n) ∈ L∞ [a, b] .
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Proof. Using the choice α2 =
we may calculate
a+b
2
=
x1 +x2
,
2
x1 =
(m−1)a+b
m
and x2 =
a+(m−1)b
m
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(4 − m) (b − a)
= (b − α3 )
(α1 − a) =
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and
(x1 − α1 ) = (α2 − x1 ) = (x2 − α2 ) = (α3 − x2 )
(m − 2) (b − a)
=
.
2m
Substituting in the inequality (6.13) we obtain the proof of (6.14).
Remark 6.5. For m = 3, we have the best estimation of (6.14) such that
(6.15) Q̂3,n Z
n
b
X
(−1)j
:= f (t) dt +
a
j!
j=1
"
j
o b − a j
b − a n (j−1)
j (j−1)
f
×
(b) − (−1) f
(a) +
6
6
n
o
2a + b
a + 2b
j
(j−1)
(j−1)
× 1 − (−1)
f
+f
2
2
(n) f ∞
≤ n
(b − a)n+1 , f (n) ∈ L∞ [a, b] .
6 (n + 1)!
Proof. From the right hand side of (6.14), consider the mapping
Mm,n :=
4−m
2m
n+1
+2
m−2
2m
n+1
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then
n n 2 (n + 1)
2
1
1
1
=
−
−
+
−
m2
m 2
2 m
and Mm,n attains its optimum when
0
Mm,n
1
1
1
2
− = − ,
m 2
2 m
in which case m = 3. Substituting m = 3 into (6.14), we obtain (6.15) and the
corollary is proved.
When n = 2, then from (6.15) we have
Z b
b
−
a
f (t) dt−
(f (b) + f (a))
Q̂3,2 := 6
a
2
1 b−a
+
(f 0 (b) − f 0 (a))
2
6
2a + b
a + 2b
b−a
f
+f
+
3
2
2
kf 00 k∞
≤
(b − a)3 , f 00 ∈ L∞ [a, b] .
216
The next corollary encapsulates the generalised Newton-Cotes inequality.
Corollary 6.11. Let f be defined as in Theorem 6.7 and choose
2a + b
a + 2b
a+b
, x2 =
, α2 =
,
3
3
2
(r − 1) a + b
a + (r − 1) b
=
and α3 =
.
r
r
x1 =
α1
Integral Inequalities of the
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Then for r a natural number, r ≥ 3, we have the inequality
(6.16) |Tr,n |
Z b
:= f (t) dt
a
n
X
"
j
o
b − a n (j−1)
f
(b) − (−1)j f (j−1) (a)
+
r
j=1
(
j
j )
(b − a) (r − 3)
b−a
j
+
− (−1)
f (j−1) (x1 )
3r
6
(
#
j
j )
b−a
(b − a) (r − 3)
j
(j−1)
+
− (−1)
f
(x2 ) 6
3r
!
(n) n+1
2 f ∞
1
r−3
1
n+1
≤
(b − a)
+
+ n+1 ,
n+1
(n + 1)!
r
3r
6
(−1)j
j!
Integral Inequalities of the
Ostrowski Type
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J
f (n) ∈ L∞ [a, b] .
Proof. From Theorem 6.7, we put x1 = 2a+b
, x2 =
3
(r−1)a+b
a+(r−1)b
and α3 =
. Then (6.16) follows.
r
r
a+2b
,
3
α2 =
a+b
,
2
α1 =
Remark 6.6. The optimum estimate of the inequality (6.16) occurs when r = 6.
from (6.16) consider the mapping
n+1
1
r−3
1
Mr,n := n+1 +
+ n+1
r
3r
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1 1 n
0
the Mr,n
= − (n + 1) r−n−2 + n+1
− r and Mr,n attains its optimum
r2
3
1
1
1
when r = 3 − r , in which case r = 6. In this case, we obtain the inequality
(6.15) and specifically for n = 1, we obtain a Simpson type inequality
Z b
b−a
(6.17)
(f (a) + f (b))
f (t) dt −
Q̂3,1 := 6
a
2a + b
a + 2b
b−a
f
+f
−
3
3
3
kf 0 k∞
≤
(b − a)2 , f 0 ∈ L∞ [a, b] ,
12
which is better than that given by (6.7).
Corollary 6.12. Let f be defined as in Theorem 6.7 and choose m = 8 such
that α1 = 7a+b
, α2 = a+b
and α3 = a+7b
with x1 = 2a+b
and x2 = a+2b
. Then
8
8
8
3
3
we have the inequality
(6.18) |T8,n |
Z b
:= f (t) dt
a
"
j
n
X
(−1)j
b − a (j−1)
+
f
(b) − (−1)j f (j−1) (a)
j!
8
j=1
!
j j
b−a
5
+
− (−1)j
6
4
i
× f (j−1) (x1 ) − (−1)j f (j−1) (x2 ) Integral Inequalities of the
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2 f (n) ∞ b − a n+1 n+1
≤
3
+ 4n+1 + 5n+1 ,
(n + 1)!
24
f (n) ∈ L∞ [a, b] .
Proof. From Theorem 6.7 we put
x1 =
and α2 =
a+b
2
2a + b
a + 2b
7a + b
a + 7b
, x2 =
, α1 =
, α3 =
3
3
8
8
and the inequality (6.18) is obtained.
When n = 1 we obtain from (6.18) the ‘three-eighths rule’ of Newton-Cotes.
Remark 6.7. From (6.16) with r = 3 we have
Z
"
j
n
b
X
b − a (j−1)
(−1)j
j (j−1)
f
(b) − (−1) f
(a)
|T3,n | := f (t) dt +
a
j!
3
j=1
#
j
b − a (j−1)
+
f
(x2 ) − f (j−1) (x1 ) 6
(n) 2 f ∞
1
1
n+1
≤
(b − a)
+
, f (n) ∈ L∞ [a, b] .
(n + 1)!
3n+1 6n+1
In particular, for n = 2, we have the inequality
Z
2 0
b
b−a
b−a
f (b) − f 0 (a)
|T3,2 | := f (t) dt −
(f (b) + f (a)) +
a
3
3
2
b−a
a + 2b
2a + b
−
f
+f
6
3
3
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+
≤
b−a
6
2
f0
a+2b
3
− f0
2
2a+b
3
!
kf 00 k∞
(b − a)3 , f 00 ∈ L∞ [a, b] .
72
The following theorem encapsulates Boole’s rule.
Theorem 6.13. Let f : [a, b] → R be an absolutely continuous mapping on
[a, b] and let a < x1 < x2 < x3 < b and α1 ∈ [a, x1) , α2 ∈ [x1 , x2 ), α3 ∈
[x2 , x3 ) and α4 ∈ [x3 , b]. Then we have the inequality
Z
n
b
X
(−1)j h
(6.19)
− (a − α1 )j f (j−1) (a)
f
(t)
dt
+
a
j!
j=1
+ (x1 − α1 )j − (x1 − α2 )j f (j−1) (x1 )
+ (x2 − α2 )j − (x2 − α3 )j f (j−1) (x2 )
j
j
+ (x3 − α3 ) − (x3 − α4 ) f (j−1) (x3 )
i
+ (b − α4 )j f (j−1) (b) (n) f ∞
≤
(α1 − a)n+1 + (x1 − α1 )n+1 + (α2 − x1 )n+1
(n + 1)!
+ (x2 − α2 )n+1 + (α3 − x2 )n+1 + (x3 − α3 )n+1
+ (α4 − x3 )n+1 + (b − α4 )n+1 if f (n) ∈ L∞ [a, b] .
Integral Inequalities of the
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Proof. Follows directly from (3.1) with the points α0 = x0 = a, x4 = α5 = b
and the division a = x0 < x1 < x2 < x3 = b, α1 ∈ [a, x1) , α2 ∈ [x1 , x2 ),
α3 ∈ [x2 , x3 ) and α4 ∈ [x3 , b].
The following inequality arises from Theorem 6.13.
Corollary 6.14. Let f be defined as in Theorem 6.7 and choose α1 = 11a+b
,
12
7a+11b
a+11b
7a+2b
2a+7b
α2 = 11a+7b
,
α
=
,
α
=
,
x
=
,
x
=
and
x
=
3
4
1
3
2
18
18
12
9
9
x1 +x3
a+b
=
,
then
we
can
state:
2
2
Z
"
j
n
j b
o
X
b − a n (j−1)
(−1)
j (j−1)
f
(b)
−
(−1)
f
(a)
f
(t)
dt
+
a
j!
12
j=1
(
)
j j
b−a
5
j
+
− (−1)
6
6
7a + 2b
2a + 7b
j (j−1)
(j−1)
× f
− (−1) f
9
9
#
j n
o
b−a
a + b +
1 − (−1)j f (j−1)
9
2
(n) 2 f ∞ b − a n+1 n+1
≤
3
+ 4n+1 + 5n+1 + 6n+1 ,
(n + 1)!
36
if f
(n)
∈ L∞ [a, b] .
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7.
Applications for Numerical Integration
In this section we utilise the particular inequalities of the previous sections and
apply them to numerical integration.
Consider the partitioning of the interval [a, b] given by ∆m : a = x0 <
x1 < · · · < xm−1 < xm = b, put hi := xi+1 − xi (i = 0, . . . , m − 1) and put
ν (h) := max (hi |i = 0, . . . , m − 1). The following theorem holds.
Theorem 7.1. Let f : [a, b] → R be AC[a, b], k ≥ 1 and m ≥ 1. Then we have
the composite quadrature formula
Z b
(7.1)
f (t) dt = Ak (∆m , f ) + Rk (∆m , f )
Integral Inequalities of the
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a
where
(7.2)
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Ak (∆m , f ) := −Tk (∆m , f ) − Uk (∆m , f ) ,
(7.3) Tk (∆m , f ) :=
m−1
n XX
i=0 j=1
hi
2k
j
i
1 h (j−1)
j (j−1)
−f
(xi ) + (−1) f
(xi+1 )
j!
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and
(7.4) Uk (∆m , f ) :=
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m−1
n XX
i=0 j=1
hi
2k
j
1
j!
" k−1
#
o
Xn
(k
−
r)
x
+
rx
i
i+1
×
(−1)j − 1 f (j−1)
k
r=1
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is a perturbed quadrature formula. The remainder Rk (∆m , f ) satisfies the estimation
(n) m−1
f X
∞
(7.5)
|Rk (∆m , f )| ≤ n
hn+1
, if f (n) ∈ L∞ [a, b] ,
i
n+1
2 (n + 1)!k
i=0
where ν (h) := max (hi |i = 0, . . . , m − 1).
Proof. We shall apply Corollary 3.3 on the interval [xi , xi+1 ], (i = 0, . . . , m − 1).
Thus we obtain
Z
j h
n xi+1
X
hi
1
f (t) dt +
−f (j−1) (xi ) + (−1)j f (j−1) (xi+1 )
xi
2k
j!
j=1
#
k−1 n
o
X
(k
−
r)
x
+
rx
i
i+1
+
(−1)j − 1 f (j−1)
k
r=1
(n) xi+1 − xi n+1
1
≤
sup f (t)
.
(n + 1)!2n t∈[xi ,xi+1 ]
k
Summing over i from 0 to m − 1 and using the generalised triangle inequality,
we have
j h
m−1
n X Z xi+1
X
hi
1
f (t) dt +
−f (j−1) (xi ) + (−1)j f (j−1) (xi+1 )
xi
2k
j!
i=0
j=1
#
k−1
n
o
X
(k − r) xi + rxi+1 j
(j−1)
+
(−1) − 1 f
k
r=1
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m−1
X hn+1
1
i
≤
(n + 1)!2n i=0 k n+1
sup
(n) f (t) .
t∈[xi ,xi+1 ]
Now,
Z
j
m−1
n b
XX
1
hi
(7.6) f (t) dt +
a
2k
j!
i=0 j=1
j
n i m−1
XX
hi
1
× −f
(xi ) + (−1) f
(xi+1 ) +
2k
j!
i=0 j=1
" k−1
#
o
Xn
(k
−
r)
x
+
rx
i
i+1
j
×
(−1) − 1 f (j−1)
k
r=1
h
As
(j−1)
j
(j−1)
≤ Rk (∆m , f ) .
sup f (n) (t) ≤ f (n) ∞ , the inequality in (7.5) follows and the theo-
t∈[xi ,xi+1 ]
rem is proved.
The following corollary holds.
Corollary 7.2. Let f : [a, b] → R be a mapping such that f (n−1) is AC[a, b],
then we have the equality
Z b
(7.7)
f (t) dt = −T2 (∆m , f ) − U2 (∆m , f ) + R2 (∆m , f ) ,
a
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where
T2 (∆m , f ) :=
m−1
n XX
i=0 j=1
hi
4
j
i
1 h (j−1)
j (j−1)
−f
(xi ) + (−1) f
(xi+1 )
j!
U2 (∆m , f ) is the perturbed midpoint quadrature rule, containing only even
derivatives
j n
m−1
n o
XX
hi
1
xi + xi+1
j
(j−1)
U2 (∆m , f ) :=
(−1) − 1 f
4
j!
2
i=0 j=1
Integral Inequalities of the
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and the remainder, R2 (∆m , f ) satisfies the estimation
(n) m−1
f X
∞
|R2 (∆m , f )| ≤ 2n+1
hn+1 , if f (n) ∈ L∞ [a, b] .
2
(n + 1)! i=0 i
Corollary 7.3. Let f and ∆m be defined as above. Then we have the equality
Z b
(7.8)
f (t) dt = −T3 (∆m , f ) − U3 (∆m , f ) + R3 (∆m , f ) ,
a
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where
T3 (∆m , f ) :=
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m−1
n XX
i=0 j=1
hi
6
j
i
1 h (j−1)
j (j−1)
−f
(xi ) + (−1) f
(xi+1 )
j!
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and
U3 (∆m , f ) :=
m−1
n
XX
(−1)j − 1 h j
i
j!
i=0 j=1
6
2xi + xi+1
xi + 2xi+1
(j−1)
(j−1)
× f
+f
3
3
and the remainder satisfies the bound
(n) m−1
f X
∞
hn+1 , if f (n) ∈ L∞ [a, b] .
|R3 (∆m , f )| ≤
3 · 6n (n + 1)! i=0 i
Theorem 7.4. Let f and ∆m be defined as in Theorem 7.1 and suppose that
ξi ∈ [xi , xi+1 ] (i = 0, . . . , m − 1). Then we have the quadrature formula:
Z
b
f (t) dt =
(7.9)
a
m−1
n
XX
i=0 j=1
(−1)j n
(ξi − xi )j f (j−1) (xi )
j!
o
− (−1)j (xi+1 − ξi )j f (j−1) (xi+1 ) + R (ξ, ∆m , f )
and the remainder, R (ξ, ∆m , f ) satisfies the inequality
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(7.10)
(n) m−1
f X
∞
|R (ξ, ∆m , f )| ≤
(ξi − xi )n+1
(n + 1)! i=0
+ (xi+1 − ξi )n+1 , if f (n) ∈ L∞ [a, b] .
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Proof. From Theorem 3.1 we put α0 = a, x0 = a, x1 = b, α2 = b and α1 =
α ∈ [a, b] such that
Z
b
f (t) dt +
a
n
X
(−1)j h
j!
j=1
− (a − α)j f (j−1) (a)
i
+ (b − α)j f (j−1) (b) = R (ξ, ∆m , f ) .
Over the interval [xi , xi+1 ] (i = 0, . . . , m − 1), we have
Z
xi+1
f (t) dt +
xi
n
X
(−1)j h
j=1
j!
and therefore, using the generalised triangle inequality
|R (ξ, ∆m , f )|
m−1
m−1
n
X Z xi+1
XX
(−1)j h
≤
(xi+1 − ξi )j f (j−1) (xi+1 )
f (t) dt +
xi
j!
i=0
i=0 j=1
i j
j (j−1)
− (−1) (ξi − xi ) f
(xi ) ≤
1
(n + 1)!
A. Sofo
(xi+1 − ξi )j f (j−1) (xi+1 )
i
− (−1)j (ξi − xi )j f (j−1) (xi ) = R (ξ, ∆m , f )
m−1
X
Integral Inequalities of the
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sup
i=0 t∈[xi ,xi+1 ]
(n) f (t) (ξi − xi )n+1 + (xi+1 − ξi )n+1 .
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The inequality in (7.11) follows, since we have
sup
t∈[xi ,xi+1 ]
(n) (n) f (t) ≤ f ,
∞
and Theorem 7.4 is proved.
The following corollary is a consequence of Theorem 7.4.
Corollary 7.5. Let f and ∆m be defined as in Theorem 7.1. The following
estimates apply.
(i) The nth order left rectangle rule
b
Z
f (t) dt =
m−1
n
XX
a
(−hi )j (j−1)
f
(xi ) + Rl (∆m , f ) .
j!
i=0 j=1
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(ii) The nth order right rectangle rule
Contents
Z
b
f (t) dt = −
m−1
n
XX
a
i=0 j=1
j
(hi ) (j−1)
f
(xi+1 ) + Rr (∆m , f ) .
j!
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(iii) The nth order trapezoidal rule
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Z
b
f (t) dt =
a
m−1
n XX
i=0 j=1
−
hi
2
j
1
j!
n
o
× f (j−1) (xi ) − (−1)j f (j−1) (xi+1 ) + RT (∆m , f ) ,
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where
(n) m−1
f X
∞
|Rl (∆m , f )| = |Rr (∆m , f )| ≤
hn+1 , if f (n) ∈ L∞ [a, b]
(n + 1)! i=0 i
and
(n) m−1
f X
∞
|RT (∆m , f )| ≤ n
hn+1 , if f (n) ∈ L∞ [a, b] .
2 (n + 1)! i=0 i
(1)
(2)
Theorem 7.6. Consider the interval xi ≤ αi ≤ ξi ≤ αi ≤ xi+1 , i =
0, . . . , m − 1, and let f and ∆m be defined as above. Then we have the equality
Z b
m−1
n
j
XX
(−1)j (i)
(7.11)
xi − αi
f (j−1) (xi )
f (t) dt =
j!
a
i=0 j=1
j j (1)
(2)
− ξi − αi
f (j−1) (ξi )
−
ξi − αi
j
(2)
(j−1)
− xi+1 − αi
f
(xi+1 )
(1)
(2)
+ R ξ, αi , αi , ∆m , f
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and the remainder satisfies the estimation
(1)
(2)
R
ξ,
α
,
α
,
∆
,
f
m
i
i
(n) m−1 n+1 n+1
f X (1)
(1)
∞
αi − xi
+ ξi − αi
≤
(n + 1)! i=0
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+
(2)
αi
− ξi
n+1
+ xi+1 −
(2)
αi
n+1 , if f (n) ∈ L∞ [a, b] .
The proof follows directly from Theorem 6.1 on the intervals [xi , xi+1 ],
(i = 0, . . . , m − 1).
The following Riemann type formula also holds.
Corollary 7.7. Let f and ∆m be defined as in Theorem 7.1 and choose ξi ∈
[xi , xi+1 ],
(i = 0, . . . , m − 1). Then we have the equality
Z
b
f (t) dt =
(7.12)
a
Integral Inequalities of the
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m−1
n
XX
i=0
(−1)j (ξi − xi )n+1
j!
j=1
− (ξi − xi+1 )n+1 f (j−1) (ξi ) + RR (ξ, ∆m , f )
and the remainder satisfies the estimation
(n) m−1
f X
∞
|RR (ξ, ∆m , f )| ≤
(ξi − xi )n+1 + (xi+1 − ξi )n+1 ,
(n + 1)! i=0
if f (n) ∈ L∞ [a, b] .
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The proof follows from (7.11) where
(1)
αi
= xi and
(2)
αi
= xi+1 .
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Remark 7.1. If in (7.12) we choose the midpoint 2ξi = xi+1 + xi we obtain the
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generalised midpoint quadrature formula
Z
(7.13)
a
b
n
m−1
XX
j
(−1)j+1 hi
f (t) dt =
j!
2
i=0 j=1
n
o
xi + xi+1
j
(j−1)
× 1 − (−1) f
+ RM (∆m , f )
2
and RM (∆m , f ) is bounded by
(n) m−1
f X
∞
|RM (∆m , f )| ≤
hn+1 , if f (n) ∈ L∞ [a, b] .
(n + 1)!2n i=0 i
Corollary 7.8. Consider a set of points
5xi + xi+1 xi + 5xi+1
ξi ∈
,
(i = 0, . . . , m − 1)
6
6
and let f and ∆m be defined as in Theorem 7.1. Then we have the equality
" Z b
m−1
n
j
XX
hi n
(−1)j
(7.14)
(−1)j f (j−1) (xi )
f (t) dt =
j!
6
a
i=0 j=1
(
j
5xi + xi+1
(j−1)
−f
(xi+1 ) −
ξi −
6
#
j )
xi + 5xi+1
− ξi −
f (j−1) (ξi ) + Rs (∆m , f )
6
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and the remainder, Rs (∆m , f ) satisfies the bound
(n) m−1 ( n+1 n+1
f X
hi
5xi + xi+1
∞
|Rs (∆m , f )| ≤
2
+ ξi −
(n + 1)! i=0
6
6
n+1 )
xi + 5xi+1
− ξi
, if f (n) ∈ L∞ [a, b] .
+
6
Remark 7.2. If in (7.14) we choose the midpoint ξi = xi+12+xi we obtain a
generalised Simpson formula:
"
j
Z b
m−1
n
j o
XX
hi n
(−1)
j (j−1)
(j−1)
(−1) f
(xi ) − f
(xi+1 )
f (t) dt =
j!
6
a
i=0 j=1
#
j n
o
x
+
x
hi
i+1
i
1 − (−1)j f (j−1)
−
+ Rs (∆m , f )
3
2
and Rs (∆m , f ) is bounded by
(n) f X hi n+1
m−1
n+1
∞
|Rs (∆m , f )| ≤
2 1+2
, if f (n) ∈ L∞ [a, b] .
(n + 1)!
6
i=0
The following is a consequence of Theorem 7.6.
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Corollary 7.9. Consider the interval
(1)
xi ≤ αi ≤
xi+1 + xi
(2)
≤ αi ≤ xi+1 (i = 0, . . . , m − 1) ,
2
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and let f and ∆m be defined as in Theorem 7.1.
The following equality is obtained:
Z b
(7.15)
f (t) dt
a
=
n
m−1
XX
i=0 j=1
(−1)j
j!
(i)
xi − αi
j
(
j
xi+1 + xi
(1)
×f
(xi ) −
− αi
2
j )
xi+1 + xi
xi+1 + xi
(2)
(j−1)
−
− αi
f
2
2
j
(2)
(1)
(2)
− xi+1 − αi
f (j−1) (xi+1 ) + RB αi , αi , ∆m , f ,
(j−1)
where the remainder satisfies the bound
(1)
(2)
RB αi , αi , ∆m , f (n) m−1 (
n+1
n+1 x + x
f X i+1
i
(1)
(1)
∞
≤
αi − xi
+
− αi
(n + 1)! i=0
2
)
n+1 n+1
xi+1 + xi
(2)
(2)
+ xi+1 − αi
.
+ αi −
2
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The following remark applies to Corollary 7.9.
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Remark 7.3. If in (7.15) we choose
(1)
αi =
3xi + xi+1
xi + 3xi+1
(2)
and αi =
,
4
4
we have the formula:
b
Z
f (t) dt
(7.16)
a
=
m−1
n
XX
i=0 j=1
j
(−1)j hi h
(−1)j f (j−1) (xi ) − f (j−1) (xi+1 )
j!
4
n
o
xi+1 + xi
j
(j−1)
− 1 − (−1) f
+ RB (∆m , f ) .
2
The remainder, RB (∆m , f ) satisfies the bound
(n) m−1
f X hi n+1
∞
×4
, if f (n) ∈ L∞ [a, b] .
|RB (∆m , f )| ≤
(n + 1)!
4
i=0
The following theorem incorporates the Newton-Cotes formula.
Theorem 7.10. Consider the interval
(1)
(1)
xi ≤ αi ≤ ξi
(2)
(2)
≤ αi ≤ ξi
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(3)
≤ αi ≤ xi+1 (i = 0, . . . , m − 1) ,
and let ∆m and f be defined as in Theorem 7.1. This consideration gives us the
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equality
Z
(7.17)
a
b
n
m−1
XX
j
(−1)j (1)
f (j−1) (xi )
f (t) dt =
xi − αi
j!
i=0 j=1
j
(3)
− xi+1 − αi
f (j−1) (xi+1 )
j j (1)
(1)
(1)
(2)
(1)
(j−1)
−
ξi − αi
− ξi − αi
f
ξi
j j (2)
(2)
(2)
(2)
(3)
(j−1)
− ξi − αi
f
ξi
−
ξi − αi
(1)
(2)
(3) (1) (2)
+ R αi , αi , αi , ξi , ξi , ∆m , f .
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The remainder satisfies the bound
(1)
(2)
(3) (1) (2)
R αi , αi , αi , ξi , ξi , ∆m , f (n) m−1 n+1 n+1
f X (1)
(1)
(1)
∞
αi − xi
+ ξi − αi
≤
(n + 1)! i=0
n+1 n+1 n+1
(2)
(1)
(2)
(2)
(3)
(2)
+ αi − ξi
+ ξi − αi
+ αi − ξi
n+1 (3)
+ xi+1 − αi
, if f (n) ∈ L∞ [a, b] .
The following is a consequence of Theorem 7.10.
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(1)
Corollary 7.11. Let f and ∆m be defined as above and make the choices αi =
(2)
(3)
(1)
(2)
7xi +xi+1
i+1
i+1
i+1
, αi = xi +x2 i+1 , αi = xi +7x
, ξi = 2xi +x
and ξi = xi +2x
,
8
8
3
3
then we have the equality:
Z b
(7.18)
f (t) dt
a
" m−1
n
j
o
XX
hi n
(−1)j
j (j−1)
(j−1)
(−1) f
(xi ) − f
(xi+1 )
=
j!
8
i=0 j=1
j n
o
hi
2xi + xi+1
j
j
(j−1)
−
5 − (−4) f
24
3
j n
#
o
x
+
2x
hi
i
i+1
j
−
4j − (−5) f (j−1)
24
3
Integral Inequalities of the
Ostrowski Type
A. Sofo
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+ RN (∆m , f ) ,
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J
where the remainder satisfies the bound
X hi n+1
m−1
2 f (n) ∞ n+1
n+1
n+1
3
+4
+5
,
|RN (∆m , f )| ≤
(n + 1)!
24
i=0
if f (n) ∈ L∞ [a, b] .
When n = 1, we obtain from (7.18) the three-eighths rule of Newton-Cotes.
Remark 7.4. For n = 2 from (7.18), we obtained a perturbed three-eighths
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Newton-Cotes formula:
Z
a
b
m−1
X
2 0
hi
f (xi ) − f 0 (xi+1 )
f (t) dt =
(f (xi ) + f (xi+1 )) +
8
2
i=0
3hi
2xi + xi+1
xi + 2xi+1
+
f
+f
8
3
3
)#
2 ( 0 2xi +xi+1 0 xi +2xi+1
f
−
f
3hi
3
3
+ RN (∆m , f ) ,
−
2
8
hi
8
Integral Inequalities of the
Ostrowski Type
where the remainder satisfies the bound
m−1
kf 00 k∞ X 3
|RN (∆m , f )| ≤
hi , if f 00 ∈ L∞ [a, b] .
192 i=0
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8.
Conclusion
This paper has extended many previous Ostrowski type results. Integral inequalities for n−times differentiable mappings have been obtained by the use
of a generalised Peano kernel. Some particular integral inequalities, including
the trapezoid, midpoint, Simpson and Newton-Cotes rules have been obtained
and further developed into composite quadrature rules.
Further work in this area may be undertaken by considering the Chebychev
and Lupaş inequalities. Similarly, the following alternate Grüss type results
may be used to examine all the interior point rules of this paper.
Let σ (h (x)) = h (x) − M (g) where
1
M (h) =
b−a
Z
Integral Inequalities of the
Ostrowski Type
A. Sofo
b
h (t) dt.
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Contents
Then from (5.2)
T (h, g) = M (hg) − M (h) M (g) .
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