Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 2, Article 21, 2002
INTEGRAL INEQUALITIES OF THE OSTROWSKI TYPE
A. SOFO
S CHOOL OF C OMMUNICATIONS AND I NFORMATICS
V ICTORIA U NIVERSITY OF T ECHNOLOGY
P.O. B OX 14428
M ELBOURNE C ITY MC
V ICTORIA 8001, AUSTRALIA .
sofo@matilda.vu.edu.au
URL: http://rgmia.vu.edu.au/sofo/
Received December, 2000; accepted 16 November, 2001.
Communicated by T. Mills
A BSTRACT. 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, NewtonCotes and left and right rectangle rules. The results obtained provide sharper bounds than those
obtained by Dragomir [5] and Cerone, Dragomir and Roumeliotis [2].
Key words and phrases: Ostrowski Integral Inequality, Quadrature Formulae.
2000 Mathematics Subject Classification. 26D15, 41A55.
1. I NTRODUCTION
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
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.
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
I wish to acknowledge the useful comments of an anonymous referee which led to an improvement of this work.
083-01
2
A. S OFO
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.
Theorem 1.2. Let f : [a, b] → R be a mapping such that f (n−1) is absolutely continuous on
[a, b], (AC[a, b] for short). Then for all x ∈ [a, b] the following bound is valid:
Z
!
n−1
j+1
j
j+1
b
X
(b − x) + (−1) (x − a)
(j)
(1.2) f (t) dt −
f (x)
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]
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 X
X
xi + xi+1
1
h2i +
αi+1 −
kf 0 k∞
≤
4 i=0
2
i=0
k−1
X
1
≤ kf 0 k∞
h2i
2
i=0
1
(b − a) kf 0 k∞ ν (h) ,
2
where hi := xi+1 − xi (i = 0, ..., k − 1) and ν (h) := max {hi |i = 0, ..., k − 1}.
The constant 41 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.
≤
2. I NTEGRAL I DENTITIES
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
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
http://jipam.vu.edu.au/
O STROWSKI T YPE I NEQUALITIES
3
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
n
k−1
X
(−1)j X n
f (t) dt +
(xi+1 − αi+1 )j f (j−1) (xi+1 )
j! i=0
j=1
j
− (xi − αi+1 ) f
(j−1)
Z b
o
n
(xi ) = (−1)
Kn,k (t) f (n) (t) dt,
a
where the Peano kernel

(t − α1 )n


,


n!




(t − α2 )n


,



n!

t ∈ [x1 , x2 )
..
.


n


(t − αk−1 )


, t ∈ [xk−2 , xk−1 )


n!



n


 (t − αk ) ,
t ∈ [xk−1 , b] ,
n!
Kn,k (t) :=
(2.2)
t ∈ [a, x1 )
n and k are natural numbers, n ≥ 1, k ≥ 1 and f (0) (x) = f (x).
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 ) ,





t ∈ [a, x1 )
t ∈ [x1 , x2 )
..
.







(t − αk−1 ) , t ∈ [xk−2 , xk−1 )






(t − αk ) ,
t ∈ [xk−1 , b] .
To prove (2.3), we integrate by parts as follows
Z b
K1,k (t) f 0 (t) dt
K1,k (t) :=
a
=
k−1 Z
X
i=0
xi+1
(t − αi+1 ) f 0 (t) dt
xi
Z
k−1 X
xi+1
=
(t − αi+1 ) f (t) xi −
i=0
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
xi+1
f (t) dt
xi
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4
A. S OFO
=
k−1
X
[(xi+1 − αi+1 ) f (xi+1 ) − (xi − αi+1 ) f (xi )] −
i=0
Z
b
i=0
Z
f (t) dt +
a
k−1 Z
X
b
K1,k (t) f 0 (t) dt =
k−1
X
a
xi+1
f (t) dt.
xi
[(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
(2.4)
a
b
k−1 X
n+1
o
X
(−1)j n
j (j−1)
j (j−1)
(xi+1 − αi+1 ) f
(xi+1 ) − (xi − αi+1 ) f
(xi )
f (t) dt +
j!
i=0 j=1
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 )
..
.
(t−αk−1 )n+1
,
(n+1)!
t ∈ [xk−2 , xk−1 )
(t−αk )n+1
,
(n+1)!
t ∈ [xk−1 , b] .
Consider
Z
b
Kn+1,k (t) f
(n+1)
a
(t) dt =
k−1 Z
X
i=0
xi+1
xi
(t − αi+1 )n+1 (n+1)
f
(t) dt
(n + 1)!
and upon integrating by parts we have
Z b
Kn+1,k (t) f (n+1) (t) dt
a
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 )n+1
(x
−
α
)
i
i+1
=
f (n) (xi+1 ) −
f (n) (xi )
(n
+
1)!
(n
+
1)!
i=0
Z b
−
Kn,k (t) f (n) (t) dt.
a
Upon rearrangement we may write
(
)
Z b
k−1
n+1
n+1
X
(x
−
α
)
(x
−
α
)
i+1
i+1
i
i+1
Kn,k (t) f (n) (t) dt =
f (n) (xi+1 ) −
f (n) (xi )
(n + 1)!
(n + 1)!
a
i=0
Z b
−
Kn+1,k (t) f (n+1) (t) dt.
a
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
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O STROWSKI T YPE I NEQUALITIES
Now substitute
(−1)n
Z
a
Rb
a
b
5
Kn,k (t) f (n) (t) dt from the induction hypothesis (2.1) such that
" n
k−1 X
X
(−1)j n
(xi+1 − αi+1 )j f (j−1) (xi+1 )
f (t) dt + (−1)n
j!
i=0
j=1
oi
− (xi − αi+1 )j f (j−1) (xi )
(
)
k−1
X
(xi − αi+1 )n+1 (n)
(xi+1 − 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
Collecting the second and third terms and rearranging, we can state
Z
a
b
k−1 X
n+1
o
X
(−1)j n
f (t) dt +
(xi+1 − αi+1 )j f (j−1) (xi+1 ) − (xi − αi+1 )j f (j−1) (xi )
j!
i=0 j=1
Z b
n+1
= (−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.
Corollary 2.2. From Theorem 2.1, the equality (2.1) may be represented as
Z
(2.5)
a
b
" k
#
n
o
X
(−1)j X n
j
j
f (t) dt +
(xi − αi ) − (xi − αi+1 ) f (j−1) (xi )
j!
j=1
i=0
Z b
n
= (−1)
Kn,k (t) f (n) (t) dt.
a
Proof. From (2.1) consider the second term and rewrite it as
(2.6) S1 + S2 :=
k−1
X
{− (xi+1 − αi+1 ) f (xi+1 ) + (xi − αi+1 ) f (xi )}
i=0
" n
k−1 X
X
(−1)j n
+
(xi+1 − αi+1 )j f (j−1) (xi+1 )
j!
i=0
j=2
j
− (xi − αi+1 ) f
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
(j−1)
o
(xi ) .
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6
A. S OFO
Now
S1 = (a − α1 ) f (a) +
k−1
X
(xi − αi+1 ) f (xi )
i=1
+
k−2
X
{− (xi+1 − αi+1 ) f (xi+1 )} − (b − αk ) f (b)
i=0
= (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
Also,
S2
#
" n
k−2 X
o
X
(−1)j n
(xi+1 − αi+1 )j f (j−1) (xi+1 )
=
j!
i=0
j=2
n
o
X
(−1)j n
j (j−1)
(xk − αk ) f
(xk )
+
j!
j=2
n
o
X
(−1)j 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
=
n
X
(−1)j
j
(b − αk ) f
(j−1)
(b) −
n
X
(−1)j
(a − α1 )j f (j−1) (a)
j!
j!
j=2
"
#
k−1 X
n
o
X
(−1)j n
+
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi ) .
j!
i=1
j=2
j=2
From (2.6)
(
S1 + S2 := − (b − αk ) f (b) + (α1 − a) f (a) +
k−1
X
)
(αi+1 − αi ) f (xi )
i=1
n
o
X
(−1)j n
+
(b − αk )j f (j−1) (b) − (a − α1 )j f (j−1) (a)
j!
j=2
"
#
k−1 X
n
j n
o
X
(−1)
+
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi )
j!
i=1
j=2
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
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O STROWSKI T YPE I NEQUALITIES
(
= − (α1 − a) f (a) +
k−1
X
7
)
(αi+1 − αi ) f (xi ) + (b − αk ) f (b)
i=1
+
n
X
(−1)j
j!
j=2
+
"
− (a − α1 )j f (j−1) (a)
k−1 n
X
j
j
(xi − αi ) − (xi − αi+1 )
o
#
j
f (j−1) (xi ) + (b − αk ) f (j−1) (b) .
i=1
Keeping in mind that x0 = a, α0 = 0, xk = b and αk+1 = b we may write
S1 + S 2 = −
k
X
(αi+1 − αi ) f (xi )
i=0
" k
#
n
j X
n
o
X
(−1)
+
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi )
j!
i=0
j=2
" k
#
n
o
X (−1)j X n
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi ) .
=
j!
j=1
i=0
And substituting S1 + S2 into the second term of (2.1) we obtain the identity (2.5).
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
Z
(2.7)
a
b
" k
#
n
o
X
1 Xn j
f (t) dt +
−hi + (−1)j hji−1 f (j−1) (xi )
j j!
2
j=1
i=0
n
b
Z
Kn,k (t) f (n) (t) dt,
= (−1)
a
where hi := xi+1 − xi , h−1 := 0 and hk := 0.
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) +
k−1
X
(αi+1 − αi ) f (xi )
i=1
1
=
2
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
(
h0 f (a) +
k−1
X
)
(hi + hi−1 ) f (xi ) + hk−1 f (b) ,
i=1
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8
A. S OFO
the term
n
o
X
(−1)j n
j (j−1)
j (j−1)
(b − αk ) f
(b) − (a − α1 ) f
(a)
j!
j=2
n
o
X
(−1)j n j
j j (j−1)
(j−1)
=
hk−1 f
(b) − (−1) h0 f
(a)
j!2j
j=2
and the term
#
" n
k−1 X
o
X
(−1)j n
j
j
(j−1)
(xi − αi ) − (xi − αi+1 ) f
(xi )
j!
i=1
j=2
" 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
a
1
f (t) dt −
2
(
h0 f (a) +
k−1
X
)
(hi + hi−1 ) f (xi ) + hk−1 f (b)
i=1
j n
n
X
(−1)
j
hjk−1 f (j−1)
hj0 f (j−1)
o
(a)
(b) − (−1)
j!2j
" 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
Z b
n
= (−1)
Kn,k (t) f (n) (t) dt.
+
j=2
a
Collecting the inner three terms of the last expression, we have
Z
a
b
Z b
n
k
o
X
(−1)j X n j
j j
n
(j−1)
f (t) dt+
hi−1 − (−1) hi f
(xi ) = (−1)
Kn,k (t) f (n) (t) dt,
j
j!2
a
j=1
i=0
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.
Corollary 2.4. Let
(2.8)
Ik : xi = a + i
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
b−a
k
, i = 0, . . . , k
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O STROWSKI T YPE I NEQUALITIES
9
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
j "
Z b
n X
b−a
1
(2.9)
f (t) dt +
− f (j−1) (a)
2k
j!
a
j=1
#
k−1 n
o
X
+
(−1)j − 1 f (j−1) (xi ) + (−1)j f (j−1) (b)
i=1
n
b
Z
Kn,k (t) f (n) (t) dt.
= (−1)
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
b−a
b−a
, hk−1 = (xk − xk−1 ) =
,
k
k
b−a
b−a
hi = xi+1 − xi =
and hi−1 = xi − xi−1 =
, (i = 1, . . . , k − 1)
k
k
and substituting into (2.7) we have
" ( j
j
Z b
n
k−1
X
X
1
b−a
b−a
(j−1)
f (t) dt +
−
f
(a) +
−
j
j!2
2k
k
a
j=1
i=0
#
j )
j
b
−
a
b−a
j
j
f (j−1) (xi ) + (−1)
f (j−1) (b)
+ (−1)
k
k
Z b
n
= (−1)
Kn,k (t) f (n) (t) dt,
h0 = x1 − x0 =
a
which simplifies to (2.9) after some minor manipulation.
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
" k
#
n
o
X
(−1)j X n
(2.11) g (y) = g (a) −
(xi − αi )j − (xi − αi+1 )j g (j) (xi )
j!
j=1
i=0
Z y
n
+ (−1)
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 .
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A. S OFO
3. I NTEGRAL I NEQUALITIES
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
j X
b
n
o
X
(−1)
j
j
(j−1)
(3.1)
f
(t)
dt
+
(x
−
α
)
−
(x
−
α
)
f
(x
)
i
i
i
i+1
i
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)!
where
(n) f :=
∞
sup f (n) (t) < ∞,
t∈[a,b]
hi := xi+1 − xi and
ν (h) := max {hi |i = 0, . . . , k − 1} .
Proof. From Corollary 2.2 we may write
Z
" k
#
n
b
o
X
(−1)j X n
j
j
(j−1)
(xi − αi ) − (xi − αi+1 ) f
(xi ) (3.2) f (t) dt +
a
j!
j=1
i=0
Z b
n
(n)
= (−1)
Kn,k (t) f (t) dt ,
a
and
Z b
Z b
(n) n
(n)
(−1)
Kn,k (t) f (t) dt ≤ f ∞
|Kn,k (t)| dt,
a
Z
a
b
|Kn,k (t)| dt =
a
k−1 Z
X
i=0
=
xi+1
xi
k−1 Z
X
i=0
|t − αi+1 |n
dt
n!
αi+1
xi
(αi+1 − t)n
dt +
n!
Z
xi+1
αi+1
(t − αi+1 )n
dt
n!
k−1
X
1
=
(αi+1 − xi )n+1 + (xi+1 − αi+1 )n+1
(n + 1)! i=0
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
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O STROWSKI T YPE I NEQUALITIES
11
and thus
Z b
n
(n)
(−1)
Kn,k (t) f (t) dt
(n) k−1
f X ∞
(αi+1 − xi )n+1 + (xi+1 − αi+1 )n+1 .
(n + 1)! i=0
≤
a
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
n+1 n+1
∞
∞
(αi+1 − xi )
+ (xi+1 − αi+1 )
≤
hn+1
,
(n + 1)! i=0
(n + 1)! i=0 i
since for 0 < B < A < C it is well known that
(A − B)n+1 + (C − A)n+1 ≤ (C − B)n+1 .
(3.3)
Also
(n) k−1
(n) (n) k−1
f X
f f X
n+1
n
∞
∞
∞
hi ≤
ν (h)
hi =
(b − a) ν n (h) ,
(n + 1)! i=0
(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.
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
j X
b
n
o
X
(−1)
j j
j
(j−1)
−h
+
(−1)
h
f
(x
)
f
(t)
dt
+
i
i
i+1
a
2j j!
i=0
j=1
(n) k−1
f X
∞
hn+1
≤
(n + 1)!2n i=0 i
for f (n) ∈ L∞ [a, b] .
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.
For the equidistant partitioning case we have the following inequality.
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12
A. S OFO
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
1
b−a
(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
h0 = x1 − x0 =
b−a
b−a
and hi = xi+1 − xi =
, i = 1, . . . , k − 1
k
k
in which case (3.5) follows.
The following inequalities for Taylor-like expansions also hold.
Corollary 3.4. Let g be defined as in Corollary 2.5. Then we have the inequality
" k
#
n
o
X
(−1)j X n
(xi − αi )j − (xi − αi+1 )j g (j) (xi ) (3.6)
g (y) − g (a) +
j!
j=1
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
for all xi ∈ [a, y] where
(n+1) g
:= sup g (n+1) (t) < ∞.
∞
t∈[a,y]
Proof. Follows directly from (2.11) and using the norm as in (3.1).
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
n
o
X
X
1
j j
j
(j)
(3.7) g (y) − g (a) +
−h
+
(−1)
h
g
(x
)
i
i
i+1
j j!
2
j=1
i=0
(n+1) k−1
g
X
∞
≤
hn+1 if g (n+1) ∈ L∞ [a, y] .
(n + 1)!2n i=0 i
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
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
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O STROWSKI T YPE I NEQUALITIES
13
be an equidistant partitioning of [a, y], then we have the inequality:
(3.8)
j
n X
1
y−a
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
n+1
∞
≤
if g (n+1) ∈ L∞ [a, y] .
n (y − a)
(n + 1)! (2k)
Proof. The proof follows directly upon using (2.9) with f 0 = g and b = y.
4. T HE C ONVERGENCE
OF A
(m)
< x1
G ENERAL Q UADRATURE F ORMULA
Let
∆m : a = x0
(m)
(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
X
(m)
(m)
wj f xj
−
" m (
n
X
(−1)r X
r=2
j=0
r!
−
(m)
xj
j−1
X
−a−
−a−
j
X
ws(m)
s=0
!r )
j=0
(m)
xj
!r
ws(m)
f (r−1)
(m)
xj
#
,
s=0
where wj (j = 0, . . . , m) are the quadrature weights and assume that
Pm
j=0
(m)
wj
=b−a.
(m)
The following theorem contains a sufficient condition for the weights wj
so that
R
b
Im f, f 0 ,. . . , f(n) , ∆m , wm approximates the integral a f (x) dx with an error expressed in
terms of f (n) ∞ .
Theorem 4.1. Let f : [a, b] → R be a continuous mapping on [a, b] such that f (n−1) is AC[a, b].
(m)
If the quadrature weights, wj satisfy the condition
(4.2)
(m)
xi
−a≤
i
X
(m)
wj
(m)
≤ xi+1 − a for all i = 0, . . . , m − 1
j=0
then we have the estimation
Z b
0
(n)
(4.3)
I
f,
f
,
.
.
.
,
f
,
∆
,
w
−
f
(t)
dt
m
m
m
a
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A. S OFO
!n+1
!n+1 
(n) m−1 
i
i
f X
X
X
(m)
(m)
(m)
(m)
∞
 a+

≤
wj − xi
− xi+1 − a −
wj
(n + 1)! i=0
j=0
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)!
Also
ν h(m)
:=
(m)
hi
max
i=0,...,m−1
(m)
n
o
(m)
hi
(m)
:= xi+1 − xi .
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 , i = 0, . . . , m.
j=0
Note that
(m)
αi+1
=a+
m
X
(m)
wj
= a + b − a = b.
j=0
By the assumption (4.2), we have
i
h
(m)
(m)
(m)
αi+1 ∈ xi , xi+1 for all i = 0, . . . , m − 1.
(m)
Define α0
= a and compute
(m)
α1
(m)
αi+1
(m)
− α0
−
(m)
αi
(m)
= w0
= a+
i
X
(m)
wj
−a−
j=0
i−1
X
(m)
(m)
= wi
wj
(i = 0, . . . , m − 1)
j=0
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)
αi+1
−
(m)
αi
i=0
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
m
X
(m)
(m)
(m)
f xi
=
wi f xi
,
i=0
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O STROWSKI T YPE I NEQUALITIES
15
and let
m
X
j=0
(m)
(m)
wj f xj
−
" m (
n
X
(−1)r X
r=2
r!
(m)
xj
−a−
−
−a−
ws(m)
s=0
j=0
!r )
j
(m)
xj
!r
j−1
X
X
ws(m)
f (r−1)
(m)
xj
#
s=0
:= Im f, f 0 , . . . , f (n) , ∆m , wm .
Applying the inequality (3.1) we obtain the estimate (4.3).
The case when the partitioning is equidistant is important in practice. Consider the equidistant
partition
b−a
(m)
Em := xi := a + i
, (i = 0, . . . , m)
m
and define the sequence of numerical quadrature formulae
Im f, f 0 , . . . , f (n) , ∆m , wm
" m (
!r
X
j−1
m
n
X
j (b − a)
(−1)r X
j (b − a) X (m)
(m)
:=
wj f a +
−
−
ws
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.
(m)
Corollary 4.2. Let f : [a, b] → R be AC[a, b]. If the quadrature weights wj
condition
i
1 X (m) i + 1
i
≤
w
≤
, i = 0, 1, . . . , n − 1,
m
b − a j=0 j
m
satisfy the
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)
(n) n+1
f b−a
∞
≤
,
(n + 1)!
m
b−a
m
−
i
X
!n+1 
(m)
wj

j=0
where f (n) ∈ L∞ [a, b] .
In particular, if f (n) ∞ < ∞, then
0
lim Im f, f , . . . , f
m→∞
(n)
Z
, wm =
b
f (t) dt
a
uniformly by the influence of the weights wm .
The proof of Corollary 4.2 follows directly from Theorem 4.1.
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
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16
A. S OFO
5. G RÜSS T YPE I NEQUALITIES
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
|T (h, g)| ≤
(5.1)
1
(Φ − φ) (Γ − γ) ,
4
where
(5.2)
1
T (h, g) :=
b−a
b
Z
a
1
h (x) g (x) dx −
b−a
Z
a
b
1
h (x) dx ·
b−a
Z
b
g (x) dx
a
and the inequality (5.1) is sharp, in the sense that the constant 14 cannot be replaced by a smaller
one.
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.
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) ,
we have the inequality
Z b
n
X
(−1)j
n
n
(5.4) (−1)
f (t) dt + (−1)
j!
a
j=1
" k−1 (
)#
j
j
X
hi
hi
(j−1)
(j−1)
×
− δi f
(xi+1 ) −
+ δi f
(xi )
2
2
i=0
#
(n−1)
k−1 n+1 "X
r
n+1 f
(b) − f (n−1) (a) X hi
n+1
2δi
r
−
×
{1 + (−1) } (b − a) (n + 1)!
2
r
hi
r=0
i=0
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
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O STROWSKI T YPE I NEQUALITIES
17
"
#
2n+1 "2n+1
k−1 X 2n + 1 2δi r
X
M −m
b−a
hi
≤
×
{1 + (−1)r }
r
2
h
(2n + 1) (n!)2 i=0 2
i
r=0
#!2  21
n+1 "X
r
k−1 n+1 X
1
hi
n+1
2δi

−
{1 + (−1)r }
(n + 1)! i=0 2
r
h
i
r=0
where
hi := xi+1 − xi and
xi+1 + xi
, i = 0, . . . , k − 1.
δi := αi+1 −
2
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
Z b
Z b
Z b
1
(n)
(n)
(5.5) Kn,k (t) f (t) dt −
f (t) dt ·
Kn,k (t) dt
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
Now we may evaluate
Z
b
f (n) (t) dt = f (n−1) (b) − f (n−1) (a)
a
and
Z
G1 :=
=
b
Kn,k (t) dt
a
k−1 Z xi+1
X
i=0
1
(t − αi+1 )n dt
n!
xi
k−1
=
X
1
(xi+1 − αi+1 )n+1 + (αi+1 − xi )n+1 .
(n + 1)! i=0
Using the definitions of hi and δi we have
xi+1 − αi+1 =
hi
− δi
2
and
αi+1 − xi =
hi
+ δi
2
such that
n+1 n+1 )
hi
hi
− δi
+
+ δi
2
2
" n+1 n+1−r X
n+1−r #
k−1 X
n+1 X
1
n+1
hi
n + 1 r hi
r
=
(−δi )
+
δi
(n + 1)! i=0 r=0
r
2
r
2
r=0
"
#
n+1 X
r
k−1 n+1 X
1
hi
n+1
2δi
=
{1 + (−1)r } .
(n + 1)! i=0 2
r
hi
r=0
k−1
X
1
G1 =
(n + 1)! i=0
(
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18
A. S OFO
Also,
Z
b
G2 :=
2
(t) dt
Kn,k
a
=
k−1 Z
X
i=0
=
=
=
xi+1
xi
(t − αi+1 )2n
dt
(n!)2
k−1
X
1
2n+1
2n+1 (x
−
α
)
+
(α
−
x
)
i+1
i+1
i+1
i
(2n + 1) (n!)2 i=0
(
2n+1 2n+1 )
k−1
X
hi
1
hi
− δi
+
+ δi
2
2
(2n + 1) (n!)2 i=0
#
2n+1 "2n+1
k−1 X
X 2n + 1 2δi r
1
hi
{1 + (−1)r } .
2
2
r
h
(2n + 1) (n!) i=0
i
r=0
From identity (2.1), we may write
Z b
Z b
n
X
(−1)j
n
n
(n)
Kn,k (t) f (t) dt = (−1)
f (t) dt + (−1)
j!
a
a
j=1
" k−1
#
o
Xn
j (j−1)
j (j−1)
×
(xi+1 − αi+1 ) f
(xi+1 ) − (αi+1 − xi ) f
(xi )
i=0
and from the left hand side of (5.5) we obtain
Z b
n
X
(−1)j
n
G3 := (−1)
f (t) dt + (−1)n
j!
a
j=1
" k−1 (
)#
j
j
X
hi
h
i
×
− δi f (j−1) (xi+1 ) −
+ δi f (j−1) (xi )
2
2
i=0
#
(n−1)
k−1 n+1 "X
r
n+1 f
(b) − f (n−1) (a) X hi
n+1
2δi
−
×
{1 + (−1)r }
(b − a) (n + 1)!
2
r
h
i
r=0
i=0
after substituting for G1 .
From the right hand side of (5.5) we substitute for G1 and G2 so that
Z b
2
Z b
2
G4 := (b − a)
Kn,k (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
1
hi
n+1
2δi
−
{1 + (−1)r }
.
(n + 1)! i=0 2
r
h
i
r=0
Hence,
|G3 | ≤
and Theorem 5.3 has been proved.
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
1
M −m
(G4 ) 2 ,
2
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O STROWSKI T YPE I NEQUALITIES
19
Corollary 5.4. Let f , Ik and αk be defined as in Theorem 5.3 and further define
δ = αi+1 −
(5.6)
xi+1 + xi
2
for all i = 0, . . . , k − 1 such that
(5.7)
|δ| ≤
1
min {hi |i = 1, . . . , k} .
2
The following inequality applies:
Z b
n
X
1
n
n
(5.8)
f (t) dt + (−1)
(−1)
j!
a
j=1
)#
" k−1 (
j
j
X
hi
h
i
j
×
(−i)
− δi f (j−1) (xi+1 ) −
+ δi f (j−1) (xi )
2
2
i=0
(n−1)
k−1 n+1 n+1−r
f
(b) − f (n−1) (a) X X r hi
{1 + (−1)r }
−
δ
(b − a) (n + 1)!
2
i=0 r=0
"
"
#
k−1 2n+1
X
X 2n + 1 hi 2n+1−r
M −m
b−a
≤
δr
{1 + (−1)r }
2
r
2
2
(2n + 1) (n!) i=0 r=0
!2  21
k−1
n+1
n+1−r
XX n + 1
hi
1
−
δr
{1 + (−1)r }  .
(n + 1)! i=0 r=0
r
2
The proof follows directly from (5.4) upon the substitution of (5.6) and some minor simplification.
Remark 5.5. If for any division Ik : a = x0 < x1 < · · · < xk−1 < xk = b of the interval [a, b],
we choose δ = 0 in (5.6), we have the inequality
(5.9)
j
Z b
n
k−1 o
X
X
1
hi n
n
n
f (t) dt + (−1)
×
(−1)j f (j−1) (xi+1 ) − f (j−1) (xi )
(−1)
j!
2
a
i=0
j=1
(n−1)
X
n+1 k−1 (n−1)
f
(b) − f
(a)
hi
−2
(b − a) (n + 1)!
2
i=0
1
"
2n+1
n+1 !2 2
k−1 k−1 X
X
M −m
2 (b − a)
hi
2
hi
 .
≤
−
2
2
(n + 1)! i=0 2
(2n + 1) (n!) i=0 2
The proof follows directly from (5.8).
Remark 5.6. Let f (n) be defined as in Theorem 5.3 and consider an equidistant partitioning Ek
of the interval [a, b], where
b−a
Ek := xi = a + i
, i = 0, . . . , k.
k
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
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20
A. S OFO
The following inequality applies
(5.10)
j
Z b
n
X
1 b−a
n
n
f (t) dt + (−1)
(−1)
j!
2k
a
j=1
a (k − i − 1) + b (i − 1)
a (k − i) + ib
j (j−1)
(j−1)
× (−1) f
−f
k
k
(n−1)
n+1
f
(b) − f (n−1) (a)
b−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
k
, i = 0, . . . , k.
6. S OME PARTICULAR I NTEGRAL I NEQUALITIES
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,
(6.1)
 1
n


 n! (t − a) , t ∈ [a, x)
Kn,2 (t) :=
 1


(t − b)n , t ∈ (x, b]
n!
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.
Theorem 6.1. Let f : [a, b] → R be an absolutely continuous mapping on [a, b] and a ≤ x1 ≤ b,
a ≤ α1 ≤ x1 ≤ α2 ≤ b. Then we have
(6.2)
Z
n
b
X
(−1)j f (t) dt +
− (a − α1 )j f (j−1) (a)
a
j!
j=1
n
o
j (j−1)
j
j
(j−1)
+ (x1 − α1 ) − (x1 − α2 ) f
(x1 ) + (b − α2 ) f
(b) J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
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O STROWSKI T YPE I NEQUALITIES
21
(n) f ∞
≤
(α1 − a)n+1 + (x1 − α1 )n+1 + (α2 − x1 )n+1 + (b − α2 )n+1
(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
(xi − αi )j − (xi − αi+1 )j f (j−1) (xi )
j! i=0
j=1
=
n
X
(−1)j j=1
j!
n
o
− (a − α1 )j f (j−1) (a) + (x1 − α1 )j − (x1 − α2 )j f (j−1) (x1 )
+ (b − α2 )j f (j−1) (b) .
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).
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 a ≤
≤ b for m a natural number, m ≥ 2, then we have the inequality
x1 ≤ a+(m−1)b
m
(6.3)
(m−1)a+b
m
≤
|Pm,n |
Z
"
j
n
j b
o
X
(−1)
b − a n (j−1)
:= f (t) dt +
f
(b) − (−1)j f (j−1) (a)
a
j!
m
j=1
(
#
j j )
b−a
b−a
(j−1)
+
(x1 − a) −
−
− (b − x1 )
f
(x1 ) m
m
(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
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22
A. S OFO
Proof. From Theorem 6.1 we note that
α1 =
(m − 1) a + b
a + (m − 1) b
and α2 =
m
m
so that
a − α1
x1 − α1
b−a
b−a
= −
, b − α2 =
,
m
m
b−a
b−a
= x1 − a −
and x1 − α2 =
− (b − x1 ) .
m
m
From the left hand side of (6.2) we have
n
o
j (j−1)
j
j
− (a − α1 ) f
(a) + (x1 − α1 ) − (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
+
−
− (b − x1 )
f (j−1) (x1 ) .
x1 − a −
m
m
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 n+1
b−a
b−a
b−a
=2
+ x1 − a −
+ b − x1 −
m
m
m
and the inequality (6.3) follows, hence the proof is complete.
The following corollary points out that the optimum of Proposition 6.2 occurs at x1 =
= a+b
in which case we have:
2
α1 +α2
2
Corollary 6.3. Let f be defined as in Proposition 6.2 and let x1 = a+b
in which case we have
2
the inequality
a
+
b
Pm,n
(6.4)
2
Z
"
j
n
j b
o
X
(−1)
b − a n (j−1)
:= f (t) dt +
f
(b) − (−1)j f (j−1) (a)
a
j!
m
j=1
j
#
(m − 2) (b − a) a
+
b
+
1 − (−1)j f (j−1)
2m
2
!
(n) n+1
2 f ∞ b − a n+1
m−2
≤
1+
if f (n) ∈ L∞ [a, b] .
(n + 1)!
m
2
.
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.
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O STROWSKI T YPE I NEQUALITIES
23
Corollary 6.4. Let the conditions of Corollary 6.3 hold and put m = 6. Then we have the
inequality
(6.5)
|P6,n |
Z
"
j
n
b
o
X
b − a n (j−1)
(−1)j
j (j−1)
:= f (t) dt +
f
(b) − (−1) f
(a)
a
j!
6
j=1
(
#
j j )
a + 5b
5a + b
+
x1 −
− x1 −
f (j−1) (x1 ) 6
6
(n) n+1 n+1
f b−a
5a + b
∞
≤
2
+ x1 −
(n + 1)!
6
6
!
n+1
a + 5b
+ −x1 +
, f (n) ∈ L∞ [a, b] ,
6
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
P6,n
(6.6)
2
Z
"
j
n
j b
o
X
(−1)
b − a n (j−1)
:= f (t) dt +
f
(b) − (−1)j f (j−1) (a)
a
j!
6
j=1
j
#
b−a a
+
b
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 6.6. Choosing n = 1 in (6.6) we have
Z b
a
+
b
b
−
a
a
+
b
:= (6.7) P6,1
f
(t)
dt
−
f
(a)
+
4f
+
f
(b)
2
6
2
a
5
≤
kf 0 k∞ (b − a)2 , f 0 ∈ L∞ [a, b] .
36
Remark 6.7. Choosing n = 2 in (6.6) we have a perturbed Simpson type inequality,
a + b (6.8)
P6,2
2
Z b
b−a
a+b
:= f (t) dt −
f (a) + 4f
+ f (b)
6
2
a
2 0
b−a
f (b) − f 0 (a) +
6
2
(b − a)3 00
≤
kf k∞ , f 00 ∈ L∞ [a, b] .
72
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24
A. S OFO
Corollary 6.8. Let f be defined as in Corollary 6.3 and let m = 4, then we have the inequality
a + b (6.9) P4,n
2
Z
j
n
b
X
(−1)j b − a
:= f (t) dt +
a
j!
4
j=1
a + b j (j−1)
j
(j−1)
(j−1)
× f
(b) − (−1) f
(a) + 1 − (−1) f
2
(n) f ∞
≤
(b − a)n+1 , f (n) ∈ L∞ [a, b] .
n
(n + 1)!4
Remark 6.9. From (6.9) we choose n = 2 and we have the inequality
a
+
b
P4,2
(6.10)
2
Z b
b−a
a+b
f (b) + f (a) + 2f
:= f (t) dt −
4
2
a
2 0
b−a
f (b) − f 0 (a) +
4
2
kf 00 k∞
≤
(b − a)3 , f 00 ∈ L∞ [a, b] .
96
Theorem 6.10. 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) f (t) dt +
− (a − α1 )j f (j−1) (a)
a
j!
j=1
+ (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) (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
n
3
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
j
j
j (j−1)
(j−1)
+ (x2 − α2 ) − (x2 − α3 ) f
(x2 ) + (b − α3 ) f
(b) .
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
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O STROWSKI T YPE I NEQUALITIES
25
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
and hence the first line of the inequality (6.11) follows and Theorem 6.10 is proved.
Corollary 6.11. Let f be defined as in Theorem 6.10 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)
Z
n
b
X
(−1)j h
|Qm,n | := f (t) dt +
− (a − α1 )j f (j−1) (a)
a
j!
j=1
(
j j )
(m − 1) a + b
(m − 1) a + b
+
− α1 −
− α2
f (j−1) (x1 )
m
m
(
j j )
a + (m − 1) b
a + (m − 1) b
+
− α2 −
− α3
f (j−1) (x2 )
m
m
i
j (j−1)
+ (b − α3 ) f
(b) (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
n+1
+ α3 −
+ (b − α3 )
, f (n) ∈ L∞ [a, b] .
m
Proof. Choose in Theorem 6.10, x1 =
proved.
(m−1)a+b
,
m
and x2 =
a+(m−1)b
,
m
hence the theorem is
Remark 6.12. For particular choices of the parameters m and n, Corollary 6.11 contains a
generalisation of the three-eighths rule of Newton and Cotes.
The following corollary is a consequence of Corollary 6.11.
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
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26
A. S OFO
2
= x1 +x
, then we
Corollary 6.13. Let f be defined as in Theorem 6.10 and choose α2 = a+b
2
2
have the inequality
Z
n
b
X
(−1)j h
Q̄m,n := (6.13)
f (t) dt +
− (a − α1 )j f (j−1) (a)
a
j!
j=1
(
j )
(m − 2) (b − a)
j
j
f (j−1) (x1 )
+ (x1 − α1 ) − (−1)
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
(m − 2)
a + (m − 1) b
+2 (b − a)
+ α3 −
2m
m
n+1 (n)
+ (b − α3 )
,
f ∈ L∞ [a, b] .
Proof. If we put α2 =
is proved.
a+b
2
=
x1 +x2
2
into (6.12), we obtain the inequality (6.13) and the corollary
The following corollary contains an optimum estimate for the inequality (6.13).
Corollary 6.14. Let f be defined as in Theorem 6.10 and make the choices
3m − 4
4−m
α1 =
a+
b and
2m
2m
4−m
3m − 4
α3 =
a+
b
2m
2m
then we have the best estimate
(6.14) Q̂m,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
+
1 − (−1)j f (j−1) (x1 ) + f (j−1) (x2 ) 2m
(n) 2 f ∞ b − a n+1
≤
(4 − m)n+1 + 2 (m − 2)n+1 , f (n) ∈ L∞ [a, b] .
(n + 1)!
2m
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O STROWSKI T YPE I NEQUALITIES
Proof. Using the choice α2 =
a+b
2
=
x1 +x2
,
2
(α1 − a) =
x1 =
(m−1)a+b
m
27
and x2 =
a+(m−1)b
m
we may calculate
(4 − m) (b − a)
= (b − α3 )
2m
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 6.15. For m = 3, we have the best estimation of (6.14) such that
Z
n
b
X
(−1)j
(6.15)
f (t) dt +
Q̂3,n := 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 ∞
≤
(b − a)n+1 , f (n) ∈ L∞ [a, b] .
n
6 (n + 1)!
Proof. From the right hand side of (6.14), consider the mapping
n+1
n+1
4−m
m−2
Mm,n :=
+2
2m
2m
then
2 (n + 1)
=
m2
and Mm,n attains its optimum when
0
Mm,n
n n 2
1
1
1
−
−
+
−
m 2
2 m
2
1
1
1
− = − ,
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
b−a
2a + b
a + 2b
+
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.
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28
A. S OFO
Corollary 6.16. Let f be defined as in Theorem 6.10 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
Then for r a natural number, r ≥ 3, we have the inequality
(6.16) |Tr,n |
Z
"
j
n
j b
o
X
(−1)
b − a n (j−1)
j (j−1)
:= f (t) dt +
f
(b) − (−1) f
(a)
a
j!
r
j=1
(
j
j )
b
−
a
(b − a) (r − 3)
+
− (−1)j
f (j−1) (x1 )
3r
6
(
#
j
j )
b−a
(b
−
a)
(r
−
3)
+
− (−1)j
f (j−1) (x2 ) 6
3r
!
(n) n+1
2 f ∞
1
r−3
1
n+1
≤
(b − a)
+
+ n+1 , f (n) ∈ L∞ [a, b] .
n+1
(n + 1)!
r
3r
6
Proof. From Theorem 6.10, we put x1 =
α3 = a+(r−1)b
. Then (6.16) follows.
r
2a+b
,
3
x2 =
a+2b
,
3
α2 =
a+b
,
2
α1 =
(r−1)a+b
r
and
Remark 6.17. The optimum estimate of the inequality (6.16) occurs when r = 6.
from (6.16) consider the mapping
Mr,n :=
1
rn+1
+
r−3
3r
n+1
+
1
6n+1
1 1 n
0
= − (n + 1) r−n−2 + n+1
− r and Mr,n attains its optimum when 1r = 13 − 1r ,
the Mr,n
r2
3
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
(6.17)
Z b
b
−
a
f (t) dt −
(f (a) + f (b))
Q̂3,1 := 6
a
b−a
2a + b
a + 2b
−
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).
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O STROWSKI T YPE I NEQUALITIES
29
Corollary 6.18. Let f be defined as in Theorem 6.10 and choose m = 8 such that α1 =
α2 = a+b
and α3 = a+7b
with x1 = 2a+b
and x2 = a+2b
. Then we have the inequality
8
8
3
3
7a+b
,
8
(6.18) |T8,n |
Z
"
j
n
j b
X
b − a (j−1)
(−1)
:= f
(b) − (−1)j f (j−1) (a)
f (t) dt +
a
j!
8
j=1
!
#
j j
b−a
5
j
j
+
− (−1)
f (j−1) (x1 ) − (−1) f (j−1) (x2 ) 6
4
(n) 2 f ∞ b − a n+1 n+1
≤
3
+ 4n+1 + 5n+1 , f (n) ∈ L∞ [a, b] .
(n + 1)!
24
Proof. From Theorem 6.10 we put
x1 =
and α2 =
a+b
2
a + 2b
7a + b
a + 7b
2a + b
, 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 6.19. From (6.16) with r = 3 we have
Z
"
j
n
b
X
b − a (j−1)
(−1)j
f
(b) − (−1)j f (j−1) (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
!
2
0 2a+b
−
f
f 0 a+2b
b−a
3
3
+
6
2
kf 00 k∞
≤
(b − a)3 , f 00 ∈ L∞ [a, b] .
72
The following theorem encapsulates Boole’s rule.
Theorem 6.20. 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
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30
A. S OFO
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
j
j
+ (x1 − α1 ) − (x1 − α2 ) f (j−1) (x1 )
+ (x2 − α2 )j − (x2 − α3 )j f (j−1) (x2 )
i
j
j
j (j−1)
(j−1)
+ (x3 − α3 ) − (x3 − α4 ) f
(x3 ) + (b − α4 ) f
(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] .
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 6.20.
Corollary 6.21. Let f be defined as in Theorem 6.10 and choose α1 = 11a+b
, α2 = 11a+7b
,
12
18
x1 +x3
a+11b
7a+2b
2a+7b
a+b
α3 = 7a+11b
,
α
=
,
x
=
,
x
=
and
x
=
=
,
then
we
can
state:
4
1
3
2
18
12
9
9
2
2
Z
"
n
j
b
o
X
(−1)j
b − a n (j−1)
j (j−1)
f (t) dt +
f
(b) − (−1) f
(a)
a
j!
12
j=1
)
j ( j
b−a
5
7a + 2b
2a + 7b
j
j (j−1)
(j−1)
+
− (−1)
f
− (−1) f
6
6
9
9
#
j
o
b−a n
a + b j
(j−1)
+
1 − (−1) f
9
2
(n) 2 f ∞ b − a n+1 n+1
≤
3
+ 4n+1 + 5n+1 + 6n+1 , if f (n) ∈ L∞ [a, b] .
(n + 1)!
36
7. A PPLICATIONS FOR N UMERICAL I NTEGRATION
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 )
a
where
(7.2)
Ak (∆m , f ) := −Tk (∆m , f ) − Uk (∆m , f ) ,
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O STROWSKI T YPE I NEQUALITIES
Tk (∆m , f ) :=
(7.3)
m−1
n XX
i=0 j=1
hi
2k
j
31
i
1 h (j−1)
−f
(xi ) + (−1)j f (j−1) (xi+1 )
j!
and
(7.4) Uk (∆m , f ) :=
m−1
n XX
i=0 j=1
hi
2k
j
" k−1
#
o
Xn
1
(k
−
r)
x
+
rx
i
i+1
×
(−1)j − 1 f (j−1)
j!
k
r=1
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] ,
2 (n + 1)!k n+1 i=0 i
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 (j−1) (xi ) + (−1)j f (j−1) (xi+1 )
f (t) dt +
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)
x
+
rx
i
i+1
+
(−1)j − 1 f (j−1)
k
r=1
m−1
X hn+1
1
i
≤
n
n+1
(n + 1)!2 i=0 k
sup
(n) f (t) .
t∈[xi ,xi+1 ]
Now,
(7.6)
As
Z
j h
m−1
n b
i
XX
hi
1
j (j−1)
(j−1)
f
(t)
dt
+
−f
(x
)
+
(−1)
f
(x
)
i
i+1
a
2k
j!
i=0 j=1
j " X
#
m−1
n k−1 n
o
XX
hi
1
(k
−
r)
x
+
rx
i
i+1
+
(−1)j − 1 f (j−1)
2k
j!
k
r=1
i=0 j=1
sup
t∈[xi ,xi+1 ]
≤ Rk (∆m , f ) .
(n) (n) f (t) ≤ f , the inequality in (7.5) follows and the theorem is proved. ∞
The following corollary holds.
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32
A. S OFO
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
f (t) dt = −T2 (∆m , f ) − U2 (∆m , f ) + R2 (∆m , f ) ,
(7.7)
a
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
U2 (∆m , f ) :=
m−1
n XX
i=0 j=1
hi
4
j
o
1n
xi + xi+1
j
(j−1)
(−1) − 1 f
j!
2
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] .
i
2
(n + 1)! i=0
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
where
T3 (∆m , f ) :=
m−1
n XX
i=0 j=1
hi
6
j
i
1 h (j−1)
−f
(xi ) + (−1)j f (j−1) (xi+1 )
j!
and
U3 (∆m , f ) :=
m−1
n
XX
(−1) − 1 h j
i=0 j=1
j
i
j!
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
∞
|R3 (∆m , f )| ≤
hn+1 , if f (n) ∈ L∞ [a, b] .
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!
j
j
− (−1) (xi+1 − ξi ) f
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
(j−1)
o
(xi+1 ) + R (ξ, ∆m , f )
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O STROWSKI T YPE I NEQUALITIES
33
and the remainder, R (ξ, ∆m , f ) satisfies the inequality
|R (ξ, ∆m , f )|
(7.10)
(n) m−1
f X
∞
(ξi − xi )n+1 + (xi+1 − ξi )n+1 , if f (n) ∈ L∞ [a, b] .
≤
(n + 1)! i=0
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=1
j!
i
− (a − α)j f (j−1) (a) + (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!
i
(xi+1 − ξi )j f (j−1) (xi+1 ) − (−1)j (ξi − xi )j f (j−1) (xi )
= R (ξ, ∆m , f )
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 ) m−1
X
1
sup f (n) (t) (ξi − xi )n+1 + (xi+1 − ξi )n+1 .
≤
(n + 1)! i=0 t∈[xi ,xi+1 ]
The inequality in (7.11) follows, since we have sup f (n) (t) ≤ f (n) ∞ , and Theorem
t∈[xi ,xi+1 ]
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
Z b
m−1
n
XX
(−hi )j (j−1)
f (t) dt =
f
(xi ) + Rl (∆m , f ) .
j!
a
i=0 j=1
(ii) The nth order right rectangle rule
Z b
m−1
n
XX
(hi )j (j−1)
f (t) dt = −
f
(xi+1 ) + Rr (∆m , f ) .
j!
a
i=0 j=1
(iii) The nth order trapezoidal rule
Z
b
f (t) dt =
a
m−1
n XX
i=0 j=1
hi
−
2
j
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
o
1 n (j−1)
j (j−1)
f
(xi ) − (−1) f
(xi+1 ) + RT (∆m , f ) ,
j!
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34
A. S OFO
where
(n) m−1
f X
∞
|Rl (∆m , f )| = |Rr (∆m , f )| ≤
hn+1
, if f (n) ∈ L∞ [a, b]
i
(n + 1)! i=0
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
− ξi − αi
f (j−1) (ξi )
j
(2)
(1)
(2)
(j−1)
− xi+1 − αi
f
(xi+1 ) + R ξ, αi , αi , ∆m , f
and the remainder satisfies the estimation
f (n) m−1
n+1 n+1
X (1)
(1)
(2)
(1)
∞
αi − xi
+ ξi − αi
R ξ, αi , αi , ∆m , f ≤
(n + 1)! i=0
n+1 n+1 (2)
(2)
+ αi − ξi
+ xi+1 − αi
, 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
m−1
n
XX
i=0 j=1
(−1)j (ξi − xi )n+1 − (ξi − xi+1 )n+1 f (j−1) (ξi )
j!
+ 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
(1)
if f (n) ∈ L∞ [a, b] .
(2)
The proof follows from (7.11) where αi = xi and αi = xi+1 .
Remark 7.8. If in (7.12) we choose the midpoint 2ξi = xi+1 + xi we obtain the generalised
midpoint quadrature formula
(7.13)
Z b
f (t) dt =
a
m−1
n
XX
i=0 j=1
(−1)j+1
j!
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
hi
2
j n
o
xi + xi+1
j
(j−1)
1 − (−1) f
+ RM (∆m , f )
2
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O STROWSKI T YPE I NEQUALITIES
35
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 7.9. Consider a set of points
5xi + xi+1 xi + 5xi+1
ξi ∈
,
6
6
(i = 0, . . . , m − 1)
and let f and ∆m be defined as in Theorem 7.1. Then we have the equality
b
Z
f (t) dt =
(7.14)
a
m−1
n
XX
i=0 j=1
(
−
ξi −
(−1)j
j!
5xi + xi+1
6
"
j
hi
6
j n
o
(−1)j f (j−1) (xi ) − f (j−1) (xi+1 )
− ξi −
xi + 5xi+1
6
j )
#
f (j−1) (ξi ) + Rs (∆m , f )
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 7.10. If in (7.14) we choose the midpoint ξi =
son formula:
Z
b
f (t) dt =
a
m−1
n
XX
i=0 j=1
(−1)j
j!
xi+1 +xi
2
we obtain a generalised Simp-
"
j
o
hi n
(−1)j f (j−1) (xi ) − f (j−1) (xi+1 )
6
j n
#
o
hi
xi+1 + xi
j
(j−1)
−
1 − (−1) f
+ 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.
Corollary 7.11. Consider the interval
(1)
xi ≤ αi ≤
xi+1 + xi
(2)
≤ αi ≤ xi+1 (i = 0, . . . , m − 1) ,
2
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36
A. S OFO
and let f and ∆m be defined as in Theorem 7.1.
The following equality is obtained:
(7.15)
m−1
n
XX
j
(−1)j (i)
f (t) dt =
xi − αi
f (j−1) (xi )
j!
a
i=0 j=1
(
j j )
xi+1 + xi
x
+
x
xi+1 + xi
i+1
i
(1)
(2)
(j−1)
−
− αi
−
− αi
f
2
2
2
j
(2)
(1)
(2)
− xi+1 − αi
f (j−1) (xi+1 ) + RB αi , αi , ∆m , f ,
Z
b
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
x
+
x
i+1
i
(2)
(2)
+ αi −
+ xi+1 − αi
.
2
The following remark applies to Corollary 7.11.
Remark 7.12. If in (7.15) we choose
(1)
αi =
3xi + xi+1
xi + 3xi+1
(2)
and αi =
,
4
4
we have the formula:
Z
b
f (t) dt =
(7.16)
m−1
n
XX
a
i=0 j=1
(−1)j
j!
j
hi h
(−1)j f (j−1) (xi ) − f (j−1) (xi+1 )
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
∞
|RB (∆m , f )| ≤
×4
, if f (n) ∈ L∞ [a, b] .
(n + 1)!
4
i=0
The following theorem incorporates the Newton-Cotes formula.
Theorem 7.13. Consider the interval
(1)
(1)
xi ≤ αi ≤ ξi
(2)
(2)
≤ αi ≤ ξi
J. Inequal. Pure and Appl. Math., 3(2) Art. 21, 2002
(3)
≤ αi ≤ xi+1 (i = 0, . . . , m − 1) ,
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O STROWSKI T YPE I NEQUALITIES
37
and let ∆m and f be defined as in Theorem 7.1. This consideration gives us the equality
Z
(7.17)
a
b
n
m−1
XX
j
(−1)j (1)
xi − αi
f (j−1) (xi )
f (t) dt =
j!
i=0 j=1
j
(3)
− xi+1 − αi
f (j−1) (xi+1 )
j j (1)
(1)
(1)
(2)
(1)
−
ξi − αi
− ξi − αi
f (j−1) ξi
j j (2)
(2)
(2)
(3)
(2)
(j−1)
ξi
−
ξi − αi
− ξi − αi
f
(1)
(2)
(3) (1) (2)
+R αi , αi , αi , ξi , ξi , ∆m , f .
The remainder satisfies the bound
(1)
(2)
(3) (1) (2)
R
α
,
α
,
α
,
ξ
,
ξ
,
∆
,
f
m
i
i
i
i
i
(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
(3)
(2)
(2)
(1)
(2)
(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 7.13.
(1)
i+1
Corollary 7.14. Let f and ∆m be defined as above and make the choices αi = 7xi +x
,
8
(2)
(3)
(1)
(2)
xi +7xi+1
2xi +xi+1
xi +2xi+1
xi +xi+1
, αi =
, ξi =
and ξi =
, then we have the equality:
αi =
2
8
3
3
Z
b
f (t) dt
(7.18)
a
=
" j
o
(−1)j
hi n
(−1)j f (j−1) (xi ) − f (j−1) (xi+1 )
j!
8
j=1
j n
o
hi
2xi + xi+1
j
j
(j−1)
−
5 − (−4) f
24
3
j n
#
o
hi
x
+
2x
i
i+1
−
4j − (−5)j f (j−1)
+ RN (∆m , f ) ,
24
3
m−1
n
XX
i=0
where the remainder satisfies the bound
X hi n+1
m−1
2 f (n) ∞ n+1
n+1
n+1
|RN (∆m , f )| ≤
3
+4
+5
,
(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.
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38
A. S OFO
Remark 7.15. For n = 2 from (7.18), we obtained a perturbed three-eighths Newton-Cotes
formula:
2 0
Z b
m−1
X hi hi
f (xi ) − f 0 (xi+1 )
f (t) dt =
(f (xi ) + f (xi+1 )) +
8
8
2
a
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
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
8. C ONCLUSION
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 NewtonCotes 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
Z b
1
M (h) =
h (t) dt.
b−a a
Then from (5.2)
T (h, g) = M (hg) − M (h) M (g) .
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