volume 3, issue 2, article 21, 2002.
Received December, 2000;
accepted 16 November, 2001.
Communicated by:T. Mills
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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/
c
2000Victoria University ISSN (electronic): 1443-5756 083-01
Integral Inequalities of the Ostrowski Type
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Abstract
Integral inequalities of Ostrowski type are developed forn−times differentiable mappings, with multiple branches, on theL∞norm. Some particular inequali- ties are also investigated, which include explicit bounds for perturbed trapezoid, midpoint, Simpson’s, Newton-Cotes and left and right rectangle rules. The re- sults obtained provide sharper bounds than those obtained by Dragomir [5] and Cerone, Dragomir and Roumeliotis [2].
2000 Mathematics Subject Classification:26D15, 41A55.
Key words: Ostrowski Integral Inequality, Quadrature Formulae.
I wish to acknowledge the useful comments of an anonymous referee which led to an improvement of this work.
Contents
1 Introduction. . . 3
2 Integral Identities . . . 6
3 Integral Inequalities . . . 20
4 The Convergence of a General Quadrature Formula. . . 27
5 Grüss Type Inequalities . . . 33
6 Some Particular Integral Inequalities . . . 43
7 Applications for Numerical Integration . . . 64
8 Conclusion. . . 80 References
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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] → Rbe a differentiable mapping on [a, b]and let
|f0(t)| ≤M for allt ∈(a, b), then the following bound is valid (1.1)
f(x)− 1 b−a
Z b a
f(t)dt
≤(b−a)M
"
1
4+ x− a+b2 2
(b−a)2
#
for allx∈[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 ex- tended 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 mono- tonic mappings.
Cerone, Dragomir and Roumeliotis [3] as well as Dedi´c, Mati´c and Peˇcari´c [4] and Pearce, Peˇcari´c, Ujevi´c and Varošanec [19] further extended the result (1.1) by consideringn−times differentiable mappings on an interior pointx∈ [a, b].
In particular, Cerone and Dragomir [1] proved the following result.
Theorem 1.2. Let f : [a, b] → Rbe a mapping such that f(n−1) is absolutely continuous on [a, b], (AC[a, b]for short). Then for allx ∈ [a, b]the following
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bound is valid:
(1.2)
Z b a
f(t)dt−
n−1
X
j=0
(b−x)j+1+ (−1)j(x−a)j+1 (j + 1)!
!
f(j)(x)
≤
f(n) ∞
(n+ 1)! (x−a)n+1+ (b−x)n+1
if f(n) ∈L∞[a, b], where
f(n)
∞:= sup
t∈[a,b]
f(n)(t) <∞.
Dragomir also generalised the Ostrowski inequality fork 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)andαk+1 =b. Iff : [a, b]→Ris AC[a, b], then we have the inequality
Z b a
f(x)dx−
k
X
i=0
(αi+1−αi)f(xi) (1.3)
≤
"
1 4
k−1
X
i=0
h2i +
k−1
X
i=0
αi+1− xi+xi+1 2
2# kf0k∞
≤ 1 2kf0k∞
k−1
X
i=0
h2i ≤ 1
2(b−a)kf0k∞ν(h) ,
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wherehi :=xi+1−xi(i= 0, ..., k−1)andν(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 forn−times differentiable mappings. The mo- tivation 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 ordern.
We begin the process by obtaining the following integral equalities.
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2. Integral Identities
Theorem 2.1. Let Ik : a = x0 < x1 < · · · < xk−1 < xk = bbe 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)andαk+1 =b. Iff : [a, b]→Ris a mapping such thatf(n−1)is AC[a, b], then for allxi ∈[a, b]we have the identity:
(2.1) Z b
a
f(t)dt+
n
X
j=1
(−1)j j!
k−1
X
i=0
n
(xi+1−αi+1)jf(j−1)(xi+1)
−(xi−αi+1)jf(j−1)(xi)o
= (−1)n Z b
a
Kn,k(t)f(n)(t)dt, where the Peano kernel
(2.2) Kn,k(t) :=
(t−α1)n
n! , t ∈[a, x1) (t−α2)n
n! , t ∈[x1, x2) ...
(t−αk−1)n
n! , t ∈[xk−2, xk−1) (t−αk)n
n! , t ∈[xk−1, b], nandkare natural numbers,n ≥1,k ≥1andf(0)(x) =f(x).
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Proof. The proof is by mathematical induction. Forn = 1, from (2.1) we have the equality
(2.3) Z b
a
f(t)dt =
k−1
X
i=0
h
(xi+1−αi+1)jf(xi+1)−(xi−αi+1)jf(xi)i
− Z b
a
K1,k(t)f0(t)dt, where
K1,k(t) :=
(t−α1), t∈[a, x1) (t−α2), 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 a
K1,k(t)f0(t)dt
=
k−1
X
i=0
Z xi+1
xi
(t−αi+1)f0(t)dt
=
k−1
X
i=0
(t−αi+1)f(t)
xi+1
xi − Z xi+1
xi
f(t)dt
=
k−1
X
i=0
[(xi+1−αi+1)f(xi+1)−(xi−αi+1)f(xi)]−
k−1
X
i=0
Z xi+1
xi
f(t)dt,
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Z b a
f(t)dt+ Z b
a
K1,k(t)f0(t)dt
=
k−1
X
i=0
[(xi+1−αi+1)f(xi+1)−(xi−αi+1)f(xi)]. 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
(2.4) Z b
a
f(t)dt+
k−1
X
i=0 n+1
X
j=1
(−1)j j!
×n
(xi+1−αi+1)jf(j−1)(xi+1)−(xi−αi+1)jf(j−1)(xi)o
= (−1)n+1 Z b
a
Kn+1,k(t)f(n+1)(t)dt, 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].
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Consider Z b a
Kn+1,k(t)f(n+1)(t)dt=
k−1
X
i=0
Z xi+1
xi
(t−αi+1)n+1
(n+ 1)! f(n+1)(t)dt and upon integrating by parts we have
Z b a
Kn+1,k(t)f(n+1)(t)dt
=
k−1
X
i=0
"
(t−αi+1)n+1
(n+ 1)! f(n)(t)
xi+1
xi
− Z xi+1
xi
(t−αi+1)n
n! f(n)(t)dt
#
=
k−1
X
i=0
((xi+1−αi+1)n+1
(n+ 1)! f(n)(xi+1)− (xi−αi+1)n+1
(n+ 1)! f(n)(xi) )
− Z b
a
Kn,k(t)f(n)(t)dt.
Upon rearrangement we may write Z b
a
Kn,k(t)f(n)(t)dt
=
k−1
X
i=0
((xi+1−αi+1)n+1
(n+ 1)! f(n)(xi+1)− (xi−αi+1)n+1
(n+ 1)! f(n)(xi) )
− Z b
a
Kn+1,k(t)f(n+1)(t)dt.
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Now substitute Rb
a Kn,k(t)f(n)(t)dt from the induction hypothesis (2.1) such that
(−1)n Z b
a
f(t)dt+ (−1)n
k−1
X
i=0
" n X
j=1
(−1)j j!
n
(xi+1−αi+1)jf(j−1)(xi+1)
− (xi−αi+1)jf(j−1)(xi)oi
=
k−1
X
i=0
((xi+1−αi+1)n+1
(n+ 1)! f(n)(xi+1)− (xi−αi+1)n+1
(n+ 1)! f(n)(xi) )
− Z b
a
Kn+1,k(t)f(n+1)(t)dt.
Collecting the second and third terms and rearranging, we can state Z b
a
f(t)dt+
k−1
X
i=0 n+1
X
j=1
(−1)j j!
×n
(xi+1−αi+1)jf(j−1)(xi+1) −(xi−αi+1)jf(j−1)(xi)o
= (−1)n+1 Z b
a
Kn+1,k(t)f(n+1)(t)dt, which is identical to (2.4), hence Theorem2.1is 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 Theorem2.1, the equality (2.1) may be represented as (2.5)
Z b a
f(t)dt +
n
X
j=1
(−1)j j!
" k X
i=0
n
(xi−αi)j−(xi−αi+1)jo
f(j−1)(xi)
#
= (−1)n Z b
a
Kn,k(t)f(n)(t)dt.
Proof. From (2.1) consider the second term and rewrite it as
(2.6) S1+S2 :=
k−1
X
i=0
{−(xi+1−αi+1)f(xi+1) + (xi−αi+1)f(xi)}
+
k−1
X
i=0
" n X
j=2
(−1)j j!
n
(xi+1−αi+1)jf(j−1)(xi+1)
− (xi−αi+1)jf(j−1)(xi)o . Now
S1 = (a−α1)f(a) +
k−1
X
i=1
(xi−αi+1)f(xi)
+
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
i=1
(xi−αi+1)f(xi)
+
k−1
X
i=1
{−(xi−αi)f(xi)} −(b−αk)f(b)
=−(α1−a)f(a)−
k−1
X
i=1
(αi+1−αi)f(xi)−(b−αk)f(b). Also,
S2 =
k−2
X
i=0
" n X
j=2
(−1)j j!
n
(xi+1−αi+1)jf(j−1)(xi+1)o
#
+
n
X
j=2
(−1)j j!
n
(xk−αk)jf(j−1)(xk)o
−
n
X
j=2
(−1)j j!
n
(x0−α1)jf(j−1)(x0) o
−
k−1
X
i=1
" n X
j=2
(−1)j j!
n
(xi−αi+1)jf(j−1)(xi)o
#
=
n
X
j=2
(−1)j
j! (b−αk)jf(j−1)(b)−
n
X
j=2
(−1)j
j! (a−α1)jf(j−1)(a)
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+
k−1
X
i=1
" n X
j=2
(−1)j j!
n
(xi −αi)j−(xi−αi+1)jo
f(j−1)(xi)
# . From (2.6)
S1+S2
:=− (
(b−αk)f(b) + (α1−a)f(a) +
k−1
X
i=1
(αi+1−αi)f(xi) )
+
n
X
j=2
(−1)j j!
n
(b−αk)jf(j−1)(b)−(a−α1)jf(j−1)(a)o
+
k−1
X
i=1
" n X
j=2
(−1)j j!
n
(xi−αi)j −(xi−αi+1)jo
f(j−1)(xi)
#
=− (
(α1−a)f(a) +
k−1
X
i=1
(αi+1−αi)f(xi) + (b−αk)f(b) )
+
n
X
j=2
(−1)j j!
"
−(a−α1)jf(j−1)(a)
+
k−1
X
i=1
n
(xi−αi)j −(xi−αi+1)jo
f(j−1)(xi) + (b−αk)jf(j−1)(b)
# .
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Keeping in mind thatx0 =a,α0 = 0,xk =bandαk+1 =bwe may write S1+S2 =−
k
X
i=0
(αi+1−αi)f(xi)
+
n
X
j=2
(−1)j j!
" k X
i=0
n
(xi−αi)j −(xi−αi+1)jo
f(j−1)(xi)
#
=
n
X
j=1
(−1)j j!
" k X
i=0
n
(xi −αi)j−(xi−αi+1)j o
f(j−1)(xi)
# . 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 divisionIk are fixed, we obtain the following corollary.
Corollary 2.3. LetIk :a =x0 < x1 <· · · < xk−1 < xk =b be a division of the interval[a, b]. Iff : [a, b]→ Ris as defined in Theorem2.1, then we have the equality
(2.7) Z b
a
f(t)dt+
n
X
j=1
1 2jj!
" k X
i=0
n−hji + (−1)jhji−1o
f(j−1)(xi)
#
= (−1)n Z b
a
Kn,k(t)f(n)(t)dt, wherehi :=xi+1−xi,h−1 := 0andhk := 0.
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Proof. Choose
α0 = a, α1 = a+x1
2 , α2 = x1+x2 2 , . . . , αk−1 = xk−2+xk−1
2 ,αk = xk−1+xk
2 and αk+1 =b.
From Corollary2.2, the term
(b−αk)f(b) + (α1−a)f(a) +
k−1
X
i=1
(αi+1−αi)f(xi)
= 1 2
(
h0f(a) +
k−1
X
i=1
(hi+hi−1)f(xi) +hk−1f(b) )
, the term
n
X
j=2
(−1)j j!
n
(b−αk)jf(j−1)(b)−(a−α1)jf(j−1)(a)o
=
n
X
j=2
(−1)j j!2j
n
hjk−1f(j−1)(b)−(−1)jhj0f(j−1)(a)o and the term
k−1
X
i=1
" n X
j=2
(−1)j j!
n
(xi−αi)j−(xi−αi+1)jo
f(j−1)(xi)
#
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=
k−1
X
i=1
" n X
j=2
(−1)j j!2j
n
hji−1−(−1)jhjio
f(j−1)(xi)
# . Putting the last three terms in (2.5) we obtain
Z b a
f(t)dt− 1 2
(
h0f(a) +
k−1
X
i=1
(hi+hi−1)f(xi) +hk−1f(b) )
+
n
X
j=2
(−1)j j!2j
n
hjk−1f(j−1)(b)−(−1)jhj0f(j−1)(a)o
+
k−1
X
i=1
" n X
j=2
(−1)j j!2j
n
hji−1−(−1)jhjio
f(j−1)(xi)
#
= (−1)n Z b
a
Kn,k(t)f(n)(t)dt.
Collecting the inner three terms of the last expression, we have Z b
a
f(t)dt+
n
X
j=1
(−1)j j!2j
k
X
i=0
n
hji−1−(−1)jhji o
f(j−1)(xi)
= (−1)n Z b
a
Kn,k(t)f(n)(t)dt, 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.
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Corollary 2.4. Let
(2.8) Ik:xi =a+i
b−a k
, i= 0, . . . , k
be an equidistant partitioning of[a, b], andf : [a, b] → Rbe a mapping such thatf(n−1)is AC[a, b],then we have the equality
(2.9) Z b
a
f(t)dt+
n
X
j=1
b−a 2k
j
1 j!
"
−f(j−1)(a)
+
k−1
X
i=1
n(−1)j −1o
f(j−1)(xi) + (−1)jf(j−1)(b)
#
= (−1)n Z b
a
Kn,k(t)f(n)(t)dt.
It is of some interest to note that the second term of (2.9) involves only even derivatives at all interior pointsxi,i= 1, . . . , k−1.
Proof. Using (2.8) we note that
h0 = x1−x0 = b−a
k , hk−1 = (xk−xk−1) = b−a k , hi = xi+1−xi = b−a
k and hi−1 = xi−xi−1 = b−a
k , (i= 1, . . . , k−1)
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and substituting into (2.7) we have Z b
a
f(t)dt+
n
X
j=1
1 j!2j
"
−
b−a 2k
j
f(j−1)(a) +
k−1
X
i=0
(
−
b−a k
j
+ (−1)j
b−a k
j)
f(j−1)(xi) + (−1)j
b−a k
j
f(j−1)(b)
#
= (−1)n Z b
a
Kn,k(t)f(n)(t)dt, which simplifies to (2.9) after some minor manipulation.
The following Taylor-like formula with integral remainder also holds.
Corollary 2.5. Letg : [a, y]→Rbe a mapping such thatg(n)is AC[a, y]. Then for allxi ∈[a, y]we have the identity
(2.10) g(y) =g(a)−
k−1
X
i=0
" n X
j=1
(−1)j j!
n
(xi+1−αi+1)jg(j)(xi+1)
− (xi−αi+1)jg(j)(xi)o
+ (−1)n Z y
a
Kn,k(y, t)g(n+1)(t)dt or
(2.11) g(y)
=g(a)−
n
X
j=1
(−1)j j!
" k X
i=0
n
(xi−αi)j −(xi−αi+1)jo
g(j)(xi)
#
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+ (−1)n Z y
a
Kn,k(y, t)g(n+1)(t)dt.
The proof of (2.10) and (2.11) follows directly from (2.1) and (2.5) respec- tively upon choosingb=yandf =g0.
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3. Integral Inequalities
In this section we utilise the equalities of Section2and 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 = bbe 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)andαk+1 =b. Iff : [a, b]→Ris a mapping such thatf(n−1)is AC[a, b], then for allxi ∈[a, b]we have the inequality:
Z b a
f(t)dt+
n
X
j=1
(−1)j j!
" k X
i=0
n
(xi −αi)j−(xi−αi+1)jo
f(j−1)(xi)
# (3.1)
≤
f(n) ∞ (n+ 1)!
k−1
X
i=0
(αi+1−xi)n+1+ (xi+1−αi+1)n+1
≤
f(n) ∞ (n+ 1)!
k−1
X
i=0
hn+1i
≤
f(n) ∞
(n+ 1)! (b−a)νn(h) if f(n) ∈L∞[a, b], where
f(n)
∞ := sup
t∈[a,b]
f(n)(t) <∞, hi :=xi+1−xi and
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ν(h) := max{hi|i= 0, . . . , k−1}. Proof. From Corollary2.2we may write
(3.2)
Z b a
f(t)dt +
n
X
j=1
(−1)j j!
" k X
i=0
n
(xi−αi)j−(xi−αi+1)jo
f(j−1)(xi)
#
=
(−1)n Z b
a
Kn,k(t)f(n)(t)dt ,
and
(−1)n Z b
a
Kn,k(t)f(n)(t)dt
≤ f(n)
∞
Z b a
|Kn,k(t)|dt,
Z b a
|Kn,k(t)|dt =
k−1
X
i=0
Z xi+1
xi
|t−αi+1|n
n! dt
=
k−1
X
i=0
Z αi+1
xi
(αi+1−t)n n! dt+
Z xi+1
αi+1
(t−αi+1)n
n! dt
= 1
(n+ 1)!
k−1
X
i=0
(αi+1−xi)n+1+ (xi+1−αi+1)n+1
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and thus
(−1)n Z b
a
Kn,k(t)f(n)(t)dt
≤
f(n) ∞
(n+ 1)!
k−1
X
i=0
(αi+1−xi)n+1+ (xi+1−αi+1)n+1 . Hence, from (3.2), the first part of the inequality (3.1) is proved. The second and third lines follow by noting that
f(n) ∞ (n+ 1)!
k−1
X
i=0
(αi+1−xi)n+1+ (xi+1−αi+1)n+1 ≤
f(n) ∞ (n+ 1)!
k−1
X
i=0
hn+1i , since for0< B < A < C it is well known that
(3.3) (A−B)n+1+ (C−A)n+1 ≤(C−B)n+1. Also
f(n) ∞ (n+ 1)!
k−1
X
i=0
hn+1i ≤
f(n) ∞ (n+ 1)!νn(h)
k−1
X
i=0
hi =
f(n) ∞
(n+ 1)! (b−a)νn(h), where ν(h) := max{hi|i= 0, . . . , k−1} and therefore the third line of the inequality (3.1) follows, hence Theorem3.1is proved.
When the points of the divisionIkare fixed, we obtain the following inequal- ity.
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Corollary 3.2. Let f : [a, b] → R be a mapping such that f(n−1) is AC[a, b], andIkbe defined as in Corollary2.3, then
(3.4)
Z b a
f(t)dt+
n
X
j=1
(−1)j 2jj!
" k X
i=0
n−hji + (−1)jhji+1o
f(j−1)(xi)
#
≤
f(n) ∞ (n+ 1)!2n
k−1
X
i=0
hn+1i forf(n) ∈L∞[a, b].
Proof. From Corollary2.3we choose α0 = a, α1 = a+x1
2 , . . . , αk−1 = xk−2+xk−1
2 , αk = xk−1+xk
2 andαk+1 =b.
Now utilising the first line of the inequality (3.1), we may evaluate
k−1
X
i=0
(αi+1−xi)n+1+ (xi+1−αi+1)n+1 =
k−1
X
i=0
2 hi
2 n+1
and therefore the inequality (3.4) follows.
For the equidistant partitioning case we have the following inequality.
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Corollary 3.3. Letf : [a, b]→Rbe a mapping such thatf(n−1) is AC[a, b]and letIkbe defined by (2.8). Then
(3.5)
Z b a
f(t)dt+
n
X
j=1
b−a 2k
j
1 j!
×
"
−f(j−1)(a) +
k−1
X
i=1
n(−1)j−1o
f(j−1)(xi) + (−1)jf(j−1)(b)
#
≤
f(n) ∞
(n+ 1)! (2k)n (b−a)n+1 forf(n) ∈L∞[a, b].
Proof. We may utilise (2.9) and from (3.1), note that
h0 =x1−x0 = b−a k and hi =xi+1−xi = b−a
k , i= 1, . . . , k−1 in which case (3.5) follows.
The following inequalities for Taylor-like expansions also hold.
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Corollary 3.4. Letgbe defined as in Corollary2.5. Then we have the inequality (3.6)
g(y)−g(a) +
n
X
j=1
(−1)j j!
×
" k X
i=0
n
(xi−αi)j−(xi−αi+1)jo
g(j)(xi)
#
≤
g(n+1) ∞ (n+ 1)!
k−1
X
i=0
(αi+1−xi)n+1+ (xi+1−αi+1)n+1 ifg(n+1)∈L∞[a, b]
for allxi ∈[a, y]where g(n+1)
∞ := sup
t∈[a,y]
g(n+1)(t) <∞.
Proof. Follows directly from (2.11) and using the norm as in (3.1).
When the points of the divisionIkare fixed we obtain the following.
Corollary 3.5. Let g be defined as in Corollary2.5 and Ik : a = x0 < x1 <
· · · < xk−1 < xk = y be a division of the interval [a, y]. Then we have the inequality
(3.7)
g(y)−g(a) +
n
X
j=1
1 2jj!
" k X
i=0
n
−hji + (−1)jhji+1 o
g(j)(xi)
#
≤
g(n+1) ∞ (n+ 1)!2n
k−1
X
i=0
hn+1i ifg(n+1) ∈L∞[a, y].
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Proof. The proof follows directly from using (2.7).
For the equidistant partitioning case we have:
Corollary 3.6. Letg be defined as in Corollary2.5and Ik :xi =a+i·
y−a k
, i= 0, . . . , k
be an equidistant partitioning of[a, y], then we have the inequality:
(3.8)
g(y)−g(a) +
n
X
j=1
y−a 2k
j
1 j!
×
"
−g(j)(a) +
k−1
X
i=1
n(−1)j−1o
g(j)(xi) + (−1)jg(j)(y)
#
≤
g(n+1) ∞
(n+ 1)! (2k)n(y−a)n+1 ifg(n+1) ∈L∞[a, y]. Proof. The proof follows directly upon using (2.9) withf0 =g andb =y.
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4. The Convergence of a General Quadrature For- mula
Let
∆m :a=x(m)0 < x(m)1 <· · ·< x(m)m−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, f0, . . . , f(n),∆m, wm :=
m
X
j=0
wj(m)f x(m)j
−
n
X
r=2
(−1)r r!
" m X
j=0
(
x(m)j −a−
j−1
X
s=0
ws(m)
!r
− x(m)j −a−
j
X
s=0
w(m)s
!r) f(r−1)
x(m)j
# ,
wherewj(j = 0, . . . , m)are the quadrature weights and assume thatPm
j=0w(m)j = b−a.
The following theorem contains a sufficient condition for the weights wj(m)
so that
Im f, f0, . . . , f(n),∆m, wm
approximates the integral Rb
af(x)dx with an er- ror expressed in terms of
f(n) ∞.
Theorem 4.1. Let f : [a, b] → Rbe a continuous mapping on[a, b] such that
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f(n−1) is AC[a, b]. If the quadrature weights,w(m)j satisfy the condition
(4.2) x(m)i −a≤
i
X
j=0
w(m)j ≤x(m)i+1 −a for alli= 0, . . . , m−1
then we have the estimation
Im f, f0, . . . , f(n),∆m, wm
− Z b
a
f(t)dt (4.3)
≤
f(n) ∞ (n+ 1)!
m−1
X
i=0
a+
i
X
j=0
wj(m)−x(m)i
!n+1
− x(m)i+1−a−
i
X
j=0
w(m)j
!n+1
≤
f(n) ∞ (n+ 1)!
m−1
X
i=0
h(m)i
n+1
≤
f(n) ∞ (n+ 1)!
ν h(m)
(b−a), where f(n) ∈L∞[a, b]. Also
ν h(m)
:= max
i=0,...,m−1
n h(m)i o h(m)i := x(m)i+1 −x(m)i .
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In particular, if f(n)
∞<∞, then lim
ν(h(m))→0Im f, f0, . . . , f(n),∆m, wm
= Z b
a
f(t)dt uniformly by the influence of the weightswm.
Proof. Define the sequence of real numbers
α(m)i+1 :=a+
i
X
j=0
w(m)j , i= 0, . . . , m.
Note that
α(m)i+1 =a+
m
X
j=0
w(m)j =a+b−a=b.
By the assumption (4.2), we have α(m)i+1 ∈h
x(m)i , x(m)i+1i
for all i= 0, . . . , m−1.
Defineα(m)0 =aand compute α(m)1 −α(m)0 = w0(m) α(m)i+1 −α(m)i = a+
i
X
j=0
w(m)j −a−
i−1
X
j=0
w(m)j =wi(m) (i= 0, . . . , m−1)