Directory UMM :Journals:Journal_of_mathematics:OTHER:
ACTA UNIVERSITATIS APULENSIS
No 16/2008
THE ESTIMATES FOR THE REMAINDER TERM OF SOME
QUADRATURE FORMULAE
Ana Maria Acu, Mugur Acu, Arif Rafiq
Abstract. A new generalization of Ostrowski ′ s integral inequality is
established. A consequence of the generalization is that we can derive new
estimates for remainder term of the midpoint, trapezoid and Simpson formulae. These estimates are improvements of some recently obtained estimates.
Applications in numerical integration are also given.
2000 Mathematics Subject Classification: 65D30 , 65D32.
Keywords and phrases: quadrature rule, Ostrowski inequality
1. Introduction
In [2] Cerone, Dragomir and Roumeliotis proved the following generalization of Ostrowski ′ s inequality:
Theorem 1.[2] Let f : [a, b] → R be continous on (a, b) and whose first derivative f ′ : (a, b) → R is bounded on (a,b). Denote kf ′ k∞ = sup |f ′ (t)| < ∞.
t∈[a,b]
Then we have:
Z b
f
(a)
+
f
(b)
f (t)dt − f (x)(1 − λ) +
λ (b − a)
2
"a
2 #
1
a
+
b
(b − a)2 (λ2 + (1 − λ)2 ) + x −
kf ′ k∞
≤
4
2
for all λ ∈ [0, 1] and a + λ
b−a
b−a
≤x≤b−λ
.
2
2
53
(1)
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
Using (1), they also obtained estimates for the remainder term of the midpoint, trapezoid and Simpson ′ s formulae. They also gave applications of the
mentioned rezults in numerical integration and for special means.
In [4], N. Ujevic proved the following generalization of Ostrowski ′ s inequality:
Theorem 2.[4] Let I ⊂ R be a open interval and a, b ∈ I, a < b. If f : I → R
is a differentiable function such that γ ≤ f ′ (t) ≤ Γ, for all t ∈ [a, b], for some
constants Γ, γ ∈ R, then we have
"
#
Z b
a+b 2
1
x−
Γ+γ
1
a
+
b
1
2
f (x)−
f (t)dt ≤ (Γ−γ)(b−a) +
x−
−
.
2
2
b−a a
2
4
(b−a)2
In this paper we give some generalizations of Ostrowski ′ s inequalities and
we obtain new estimations for the remainder term of some quadrature formulae.
2. Main results
We denote [x] the integer part of x, x ∈ R.
Theorem 3. If f ∈ C n−1 [a, b], f (n−1) , n > 1 is absolutely continuous and
there exist real numbers γ, Γ such that γ ≤ f (n) (t) ≤ Γ, t ∈ [a, b], then
Z b
n−1 X
m
X
1
(−1)k
k
(k)
f (t)dt
(b − a) Ak,k−2j f (x) −
2
(k + 1)!
b−a a
k=0 j=0
m
˜
n X
(b
−
a)
+ (−1)n (Γ + γ)
An,n−2j
(n + 1)! j=0
p˜
) "
( p
2 #
X
x− a+b
Γ−γ (b−a)n X
1
2
·
+
An−2,2j +
, (2)
An−2,2j+1 ·
≤
2
2
n!
4
(b−a)
j=0
j=0
where
i
a+b
1
k+1
x−
Ak,i = k+1−i
, k = 0, n − 1, i = 0, k + 1,
i
2
(b − a)i
2
hni
n−1
n−2
k
, m
˜ =
, p=
, p˜ =
.
and m =
2
2
2
2
54
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
Proof. Let P : [a, b]2 → R a mapping given by
(t − a)n
, if t ∈ [a, x)
n!
P (x, t) =
n
(t − b) , if t ∈ [x, b].
n!
We observe that successive integration by parts yields the relation:
x
Z b
Z x
n−1
n (n−1−k)
X
(n)
n−1−k (t−a)
(k)
n
P (x, t)f (t)dt =
(−1)
· f (t) +(−1)
f (t)dt
n!
a
a
k=0
a
b
Z b
n−1
n (n−1−k)
X
(t
−
b)
(k)
n−1−k
n
f (t)dt
· f (t) + (−1)
+
(−1)
n!
x
k=0
x
Z b
n−1
n−1
k+1
X
X
(x−b)k+1 (k)
(k)
n−1−k (x−a)
n
f (x)− (−1)n−1−k
f (x)
f (t)dt+ (−1)
= (−1)
(k+1)!
(k+1)!
a
k=0
k=0
( Z
)
n−1
b
k
X
(−1)
f (t)dt+
= (−1)n−1 −
(x−a)k+1 −(x−b)k+1 f (k) (x)
(k+1)!
k=0
( aZ
n−1
k+1
b
X
(−1)k X
n−1
f (t)dt +
−
= (−1)
1 + (−1)k−i
(k + 1)! i=0
a
k=0
)
i
1
a
+
b
k+1
(b − a)k+1−i f (k) (x)
· x−
·
i
2
2k+1−i
)
( Z
n−1 X
k+1
b
k
X
(−1)
1+(−1)k−i (b−a)k+1 Ak,i f (k) (x)
f (t)dt +
= (−1)n−1 −
(k + 1)!
a
Zk=0b i=0
1
= (−1)n−1 (b − a) −
f (t)dt
b−a a
)
k+1
n−1 X
X
(−1)k
(b − a)k 1 + (−1)i+1 Ak,k+1−i f (k) (x)
+
(k
+
1)!
k=0 i=0
)
(
Z b
m
n−1 X
k
X
(−1)
1
f (t)dt+2
(b−a)k Ak,k−2j · f (k) (x) .
= (−1)n−1 (b−a) −
b−a a
(k+1)!
k=0 j=0
55
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
We have
Z b
(t−b)n
1
(t−a)n
(x−a)n+1 −(x−b)n+1
dt+
dt =
P (x, t)dt =
n!
n!
(n + 1)!
x
a
a
n+1
n+1
X
1
(b−a)n+1 X
n+1
n−i
(b−a) An,i =
1+(−1)
1+(−1)i+1 An,n+1−i
=
(n+1)! i=0
(n + 1)! i=0
m
˜
(b − a)n+1 X
An,n−2j .
=2
(n + 1)! j=0
Z
b
Z
x
Since
b
Z b
Γ+γ
1
(n)
n−1
P (x, t) f (t) −
f (t)dt
dt = (−1) (b − a) −
2
b−a a
a
)
n−1 X
m
m
˜
n X
X
(b
−
a)
(−1)k
An,n−2j
(b−a)k · Ak,k−2j · f (k) (x)+(−1)n (Γ+γ)
+2
(k+1)!
(n+1)! j=0
k=0 j=0
Z
we have
Z b
n−1 X
m
X
1
(−1)k
k
(k)
f (t)dt
(b − a) · Ak,k−2j · f (x) −
2
(k
+
1)!
b
−
a
a
k=0 j=0
Z b
m
˜
n X
Γ
+
γ
(b
−
a)
1
dt
+(−1)n (Γ + γ)
|P (x, t)| f (n) (t) −
An,n−2j ≤
b−a a
(n + 1)! j=0
2
Z b
(n)
1
Γ
+
γ
|P (x, t)| dt.
max f (t) −
≤
b − a t∈[a,b]
2 a
By using the relation
1
max {(x−a)n−1 , (b−x)n−1 } =
(x−a)n−1 +(b−x)n−1 + (x−a)n−1 −(b−x)n−1
2
( n−1
n−1−i
i
1 X
b−a
a+b
n−1
i
=
x−
1 + (−1)
i
2 i=0
2
2
i )
n−1
n−1−i
X
b
−
a
a
+
b
n−1
1 + (−1)i+1
x−
+
i
2
2
i=0
n−1
)
( n−1
n−1
X
X
(b − a)
i
i+1
=
1 + (−1) An−2,i +
An−2,i
1 + (−1)
2
i=0
p˜
)
( i=0
p
X
X
An−2,2j +
= (b − a)n−1
An−2,2j+1
j=0
j=0
56
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
we obtain
Z x
Z b
Z b
(t − a)n
(b − t)n
|P (x, t)| dt =
dt +
dt
n!
a
x Z n!
Z x a
b
t−a
b−t
dt + (b − x)n−1
dt
≤ (x − a)n−1
n!
a
Zx x n!
Z b
1
n−1
n−1
≤
(b − t)dt
(t − a)dt +
·
max (x − a) , (b − x)
n!
x
a
"
#
a+b 2
x
−
1
1
2
max (x − a)n−1 , (b − x)n−1 · (b − a)2 ·
+
=
n!
4
(b − a)2
)
"
#
(
2
p˜
p
X
X
x − a+b
1
(b − a)n+1
2
·
+
An−2,2j +
=
·
An−2,2j+1 .
2
n!
4
(b − a)
j=0
j=0
Since γ ≤ f (n) (t) ≤ Γ, we obtain
(n)
f (t) − Γ + γ ≤ Γ − γ
2
2
(∀) t ∈ [a, b].
From the above relations we get the desired inequality (3).
From Theorem 3 we obtain a new quadrature formula and we give a estimation of the remainder term.
Theorem 4. If f ∈ C n−1 [a, b], f (n−1) , n > 1 is absolutely continuous and there
exist real numbers γ, Γ such that γ ≤ f (n) (t) ≤ Γ, t ∈ [a, b] then
Z b
n−1 X
m
X
(−1)k
(b − a)k+1 Ak,k−2j f (k) (x)
f (t)dt = 2
(k
+
1)!
a
k=0 j=0
m
˜
(b − a)n+1 X
+ (−1) (Γ + γ)
An,n−2j + Rn [f ]
(n + 1)! j=0
n
and the remainder term Rn [f ] satisfies the estimation
p˜
) "
( p
#
a+b 2
n+1
X
X
x
−
Γ − γ (b − a)
1
2
An−2,2j+1 · +
|Rn [f ]| ≤
An−2,2j +
·
,
2
n!
4
(b − a)2
j=0
j=0
where
Ak,i
1
= k+1−i
2
(b − a)i
k+1
i
a+b
x−
2
57
i
, k = 0, n − 1, i = 0, k + 1,
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
hni
k
n−1
n−2
and m =
, m
˜ =
, p=
, p˜ =
.
2
2
2
2
a+b
, n = 2, we obtain the following quadrature
Remark 1. If we choose x =
2
formula:
Z b
a+b
Γ+γ
f (t)dt = (b − a)f
+
(b − a)3 + R2 [f ]
2
48
a
and the remainder therm R2 [f ] satisfies the estimation
|R2 [f ]| ≤
Γ−γ
(b − a)3 .
32
Remark 2. If in Theorem 4, for n = 2, we choose x = a, x = b, respectively,
and we sum the results, to obtain the following quadrature formula
Z b
b−a
(b−a)2 ′
1
f (t)dt =
[f (a)+f (b)]+
[f (a)−f ′ (b)]+ (Γ + γ)(b − a)3 + R2 [f ]
2
4
12
a
and the remainder therm R2 [f ] satisfies the estimation
1
|R2 [f ]| ≤ (Γ − γ)(b − a)3 .
8
Theorem 5. If f ∈ C 1 [a, b], f ′ is absolutely continuous and there exist real
numbers γ, Γ such that γ ≤ f ′′ (t) ≤ Γ, t ∈ [a, b] then
Z b
(b − a)3
a+b
w(t)f (t)dt =
f (x) − x −
f ′ (x)
6
2
a
2
2 !#
b−a
Γ+γ
a+b
+
+5 x−
+R[f ],
20
2
2
where w(t) = (b−t)(t−a) and the remainder term R[f ] satisfies the estimation
"
# "
#
a+b 2
a+b 2
x− a+b
x−
x−
7
1
Γ−γ
2
2
2
·
+
(b−a)5
+
+
. (3)
|R[f ]| ≤
8
4
(b−a)2
b−a
36
(b−a)2
58
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
Proof. Let P : [a, b]2 → R a mapping given by
3
4
(b − a) (t − a) − (t − a) , t ∈ [a, x)
6 3
12 4
P (x, t) =
(t
−
b)
(t
−
b)
(a − b)
−
, t ∈ [x, b].
6
12
We observe that successive integration by parts yields the relation:
b
x
(t − a)3 (t − a)4 ′′
(b − a)
f (t)dt
P (x, t)f (t)dt =
−
6
12
a
aZ
b
(t − b)3 (t − b)4 ′′
(a − b)
+
f (t)dt
−
6
12
Zx b
1
b−a
2
2
3
3
f (x)
(x−a) +(b − x) + (x−a) +(b−x)
w(t)f (t)dt+ −
=
2
3
a
b−a
1
+
(x − a)3 − (b − x)3 −
(x − a)4 − (b − x)4 f ′ (x)
12
Z b6
(b − a)3
a+b
(b − a)3
f (x) +
x−
f ′ (x).
w(t)f (t)dt −
=
6
6
2
a
Z
′′
Z
We have
Z
b
a
Z x
Z b
(t−a)3 (t−a)4
(t−b)3 (t−b)4
(b−a)
P (x, t)dt =
(a−b)
−
dt+
−
dt
6
12
6
12
a
x
1
b−a
(x − a)4 + (b − x)4 −
(x − a)5 + (b − x)5
=
24
60
(
2
2 )
3
(b − a)
b−a
a+b
.
=
+5 x−
60
2
2
Since
Z
a
b
Z b
Γ+γ
(b−a)3
a+b
′′
P (x, t) f (t)−
w(t)f (t)dt−
dt =
f (x)− x−
f ′ (x)
2
6
2
a
"
2
2 #)
b−a
a+b
Γ+γ
+5 x−
,
+
20
2
2
we have
59
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
Z b
3
a
+
b
(b
−
a)
f (x) − x −
f ′ (x)
w(t)f (t)dt −
6
2
a
"
2
2 #) Z b
′′
Γ
+
γ
b−a
Γ+γ
a+b
dt
+
|P (x, t)| · f (t) −
+5 x−
≤
20
2
2
2
a
Z b
′′
Γ + γ
≤ max f (t) −
|P (x, t)| dt.
·
t∈[a,b]
2 a
Using the relation
1
(x − a)2 + (b − x)2 + (x − a)2 − (b − x)2
max (x − a)2 , (b − x)2 =
2
"
#
a+b 2
x − a+b
x
−
1
2
2
= (b − a)2
+
+
4
(b − a)2
b−a
we obtain
Z
b
|P (x, t)|dt =
a
≤
+
+
≤
+
=
=
Z x
Z b
3
3
4
4
(a−b) (t−b) − (t−b) dt
(b−a) (t−a) − (t−a) dt+
6
12
6
12
a
x
Z
Z
Z x
x
b
(b − t)3
(t − a)4
(t − a)3
(b − a)
dt +
dt +
dt
(b − a)
6
12
6
a
x
a
Z x
Z x
Z b
(b−t)4
(t−a)2
1
2
2
(b−a)(t−a)dt+(x−a)
(x−a)
dt ≤
dt
12
6
2
a
a
x
Z b
Z b
(b − t)2
2
2
(b − a)(b − t)dt + (b − x)
(b − x)
dt
2
x
x
Z x
Z x
(t−a)2
1
2
2
(b−a)(t−a)dt+
max (x−a) , (b−x) ·
dt
6
2
a
a
Z b
Z b
(b − t)2
(b − a)(b − t)dt +
dt
2
x
x
"
#
a+b 2
3
x
−
7
(b
−
a)
2
·
+
max (x − a)2 , (b − x)2 ·
4
36
(b − a)2
"
# "
#
a+b 2
a+b 2
x− a+b
x−
x−
1
1
7
2
2
2
+
(b−a)5 ·
+
+
.
·
4
4
(b−a)2
b−a
36
(b−a)2
60
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
Since γ ≤ f ′′ (t) ≤ Γ we obtain
′′
f (t) − Γ + γ ≤ Γ − γ
2
2
(∀) t ∈ [a, b].
From the above relations we get the relation (3).
a+b
Remark 3. If we choose x =
, we obtain the following quadrature for2
mula:
Z b
Γ+γ
a+b
(b − a)3
+
f
(b − a)2 + R[f ]
w(t)f (t)dt =
6
2
80
a
and the remainder therm R[f ] satisfies the estimation
|R[f ]| ≤
7
(Γ − γ)(b − a)5 .
1152
Remark 4. If in Theorem 5, for n = 2, we choose x = a, x = b, respectively,
and we sum the results, to obtain the following quadrature formula
Z b
b−a ′
3
(b−a)3
′
2
f (a)+f (b)+
(f (a)−f (b))+ (Γ+γ) (b−a) +R[f ]
w(t)f (t)dt =
12
2
20
a
and the remainder therm R[f ] satisfies the estimation
|R[f ]| ≤
1
(Γ − γ)(b − a)5 .
18
References
[1] P. Cerone, S.S. Dragomir, Midpoint type rules from an inequalities point
of view, in Analytic-Compuational Methods in Applied Mathematics, G.A.
Anastassiou (Ed), CRC Press, New York, 2000,135-200.
[2] P. Cerone, S.S. Dragomir, Trapezoidal type rules from an inequalities
point of view, in Analytic-Compuational Methods in Applied Mathematics,
G.A. Anastassiou (Ed), CRC Press, New York, 2000, 65-134.
61
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
[3]S.S. Dragomir, P. Cerone, J. Roumeliotis, A new generalization of Ostrowski ′ s
integral inequality for mappings whose derivatives are bounded and applications
in numerical integration, submitted, Appl. Math. Lett. (20/04/1999).
[4] N.Ujevi´
c, A generalization of Ostrovski ′ s inequality and applications in
numerical integration, Appl. Math. Lett., 17(2),133-137, 2004.
Authors:
Ana Maria Acu
Department of Mathematics
”Lucian Blaga” University
Sibiu, Romania
e-mail:[email protected]
Arif Rafiq
Department of Mathematics
COMSATS Institute of Information Technology
Islamabad, Pakistan
e-mail:[email protected]
Mugur Acu
”Lucian Blaga” University
Department of Mathematics
Sibiu, Romania
e-mail:acu [email protected]
62
No 16/2008
THE ESTIMATES FOR THE REMAINDER TERM OF SOME
QUADRATURE FORMULAE
Ana Maria Acu, Mugur Acu, Arif Rafiq
Abstract. A new generalization of Ostrowski ′ s integral inequality is
established. A consequence of the generalization is that we can derive new
estimates for remainder term of the midpoint, trapezoid and Simpson formulae. These estimates are improvements of some recently obtained estimates.
Applications in numerical integration are also given.
2000 Mathematics Subject Classification: 65D30 , 65D32.
Keywords and phrases: quadrature rule, Ostrowski inequality
1. Introduction
In [2] Cerone, Dragomir and Roumeliotis proved the following generalization of Ostrowski ′ s inequality:
Theorem 1.[2] Let f : [a, b] → R be continous on (a, b) and whose first derivative f ′ : (a, b) → R is bounded on (a,b). Denote kf ′ k∞ = sup |f ′ (t)| < ∞.
t∈[a,b]
Then we have:
Z b
f
(a)
+
f
(b)
f (t)dt − f (x)(1 − λ) +
λ (b − a)
2
"a
2 #
1
a
+
b
(b − a)2 (λ2 + (1 − λ)2 ) + x −
kf ′ k∞
≤
4
2
for all λ ∈ [0, 1] and a + λ
b−a
b−a
≤x≤b−λ
.
2
2
53
(1)
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
Using (1), they also obtained estimates for the remainder term of the midpoint, trapezoid and Simpson ′ s formulae. They also gave applications of the
mentioned rezults in numerical integration and for special means.
In [4], N. Ujevic proved the following generalization of Ostrowski ′ s inequality:
Theorem 2.[4] Let I ⊂ R be a open interval and a, b ∈ I, a < b. If f : I → R
is a differentiable function such that γ ≤ f ′ (t) ≤ Γ, for all t ∈ [a, b], for some
constants Γ, γ ∈ R, then we have
"
#
Z b
a+b 2
1
x−
Γ+γ
1
a
+
b
1
2
f (x)−
f (t)dt ≤ (Γ−γ)(b−a) +
x−
−
.
2
2
b−a a
2
4
(b−a)2
In this paper we give some generalizations of Ostrowski ′ s inequalities and
we obtain new estimations for the remainder term of some quadrature formulae.
2. Main results
We denote [x] the integer part of x, x ∈ R.
Theorem 3. If f ∈ C n−1 [a, b], f (n−1) , n > 1 is absolutely continuous and
there exist real numbers γ, Γ such that γ ≤ f (n) (t) ≤ Γ, t ∈ [a, b], then
Z b
n−1 X
m
X
1
(−1)k
k
(k)
f (t)dt
(b − a) Ak,k−2j f (x) −
2
(k + 1)!
b−a a
k=0 j=0
m
˜
n X
(b
−
a)
+ (−1)n (Γ + γ)
An,n−2j
(n + 1)! j=0
p˜
) "
( p
2 #
X
x− a+b
Γ−γ (b−a)n X
1
2
·
+
An−2,2j +
, (2)
An−2,2j+1 ·
≤
2
2
n!
4
(b−a)
j=0
j=0
where
i
a+b
1
k+1
x−
Ak,i = k+1−i
, k = 0, n − 1, i = 0, k + 1,
i
2
(b − a)i
2
hni
n−1
n−2
k
, m
˜ =
, p=
, p˜ =
.
and m =
2
2
2
2
54
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
Proof. Let P : [a, b]2 → R a mapping given by
(t − a)n
, if t ∈ [a, x)
n!
P (x, t) =
n
(t − b) , if t ∈ [x, b].
n!
We observe that successive integration by parts yields the relation:
x
Z b
Z x
n−1
n (n−1−k)
X
(n)
n−1−k (t−a)
(k)
n
P (x, t)f (t)dt =
(−1)
· f (t) +(−1)
f (t)dt
n!
a
a
k=0
a
b
Z b
n−1
n (n−1−k)
X
(t
−
b)
(k)
n−1−k
n
f (t)dt
· f (t) + (−1)
+
(−1)
n!
x
k=0
x
Z b
n−1
n−1
k+1
X
X
(x−b)k+1 (k)
(k)
n−1−k (x−a)
n
f (x)− (−1)n−1−k
f (x)
f (t)dt+ (−1)
= (−1)
(k+1)!
(k+1)!
a
k=0
k=0
( Z
)
n−1
b
k
X
(−1)
f (t)dt+
= (−1)n−1 −
(x−a)k+1 −(x−b)k+1 f (k) (x)
(k+1)!
k=0
( aZ
n−1
k+1
b
X
(−1)k X
n−1
f (t)dt +
−
= (−1)
1 + (−1)k−i
(k + 1)! i=0
a
k=0
)
i
1
a
+
b
k+1
(b − a)k+1−i f (k) (x)
· x−
·
i
2
2k+1−i
)
( Z
n−1 X
k+1
b
k
X
(−1)
1+(−1)k−i (b−a)k+1 Ak,i f (k) (x)
f (t)dt +
= (−1)n−1 −
(k + 1)!
a
Zk=0b i=0
1
= (−1)n−1 (b − a) −
f (t)dt
b−a a
)
k+1
n−1 X
X
(−1)k
(b − a)k 1 + (−1)i+1 Ak,k+1−i f (k) (x)
+
(k
+
1)!
k=0 i=0
)
(
Z b
m
n−1 X
k
X
(−1)
1
f (t)dt+2
(b−a)k Ak,k−2j · f (k) (x) .
= (−1)n−1 (b−a) −
b−a a
(k+1)!
k=0 j=0
55
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
We have
Z b
(t−b)n
1
(t−a)n
(x−a)n+1 −(x−b)n+1
dt+
dt =
P (x, t)dt =
n!
n!
(n + 1)!
x
a
a
n+1
n+1
X
1
(b−a)n+1 X
n+1
n−i
(b−a) An,i =
1+(−1)
1+(−1)i+1 An,n+1−i
=
(n+1)! i=0
(n + 1)! i=0
m
˜
(b − a)n+1 X
An,n−2j .
=2
(n + 1)! j=0
Z
b
Z
x
Since
b
Z b
Γ+γ
1
(n)
n−1
P (x, t) f (t) −
f (t)dt
dt = (−1) (b − a) −
2
b−a a
a
)
n−1 X
m
m
˜
n X
X
(b
−
a)
(−1)k
An,n−2j
(b−a)k · Ak,k−2j · f (k) (x)+(−1)n (Γ+γ)
+2
(k+1)!
(n+1)! j=0
k=0 j=0
Z
we have
Z b
n−1 X
m
X
1
(−1)k
k
(k)
f (t)dt
(b − a) · Ak,k−2j · f (x) −
2
(k
+
1)!
b
−
a
a
k=0 j=0
Z b
m
˜
n X
Γ
+
γ
(b
−
a)
1
dt
+(−1)n (Γ + γ)
|P (x, t)| f (n) (t) −
An,n−2j ≤
b−a a
(n + 1)! j=0
2
Z b
(n)
1
Γ
+
γ
|P (x, t)| dt.
max f (t) −
≤
b − a t∈[a,b]
2 a
By using the relation
1
max {(x−a)n−1 , (b−x)n−1 } =
(x−a)n−1 +(b−x)n−1 + (x−a)n−1 −(b−x)n−1
2
( n−1
n−1−i
i
1 X
b−a
a+b
n−1
i
=
x−
1 + (−1)
i
2 i=0
2
2
i )
n−1
n−1−i
X
b
−
a
a
+
b
n−1
1 + (−1)i+1
x−
+
i
2
2
i=0
n−1
)
( n−1
n−1
X
X
(b − a)
i
i+1
=
1 + (−1) An−2,i +
An−2,i
1 + (−1)
2
i=0
p˜
)
( i=0
p
X
X
An−2,2j +
= (b − a)n−1
An−2,2j+1
j=0
j=0
56
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
we obtain
Z x
Z b
Z b
(t − a)n
(b − t)n
|P (x, t)| dt =
dt +
dt
n!
a
x Z n!
Z x a
b
t−a
b−t
dt + (b − x)n−1
dt
≤ (x − a)n−1
n!
a
Zx x n!
Z b
1
n−1
n−1
≤
(b − t)dt
(t − a)dt +
·
max (x − a) , (b − x)
n!
x
a
"
#
a+b 2
x
−
1
1
2
max (x − a)n−1 , (b − x)n−1 · (b − a)2 ·
+
=
n!
4
(b − a)2
)
"
#
(
2
p˜
p
X
X
x − a+b
1
(b − a)n+1
2
·
+
An−2,2j +
=
·
An−2,2j+1 .
2
n!
4
(b − a)
j=0
j=0
Since γ ≤ f (n) (t) ≤ Γ, we obtain
(n)
f (t) − Γ + γ ≤ Γ − γ
2
2
(∀) t ∈ [a, b].
From the above relations we get the desired inequality (3).
From Theorem 3 we obtain a new quadrature formula and we give a estimation of the remainder term.
Theorem 4. If f ∈ C n−1 [a, b], f (n−1) , n > 1 is absolutely continuous and there
exist real numbers γ, Γ such that γ ≤ f (n) (t) ≤ Γ, t ∈ [a, b] then
Z b
n−1 X
m
X
(−1)k
(b − a)k+1 Ak,k−2j f (k) (x)
f (t)dt = 2
(k
+
1)!
a
k=0 j=0
m
˜
(b − a)n+1 X
+ (−1) (Γ + γ)
An,n−2j + Rn [f ]
(n + 1)! j=0
n
and the remainder term Rn [f ] satisfies the estimation
p˜
) "
( p
#
a+b 2
n+1
X
X
x
−
Γ − γ (b − a)
1
2
An−2,2j+1 · +
|Rn [f ]| ≤
An−2,2j +
·
,
2
n!
4
(b − a)2
j=0
j=0
where
Ak,i
1
= k+1−i
2
(b − a)i
k+1
i
a+b
x−
2
57
i
, k = 0, n − 1, i = 0, k + 1,
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
hni
k
n−1
n−2
and m =
, m
˜ =
, p=
, p˜ =
.
2
2
2
2
a+b
, n = 2, we obtain the following quadrature
Remark 1. If we choose x =
2
formula:
Z b
a+b
Γ+γ
f (t)dt = (b − a)f
+
(b − a)3 + R2 [f ]
2
48
a
and the remainder therm R2 [f ] satisfies the estimation
|R2 [f ]| ≤
Γ−γ
(b − a)3 .
32
Remark 2. If in Theorem 4, for n = 2, we choose x = a, x = b, respectively,
and we sum the results, to obtain the following quadrature formula
Z b
b−a
(b−a)2 ′
1
f (t)dt =
[f (a)+f (b)]+
[f (a)−f ′ (b)]+ (Γ + γ)(b − a)3 + R2 [f ]
2
4
12
a
and the remainder therm R2 [f ] satisfies the estimation
1
|R2 [f ]| ≤ (Γ − γ)(b − a)3 .
8
Theorem 5. If f ∈ C 1 [a, b], f ′ is absolutely continuous and there exist real
numbers γ, Γ such that γ ≤ f ′′ (t) ≤ Γ, t ∈ [a, b] then
Z b
(b − a)3
a+b
w(t)f (t)dt =
f (x) − x −
f ′ (x)
6
2
a
2
2 !#
b−a
Γ+γ
a+b
+
+5 x−
+R[f ],
20
2
2
where w(t) = (b−t)(t−a) and the remainder term R[f ] satisfies the estimation
"
# "
#
a+b 2
a+b 2
x− a+b
x−
x−
7
1
Γ−γ
2
2
2
·
+
(b−a)5
+
+
. (3)
|R[f ]| ≤
8
4
(b−a)2
b−a
36
(b−a)2
58
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
Proof. Let P : [a, b]2 → R a mapping given by
3
4
(b − a) (t − a) − (t − a) , t ∈ [a, x)
6 3
12 4
P (x, t) =
(t
−
b)
(t
−
b)
(a − b)
−
, t ∈ [x, b].
6
12
We observe that successive integration by parts yields the relation:
b
x
(t − a)3 (t − a)4 ′′
(b − a)
f (t)dt
P (x, t)f (t)dt =
−
6
12
a
aZ
b
(t − b)3 (t − b)4 ′′
(a − b)
+
f (t)dt
−
6
12
Zx b
1
b−a
2
2
3
3
f (x)
(x−a) +(b − x) + (x−a) +(b−x)
w(t)f (t)dt+ −
=
2
3
a
b−a
1
+
(x − a)3 − (b − x)3 −
(x − a)4 − (b − x)4 f ′ (x)
12
Z b6
(b − a)3
a+b
(b − a)3
f (x) +
x−
f ′ (x).
w(t)f (t)dt −
=
6
6
2
a
Z
′′
Z
We have
Z
b
a
Z x
Z b
(t−a)3 (t−a)4
(t−b)3 (t−b)4
(b−a)
P (x, t)dt =
(a−b)
−
dt+
−
dt
6
12
6
12
a
x
1
b−a
(x − a)4 + (b − x)4 −
(x − a)5 + (b − x)5
=
24
60
(
2
2 )
3
(b − a)
b−a
a+b
.
=
+5 x−
60
2
2
Since
Z
a
b
Z b
Γ+γ
(b−a)3
a+b
′′
P (x, t) f (t)−
w(t)f (t)dt−
dt =
f (x)− x−
f ′ (x)
2
6
2
a
"
2
2 #)
b−a
a+b
Γ+γ
+5 x−
,
+
20
2
2
we have
59
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
Z b
3
a
+
b
(b
−
a)
f (x) − x −
f ′ (x)
w(t)f (t)dt −
6
2
a
"
2
2 #) Z b
′′
Γ
+
γ
b−a
Γ+γ
a+b
dt
+
|P (x, t)| · f (t) −
+5 x−
≤
20
2
2
2
a
Z b
′′
Γ + γ
≤ max f (t) −
|P (x, t)| dt.
·
t∈[a,b]
2 a
Using the relation
1
(x − a)2 + (b − x)2 + (x − a)2 − (b − x)2
max (x − a)2 , (b − x)2 =
2
"
#
a+b 2
x − a+b
x
−
1
2
2
= (b − a)2
+
+
4
(b − a)2
b−a
we obtain
Z
b
|P (x, t)|dt =
a
≤
+
+
≤
+
=
=
Z x
Z b
3
3
4
4
(a−b) (t−b) − (t−b) dt
(b−a) (t−a) − (t−a) dt+
6
12
6
12
a
x
Z
Z
Z x
x
b
(b − t)3
(t − a)4
(t − a)3
(b − a)
dt +
dt +
dt
(b − a)
6
12
6
a
x
a
Z x
Z x
Z b
(b−t)4
(t−a)2
1
2
2
(b−a)(t−a)dt+(x−a)
(x−a)
dt ≤
dt
12
6
2
a
a
x
Z b
Z b
(b − t)2
2
2
(b − a)(b − t)dt + (b − x)
(b − x)
dt
2
x
x
Z x
Z x
(t−a)2
1
2
2
(b−a)(t−a)dt+
max (x−a) , (b−x) ·
dt
6
2
a
a
Z b
Z b
(b − t)2
(b − a)(b − t)dt +
dt
2
x
x
"
#
a+b 2
3
x
−
7
(b
−
a)
2
·
+
max (x − a)2 , (b − x)2 ·
4
36
(b − a)2
"
# "
#
a+b 2
a+b 2
x− a+b
x−
x−
1
1
7
2
2
2
+
(b−a)5 ·
+
+
.
·
4
4
(b−a)2
b−a
36
(b−a)2
60
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
Since γ ≤ f ′′ (t) ≤ Γ we obtain
′′
f (t) − Γ + γ ≤ Γ − γ
2
2
(∀) t ∈ [a, b].
From the above relations we get the relation (3).
a+b
Remark 3. If we choose x =
, we obtain the following quadrature for2
mula:
Z b
Γ+γ
a+b
(b − a)3
+
f
(b − a)2 + R[f ]
w(t)f (t)dt =
6
2
80
a
and the remainder therm R[f ] satisfies the estimation
|R[f ]| ≤
7
(Γ − γ)(b − a)5 .
1152
Remark 4. If in Theorem 5, for n = 2, we choose x = a, x = b, respectively,
and we sum the results, to obtain the following quadrature formula
Z b
b−a ′
3
(b−a)3
′
2
f (a)+f (b)+
(f (a)−f (b))+ (Γ+γ) (b−a) +R[f ]
w(t)f (t)dt =
12
2
20
a
and the remainder therm R[f ] satisfies the estimation
|R[f ]| ≤
1
(Γ − γ)(b − a)5 .
18
References
[1] P. Cerone, S.S. Dragomir, Midpoint type rules from an inequalities point
of view, in Analytic-Compuational Methods in Applied Mathematics, G.A.
Anastassiou (Ed), CRC Press, New York, 2000,135-200.
[2] P. Cerone, S.S. Dragomir, Trapezoidal type rules from an inequalities
point of view, in Analytic-Compuational Methods in Applied Mathematics,
G.A. Anastassiou (Ed), CRC Press, New York, 2000, 65-134.
61
A.M. Acu, M.Acu, A. Rafiq - The estimates for the remainder term...
[3]S.S. Dragomir, P. Cerone, J. Roumeliotis, A new generalization of Ostrowski ′ s
integral inequality for mappings whose derivatives are bounded and applications
in numerical integration, submitted, Appl. Math. Lett. (20/04/1999).
[4] N.Ujevi´
c, A generalization of Ostrovski ′ s inequality and applications in
numerical integration, Appl. Math. Lett., 17(2),133-137, 2004.
Authors:
Ana Maria Acu
Department of Mathematics
”Lucian Blaga” University
Sibiu, Romania
e-mail:[email protected]
Arif Rafiq
Department of Mathematics
COMSATS Institute of Information Technology
Islamabad, Pakistan
e-mail:[email protected]
Mugur Acu
”Lucian Blaga” University
Department of Mathematics
Sibiu, Romania
e-mail:acu [email protected]
62