Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 9 (2016), 537–552
Research Article
Companion of Ostrowski-type inequality based on
5-step quadratic kernel and applications
Ather Qayyuma,∗, Muhammad Shoaibb , Ibrahima Fayea
a
Department of Fundamental and Applied Sciences, Universiti Teknologi PETRONAS, 32610 Bandar Seri Iskandar, Perak Darul
Ridzuan, Malaysia.
b
Abu Dhabi Mens College, Higher Colleges of Technology, P. O. Box 25035, Abu Dhabi, United Arab Emirates.
Communicated by Nawab Hussain
Abstract
The purpose of this paper is to establish an improved version of companion of Ostrowski’s type integral
inequalities. The inequalities are obtained by using a newly developed special type of five steps quadratic
kernel. The introduction of this new Kernel gives some new error bounds for various quadrature rules.
Applications for composite quadrature rules and Cumulative Distributive Functions are considered. c 2016
All rights reserved.
Keywords: Ostrowski inequality, numerical integration, composite quadrature rule, cumulative
distributive function.
2010 MSC: 26D15, 26D20, 26D99.
1. Introduction
The field of inequalities have applications in most of the domains of Mathematics. Their importance
have increased during the past few decades and is now studied as a separate branch of Mathematics. A
number of research papers and books have been written on inequalities and their applications (see for
instance [5], [6], [8]-[20] and [14]-[19]). In many practical problems, it is important to bound one quantity
by another quantity. The classical inequalities such as Ostrowski are very useful for this purpose. Ostrowski
type inequalities have immediate applications in numerical integration, optimization theory, statistics, and
integral operator theory.
In 1938, Ostrowski [13] discovered the following useful integral inequality.
∗
Corresponding author
Email addresses: atherqayyum@gmail.com (Ather Qayyum), safridi@gmail.com (Muhammad Shoaib),
ibrahima faye@petronas.com.my (Ibrahima Faye)
Received 2015-05-10
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
538
Theorem 1.1. Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) , whose derivative
f 0 : (a, b) → R is bounded on (a, b) , i.e.
f0
∞
= sup f 0 (t) < ∞
t∈[a,b]
then for all x ∈ [a, b]
1
f (x) −
b−a
Zb

1
f (t)dt ≤  +
4
x − a+b
2
b−a
!2 
 (b − a) f 0
∞
.
(1.1)
a
We mention another inequality called Grüss inequality [12] which is stated as the integral inequality
that establishes a connection between the integral of the product of two functions and the product of the
integrals, which is given below.
1
b−a
Zb
1
f (x)g(x)dx −
b−a
a
Zb
1
f (x)dx.
b−a
a
Zb
1
g(x)dx ≤ (Φ − ϕ)(Γ − γ),
4
(1.2)
a
where ϕ ≤ f (x) ≤ Φ and γ ≤ g (x) ≤ Γ, for all x ∈ [a, b] . The constant 14 is sharp in (1.2).
In [5], Dragomir and Wang combined Ostrowski and Grüss inequality to give a new inequality which
they named Ostrowski-Grüss type inequalities.
In [6], Guessab and Schmeisser proved the following Ostrowski’s inequality:
Theorem
1.2. Let f : [a, b] → R satisfy the Lipschitz condition i.e., |f (t) − f (s)| ≤ M |t − s| . Then for all
x ∈ a, a+b
2 , we have

!2 
Zb
x − 3a+b
1
f (x) + f (a + b − x)
1
4
 (b − a) M.
−
f (t)dt ≤  + 2
(1.3)
2
b−a
8
b−a
a
In (1.3), the point x =
3a+b
4
yields the following trapezoid type inequality.
f
3a+b
4
+f
2
a+3b
4
1
−
b−a
Zb
f (t)dt ≤
b−a
M.
8
(1.4)
a
The constant
1
8
is sharp in (1.4).
In [3], Barnett et al. proved some Ostrowski and generalized trapezoid inequalities. Dragomir [4] and
Liu [8] established some companions of Ostrowski type integral inequalities. Alomari [1] proved the following
inequality:
Let f : [a, b] → R be a differentiable mapping on (a, b). If f 0 ∈ L1 [a, b] and γ ≤ f 0 (t) ≤ Γ, for all
t ∈ [a, b] , then
Zb
f (x) + f (a + b − x)
1
1
−
f (t) dt ≤ (b − a) (Γ − γ) .
(1.5)
2
b−a
8
a
Recently, Liu [9] and Liu et al. [10] proved some Ostrowski type inequalities. In all references mentioned
above, authors proved their results by using kernels with two or three steps.
2. Main Results
Before we prove our results for the 5-step quadratic kernel, we give the following lemma for 5-step linear
kernel.
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
539
Lemma 2.1. Consider the kernel

t − a,



3a+b

 t− 4 ,
t − a+b
K(x, t) =
2 ,

a+3b


t
−

4 ,

t − b,
t ∈ a, a+x
2 ,
t ∈ a+x
2 ,x ,
t ∈ (x, a + b − x] ,
,
t ∈ a + b − x, a+2b−x
2
a+2b−x
t∈
,b
2
(2.1)
, then the following identity holds.
for all x ∈ a, a+b
2
1
b−a
Zb
K(x, t)f 0 (t) dt
a
(2.2)
Zb
a+x
a + 2b − x
1
1
f (x) + f (a + b − x) + f
+f
−
f (t) dt.
=
4
2
2
b−a
a
Proof. From (2.1), we have
Zb
a+x
Z2
0
Zx 0
(t − a) f (t) dt +
K(x, t)f (t) dt =
a
a
3a + b
t−
4
a+b−x
Z 0
Z
+
a+b
2
f 0 (t) dt
x
a+x
2
a+2b−x
2
t−
f (t) dt +
a + 3b
t−
f 0 (t) dt +
4
Zb
(t − b) f 0 (t) dt
a+2b−x
2
a+b−x
1
a+x
3a + b
1
a+b
a+x
= (x − a) f
+ x−
f (x) −
x−
f
2
2
4
2
2
2
a+b
a+b
f (a + b − x) − x −
f (x)
− x−
2
2
1
a+b
a + 2b − x
3a + b
−
x−
f
+ x−
f (a + b − x)
2
2
2
4
Zb
1
a + 2b − x
+ (x − a) f
− f (t) dt.
2
2
a
Hence after simplification, we get (2.2).
Now with the help of above identity, we state and prove the following theorem.
Theorem 2.2. Let f : [a, b] → R be differentiable on (a, b). If f
t ∈ [a, b] , then the inequality
0
∈ L1 [a, b] and γ ≤ f 0 (t) ≤ Γ, for all
Zb
1
1
a+x
a + 2b − x
1
f (x) + f (a + b − x) + f
+f
−
f (t) dt ≤
(b − a) (Γ − γ) (2.3)
4
2
2
b−a
16
a
holds for all x ∈ a,
a+b
2
.
Proof. As we know that for all t ∈ [a, b] and x ∈ a, a+b
2 , we have
x−
3a + b
≤ K(x, t) ≤ x − a.
4
Applying Grüss inequality to the mappings K(x, .) and f 0 (.), for all x ∈ a, a+b
, we obtain
2
(2.4)
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
1
b−a
Zb
a
1
K(x, t)f (t)dt −
(b − a)2
0
We also have
1
b−a
Zb
Zb
K(x, t)dt
a
f 0 (t)dt ≤
540
1
(b − a) (Γ − γ) .
16
(2.5)
a
Zb
K(x, t)dt = 0
(2.6)
a
and
1
b−a
Zb
f 0 (t)dt =
f (b) − f (a)
.
b−a
(2.7)
a
Hence from (2.4)-(2.7), we get our required result (2.3).
We now introduce the quadratic version of 5-step kernel that further generalize (2.3) and various previous
results contained in [1], [2], [6], and [10]. With the help of this special 5-step quadratic kernel, a number
of different types of useful and interesting results are obtained. Moreover, we describe new results by using
Grüss inequality, Cauchy inequality and Diaz-Metcalf inequality. At the end, some obtained inequalities are
applied for quadrature rules and cumulative distributive function.
Now we start our main result for the 5-step quadratic kernel. Firstly, we need to state the following
lemma.
Lemma 2.3. Let f : [a, b] → R be such that f 0 is absolutely continuous on [a, b] . Define the kernel P (x, t)
as:
 1
2
a+x
(t
−
a)
,
t
∈
a,

2
2 ,



1
3a+b 2
a+x

 2 t − 4 , t ∈ 2 ,x ,
a+b 2
1
P (x, t) =
(2.8)
t ∈ (x, a + b − x] ,
2 t− 2 ,

2

1
a+3b
a+2b−x

 2 t− 4
, t ∈ a + b − x, 2
,

 1
2
a+2b−x
(t
−
b)
,
t
∈
,
b
2
2
a+b for all x ∈ a, 2 . Then the following identity
1
b−a
Zb
a
Zb
1
1
a+x
a + 2b − x
P (x, t)f dt =
f (t) dt −
f (x) + f (a + b − x) + f
+f
b−a
4
2
2
a
5a + 3b 0
f (a + b − x) − f 0 (x)
(2.9)
+ x−
8
1
3a + b
0 a + 2b − x
0 a+x
+
x−
f
−f
2
4
2
2
00
holds.
Now we present our results by imposing three different types of conditions on f
2.1. f
00
00
and f
000 .
∈ L1 [a, b] :
0
Theorem 2.4. Let f : [a, b] → R be differentiable
a+b on (a, b), f is absolutely continuous on [a, b] and
00
γ ≤ f (t) ≤ Γ, ∀ t ∈ [a, b] , then for all x ∈ a, 2 , we have
1
a+x
a + 2b − x
f (x) + f (a + b − x) + f
+f
(2.10)
4
2
2
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
541
5a + 3b 0
3a + b
1
0
+ x−
x−
f (a + b − x) − f (x) +
8
2
4
0 (b) − f 0 (a)
a
+
x
a
+
2b
−
x
f
−f 0
+
× f0
2
2
(b − a)2
(
3
3 )
Zb
1
3a
+
b
3
a
+
b
1
1
3
×
x−
−
x−
−
(x − a) +
f (t) dt
24
3
4
8
2
b−a
a
≤υ (x) (b − a) (S − γ)
and
a+x
a + 2b − x
1
f (x) + f (a + b − x) + f
+f
4
2
2
5a + 3b 0
1
3a + b
0
+ x−
f (a + b − x) − f (x) +
x−
8
2
4
0
0
a + 2b − x
a+x
f (b) − f (a)
× f0
−f 0
+
2
2
(b − a)2
(
)
Zb
1
1
3a + b 3 3
a+b 3
1
3
(x − a) +
x−
−
x−
−
f (t) dt
×
24
3
4
8
2
b−a
(2.11)
a
≤υ (x) (b − a) (Γ − S) ,
where
S=
f 0 (b) − f 0 (a)
b−a
and
υ (x) =
1
max −a2 − 13ab + 15ax − 4b2 + 21bx − 18x2 ,
96
14a2 + 5ab − 33ax − b2 − 3bx + 18x2 , −a2 + 11ab − 9ax + 8b2 − 27bx + 18x2 ,
−10a2 − 7ab − b2 + 27ax + 9bx − 18x2 , 13a2 + 13a(b − 3x) + 4b2 − 21bx + 30x2
.
Proof. The use of
1
b−a
Zb
f 00 (t)dt =
f 0 (b) − f 0 (a)
,
b−a
(2.12)
a
1
b−a
Zb
"
1
1
1
P (x, t)dt =
(x − a)3 +
b − a 24
3
#
3a + b 3 3
a+b 3
x−
−
x−
4
8
2
(2.13)
a
and (2.8) implies that
1
b−a
Zb
a
1
P (x, t)f (t)dt −
(b − a)2
00
Zb
=
Zb
Zb
P (x, t)dt
a
f 00 (t)dt
a
1
1
a+x
a + 2b − x
f (t) dt −
f (x) + f (a + b − x) + f
+f
b−a
4
2
2
a
5a + 3b 0
1
3a + b
0
+ x−
f (a + b − x) − f (x) +
x−
8
2
4
(2.14)
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
542
a+x
f 0 (b) − f 0 (a)
× f
−f
−
2
(b − a)2
(
3
)
3a + b
3
a+b 3
1
1
3
x−
−
x−
.
×
(x − a) +
24
3
4
8
2
0
a + 2b − x
2
0
We suppose that
1
Rn (x) =
b−a
Zb
a
1
P (x, t)f 00 (t)dt −
(b − a)2
Zb
Zb
P (x, t)dt.
a
f 00 (t)dt.
(2.15)
a
If C ∈ R is an arbitrary constant, then we have
Rn (x) =
1
b−a
Zb

f 00 (t) − C P (x, t) −
1
b−a
Zb
a

P (x, s)ds dt.
(2.16)
a
Furthermore, we have
1
1
max P (x, t) −
|Rn (x)| ≤
b − a t∈[a,b]
b−a
Zb
Zb
P (x, s)ds
a
f 00 (t) − C dt.
(2.17)
a
Now
Zb
1
P (x, s)ds
max P (x, t) −
b−a
a
( λ(x)
1
3a + b 2
λ(x)
1 x−a 2
= max
−
,
x−
−
,
2
2
b−a 2
4
b−a
)
1
a+b 2
λ(x)
1
a+b 2
λ(x) λ(x)
x−
−
,
x−
−
,
,
2
2
b−a 8
2
b−a b−a
where
λ(x) =
1
1
(x − a)3 +
24
3
x−
3a + b
4
3
−
3
8
x−
a+b
2
(2.18)
3
and
υ (x) =
1
(2.19)
max −a2 − 13ab + 15ax − 4b2 + 21bx − 18x2 ,
96
14a2 + 5ab − 33ax − b2 − 3bx + 18x2 , −a2 + 11ab − 9ax + 8b2 − 27bx + 18x2 ,
−10a2 − 7ab − b2 + 27ax + 9bx − 18x2 , 13a2 + 13a(b − 3x) + 4b2 − 21bx + 30x2
We also have
Zb
.
f 00 (t) − γ dt = (S − γ) (b − a)
(2.20)
f 00 (t) − Γ dt = (Γ − S) (b − a) .
(2.21)
a
and
Zb
a
Therefore, we obtain (2.10) and (2.11) by using (2.12) to (2.21) and choosing C = γ and C = Γ in (2.17),
respectively.
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
543
Corollary 2.5. Substitution of x = a in (2.10) and (2.11) gives
f (a) + f (b)
f 0 (b) − f 0 (a)
1
− (b − a)
−
2
12
b−a
Zb
f (t) dt ≤
1
(b − a)3 (S − γ) ,
12
(2.22)
f (t) dt ≤
1
(b − a)3 (Γ − S) .
12
(2.23)
a
f (a) + f (b)
f 0 (b) − f 0 (a)
1
− (b − a)
−
2
12
b−a
Zb
a
Corollary 2.6. Substitution of x = a+b
2 , in (2.10) and (2.11) gives
1
3a + b
a + 3b
1
1
a+b
0 3a + b
0 a + 3b
+
f
+f
+
−f
f
(b − a) f
2
2
4
4
4
32
4
4
b
Z
0
1
1
1
0
+ (b − a) f (b) − f (a) −
f (t) dt ≤
(b − a)3 (S − γ) ,
(2.24)
96
b−a
48
a
3a + b
a + 3b
f
+f
+
4
4
Zb
0
1
1
0
f (t) dt ≤
+ (b − a) f (b) − f (a) −
96
b−a
1
f
2
a+b
2
1
+
4
1
0 a + 3b
0 3a + b
(b − a) f
−f
32
4
4
1
(b − a)3 (Γ − S) .
48
(2.25)
a
Corollary 2.7. Substitution of x = 3a+b
4 , in (2.10) and (2.11) gives
3a + b
a + 3b
7a + b
a + 7b
1
f
+f
+f
+f
4
4
4
8
8
1
a
+
3b
3a
+
b
5
−
(b − a) f 0
−f 0
−
(b − a)
32
4
4
768
Zb
0
1
19
× f (b) − f 0 (a) −
f (t) dt ≤
(b − a)3 (S − γ) ,
b−a
768
(2.26)
a
3a + b
a + 3b
7a + b
a + 7b
f
+f
+f
+f
4
4
8
8
1
a + 3b
3a + b
5
−
(b − a) f 0
−f 0
−
(b − a)
32
4
4
768
Zb
0
19
1
0
× f (b) − f (a) −
f (t) dt ≤
(b − a)3 (Γ − S) .
b−a
768
1
4
(2.27)
a
2.2. f
000
∈ L2 [a, b]
Theorem
2.8. Let f : [a, b] → R be three times differentiable function on (a, b) . If f
all x ∈ a, a+b
, we have
2
1
a+x
a + 2b − x
f (x) + f (a + b − x) + f
+f
4
2
2
000
∈ L2 [a, b] , then for
(2.28)
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
544
5a + 3b 0
3a + b
1
0
+ x−
x−
f (a + b − x) − f (x) +
8
2
4
0 (b) − f 0 (a)
a
+
x
a
+
2b
−
x
f
−f 0
+
× f0
2
2
(b − a)2
(
3
3 )
Zb
1
3a
+
b
3
a
+
b
1
1
3
×
x−
−
x−
−
(x − a) +
f (t) dt
24
3
4
8
2
b−a
a
"
5
5
1
3a + b
a+b
1
33
1
f 000 2
x−
x−
(x − a)5 +
−
≤
π
320
10
4
320
2
1
(
3
3 )2 2
1
1
3a + b
3
a+b
1
 .
x−
−
x−
−
(x − a)3 +
b − a 24
3
4
8
2
(2.29)
Proof. Let Rn (x) be defined by (2.15). From (2.14), we get
Zb
1
a+x
a + 2b − x
1
f (t) dt −
f (x) + f (a + b − x) + f
+f
Rn (x) =
b−a
4
2
2
a
5a + 3b 0
1
3a + b
+ x−
f (a + b − x) − f 0 (x) +
x−
8
2
4
0
0
a + 2b − x
a+x
f (b) − f (a)
× f0
−f 0
−
2
2
(b − a)2
(
3
)
1
1
3a + b
3
a+b 3
3
×
(x − a) +
x−
−
x−
.
24
3
4
8
2
If we choose C = f
00
1
|Rn (x)| ≤
b−a
a+b
2
Zb
(2.30)
in (2.16) and use the Cauchy Inequality, then we get
00
f (t) − f
00
a+b
2
1
P (x, t) −
b−a
a
Zb
P (x, s)ds dt
(2.31)
a
 b
 21  b 
2  12
2
Z Z
Zb
1 
a+b
1


≤
f 00 (t) − f 00
dt ×  P (x, t) −
P (x, s)ds dt .
b−a
2
b−a
a
a
a
We use the Diaz-Metcalf inequality from [11] or [20] to get
2
Zb (b − a)2
00
00 a + b
dt ≤
f
f (t) − f
2
π2
000 2
2
a
and
Zb

P (x, t) −
a
1
b−a
2
Zb
P (x, s)ds dt
(2.32)
a
Zb
=
a
1
P (x, t)2 dt −
b−a
(
1
1
(x − a)3 +
24
3
)2
3a + b 3 3
a+b 3
x−
−
x−
4
8
2
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
545
1
3a + b 5
a+b 5
1
33
5
=
x−
x−
(x − a) +
−
320
10
4
320
2
(
3
3 )2
1
3a
+
b
3
a
+
b
1
1
x−
−
x−
(x − a)3 +
.
−
b − a 24
3
4
8
2
Therefore, using the above relations (2.30)-(2.32), we obtain (2.28).
Corollary 2.9. Substitution of x = a in (2.28) gives
f (a) + f (b)
f 0 (b) − f 0 (a)
1
− (b − a)
−
2
12
b−a
Zb
f (t) dt ≤
kf
000 k
2
π
1
(b − a)
64
5
2
r
89
.
45
a
Corollary 2.10. Substitution of x = a+b
2 in (2.28) gives
3a + b
1
3a + b
a + 3b
1
a + 3b
1
a+b
+
f
+f
+
−f 0
f
(b − a) f 0
2
2
4
4
4
32
4
4
Zb
0
5
kf 000 k2
1
1
1
0
− (b − a) f (b) − f (a) −
f (t) dt ≤
(b − a) 2 √
.
(2.33)
96
b−a
π
16645
a
Corollary 2.11. Substitution of x = 3a+b
in (2.28) gives
4
1
3a + b
a + 3b
7a + b
a + 7b
b−a
0 a + 3b
0 3a + b
f
+f
+f
+f
−
f
−f
4
4
4
8
8
32
4
4
√
Zb
5
kf 000 k2
5
1
97
−
(b − a) f 0 (b) − f 0 (a) −
(b − a) 2
.
(2.34)
f (t) dt ≤
768
b−a
π
768
a
2.3. f 00 ∈ L2 [a, b]
Theorem 2.12. Let f : [a, b] → R be an absolutely continuous function on (a, b) with f 00 ∈ L2 [a, b]. Then,
we have
1
a+x
a + 2b − x
f (x) + f (a + b − x) + f
+f
(2.35)
4
2
2
1
3a + b
5a + 3b 0
0
+ x−
f (a + b − x) − f (x) +
x−
8
2
4
0
0
a + 2b − x
a+x
f (b) − f (a)
× f0
−f 0
+
2
2
(b − a)2
(
3
3 )
Zb
1
1
3a
+
b
3
a
+
b
1
3
×
(x − a) +
x−
−
x−
−
f (t) dt
24
3
4
8
2
b−a
a
"
p
5
5
00
σ (f ) 1
1
3a + b
33
a+b
≤
(x − a)5 +
x−
−
x−
b−a
320
10
320
2
1
(
3
3 )2 2
1
1
1
3a + b
3
a+b
 .
−
(x − a)3 +
x−
−
x−
b − a 24
3
4
8
2
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
546
for all x ∈ a, a+b
2 , where
σ f
00
= f
00 2
2
−
(f 0 (b) − f 0 (a))2
= f
b−a
00 2
2
− S 2 (b − a) ,
(2.36)
and S is defined in Theorem 2.4.
Proof. Let Rn (x) is defined as in (2.15). If we choose C =
1
b−a
Rb
f 00 (s) ds in (2.16) and use the Cauchy
a
inequality and (2.32), then we get
1
|Rn (x)| ≤
b−a
Zb
1
f (t) −
b−a
00
a
1
≤
b−a
Zb
1
f (s) ds P (x, t) −
b−a
00
a
 
Zb
  00
f (t) −

a
Zb
P (x, s)ds dt
a
1
1
b−a
 
2
2
Zb
Zb
 

00

f (s) ds dt × 
P (x, t) −
1
b−a
a
a
2
Zb
1
2

P (x, s)ds dt
a
"
p
σ (f 00 ) 1
3a + b 5
a+b 5
1
33
5
x−
x−
=
(x − a) +
−
b−a
320
10
4
320
2
1
(
3
3 )2 2
1
1
1
3a + b
3
a+b
 .
−
(x − a)3 +
x−
−
x−
b − a 24
3
4
8
2
Hence proved (2.35).
Corollary 2.13. Substitution of x = a, in (2.35) gives
f (a) + f (b) 1
1
− (b − a) f 0 (b) − f 0 (a) −
2
6
b−a
Zb
p
3 1
f (t) dt ≤ σ (f 00 ) (b − a) 2
64
r
89
.
45
(2.37)
a
Corollary 2.14. Substitution of x = a+b
2 , in (2.35) gives
1
a+b
1
3a + b
a + 3b
1
0 a + 3b
0 3a + b
f
+
f
+f
+
(b − a) f
−f
2
2
4
4
4
32
4
4
r
b
Z
p
0
3
1
1
1
0
− (b − a) f (b) − f (a) −
f (t) dt ≤ σ (f 00 ) (b − a) 2
.
(2.38)
96
b−a
16645
a
3a+b
4 ,
Corollary 2.15. Substitution of x =
in (2.35) gives
1
3a + b
a + 3b
7a + b
a + 7b
b−a
a + 3b
3a + b
f
+f
+f
+f
−
f0
−f 0
4
4
4
8
8
32
4
4
√
Zb
p
0
3
5
1
97
0
00
−
(b − a) f (b) − f (a) −
f (t) dt ≤ σ (f ) (b − a) 2
.
(2.39)
768
b−a
768
a
3. An application to composite quadrature rules
Let In : a = x0 < x1 < x2 < . . . < xn−1 < xn = b be a division of the interval [a, b] and ξ =
(ξ0 , ξ1 , . . . , ξn−1 ) a sequence of intermediate points ξi ∈ [xi , xi+1 ] and h = xi+1 − xi (i = 0, 1, . . . , n − 1).
We have the following quadrature formula:
Consider the perturbed composite quadrature rules
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
0
An f, f , ξ, In
547
1
x i + ξi
xi + 2xi+1 − ξi
= hi f (ξi ) + f (xi + xi+1 − ξi ) + f
+f
4
2
2
5xi + 3xi+1 0
+ ξi −
f (xi + xi+1 − ξi ) − f 0 (ξi )
8
1
3xi + xi+1
0 xi + 2xi+1 − ξi
0 xi + ξi
+
ξi −
× f
−f
2
4
2
2
f 0 (xi+1 ) − f 0 (xi )
+
hi
(
)
3xi + xi+1 3 3
xi + xi+1 3
1
1
3
ξi −
ξi −
×
(ξi − xi ) +
−
24
3
4
8
2
(3.1)
h
i
i+1
for all ξi ∈ xi , xi +x
.
2
Theorem 3.1. Let f : [a, b] → R be such that f 0 is absolutely continuous function. If f
γ ≤ f 00 (x) ≤ Γ for all x ∈ a, a+b
, then we have the following quadrature formulae:
2
Zb
00
∈ L1 [a, b] and
f (t)dt = An f, f 0 , ξ, In + Rn1 f, f 0 , ξ, In ,
(3.2)
f (t)dt = An f, f 0 , ξ, In + Rn2 f, f 0 , ξ, In ,
(3.3)
a
Zb
a
where An (f, f
0 , ξ, I
n)
is defined in (3.1) and remainder satisfies the estimation
Rn1 f, f 0 , ξ, In
≤ (S − γ)
n−1
X
h2i υ (ξi )
(3.4)
h2i υ (ξi ) .
(3.5)
i=0
and
Rn2 f, f 0 , ξ, In
≤ (Γ − S)
n−1
X
i=0
Proof. Apply (2.10) and (2.11) on the interval [xi , xi+1 ] to get
Rn1
0
0
f, f , ξ, In
xZ
i+1
xi + 2xi+1 − ξi
xi + ξi
1
+f
=
f (t)dt − hi f (ξi ) + f (xi + xi+1 − ξi ) + f
4
2
2
xi
5xi + 3xi+1 0
+ ξi −
f (xi + xi+1 − ξi ) − f 0 (ξi )
8
1
3xi + xi+1
xi + 2xi+1 − ξi
xi + ξi
f 0 (xi+1 ) − f 0 (xi )
+
ξi −
× f0
−f 0
−
2
4
2
2
hi
(
3
3 )
1
1
3xi + xi+1
3
xi + xi+1
≤ (S − γ) h2i υ (ξi )
×
(ξi − xi )3 +
ξi −
−
ξi −
24
3
4
8
2
and
Rn2
f, f , ξ, In
xZ
i+1
=
xi
1
xi + 2xi+1 − ξi
xi + ξi
f (t)dt − hi f (ξi ) + f (xi + xi+1 − ξi ) + f
+f
4
2
2
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
548
5xi + 3xi+1 0
+ ξi −
f (xi + xi+1 − ξi ) − f 0 (ξi )
8
3xi + xi+1
f 0 (xi+1 ) − f 0 (xi )
1
0 xi + 2xi+1 − ξi
0 xi + ξi
ξi −
× f
−f
−
+
2
4
2
2
hi
(
3
3 )
3xi + xi+1
3
xi + xi+1
1
1
ξi −
−
ξi −
≤ (Γ − S) h2i υ (ξi )
×
(ξi − xi )3 +
24
3
4
8
2
for (i = 1, 2, . . . , n − 1) .
Summing over i from 0 to n − 1, and using the triangle inequality, we get (3.4) and (3.5).
Theorem 3.2. Let
f : [a,
b] → R be a thrice continuously differentiable mappingin (a, b) with f
Then for all x ∈ a, a+b
, we have
2
Zb
000
∈ L2 [a, b].
f (t)dt = An f, f 0 , ξ, In + Rn3 f, f 0 , ξ, In ,
(3.6)
a
where An (f, f 0 , ξ, In ) is defined by formula (3.1) and the remainder Rn3 (f, f 0 , ξ, In ) satisfies the estimation
"
n−1
X
1
1
3xi + xi+1 5
33
xi + xi+1 5
1
5
000
3
0
f
hi
(ξi − xi ) +
ξi −
−
ξi −
Rn f, f , ξ, In ≤
2
π
320
10
4
320
2
i=0
1
(
3
3 )2 2
1
1
3xi + xi+1
3
xi + xi+1
1
 .
(ξi − xi )3 +
ξi −
−
ξi −
(3.7)
−
hi 24
3
4
8
2
Proof. Applying (2.28) to the interval [xi , xi+1 ], then we get
xZ
i+1
1
xi + 2xi+1 − ξi
x i + ξi
f (t)dt − hi f (ξi ) + f (xi + xi+1 − ξi ) + f
+f
4
2
2
xi
5xi + 3xi+1 0
+ ξi −
f (xi + xi+1 − ξi ) − f 0 (ξi )
8
3xi + xi+1
f 0 (xi+1 ) − f 0 (xi )
1
0 xi + 2xi+1 − ξi
0 x i + ξi
+
ξi −
× f
−f
−
2
4
2
2
hi
(
3
3 )
1
1
3xi + xi+1
3
xi + xi+1
×
(ξi − xi )3 +
ξi −
−
ξi −
24
3
4
8
2
"
1
1
3xi + xi+1 5
33
xi + xi+1 5
1
5
000
≤
f
h
(ξi − xi ) +
ξi −
−
ξi −
2 i 320
π
10
4
320
2
1
(
3
3 )2 2
1
1
1
3xi + xi+1
3
xi + xi+1
 .
−
(ξi − xi )3 +
ξi −
−
ξi −
hi 24
3
4
8
2
for (i = 0, 1, . . . , n − 1).
Now summing over i from 0 to n − 1, using the triangle inequality, we get (3.7).
0
Theorem 3.3. Let
f : [a, b] → R be such that f is absolutely continuous on [a, b] with f
a+b
for all x ∈ a, 2 , we have
Zb
a
f (x)dx = An f, f 0 , ξ, In + Rn4 f, f 0 , ξ, In
00
∈ L2 [a, b]. Then
(3.8)
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
549
where An (f, f 0 , ξ, In ) is defined by formula (3.1) and the remainder Rn4 (f, f 0 , ξ, In ) satisfies the estimation
"
n−1
X 1
p
3xi + xi+1 5
33
xi + xi+1 5
1
5
4
0
00
Rn (f, f , ξ, In ) ≤ σ (f )
ξi −
−
ξi −
(ξi − xi ) +
320
10
4
320
2
i=0
1
(
3
3 )2 2
1
1
3x
+
x
x
+
x
1
3
i
i+1
i
i+1
 .
−
(3.9)
ξi −
ξi −
(ξi − xi )3 +
−
hi 24
3
4
8
2
Proof. Applying (2.35) to the interval [xi , xi+1 ], then we get
xZ
i+1
x i + ξi
xi + 2xi+1 − ξi
1
+f
f (t)dt − hi f (ξi ) + f (xi + xi+1 − ξi ) + f
4
2
2
xi
5xi + 3xi+1 0
+ ξi −
f (xi + xi+1 − ξi ) − f 0 (ξi )
8
f 0 (xi+1 ) − f 0 (xi )
1
3xi + xi+1
0 xi + 2xi+1 − ξi
0 x i + ξi
ξi −
× f
−f
−
+
2
4
2
2
hi
(
3
3 )
1
1
3xi + xi+1
3
xi + xi+1
×
(ξi − xi )3 +
ξi −
−
ξi −
24
3
4
8
2
"
p
1
1
3xi + xi+1 5
33
xi + xi+1 5
≤ σ (f 00 )
(ξi − xi )5 +
ξi −
−
ξi −
320
10
4
320
2
1
(
3
3 )2 2
1
1
1
3xi + xi+1
3
xi + xi+1
 .
−
(ξi − xi )3 +
ξi −
−
ξi −
hi 24
3
4
8
2
for (i = 0, 1, . . . , n − 1) .
Now summing over i from 0 to n − 1, using the triangle inequality, we get (3.9).
4. An application to cumulative distribution function
Let X be a random variable taking values in the finite interval [a, b] with the probability density function
f : [a, b] → [0, 1] and cumulative distributive function
Z x
F (x) = Pr (X ≤ x) =
f (t) dt,
(4.1)
a
Zb
F (b) = Pr (X ≤ b) =
f (u) du = 1.
(4.2)
a
Theorem 4.1. With the assumptions of Theorem 2.4, we have the following inequality which holds
b − E (X) 1
a+x
a + 2b − x
−
F (x) + F (a + b − x) + F
+F
b−a
4
2
2
5a + 3b
+ x−
{f (a + b − x) − f (x)}
8
1
3a + b
a + 2b − x
a+x
f (b) − f (a)
+
x−
f
−f
−
2
4
2
2
(b − a)2
(4.3)
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
(
×
1
1
(x − a)3 +
24
3
550
)
3a + b 3 3
a+b 3
x−
−
x−
≤ υ (x) (b − a) (S − γ) ,
4
8
2
a+x
a + 2b − x
b − E (X) 1
F (x) + F (a + b − x) + F
+F
−
b−a
4
2
2
5a + 3b
+ x−
{f (a + b − x) − f (x)}
8
3a + b
f (b) − f (a)
1
a + 2b − x
a+x
x−
f
−f
−
+
2
4
2
2
(b − a)2
(
)
3a + b 3 3
a+b 3
1
1
3
x−
x−
×
(x − a) +
−
≤ υ (x) (b − a) (Γ − S) ,
24
3
4
8
2
for all x ∈ a, a+b
2 . Where E (X) is the expectation of X.
Proof. By (2.10) and (2.11) on choosing f = F and using the fact
Z b
Z b
F (t) dt,
tdF (t) = b −
E (X) =
a
(4.4)
(4.5)
a
we obtain (4.3) and (4.4).
Corollary 4.2. Under the assumptions of Theorem 4.1, if we put x = 3a+b
in (4.3) and (4.4) then we get
4
3a + b
a + 3b
7a + b
a + 7b
b − E (X) 1
−
F
+F
+F
+F
b−a
4
4
4
8
8
b−a
a + 3b
3a + b
5
19
−
f
−f
−
(b − a) {f (b) − f (a)} ≤
(b − a)3 (S − γ) ,
(4.6)
8
4
4
768
768
b − E (X) 1
3a + b
a + 3b
7a + b
a + 7b
−
F
+F
+F
+F
b−a
4
4
4
8
8
b−a
a + 3b
3a + b
5
19
−
f
−f
−
(b − a) {f (b) − f (a)} ≤
(b − a)3 (Γ − γ) .
8
4
4
768
768
Theorem 4.3. With the assumptions of Theorem 2.8, we have the following inequality which holds
b − E (X) 1
a+x
a + 2b − x
−
F (x) + F (a + b − x) + F
+F
b−a
4
2
2
5a + 3b
+ x−
{f (a + b − x) − f (x)}
8
1
3a + b
a + 2b − x
a+x
f (b) − f (a)
+
x−
f
−f
−
2
4
2
2
(b − a)2
)
(
1
1
3a + b 3 3
a+b 3
3
×
(x − a) +
x−
−
x−
24
3
4
8
2
"
5
1
1
1
3a + b
33
a+b 5
5
000
≤
f
(x − a) +
x−
−
x−
2 320
π
10
4
320
2

(
3
3 )2
1
1
1
3a + b
3
a+b

−
(x − a)3 +
x−
−
x−
b − a 24
3
4
8
2
for all x ∈ a, a+b
2 , where E (X) is the expectation of X.
(4.7)
(4.8)
A. Qayyum, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 537–552
551
Proof. Using (2.28) and the same conditions that we used in above theorem, we get the required inequality
(4.8).
Corollary 4.4. Under the assumptions of Theorem 4.3, if we put x = 3a+b
in (4.8), then we get
4
3a + b
a + 3b
b − E (X) 1
F
+F
−
b−a
4
4
4
7a + b
a + 7b
b−a
a + 3b
3a + b
+F
+F
−
f
−f
8
8
8
4
4
√
000
5
kf k2
97
5
−
(b − a) {f (b) − f (a)} ≤
(b − a) 2
.
768
π
768
(4.9)
Theorem 4.5. With the assumptions of Theorem 2.12, we have the following inequality which holds
b − E (X) 1
a+x
a + 2b − x
F (x) + F (a + b − x) + F
+F
−
b−a
4
2
2
5a + 3b
+ x−
{f (a + b − x) − f (x)}
8
1
3a + b
a + 2b − x
a+x
f (b) − f (a)
+
x−
f
−f
−
(4.10)
2
4
2
2
(b − a)2
(
3
3 )
1
3a
+
b
3
a
+
b
1
(x − a)3 +
x−
−
x−
×
24
3
4
8
2
"
p
σ (f 00 ) 1
1
3a + b 5
33
a+b 5
5
≤
(x − a) +
x−
−
x−
b−a
320
10
4
320
2
1
(
3
3 )2 2
1
1
1
3a
+
b
3
a
+
b
 .
−
(x − a)3 +
x−
−
x−
b − a 24
3
4
8
2
Proof. Applying (2.35) and using the same conditions that we used in Theorem 4.1, we get the required
inequality (4.10).
Corollary 4.6. Under the assumptions of Theorem 4.5, if we put x = 3a+b
in (4.10), then we get
4
b − E (X) 1
3a + b
a + 3b
7a + b
a + 7b
−
F
+F
+F
+F
b−a
4
4
4
8
8
1
5
a + 3b
3a + b
− (b − a) f
−f
−
(b − a) {f (b) − f (a)}
8
4
4
768
√
p
3
97
≤ σ (f 00 ) (b − a) 2
.
768
(4.11)
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