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J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
Research Article
Derivation and applications of inequalities of
Ostrowski type for n-times differentiable mappings
for cumulative distribution function and some
quadrature rules
Ather Qayyuma,∗, Abdul Rehman Kashifb , Muhammad Shoaibc , Ibrahima Fayea
a
Department of Fundamental and Applied Sciences, Universiti Teknologi PETRONAS, 32610 Bandar Seri Iskandar, Perak Darul
Ridzuan, Malaysia.
b
c
Department of Mathematics, University of Hail, 2440, Saudi Arabia.
Abu Dhabi Mens College, Higher Colleges of Technology, P.O. Box 25035, Abu Dhabi, United Arab Emirates.
Communicated by M. Bohner
Abstract
In this paper new integral inequalities of Ostrowski type are developed for n-times differentiable mappings. Some well known inequalities become special cases of the inequalities obtained in this paper. With
the help of obtained inequalities, we will derive new and efficient quadrature rules which are analyzed with
c
the help of specific examples. We also give applications for cumulative distribution function. 2016
All
rights reserved.
Keywords: Ostrowski inequality, numerical integration, composite quadrature rule, cumulative
distributive function.
2010 MSC: 26D15, 26D20, 26D99.
1. Introduction
In 1938, Ostrowski [14] obtained an integral inequality:
∗
Corresponding author
Email addresses: atherqayyum@gmail.com (Ather Qayyum), kashmology@gmail.com (Abdul Rehman Kashif),
safridi@gmail.com (Muhammad Shoaib), ibrahima faye@petronas.com.my (Ibrahima Faye)
Received 2015-07-26
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
1845
Theorem 1.1. Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) , whose derivative is
bounded on (a, b) , then
"
#
b−a 2
a+b 2 M
|S (f ; a, b)| ≤
+ x−
(1.1)
2
2
b−a
for all x ∈ [a, b], where the functional S (f ; a, b) represent the deviation of f (x) from its integral mean over
[a, b],
S (f ; a, b) = f (x) − M (f ; a, b)
(1.2)
and
1
M (f ; a, b) =
b−a
Zb
f (x) dx
(1.3)
a
is the integral mean.
If we assume that f 0 ∈ L∞ [a, b] and kf 0 k∞ = ess |f 0 (t)| then M in (1.1) may be replaced by kf 0 k∞ .
t∈[a,b]
The inequality of Ostrowski type is considered the most useful inequality in mathematical analysis. Some
of the classical inequalities for means can be derived from [2] for particular choices of the function f. Many
researchers have given considerable attention to the inequality (1.1) and its various generalizations, have
appeared in the literature, to mention a few, see [1, 4, 5, 6, 8, 9, 10, 11, 13, 16, 20, 21] and the references
cited therein.
In the recent year, many new inequalities similar to (1.1) have been established; see [7, 15, 17, 18, 19].
In this current work we will extend, subsume and generalize Ostrowski’s inequality for n-times differentiable
mappings by using a more generalized kernel. The generalized kernel is expected to reduce approximation
errors.
2. Main Results
The following lemma is required to prove the main theorem.
Lemma 2.1. Let f : [a, b] → R be an n-time differentiable function such that f (n−1) (x) for n ∈ N is
absolutely continuous on [a, b], then we have the identity
Zb
Pn (x, t)f
a
(n)
n−1
X
h
(k)
1
(−1)k A (x − a)k+1 + B (b − x)k+1 f (x)
(k + 1)!
k=0

Zx
Zb
+ (−1)n A f (t)dt + B f (t)dt
n+1
(t)dt = (−1)
a
(2.1)
x
for all x ∈ [a, b], where α, β ∈ R are non negative and not both zero. Consider P (x, .): [a, b] → R, the peano
kernel
 A
n
 n! (t − a) , a ≤ t ≤ x,
Pn (x, t) =
(2.2)
 B
n
n! (t − b) , x < t ≤ b,
where,
α
,
(α + β) (x − a)
β
B=
,
(α + β) (b − x)
A=
n ≥ 1, k ≥ 1 are natural numbers and f (0) (x) = f (x) .
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
1846
Proof. The proof of (2.1) is established using mathematical induction.
Take n = 1,
Zb
L.H.S of (2.1) = P1 (x, t)f 0 (t)dt.
(2.3)
a
After integration by parts, we get
Zb

1  α
α+β x−a
P1 (x, t)f 0 (t)dt = f (x) −
a
Zx
f (t)dt +
β
b−x
a
Zb

f (t)dt .
x
Equation (2.3) is identical to the R.H.S of (2.1) for n = 1.
Assume that (2.1) is true for n and let us prove it for n + 1. That is, we have to prove the equality
Zb
n
X
h
i
1
(−1)k A (x − a)k+1 + B (b − x)k+1 f (k) (x)
(k + 1)!
k=0
 x

Z
Zb
+ (−1)n+1 A f (t)dt + B f (t)dt ,
Pn+1 (x, t)f (n+1) (t)dt = (−1)n+2
a
a
where
Pn+1 (x, t) =
(2.4)
x



A
(n+1)!
(t − a)n+1 , a ≤ t ≤ x,


B
(n+1)!
(t − b)n+1 , x < t ≤ b.
We integrate the left hand side of (2.4)
Zb

Zx
Zb

1
A (t − a)n+1 f (n+1) (t)dt + B (t − b)n+1 f (n+1) (t)dt
(n + 1)!
a
x
h
i
1
n+1
n+1
A (x − a)
+ B (−1) (x − b)
f (n) (x)
=
(n + 1)!
 x

Z
Zb
(n + 1) 
−
A (t − a)n f (n) (t)dt + B (t − b)n f (n) (t)dt
(n + 1)!
a
x
h
i
1
=
A (x − a)n+1 + B (−1)n+2 (b − x)n+1 f (n) (x)
(n + 1)!
 x

Z
Zb
1
− A (t − a)n f (n) (t)dt + B (t − b)n f (n) (t)dt .
n!
Pn+1 (x, t)f (n+1) (t)dt =
a
a
x
Multiplying both sides with (−1)n , we get
n
Zb
(−1)
Pn+1 (x, t)f (n+1) (t)dt =
h
i
1
A (−1)n (x − a)n+1 + B (b − x)n+1 f (n) (x)
(n + 1)!
a
n
Zb
− (−1)
a
Pn (x, t)f (n) (t)dt
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
=
h
i
1
A (−1)n (x − a)n+1 + B (b − x)n+1 f (n) (x)
(n + 1)!
"
#
n−1
X B (b − x)k+1 + (−1)k A (x − a)k+1
+
f (k) (x)
(k + 1)!
k=0
Zx
−A
Zb
f (t)dt − B
a
=
1847
n
X
k=0
x
1
(k + 1)!
h
i
(−1)k A (x − a)k+1 + B (b − x)k+1 f (k) (x)
Zx
−A
f (t)dt
Zb
f (t)dt − B
a
f (t)dt.
x
So,
Zb
n
X
h
i
1
(−1)k A (x − a)k+1 + B (b − x)k+1 f (k) (x)
(k + 1)!
k=0
 x

Z
Zb
+ (−1)n+1 A f (t)dt + B f (t)dt .
Pn+1 (x, t)f (n+1) (t)dt = (−1)n+2
a
a
x
Hence (2.4) is proved.
Theorem 2.2. Let f : [a, b] → R be an n-time differentiable function such that f (n−1) (x) for n ∈ N is
absolutely continuous on [a, b]. Using (2.1), we define
τ (x; α, β) = (−1)n+1
n−1
X
k=0
h
i (k)
1
(−1)k A (x − a)k+1 + B (b − x)k+1 f (x)
(k + 1)!
(−1)n
+
[α (M (a, x)) + βM (x, b)] ,
α+β
where M (f ; a; b) is the integral















|τ (x; α, β)| ≤














mean as defined by (1.3), then
h
h
A (x − a)n+1 + B (b − x)n+1
i
kf (n) k∞
(n+1)!
Aq (x − a)nq+1 + B q (b − x)nq+1
, f (n) ∈ L∞ [a, b] ,
i 1 kf (n) k
q
p
1
,
n!(nq+1) q
f (n) ∈ Lp [a, b] , p > 1, p1 +
1
q
(2.6)
= 1,
[A (x − a)n + B (b − x)n + |A (x − a)n − B (b − x)n |]
f (n) ∈ L1 [a, b] .
kf (n) k1
2n!
,
Proof. Taking the modulus of (2.1) and using (1.3), we have from (2.5)
b
Z
Zb
Zb Zb
|τ (x; α, β)| = Pn (x, t)f (n) (t)dt ≤ |Pn (x, t)| dt f (n) (t) dt ≤ f (n) (t)
|Pn (x, t)| dt.
∞
a
(2.5)
a
a
a
(2.7)
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
1848
Using (2.2), we get
Zb
|Pn (x, t)| dt =
h
i
1
A (x − a)n+1 + B (b − x)n+1 .
(n + 1)!
(2.8)
a
Using (2.8) in (2.7), we get first inequality of (2.6).
Using Hölder’s integral inequality, from (2.7)
 1q
 b
Z
|τ (x; α, β)| ≤  |Pn (x, t)|q dt f (n) .
p
a
Now
 1q
 b
Z
q
(α + β)  |Pn (x, t)| dt =
1
n! (nq + 1)
α
x−a
q
nq+1
(x − a)
1
+
n! (nq + 1)
β
b−x
q
nq+1
1
q
(b − x)
,
a
This gives
(n) f p
|τ (x; α, β)| ≤
n! (nq + 1)
h
1
q
Aq (x − a)nq+1 + B q (b − x)nq+1
i1
q
.
This completes the proof of second inequality.
Finally, by using the definition of k.k1 − norm, we get
|τ (x; α, β)| ≤ sup |Pn (x, t)| f (n) ,
1
t∈[a,b]
so
(α + β) sup |Pn (x, t)| =
t∈[a,b]
1 1
n−1
[α (x − a)n−1 + β (b − x)n−1 ] +
− β (b − x)n−1 .
α (x − a)
2n!
2n!
Therefore, by using (2.7), we get last inequality. This completes the proof of Theorem 2.2.
Remark 2.3. If we put n = 1 and n = 2 in (2.6), we get the results of Cerone [4] and Qayyum et al [19]
respectively.
3. Perturbed Results using Čebyŝev functional
In this section, we will establish some new results by using Grüss inequality and the Čebyŝev functional.
In 1882, Čebyŝev [3] gave the following inequality.
|T (f, g)| ≤
1
(b − a)2 f 0 ∞ g 0 ∞ ,
12
(3.1)
where f, g : [a, b] → R are absolutely continuous functions, which have bounded first derivatives, and
1
T (f, g) :=
b−a
Zb
a
1
f (x) g (x) dx −
(b − a)2
Zb
Zb
f (x) dx. g (x) dx
a
a
=M (f g; a, b) − M (f ; a, b) M (g; a, b) .
(3.2)
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
In 1935, Grüss [12] proved the following inequality:
Zb
Zb
Zb
1
1
1
1
b − a f (x) g (x) dx − b − a f (x) dx. b − a g (x) dx ≤ 4 (Φ − ϕ) (Γ − γ) ,
a
a
1849
(3.3)
a
provided that f and g are two integrable functions on [a, b] and satisfy the condition
ϕ ≤ f (x) ≤ Φ and γ ≤ g (x) ≤ Γ for all x ∈ [a, b] .
(3.4)
The constant 14 is best possible. The perturbed version of the results of Theorem 3.1 can be obtained by
using Grüss type results involving the Čebyŝev functional
T (f, g) = M (f g; a, b) − M (f ; a, b) M (g; a, b) ,
where M is the integral mean defined in (1.3).
Theorem 3.1. Let f : [a, b] → R be an n-time differentiable function such that f (n−1) (x) for n ∈ N is
absolutely continuous on [a, b] and α, β are non-negative real numbers. Then
τ (x; α, β) −
1
2
1
κ
1 (n) 2
n
n
2
[α (x − a) + β (x − b) ]
f
≤
(b
−
a)
N
(x)
−
κ
(α + β)
(n + 1)! b−a
2
(b − a)
(Φ − ϕ) (Γ − γ) ,
≤
2 (α + β)
(3.5)
where τ (x; α, β) is as given by (2.5),
κ :=
2
"
α
α+β
N (x) :=
−
2
f (n−1) (b) − f (n−1) (a)
,
b−a
#
1
1
2n−1
×
2 (x − b)
b
−
a
(2n + 1) (n!)
2
[α (x − a)n + β (x − b)n ] ,
(x − a)2n−1
−
(n!)2 (2n + 1)
1
(b − a) (α + β) (n + 1)!
β
α+β
2
and
Φ = sup Pn (x, t) =
t∈[a,b]
h
i
1
α (x − a)n−1 + β (x − b)n−1 ,
(α + β) n!
ϕ = inf Pn (x, t) = 0,
t∈[a,b]
if n is even
Φ = sup Pn (x, t) =
t∈[a,b]
1
α (x − a)n−1
(α + β) n!
and
ϕ = inf Pn (x, t) =
t∈[a,b]
if n is odd.
(3.6)
−β (b − x)n−1
,
(α + β) n!
(3.7)
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
1850
Proof. Associating f (t) with Pn (x, t) and g (t) with f (n) (t), we obtain
T Pn (x, .) , f (n) (.); a, b = M Pn (x, .) f (n) (.); a, b − M (Pn (x, .) ; a, b) M f (n) (.); a, b .
Now using identity (2.1),
(b − a) T Pn (x, .) , f (n) (.); a, b = τ (x; α, β) − (b − a) M (Pn (x, .) ; a, b) κ,
(3.8)
where κ is the secant slope of f 0 over [a, b] as given in (3.6). Now, from (2.1) and (3.2),
Zb
(b − a) M (Pn (x, .) ; a, b) =
Pn (x, t)dt
a

=
1
 α
(α + β) n! x − a
Zx
(t − a)n dt +
a
β
b−x
Zb

(t − b)n  dt
(3.9)
x
1
=
[α (x − a)n + β (x − b)n ] .
(α + β) (n + 1)!
Now combining (3.9) with (3.7), the left hand side of (3.5) is obtained.
Let f, g : [a, b] → R and f g : [a, b] → R be integrable on [a, b], then [4]
1
1
|T (f, g)| ≤ T 2 (f, f ) T 2 (g, g)
(Γ − γ) 1
≤
T 2 (f, f )
2
1
≤ (Φ − ϕ) (Γ − γ)
4
(f, g ∈ L2 [a, b])
(γ ≤ g (x) ≤ Γ, t ∈ [a, b])
(3.10)
(ϕ ≤ f (x) ≤ Φ, t ∈ [a, b]) .
Also, note that
i 1
h 1
2
0 ≤ T 2 f (n) (.), f (n) (.) = M f (n) (.)2 ; a, b − M 2 f (n) (.); a, b

2  12

Rb (n)

Zb  f (t) dt  
 1
(n) 2
 
a
=
f
(t)
dt
−
 

b − a  b−a  


a
(3.11)
1
2
1 (n) 2
2
=
f − κ
b−a
2
(Γ − γ)
≤
,
2
1
where γ ≤ f 00 (t) ≤ Γ, t ∈ [a, b] . Now, for the bounds on (3.8), we have to determine T 2 (Pn (x, .) , Pn (x, .)) ,
and φ and Φ such that ϕ ≤ Pn (x, .) ≤ Φ.
Now from (2.2), the definition of Pn (x, t), we have
T (Pn (x, .) , Pn (x, .)) = M P 2 (x, .) ; a, b − M 2 (Pn (x, .) ; a, b) .
(3.12)
From (3.9) we obtain
M (Pn (x, .) ; a, b) =
1
[α (x − a)n + β (x − b)n ]
(b − a) (α + β) (n + 1)!
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
1851
and
2
(b − a) M P (x, .) ; a, b =
=
α
α+β
2
α
α+β
2
1
(n! (x − a))2
Zx
(t − a)
2n
dt +
β
α+β
2
a
2n−1
(x − a)
−
(n!)2 (2n + 1)
β
α+β
2
1
(n! (b − x))2
Zb
(t − b)2n dt
x
2n−1
(x − b)
.
(2n + 1) (n!)2
Thus, substituting the above results into (3.12) gives
1
0 ≤ N (x) := T 2 (Pn (x, .) , Pn (x, .)) ,
which is given explicitly by (3.7). Combining (3.8), (3.11) and (3.12) gives from the first inequality in (3.10),
the first inequality in (3.5). Now utilizing the inequality in (3.11) produces the second result in (3.5). Further,
it may be noticed from the definition of P (x, t) in (2.2) that for α, β ≥ 0,
Φ = sup Pn (x, t) and ϕ = inf Pn (x, t) .
t∈[a,b]
t∈[a,b]
Remark 3.2. Similarly, If we put n = 1 and n = 2, in Theorem 3.1, we get the results proved in [4] and [19]
respectively.
4. An Application to the Cumulative Distributive Function
Let X ∈ [a, b] be a random variable with the cumulative distributive function
Zx
F (x) = Pr (X ≤ x) =
f (u) du,
a
where f is the probability density function. In particular,
Zb
f (u) du = 1.
a
The following theorem holds.
Theorem 4.1. Let X and F be as above, then
"
#
n−1
n+1
k
k+1
X
(−1)
(k)
(−1)
α
(x
−
a)
(b
−
x)
f (x) + (−1)n [{α (b − x) − β (x− a)}F (x) − β (x − a)]|
k+1
(k + 1)!
+β (x − a) (b − x)
k=0

h
i (n)

n+1
n+1 kf k∞
(n) ∈ L [a, b] ,

(α
+
β)
(b
−
x)
(x
−
a)
A
(x
−
a)
+
B
(b
−
x)

∞
(n+1)! , f





(4.1)


h
i1
(n)

nq+1
nq+1 q kf kp
q
q
(n)
(α + β) (b − x) (x − a) A (x − a)
+ B (b − x)
∈ Lp [a, b] ,
1 ,f
≤
n!(nq+1) q





(n)

n
n


A
(x
−
a)
+
B
(b
−
x)
kf k1 (n)


∈ L1 [a, b] .
 (α + β) (b − x) (x − a) + |A (x − a)n − B (b − x)n |
2n! , f
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
1852
Proof. From (2.5) and by using the definition of probability density function, we have
τ (x; α, β) =
n
i
1 X (−1)n+k h
(k−1)
(k−1)
α (x − a)k−1 f
(x) + β (x − b)k−1 f
(x)
α+β
k!
k=1
n
(−1)
[α (M (a, x)) + βM (x, b)]
α+β
n
i
1 X (−1)n+k h
(k−1)
(k−1)
α (x − a)k−1 f
(x) + β (x − b)k−1 f
(x)
=
α+β
k!
k=1
(−1)n {α (b − x) − β (x − a)}F (x)
β
+
.
+
α+β
(x − a) (b − x)
(b − x)
+
(4.2)
Now using (2.6) and (4.2), we get our required result (4.1).
Putting α = β =
1
2
in Theorem 3.1 gives the following result.
Corollary 4.2. Let X be a random variable with the cumulative distributive function F (x) and the probability density function f (x). Then
(−1)n+1 n−1
i
X f (k) (x) h
(−1)k (x − a)k+1 (b − x) + (x − a) (b − x)k+1
2
(k + 1)!
k=0
(4.3)
a+b
n
+ (−1)
− x 2F (x) − (x − a) 2

kf (n) k


(b − x) (x − a) [(x − a)n + (b − x)n ] 2(n+1)!∞ ,



h
i 1 kf (n) k

q

p
(b − x) (x − a) (x − a)nq+1−q + (b − x)nq+1−q
1 ,
≤
q
2n!(nq+1)
"
#

n−1
n−1


(n)
(x
−
a)
+
(b
−
x)

kf k1 .


 (b − x) (x − a) + (x − a)n−1 − B (b − x)n−1 4n!
Remark 4.3. The above result allow the approximation of F (x) in terms of f (x). The approximation of
R (x) = 1 − F (x)
could also be obtained by a simple substitution. R (x) is of importance in reliability theory where f (x) is
the probability density function of failure.
Remark 4.4. We put β = 0 in (4.1), assuming that α 6= 0 to obtain
(k)
n−1
h
i
X
f
(x)
n+1
(−1)k (x − a)k+1 (b − x) + (−1)n (b − x) F (x)
(−1)
(k + 1)!
k=0

kf (n) k

(b − x) (x − a)n+1 (n+1)!∞ ,




h
i 1 kf (n) k
q
p
(b − x) (x − a) (x − a)nq+1−q
≤
1 ,
q

n!(nq+1)



 (b − x) (x − a) (x − a)n−1 + (x − a)n−1 kf (n) k1 .
2n!
Further we note that
Zb
F (u) du = uF
a
(u)|ba
Zb
−
xf (x) dx = b − E (X) .
a
Remark 4.5. By choosing n = 1 and n = 2 in (4.1)-(4.4), we can get results proved in [4] and [19].
(4.4)
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
1853
5. Some new Ostrowski Type Inequalities for α = x − a and β = b − x
(n−1)
is absolutely continuous on [a, b] and
|X| ≤
Y
(S − γ)
2 (b − a) (n + 1)!
(5.1)
|X| ≤
Y
(Γ − S) ,
2 (b − a) (n + 1)!
(5.2)
|X| ≤
Z
(S − γ)
2 (b − a) (n + 1)!
(5.3)
|X| ≤
Z
(Γ − S)
2 (b − a) (n + 1)!
(5.4)
Theorem 5.1. Let f : [a, b] → R be differentiable
on (a, b), f
γ ≤ f (n) (t) ≤ Γ ∀ t ∈ [a, b] , then for all x ∈ a, a+b
, we have
2
and
if n is even and
and
if n is odd,
where
n+1
X := (−1)
n−1
X
h
i (k)
1
k
k+1
k+1
(−1) (x − a)
+ (b − x)
f (x)
(b − a)2 (k + 1)!
k=0
(−1)n
+
(b − a)2
Zb
f (t)dt −
a
h
i
1
n+1
n+1
(x
−
a)
−
(x
−
b)
× f (n−1) (b) − f (n−1) (a),
(b − a)2 (n + 1)!
Y := (x − a)n [n + 1 + 2 (a − x)] + (x − b)n [n + 1 + 2 (x − b)] + (n + 1) [(x − a)n − (x − b)n ]
and
Z := (x − a)n [n + 1 + 2 (a − x)] + 2 (x − b)n+1 + (n + 1) (x − a)n .
Proof. From (2.2) and using α = x − a and β = b − x and
1
b−a
Zb
f (n−1) (b) − f (n−1) (a)
b−a
(5.5)
h
i
1
(x − a)n+1 − (x − b)n+1 ,
(b − a) (n + 1)!
(5.6)
f (n) (t)dt =
a
and
1
b−a
Zb
Pn (x, t)dt =
a
we get
1
b−a
Zb
Pn (x, t)f
(n)
a
= (−1)n+1
1
(t)dt −
(b − a)2
×f
Zb
Pn (x, t)dt
a
f (n) (t)dt
a
n−1
X
h
i (k)
1
k
k+1
k+1
(−1)
(x
−
a)
+
(b
−
x)
f (x)
(b − a)2 (k + 1)!
k=0
(−1)n
+
(b − a)2
(n−1)
Zb
Zb
f (t)dt −
a
h
i
1
n+1
n+1
(x
−
a)
−
(x
−
b)
(b − a)2 (n + 1)!
(b) − f (n−1) (a).
(5.7)
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
1854
We suppose that
1
Rn (x) =
b−a
Zb
Pn (x, t)f
(n)
a
1
(t)dt −
(b − a)2
Zb
Zb
Pn (x, t)dt
a
f (n) (t)dt.
(5.8)
a
If C ∈ R is an arbitrary constant, then we have
Rn (x) =
1
b−a
Zb 
f (n) (t) − C Pn (x, t) −
a
1
b−a
Zb

P (n) (x, s)ds dt.
(5.9)
a
Furthermore, we have
b
Z Zb
1
1
(n)
|Rn (x)| ≤
max Pn (x, t) −
P (x, s)ds f (n) (t) − C dt.
b − a t∈[a,b] b−a
a
(5.10)
a
Now, if n is even
Pn (x, t) −
1
b−a
Zb
P
(n)
(x, s)ds
a
hn
o
1
=
max (x − a)n {n + 1 − x + a} + (x − b)n+1 ,
(b − a) (n + 1)!
n
oi
n
n+1 (x − b) {n + 1 + x − b} − (x − a)
=
(5.11)
1
[(x − a)n {n + 1 + 2 (a − x)}
2 (b − a) (n + 1)!
+ (x − b)n {n + 1 + 2 (x − b)}
+ (n + 1) |{(x − a)n − (x − b)n }|]
and if n is odd
Pn (x, t) −
1
b−a
Zb
P (n) (x, s)ds
a
i (5.12)
i hh
1
=
max (x − a)n (n + 1 − x + a) + (x − b)n+1 , (x − b)n+1 − (x − a)n+1 (b − a) (n + 1)!
h
1
=
(x − a)n [n + 1 + 2 (a − x)] + 2 (x − b)n+1 + (n + 1) |(x − a)n |] .
2 (b − a) (n + 1)!
We also have
Zb (n)
f (t) − γ dt = (S − γ) (b − a)
(5.13)
a
and
Zb (n)
f (t) − Γ dt = (Γ − S) (b − a) .
(5.14)
a
Therefore, we can obtain (5.1) and (5.4) by using (5.5) to (5.14) taking C = γ and C = Γ in (5.10),
respectively.
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
1855
5.1. Derivation of Numerical Quadrature Rules
We propose some new quadrature rules involving higher order derivatives of the function f . In fact, the
following new quadrature rules can be obtained while investigating error bounds using Theorem 5.1.
Zb
n−1
X
f (t)dt ≈
Qn,1 (f ) :=
k=0
a
Zb
f (t)dt ≈
Qn,2 (f ) :=
n−1
X
k=0
a
n+1
+
(b − a)
+ 1)!
2n+1 (n
Zb
f (t)dt ≈
Qn,3 (f ) :=
n−1
X
k=0
a
n
+
i
(b − a)k+1 (k)
(b − a)n+1 h (n−1)
f (a) +
f
(b) − f (n−1) (a) ,
(k + 1)!
(n + 1)!
i (k)
(b − a)k+1 h
k
1
+
(−1)
f
2k+1 (k + 1)!
a+b
2
(5.15)
(5.16)
h
i
[1 + (−1)n ] f (n−1) (b) − f (n−1) (a) ,
i (k)
(b − a)k+1 h
k
k+1
(−1)
+
(3)
f
4k+1 (k + 1)!
n+1
(−1) (b − a)
4n+1 (n + 1)!
h
3a + b
4
(5.17)
ih
i
1 + (−1)n (3)n+1 f (n−1) (b) − f (n−1) (a) .
To investigate the efficiency of these quadrature rules we integrate various types of functions including
Polynomial, Exponential, Logarithmic. The results of the numerical integrations are given in Table 1 with
absolute error and the value of n used to obtain the mentioned accuracy.
f1 (x) = x2 + x + 2,
x
f3 (x) = e sin x ,
2
f5 (x) = ex ,
f7 (x) = x + cos x,
f2 (x) = x sin x,
f4 (x) = x2 + sin x,
f6 (x) = ex cos (ex − 2x) ,
f8 (x) = log x2 + 2 sin log x2 + 2 .
(5.18)
The errors shown in the Table 1 are the absolute value of the difference of the exact value of the integral
and its numerical value. All three quadrature rules show exact value of the integral of f1 for n = 2. For a
polynomial of degree k, n = k + 1 will give exact value of the integral f1 . Though acceptable error estimates
can be obtained for smaller values of n. to save computational time.
The integral of f5 , Qn,3 (f ) give an error of the order of 10−5 for n = 7 while the other two quadrature
rules give a similar error for n = 13 and n = 16. Similarly for all other functions Qn,3 (f ) report errors
of the order of 10−5 or 10−6 for relatively smaller values of n as compared to the other two quadrature
rules. Specifically, Qn,3 (f ) give an excellent estimate for the integrals of f6 , f8 and f5 at n = 6 and n = 7
respectively. In general Qn,3 (f ) gave better results as compared to the rest of the quadrature rules for much
smaller values of n. Therefore we can conjecture that Qn,3 (f ) is computationally more efficient
both in terms
of error approximation, simplicity and time. As a rough estimate we integrated log x2 + 2 sin log x2 + 2
using the builtin algorithms of Mathematica 10.0 which took 26.30 seconds to give its approximate answer.
To obtain similar approximation for the integral of f8 , Qn,3 (f ) took less than a second. The performance
of some the quadrature rules can be seen to be poor for the integrals f5 , f6 and f8 where we had to take n
around 15, 16 and 18 for Qn,1 (f ) to achieve a reasonable approximation error.
A. Qayyum, A. R. Kashif, M. Shoaib, I. Faye, J. Nonlinear Sci. Appl. 9 (2016), 1844–1857
1856
Table 1: Performance of the proposed quadrature rules
Sr. No.
1.
Method
R1
f1 (x)dx
n : Qn,1 (f )
2:
2.83333
n:
Qn,2 (f )
n:
Qn,3 (f )
Exact Value
2:
2.83333
2:
2.83333
2.83333
0
Error:
2.
R1
f2 (x)dx
0
7:
0.301101
0
6:
0.301149
0
5:
0.301153
0.301169
0
Error:
3.
R1
0.00006748
0.0000201585
0.0000155081
6:
5:
f3 (x)dx
8:
Error:
0.0000348171
0.000011471
0.0000357479
f4 (x)dx
6:
5:
4:
Error:
0.0000666818
0.0000225644
f5 (x)dx
16:
13:
Error:
0.0000562577
0.000015241
0.0000864484
f6 (x)dx
18:
12:
6:
0.909365
0.909319
0.909366
0.909331
0
4.
R1
0.792964
0.793008
0
5.
R1
1.46271
1.46267
0.793022
0.793031
8.79×10−6
7:
1.46257
1.46265
0
6.
R1
0
Error
7.
R1
f7 (x)dx
1.31384
5.3×10−6
6:
1.3415
1.31385
1.31386
0.0000229733
0.0000308985
5:
4:
1.34144
1.34145
1.31383
1.34147
0
8.
R1
Error:
0.0000287233
0.000027542
0.0000160969
f8 (x)dx
15:
9:
6:
Error:
0.0000532107
0.629715
0.629866
0.629677
0.629769
0
0.0000973014
0.0000916031
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