Applied Mathematics Letters 22 (2009) 1345–1350

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Applied Mathematics Letters
journal homepage: www.elsevier.com/locate/aml

New inequalities of Ostrowski-like type involving n knots and the
Lp -norm of the m-th derivative
Vu Nhat Huy a , Quô´c-Anh Ngô a,b,∗
a

Department of Mathematics, College of Science, Viêt Nam National University, Hà Nôi, Viêt Nam

b

Department of Mathematics, National University of Singapore, 2 Science Drive 2, Singapore 117543, Singapore

article

abstract

info

Article history:
Received 23 April 2008
Received in revised form 3 March 2009
Accepted 30 March 2009

On the basis of recent results due to Nenad Ujević, we obtain some new inequalities of
Ostrowski-like type involving n knots and the Lp -norm of the m-th derivative where n, m,


f

a+b
2

√
− 2−

3 ( b − a)

√

√
+ 2−

7−4 3

3 (b − a)

8

f

∞

( b − a) 3 .

(1)


E-mail addresses: (V.N. Huy), (Q.-A. Ngô).

0893-9659/$ – see front matter © 2009 Elsevier Ltd. All rights reserved.
doi:10.1016/j.aml.2009.03.002


1346

V.N. Huy, Q.-A. Ngô / Applied Mathematics Letters 22 (2009) 1345–1350

√
a+b

+f

2

+

3−

6

49

(b − a)

2

80


√
6 in (2) are sharp in the sense that these cannot

be replaced by smaller ones. This leads us to strengthen (1) and (2) by enlarging the number of knots (two knots in both (1)
and (2)) and replacing the norms · ∞ in (1) and · 2 in (2).
Before stating our main result, let us introduce the following notation.
b

f (x)dx.

I (f ) =
a

Let 1

m, n < ∞ and 1 p ∞. For each i = 1, n, we assume 0 < xi < 1 such that

n

x1 + x2 + · · · + xn = ,


2




· · ·
n

f (a + xi (b − a)) .

n

i=1

Remark 3. With the above notation, (1) reads as follows:
I (f ) − Q

f , 2, 2,

1
2

√

1

3 ,

− 2−

2

√

√
+ 2−

7−4 3


2

,

2

+

3−

6

49

2

80

−

1√
4

6 f

5

2


f (m)

1

p

(b − a)m+ q

(5)

where
1
p

+

1
q

=1

and

f

r

:=



1347

2. Proofs
Before proving our main theorem, we need an essential lemma below. It is well-known in the literature as Taylor’s
formula or Taylor’s theorem with the integral remainder.
Lemma 6 (See [3]). Let f : [a, b] → R and let r be a positive integer. If f is such that f (r −1) is absolutely continuous on [a, b],
x0 ∈ (a, b), then for all x ∈ (a, b) we have
f (x) = Tr −1 (f , x0 , x) + Rr −1 (f , x0 , x)
where Tr −1 (f , x0 , ·) is Taylor’s polynomial of degree r − 1, that is,
r −1

Tr −1 (f , x0 , x) =

f (k) (x0 ) (x − x0 )k
k!

k=0

and the remainder can be given by
x

Rr −1 (f , x0 , x) =
x0

(x − t )r −1 f (r ) (t )
dt .
(r − 1)!

(6)


x

F (x) =

f (x)dx.
a

By the Fundamental Theorem of Calculus
I (f ) = F (b) − F (a) .
Applying Lemma 6 to F (x) with x = a and u = b − a, we get
m

F (b) = F (a) +
k=1

(b − a)k (k)
F ( a) +
k!

b−a
0

(b − a − t )m (m+1)
F
(a + t ) dt
m!

which yields

(b − a)k (k)

b−a
0

(b − a − t )m (m)
f
(a + t ) dt .
m!

n, applying Lemma 6 to f (x) with x = a and u = xi (b − a), we get
xki (b − a)k (k)
f (a) +
k!
k=0

xi (b−a)

m−1

f (a + xi (b − a)) =

xki (b − a)k (k)
f (a) +
k!
k=0

0

m−1

=

n

n

f (a + xi (b − a)) =
i=1

n

n

xki (b − a)k

m−1
i=1

=

k!

k=0

0

n

f (k) (a) +

n (b − a)k (k)
f (a) +

b−a

m−1
xm
i ( b − a − u)
f (m) (a + xi u) du.
(m − 1)!

(9)

Thus,
m−1

Q (f , n, m, x1 , . . . , xn ) =
k=0

b−a
(b − a)k+1 (k)
f ( a) +
n
(k + 1)!

n

b−a
0

i =1

m−1

b

=
a

n

b−a
(b − x) (m)
f (x)dx +
m!
n

n

(b − ·) (m)
f
m!
m

+

n

n

1

a



b−a

i=1

b

0

i =1

m

m

b−a

xm
i

i=1

.
1

Using the Hölder inequality, we get

(b − ·)m (m)
f
m!


p

f (m) ∞ while, for 1
b

((1 − xi ) a + xi ·)

1

(m)

f

(m)

m
q

=

1
m!

(b − a)mq+1
mq + 1

p < ∞, we have
p


1
1
p

1

f (m) (t )

a
b

f

(m)

a

f (m)

p

xi
xi

((1 − xi ) a + xi x) d ((1 − xi ) a + xi x)

(1−xi )a+xi b

1
xi

1
p

1
q

f (m)

p

.

(10)


V.N. Huy, Q.-A. Ngô / Applied Mathematics Letters 22 (2009) 1345–1350

1349

This helps us to deduce that

n

m−1
xm
i (b − ·)
f (m) ((1 − xi )a + xi ·)
(m − 1)!

n


−1
xm
i
i =1

f (m)

f (m)

n

b−a

f (m) ((1 − xi ) a + xi ·)

(m − 1)!

n

b−a

=

xm
i

i =1

(m − 1)!

(b − a)mq+1
(m − 1) q + 1

p

1

(b − ·)m−1 f (m) ((1 − xi ) a + xi ·)

(m − 1)!

i =1

b−a

=

xm
i

.

It follows that
1
q

(b − a)mq+1
mq + 1

1

√
0

−

21
12

√

√
+f

a+

1
4

+

(b − a) + f

21

1
2

(b − a)

√

7+


1

2 3

2

− √



1350

V.N. Huy, Q.-A. Ngô / Applied Mathematics Letters 22 (2009) 1345–1350

Example 9. When m = 2 and 0 < xi < 1 such that
b

f (x)dx −
a

b−a
n

n

f (a + xi (b − a))
i=1

n
i =1

5
12

xi =

f



i=1

n

n

f (a + xi (b − a))
i=1

3

√

Example 10. When m = 3 and 0 < xi < 1 such that
b

√

√

n

n
i =1

7
72

f

− a)4 .

(17)

Acknowledgments
The authors would like to express sincere gratitude to the two anonymous referees for their constructive suggestions on
the manuscript.
References
[1] N. Ujević, Error inequalities for a quadrature formula of open type, Rev. Colombiana Mat. 37 (2003) 93–105.
[2] N. Ujević, Error inequalities for a quadrature formula and applications, Comput. Math. Appl. 48 (2004) 1531–1540.
[3] G.A. Anastassiou, S.S. Dragomir, On some estimates of the remainder in Taylor’s formula, J. Math. Anal. Appl. 263 (2001) 246–263.