DSpace at VNU: A new way to think about Ostrowski-like type inequalities

Computers and Mathematics with Applications 59 (2010) 3045–3052

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Computers and Mathematics with Applications
journal homepage: www.elsevier.com/locate/camwa

A new way to think about Ostrowski-like type inequalities
´
Ngô a,b,∗
Vu Nhat Huy a , Quôc-Anh
a

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

b

Department of Mathematics, National University of Singapore, Block S17 (SOC1), 10 Lower Kent Ridge Road, Singapore 119076, Singapore

article

abstract

info

Article history:
Received 5 May 2009
Received in revised form 17 February 2010
Accepted 17 February 2010

In this present paper, by considering some known inequalities of Ostrowski-like type, we

propose a new way to treat a class of Ostrowski-like type inequalities involving n points
and m-th derivative. To be precise, the following inequality
b

1
Keywords:
Inequality
Error
Integral
Taylor
Ostrowski
New estimations
Numerical integration

b−a

f (x) dx −

b−a

a

n

n

f (a + xi (b − a))
i=1

2m + 5 (b − a)m+1

4

(m + 1)!

(S − s)
( )

(m)

(m)

holds, where S := supa x b f (x), s := infa x b f (x) and for suitable x1 , x2 , . . . , xn . It is
worth noticing that n, m are arbitrary numbers. This means that the estimate in ( ) is more
accurate when m is large enough. Our approach is also elementary.
© 2010 Elsevier Ltd. All rights reserved.

1. Introduction
In recent years, a number of authors have considered error inequalities for some known and some new quadrature
formulas. Sometimes they have considered generalizations of these formulas, see [1–5] and their references therein where
the midpoint and trapezoid quadrature rules are considered.
In [6, Corollary 3] the following Simpson–Grüss type inequalities have been proved. If f : [a, b] → R is such that f (n−1)
is an absolutely continuous function and
f (n) (t )

γn

Γn ,

(a.e.) on [a, b]

for some real constants γn and Γn , then for n = 1, 2, 3, we have
b

f (t ) dt −
a

b−a

f (a) + 4f

6

a+b

+ f (b)

2

Cn (Γn − γn ) (b − a)n+1 ,

where
C1 =

∗

5
72

,

C2 =

1
162

,

C3 =

1
1152

.

Corresponding author at: Department of Mathematics, College of Science, Viêt Nam National University, Hà Nôi, Viet Nam.
E-mail addresses: (V.N. Huy), (Q.-A. Ngô).

0898-1221/$ – see front matter © 2010 Elsevier Ltd. All rights reserved.
doi:10.1016/j.camwa.2010.02.024

(1)


3046

V.N. Huy, Q.-A. Ngô / Computers and Mathematics with Applications 59 (2010) 3045–3052

In [2, Theorem 3], the following results obtained: Let I ⊂ R be an open interval such that [a, b] ⊂ I and let f : I → R be
a twice differentiable function such that f is bounded and integrable. Then we have
b

b−a

f (x) dx −
a

a+b

f

2

2

√
− 2−

3 (b − a)

+f

a+b
2

√
+ 2−

3 (b − a)

√

7−4 3

f

8

∞

( b − a) 3 .

(2)
√

In the above mentioned results, constants Cn in (1) and 7−48

3

in (2) are sharp in the sense that these cannot be replaced

by smaller ones. We may think the estimate in (1) involves the following six points xi , i = 1, 6 which will be called knots in
the sequel
a+

0 × ( b − a) < a +
x1

1

1

(b − a) = · · · = a +

2

(b − a) < a + 1 × (b − a) .

2

x2

x6

x5

While in (2), we have two knots x1 < x2 as following
a+

1
2

√
− 2−

× ( b − a) < a +

3

1
2

√
+ 2−

x1

( b − a) .

3

x2

On the other hand, as can be seen in both (1) and (2) the number of knots in the left hand side reflects the exponent of
b − a in the right hand side. This leads us to strengthen (1)–(2) by enlarging the number of knots (six knots in (1) and two
knots in (2)).
Before stating our main result, let us introduce the following notation
b

f (x) dx.

I (f ) =
a

Let 1

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


2



···


n


xj1 + xj2 + · · · + xjn =
,
j+1
···



n

−1
−1
−1

xm
+ xm
+ · · · + xm
= ,

n

1
2


m


xm + xm + · · · + xm = n .
1
2
n
m+1

Put
Q (f , n, m, x1 , . . . , xn ) =

n

b−a
n

f (a + xi (b − a)) .
i=1

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

f , 6, m, 0,

1 1 1 1

, , , ,1

Cm (Γm − γm ) (b − a)m+1 ,

2 2 2 2

m = 1, 3,

(3)

while (2) reads as follows
I (f ) − Q

f , 2, 2,

1
2

√
− 2−

3 ,

1
2

√

√

+ 2−

3

7−4 3
8

f

∞

(b − a)3 .

(4)

We are now in a position to state our main result.
Theorem 2. Let I ⊂ R be an open interval such that [a, b] ⊂ I and let f : I → R be a m-th differentiable function. Then we
have

|I (f ) − Q (f , n, m, x1 , . . . , xn )|
where S := supa

x b

2m + 5 (b − a)m+1

f (m) (x) and s := infa

4
x b

(m + 1)!
f (m) (x).

(S − s)

(5)


V.N. Huy, Q.-A. Ngô / Computers and Mathematics with Applications 59 (2010) 3045–3052

3047

Remark 3. It is worth noticing that the right hand side of (5) does not involve xi , i = 1, n and that m can be chosen arbitrarily.
This means that our inequality (5) is better in some sense, especially when b − a
1.
This work can be considered as a continued and complementary part to a recent paper [7]. More specifically, [7, Theorem
4] provides a similar estimate as (5). However, in contrast to the result presented here our estimate in (5) depends only on
the Lp -norm of f (m) (x). There is one thing we should mention here; both Theorem 2 presented here and Theorem 4 in [7]
are not optimal. This is because of the restriction of the technique that we use. It is better if we leave these to be solved by
the interested reader.
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 4 (See [8]). 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)

By a simple calculation, the remainder in (6) can be rewritten as
x −x 0

Rr −1 (f , x0 , x) =
0

(x − x0 − t )r −1 f (r ) (x0 + t )
dt
(r − 1)!

which helps us to deduce a similar representation of f as follows

u
uk (k)
(u − t )r −1 (r )
f (x + t ) dt .
f (x) +
k!
(r − 1)!
0
k=0
Before proving Theorem 2, we see that
r −1

f ( x + u) =

b

1
b−a

a

(b − x)m
dx
m!

b

f (m) (x) dx

=

a

(7)

(b − a)m (m−1)
(f
(b) − f (m−1) (a)).
(m + 1)!

Since
f (m−1) (b) − f (m−1) (a)

( b − a) s

( b − a) S

then

(b − a)m+1
s
(m + 1)!

b

1
b−a

a

(b − x)m
dx
m!

b

(b − a)m+1
S.
(m + 1)!

f (m) (x) dx

a

Besides,
1

n

b

n i=1

=

1

m−1
xm
i ( b − x)

(m − 1)!

a
n

xm
i

n i =1

b

dx

f (m) ((1 − xi )a + xi x) dx

a

(b − a)
m!

b

m

f (m) ((1 − xi )a + xi x) dx

.

a

Clearly,
b

( b − a) s

f (m) ((1 − xi )a + xi x) dx

(b − a) S

a

which implies that

(b − a)m+1
s
(m + 1)!

1

n

n i=1

b
a

m−1
xm

i ( b − x)

(m − 1)!

b

dx
a

f (m) ((1 − xi )a + xi x) dx

(b − a)m+1
S.
(m + 1)!


3048

V.N. Huy, Q.-A. Ngô / Computers and Mathematics with Applications 59 (2010) 3045–3052

Lemma 5 (Grüss Inequality, See [9]). Let f and g be two functions defined and integrable over [a, b]. Then we have
b

1
b−a
where s1

( b − a)

f ( x)

4

a

a

(S1 − s1 ) (S2 − s2 )

S2 for all x ∈ [a, b].

g ( x)

S1 and s2

1

g (x) dx

f (x) dx

2

a

b

b

1

f (x) g (x) dx −

Proof of Theorem 2. Denote
x

f (x) dx.

F (x) =
a

By the Fundamental Theorem of Calculus
I ( f ) = F ( b ) − F ( a) .
Applying Lemma 4 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

(b − t )m (m+1)
F
(t ) dt
m!

which yields
m

I (f ) =
k=1

(b − a)k (k)
F (a) +
k!

(b − t )m (m+1)
F
(t ) dt .
m!

b
a

Equivalently,
m−1

I (f ) =
k=0

For each 1

i

(b − a)k+1 (k)

f (a) +
(k + 1)!

b
a

(b − t )m (m)
f
(t ) dt .
m!

(8)

n, applying Lemma 4 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)) =

m−1

=
k=0

xki

(b − a) (k)
f (a) +
k!

0

k

xki (b − a)k (k)
f (a) +
k!
k=0

b −a
0

m−1

=

b
a

(xi (b − a) − t )m−1 (m)
f
(a + t ) dt
(m − 1)!

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

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

(9)

By applying (9) to i = 1, n and then summing up, we deduce that
n

xki (b − a)k (k)
f (a) +
k!
i=1
i=1 k=0
n

f (a + xi (b − a)) =
i=1

m−1

n

m−1

n

xki (b − a)k

i=1

=

k!

k=0

a

n

b
a

i =1

n (b − a)k (k)
f (a) +
(k + 1)!
k=0
i =1

m−1

=

f (k) (a) +

b

n

b
a

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

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

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

(10)

Thus,
m−1

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

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

n

i =1

b
a

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

Therefore, by combining (8) and (11), we get
b

I (f ) − Q (f , n, m, x1 , . . . , xn ) =
a

(b − x)m (m)
b−a
f (x) dx −
m!
n

n

i=1

b
a

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

(11)


V.N. Huy, Q.-A. Ngô / Computers and Mathematics with Applications 59 (2010) 3045–3052

=

(b − a)m (m−1)

(f
(b) − f (m−1) (a)) +
(m + 1)!
−
−
−

=

n

b−a
n

b
a

i=1
b

1
b−a
n

1

(m − 1)!
b

n i =1

b

dx

(m − 1)!

b

+
a

−
−

n

n

b
a

i=1
b

1
b−a

f (m) (x) dx

a

f (m) ((1 − xi )a + xi x) dx

n

b

a

(m − 1)!

a

i =1
b

m−1
xm
i (b − x)

(b − x)m
dx
m!

b

b

dx

f (m) ((1 − xi )a + xi x) dx

a

f (m) (x) dx

a

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

m−1
xm
i ( b − x)

(m − 1)!

a

b

a

(b − x)m (m)
1
f (x) dx −

m!
b−a

b−a

a

(b − x)m
dx
m!

f (m) ((1 − xi )a + xi x) dx

dx

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

b

a

m−1
xm
i (b − x)

a

a

(b − x)m (m)
1
f (x) dx −
m!
b−a

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

m−1
xm
i ( b − x)

a

b

3049

b

dx

f (m) ((1 − xi )a + xi x) dx

.

a

Then it follows from using the Grüss inequality that
b
a

(b − x)m (m)
1
f (x) dx −
m!
b−a

b
a

(b − x)m
dx
m!

b

f (m) (x) dx

a

1 (b − a)m+1
m!

4

(S − s)

and
n

b
a

i =1

1

=

m−1
xm
1
i ( b − x)
f (m) ((1 − xi )a + xi x) dx −
(m − 1)!
b−a
n

b

m−1
xm
i (b − x)

a

(m − 1)!

b

dx

f (m) ((1 − xi )a + xi x) dx

a

(b − a)m xm
i
(S − s)
(m − 1)!

4 i =1
n(b − a)m
1

4 (m − 1)!(m + 1)

(S − s).

We know that

(b − a)m+1
s
(m + 1)!

(b − a)m (m−1)
f
(b) − f (m−1) (a)
(m + 1)!

(b − a)m+1
s
(m + 1)!

1

(b − a)m+1
S
(m + 1)!

and
n

n i=1

b

m−1
xm
i ( b − x)

a

(m − 1)!

b

dx

f (m) ((1 − xi )a + xi x) dx

a

(b − a)m+1
S.
(m + 1)!

Therefore,
1
(b − a)m (m−1)
f
(b) − f (m−1) (a) −
n
(m + 1)!

n

i=1

b

a

m−1
xm
i (b − x)

(m − 1)!

b

dx

f (m) ((1 − xi )a + xi x) dx

a

(b − a)m+1
(S − s) .
(m + 1)!
Thus,

|I (f ) − Q (f , n, m, x1 , . . . , xn )|

1 (b − a)m+1

(b − a)m+1
(b − a)m+1
(S − s) +
(S − s)
4

m!
4 (m − 1)!(m + 1)
(m + 1)!
1
1
1
(b − a)m+1
+
+
=
(S − s)
4m
4 (m + 1)
m (m + 1)
(m − 1)!
2m + 5 (b − a)m+1
=
( S − s)
4
(m + 1)!
(S − s) +

1


3050

V.N. Huy, Q.-A. Ngô / Computers and Mathematics with Applications 59 (2010) 3045–3052

where

S := sup f (m) (x)

and s := inf f (m) (x)
a x b

a x b

which completes our proof.
3. Examples
In this section, by applying our main theorem, we will obtain some new inequalities which cannot be easy obtained
by [2,3]. Actually, our result covers several known results in the numerical integration.
Example 6. Assume n = 6, m = 1, 2, or 3. Clearly x1 = 0, x2 = x3 = x4 = x5 =
system

1
,
2

and x6 = 1 satisfy the following linear


6


x1 + x2 + · · · + x6 = ,


2




···


6
j
j
j
,
x1 + x2 + · · · + x6 =

j
+
1



···




xm + xm + · · · + xm = 6 .
1
2
6
m+1
Therefore, we obtain the following inequalities
b

b−a

f (t ) dt −
a

where Sm = supa
C1 =

7
8

x b

,

a+b

f (a) + 4f

6

f (m) (x) and sm = infa

C2 =

9
24

,

C3 =

+ f (b)

2

11
96

x b

Cm (Sm − sm ) (b − a)m+1

(12)

f (m) (x) and

.

Clearly, the left hand side of (12) is similar to the Simpson rule.
Example 7. Assume n = 3, m = 3. By solving the following linear system


3

x1 + x2 + x3 = ,



2



3
2
2
2
x1 + x2 + x3 = ,

3




x3 + x3 + x3 = 3 ,
1
2
3
4

we obtain {x1 , x2 , x3 } is a permutation of

√
1
2

,1 −

1
2

1±

2
2

√
,

1

2

1±

2

.

2

Therefore, we obtain the following inequalities

√

b

f (x) dx −
a

b−a

3

f

a+

1−

1
2

2

1±

2

( b − a)

√
+ f

a+

where S = supa

1
2

x b

(b − a) + f

a+

f (x) and s = infa

1
2

x b

1±

2
2

( b − a)

f (x).

Example 8. If n = 2, m = 2, then by solving the following system


2

x1 + x2 = ,
2



 x2 + x2 = 2 ,
1
2
3

11(b − a)4
96

(S − s)

(13)


V.N. Huy, Q.-A. Ngô / Computers and Mathematics with Applications 59 (2010) 3045–3052

3051

we obtain

√
(x1 , x2 ) =

1

±

2

√

3 1
6

,

2

3

∓

6

.

We then obtain a similar 2-point Gaussian quadrature rule

√

b

f (x) dx −

b−a
2

a

where S = supa

x b

f

a+

f (x) and s = infa

1
2

−

x b

√

3

(b − a) + f

6

a+

1
2

+

3
6

(b − a)

9(b − a)3
24

( S − s)

(14)

f (x).

Remark 9. Note that using (13) provides a better result than using (14) (or the 2-point Gaussian quadrature rule). For
example, let us consider the following function f (x) = xesin x . Then
1

f (x) dx ≈ 0.9291567730.
0

If we use (13), we then have

√

1

1

f (x) dx ≈

f

3

0

1

−

2

√
2

4

1

+f

+f

2

1

f

2

+

2
4

≈ 0.9301849429.

If we use (14), we then have

√

1

1

f (x) dx ≈

f

2

0

1

−

2

√
3

6

+f

1

f

2

+

3

≈ 0.9319357678.

6

Example 10. If m = 2 and n = 3, then by solving the following system


3

 x1 + x2 + x3 = ,
2


 x2 + x2 + x2 = 3 ,
1
2
3
3

we obtain {x1 , x2 , x3 } is a permutation of t ,

3
2

− t − k, k , where k is a solution of the following algebraic equation

8x2 + (8t − 12) x + 8t 2 − 12t + 5 = 0
with

√
1

t ∈

2

−

√

6 1

,

6

2

+

6
6

.

We then obtain
b

f (x) dx −
a

where S = supa

x b

b−a
3

f (a + t (b − a)) + f

f (x) and s = infa

x b

a+

3
2

− t − k (b − a) + f (a + k (b − a))

9(b − a)3
24

(S − s)

f (x).

Acknowledgements
The authors wish to express their gratitude to the anonymous referees for a number of valuable comments and
suggestions.
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