DSpace at VNU: New inequalities of Simpson-like type involving n knots and the mth derivative

Mathematical and Computer Modelling 52 (2010) 522–528

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Mathematical and Computer Modelling
journal homepage: www.elsevier.com/locate/mcm

New inequalities of Simpson-like type involving n knots and
the mth derivative
´
Vu Nhat Huy a , Quôc-Anh
Ngô a,b,∗
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 9 May 2009
Received in revised form 4 March 2010
Accepted 22 March 2010

Based on recent results due to Nenad Ujević, we obtain some new inequalities of Simpsonlike type involving n knots and the mth derivative where n, m are arbitrary numbers. Our
method is also elementary.
© 2010 Elsevier Ltd. All rights reserved.

Keywords:
Inequality
Error
Integral
Taylor
Simpson

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, for example, the Simpson inequality (which
gives an error bound for the well-known Simpson rule) is considered in [1–10]. In [5], we can find the inequality,
b

f (t ) dt −

b−a

a

f (a) + 4f

6

a+b
2

+ f ( b)

Γ −γ
12

( b − a) 2 ,

(1)

where Γ , γ are real numbers, such that γ < f (t ) < Γ , t ∈ [a, b]. We define the Chebyshev functional,
T (f , g ) =

b

1
b−a

f (t ) g (t ) dt −
a

b

1

( b − a) 2

b

g (t ) dt .

f (t ) dt
a

a

Then
T (f , f ) =

1
b−a

f

2
L2

−

(b − a)2

2

b

1

f (t ) dt

.

a

We also define

σ (f ) = (b − a)T (f , f ).
In [10], the author proved the following result.

∗

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).

0895-7177/$ – see front matter © 2010 Elsevier Ltd. All rights reserved.
doi:10.1016/j.mcm.2010.03.049

(2)


V.N. Huy, Q.-A. Ngô / Mathematical and Computer Modelling 52 (2010) 522–528

523

Theorem 1 (See [10], Theorem 1). Let f : [a, b] → R be an absolutely continuous function, whose derivative f ∈ L2 [(a, b)].
Then
b

f (t ) dt −

b−a
6

a

3

a+b

f (a) + 4f

( b − a) 2

+ f (b)

2

6

σ (f ),

where σ (·) is defined by (2). Inequality (3) is sharp in the sense that the constant

1
6

(3)

cannot be replaced by a smaller one.

Since the constant 16 in (3) is sharp, in order to strengthen (3) we have to replace the exponent 32 on the right-hand side
of (3). This leads us to strengthen (3) by enlarging the number of knots (6 knots in (3)) and replacing f in (3) (see [11] for

more details). 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


xj1 + xj2 + · · · + xjn =
,
j+1

···



n

m−1
m−1
m−1


 x1 + x2 + · · · + xn = m ,



 xm + xm + · · · + xm = n .
1
2
n
m+1

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

b−a
n

n

f (a + xi (b − a)) .

i =1

Remark 2. With the above notations, inequality (3) reads as follows
I (f ) − Q

f , 6, 1, 0,

3

1 1 1 1

(b − a) 2

2 2 2 2

6

, , , ,1

σ (f ).

(4)

We are now in a position to state our main result. Precisely, we shall apply the Fundamental Theorem of Calculus, Taylor’s
formula and the Hölder inequality to establish the following result.
Theorem 3. Let I ⊂ R be an open interval such that [a, b] ⊂ I and let f : I → R be an m-times differentiable function such
that f (m) ∈ L2 (a, b). Then we have

|I (f ) − Q (f , n, m, x1 , ., xn ) |

√

1

2m + 1

+√

1

2m − 1

1

(b − a)m+ 2
m!

σ f (m) .

(5)

This work can be considered as a continued and complementary part to our recent papers [11–13].
Remark 4. 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. However, the constant

√

1

2m + 1

+√

1

2m − 1

1
m!

in the inequality (5) is not sharp. 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.


524

V.N. Huy, Q.-A. Ngô / Mathematical and Computer Modelling 52 (2010) 522–528

Lemma 5 (See [14]). 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−x0

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
r −1

f (x + u) =

uk (k)
f (x) +
k!
k=0

u
0

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

(7)

Proof of Theorem 3. Define
x

f (t ) dt .

F (x) =
a

By Fundamental Theorem of Calculus
I ( f ) = F ( b ) − F ( a) .
Applying Lemma 5 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
m

I (f ) =
k=1

(b − a)k (k)
F (a) +
k!

b −a
0

(b − a − t )m (m+1)
F
(a + t ) dt .

m!

Equivalently,
m−1

I (f ) =
k=0

For each 1

i

(b − a)k+1 (k)
f (a) +
(k + 1)!

b −a
0

(b − a − t )m (m)
f
(a + t ) dt .
m!

n, applying Lemma 5 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

=

b −a
0

(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)!

(8)


V.N. Huy, Q.-A. Ngô / Mathematical and Computer Modelling 52 (2010) 522–528

525

By applying (8) to i = 1, n and then summing, we deduce that
xki (b − a)k (k)
f ( a) +
k!
i=1 k=0
i=1
m−1

n

n

f (a + xi (b − a)) =
i =1

n

n

xki (b − a)k

m−1

i =1

=

k!

k=0
m−1

=
k=0

f (k) (a) +

n ( b − a)

(k + 1)!

n

f (k) (a) +

0

n
0

b−a

i=1

0

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

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

b−a

i=1
k

b −a

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

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

Therefore,
b −a

|I (f ) − Q (f , n, m, x1 , . . . , xn )| =
0

b−a

(b − a − t )m (m)
f
(a + t ) dt −
m!
n

n

b−a

×
i=1

0

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

b−a
(b − x)m (m)
f
(x) dx −
m!
n

b

=
a
n

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

b

×
i=1

a

which yields
b

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

−

b−a
n

b−a
b

a

−

n

b

m−1
xm
i (b − x)

(m − 1)!

a

i =1
b

1

−

+

1
(b − x)m (m)
f
( x) −
m!

b−a

n

f (m) ((1 − xi ) a + xi x)

a

n

b
a

i =1
b

1
b−a

f (m) (t ) dt dx

a

f (m) (t ) dt dx

1
(b − x)m (m)
f
( x) −
m!

b−a

b−a

b

m−1
xm
i ( b − x)

(m − 1)!

b

f (m) (t ) dt dx

a

f (m) ((1 − xi ) a + xi x)

f (m) (t ) dt dx .

a

We note by the Hölder inequality that
b
a

1
(b − x)m (m)

f
(x) −
m!
b−a
b
a

(b − x)m
m!

b

1
2

2

f (m) (t ) dt dx

a
b

f

dx
a

(m)

( x) −

b−a

f
a

1
2

2

b

1

(m)

(t ) dt

dx

.

(9)


526

V.N. Huy, Q.-A. Ngô / Mathematical and Computer Modelling 52 (2010) 522–528

We now compare the last integral on the right-hand side of the above inequality with
b

(m)

f

(x) −

a
b

f (m) (x) −

=

f

b−a

(m)

=

f

(m)

(x)

2

(t ) dt

dx

a

− 2f

f (m−1) (b) − f (m−1) (a)

(x)

=

f

2

(m)

(x) dx − 2

f (m−1) (b) − f (m−1) (a)

=

f

f (m−1) (b) − f (m−1) (a)

2

(m)

(x) dx − 2

b

f (m) (x)

=

2

dx −

f (m−1) (b) − f (m−1) (a)

=

b−a

b−a

√
=

b

1

b−a

f

(m)

f

(m)

b

b−a

(x) dx +

f (m−1) (b) − f (m−1) (a)
b−a

2

f (m−1) (b) − f (m−1) (a)
b−a

1
2

2

dx

1
2

2

(b − a)

1
2

2

f (m−1) (b) − f (m−1) (a)

2

(x) dx −

2

1
2

2

1

2

b−a

a

1

dx

b−a

a

√

b
a

+

1
2

2

b−a

a

b−a

a

(m)

f

b−a

a
b

b

f (m−1) (b) − f (m−1) (a)

+

b−a

a
b

dx

b−a
(m)

1

2

2

f (m−1) (b) − f (m−1) (a)

a
b

1
2

2

b

1

σ (f (m) ). More precisely, one has

b

1

2

(x) dx −

f

(b − a)2

a

(m)

(x) dx

a

σ f (m) .

=
Hence,

1
(b − x)m (m)
f
(x) −
m!
b−a

b
a

1

b

(b − a)m+ 2

σ f (m) .
√
m! 2m + 1

f (m) (t ) dt dx

a

(10)

Again by the Hölder inequality, one obtains
b

m−1
xm
i (b − x)

(m − 1)!

a

f (m) ((1 − xi ) a + xi x) −

m−1
xm
i ( b − x)

b

b

dx

(m − 1)!

a

1
2

2

f

(m)

b

1
b−a

f (m) (t ) dt dx

a

((1 − xi ) a + xi x) −

a

b−a

1
2

2

b

1

f

(m)

(t ) dt

dx

a

Clearly,
b

m−1
xm
i (b − x)

1
2

2

(m − 1)!

a

=

dx

1
2

b

xm
i

(b − x)

2m−2

(m − 1)!

dx

a

and
b

f

(m)

((1 − xi ) a + xi x) −

a

=

(1−xi )a+xi b

1
xi

f

(y) −

a
b

1
xi

(m)

b−a

f
a

(m)

(y) −

b−a

f
a

(t ) dt

(m)

dx
1
2

2

b

f

(m)

(t ) dt

a
1
2

2

b

1

f

(m)

a

1
b−a

1
2

2

b

1

(t ) dt

dy

dy

=

1

(b − a)m− 2
,
√
(m − 1)! 2m − 1
xm
i

.


V.N. Huy, Q.-A. Ngô / Mathematical and Computer Modelling 52 (2010) 522–528
b

1

f (m) (y) −

= √

xi

a

1

1

σ f (m)

= √

xi

xi

b

1
b−a

2

f (m) (t ) dt

527

1
2

dy

a

σ f (m) .

Therefore,
b

m−1
xm
i (b − x)

f (m) ((1 − xi ) a + xi x) −

(m − 1)!

a

b

1
b−a

f (m) (t ) dt dx

a

1

−1
xm
i

(b − a)m− 2
σ f (m) .
√
(m − 1)! 2m − 1

Thus,
n

b

i =1

m−1
xm
i ( b − x)

(m − 1)!

a

f (m) ((1 − xi ) a + xi x) −

b

1
b−a

f (m) (t ) dt dx

a

1

n (b − a)m− 2

σ f (m) .

√

m! 2m − 1

(11)

Combining (10) and (11) gives
1

|I (f ) − Q (f , n, m, x1 , ., xn ) |

1

b − a n (b − a)m− 2
(b − a)m+ 2
+
√
√
n m! 2m − 1
m! 2m + 1

σ f (m) ,

or equivalently,

|I (f ) − Q (f , n, m, x1 , ., xn ) |

√

1

2m + 1

+√

1

2m − 1

1

(b − a)m+ 2
m!

σ f (m)

which completes the proof.
3. Examples
In this section, by applying our main theorem, we will obtain some new inequalities which cannot be easy obtained
from [10].
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

f (t ) dt −
a

b−a
6

f (a) + 4f

a+b
2

+ f (b)

√

1

2m + 1

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

+√

1

2m − 1

1

(b − a)m+ 2
m!

σ f (m) .


528

V.N. Huy, Q.-A. Ngô / Mathematical and Computer Modelling 52 (2010) 522–528

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

√
1
2

,1 −

1
2

√

2

1±

,

2

1
2

2

1±

.

2

Therefore, we obtain the following inequalities

√

b

f (x) dx −

b−a

f

3

a

a+

1−

1

1±

2

2
2

( b − a)
√

+ f

a+

1
2

(b − a) + f

a+

1−

1
2

1±

2
2

√

√

7+

( b − a)

√

5

6 35

7

(b − a) 2

σ (f ).

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


2

x1 + x2 = ,
2


 x2 + x2 = 2 ,
1
2
3

we obtain

√
1

(x1 , x2 ) =

2

√

3 1

±

6

,

2

∓

3
6

.

We then obtain

√

b

f (x) dx −
a

√

b−a
2

f

a+

1
2

−

3
6

√
(b − a) + f

a+

1
2

+

3
6

( b − a)

√

3+

√

5

2 15

5

(b − a) 2

σ (f ).

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