General Mathematics Vol. 13, No. 4 (2005), 51–64
Error Inequalities for a Generalized
Quadrature Rule
Nenad Ujević
Dedicated to Professor Dumitru Acu on his 60th anniversary
Abstract
Error inequalities for a generalized quadrature rule are derived.
A summation formula for the special function Si(x) is given.
2000 Mathematics Subject Classification: 26D15, 65D30.
Keywords: quadrature rule, generalization, numerical integration, error
bounds.
1
Introduction
In recent years a number of authors have considered generalizations of some
known and some new quadrature rules. For example, generalizations of the
trapezoid, mid-point and Simpson’s quadrature rules are considered in [1],
51
52
Nenad Ujević
[2], [3], [5] and [9]. As an illustration we give a generalization of the midpoint quadrature rule (see [3]),
Z
b
f (t)dt =
a
n−1
X
k=0
(b − a)k+1 (k) a + b
1 + (−1)
f (
)+(−1)n
k+1
2 (k + 1)!
2
Z
b
k
where


Kn (t) =

(t−a)n
,
n!
n
(t−b)
,
n!
Kn (t)f (n) (t)dt,
a
t ∈ a, a+b
2
.
t ∈ a+b
,
b
2
For n = 1 we get the mid-point rule
Z b
Z b
a+b
f (t)dt = (b − a)f (
K1 (t)f 0 (t)dt.
)−
2
a
a
In this paper we consider a generalization of a simple quadrature rule of
open type which has the form
Zb
(1)
3a + b
a + 3b
b−a
f(
) + f(
) + R(f ).
f (t)dt =
2
4
4
a
In [14] it is shown that the above 2-point quadrature rule of open type is
optimal with respect to a given way of estimation of the remainder term
(error) R(f ). We have (see [14])
(2)
|R(f )| ≤
Γ−γ
(b − a)2 ,
16
where γ ≤ f 0 (t) ≤ Γ, t ∈ [a, b].
On the other hand, the well-known 2-point Gauss quadrature rule
(3)
"
#
√
√
Zb
b−a
3
a+b
3
a+b
f (t)dt =
−
(b − a)) + f (
+
(b − a)) +R1 (f )
f(
2
2
6
2
6
a
Error Inequalities for a Generalized Quadrature Rule
53
has the estimation of the error (see [10])
(4)
√
Γ−γ
(5 − 2 3)(b − a)2 .
24
√
1
= 0.062 5 < 24 (5 − 2 3) = 0.063 99 we conclude that (2) is
|R1 (f )| ≤
Since
1
16
better than (4).
In [14] we can also find various error inequalities for this rule. Here we
also give various error bounds for the generalization of this rule. These error
bounds are generalizations of the error bounds obtained in [14] and they are
similar to error bounds obtained in [15].
Finally, we give a numerical example. In fact, we derive a summation
Rx
formula for the special function Si(x) = 0 sint t dt.
2
Main results
Lemma 1. Let f : [a, b] → R be a function such that f (n−1) is absolutely
continuous. Then
Z b
b−a
3a + b
a + 3b
f (x)dx =
(5)
f(
) + f(
) −
2
4
4
a
m
X
(b − a)2i+1
(2i) 3a + b
(2i) a + 3b
f (
)+f (
) + R(f ),
−2
42i+1 (2i + 1)!
4
4
i=1
where m = n−1
, the integer part of (n − 1)/2,
2
Z b
n
Sn (t)f (n) (t)dt
(6)
R(f ) = (−1)
a
and
(7)





Sn (t) =




t ∈ a, 3a+b
4
a+b n
1
3a+b a+3b
.
t
−
,
,
t
∈
n!
2
4
4
1
(t − b)n ,
t ∈ a+3b
,b
n!
4
1
(t
n!
− a)n ,
54
Nenad Ujević
Proof. We prove (5) by induction. First we note that





S1 (t) =
t − a,
t−




t ∈ a, 3a+b
4
t ∈ 3a+b
, a+3b
4
4
a+3b t ∈ 4 ,b
a+b
,
2
t − b,
is a Peano kernel for the quadrature rule of open type, that is, we have
Z
b
−
a
f ( 3a+b
) + f ( a+3b
)
4
4
S1 (t)f (t)dt = −
(b − a) +
2
Z
b
0
f (t)dt.
a
We easily show that (5) holds for n = 2. Now suppose that (5) holds for an
arbitrary n. We have to prove that (5) holds for n → n + 1. To simplify
the proof we introduce the notations
(8)
(9)
Pn (t) =
1
Qn (t) =
n!
(t − a)n
,
n!
a+b
t−
2
n
,
(t − b)n
Rn (t) =
.
n!
(10)
We see that Pn , Qn and Rn form Appell sequences of polynomials, that is
Pn0 (t) = Pn−1 (t), Q0n (t) = Qn−1 (t), Rn0 (t) = Rn−1 (t),
P0 (t) = Q0 (t) = R0 (t) = 1.
We have
Z
b
n+1
(−1)
a
Sn+1 (t)f (n+1) (t)dt
Error Inequalities for a Generalized Quadrature Rule
Z
= (−1)
Z
(−1)n+1
Z
3a+b
4
n+1
Pn+1 (t)f
(n+1)
n+1
(t)dt + (−1)
Qn+1 (t)f (n+1) (t)dt
Rn+1 (t)f (n+1) (t)dt
a+3b
4
= (−1)
a+3b
4
3a+b
4
a
b
55
n+1
3a + b (n) 3a + b
(n)
)f (
) − Pn+1 (a)f (a)
Pn+1 (
4
4
a + 3b (n) a + 3b
3a + b (n) 3a + b
+(−1)
Qn+1 (
)f (
) − Qn+1 (
)f (
)
4
4
4
4
a + 3b (n) a + 3b
n+1
(n)
+(−1)
Rn+1 (b)f (b) − Rn+1 (
)f (
)
4
4
n+1
Z
+(−1)
Z
+(−1)
3a+b
4
n
Z
Pn (t)f
(n)
(t)dt + (−1)
a
b
n
a+3b
4
Z
n
n
a+3b
4
3a+b
4
Qn (t)f (n) (t)dt
Rn (t)f (n) (t)dt
b
Sn (t)f (n) (t)dt
a
3a + b
3a + b
3a + b
n+1
+(−1)
Pn+1 (
) − Qn+1 (
) f (n) (
)
4
4
4
a + 3b
a + 3b
a + 3b
n+1
+(−1)
Qn+1 (
) − Rn+1 (
) f (n) (
)
4
4
4
= (−1)
Z b
f ( 3a+b
) + f ( a+3b
)
4
4
= −
(b − a) +
f (t)dt
2
a
m
X
2(b − a)2i+1
(2i) a + 3b
(2i) 3a + b
)+f (
)
+
f (
2i+1 (2i + 1)!
4
4
4
i=1
+(−1)
n+1
3a + b
3a + b
3a + b
) − Qn+1 (
) f (n) (
)
Pn+1 (
4
4
4
56
Nenad Ujević
n+1
+(−1)
a + 3b
a + 3b
a + 3b
) − Rn+1 (
) f (n) (
)
Qn+1 (
4
4
4
Z b
f ( 3a+b
) + f ( a+3b
)
4
4
= −
(b − a) +
f (t)dt
2
a
m1
X
2(b − a)2i+1
(2i) 3a + b
(2i) a + 3b
+
f (
)+f (
)
42i+1 (2i + 1)!
4
4
i=1
where m1 =
n
, since
3a + b
3a + b
3a + b
n+1
+(−1)
Pn+1 (
) − Qn+1 (
) f (n) (
)
4
4
4
2
a + 3b
a + 3b
a + 3b
+(−1)
Qn+1 (
) − Rn+1 (
) f (n) (
)
4
4
4
(n) 3a + b
(b − a)n+1 n+1
(n) a + 3b
= n+1
1 − (−1)
f (
)+f (
) .
4 (n + 1)!
4
4
n+1
This completes the proof.
Lemma 2. The Peano kernels Sn (t), n > 1, satisfy:
Z
b
(11)
Sn (t)dt = 0, if n is odd,
a
Z
b
|Sn (t)| dt =
(b − a)n+1
,
4n (n + 1)!
max |Sn (t)| =
(b − a)n
.
4n n!
(12)
a
(13)
t∈[a,b]
Proof. A simple calculation gives
Z
b
Sn (t)dt =
a
2(b − a)n+1 1 − (−1)n+1 .
n+1
4 (n + 1)!
Error Inequalities for a Generalized Quadrature Rule
57
From the above relation we see that (11) holds, since 1 − (−1)n+1 = 0 if n
is odd.
We now consider some properties of the Appell sequences of polynomials
Pn (t), Qn (t) and Rn (t), given by (8), (9) and (10), respectively. We have
that (t−a)n ≥ 0, for each n, such that Pn (t) ≥ 0, ∀n and t ∈ a, 3a+b
. Since
2
Pn0 (t) = Pn−1 (t) we conclude that Pn (t) are increasing functions. If n is even
n
n
a+3b
then t − a+b
≥ 0. If n is odd then t − a+b
≥ 0, for t ∈ a+b
and
2
2
2
4
n
t − a+b
≤ 0, for t ∈ 3a+b
, a+b
.
2
4
2
Since Q0n (t) = Qn−1 (t) we conclude that Qn (t) is increasing function if
, a+b
, while Qn (t) is
n is even and Qn (t) is decreasing function for t ∈ 3a+b
4
2
a+b a+3b increasing function for t ∈ 2 4 , if n is odd.
We have that (t − b)n ≤ 0 if n ≥ 1, n is odd and (t − b)n ≥ 0 if n ≥ 0,
n is even. Thus, we have that Rn (t) ≤ 0 if n is odd and Rn (t) ≥ 0 if n is
even. As we know Rn0 (t) = Rn−1 (t) such that Rn (t) are decreasing functions
if n is even and Rn (t) are increasing functions if n is odd. We use these
properties to prove (12) and (13).
We have
Z
Z
b
3a+b
4
|Sn (t)| dt =
a
Z
|Pn (t)| dt +
a
=
(b − a)n+1
.
4n (n + 1)!
a+3b
4
3a+b
4
Z
|Qn (t)| dt +
b
a+3b
4
|Rn (t)| dt
58
Nenad Ujević
Finally, we have
(
|Qn (t)| , max |Rn (t)|
t∈[ a+3b
,b]
4
3a + b
3a + b
a + 3b
a + 3b
= max Pn (
) , Qn (
) , Qn (
) , Rn (
)
4
4
4
4
(b − a)n
=
.
4n n!
max |Sn (t)| = max
t∈[a,b]
)
max |Pn (t)| ,
t∈[a, 3a+b
4 ]
max
t∈[ 3a+b
, a+3b
4
4 ]
We introduce the notations
Z
b
I=
f (t)dt,
a
) + f ( a+3b
)
f ( 3a+b
4
4
(b − a)
2
m
X
2(b − a)2i+1
(2i) 3a + b
(2i) a + 3b
+
f (
)+f (
) .
42i+1 (2i + 1)!
4
4
i=1
F = −
Theorem 3. Let f : [a, b] → R be a function such that f (n−1) , n > 1, is
absolutely continuous and there exist real numbers γn , Γn such that γn ≤
f (n) (t) ≤ Γn , t ∈ [a, b]. Then
(14)
|I − F | ≤
1 Γn − γn 1
(b − a)n+1 if n is odd
2 (n + 1)! 4n
and
(b − a)n+1 n (n)
|I − F | ≤ n
f
4 (n + 1)!
(15)
∞
if n is even.
Proof. Let n be odd. From (6) and (11) we get
Z
Z
b
n
R(f ) = (−1)
Sn (t)f
a
(n)
b
n
(t)dt = (−1)
a
γn + Γ n
(n)
Sn (t) f (t) −
dt
2
Error Inequalities for a Generalized Quadrature Rule
59
such that we have
(16)
|R(f )| = |I − F | ≤ max f
(n)
t∈[a,b]
γn + Γ n
(t) −
2
Z
b
|Sn (t)| dt.
a
We also have
max f (n) (t) −
(17)
t∈[a,b]
γn + Γ n
Γ n − γn
≤
.
2
2
From (16), (17) and (12) we get
|I − F | ≤
1 Γ n − γn 1
(b − a)n+1 .
2 (n + 1)! 4n
Let n be even. Then we have
Z b
|R(f )| = |I − F | ≤
|Sn (t)| dt f (n)
a
∞
=
(b − a)n+1 n (n)
f
4n (n + 1)!
∞
.
Theorem 4. Let f : [a, b] → R be a function such that f (n−1) , n > 1, is
absolutely continuous and let n be odd. If there exists a real number γn such
that γn ≤ f (n) (t), t ∈ [a, b] then
(18)
|I − F | ≤ (Tn − γn )
(b − a)n+1
,
4n n!
where
Tn =
f (n−1) (b) − f (n−1) (a)
.
b−a
If there exists a real number Γn such that f (n) (t) ≤ Γn , t ∈ [a, b] then
(b − a)n+1
.
|I − F | ≤ (Γn − Tn )
4n n!
(19)
Proof. We have
Z
b
|R(f )| = |I − F | =
a
(f (n) (t) − γn )Sn (t)dt ,
60
Nenad Ujević
since (11) holds. Then we have
Z
Z
b
(f
(n)
(t) − γn )Sn (t)dt
b
≤ max |Sn (t)|
t∈[a,b]
a
(f (n) (t) − γn )dt
a
(b − a)n (n−1)
(n−1)
=
f
(b)
−
f
(a)
−
γ
(b
−
a)
n
4n n!
(b − a)n+1
=
(Tn − γn ) .
4n n!
In a similar way we can prove that (19) holds.
Remark 5. Note that we can apply the estimations (14) and (15) only
if f (n) is bounded. On the other hand, we can apply the estimation (18)
if f (n) is unbounded above and we can apply the estimation (19) if f (n) is
unbounded below.
3
A numerical example
Here we consider the integral (special function) Si(x) =
Rx
0
sin t
dt
t
and apply
the summation formula (5) to this integral. We get the summation formula
Si(x) = F (x) + R(x), where
m
2x2i+1
3x X
x 2
(2i) x
(2i) 3x
(20) F (x) = 2 sin + sin
+
f ( )+f ( )
4 3
4
42i+1 (2i + 1)!
4
4
i=1
and f (t) = (sin t)/t. We calculate the derivatives f (j) (t) as follows. We
have
(j)
(g(t)h(t))
=
j
X
k=0
j
k
g (k) (t)h(j−k) (t).
Error Inequalities for a Generalized Quadrature Rule
61
If we choose g(t) = sin t and h(t) = 1/t then we get
x
f (j) ( ) =
4
j−1
[X
2 ]
j
2i+1
(−1)j−i+1
i=0
+
[ 2j ]
X
j
2i
(−1)j−i
i=0
f
(j)
3x
( ) =
4
j−1
[X
2 ]
j
2i+1
(j − 2i)!4j−2i+1
x
sin ,
j−2i+1
x
4
j−i+1 (j
(−1)
i=0
+
[ 2j ]
X
i=0
j
2i
x
(j − 2i − 1)!4j−2i
cos
xj−2i
4
(−1)j−i
− 2i − 1)!4j−2i
3x
cos
j−2i
j−2i
3
x
4
3x
(j − 2i)!4j−2i+1
sin .
j−2i+1
j−2i+1
3
x
4
We now compare the summation formula (20) with the known compound
formula (for the given quadrature rule of open type),
Z x
n−1 hX
3xi + xi+1
xi + 3xi+1
(21)
f (t)dt =
f(
) + f(
) + R(x),
2 i=0
4
4
0
where xi = ih, h = x/n, f (t) = (sin t)/t.
Let us choose x = 1. The ”exact” value is Si(1) = 0.946083070367. If
we use (20) with m = 5 then we get Si(1) ≈ 0.946083070363. If we use (21)
with n = 40000 then we get Si(1) ≈ 0.946083070369. All calculations are
done in double precision arithmetic. The first approximate result is obtained
much faster than the second approximate result. The same is valid if we use
some quadrature rule of higher order, for example Simpson’s rule. This is
a consequence of the fact that we have to calculate the function sin t many
times when we apply the compound formula and we have only to calculate
sin(x/4), cos(x/4), sin(3x/4) and cos(3x/4) when we apply the summation
formula.
62
Nenad Ujević
Similar summation formulas can be obtained for the integrals (special
Rx
Rx
Rx
functions): 0 [(et − 1)/t] dt, 0 [(cos t − 1)/t] dt, 0 exp(−t2 )dt, etc.
References
[1] G. A. Anastassiou, Ostrowski Type Inequalities, Proc. Amer. Math.
Soc., Vol 123, No 12, (1995), 3775–3781.
[2] P. Cerone and S. S. Dragomir, Midpoint-type Rules from an Inequalities Point of View, Handbook of Analytic-Computational Methods in
Applied Mathematics, Editor: G. Anastassiou, CRC Press, New York,
(2000), 135–200.
[3] P. Cerone and S. S. Dragomir, Trapezoidal-type Rules from an Inequalities Point of View, Handbook of Analytic-Computational Methods in
Applied Mathematics, Editor: G. Anastassiou, CRC Press, New York,
(2000), 65–134.
[4] D. Cruz-Uribe and C. J. Neugebauer, Sharp error bounds for the trapezoidal rule and Simpson’s rule, J. Inequal. Pure Appl. Math., 3(4),
Article 49, (2002), 1–22.
[5] Lj. Dedić, M. Matić and J. Pečarić, On Euler trapezoid formulae, Appl.
Math. Comput., 123 (2001), 37–62.
[6] S. S. Dragomir, P. Cerone and J. Roumeliotis, A new generalization of Ostrowski’s integral inequality for mappings whose derivatives
Error Inequalities for a Generalized Quadrature Rule
63
are bounded and applications in numerical integration and for special
means, Appl. Math. Lett., 13 (2000), 19–25.
[7] A. R. Krommer, C. W. Ueberhuber, Computational integration, SIAM,
Philadelphia, 1998.
[8] D. S. Mitrinović, J. Pečarić and A. M. Fink, Classical and new inequalities in analysis, Kluwer Acad. Publ., Dordrecht/Boston/London,
1993.
[9] C. E. M. Pearce, J. Pečarić, N. Ujević and S. Varošanec, Generalizations of some inequalities of Ostrowski-Grüss type, Math. Inequal.
Appl., 3(1), (2000), 25–34.
[10] N. Ujević, Inequalities of Ostrowski-Grüss type and applications, Appl.
Math., 29(4), (2002), 465–479.
[11] N. Ujević, A generalization of Ostrowski’s inequality and applications
in numerical integration, Appl. Math. Lett., 17(2), (2004), 133-137.
[12] N. Ujević, New bounds for the first inequality of Ostrowski-Gruss type
and applications, Comput. Math. Appl., 46, (2003), 421-427.
[13] N. Ujević and A. J. Roberts, A corrected quadrature formula and applications, ANZIAM J., 45(E), (2004), E41–E56.
[14] N. Ujević, Error inequalities for an optimal 2-point quadrature formula
of open type, (to appear in ”Inequality Theory and Applications”, Nova
Science Publishers, Inc., New York).
64
Nenad Ujević
[15] N. Ujević, Error inequalities for a generalized trapezoid rule, Appl.
Math. Lett, (in press).
Department of Mathematics
University of Split
Teslina 12/III, 21000 Split
CROATIA