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Error bounds for weighted 2-point and 3-point Radau and Lobatto quadrature rules for functions of bounded variation
Mathematical and Computer Modelling 54 (2011) 1365–1379
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Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm
Error bounds for weighted 2-point and 3-point Radau and Lobatto quadrature rules for functions of bounded variation A. Aglić Aljinović a,∗ , S. Kovač b , J. Pečarić c a
Department of Applied Mathematics, Faculty of Electrical Engineering and Computing, University of Zagreb, Unska 3, 10000 Zagreb, Croatia
b
Faculty of Geotechnical Engineering, University of Zagreb, Hallerova aleja 7, 42000 Varaždin, Croatia
c
Faculty of Textile Technology, University of Zagreb, Pierottijeva 6, 10000 Zagreb, Croatia
article
abstract
info
Article history: Received 6 February 2010 Received in revised form 12 December 2010 Accepted 6 April 2011
We present a weighted generalization of Montgomery identity for Riemann–Stieltjes integral and use it to obtain weighted generalization of a recently obtained inequality, as well as weighted 2-point and 3-point quadrature formulae of closed and semi-closed type for functions of bounded variation. © 2011 Elsevier Ltd. All rights reserved.
Keywords: 2-point and 3-point Radau and Lobatto quadrature formulae Montgomery identity
1. Introduction Dragomir et al. in [1] established the following identity: Theorem 1. Let f : [a, b] → R be a bounded function on [a, b], and x1 , x2 , x3 ∈ [a, b] such that x1 ≤ x2 ≤ x3 . Then the following identity holds x1 − a b−a
f ( a) +
x3 − x1 b−a
f ( x2 ) +
b − x3 b−a
f ( b) =
1 b−a
b
∫
f (t )dt + a
1 b−a
∫
b
S (x1 , x2 , x3 , t )df (t )
(1.1)
a
where S (x1 , x2 , x3 , t ) is defined by S (x1 , x2 , x3 , t ) =
t − x1 , t − x3
a ≤ t ≤ x2 , x2 < t ≤ b.
If we take x1 = a, x3 = b the identity (1.1) reduces to a Montgomery identity for Riemann–Stieltjes integral (see for instance [2]) f ( x) =
∗
1 b−a
b
∫
f (t )dt + a
b
∫
P (x, t )df (t ) a
Corresponding author. Tel.: +385 16129965; fax: +385 16129946. E-mail addresses: [email protected] (A. Aglić Aljinović), [email protected] (S. Kovač), [email protected] (J. Pečarić).
0895-7177/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2011.04.006
(1.2)
1366
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
where P (x, t ) is the Peano kernel, defined by
t −a , −a P (x, t ) = bt − b b−a
a ≤ t ≤ x, x < t ≤ b.
The identity (1.1) was used in the same paper to obtain the following inequality: Theorem 2. Let f : [a, b] → R be a monotonic non-decreasing function on [a, b], and x1 , x2 , x3 ∈ [a, b] be such that x1 ≤ x2 ≤ x3 . Then the following inequality holds
∫ (x1 − a)f (a) + (x3 − x1 )f (x2 ) + (b − x3 )f (b) −
b a
f (t )dt
≤ (x1 − a)(f (x1 ) − f (a)) + (x2 − x1 )(f (x2 ) − f (x1 )) + (x3 − x2 )(f (x3 ) − f (x2 )) + (b − x3 )(f (b) − f (x3 )) ≤ max{x1 − a, x2 − x1 , x3 − x2 , b − x3 }(f (b) − f (a)). The aim of this paper is to generalize the result from the Theorem 2, as well as to give the result of the same type for a wider class of functions, namely for the functions of bounded variation. Also, we shall consider the more general integrals of the form b
∫
w(t )f (t )dt a
where w : [a, b] → [0, ∞) is a normalized weight function. These integrals are useful because they split the integrand into the product of a function with possible singularities (w ) and a continuous function (f ). In Section 2, we give a weighted generalization of the identity (1.1). In Section 3, we use this identity to obtain a weighted generalization of the inequality from the Theorem 2, which holds for a wider class of functions. In Section 4, we use this generalization to obtain bounds of the reminder of the general weighted 3-point quadrature formula of semi-closed type. In Section 5. the same is done for general weighted 3-point quadrature formula of closed type and in Section 6, for general weighted 2-point quadrature formulae of semi-closed type. In Section 7, we apply all the results with some well-known weight functions. For other weighted generalizations of the quadreature rules see e.g. [3–5]. 2. Weighted generalization of identity (1.1) In this section we establish a new weighted generalization of the identity (1.1). The following two well-known lemmas are a basic result for the Riemann–Stieltjes integral and the integration by parts for Riemann–Stieltjes integral. Lemma 1. Let g : [a, b] → R be a continuous function and ϕ : [a, b] → R a function of bounded variation on [a, b]. Then it holds
∫ where
b a
b a
g (t )dϕ(t ) ≤
b (ϕ) · max |g (t )| t ∈[a,b]
a
(f ) is total variation of ϕ on [a, b].
Lemma 2. Let ϕ and g be functions of bounded variation on [a, b] which do not have common discontinuity points on [a, b]. Then b
∫
g (t )dϕ(t ) = g (b)ϕ(b) − g (a)ϕ(a) − a
b
∫
ϕ(t )dg (t ). a
Theorem 3. Let f : [a, b] → R be a function of bounded variation and w : [a, b] → [0, ∞⟩ some integrable weight function. Let also x1 , x2 , x3 , y ∈ [a, b], y ≤ x2 . If f is continuous at x2 and y then the following identity holds W (x1 )f (y) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) =
b
∫
w(t )f (t )dt + a
b
∫
Pw (x1 , x2 , x3 , y, t )df (t )
(2.1)
a
where Pw (x1 , x2 , x3 , y, t ) is the generalized weighted Peano kernel, defined by W (t ), W (t ) − W (x1 ), W (t ) − W (x3 ),
Pw (x1 , x2 , x3 , y, t ) =
a ≤ t ≤ y, y < t ≤ x2 , x2 < t ≤ b.
(2.2)
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
and W (t ) =
t a
1367
w(x)dx for t ∈ [a, b].
Proof. We have b
∫
Pw (x1 , x2 , x3 , y, t )df (t ) =
∫
y
W (t )df (t ) +
(W (t ) − W (x1 ))df (t ) +
b
∫
(W (t ) − W (x3 ))df (t ). x2
y
a
a
x2
∫
Since f and Pw do not have common discontinuity points, we may use the integration by parts formula for the Riemann–Stieltjes integral to obtain y
∫
W (t )df (t ) = W (y)f (y) −
∫a x2
y
∫
f (t )dW (t ), a
(W (t ) − W (x1 ))df (t ) = (W (x2 ) − W (x1 ))f (x2 ) − (W (y) − W (x1 ))f (y) −
x2
∫
f (t )dW (t ),
y
y
and
∫
b
(W (t ) − W (x3 ))df (t ) = (W (b) − W (x3 ))f (b) − (W (x2 ) − W (x3 ))f (x2 ) −
b
∫
f (t )dW (t ). x2
x2
By adding these formulae together and by using dW (t ) = w(t )dt we obtain (2.3).
Remark 1. Identity (2.1) is a weighted generalization of the identity (1.1). Indeed, if we first take y = a in (2.1) we obtain W (x1 )f (a) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) =
b
∫
w(t )f (t )dt + a
b
∫
Pw (x1 , x2 , x3 , t )df (t )
(2.3)
a
where Pw (x1 , x2 , x3 , t ) is given by Pw (x1 , x2 , x3 , t ) =
W (t ) − W (x1 ), W (t ) − W (x3 )
a ≤ t ≤ x2 , x2 < t ≤ b.
(2.4)
1 a Further, if we take uniform normalized weight function w(t ) = b− , t ∈ [a, b] and thus W (t ) = bt − , t ∈ [a, b], identity a −a (2.3) reduces to (1.1).
Remark 2. Identity (2.3) has also been obtained as a special case of other weighted generalizations of the Montgomery identity by Čuljak et al. in [6]. 3. Inequalities for functions of bounded variation on [a, b] In this section we give a weighted generalization of inequality from the Theorem 2 which holds for a wider class of functions. Theorem 4. Let f : [a, b] → R be a function of bounded variation on [a, b], w : [a, b] → [0, ∞⟩ some integrable weight function. Let also x1 , x2 , x3 , y ∈ [a, b], y ≤ x2 . If f is continuous at x2 and y then the following identity holds
∫ W (x1 )f (y) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) −
b a
w(t )f (t )dt
≤ max{W (y), |W (x2 ) − W (x1 )|, |W (y) − W (x1 )|, |W (x2 ) − W (x3 )|, W (b) − W (x3 )}
b (f ). a
Proof. By using the (2.3) and triangle inequality, we have
∫ b W (x1 )f (y) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) − w(t )f (t )dt a ∫ b ∫ y = Pw (x1 , x2 , x3 , y, t )df (t ) ≤ Pw (x1 , x2 , x3 , y, t )df (t ) a a ∫ x ∫ b 2 + Pw (x1 , x2 , x3 , y, t )df (t ) + Pw (x1 , x2 , x3 , y, t )df (t ) y
y
∫
W (t )|df (t )| +
= a
x2
x2
∫ y
|W (t ) − W (x1 )| |df (t )| +
∫
b
|W (t ) − W (x3 )||df (t )| x2
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≤ W (y)
x2 y (f ) + max {|W (x2 ) − W (x1 )| , |W (y) − W (x1 )|} (f ) y
a
+ max {|W (x2 ) − W (x3 )| , W (b) − W (x3 )}
b
(f )
x2
≤ max{W (y), |W (x2 ) − W (x1 )|, |W (y) − W (x1 )|, |W (x2 ) − W (x3 )|, W (b) − W (x3 )}
b (f ) a
and the proof follows.
The following corollary is the special case of the previous Theorem for y = a. Corollary 1. Let f : [a, b] → R be a function of bounded variation on [a, b], w : [a, b] → [0, ∞⟩ some integrable weight function. Let also x1 , x2 , x3 ∈ [a, b]. If f is continuous at x2 and a then the following identity holds
∫ W (x1 )f (a) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) −
b
a
w(t )f (t )dt
≤ max{W (x1 ), |W (x2 ) − W (x1 )|, |W (x2 ) − W (x3 )|, W (b) − W (x3 )}
b (f ). a
Proof. We apply Theorem 4 with y = a.
Remark 3. If we additionally assume that x1 ≤ x2 ≤ x3 we have
∫ W (x1 )f (a) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) −
b
a
w(t )f (t )dt
x3 x1 x2 b ≤ W (x1 ) (f ) + (W (x2 ) − W (x1 )) (f ) + (W (x3 ) − W (x2 )) (f ) + (W (b) − W (x3 )) (f ) a
x1
x2
x3
b ≤ max{W (x1 ), W (x2 ) − W (x1 ), W (x2 ) − W (x3 ), W (b) − W (x3 )} (f ). a
Corollary 2. Let f : [a, b] → R be a monotonic non-decreasing function on [a, b], w : [a, b] → [0, ∞⟩ is some integrable weight function. Let also x1 , x2 , x3 ∈ [a, b] be such that x1 ≤ x2 ≤ x3 . Then the following inequality holds
∫ W (x1 )f (a) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) −
a
b
w(t )f (t )dt
≤ W (x1 )(f (x1 ) − f (a)) + (W (x2 ) − W (x1 ))(f (x2 ) − f (x1 )) + (W (x3 ) − W (x2 ))(f (x3 ) − f (x2 )) + (W (b) − W (x3 ))(f (b) − f (x3 )) ≤ max{W (x1 ), W (x2 ) − W (x1 ), W (x3 ) − W (x2 ), W (b) − W (x3 )}(f (b) − f (a)). Proof. Since f : [a, b] → R is a monotonic non-decreasing function on [a, b], we have result from the Remark 3 the proof follows.
b a
(f ) = f (b) − f (a). By using the
1 Remark 4. If we take w(t ) = b− , t ∈ [a, b] to be a uniform normalized weight function, inequalities from the Corollary 2 a reduce to inequalities from the Theorem 2.
4. Weighted 3-point quadrature formulae of semi-closed type In this section we apply all the results to establish bounds of the remainder E (f ) of the general weighted 3-point quadrature formulae of semi-closed type: b
∫
w(t )f (t )dt = A1 f (y) + A2 f (x) + A3 f (b) + E (f ) a
where y, x ∈ [a, b], y ≤ x and
∑3
k=1
Ak = 1.
(4.1)
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
1369
Note that in our case we have A1 = W (x1 ),
A2 = W (x3 ) − W (x1 ),
A3 = W (b) − W (x3 ),
so if we take a normalized weight function w(t ), we have 3 −
Ak = W (b) = 1.
k=1
In case of non-symmetric weight function we can formulate the following result. Theorem 5. Let f : [a, b] → R be a function of bounded variation on [a, b], w : [a, b] → [0, ∞) normalized weight function. b I b −I I y−I I b−I Further, let a ≤ y < 1b−I 2 and 1b−I 2 ≤ x ≤ 1y−I 2 where Ij = a w(t )t j dt, for j = 1, 2. If f is continuous at y and x, then the 1 1 1 following inequality holds:
I − I (b + x) + xb −I2 + I1 (b + y) − by I2 − I1 (x + y) + xy 1 2 f (y) + f (x) + f (b) (b − y)(x − y) (x − y)(b − x) (b − y)(b − x) ∫ b b − w(t )f (t )dt ≤ Kw (y, x) · (f ), a a
(4.2)
where
, W (y) − I2 − I1 (b + x) + xb , (x − y)(b − y) (x − y)(b − y) I2 − I1 (x + y) + xy I2 − I1 (x + y) + xy W (x) − 1 + , . (b − y)(b − x) (b − y)(b − x)
Kw (y, x) = max W (y), W (x) −
Proof. Obviously, a, x∈
I1 b−I2 b−I1
I1 b−I2 b−I1
I2 − I1 (b + x) + xb
(4.3)
̸= ∅. For fixed y ∈ a, I1bb−−I1I2 , it is easy to check that I1bb−−I1I2 , I1yy−−I1I2 ̸= ∅, so we choose
, I1yy−−I1I2 . The quadrature formula (4.1) will be accurate for all polynomials of degree ≤ 2 when
A1 =
I2 − I1 (b + x) + xb
(b − y)(x − y) −I2 + I1 (b + y) − by A2 = (x − y)(b − x) I2 − I1 (x + y) + xy A3 = . (b − y)(b − x) , (x − y)(b − y) > 0 we have A1 ≥ 0. Further, since I12 ≤ I2 we have y <
I1 b−I2 b−I1 bx+I2 −I1 (x+b) (x−y)(b−y)
Since x ≥
I1 b−I12 I1 b−I2 b−I1 b−I1 I2 −I1 (b+x)+xb . (b−y)(x−y)
≤
= I1 , so we have
≤ 1. Therefore, A1 ∈ [0, 1], so there exist x1 ∈ [a, b] such that W (x1 ) = I b −I On the other hand, since y ≤ 1b−I 2 and (b − y)(b − x) > 0 we have A3 ∈ [0, 1], so there exist x3 ∈ [a, b] such that 1
A1 =
I −I (x+y)+xy
W (x3 ) = 1 − 2 (b−1 y)(b−x) . It is easy to check that W (x1 ) ≤ W (x3 ), which implies x1 ≤ x3 . Apply Theorem 4 with y, x2 = x, x1 and x3 as above.
Now, we consider the special case of symmetric normalized weight function on [−1, 1] Analog results for interval [a, b] can easily be obtained with linear transformation t =
b−a 2
x+
b+a 2
.
Corollary 3. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1], w : [−1, 1] → [0, ∞) symmetric normalized 1 I weight function. Further, let −1 ≤ y < −I2 and −I2 ≤ x ≤ − y2 , where I2 = −1 w(t )t 2 dt. If f is continuous at y and x, then the following inequality holds:
∫ 1 1 x + I2 −y − I 2 xy + I2 ≤ Kw (y, x) · (f ), f ( y ) + f ( x ) + f ( 1 ) − w( t ) f ( t ) dt (x − y)(1 − y) (x − y)(1 − x) (1 − y)(1 − x) −1 −1
(4.4)
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A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
where
x + I2 , W (y) − , (x − y)(1 − y) (x − y)(1 − y) xy + I2 xy + I2 W (x) − 1 + , . (1 − y)(1 − x) (1 − y)(1 − x) x + I2
Kw (y, x) = max W (y), W (x) −
Proof. Apply Theorem 5 with a = −1, b = 1, and I1 = 0.
(4.5)
Remark 5. If we put y = −1 in (4.4), then −I2 ≤ x ≤ I2 . Specially, for x = −I2 or x = I2 , the Radau 2-point inequalities are obtained. For x = 0, the Lobatto 3-point inequality is obtained. Remark 6. If we put x = − y2 , for some y ∈ (−1, −I2 ), the inequality related to the open 2-point quadrature formula. I
√
Specially, for x = −y =
I2 , the Gauss 2-point inequality is obtained:
∫ 1 f (− I2 ) + 1 f ( I2 ) − 2 2
1
−1
1 w(t )f (t )dt ≤ Kw (− I2 , I2 ) · (f ),
(4.6)
−1
where
Kw (− I2 , I2 ) = max W (− I2 ), W ( I2 ) −
, W (− I2 ) − 2 1
, 1 − W ( I2 ) . 2 1
(4.7)
5. Weighted 3-point quadrature formulae of closed type Now, we will apply all the results to obtain bounds of the reminder E (f ) of the general weighted 3-point quadrature formulae of closed type b
∫
w(t )f (t )dt = A1 f (a) + A2 f (x) + A3 f (b) + E (f )
(5.1)
a
∑3
where k=1 Ak = 1. Note that in our case we have A1 = W (x1 ),
A2 = W (x3 ) − W (x1 ),
A3 = W (b) − W (x3 ),
so if we take a normalized weight function w(t ), we have 3 −
Ak = W (b) = 1.
k=1
In the case of a symmetric normalized weight function we can formulate the following result. Theorem 6. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1], w : [−1, 1] → [0, ∞⟩ be a normalized symmetric weight function and x1 ∈ [−1, 0]. If f is continuous at − 1 and 0, then the following inequality holds:
∫ A1 (f (−1) + f (1)) + A2 f (0) −
1
−1
1 w(t )f (t )dt ≤ Kw (x1 ) (f ),
(5.2)
−1
t
where A1 = W (x1 ), A2 = 1 − 2W (x1 ), W (t ) = −1 w(s)ds and Kw (x1 ) = max W (x1 ),
Proof. Apply Corollary 1 with x1 ∈ [−1, 0], x2 = 0 and x3 = −x1 .
1 2
− W (x1 ) .
Remark 7. The quadrature formula related to the inequality (5.2) is exact for all polynomials of degree ≤ 1. Specially, for x1 such that W (x1 ) = 16 , the weighted version of famous Simpson’s rule is obtained
∫ 1 (f (−1) + f (1)) + 2 f (0) − 6 3
1
−1
1 1 w(t )f (t )dt ≤ (f ).
This result has been obtained by Dragomir in [7].
3 −1
(5.3)
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
1371
In the next result the inequality with the smallest constant Kw (x1 ) is derived. Corollary 4. Let f : [−1, 1] → R be a function of the bounded variation on [−1, 1] and let w : [−1, 1] → [0, ∞⟩ be some symmetric normalized weight function. If f is continuous at − 1 and 0, then the following inequality holds
∫ 1 f (−1) + 1 f (0) + 1 f (1) − 4 2 4
1
−1
1 1 w(t )f (t )dt ≤ (f ).
Proof. The constant Kw (x1 ) = max W (x1 ),
1 − 2W (x1 ) =
1 2
and Kw (x1 ) =
1 . 4
(5.4)
4 −1
1 2
− W (x1 ) will be minimal if W (x1 ) =
1 2
− W (x1 ) = 14 . Thus A1 = 41 , A2 =
Corollary 5. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1], w : [−1, 1] → [0, ∞⟩ be a normalized x 1 symmetric weight function, x1 such that −11 w(t )dt = 0 t 2 w(t )dt and x3 = −x1 . If f is continuous at − 1 and 0, then the following inequality holds:
∫ A1 (f (−1) + f (1)) + A2 f (0) − where A1 =
1 0
1
−1
1 w(t )f (t )dt ≤ Kw (x1 ) (f ),
(5.5)
−1
t 2 w(t )dt , A2 = 1 − 2A1 and Kw (x1 ) = max
x1 −1
w(t )dt ,
w( t ) dt . x1
0
Proof. Since
∫
x1
−1
w(t )dt =
1
∫
1
∫
t 2 w(t )dt ≤ 0
w(t )dt =
∫
0
w(t )dt ,
−1
0
t
it is obvious that −1 ≤ x1 ≤ 0 ≤ x3 ≤ 1. Therefore we can apply Corollary 1 with W (t ) = −1 w(s)ds and x = 0. Because of x the symmetry of the weight function w , we have W (x1 ) = 1 − W (x3 ) = −11 w(t )dt and W (x) − W (x1 ) = W (x3 ) − W (x) =
0 x1
w(t )dt, so the assertion follows easily.
Remark 8. The quadrature rule related to the inequality (5.5)
∫
1
w(t )f (t )dt ≈ A1 (f (−1) + f (1)) + A2 f (0) −1
is accurate for all polynomials of degree ≤3 and is called the weighted Lobatto quadrature formula. In case of non-symmetric weight function we can formulate the following result. Theorem 7. Let f : [a, b] → R be a function of bounded variation on [a, b], w : [a, b] → [0, ∞⟩ be a normalized weight function and a ≤ x1 ≤ x ≤ x3 ≤ b. If f is continuous at a and x, then the following inequality holds:
∫ A1 f (a) + A2 f (x) + A3 f (b) −
b a
b w(t )f (t )dt ≤ Kw (x1 , x, x3 ) (f ),
(5.6)
a
where A1 = W (x1 ), A2 = W (x3 ) − W (x1 ), A3 = 1 − W (x3 ), W (t ) =
t a
w(s)ds and
Kw (x1 , x, x3 ) = max{W (x1 ), W (x) − W (x1 ), W (x3 ) − W (x), 1 − W (x3 )}. Proof. The proof follows directly from Remark 3.
b Remark 9. If we take x = a+ , x1 and x3 such that W (x1 ) = 2 formula is obtained. Thus
1 6
and W (x3 ) =
5 , 6
then the weighted version of Simpson’s
∫ b b 1 f (a) + 2 f a + b + 1 f (b) − ≤ max 1 , W a + b − 1 , 5 − W a + b w( t ) f ( t ) dt (f ). 6 3 2 6 6 2 6 6 2 a a Remark 10. The minimal constant Kw (x1 , x, x3 ) is achieved when W (x1 ) =
∫ 1 f (a) + 1 f (x) + 1 f (b) − 4 2 4
a
b b 1 w(t )f (t )dt ≤ (f ). 4 a
1 4
, W (x) =
1 2
and W (x3 ) =
3 . 4
(5.7)
Thus we get (5.8)
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6. Weighted 2-point quadrature formulae of semi-closed type In this section we apply all the results to establish bounds of the remainder E (f ) of the general weighted 2-point quadrature formulae of semi-closed type. Such a formula is obtained by putting A3 = 0 in (5.1), that is b
∫
w(t )f (t )dt = A1 f (a) + A2 f (x) + E (f ),
(6.1)
a
where A1 + A2 = 1. Note that in this case we have x3 = b, A1 = W (x1 ) and A2 = W (b) − W (x1 ), so if w(t ) is normalized weight function, then A1 + A2 = W (b) = 1. In the case of non-symmetric weight function, we formulate the following result concerning quadrature formulae with degree of exactness 1. Theorem 8. Let f : [a, b] → R be a function of bounded variation on [a, b], w : [a, b] → [0, ∞⟩ is some normalized weight b function, I ≤ x ≤ b, where I = a w(t )tdt. If f is continuous at x and a, then the following inequality holds
∫ b b I −x a−I ≤ Kw (x) (f ), w( t ) f ( t ) dt f ( a ) + f ( x ) − a − x a−x a a
(6.2)
where Kw (x) = max
I −x a−x
, W (x) −
I −x a−x
, 1 − W (x) ,
t
w(s)ds. Proof. Since a < I < b, we have [I , b] ̸= ∅. The quadrature formula (6.1) will be accurate for all polynomials of degree ≤ 1 and W (t ) =
a
when A1 =
I −x a−x
,
A2 =
a−I a−x
.
x ∈ [0, 1]. Apply Theorem 4 with y = a, x2 = x, x3 = b and x1 such that Since x ∈ [I , b], we have A1 = aI − −x I −x It is easy to check that a−x ≤ W (x), so the assertion follows directly.
x1 a
w(t )dt =
I −x . a −x
In case of symmetric normalized weight function we can formulate the following result. Corollary 6. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1], w : [−1, 1] → [0, ∞⟩ is some symmetric normalized weight function and 0 ≤ x ≤ 1. If f is continuous at x and − 1, then the following inequality holds
∫ 1 1 x 1 ≤ Kw (x) (f ), f (− 1 ) + f ( x ) − w( t ) f ( t ) dt 1 + x 1+x −1 −1
(6.3)
where Kw (x) = max
x 1+x
, W ( x) −
x 1+x
, 1 − W (x)
t
and W (t ) = −1 w(s)ds. Proof. Apply Theorem 8 with a = −1, b = 1 and I1 = 0. Remark 11. If we take x = 1, then Kw (1) =
∫ 1 f (−1) + 1 f (1) − 2 2
1
−1
1
−1
so the inequality (6.3) reduces to the weighted trapezoid inequality
1 1 w(t )f (t )dt ≤ (f ). 2
(6.4)
−1
Remark 12. If we take x = 0, then Kw (0) =
∫ f (0) −
1 , 2
1 1 w(t )f (t )dt ≤ (f ). 2 −1
1 , 2
so the inequality (6.3) reduces to the weighted midpoint inequality (6.5)
Remark 13. In order to enlarge degree of exactness, we have to consider quadrature formulae of Radau type which are 1 accurate for all polynomials of degree ≤ 2. Therefore we have to take x = −1 w(t )t 2 dt in (6.3).
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
1373
The next result establishes the inequality related to the 2-point semi-closed quadrature formula with the minimal constant K (x). Corollary 7. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1], w : [−1, 1] → [0, ∞⟩ is some symmetric 1
normalized weight function such that −21 w(t )dt =
∫ 1 f (−1) + 2 f 1 − 3 3 2
1
−1
2 . 3
If f is continuous at
1 2
and − 1, then the following inequality holds
1 1 w(t )f (t )dt ≤ (f ). 3
(6.6)
−1
Proof. Obviously, constant Kw (x) is minimal if x
x
= W (x) −
1+x
1+x
= 1 − W (x) =
t
where W (t ) = −1 w(s)ds. That is x =
1 2
1 3
,
and W (x) =
2 . 3
Therefore inequality (6.3) becomes (6.6).
7. Applications for some special weight functions 7.1. Case w(t ) =
1 2
, t ∈ [−1, 1]
Corollary 8. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1] and −1 ≤ y < − 13 ≤ x < 1. If f is continuous at y and x, then the following inequality holds:
∫ 1 x + 1/3 −y − 1/3 xy + 1/3 1 1 ≤ Kw (y, x) · (f ), f ( y ) + f ( x ) + f ( 1 ) − f ( t ) dt (x − y)(1 − y) (x − y)(1 − x) (1 − y)(1 − x) 2 −1 −1
(7.1)
where
x + 1/3
, 2 2
, Kw (y, x) = max − (x − y)(1 − y) y + 1 x − 1 x + 1/3 xy + 1/3 xy + 1/3 2 − (x − y)(1 − y) , 2 + (1 − y)(1 − x) , (1 − y)(1 − x) . y + 1 x + 1
Proof. Apply Corollary 3 with w(t ) = √
Remark 14. For y = − 1+5
6
1 2
and W (t ) = t +2 1 . Obviously, I2 =
1 , 3
so the assertion follows easily.
√
and x =
6−1 5
we get Radau 3-point inequality:
√ √ √ ∫ 1 16 − √6 23√6 − 8 1+ 6 16 + 6 6−1 1 1 1 − + + f f f ( 1 ) − f ( t ) dt · (f ). ≤ 36 5 36 5 9 2 −1 180 −1
(7.2)
For y = −1 and x = 0 we get the Simpson inequality:
∫ 1 f (−1) + 2 f (0) + 1 f (1) − 1 6 3 6 2
1
−1
1 1 f (t )dt ≤ · (f ). 3
(7.3)
−1
For x = − 13 we get Radau 2-point inequality:
∫ 3 f − 1 + 1 f (1) − 1 4 3 4 2
1
−1
1 5 f (t )dt ≤ · (f ). 12
(7.4)
−1
For y = − √1 and x = √1 we get Legendre–Gauss 2-point inequality: 3
3
∫ 1 1 1 1 f − √1 + 1 f √1 − 1 ≤ √ f ( t ) dt · (f ). 2 2 3 2 2 −1 3 3 −1 Obviously, the best estimate is obtained for Radau 3-point formula.
(7.5)
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Corollary 9. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1] and 0 ≤ x ≤ 1. If f is continuous at x, then the following inequality holds
∫ 1 2 x 1 1 1 ≤ 1+x f ( t ) dt f (− 1 ) + f ( x ) − (f ). 2(1 + x) 1 + x 1+x 2 −1 −1 Proof. Apply Corollary 6 with w(t ) = 1+x2 . 2(1+x)
and W (t ) =
1 2
t +1 . 2
(7.6)
It is easy to check that Kw (x) = max
2
, 1+x , 1−2 x 1+x 2(1+x) x
=
Remark 15. For x = 0, 31 , 1 we get the midpoint, Radau 2-point and trapezoid inequality, respectively. Namely,
∫ f (0) − 1 2
1
−1
f (t )dt ≤
1 1
2 −1
∫ 1 f (−1) + 3 f 1 − 1 4 4 3 2
(f ),
1
−1
f (t )dt ≤
1 5
12 −1
(f )
and
∫ 1 f (−1) + 1 f (1) − 1 2 2 2
1
−1
1 1 f (t )dt ≤ (f ). 2 −1
Obviously, the best estimate is obtained for Radau 2-point formula. 7.2. Case w(t ) =
√1 π
1−t 2
, t ∈ ⟨−1, 1⟩
Corollary 10. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1] and let −1 ≤ y ≤ − 12 ≤ x ≤ 1. If f is continuous at y and x, then the following inequality holds:
∫ 1 1 x + 1/2 −y − 1/2 xy + 1/2 1 f (y) + f (x) + f (1) − (f ), (7.7) f (t )dt ≤ Kw (y, x) · √ (x − y)(1 − y) (x − y)(1 − x) (1 − y)(1 − x) −1 π 1 − t 2 −1 where
Kw (y, x) = max
1
+
, +
1 + 2x
arcsin y 1
+
,
arcsin x
2 π 2 2(x − y)(y − 1) π 1 1 + 2x arcsin y 1 x + y + xy arcsin x 1 + 2xy + , + , + . 2 2(x − y)(y − 1) + π 2 2(1 − x)(1 − y) π 2(1 − y)(1 − x)
Proof. Apply Corollary 3 with w(t ) = √
Remark 16. For y = − 1+4
√1 π
1−t 2
, t ∈ ⟨−1, 1⟩ and W (t ) =
arcsin t
π
+ 12 , t ∈ [−1, 1].
√ 5
5 −1 4
and x =
we get the Radau 3-point inequality:
√ √ ∫ 1 1 2 1 1+ 5 2 5−1 1 f (t ) + f + f (1) − dt ≤ (f ). √ f − 5 −1 5 4 5 4 5 −1 π 1 − t 2
(7.8)
For y = −1 and x = 0 we get the Lobatto inequality
∫ 1 f (−1) + 1 f (0) + 1 f (1) − 4 2 4
f (t )
1
√
−1
π 1 − t2
dt ≤
1 1
4 −1
(f ).
(7.9)
For x = − 12 we get the Radau 2-point inequality:
∫ 2 f − 1 + 1 f (1) − 3 2 3
1
f (t )
√
−1
π 1 − t2
dt ≤
1 1
3 −1
(f ).
(7.10)
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
√
For y = −
2 2
1375
√ 2 2
and x =
we get the Gauss–Chebyshev 2-point inequality:
√ √ ∫ 1 1 1 1 2 2 f (t ) 1 + f − dt ≤ (f ). √ f − 2 4 −1 2 2 2 −1 π 1 − t 2
(7.11)
Corollary 11. Let f : [−1, 1] → R be a function of a bounded variation on [−1, 1]. If 0 ≤ x ≤ then the following inequality holds
1 2
and if f is continuous at x,
∫ 1 1 x f (t ) 1 ≤ 1 − arcsin x f (− 1 ) + f ( x ) − dt (f ). √ x + 1 x+1 2 π −1 π 1 − t 2 −1 If
1 2
(7.12)
≤ x ≤ 1, then the following inequality holds ∫ 1 1 x 1 f (t ) x dt ≤ (f ). √ x + 1 f (−1) + x + 1 f (x) − x + 1 −1 −1 π 1 − t 2
Proof. Apply Corollary 6 with w(t ) =
√1 π
1 −t 2
and W (t ) =
1 2
+
arcsin t
π
.
(7.13)
Remark 17. For x = 0, 12 , 1 we get the midpoint, Radau 2-point and trapezoid inequality, respectively. Namely,
∫ f (0) −
f (t )
1
√
dt ≤
π 1 − t2 ∫ 1 f (−1) + 2 f 1 − −1
3
3
2
1
−1
1 1
2 −1
(f ),
(7.14)
f (t )
1 1 dt ≤ (f ) √ 3 π 1 − t2
(7.15)
−1
and
∫ 1 f (−1) + 1 f (1) − 2 2
1
f (t )
√
−1
π 1 − t2
dt ≤
1 1
2 −1
(f ).
(7.16)
Obviously, the best estimate for 2-point semi-closed quadrature rule is obtained for the Radau quadrature formula (7.15). 7.3. Case w(t ) = π2
√
1 − t 2 , t ∈ [−1, 1]
Corollary 12. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1] and let −1 ≤ y ≤ − 14 ≤ x ≤ 1. If f is continuous at y and x, then the following inequality holds:
∫ 1 1 x + 1/4 −y − 1/4 xy + 1/4 2 f (y) + f (x) + f (1) − 1 − t 2 f (t )dt ≤ Kw (y, x) · (f ), (x − y)(1 − y) (x − y)(1 − x) (1 − y)(1 − x) −1 π −1 where
√ π + y 1 − y2 − arccos y 1 + 4x π + x 1 − x2 − arccos x Kw (y, x) = max , + 4(x − y)(y − 1) π π 1 + 4x π + y 1 − y2 − arccos y × + , 4(x − y)(y − 1) π √ 4x + 4y − 3 π + x 1 − x2 − arccos x 1 + 4xy + . , 4(1 − x)(1 − y) 4(1 − x)(1 − y) π
Proof. Apply Corollary 3 with w(t ) = π2
√
√ 1 − t 2 , t ∈ [−1, 1] and W (t ) =
π+t
1−t 2 −arccos t
π
, t ∈ [−1, 1].
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A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
√
Remark 18. For y = − 1+6
√ 7
7 −1 6
and x =
we get the Radau 3-point inequality:
√ √ √ ∫ 1 1 1 13 − √7 2 1+ 7 7−1 13 + 7 1 1 − t 2 f (t )dt ≤ f − f + + f (1) − (f ). 5 −1 28 6 28 6 14 −1 π
(7.17)
For y = −1 and x = 0 we get the Lobatto inequality
∫ 1 f (−1) + 3 f (0) + 1 f (1) − 8 4 8
1 3 2 1 − t 2 f (t )dt ≤ (f ). 8 −1 −1 π 1
(7.18)
For x = − 14 we get the Radau 2-point inequality:
∫ 4 f − 1 + 1 f (1) − 5 4 5 For y = − 12 and x =
1 2
1
−1
2
π
1 − t 2 f (t )dt ≤
√
1 5 15 − 16π + 80 arccos(1/4)
(f ).
80π
(7.19)
−1
we get the Gauss–Chebyshev 2-point inequality of the second:
∫ 1 f −1 + 1f 1 − 2 2 2 2
1
−1
2
π
√ 1 1 3 1 − t 2 f (t )dt ≤ + (f ). 6 4π −1
(7.20)
Corollary 13. Let f : [−1, 1] → R be a function of a bounded variation on [−1, 1]. If 0 ≤ x ≤ 1 and if f is continuous at x, then the following inequality holds
√ ∫ 1 1 x 1 2 1 x 1 − x2 − arccos x 2 (f ). f (− 1 ) + f ( x ) − 1 − t f ( t ) dt ≤ + x + 1 x+1 1+x π −1 π −1 Proof. Apply Corollary 6 with w(t ) = π2
√
1 − t 2 and W (t ) =
(7.21)
√ π+t
1−t 2 −arccos t
π
.
Remark 19. For x = 0, 41 , 1 we get the midpoint, Radau 2-point and trapezoid inequality, respectively. Namely,
∫ 1 √ 1 1 2 1 − t 2 f (t ) (f ), dt ≤ f (0) − 2 −1 π −1 ∫ 1 √ 1 1 5√15 + 64π − 80 arccos 1 4 1 2 1 − t 2 f (t ) 4 − dt ≤ (f ) f (−1) + f 5 5 4 π 80π −1 −1
(7.22)
(7.23)
and
∫ 1 √ 1 1 1 1 2 1 − t 2 f (t ) dt ≤ (f ). f (−1) + f (1) − 2 2 −1 2 π −1
(7.24)
√ 5 15+64π −80 arccos 1
4 Since ≈ 0.457481, the best estimate for 2-point semi-closed quadrature rule is obtained for the Radau 80π quadrature formula (7.23).
7.4. Case w(t ) =
3 2
√
t , t ∈ [0, 1]
Corollary 14. Let f : [0, 1] → R be a function of a bounded variation on [0, 1] and let 0 ≤ y ≤ at y and x, then the following inequality holds:
3 7
≤ x ≤ 1. If f is continuous
∫ 1 √ 1 2(3 − 7y) 15 + 35xy − 21(x + y) 3 tf (t ) 2(7x − 3) f (y) + f (x) + f (1) − dt ≤ Kw (y, x) (f ), 35(x − y)(1 − y) 35(1 − x)(x − y) 35(x − 1)(y − 1) 2 0 0
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
1377
where
2(7x − 3)
3/2 2(7x − 3) , y + 7(x − y)(y − 1) 7(x − y)(y − 1) 3/2 14 + 34xy − 20(x + y) 15 + 35xy − 21(x + y) × x + , 35(x − y)(1 − y) 35(x − 1)(y − 1)
Kw (y, x) = max y3/2 , x3/2 +
Proof. Apply Theorem 5 with w(t ) =
3 2
√
t , t ∈ [0, 1] and W (t ) = t 3/2 , t ∈ [0, 1].
√
Remark 20. For y =
15−2 15 33
√
and x =
15+2 15 33
the Radau 3-point inequality is obtained
√ √ √ 150 − 13√15 15 − 2 15 15 + 2 15 150 + 13 15 f f + 350 33 350 33 ∫ 1 √ 1 3 tf (t ) 1 (f ). dt ≤ 0.287532 + f (1) − 7 2 0 0 For y = 0 and x =
5 9
(7.25)
the Lobatto inequality is obtained:
∫ 1 √ 1 16 3 tf (t ) 243 5 3 (f ). f (0) + f + f (1) − dt ≤ 0.371628 175 350 9 14 2 0 0 For x =
3 7
(7.26)
the Radau 2-point inequality is obtained
∫ 1 √ 1 7 3 3 3 tf (t ) (f ). + f (1) − dt ≤ 0.419434 f 10 7 10 2 0 0 √
For y =
35−2 70 63
(7.27)
√
and x =
35+2 70 63
the inequality of Gauss–Jacobi type is obtained:
√ √ √ ∫ 1 √ 1 50 + √70 35 − 2 70 50 − 70 35 + 2 70 3 tf (t ) (f ). f + f − dt ≤ 0.327786 100 63 100 63 2 0 0 Corollary 15. Let f : [0, 1] → R be a function of bounded variation on [0, 1]. If then the following inequality holds
3 7
≤x≤
3 2/5 5
and if f is continuous at x,
∫ 1 3 1 x − 3 3√ 5 5 f (0) + f (x) − tf (t )dt ≤ (1 − x3/2 ) (f ). x x 0 2 0 If
(7.28)
(7.29)
3 2/5 5
≤ x ≤ 1, then the following inequality holds ∫ 1 1 x − 3 x − 35 3/5 3√ 5 f (0) + f (x) − tf (t )dt ≤ (f ). x x x 0 2 0
Proof. Apply Theorem 8 with w(t ) =
3 2
√
t and W (t ) = t 3/2 .
, 57 , 1 we get midpoint, Radau 2-point and trapezoid inequality, respectively. Namely, ∫ √ 3/2 1 3 1 3 tf (t ) 3 − dt ≤ 1 − (f ), f 5 2 5 0 0 √ 1 ∫ 1 √ 4 21 5 3 tf (t ) 5 35 − dt ≤ 1 − (f ) f (0) + f 25 25 7 2 49 0 0
Remark 21. For x =
(7.30)
3 5
(7.31)
(7.32)
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A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
and
∫ 1 √ 1 2 2 3 tf (t ) 3 dt ≤ (f ). f (0) + f (1) − 5 5 0 5 2 0 √
Since 1 −
5 35 49
≈ 0.396318 and 1 −
3 3/2 5
(7.33)
≈ 0.535242 the best estimate is obtained for the Radau 2-point formula.
1 , t ∈ ⟨0, 1] 7.5. Case w(t ) = √ 2 t
Corollary 16. Let f : [0, 1] → R be a function of bounded variation on [0, 1] and let 0 ≤ y ≤ y and x, then the following inequality holds:
1 5
≤ x ≤ 1. If f is continuous at
1 2(5y − 1) 3 + 15xy − 5(x + y) 2(5x − 1) ≤ Kw (y, x) (f ), f ( y ) + f ( x ) + f ( 1 ) 15(x − y)(1 − y) 15(x − 1)(x − y) 15(1 − x)(1 − y) 0 where
√ √ √ 2(5x − 1) 2(5x − 1) , y− Kw (y, x) = max y, x − 15(x − y)(1 − y) 15(x − y)(1 − y) √ 3 + 10(x + y) 3 + 15xy − 5(x + y) , × x + . 15(x − 1)(y − 1) 15(1 − x)(1 − y) 1 Proof. Apply Theorem 5 with w(t ) = √ , t ∈ ⟨0, 1] and W (t ) =
√
2 t
√ 2 Remark 22. For y = 7−21
7
t , t ∈ [0, 1].
√ 2 and x = 7+21
7
, the Radau 3-point inequality is obtained:
√ √ √ √ 1 ∫ 1 14 + √7 7−2 7 14 − 7 7+2 7 1 f (t ) 7−2 7 (f ). f + f + f (1) − √ dt ≤ 30 21 30 21 15 21 0 2 t 0 For y = 0 and x =
3 , 7
the Lobatto type inequality is obtained:
∫ 16 f (0) + 49 f 3 + 1 f (1) − 45 90 7 10 For x =
1 5
1 0
f (t )
1 16 (f ). √ dt ≤ 45 2 t
(7.35)
0
the Radau 2-point inequality is obtained:
∫ 5 f 1 + 1 f (1) − 6 5 6 √
For y =
15−2 30 35
1 0
f (t )
1 1 (f ). √ dt ≤ √ 2 t 5
(7.36)
0
√
and x =
15+2 30 35
we get the inequality of Gauss–Jacobi type:
√ √ √ ∫ 1 √ 1 18 + √30 15 − 2 30 18 − 30 15 + 2 30 f (t ) 15 − 2 30 f + f − (f ). √ dt ≤ 36 35 36 35 35 0 2 t 0 Corollary 17. Let f : [0, 1] → R be a function of bounded variation on [0, 1]. If then the following inequality holds
∫ 1 1 x − 1 √ 1/3 1 3 f (0) + f (x) − √ f (t )dt ≤ (1 − x) (f ). x x 0 2 t 0 If
(7.34)
1 3
≤x≤
1 2/3 3
(7.37)
and if f is continuous at x,
(7.38)
1 2/3 3
≤ x ≤ 1, then the following inequality holds ∫ 1 1 x − 1 x − 13 1/3 1 3 f (0) + f (x) − (f ). √ f (t )dt ≤ x x x 0 2 t 0
1 Proof. Apply Theorem 8 with w(t ) = √ and W (t ) = 2 t
√ t.
(7.39)
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
Remark 23. For x =
∫ 1 f 3 −
1 0
, 35 , 1 we get midpoint, Radau 2-point and trapezoid inequalities. Namely, 1 f (t ) 1 (f ), √ dt ≤ 1 − 3 2 t
1379
1 3
(7.40)
0
∫ 4 f (0) + 5 f 3 − 9 9 5
1 0
f (t )
1 4 (f ) √ dt ≤ 9 2 t
(7.41)
0
and
∫ 2 f (0) + 1 f (1) − 3 3 Since 1 −
1 3
1 0
≈ 0.42265 and
f (t )
1 2 (f ). √ dt ≤ 3 2 t
(7.42)
0
4 9
≈ 0.444444 the best estimate is obtained for the midpoint formula.
References [1] S.S. Dragomir, J.E. Pečarić, S. Wang, The unified treatment of trapezoid, Simpson and Ostrowski type inequalities for monotonic mappings and applications, Math. Comput. Modelling 31 (6–7) (2001) 61–70. [2] D.S. Mitrinović, J.E. Pečarić, A.M. Fink, Inequalities for Functions and their Integrals and Derivatives, Kluwer Academic Publishers, Dordrecht, 1994. [3] S. Kovač, J. Pečarić, Weighted version of general integral formula of the Euler type, Math. Inequal. Appl. 13 (3) (2010) 579–599. [4] S. Kovač, J. Pečarić, A. Vukelić, A generalization of general two-point formula with applications in numerical integration, Nonlinear Anal. Ser. A: TMA 68 (8) (2008) 2445–2463. [5] S. Kovač, J. Pečarić, A. Vukelić, Weighted generalization of the trapezoidal rule via Fink identity, Aust. J. Math. Anal. Appl. 4 (1) (2007) Article 8, 1–12 (electronic). [6] V. Čuljak, J. Pečarić, L.E. Persson, Weighted Simpson type inequalities and an extension of Montgomery identity, J. Math. Inequal. (manuscript). [7] S.S. Dragomir, On Simpson’s quadrature formula for mappings of bounded variation and applications, Tamkang J. Math. 30 (1999) 53–58.
Contents lists available at ScienceDirect
Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm
Error bounds for weighted 2-point and 3-point Radau and Lobatto quadrature rules for functions of bounded variation A. Aglić Aljinović a,∗ , S. Kovač b , J. Pečarić c a
Department of Applied Mathematics, Faculty of Electrical Engineering and Computing, University of Zagreb, Unska 3, 10000 Zagreb, Croatia
b
Faculty of Geotechnical Engineering, University of Zagreb, Hallerova aleja 7, 42000 Varaždin, Croatia
c
Faculty of Textile Technology, University of Zagreb, Pierottijeva 6, 10000 Zagreb, Croatia
article
abstract
info
Article history: Received 6 February 2010 Received in revised form 12 December 2010 Accepted 6 April 2011
We present a weighted generalization of Montgomery identity for Riemann–Stieltjes integral and use it to obtain weighted generalization of a recently obtained inequality, as well as weighted 2-point and 3-point quadrature formulae of closed and semi-closed type for functions of bounded variation. © 2011 Elsevier Ltd. All rights reserved.
Keywords: 2-point and 3-point Radau and Lobatto quadrature formulae Montgomery identity
1. Introduction Dragomir et al. in [1] established the following identity: Theorem 1. Let f : [a, b] → R be a bounded function on [a, b], and x1 , x2 , x3 ∈ [a, b] such that x1 ≤ x2 ≤ x3 . Then the following identity holds x1 − a b−a
f ( a) +
x3 − x1 b−a
f ( x2 ) +
b − x3 b−a
f ( b) =
1 b−a
b
∫
f (t )dt + a
1 b−a
∫
b
S (x1 , x2 , x3 , t )df (t )
(1.1)
a
where S (x1 , x2 , x3 , t ) is defined by S (x1 , x2 , x3 , t ) =
t − x1 , t − x3
a ≤ t ≤ x2 , x2 < t ≤ b.
If we take x1 = a, x3 = b the identity (1.1) reduces to a Montgomery identity for Riemann–Stieltjes integral (see for instance [2]) f ( x) =
∗
1 b−a
b
∫
f (t )dt + a
b
∫
P (x, t )df (t ) a
Corresponding author. Tel.: +385 16129965; fax: +385 16129946. E-mail addresses: [email protected] (A. Aglić Aljinović), [email protected] (S. Kovač), [email protected] (J. Pečarić).
0895-7177/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2011.04.006
(1.2)
1366
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
where P (x, t ) is the Peano kernel, defined by
t −a , −a P (x, t ) = bt − b b−a
a ≤ t ≤ x, x < t ≤ b.
The identity (1.1) was used in the same paper to obtain the following inequality: Theorem 2. Let f : [a, b] → R be a monotonic non-decreasing function on [a, b], and x1 , x2 , x3 ∈ [a, b] be such that x1 ≤ x2 ≤ x3 . Then the following inequality holds
∫ (x1 − a)f (a) + (x3 − x1 )f (x2 ) + (b − x3 )f (b) −
b a
f (t )dt
≤ (x1 − a)(f (x1 ) − f (a)) + (x2 − x1 )(f (x2 ) − f (x1 )) + (x3 − x2 )(f (x3 ) − f (x2 )) + (b − x3 )(f (b) − f (x3 )) ≤ max{x1 − a, x2 − x1 , x3 − x2 , b − x3 }(f (b) − f (a)). The aim of this paper is to generalize the result from the Theorem 2, as well as to give the result of the same type for a wider class of functions, namely for the functions of bounded variation. Also, we shall consider the more general integrals of the form b
∫
w(t )f (t )dt a
where w : [a, b] → [0, ∞) is a normalized weight function. These integrals are useful because they split the integrand into the product of a function with possible singularities (w ) and a continuous function (f ). In Section 2, we give a weighted generalization of the identity (1.1). In Section 3, we use this identity to obtain a weighted generalization of the inequality from the Theorem 2, which holds for a wider class of functions. In Section 4, we use this generalization to obtain bounds of the reminder of the general weighted 3-point quadrature formula of semi-closed type. In Section 5. the same is done for general weighted 3-point quadrature formula of closed type and in Section 6, for general weighted 2-point quadrature formulae of semi-closed type. In Section 7, we apply all the results with some well-known weight functions. For other weighted generalizations of the quadreature rules see e.g. [3–5]. 2. Weighted generalization of identity (1.1) In this section we establish a new weighted generalization of the identity (1.1). The following two well-known lemmas are a basic result for the Riemann–Stieltjes integral and the integration by parts for Riemann–Stieltjes integral. Lemma 1. Let g : [a, b] → R be a continuous function and ϕ : [a, b] → R a function of bounded variation on [a, b]. Then it holds
∫ where
b a
b a
g (t )dϕ(t ) ≤
b (ϕ) · max |g (t )| t ∈[a,b]
a
(f ) is total variation of ϕ on [a, b].
Lemma 2. Let ϕ and g be functions of bounded variation on [a, b] which do not have common discontinuity points on [a, b]. Then b
∫
g (t )dϕ(t ) = g (b)ϕ(b) − g (a)ϕ(a) − a
b
∫
ϕ(t )dg (t ). a
Theorem 3. Let f : [a, b] → R be a function of bounded variation and w : [a, b] → [0, ∞⟩ some integrable weight function. Let also x1 , x2 , x3 , y ∈ [a, b], y ≤ x2 . If f is continuous at x2 and y then the following identity holds W (x1 )f (y) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) =
b
∫
w(t )f (t )dt + a
b
∫
Pw (x1 , x2 , x3 , y, t )df (t )
(2.1)
a
where Pw (x1 , x2 , x3 , y, t ) is the generalized weighted Peano kernel, defined by W (t ), W (t ) − W (x1 ), W (t ) − W (x3 ),
Pw (x1 , x2 , x3 , y, t ) =
a ≤ t ≤ y, y < t ≤ x2 , x2 < t ≤ b.
(2.2)
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
and W (t ) =
t a
1367
w(x)dx for t ∈ [a, b].
Proof. We have b
∫
Pw (x1 , x2 , x3 , y, t )df (t ) =
∫
y
W (t )df (t ) +
(W (t ) − W (x1 ))df (t ) +
b
∫
(W (t ) − W (x3 ))df (t ). x2
y
a
a
x2
∫
Since f and Pw do not have common discontinuity points, we may use the integration by parts formula for the Riemann–Stieltjes integral to obtain y
∫
W (t )df (t ) = W (y)f (y) −
∫a x2
y
∫
f (t )dW (t ), a
(W (t ) − W (x1 ))df (t ) = (W (x2 ) − W (x1 ))f (x2 ) − (W (y) − W (x1 ))f (y) −
x2
∫
f (t )dW (t ),
y
y
and
∫
b
(W (t ) − W (x3 ))df (t ) = (W (b) − W (x3 ))f (b) − (W (x2 ) − W (x3 ))f (x2 ) −
b
∫
f (t )dW (t ). x2
x2
By adding these formulae together and by using dW (t ) = w(t )dt we obtain (2.3).
Remark 1. Identity (2.1) is a weighted generalization of the identity (1.1). Indeed, if we first take y = a in (2.1) we obtain W (x1 )f (a) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) =
b
∫
w(t )f (t )dt + a
b
∫
Pw (x1 , x2 , x3 , t )df (t )
(2.3)
a
where Pw (x1 , x2 , x3 , t ) is given by Pw (x1 , x2 , x3 , t ) =
W (t ) − W (x1 ), W (t ) − W (x3 )
a ≤ t ≤ x2 , x2 < t ≤ b.
(2.4)
1 a Further, if we take uniform normalized weight function w(t ) = b− , t ∈ [a, b] and thus W (t ) = bt − , t ∈ [a, b], identity a −a (2.3) reduces to (1.1).
Remark 2. Identity (2.3) has also been obtained as a special case of other weighted generalizations of the Montgomery identity by Čuljak et al. in [6]. 3. Inequalities for functions of bounded variation on [a, b] In this section we give a weighted generalization of inequality from the Theorem 2 which holds for a wider class of functions. Theorem 4. Let f : [a, b] → R be a function of bounded variation on [a, b], w : [a, b] → [0, ∞⟩ some integrable weight function. Let also x1 , x2 , x3 , y ∈ [a, b], y ≤ x2 . If f is continuous at x2 and y then the following identity holds
∫ W (x1 )f (y) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) −
b a
w(t )f (t )dt
≤ max{W (y), |W (x2 ) − W (x1 )|, |W (y) − W (x1 )|, |W (x2 ) − W (x3 )|, W (b) − W (x3 )}
b (f ). a
Proof. By using the (2.3) and triangle inequality, we have
∫ b W (x1 )f (y) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) − w(t )f (t )dt a ∫ b ∫ y = Pw (x1 , x2 , x3 , y, t )df (t ) ≤ Pw (x1 , x2 , x3 , y, t )df (t ) a a ∫ x ∫ b 2 + Pw (x1 , x2 , x3 , y, t )df (t ) + Pw (x1 , x2 , x3 , y, t )df (t ) y
y
∫
W (t )|df (t )| +
= a
x2
x2
∫ y
|W (t ) − W (x1 )| |df (t )| +
∫
b
|W (t ) − W (x3 )||df (t )| x2
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≤ W (y)
x2 y (f ) + max {|W (x2 ) − W (x1 )| , |W (y) − W (x1 )|} (f ) y
a
+ max {|W (x2 ) − W (x3 )| , W (b) − W (x3 )}
b
(f )
x2
≤ max{W (y), |W (x2 ) − W (x1 )|, |W (y) − W (x1 )|, |W (x2 ) − W (x3 )|, W (b) − W (x3 )}
b (f ) a
and the proof follows.
The following corollary is the special case of the previous Theorem for y = a. Corollary 1. Let f : [a, b] → R be a function of bounded variation on [a, b], w : [a, b] → [0, ∞⟩ some integrable weight function. Let also x1 , x2 , x3 ∈ [a, b]. If f is continuous at x2 and a then the following identity holds
∫ W (x1 )f (a) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) −
b
a
w(t )f (t )dt
≤ max{W (x1 ), |W (x2 ) − W (x1 )|, |W (x2 ) − W (x3 )|, W (b) − W (x3 )}
b (f ). a
Proof. We apply Theorem 4 with y = a.
Remark 3. If we additionally assume that x1 ≤ x2 ≤ x3 we have
∫ W (x1 )f (a) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) −
b
a
w(t )f (t )dt
x3 x1 x2 b ≤ W (x1 ) (f ) + (W (x2 ) − W (x1 )) (f ) + (W (x3 ) − W (x2 )) (f ) + (W (b) − W (x3 )) (f ) a
x1
x2
x3
b ≤ max{W (x1 ), W (x2 ) − W (x1 ), W (x2 ) − W (x3 ), W (b) − W (x3 )} (f ). a
Corollary 2. Let f : [a, b] → R be a monotonic non-decreasing function on [a, b], w : [a, b] → [0, ∞⟩ is some integrable weight function. Let also x1 , x2 , x3 ∈ [a, b] be such that x1 ≤ x2 ≤ x3 . Then the following inequality holds
∫ W (x1 )f (a) + (W (x3 ) − W (x1 ))f (x2 ) + (W (b) − W (x3 ))f (b) −
a
b
w(t )f (t )dt
≤ W (x1 )(f (x1 ) − f (a)) + (W (x2 ) − W (x1 ))(f (x2 ) − f (x1 )) + (W (x3 ) − W (x2 ))(f (x3 ) − f (x2 )) + (W (b) − W (x3 ))(f (b) − f (x3 )) ≤ max{W (x1 ), W (x2 ) − W (x1 ), W (x3 ) − W (x2 ), W (b) − W (x3 )}(f (b) − f (a)). Proof. Since f : [a, b] → R is a monotonic non-decreasing function on [a, b], we have result from the Remark 3 the proof follows.
b a
(f ) = f (b) − f (a). By using the
1 Remark 4. If we take w(t ) = b− , t ∈ [a, b] to be a uniform normalized weight function, inequalities from the Corollary 2 a reduce to inequalities from the Theorem 2.
4. Weighted 3-point quadrature formulae of semi-closed type In this section we apply all the results to establish bounds of the remainder E (f ) of the general weighted 3-point quadrature formulae of semi-closed type: b
∫
w(t )f (t )dt = A1 f (y) + A2 f (x) + A3 f (b) + E (f ) a
where y, x ∈ [a, b], y ≤ x and
∑3
k=1
Ak = 1.
(4.1)
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
1369
Note that in our case we have A1 = W (x1 ),
A2 = W (x3 ) − W (x1 ),
A3 = W (b) − W (x3 ),
so if we take a normalized weight function w(t ), we have 3 −
Ak = W (b) = 1.
k=1
In case of non-symmetric weight function we can formulate the following result. Theorem 5. Let f : [a, b] → R be a function of bounded variation on [a, b], w : [a, b] → [0, ∞) normalized weight function. b I b −I I y−I I b−I Further, let a ≤ y < 1b−I 2 and 1b−I 2 ≤ x ≤ 1y−I 2 where Ij = a w(t )t j dt, for j = 1, 2. If f is continuous at y and x, then the 1 1 1 following inequality holds:
I − I (b + x) + xb −I2 + I1 (b + y) − by I2 − I1 (x + y) + xy 1 2 f (y) + f (x) + f (b) (b − y)(x − y) (x − y)(b − x) (b − y)(b − x) ∫ b b − w(t )f (t )dt ≤ Kw (y, x) · (f ), a a
(4.2)
where
, W (y) − I2 − I1 (b + x) + xb , (x − y)(b − y) (x − y)(b − y) I2 − I1 (x + y) + xy I2 − I1 (x + y) + xy W (x) − 1 + , . (b − y)(b − x) (b − y)(b − x)
Kw (y, x) = max W (y), W (x) −
Proof. Obviously, a, x∈
I1 b−I2 b−I1
I1 b−I2 b−I1
I2 − I1 (b + x) + xb
(4.3)
̸= ∅. For fixed y ∈ a, I1bb−−I1I2 , it is easy to check that I1bb−−I1I2 , I1yy−−I1I2 ̸= ∅, so we choose
, I1yy−−I1I2 . The quadrature formula (4.1) will be accurate for all polynomials of degree ≤ 2 when
A1 =
I2 − I1 (b + x) + xb
(b − y)(x − y) −I2 + I1 (b + y) − by A2 = (x − y)(b − x) I2 − I1 (x + y) + xy A3 = . (b − y)(b − x) , (x − y)(b − y) > 0 we have A1 ≥ 0. Further, since I12 ≤ I2 we have y <
I1 b−I2 b−I1 bx+I2 −I1 (x+b) (x−y)(b−y)
Since x ≥
I1 b−I12 I1 b−I2 b−I1 b−I1 I2 −I1 (b+x)+xb . (b−y)(x−y)
≤
= I1 , so we have
≤ 1. Therefore, A1 ∈ [0, 1], so there exist x1 ∈ [a, b] such that W (x1 ) = I b −I On the other hand, since y ≤ 1b−I 2 and (b − y)(b − x) > 0 we have A3 ∈ [0, 1], so there exist x3 ∈ [a, b] such that 1
A1 =
I −I (x+y)+xy
W (x3 ) = 1 − 2 (b−1 y)(b−x) . It is easy to check that W (x1 ) ≤ W (x3 ), which implies x1 ≤ x3 . Apply Theorem 4 with y, x2 = x, x1 and x3 as above.
Now, we consider the special case of symmetric normalized weight function on [−1, 1] Analog results for interval [a, b] can easily be obtained with linear transformation t =
b−a 2
x+
b+a 2
.
Corollary 3. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1], w : [−1, 1] → [0, ∞) symmetric normalized 1 I weight function. Further, let −1 ≤ y < −I2 and −I2 ≤ x ≤ − y2 , where I2 = −1 w(t )t 2 dt. If f is continuous at y and x, then the following inequality holds:
∫ 1 1 x + I2 −y − I 2 xy + I2 ≤ Kw (y, x) · (f ), f ( y ) + f ( x ) + f ( 1 ) − w( t ) f ( t ) dt (x − y)(1 − y) (x − y)(1 − x) (1 − y)(1 − x) −1 −1
(4.4)
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A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
where
x + I2 , W (y) − , (x − y)(1 − y) (x − y)(1 − y) xy + I2 xy + I2 W (x) − 1 + , . (1 − y)(1 − x) (1 − y)(1 − x) x + I2
Kw (y, x) = max W (y), W (x) −
Proof. Apply Theorem 5 with a = −1, b = 1, and I1 = 0.
(4.5)
Remark 5. If we put y = −1 in (4.4), then −I2 ≤ x ≤ I2 . Specially, for x = −I2 or x = I2 , the Radau 2-point inequalities are obtained. For x = 0, the Lobatto 3-point inequality is obtained. Remark 6. If we put x = − y2 , for some y ∈ (−1, −I2 ), the inequality related to the open 2-point quadrature formula. I
√
Specially, for x = −y =
I2 , the Gauss 2-point inequality is obtained:
∫ 1 f (− I2 ) + 1 f ( I2 ) − 2 2
1
−1
1 w(t )f (t )dt ≤ Kw (− I2 , I2 ) · (f ),
(4.6)
−1
where
Kw (− I2 , I2 ) = max W (− I2 ), W ( I2 ) −
, W (− I2 ) − 2 1
, 1 − W ( I2 ) . 2 1
(4.7)
5. Weighted 3-point quadrature formulae of closed type Now, we will apply all the results to obtain bounds of the reminder E (f ) of the general weighted 3-point quadrature formulae of closed type b
∫
w(t )f (t )dt = A1 f (a) + A2 f (x) + A3 f (b) + E (f )
(5.1)
a
∑3
where k=1 Ak = 1. Note that in our case we have A1 = W (x1 ),
A2 = W (x3 ) − W (x1 ),
A3 = W (b) − W (x3 ),
so if we take a normalized weight function w(t ), we have 3 −
Ak = W (b) = 1.
k=1
In the case of a symmetric normalized weight function we can formulate the following result. Theorem 6. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1], w : [−1, 1] → [0, ∞⟩ be a normalized symmetric weight function and x1 ∈ [−1, 0]. If f is continuous at − 1 and 0, then the following inequality holds:
∫ A1 (f (−1) + f (1)) + A2 f (0) −
1
−1
1 w(t )f (t )dt ≤ Kw (x1 ) (f ),
(5.2)
−1
t
where A1 = W (x1 ), A2 = 1 − 2W (x1 ), W (t ) = −1 w(s)ds and Kw (x1 ) = max W (x1 ),
Proof. Apply Corollary 1 with x1 ∈ [−1, 0], x2 = 0 and x3 = −x1 .
1 2
− W (x1 ) .
Remark 7. The quadrature formula related to the inequality (5.2) is exact for all polynomials of degree ≤ 1. Specially, for x1 such that W (x1 ) = 16 , the weighted version of famous Simpson’s rule is obtained
∫ 1 (f (−1) + f (1)) + 2 f (0) − 6 3
1
−1
1 1 w(t )f (t )dt ≤ (f ).
This result has been obtained by Dragomir in [7].
3 −1
(5.3)
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
1371
In the next result the inequality with the smallest constant Kw (x1 ) is derived. Corollary 4. Let f : [−1, 1] → R be a function of the bounded variation on [−1, 1] and let w : [−1, 1] → [0, ∞⟩ be some symmetric normalized weight function. If f is continuous at − 1 and 0, then the following inequality holds
∫ 1 f (−1) + 1 f (0) + 1 f (1) − 4 2 4
1
−1
1 1 w(t )f (t )dt ≤ (f ).
Proof. The constant Kw (x1 ) = max W (x1 ),
1 − 2W (x1 ) =
1 2
and Kw (x1 ) =
1 . 4
(5.4)
4 −1
1 2
− W (x1 ) will be minimal if W (x1 ) =
1 2
− W (x1 ) = 14 . Thus A1 = 41 , A2 =
Corollary 5. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1], w : [−1, 1] → [0, ∞⟩ be a normalized x 1 symmetric weight function, x1 such that −11 w(t )dt = 0 t 2 w(t )dt and x3 = −x1 . If f is continuous at − 1 and 0, then the following inequality holds:
∫ A1 (f (−1) + f (1)) + A2 f (0) − where A1 =
1 0
1
−1
1 w(t )f (t )dt ≤ Kw (x1 ) (f ),
(5.5)
−1
t 2 w(t )dt , A2 = 1 − 2A1 and Kw (x1 ) = max
x1 −1
w(t )dt ,
w( t ) dt . x1
0
Proof. Since
∫
x1
−1
w(t )dt =
1
∫
1
∫
t 2 w(t )dt ≤ 0
w(t )dt =
∫
0
w(t )dt ,
−1
0
t
it is obvious that −1 ≤ x1 ≤ 0 ≤ x3 ≤ 1. Therefore we can apply Corollary 1 with W (t ) = −1 w(s)ds and x = 0. Because of x the symmetry of the weight function w , we have W (x1 ) = 1 − W (x3 ) = −11 w(t )dt and W (x) − W (x1 ) = W (x3 ) − W (x) =
0 x1
w(t )dt, so the assertion follows easily.
Remark 8. The quadrature rule related to the inequality (5.5)
∫
1
w(t )f (t )dt ≈ A1 (f (−1) + f (1)) + A2 f (0) −1
is accurate for all polynomials of degree ≤3 and is called the weighted Lobatto quadrature formula. In case of non-symmetric weight function we can formulate the following result. Theorem 7. Let f : [a, b] → R be a function of bounded variation on [a, b], w : [a, b] → [0, ∞⟩ be a normalized weight function and a ≤ x1 ≤ x ≤ x3 ≤ b. If f is continuous at a and x, then the following inequality holds:
∫ A1 f (a) + A2 f (x) + A3 f (b) −
b a
b w(t )f (t )dt ≤ Kw (x1 , x, x3 ) (f ),
(5.6)
a
where A1 = W (x1 ), A2 = W (x3 ) − W (x1 ), A3 = 1 − W (x3 ), W (t ) =
t a
w(s)ds and
Kw (x1 , x, x3 ) = max{W (x1 ), W (x) − W (x1 ), W (x3 ) − W (x), 1 − W (x3 )}. Proof. The proof follows directly from Remark 3.
b Remark 9. If we take x = a+ , x1 and x3 such that W (x1 ) = 2 formula is obtained. Thus
1 6
and W (x3 ) =
5 , 6
then the weighted version of Simpson’s
∫ b b 1 f (a) + 2 f a + b + 1 f (b) − ≤ max 1 , W a + b − 1 , 5 − W a + b w( t ) f ( t ) dt (f ). 6 3 2 6 6 2 6 6 2 a a Remark 10. The minimal constant Kw (x1 , x, x3 ) is achieved when W (x1 ) =
∫ 1 f (a) + 1 f (x) + 1 f (b) − 4 2 4
a
b b 1 w(t )f (t )dt ≤ (f ). 4 a
1 4
, W (x) =
1 2
and W (x3 ) =
3 . 4
(5.7)
Thus we get (5.8)
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6. Weighted 2-point quadrature formulae of semi-closed type In this section we apply all the results to establish bounds of the remainder E (f ) of the general weighted 2-point quadrature formulae of semi-closed type. Such a formula is obtained by putting A3 = 0 in (5.1), that is b
∫
w(t )f (t )dt = A1 f (a) + A2 f (x) + E (f ),
(6.1)
a
where A1 + A2 = 1. Note that in this case we have x3 = b, A1 = W (x1 ) and A2 = W (b) − W (x1 ), so if w(t ) is normalized weight function, then A1 + A2 = W (b) = 1. In the case of non-symmetric weight function, we formulate the following result concerning quadrature formulae with degree of exactness 1. Theorem 8. Let f : [a, b] → R be a function of bounded variation on [a, b], w : [a, b] → [0, ∞⟩ is some normalized weight b function, I ≤ x ≤ b, where I = a w(t )tdt. If f is continuous at x and a, then the following inequality holds
∫ b b I −x a−I ≤ Kw (x) (f ), w( t ) f ( t ) dt f ( a ) + f ( x ) − a − x a−x a a
(6.2)
where Kw (x) = max
I −x a−x
, W (x) −
I −x a−x
, 1 − W (x) ,
t
w(s)ds. Proof. Since a < I < b, we have [I , b] ̸= ∅. The quadrature formula (6.1) will be accurate for all polynomials of degree ≤ 1 and W (t ) =
a
when A1 =
I −x a−x
,
A2 =
a−I a−x
.
x ∈ [0, 1]. Apply Theorem 4 with y = a, x2 = x, x3 = b and x1 such that Since x ∈ [I , b], we have A1 = aI − −x I −x It is easy to check that a−x ≤ W (x), so the assertion follows directly.
x1 a
w(t )dt =
I −x . a −x
In case of symmetric normalized weight function we can formulate the following result. Corollary 6. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1], w : [−1, 1] → [0, ∞⟩ is some symmetric normalized weight function and 0 ≤ x ≤ 1. If f is continuous at x and − 1, then the following inequality holds
∫ 1 1 x 1 ≤ Kw (x) (f ), f (− 1 ) + f ( x ) − w( t ) f ( t ) dt 1 + x 1+x −1 −1
(6.3)
where Kw (x) = max
x 1+x
, W ( x) −
x 1+x
, 1 − W (x)
t
and W (t ) = −1 w(s)ds. Proof. Apply Theorem 8 with a = −1, b = 1 and I1 = 0. Remark 11. If we take x = 1, then Kw (1) =
∫ 1 f (−1) + 1 f (1) − 2 2
1
−1
1
−1
so the inequality (6.3) reduces to the weighted trapezoid inequality
1 1 w(t )f (t )dt ≤ (f ). 2
(6.4)
−1
Remark 12. If we take x = 0, then Kw (0) =
∫ f (0) −
1 , 2
1 1 w(t )f (t )dt ≤ (f ). 2 −1
1 , 2
so the inequality (6.3) reduces to the weighted midpoint inequality (6.5)
Remark 13. In order to enlarge degree of exactness, we have to consider quadrature formulae of Radau type which are 1 accurate for all polynomials of degree ≤ 2. Therefore we have to take x = −1 w(t )t 2 dt in (6.3).
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1373
The next result establishes the inequality related to the 2-point semi-closed quadrature formula with the minimal constant K (x). Corollary 7. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1], w : [−1, 1] → [0, ∞⟩ is some symmetric 1
normalized weight function such that −21 w(t )dt =
∫ 1 f (−1) + 2 f 1 − 3 3 2
1
−1
2 . 3
If f is continuous at
1 2
and − 1, then the following inequality holds
1 1 w(t )f (t )dt ≤ (f ). 3
(6.6)
−1
Proof. Obviously, constant Kw (x) is minimal if x
x
= W (x) −
1+x
1+x
= 1 − W (x) =
t
where W (t ) = −1 w(s)ds. That is x =
1 2
1 3
,
and W (x) =
2 . 3
Therefore inequality (6.3) becomes (6.6).
7. Applications for some special weight functions 7.1. Case w(t ) =
1 2
, t ∈ [−1, 1]
Corollary 8. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1] and −1 ≤ y < − 13 ≤ x < 1. If f is continuous at y and x, then the following inequality holds:
∫ 1 x + 1/3 −y − 1/3 xy + 1/3 1 1 ≤ Kw (y, x) · (f ), f ( y ) + f ( x ) + f ( 1 ) − f ( t ) dt (x − y)(1 − y) (x − y)(1 − x) (1 − y)(1 − x) 2 −1 −1
(7.1)
where
x + 1/3
, 2 2
, Kw (y, x) = max − (x − y)(1 − y) y + 1 x − 1 x + 1/3 xy + 1/3 xy + 1/3 2 − (x − y)(1 − y) , 2 + (1 − y)(1 − x) , (1 − y)(1 − x) . y + 1 x + 1
Proof. Apply Corollary 3 with w(t ) = √
Remark 14. For y = − 1+5
6
1 2
and W (t ) = t +2 1 . Obviously, I2 =
1 , 3
so the assertion follows easily.
√
and x =
6−1 5
we get Radau 3-point inequality:
√ √ √ ∫ 1 16 − √6 23√6 − 8 1+ 6 16 + 6 6−1 1 1 1 − + + f f f ( 1 ) − f ( t ) dt · (f ). ≤ 36 5 36 5 9 2 −1 180 −1
(7.2)
For y = −1 and x = 0 we get the Simpson inequality:
∫ 1 f (−1) + 2 f (0) + 1 f (1) − 1 6 3 6 2
1
−1
1 1 f (t )dt ≤ · (f ). 3
(7.3)
−1
For x = − 13 we get Radau 2-point inequality:
∫ 3 f − 1 + 1 f (1) − 1 4 3 4 2
1
−1
1 5 f (t )dt ≤ · (f ). 12
(7.4)
−1
For y = − √1 and x = √1 we get Legendre–Gauss 2-point inequality: 3
3
∫ 1 1 1 1 f − √1 + 1 f √1 − 1 ≤ √ f ( t ) dt · (f ). 2 2 3 2 2 −1 3 3 −1 Obviously, the best estimate is obtained for Radau 3-point formula.
(7.5)
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Corollary 9. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1] and 0 ≤ x ≤ 1. If f is continuous at x, then the following inequality holds
∫ 1 2 x 1 1 1 ≤ 1+x f ( t ) dt f (− 1 ) + f ( x ) − (f ). 2(1 + x) 1 + x 1+x 2 −1 −1 Proof. Apply Corollary 6 with w(t ) = 1+x2 . 2(1+x)
and W (t ) =
1 2
t +1 . 2
(7.6)
It is easy to check that Kw (x) = max
2
, 1+x , 1−2 x 1+x 2(1+x) x
=
Remark 15. For x = 0, 31 , 1 we get the midpoint, Radau 2-point and trapezoid inequality, respectively. Namely,
∫ f (0) − 1 2
1
−1
f (t )dt ≤
1 1
2 −1
∫ 1 f (−1) + 3 f 1 − 1 4 4 3 2
(f ),
1
−1
f (t )dt ≤
1 5
12 −1
(f )
and
∫ 1 f (−1) + 1 f (1) − 1 2 2 2
1
−1
1 1 f (t )dt ≤ (f ). 2 −1
Obviously, the best estimate is obtained for Radau 2-point formula. 7.2. Case w(t ) =
√1 π
1−t 2
, t ∈ ⟨−1, 1⟩
Corollary 10. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1] and let −1 ≤ y ≤ − 12 ≤ x ≤ 1. If f is continuous at y and x, then the following inequality holds:
∫ 1 1 x + 1/2 −y − 1/2 xy + 1/2 1 f (y) + f (x) + f (1) − (f ), (7.7) f (t )dt ≤ Kw (y, x) · √ (x − y)(1 − y) (x − y)(1 − x) (1 − y)(1 − x) −1 π 1 − t 2 −1 where
Kw (y, x) = max
1
+
, +
1 + 2x
arcsin y 1
+
,
arcsin x
2 π 2 2(x − y)(y − 1) π 1 1 + 2x arcsin y 1 x + y + xy arcsin x 1 + 2xy + , + , + . 2 2(x − y)(y − 1) + π 2 2(1 − x)(1 − y) π 2(1 − y)(1 − x)
Proof. Apply Corollary 3 with w(t ) = √
Remark 16. For y = − 1+4
√1 π
1−t 2
, t ∈ ⟨−1, 1⟩ and W (t ) =
arcsin t
π
+ 12 , t ∈ [−1, 1].
√ 5
5 −1 4
and x =
we get the Radau 3-point inequality:
√ √ ∫ 1 1 2 1 1+ 5 2 5−1 1 f (t ) + f + f (1) − dt ≤ (f ). √ f − 5 −1 5 4 5 4 5 −1 π 1 − t 2
(7.8)
For y = −1 and x = 0 we get the Lobatto inequality
∫ 1 f (−1) + 1 f (0) + 1 f (1) − 4 2 4
f (t )
1
√
−1
π 1 − t2
dt ≤
1 1
4 −1
(f ).
(7.9)
For x = − 12 we get the Radau 2-point inequality:
∫ 2 f − 1 + 1 f (1) − 3 2 3
1
f (t )
√
−1
π 1 − t2
dt ≤
1 1
3 −1
(f ).
(7.10)
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
√
For y = −
2 2
1375
√ 2 2
and x =
we get the Gauss–Chebyshev 2-point inequality:
√ √ ∫ 1 1 1 1 2 2 f (t ) 1 + f − dt ≤ (f ). √ f − 2 4 −1 2 2 2 −1 π 1 − t 2
(7.11)
Corollary 11. Let f : [−1, 1] → R be a function of a bounded variation on [−1, 1]. If 0 ≤ x ≤ then the following inequality holds
1 2
and if f is continuous at x,
∫ 1 1 x f (t ) 1 ≤ 1 − arcsin x f (− 1 ) + f ( x ) − dt (f ). √ x + 1 x+1 2 π −1 π 1 − t 2 −1 If
1 2
(7.12)
≤ x ≤ 1, then the following inequality holds ∫ 1 1 x 1 f (t ) x dt ≤ (f ). √ x + 1 f (−1) + x + 1 f (x) − x + 1 −1 −1 π 1 − t 2
Proof. Apply Corollary 6 with w(t ) =
√1 π
1 −t 2
and W (t ) =
1 2
+
arcsin t
π
.
(7.13)
Remark 17. For x = 0, 12 , 1 we get the midpoint, Radau 2-point and trapezoid inequality, respectively. Namely,
∫ f (0) −
f (t )
1
√
dt ≤
π 1 − t2 ∫ 1 f (−1) + 2 f 1 − −1
3
3
2
1
−1
1 1
2 −1
(f ),
(7.14)
f (t )
1 1 dt ≤ (f ) √ 3 π 1 − t2
(7.15)
−1
and
∫ 1 f (−1) + 1 f (1) − 2 2
1
f (t )
√
−1
π 1 − t2
dt ≤
1 1
2 −1
(f ).
(7.16)
Obviously, the best estimate for 2-point semi-closed quadrature rule is obtained for the Radau quadrature formula (7.15). 7.3. Case w(t ) = π2
√
1 − t 2 , t ∈ [−1, 1]
Corollary 12. Let f : [−1, 1] → R be a function of bounded variation on [−1, 1] and let −1 ≤ y ≤ − 14 ≤ x ≤ 1. If f is continuous at y and x, then the following inequality holds:
∫ 1 1 x + 1/4 −y − 1/4 xy + 1/4 2 f (y) + f (x) + f (1) − 1 − t 2 f (t )dt ≤ Kw (y, x) · (f ), (x − y)(1 − y) (x − y)(1 − x) (1 − y)(1 − x) −1 π −1 where
√ π + y 1 − y2 − arccos y 1 + 4x π + x 1 − x2 − arccos x Kw (y, x) = max , + 4(x − y)(y − 1) π π 1 + 4x π + y 1 − y2 − arccos y × + , 4(x − y)(y − 1) π √ 4x + 4y − 3 π + x 1 − x2 − arccos x 1 + 4xy + . , 4(1 − x)(1 − y) 4(1 − x)(1 − y) π
Proof. Apply Corollary 3 with w(t ) = π2
√
√ 1 − t 2 , t ∈ [−1, 1] and W (t ) =
π+t
1−t 2 −arccos t
π
, t ∈ [−1, 1].
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√
Remark 18. For y = − 1+6
√ 7
7 −1 6
and x =
we get the Radau 3-point inequality:
√ √ √ ∫ 1 1 1 13 − √7 2 1+ 7 7−1 13 + 7 1 1 − t 2 f (t )dt ≤ f − f + + f (1) − (f ). 5 −1 28 6 28 6 14 −1 π
(7.17)
For y = −1 and x = 0 we get the Lobatto inequality
∫ 1 f (−1) + 3 f (0) + 1 f (1) − 8 4 8
1 3 2 1 − t 2 f (t )dt ≤ (f ). 8 −1 −1 π 1
(7.18)
For x = − 14 we get the Radau 2-point inequality:
∫ 4 f − 1 + 1 f (1) − 5 4 5 For y = − 12 and x =
1 2
1
−1
2
π
1 − t 2 f (t )dt ≤
√
1 5 15 − 16π + 80 arccos(1/4)
(f ).
80π
(7.19)
−1
we get the Gauss–Chebyshev 2-point inequality of the second:
∫ 1 f −1 + 1f 1 − 2 2 2 2
1
−1
2
π
√ 1 1 3 1 − t 2 f (t )dt ≤ + (f ). 6 4π −1
(7.20)
Corollary 13. Let f : [−1, 1] → R be a function of a bounded variation on [−1, 1]. If 0 ≤ x ≤ 1 and if f is continuous at x, then the following inequality holds
√ ∫ 1 1 x 1 2 1 x 1 − x2 − arccos x 2 (f ). f (− 1 ) + f ( x ) − 1 − t f ( t ) dt ≤ + x + 1 x+1 1+x π −1 π −1 Proof. Apply Corollary 6 with w(t ) = π2
√
1 − t 2 and W (t ) =
(7.21)
√ π+t
1−t 2 −arccos t
π
.
Remark 19. For x = 0, 41 , 1 we get the midpoint, Radau 2-point and trapezoid inequality, respectively. Namely,
∫ 1 √ 1 1 2 1 − t 2 f (t ) (f ), dt ≤ f (0) − 2 −1 π −1 ∫ 1 √ 1 1 5√15 + 64π − 80 arccos 1 4 1 2 1 − t 2 f (t ) 4 − dt ≤ (f ) f (−1) + f 5 5 4 π 80π −1 −1
(7.22)
(7.23)
and
∫ 1 √ 1 1 1 1 2 1 − t 2 f (t ) dt ≤ (f ). f (−1) + f (1) − 2 2 −1 2 π −1
(7.24)
√ 5 15+64π −80 arccos 1
4 Since ≈ 0.457481, the best estimate for 2-point semi-closed quadrature rule is obtained for the Radau 80π quadrature formula (7.23).
7.4. Case w(t ) =
3 2
√
t , t ∈ [0, 1]
Corollary 14. Let f : [0, 1] → R be a function of a bounded variation on [0, 1] and let 0 ≤ y ≤ at y and x, then the following inequality holds:
3 7
≤ x ≤ 1. If f is continuous
∫ 1 √ 1 2(3 − 7y) 15 + 35xy − 21(x + y) 3 tf (t ) 2(7x − 3) f (y) + f (x) + f (1) − dt ≤ Kw (y, x) (f ), 35(x − y)(1 − y) 35(1 − x)(x − y) 35(x − 1)(y − 1) 2 0 0
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
1377
where
2(7x − 3)
3/2 2(7x − 3) , y + 7(x − y)(y − 1) 7(x − y)(y − 1) 3/2 14 + 34xy − 20(x + y) 15 + 35xy − 21(x + y) × x + , 35(x − y)(1 − y) 35(x − 1)(y − 1)
Kw (y, x) = max y3/2 , x3/2 +
Proof. Apply Theorem 5 with w(t ) =
3 2
√
t , t ∈ [0, 1] and W (t ) = t 3/2 , t ∈ [0, 1].
√
Remark 20. For y =
15−2 15 33
√
and x =
15+2 15 33
the Radau 3-point inequality is obtained
√ √ √ 150 − 13√15 15 − 2 15 15 + 2 15 150 + 13 15 f f + 350 33 350 33 ∫ 1 √ 1 3 tf (t ) 1 (f ). dt ≤ 0.287532 + f (1) − 7 2 0 0 For y = 0 and x =
5 9
(7.25)
the Lobatto inequality is obtained:
∫ 1 √ 1 16 3 tf (t ) 243 5 3 (f ). f (0) + f + f (1) − dt ≤ 0.371628 175 350 9 14 2 0 0 For x =
3 7
(7.26)
the Radau 2-point inequality is obtained
∫ 1 √ 1 7 3 3 3 tf (t ) (f ). + f (1) − dt ≤ 0.419434 f 10 7 10 2 0 0 √
For y =
35−2 70 63
(7.27)
√
and x =
35+2 70 63
the inequality of Gauss–Jacobi type is obtained:
√ √ √ ∫ 1 √ 1 50 + √70 35 − 2 70 50 − 70 35 + 2 70 3 tf (t ) (f ). f + f − dt ≤ 0.327786 100 63 100 63 2 0 0 Corollary 15. Let f : [0, 1] → R be a function of bounded variation on [0, 1]. If then the following inequality holds
3 7
≤x≤
3 2/5 5
and if f is continuous at x,
∫ 1 3 1 x − 3 3√ 5 5 f (0) + f (x) − tf (t )dt ≤ (1 − x3/2 ) (f ). x x 0 2 0 If
(7.28)
(7.29)
3 2/5 5
≤ x ≤ 1, then the following inequality holds ∫ 1 1 x − 3 x − 35 3/5 3√ 5 f (0) + f (x) − tf (t )dt ≤ (f ). x x x 0 2 0
Proof. Apply Theorem 8 with w(t ) =
3 2
√
t and W (t ) = t 3/2 .
, 57 , 1 we get midpoint, Radau 2-point and trapezoid inequality, respectively. Namely, ∫ √ 3/2 1 3 1 3 tf (t ) 3 − dt ≤ 1 − (f ), f 5 2 5 0 0 √ 1 ∫ 1 √ 4 21 5 3 tf (t ) 5 35 − dt ≤ 1 − (f ) f (0) + f 25 25 7 2 49 0 0
Remark 21. For x =
(7.30)
3 5
(7.31)
(7.32)
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and
∫ 1 √ 1 2 2 3 tf (t ) 3 dt ≤ (f ). f (0) + f (1) − 5 5 0 5 2 0 √
Since 1 −
5 35 49
≈ 0.396318 and 1 −
3 3/2 5
(7.33)
≈ 0.535242 the best estimate is obtained for the Radau 2-point formula.
1 , t ∈ ⟨0, 1] 7.5. Case w(t ) = √ 2 t
Corollary 16. Let f : [0, 1] → R be a function of bounded variation on [0, 1] and let 0 ≤ y ≤ y and x, then the following inequality holds:
1 5
≤ x ≤ 1. If f is continuous at
1 2(5y − 1) 3 + 15xy − 5(x + y) 2(5x − 1) ≤ Kw (y, x) (f ), f ( y ) + f ( x ) + f ( 1 ) 15(x − y)(1 − y) 15(x − 1)(x − y) 15(1 − x)(1 − y) 0 where
√ √ √ 2(5x − 1) 2(5x − 1) , y− Kw (y, x) = max y, x − 15(x − y)(1 − y) 15(x − y)(1 − y) √ 3 + 10(x + y) 3 + 15xy − 5(x + y) , × x + . 15(x − 1)(y − 1) 15(1 − x)(1 − y) 1 Proof. Apply Theorem 5 with w(t ) = √ , t ∈ ⟨0, 1] and W (t ) =
√
2 t
√ 2 Remark 22. For y = 7−21
7
t , t ∈ [0, 1].
√ 2 and x = 7+21
7
, the Radau 3-point inequality is obtained:
√ √ √ √ 1 ∫ 1 14 + √7 7−2 7 14 − 7 7+2 7 1 f (t ) 7−2 7 (f ). f + f + f (1) − √ dt ≤ 30 21 30 21 15 21 0 2 t 0 For y = 0 and x =
3 , 7
the Lobatto type inequality is obtained:
∫ 16 f (0) + 49 f 3 + 1 f (1) − 45 90 7 10 For x =
1 5
1 0
f (t )
1 16 (f ). √ dt ≤ 45 2 t
(7.35)
0
the Radau 2-point inequality is obtained:
∫ 5 f 1 + 1 f (1) − 6 5 6 √
For y =
15−2 30 35
1 0
f (t )
1 1 (f ). √ dt ≤ √ 2 t 5
(7.36)
0
√
and x =
15+2 30 35
we get the inequality of Gauss–Jacobi type:
√ √ √ ∫ 1 √ 1 18 + √30 15 − 2 30 18 − 30 15 + 2 30 f (t ) 15 − 2 30 f + f − (f ). √ dt ≤ 36 35 36 35 35 0 2 t 0 Corollary 17. Let f : [0, 1] → R be a function of bounded variation on [0, 1]. If then the following inequality holds
∫ 1 1 x − 1 √ 1/3 1 3 f (0) + f (x) − √ f (t )dt ≤ (1 − x) (f ). x x 0 2 t 0 If
(7.34)
1 3
≤x≤
1 2/3 3
(7.37)
and if f is continuous at x,
(7.38)
1 2/3 3
≤ x ≤ 1, then the following inequality holds ∫ 1 1 x − 1 x − 13 1/3 1 3 f (0) + f (x) − (f ). √ f (t )dt ≤ x x x 0 2 t 0
1 Proof. Apply Theorem 8 with w(t ) = √ and W (t ) = 2 t
√ t.
(7.39)
A. Aglić Aljinović et al. / Mathematical and Computer Modelling 54 (2011) 1365–1379
Remark 23. For x =
∫ 1 f 3 −
1 0
, 35 , 1 we get midpoint, Radau 2-point and trapezoid inequalities. Namely, 1 f (t ) 1 (f ), √ dt ≤ 1 − 3 2 t
1379
1 3
(7.40)
0
∫ 4 f (0) + 5 f 3 − 9 9 5
1 0
f (t )
1 4 (f ) √ dt ≤ 9 2 t
(7.41)
0
and
∫ 2 f (0) + 1 f (1) − 3 3 Since 1 −
1 3
1 0
≈ 0.42265 and
f (t )
1 2 (f ). √ dt ≤ 3 2 t
(7.42)
0
4 9
≈ 0.444444 the best estimate is obtained for the midpoint formula.
References [1] S.S. Dragomir, J.E. Pečarić, S. Wang, The unified treatment of trapezoid, Simpson and Ostrowski type inequalities for monotonic mappings and applications, Math. Comput. Modelling 31 (6–7) (2001) 61–70. [2] D.S. Mitrinović, J.E. Pečarić, A.M. Fink, Inequalities for Functions and their Integrals and Derivatives, Kluwer Academic Publishers, Dordrecht, 1994. [3] S. Kovač, J. Pečarić, Weighted version of general integral formula of the Euler type, Math. Inequal. Appl. 13 (3) (2010) 579–599. [4] S. Kovač, J. Pečarić, A. Vukelić, A generalization of general two-point formula with applications in numerical integration, Nonlinear Anal. Ser. A: TMA 68 (8) (2008) 2445–2463. [5] S. Kovač, J. Pečarić, A. Vukelić, Weighted generalization of the trapezoidal rule via Fink identity, Aust. J. Math. Anal. Appl. 4 (1) (2007) Article 8, 1–12 (electronic). [6] V. Čuljak, J. Pečarić, L.E. Persson, Weighted Simpson type inequalities and an extension of Montgomery identity, J. Math. Inequal. (manuscript). [7] S.S. Dragomir, On Simpson’s quadrature formula for mappings of bounded variation and applications, Tamkang J. Math. 30 (1999) 53–58.