Gauss Hermite interval quadrature rule

Transcription

1 Computers and Mathematics with Applications 54 (2007) Gauss Hermite interval quadrature rule Gradimir V. Milovanović, Alesandar S. Cvetović Department of Mathematics, Faculty of Electronic Engineering, University of Niš, P.O. Box 73, 8000 Niš, Serbia Received 9 December 2006; received in revised form 2 January 2007; accepted 3 January 2007 Abstract The existence and uniqueness of the Gaussian interval quadrature formula with respect to the Hermite weight function on R is proved. Similar results have been recently obtained for the Jacobi weight on [, ] and for the generalized Laguerre weight on [0, + ). Numerical construction of the Gauss Hermite interval quadrature rule is also investigated, and a suitable algorithm is proposed. A few numerical examples are included. c 2007 Elsevier Ltd. All rights reserved. Keywords: Interval Gaussian quadrature rule; Hermite weight function; Nodes; Weights. Introduction By the Gaussian interval quadrature formula for the positive weight function w, we assume a quadrature formula of the following form b a f w dx µ x +h x h f w dx, (.) which integrates exactly all polynomials of degree less than 2n, and where the subintervals (x h, x + h ), =,..., n, do not overlap. Several results have been published on the existence of the previous inds of quadratures in the last thirty years (cf. [ 5]). The question of the existence of bounded a, b was proved in [6] in a much wider context. Suppose that w is a weight function on [, ], i.e., a nonnegative Lebesgue integrable function, such that for I = (α, β) [, ], α β, we have I w(x) dx 0. In [6], the following result is proved: Given the ordered set of odd integers {ν,..., ν n }, with the property n + n ν = N +, the Chebyshev system of functions {u 0,..., u N } on [, ], the Marov system The wor was supported by the Swiss National Science Foundation (SCOPES Joint Research Project No. IB ) and the Serbian Ministry of Science and Environmental Protection (Project #44004G). Corresponding author. Tel.: ; fax: addresses: grade@junis.ni.ac.yu, grade@elfa.ni.ac.yu (G.V. Milovanović), aca@elfa.ni.ac.yu (A.S. Cvetović) /$ - see front matter c 2007 Elsevier Ltd. All rights reserved. doi:0.06/j.camwa

2 G.V. Milovanović, A.S. Cvetović / Computers and Mathematics with Applications 54 (2007) of functions {v 0,..., v m } on [, ], where m = max{ν,..., ν n }, and a set of the lengths h 0,..., h n 0, with h <, there exists an interpolatory quadrature formula of the form f (x)w(x) dx ν ν=0 µ,ν I f (x)v ν (x)w(x) dx, where intervals I [, ], =,..., n, are non-overlapping, with the length of I equals, which integrates exactly every element of the linear span {u 0,..., u N }. In [7], it was proved that for the Legendre weight w(x) = on [, ], the Gaussian interval quadrature rule is unique. The uniqueness for the corresponding formula with respect to the Jacobi weight on [, ] and its numerical construction were given in [8]. The existence and uniqueness of the Gauss Lobatto and Gauss Radau interval quadratures for the Jacobi weight was proved in [9]. For the special case of the Chebyshev weight of the first ind and a special set of lengths, analytic solutions were derived. Recently, Bojanov and Petrov [0] proved the existence and uniqueness of the weighted Gaussian interval quadrature formula for a given system of continuously differentiable functions, which constitute an ET system of order two on a finite interval [a, b]. The case of interval quadratures of the Gaussian type on unbounded intervals was for the first time treated in [], where the existence and uniqueness of the Gaussian interval quadrature formula with respect to the generalized Laguerre weight function have been presented, including an algorithm for the numerical construction of such a formula. In this paper, we complete our investigation for all classical weight functions (cf. [2,3]), i.e., we prove the corresponding results for the Gaussian interval quadrature on R with respect to the Hermite weight function w(x) = exp( x 2 ). The paper is organized as follows. In Section 2 we introduce some definitions and formulate the main result. In Section 3 we give some preliminary results. The proof of the main result is presented in Section 3. Finally, a few numerical examples are given in Section The main result and Let h = (h,..., h n ) and H, M, ε 0 > 0. We denote } Hn {h H = R n h 0, =,..., n, h < H, X n (h) = { x R n < x h x + h < < x n h n x n + h n < + }, X n (h) = { x R n < x h x + h x n h n x n + h n < + }, X M,ε 0 n (h) = { x R n M < x h, x + h + x h > ε 0, =,..., n, x n + h n < M }, dµ = w(x) dx = exp( x 2 ) dx, φ =, ψ = 2x, such that (φw) = ψw. Let P m be the set of all algebraic polynomials of degree at most m. Definition 2.. Given h Hn H, the Gauss Hermite interval quadrature rule is an interpolation quadrature rule of the form p dµ = µ where x X n (h), and it exists. I p dµ, p P 2n, (2.)

3 546 G.V. Milovanović, A.S. Cvetović / Computers and Mathematics with Applications 54 (2007) Theorem 2.. Let w be the Hermite weight and a =, b = +. For every H > 0, there exist ε 0 > 0 and M > 0, such that for every h Hn H, the Gauss Hermite quadrature rule (.), with the nodes x Xε 0,M n and positive weights µ, =,..., n, exists uniquely. 3. Preliminary results In order to express our results in a more condensed form, we adopt the following definitions for the intervals I = (x h, x + h ), I = [x h, x + h ], =,..., n, O = (, x h ), O + = (x + h, x + h + ), =,..., n, O n+ = (x n + h n, + ), O = (+, x h ], O n+ = [x n + h n, + ), O + = [x + h, x + h + ], =,..., n, n I = I, O = R \ I. In order to prove the main theorem, we first present some auxiliary results (for similar results see [7,8,]). Lemma 3.. (i) Assume j 2, =,..., n, with n j = N +, h Hn H, x X n (h) and f m,, m =, j, =,..., n, are arbitrary numbers, then the interpolation problem p (m ) dµ = f m,, m =, j, =,..., n, (3.) I has a unique solution in P N, where for m = 2 we tae µ as the Lebesgue measure. (ii) Assume that j 2, =,..., n, with n j = N +, h Hn H, x X n (h); then, for every c C, there exists the unique q c P N, such that p = cx N+ + q c solves the following interpolation problem p (m ) dµ = 0, m =, j, =,..., n, I and there holds q c = cq, where for m = 2 the µ is the Lebesgue measure. Proof. To prove this lemma, we show that the corresponding homogenous system has only the trivial solution. Note that the conditions can be expressed as a system of linear equations for the coefficients of p. For the first part, we can simply count zeros to see that in every subinterval I, there are j zeros. So, in total we have j = N + zeros, which means that if the solution is not trivial, its degree is at least N +, and therefore it is not a solution in P N. For the second part, we can rewrite the interpolation problem in the following form q c (m ) dµ = c I Now, we can apply the first part of this lemma with f m, = c (x N+ ) (m ) dµ, m =, j, =,..., n, I I (x N+ ) (m ) dµ, m =, j, =,..., n. (3.2) to the interpolation problem (3.2), and denote the unique solution by q c. Obviously, the linear system of equations which defines q c, has a free vector multiplied by c, so that q c = cq. Lemma 3.2. Suppose h Hn H x X n (h). and there exists a Gauss Hermite quadrature rule with nodes x X n (h), then This part of the sentence for m = 2 is missing in [8,9], and [].

4 Proof. Let h H H n G.V. Milovanović, A.S. Cvetović / Computers and Mathematics with Applications 54 (2007) and x X n (h), but x X n (h). Then at least one of the equalities x + h = x + h +, =,..., n, holds. Suppose we have x + h = x + h + ; then according to the interpolation Lemma 3., there exists a monic polynomial p P 2n 2, such that p dµ = p dµ = 0, I + I + p (m ) dµ = 0, m =, 2, ν =,...,, + 2,..., n. 2h ν I ν Obviously such a p annihilates the Gauss Hermite interval quadrature sum, and it has a constant sign on O. This means that O p dµ = p dµ 0, which is a contradiction. The following lemma is crucial, since it allows us to treat the problem for an unbounded interval with tools which are designed for compact supporting sets. Lemma 3.3. There exists an M > 0, such that for every h Hn H Gauss Hermite quadrature rule (2.), there holds x < M, =,..., n. Proof. Suppose it is not the case. Then, for every M > 0, there exists h M Hn H x M X n (h), such that we have x M > M and nodes x X n(h) of the corresponding and a respective set of nodes at least for one {,..., n}. Let us assume that we have and + negative and positive nodes, respectively, that are not bounded as M increases. Then, in total we have n + 0 nodes which remain bounded as M increases. Suppose those nodes are bounded by some L, i.e., x M < L, = +,..., n +. We can always choose L sufficiently large such that n + + = + h M < L, since all other nodes are unbounded; actually it is enough to choose L bigger than H. Since the nodes x M and the lengths h M, = +,..., n +, are bounded, we can always extract the convergent sequences. Denote by x j and h j j, =,..., n, the sequences of nodes and lengths such that the nodes x, =,..., and = n + +,..., n, are unbounded and that the nodes x j and the lengths h j, = +,..., n +, j N 0, are convergent. Now consider the sets ( L, L) \ n + I j = +, j N 0, for every j N 0, which are not empty. Moreover, since H < L and the sequences of the nodes x j and the lengths h j, = +,..., n +, are convergent, there exists an interval (a, b) of positive length, such that n + + (a, b) ( L, L) \ I j, j > j 0, 2h j = + for a j 0 sufficiently large. Now, consider a polynomial p j of degree 2n, given by p (m ) j dµ = 0, m =, 2, =,..., n, I j 2h j n I j n p j dµ = 0,

5 548 G.V. Milovanović, A.S. Cvetović / Computers and Mathematics with Applications 54 (2007) with the leading coefficient, which exists, according to Lemma 3., part (ii). Obviously, this polynomial p j annihilates the quadrature formula: since it is of degree 2n it must be p j dµ = O p j dµ = 0. p j has positive sign on the set O j \ O j j n+, and negative on the set On+, so that, it must be p j dµ ( p j ) dµ = 0. O\O n+ O n+ We can give the following estimation for the first integral b p j dµ (x a) n (b x) n 2 dµ = J > 0, O\O n+ a where we used the fact that (a, b) O, and that polynomial p j has n zeros smaller than a and n 2 zeros bigger than b. As we can see quantity J does not depend on j, and it is constant. On the other hand, for the second integral we have the bound O n+ ( p j ) dµ so that we have p j dµ O\O n+ + x j n +h j n (x x j + h j )2n dµ(x) = J j 2 > 0 O n+ ( p j ) dµ J J j 2, where J j 2 depends on j, and J does not. Consider now the following monic polynomial 2h j I j p j dµ = 0, 2h j I j p (m ) j dµ = 0, m =, 2, = 2,..., n, which exists according to the Lemma 3., part (ii), it is of degree 2n, and it annihilates quadrature formula, so it must be p j dµ = O p j dµ = 0. p j is of positive sign on O j \ O j j and negative on O, so that it must be ( p j ) dµ + p j dµ = 0, O j O j \O j using the same reasoning we can conclude that b p j dµ (x a) n 3 (b x) n 4 dµ = J 3 > 0, O j \O j which does not depend on j, and O j ( p j ) dµ a x j h j so that we have ( p j ) dµ + O j O j \O j (x j n + h j n x) 2n dµ = J j 4 > 0, p j dµ J 3 J j 4. We are going to prove that at least one of the quantities J 3 and J j 4 tends to zero as x j h j and x j n + h j n tend to and +, respectively. According to this fact, we have that at least one of the quantities J J j 2 or J 3 J j 4, is positive, which produces a contradiction. For reasons of brevity, we introduce the following notation x j h j = M j and x j n + h j n = M j +.

6 G.V. Milovanović, A.S. Cvetović / Computers and Mathematics with Applications 54 (2007) At first, we see that for w(x) = e x2, m N, and M m, as well as sup x m w /2 (x) = M m w /2 (M), x M, x M + M x m w(x) dx M m w /2 (M)ε(M), ε(m) = + M w /2 (x) dx, where obviously ε(m) 0 as M tends to +. According to these facts, for M j + and M j sufficiently large, we have J j + 2 = (x + M j )2n w(x) dx = Also, we get i.e., J j M j + 2n ν=0 ( ) 2n + (M j ν )2n ν x ν w(x) dx M j + + ( + M j )2n x 2n w(x) dx M j + + (M j + ( + M j ))2n w /2 (M j + ) w /2 (x) dx (2M j + M j )2n w /2 (M j + )ε(m j + ). j M 4 = (M j + x)2n w(x) dx = J j 4 (2M j M j + )2n w /2 (M j )ε(m j ). M j + + M j (x + M j + )2n w(x) dx, Now suppose that (2M j + M j )2n w /2 (M j + ) tends to some C > 0 as M j and M j + tend to infinity. Then C /(2n ) M j 2M j + w/(2(2n )) (M j + ). Using the fact that lim x + x µ w λ (x) = 0, for each λ, µ > 0, we conclude that J j ) 4 ((M ( ( )) ε(m j ) j )2n w /4 (M j ) C /(2n ) C /(2n ) 2M j + w/(2(2n )) (M j + )w/4 2M j + w/(2(2n )) (M j + ) ) ((M j + )2n w /(2(2n )) (M j + ) 2 2n 0. C/(2n ) In the case that (2M j + M j )2n w /2 (M j j + ) tends to +, using some similar reasoning we get J4 0. Lemma 3.4. In the Gauss Hermite interval quadrature rule (2.), we have µ 0, =,..., n. Proof. Suppose there is some {,..., n}, such that µ = 0. According to the interpolation Lemma 3., there exists p P 2n 2 such that p (m ) dµ = 0, m =, 2, ν =,...,, +,..., n. 2h ν I ν This p annihilates the Gauss Hermite interval quadrature sum, but it has a constant sign on O, such that we have O p dµ = p dµ 0, which is a contradiction.

7 550 G.V. Milovanović, A.S. Cvetović / Computers and Mathematics with Applications 54 (2007) Definition 3.. We denote n Ω = (x x h )(x x + h ), and and Ω = Ω, =,..., n, (x x h )(x x + h ) (Ω φw) = (Ω φw)(x + h ) (Ω φw)(x h ), h 0, (Ω φw) = x [(Ω φw)(x )], h = 0, for =,..., n, where we use the short notation x = / x. Theorem 3.. For every h Hn H, the nodes of the quadrature rule (2.) satisfy the system of equations (Ω φw) = 0, =,..., n. For h H H n, every solution x X n(h) of the system (3.3) defines the nodes for the Gauss Hermite quadrature rule (2.). Proof. Applying the Gauss Hermite interval quadrature rule (2.) to the polynomial (Ω ν φw) /w of degree 2n, we get (Ων φw) µ (Ω ν φw) 0 = dµ = w I dµ = µ ν ν (Ω ν φw), w i.e., if x are nodes of the Gauss Hermite quadrature rule they must satisfy (3.3), since according to Lemma 3.4, we have µ ν 0, ν =,..., n. For any p P 2n 2, we have ( p(x + h ) Ω (x + h ) + p(x ) h ) Ω = 0. (3.4) (x h ) This can be proved by applying the Cauchy Residue Theorem to the rational function p/ω, over the contour Λ = {x x = R} in the complex x-plane, where R is sufficiently large. Note that for p P 2n 2, we have p/ω x 2 as x, which gives p(x) dx = 0. Λ Ω(x) Now, suppose that for x X n (h), we have (Ω φw)(x + h ) + (Ω = 0. (3.5) φw)(x h ) Then obviously, according to (3.4), we have for any p P 2n 2 0 = (Ω φw)(x + h ) ((pφw)(x + h ) (pφw)(x h )) = (Ω φw)(x + h ) I (pφw) dx. But also (pφw) dµ = (pφw) + w = 0, (3.3)

8 G.V. Milovanović, A.S. Cvetović / Computers and Mathematics with Applications 54 (2007) so that for every r P 2n of the form r = p φ + pψ, p P 2n 2, we have r dµ = C (Ω r dµ = 0, φw)(x + h ) I for an arbitrary C. All we need to do is to choose C, such that the formula is exact for all r P 2n. Thus, an element of P 2n, which is not of the form p φ + pψ, p P 2n is the constant polynomial, so that we can choose C = m 0 m 0, (Ω φw)(x +h ), m 0 = µ(r), m 0, = µ(i ). The system of equations (3.5) defines the Gauss Hermite quadrature rule, and it is equivalent to (3.3). It is enough to note that Ω (x ± h ) = ± Ω (x ± h ), =,..., n. Using these equations, we can conclude by definition of Ω, that (Ω φw)(x + h ) = (Ω φw)(x + h ) > 0, for h Hn H and x X n(h), which gives C > 0. Thus, all weights in the constructed quadrature rule are positive. To be completely fair, we need to give an explanation for the case h = 0 for some. Then, the corresponding term of (3.4) is given by p Ω pω (x ), Ω 2 and it can be transformed to the form p Ω pω Ω 2 = p Ω φw + pω (φw) pω (φw) pω φw Ω 2 φw = (pφw) Ω p(ω φw) Ω 2 φw. We require that the term with p vanish, so that we have (Ω φw) (x ) = x [(Ω φw)(x )] = 0. This is exactly what equation of the system (3.3) is when h = 0. Lemma 3.5. The weights in the Gauss Hermite interval quadrature rule (2.) are positive. Proof. Actually, we have µ = ν= m 0 m 0,ν (Ω ν φw)(x ν +h ν ) where m 0 = µ(r), m 0, = µ(i )/( ), =,..., n. Since x X n (h), all terms are positive., (Ω φw)(x + h ) =,..., n, (3.6) Lemma 3.6. There exist ε 0 > 0 and M > 0, such that for all h Hn H and all nodes x X n(h) of the Gauss Hermite quadrature rule (2.), we have x X ε 0,M n. Proof. The existence of M is already proved, so that we prove now only the existence of ε 0. Assume the contrary; then for every ε 0 > 0, there exists h ε 0 Hn H and the corresponding set of nodes xε 0 X n (h ε 0), for which the interpolation quadrature rule (2.) is Gaussian (i.e., exact on P 2n ), with the property that at least one of the following equalities holds: x ε 0 + h ε 0 + ε 0 = x ε 0 + hε 0 +, =,..., n,

9 552 G.V. Milovanović, A.S. Cvetović / Computers and Mathematics with Applications 54 (2007) Since the sets h ε 0 and x ε 0 are bounded, there are convergent sequences h, x, N, with limits h 0 and x 0, such that at least one of the equalities x 0 + h0 = x0 + h0 +, =,..., n, holds. Since weights µ ν, ν =,..., n, are continuous functions of h and x, according to (3.6), for h (0) and respective set of nodes x (0), we have that the rule µ 0 p dµ, 2h 0 I 0 is exact for p P 2n, because of continuity. Since, for this Gauss Hermite interval quadrature rule, we have at least two intervals which have the boundary point in common, we can apply the same argument as in the proof of Lemma 3.2 to produce a contradiction. This means that the statement of Lemma 3.6 is correct. 4. Proof of the main result To prove the main result, we are going to need the following topological result, which can be found in [4,5] and [6]. Assume that D is a bounded open set in R n, with the closure D and the boundary D, and Φ : D R n is a continuous mapping. By deg(φ, D, c), we denote the topological degree of Φ with respect to D and c Φ( D). Lemma 4.. (i) If deg(φ, D, c) 0, the equation Φ(x) = c has a solution in D. (ii) Let Φ(x, λ) be a continuous map Φ : D [0, ] R n, such that c Φ( D, [0, ]), then deg(φ(x, λ), D, c) is a constant independent of λ. (iii) Suppose Φ C (D), c Φ( D) and det(φ (x)) 0 for any x D such that Φ(x) = c. Then, the equation Φ(x) = c has only finitely many solutions x ν in D, and there holds deg(φ, D, c) = x ν sgn ( det(φ (x ν )) ). Now we are ready to prove the main result. Proof of Theorem 2.. We are solving the system of equations Ψ = (Ω φw) = 0, =,..., n. (4.) Suppose x X ε 0,M n (h) is a solution of (4.). Then we have ( x Ψ = (Ω φw)(x + h ) (x ν + h x ν h ν )(x h x ν h ν ) ) + (x + h x ν + h ν )(x h x ν + h ν ) + 2 > 0, and the inequality is obvious. Similarly, ( xm Ψ = (Ω φw)(x + h ) (x + h x m h m )(x + h x m + h m ) ) + < 0, (x h x m h m )(x h x m + h m ) from where the inequality is obvious. Also it is clear that x Ψ + m xm Ψ = 2(Ω φw)(x + h ) > 0,

10 which gives G.V. Milovanović, A.S. Cvetović / Computers and Mathematics with Applications 54 (2007) x Ψ > xm Ψ = xm Ψ. m m This means that the Jacobian is diagonally dominant, with positive elements on the main diagonal and negative elsewhere. Thus, sgn( xm Ψ m,,...,n ) =. The rest of the proof goes exactly as it is given in [7]. The proof has N steps, where N is defined by h = (N +η) ε 0 4, 0 < η, with h = max{h,..., h n }. At the j-th step, uniqueness is proved for the h ( j) = ( j + η) ε 0 4h h, j = 0,,..., N. In the first step, the mappings ( ) Φ (0) (x, λ) = (x, λh (0) ),..., n (x, λh (0) ) are considered on X ε 0,M n (0) for each 0 λ. It is obvious Φ (0) (x, 0) = 0 has a solution for λ = 0, and that solution is unique. Namely, this solution is, really, the classical Gauss Hermite quadrature rule. Since the sign of the determinant of the Jacobian is positive, using Lemma 4., we conclude that deg(φ (0) (x, 0), X ε 0,M n (0), 0) =. For x X ε 0,M n (0) and 0 λ, we have 0 < x λh, x + λh < x + λh +, =,..., n. Then, for any solution x of the system Φ (0) (x, λ) = 0, we have that the sign of det(j(x, λh (0) )) is positive. Hence, according to Lemma 4. part (ii), we have deg(φ (0) (x, λ), X ε 0,M n (0), 0) =, for all λ [0, ], and in particular for λ =. This means that the system Φ (0) (x, ) = 0 has a unique solution in X ε 0,M n (0). It is also the unique solution on the smaller set X ε 0,M n (h (0) ), according to Lemma 3.6. In the case N 0, we proceed with the same arguments to the mappings Φ () (x, λ) = ( (x, λh () + ( λ)h (0) ),..., n (x, λh () + ( λ)h (0) )), to prove that there is the unique solution in X ε 0,M n (h (0) ), which is also unique in the set X ε 0,M n (h (0) ), according to Lemma 3.6. After that, the same arguments are iterated to the mappings Φ ( j) (x, λ) = ( (x, λh (j) + ( λ)h (j ) ),..., n (x, λh (j) + ( λ)h (j ) )), until j reaches N. Note that we have proved existence and uniqueness. 5. Numerical examples In this section, we present some numerical results. At first it is clear that formula (3.6) is perfect for the numerical construction of the weight coefficients µ, =,..., n, since all terms involved are positive. The only problem which may occur is that the points x + h and x + h + can be very close. In that case, we can calculate Ω only with a limited precision. For the calculation of nodes x, =,..., n, we find an inspiration in the proof of Theorem 2.. It is clear that system of equations (3.3) defines nodes x, =,..., n, as the implicit functions of h Hn H, and those functions are continuous. This means that we can start with nodes of the classical Gauss Hermite quadrature rule, for which h = 0, and then increase h for small amounts, and hope that Newton Kantorovich method for the system (3.3) will converge. It might happen that x + h > x + h +, in which case we should restart the Newton Kantorovich process with a smaller increment of h.

11 554 G.V. Milovanović, A.S. Cvetović / Computers and Mathematics with Applications 54 (2007) Table 5. Nodes and weights in (2.) for n = 50 and h = (2 3,..., 2 3 ) x µ ± ( ) ± ( ) ± ( ) ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± Table 5.2 Nodes and weights in (2.) for n = 0 and h = (3/4, 3/4, 3/4, 3/4, 3/4, /8, /8, /8, /8, /8) x µ ( ) ( ) ( ) ( ) ( ) ( ) It is interesting, since there is no bound on the real line for the placement of nodes, that we can increase h for quite large amounts. For example, in Table 5., we present nodes and weights for an interval quadrature (2.), when n = 50 and h = (2 3,..., 2 3 ). Starting with nodes for the ordinary Gauss Hermite quadrature rule, using the Mathematica pacage OrthogonalPolynomials [7], we need just 8 iterations in the numerical construction. This is quite an amazing performance, since in similar situations for the Laguerre and Jacobi measures, we have much worse performance, due to the boundness of the support for the Laguerre and Jacobi measures (see [8,]). Table 5.2 presents nodes and weights for the Gauss Hermite interval quadrature rule (2.) for n = 0 and h = (3/4, 3/4, 3/4, 3/4, 3/4, /8, /8, /8, /8, /8). The construction is performed first for the vector of lengths h = (/8,..., /8), using as starting values nodes for the ordinary Gauss Hermite quadrature rule (h = 0). Then the first five components of h are increased by the amounts In all computations, we needed at most 0 iterations. However, we have 25 intermediate steps, which maes it a really painful process.

12 G.V. Milovanović, A.S. Cvetović / Computers and Mathematics with Applications 54 (2007) References [] Fr. Pitnauer, M. Reimer, Interpolation mit intervallfuntionalen, Math. Z. 46 (976) 7 5. [2] R.N. Sharipov, Best interval quadrature formulae for Lipschitz classes, in: Constructive Function Theory and Functional Analysis, Kazan University, Kazan, 983, Issue 4, pp (in Russian). [3] A.L. Kuzmina, Interval quadrature formulae with multiple node intervals, Izv. Vuzov 7 (28) (980) (in Russian). [4] V.F. Babeno, On a certain problem of optimal integration, in: Studies on Contemporary Problems of Integration and Approximation of Functions and Their Applications, Collection of Research Papers, Dnepropetrovs State University, Dnepropetrovs, 984, pp. 3 3 (in Russian). [5] V.P. Motornyi, On the best quadrature formulae in the class of functions with bounded r-th derivative, East J. Approx. 4 (998) [6] B. Bojanov, P. Petrov, Gaussian interval quadrature formula, Numer. Math. 87 (200) [7] B. Bojanov, P. Petrov, Uniqueness of the Gaussian interval quadrature formula, Numer. Math. 95 (2003) [8] G.V. Milovanović, A.S. Cvetović, Uniqueness and computation of Gaussian interval quadrature formula for Jacobi weight function, Numer. Math. 99 (2004) [9] G.V. Milovanović, A.S. Cvetović, Gauss Radau and Gauss Lobatto interval quadrature rules for Jacobi weight function, Numer. Math. 02 (2006) [0] B. Bojanov, P. Petrov, Gaussian interval quadrature formula for a Tchebycheff system, SIAM J. Numer. Anal. 43 (2005) [] G.V. Milovanoivć, A.S. Cvetović, Gauss Laguerre interval quadrature rule, J. Comput. Appl. Math. 82 (2005) [2] R.P. Agarwal, G.V. Milovanović, A characterization of the classical orthogonal polynomials, in: P. Nevai, A. Pinus (Eds.), Progress in Approximation Theory, Academic Press, New Yor, 99, pp. 4. [3] R.P. Agarwal, G.V. Milovanović, Extremal problems, inequalities, and classical orthogonal polynomials, Appl. Math. Comput. 28 (2002) [4] R.P. Agarwal, M. Meehan, D. O Regan, Fixed Point Theory and Applications, in: Cambridge Tracts in Mathematics, vol. 4, Cambridge University Press, Cambridge, 200. [5] J.M. Ortega, W.C. Rheinboldt, Iterative Solution of Nonlinear Equations in Several Variables, Academic Press, New Yor, 970. [6] J.T. Schwartz, Nonlinear Functional Analysis, Gordon and Breach, New Yor, 969. [7] A.S. Cvetović, G.V. Milovanović, The mathematica pacage OrthogonalPolynomials, Facta Univ. Ser. Math. Inform. 9 (2004) 7 36.