Quadrature Rules With an Even Number of Multiple Nodes and a Maximal Trigonometric Degree of Exactness

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1 Filomat 9:10 015), DOI 10.98/FIL151039T Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: Quadrature Rules With an Even Number of Multiple Nodes and a Maximal Trigonometric Degree of Exactness Tatjana V. Tomović a, Marija P. Stanić a a Department of Mathematics and Informatics, Faculty of Science, University of Kragujevac, Kragujevac, Serbia Abstract. This paper is devoted to the interpolatory quadrature rules with an even number of multiple nodes, which have the maximal trigonometric degree of exactness. For constructing of such quadrature rules we introduce and consider the so called s and σ orthogonal trigonometric polynomials. We present a numerical method for construction of mentioned quadrature rules. Some numerical examples are also included. 1. Introduction For a nonnegative integer n, by T n we denote the linear space of all trigonometric polynomials of degree less than or equal to n, by T n the linear space T n span{sin nx} or T n span{cos nx}, and by T the set of all trigonometric polynomials. Let us suppose that a function w is integrable and nonnegative on the interval [, π), vanishing there only on a set of a measure zero. We consider a quadrature rule of the following type f x) wx) dx = n s ν where s ν, ν = 1,,..., n, are nonnegative integers. A j,ν f j) x ν ) + R n f ), 1) j=0 Definition 1.1. The quadrature rule of the form 1) has trigonometric degree of exactness equal to d if R d f ) = 0 for every f T d, and if there exists T d+1 for which R d ) 0. We are interested in quadrature rule of the form 1) which has the maximal trigonometric degree of exactness. We consider the questions of existence and uniqueness of such quadrature rules, as well as their numerical construction. Let us notice that the quadrature rule 1) has an even number of nodes. The corresponding quadrature rules with an odd number of nodes were first considered in [8] for constant 010 Mathematics Subject Classification. Primary 65D30, 65D3 Keywords. Quadrature Rules, Trigonometric Degree Received: 08 August 014; Accepted: 16 March 015 Communicated by Marko Petković The authors were supported in part by the Serbian Ministry of Education, Science and Technological Development grant numbers #174015) addresses: tomovict@kg.ac.rs Tatjana V. Tomović), stanicm@kg.ac.rs Marija P. Stanić)

2 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), weight function wx) = 1 and the case when all of the nodes have the same multiplicity. Quadrature rules with fixed number of free nodes of fixed different multiplicities at nodes, but again only for the constant weight function wx) = 1, were considered in [4]. Finally, the general case of quadrature rules with an odd number of nodes with different multiplicities and with respect to arbitrary weight function, were considered in [9]. In this paper we pay our attention to quadrature rules with an even number of nodes and the maximal trigonometric degree of exactness. We finish this Section with some known results for a generalized Gaussian problem, given in [8]. Section is devoted to quadrature rules with simple nodes. The case when all of the nodes have the same multiplicity is considered in Section 3, while the case of different multiplicities at nodes is considered in Section 4. Finally, the numerical method for constructing of considered quadrature rules is given in Section 5, where one numerical example is given, too A generalized Gaussian problem Let the quadrature formula b a f x)wx) dx = m N 1 A j,ν f j) x ν ) + R f ) ) j=0 be such that E f ) = 0 R f ) = 0, where E is a linear differential operator of order N. According to [8, p. 8] quadrature rules which are exact for all trigonometric polynomials of degree less than or equal to ν are relative to the differential operator of order ν + 1): E = d ν ) d dx dx + k. Ghizzeti and Osicini in [8] considered whether there can exist a rule of the form b a f x)wx) dx = m N p ν 1 j=0 A j,ν f j) x ν ) + R f ), 3) with fixed integers p ν, 0 p ν N 1, ν = 1,,..., m, such that at least one of the integers p ν is greater than or equal to 1, satisfying E f ) = 0 R f ) = 0, too. They proved the following theorem see [8, p. 45]). Theorem 1.. For the given nodes x 1, x,..., x m, the linear differential operator E of order N and the nonnegative integers p 1, p,..., p m, 0 p ν N 1, ν = 1,,..., m ν {1,,..., m}) p ν 1), consider the following homogenous boundary differential problem E f ) = 0, f j) x ν ) = 0, j = 0, 1,..., N p ν 1, ν = 1,,..., m. 4) If this problem has no non trivial solutions whence N mn m p ν) it is possible to write a quadrature rule of the type 3) with mn m p ν N parameters chosen arbitrarily. If, on the other hand, the problem 4) has q linearly independent solutions U r, r = 1,,..., q, with N mn + m p ν q p ν for all ν = 1,,..., m, then the formula 3) may apply only if the following q conditions b U r x) wx) dx = 0, r = 1,,..., q, 5) a are satisfied; if so, mn m p ν N + q parameters in the formula 3) can be chosen arbitrary.. Quadrature rules with simple nodes First we consider the special case when s 1 = s = = s n = 0, i.e., the quadrature rule with simple nodes f x) wx) dx = n w ν f x ν ) + R n f ). 6)

3 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), For a given set of different nodes x ν [, π)), ν = 1,,..., n, quadrature rule 6) of the interpolatory type is exact for every t T n 1, and in addition for t T n, but not for all t T n see []). If the nodes x ν [, π)), ν = 1,,..., n, are not specified in advanced, one can try to find them such that quadrature rule 6) has the maximal trigonometric degree of exactness, i.e., such that quadrature rule 6) is exact for every t T n 1. Lemma.1. The trigonometric polynomial of degree n, T n x) = c 0 + n c k cos kx + d k sin kx), c k R k = 0, 1,..., n), d k R k = 1,..., n), c n + d n 0, 7) which is orthogonal on [, π) with respect to the weight function w to every trigonometric polynomial from T n 1 is determined uniquely, if the values of its leading coefficients, c n and d n, are given. Proof. The orthogonality conditions can be written in the form T n x) cos νx wx) dx = 0, ν = 0, 1,..., n 1, T n x) sin νx wx) dx = 0, ν = 1,,..., n 1. By changing T n x) in these equations and introducing the following notations cos νx cos kx wx) dx = α k,ν, k, ν = 0, 1,..., n, cos νx sin kx wx) dx = β k,ν, ν = 0, 1,..., n, k = 1,,..., n, sin νx sin kx wx) dx = γ k,ν, k, ν = 1,,..., n, we obtain the following system of linear equations for determining the unknown coefficients c 0, c 1, d 1,..., c n 1, d n 1 : ) c 0 α 0,ν + ck α k,ν + d k β k,ν = cn α n,ν d n β n,ν, ν = 0, 1,..., n 1, 8) ) c 0 β ν,0 + ck β ν,k + d k γ ν,k = cn β ν,n d n γ ν,n, ν = 1,,..., n 1. The determinant of this system is equal to α 0,0 α 1,0 β 1,0 α,0 β,0 α n 1,0 β n 1,0 α 0,1 α 1,1 β 1,1 α,1 β,1 α n 1,1 β n 1,1 β 1,0 β 1,1 γ 1,1 β 1, γ 1, β 1,n 1 γ 1,n 1 α 0, α 1, β 1, α, β, α n 1, β n 1, = β,0 β,1 γ,1 β, γ, β,n 1 γ,n α 0,n 1 α 1,n 1 β 1,n 1 α,n 1 β,n 1 α n 1,n 1 β n 1,n 1 β n 1,0 β n 1,1 γ n 1,1 β n 1, γ n 1, β n 1,n 1 γ n 1,n 1 Since α k,ν = α ν,k, k, ν = 0, 1,..., n 1, and γ k,ν = γ ν,k, k, ν = 1,,..., n 1, the determinant is symmetric.

4 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), Now, we consider the following quadratic form in variables ξ 0, ξ 1, η 1, ξ, η,..., ξ n 1, η n 1 : F = α k,ν ξ k ξ ν + β k,ν ξ ν η k + γ k,ν η k η ν = ν=0 k=0 + = ν=0 π wx) ξ ν cos νx ξ k cos kx dx + wx) ξ ν cos νx η k sin kx dx ν=0 k=0 wx) η ν sin νx η k sin kx dx wx) ξ ν cos νx + η k sin kx ν=0 dx > 0. The quadratic form F is positive and its determinant is equal to, thus, > 0, and the system of linear equations 8) has the unique solution for the unknown coefficients c 0, c 1, d 1,..., c n 1, d n 1. Obviously, T n x) = A n, A 0 is a constant, 9) is a trigonometric polynomial of degree n. To the contrary, every trigonometric polynomial of degree n of the form 7) can be represented in the form 9) with A = 1) n n 1 ic n id n )e i/ n x ν, where x 1, x,..., x n are the zeros of the trigonometric polynomial 7), that lie in the strip Re x < π see [10]). By using similar arguments as in [15], it is easy to prove the following result. Lemma.. Every trigonometric polynomial of degree n 1, n 1 T n 1 x) = a 0 + a k cos kx + b k sin kx), can be uniquely represented in the form T n 1 x) = A n x)b n 1 x) + R n x), where A n x) is a certain trigonometric polynomial of degree n and B n 1 x), R n x) are wanted polynomials from T n 1 and T n, respectively. By virtue of Lemma., simulating the development of the famous Gaussian quadrature rules for algebraic polynomials, the following result can be easily proved. Theorem.3. The interpolatory type quadrature rule 6) is of Gaussian type, i.e., it is exact for every t T n 1, if and only if the nodes x ν, ν = 1,,..., n, are the zeros of trigonometric polynomial T n x), which is orthogonal on [, π) with respect to the weight function wx) to every trigonometric polynomial from T n 1. It is well known that trigonometric polynomial of degree n can not have more than n distinct zeros in [, π) see [10]). Now we prove that the zeros of orthogonal trigonometric polynomials are all simple. Theorem.4. The trigonometric polynomial T n T n which is orthogonal on [, π) with respect to the weight function wx) to every trigonometric polynomial from T n 1, has in [, π) exactly n distinct simple zeros. ν=0

5 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), Proof. The trigonometric polynomial T n must have at least one zero of odd multiplicity in [, π). Indeed, if we assume the contrary, then for n N we obtain that T n x) cos 0x wx) dx = 0, which is impossible, because the integrand does not change its sign on [, π). Also, T n must change its sign an even number of times see [1, ]). Let us now suppose that the number of zeros of T n on [, π) is m, m < n. We denote these zeros by y 1, y,..., y m, and set tx) = m sin x y k. Since t T m, m < n, we have T nx)tx) wx) dx = 0, which again gives a contradiction, since the integrand does not change its sign on [, π). Therefore, T n must have exactly n different simple zeros on [, π). 3. Quadrature rules with multiple nodes with the same multiplicity and s orthogonal trigonometric polynomials In this section we consider quadrature rule of the form 1) where s 1 = s = = s n = s > 0, i.e., f x) wx) dx = n s j=0 A j,ν f j) x ν ) + R n f ), 10) which has the maximal trigonometric degree of exactness, i.e., such that R n f ) = 0 for all f T ns+1) 1. The boundary differential problem 4) in this case has the following form where E f ) = 0, f j) x ν ) = 0, j = 0, 1,..., s, ν = 1,,..., n, 11) E = d dx ns+1) 1 d dx + k ) is the differential operator of order N = 4ns + 1) 1. For n > 0 the boundary problem 11) has n 1 linear independent non trivial solutions see [8, p. 141]): U l x) = V l x) = n n s+1 s+1 cos lx, l = 0, 1,..., n 1, sin lx, l = 1,,..., n 1. According to Theorem 1., for s 1 = s = = s n = s, the nodes x 1, x,..., x n of the quadrature rule 10) satisfy the following conditions n n s+1 s+1 cos lx wx) dx = 0, l = 0, 1,..., n 1, sin lx wx) dx = 0, l = 1,,..., n 1,

6 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), i.e., x 1, x,..., x n are the zeros of the trigonometric polynomial T s,n which satisfies the following equation T s,n x)) s+1 t n 1 x) wx) dx = 0, for arbitrary t n 1 T n 1. 1) We call such trigonometric polynomials as s orthogonal trigonometric polynomials with respect to the weight function wx) on [, π). In order to prove the existence and uniqueness of T s,n x) we use the following well known facts about the best approximation see [3, p ]). Remark 3.1. Let X be a Banach space and Y be a closed linear subspace of X. For each f X, the error of approximation of f by elements from Y is defined as inf Y f. If there exists some = 0 Y for which that infimum is attained, then 0 is called the best approximation to f from Y. For each finite dimensional subspace X n of X and each f X, there exists the best approximation to f from X n. In addition, if X is a strictly convex space, then each f X has at most one element of the best approximation in each closed linear subspace Y X. Theorem 3.. Trigonometric polynomial T s,n x), with given leading coefficients, which is s orthogonal on [, π) with respect to a given weight function wx) is determined uniquely. Proof. Let us set X = L s+ [, π], u = wx) 1/s+) c n cos nx + d n sin nx) L s+ [, π], and fix the following n 1 linearly independent elements in L s+ [, π]: u j = wx) 1/s+) cos jx, j = 0, 1,..., n 1, v k = wx) 1/s+) sin kx, k = 1,,..., n 1. Here, Y = span{u 0, u 1, v 1, u, v,..., u n 1, v n 1 } is a finite dimensional subspace of X and, according to Remark 3.1, for each element from X there exists the best approximation from Y, i.e., there exist n 1 constants α j, j = 0, 1,..., n 1, β k, k = 1,,..., n 1, such that the error α u 0 u 0 + α k u k +β k v k )) = c n cos nx+d n sin nx s+ 1/s+) α 0 + α k cos kx+β k sin kx))) wx) dx), is minimal, i.e., for every n and for every choice of the leading coefficients c n, d n, c n + d n 0, there exists a trigonometric polynomial of degree n T s,n x) = c n cos nx + d n sin nx α 0 + such that T s,n x)) s+ wx) dx ) α k cos kx + β k sin kx), is minimal. Since the space L s+ [, π] is strictly convex, according to Remark 3.1, the problem of the best approximation has the unique solution, i.e., the trigonometric polynomial T s,n is unique. It follows that for each of the following n 1 functions F C k λ) = T s,n x) + λ cos kx) s+ wx) dx, k = 0, 1,..., n 1, F S k λ) = T s,n x) + λ sin kx) s+ wx) dx, k = 1,,..., n 1, its derivative must be equal to zero for λ = 0. Therefore, we get T s,n x)) s+1 cos kx wx) dx = 0, k = 0, 1,..., n 1, T s,n x)) s+1 sin kx wx) dx = 0, k = 1,,..., n 1, which means that the polynomial T s,n x) satisfies s orthogonality conditions 1).

7 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), Theorem 3.3. Trigonometric polynomial T s,n x) which is s orthogonal on [, π) with respect to a given weight function wx) has in [, π) exactly n distinct simple zeros. Proof. The trigonometric polynomial T s,n x) has on [, π) at least one zero of odd multiplicity. If we assume the contrary, for n 1 we obtain the following contradiction to 1) T s,n x)) s+1 cos 0x wx) dx 0, since T s,n x)) s+1 does not change its sign on [, π). Also, T s,n must change its sign on [, π) an even number of times see [1, ]). Let us now suppose that the number of zeros of T s,n on [, π) of odd multiplicities is m, m < n. Let us denote these zeros by y 1, y,..., y m and set tx) = m sin x y k. Since t T m, m < n, we get T s,n x)) s+1 tx) wx) dx = 0, which is a contradiction, because the integrand does not change its sign on [, π). Therefore, T s,n must have exactly n different simple zeros on [, π). 4. Quadrature rules with multiple nodes with different multiplicities and σ orthogonal trigonometric polynomials Let us denote σ = s 1, s,..., s n ) and N 1 = n s ν + 1) 1. We study quadrature rules of the form 1), which have maximal trigonometric degree of exactness, i.e., for which R n f ) = 0 for all f T N1. In this case boundary differential problem 4) has the following form where E f ) = 0, f j) x ν ) = 0, j = 0, 1,..., s ν, ν = 1,,..., n, 13) E = d dx N 1 d dx + k ) is a differential operator of order N = N The boundary problem 13) has the following n 1 linear independent nontrivial solutions U l x) = V l x) = n n ) sν +1 cos lx, l = 0, 1,..., n 1, ) sν +1 sin lx, l = 1,,..., n 1. According to Theorem 1., the nodes x 1, x,..., x n of the quadrature rule 1) satisfy conditions n n ) sν +1 cos lx wx) dx = 0, l = 0, 1,..., n 1, ) sν +1 sin lx wx) dx = 0, l = 1,,..., n 1,

8 i.e., n Trigonometric polynomial T σ,n x) = n T.V. Tomović, M.P. Stanić / Filomat 9:10 015), ) sν +1 tx) wx) dx = 0, for all t T n 1. 14) which satisfies condition 14) will be called σ orthogonal trigonometric polynomial with respect to the weight function wx) on [, π). Since the dimension of T n 1 is n 1, we have n 1 orthogonality conditions. The σ orthogonal trigonometric polynomial of degree n has n + 1 coefficients, which means that two of them can be fixed in advanced. Alternatively, if we directly compute the nodes of the σ orthogonal trigonometric polynomial of degree n, we can fix one of them in advance since for n nodes we have n 1 orthogonality conditions. Using the notation from the Theorem 1., we have mn m p ν N + q = nn n N s ν 1) n ) s ν + 1) n 1 = 0. Therefore, if one of the nodes is fixed, the quadrature rule 1) is unique. By the same arguments as in Theorem 3.3 and using conditions 14) one can prove the following theorem. Theorem 4.1. The σ orthogonal trigonometric polynomial T σ,n with respect to the weight function wx) on [, π) has exactly n distinct simple zeros on [, π). The existence of σ orthogonal trigonometric polynomials will be proved using theory of implicitly defined orthogonality. The existence of implicity defined orthogonal algebraic polynomials was proved in [6], while the existence of implicitly defined orthogonal trigonometric polynomials of semi integer degree was proved in [9]. Here we prove the existence of implicitly defined orthogonal trigonometric polynomials. Theorem 4.. Let p be a nonnegative continuous function, vanishing only on a set of a measure zero. Then there exists a trigonometric polynomial T n, of degree n, orthogonal on [, π) to every trigonometric polynomial of degree less than or equal to n 1 with respect to the weight function pt n x))wx). Proof. Let us denote by T n the set of all trigonometric polynomials of degree n which have n real distinct zeros and as one of the zeros, i.e., which have the zeros = x 1 < x < < x n < π, and S n 1 = {x = x, x 3,..., x n ) R n 1 : < x < x 3 < < x n < π}. For a given function p and an arbitrary Q n T n, we introduce the inner product as follows f, Qn = f x) x) pq n x))wx) dx, f, T. It is obvious that there is one to one correspondence between the sets T n and S n 1, which means that for every element x = x, x 3,..., x n ) S n 1 and for Q n x) = cos x n ν= 15) there exists a unique system of orthogonal trigonometric polynomials U k T k, k = 0, 1,..., n, such that U n, cos kx Qn = U n, sin jx Qn = 0, k = 0, 1,..., n 1, j = 1,,..., n 1, U n, U n Qn 0.

9 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), In such a way, we introduce a mapping F n : S n 1 S n 1, defined in the following way: for any x = x, x 3,..., x n ) S n 1 we have F n x) = y, where y = y, y 3,..., y n ) S n 1 is such that, y, y 3,..., y n are the zeros of the orthogonal trigonometric polynomial of degree n with respect to the weight function pq n x))wx), where Q n x) is given by 15). For an arbitrary x = x, x 3,..., x n ) S n 1 \S n 1, the function p cosx/) n ν= sinx x ν )/) ) wx) is an admissible weight function, too. We are going to prove that F n is continuous mapping on S n 1. Let x S n 1 be an arbitrary point, {x m) }, m N, a convergent sequence of points from S n 1, which converges to x, y = F n x), and y m) = F n x m) ), m N. Let y S n 1 be an arbitrary limit point of the sequence {y m) } when m. Thus, cos x cos x n ν= n ν= sin x ym) ν sin x ym) ν cos kx p cos x sin jx p cos x n ν= n ν= sin x xm) ν sin x xm) ν wx) dx = 0, k = 0, 1,..., n 1, wx) dx = 0, j = 1,,..., n 1. According to Lebesgue Theorem of dominant convergence see [13, p. 83]), when m + we obtain cos x cos x n ν= n ν= sin x y ν sin x y ν cos kx p cos x sin jx p cos x n ν= n ν= wx) dx = 0, k = 0, 1,..., n 1, wx) dx = 0, j = 1,,..., n 1. Hence, cosx/) n ν= sinx y ν)/) is the trigonometric polynomial of degree n which is orthogonal to all trigonometric polynomials of degree less than or equal to n 1 with respect to the weight function p cosx/) n ν= sinx x ν )/) ) wx) on [, π). According to Theorem.4, such trigonometric polynomial has n distinct simple zeros in [, π). Therefore, y S n 1 and y = F n x). Since y = F n x), because of uniqueness we have y = y, i.e., the mapping F n is continuous on S n 1. Now, we prove that the mapping F n has a fixed point. The mapping F n is continuous on the bounded, convex and closed set S n 1 R n 1. Applying the Brouwer fixed point theorem see [14]) we conclude that there exists a fixed point of F n. Since F n x) S n 1 for all x S n 1 \S n 1, the fixed point of F n belongs to S n 1. If we denote the fixed point of F n by x = x, x 3,..., x n ), then T n x) cos kx pt n x))wx) dx = 0, k = 0, 1,..., n 1, T n x) sin jx pt n x))wx) dx = 0, j = 1,,..., n 1, where T n x) = cosx/) n ν= sinx x ν )/), and we get what is stated. Now, for a [0, 1], n N, and π Fx, a) = cos x ) as1 +1 n as ν +1 n sgn t n 1x) wx) dx, t n 1 T n 1, 16) ν= we consider the following problem ν= Fx, a) = 0 for all t n 1 T n 1, 17) with unknowns x, x 3,..., x n.

10 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), The σ orthogonality conditions 14) with x 1 = ) are equivalent to the problem 17) with a = 1. Therefore, the nodes of the quadrature rule 1) can be obtained as a solution of the problem 17) for a = 1. From Theorem 4. we conclude that the problem 17) has solutions in the simplex S n 1 for every a [0, 1]. In Section 1 we proved that for a = 0 the problem 17) has the unique solution in the simplex S n 1, and, as we have already seen, the solution is also unique in the simplex S n 1 for a = 1. Our aim is to prove the uniqueness of the solution x S n 1 of the problem 17) for all a 0, 1). For this purpose we use the mathematical induction on n. Let us introduce the following notations: Wx, a, x) = n ν= as ν + and φ k x, a) = cos x ) as1 +1 Wx, a, x) sin x x wx) dx, for k =, 3,..., n. 18) k Applying the same arguments as in [4, Lemma 3.], one can prove the following auxiliary results. Lemma 4.3. There exists ε > 0 such that for every a [0, 1] the solutions x of the problem 17) belong to the simplex S ε = {y : ε y + π, ε y 3 y,..., ε y n y n 1, ε π y n }. Lemma 4.4. The problem 17) and the following problem φ k x, a) = 0, k =, 3,..., n, 19) where φ k x, a) are given by 18), are equivalent in the simplex S n 1. Proof. First we assume that x S n 1 is a solution of the problem 17). Obviously, n ν= ν k T n 1, k =, 3,..., n, and for all k =, 3,..., n we have cos x ) as1 +1 n as ν +1 n sgn ν= ν= n ν= ν k wx) dx = 0, i.e., x is a solution of 19). Conversely, let x be a solution of 19). By using the fact that every trigonometric polynomial t T n 1 can be represented in the following way see [4]) tx) = n k= n ν= tx k ) sin x x k n ν= ν k sin x k x ν,

11 we have = = = n k= n k= T.V. Tomović, M.P. Stanić / Filomat 9:10 015), cos x ) as1 +1 n as ν +1 n sgn ν= ν= ν= ν= cos x ) as1 +1 n as ν +1 n sgn n ν= ν k n ν= ν k tx k ) sin x k x ν tx k ) sin x k x ν ν= tx) wx) dx n k= n ν= tx k ) sin x x k n cos x ) as1 +1 n as ν +1 n sgn φ k x, a) = 0, for all t T n 1. Therefore, x is a solution of the problem 17). ν= ν= ν k n ν= sin x k x ν sin x x k wx) dx wx) dx According to Lemmas 4.3 and 4.4 we conclude that the problems 17) and 19) are equivalent in the simplex S ε, for some ε > 0. Also, these problems are equivalent in the simplex S ε1, for all 0 < ε 1 < ε, but they are not equivalent in S n 1. By using the same arguments as in [9, Lemma.4], one can prove the following auxiliary results. Lemma 4.5. Let p ξ,η x) be a continuous function on [, π], which depends continuously on parameters ξ, η [c, d]. i.e., if ξ m, η m ) approaches ξ 0, η 0 ) then the sequences p ξm,η m x) tends to p ξ0,η 0 x) for every fixed x. If the solution xξ, η) of the problem 19) with the weight function p ξ,η x)wx) is always unique for every ξ, η) [c, d], then the solution xξ, η) depends continuously on ξ, η) [c, d]. Now, we are going to prove the main theorem. Theorem 4.6. The problem 19) has a unique solution in the simplex S n 1 for all a [0, 1]. Proof. Let us call the problem 19) as a; s, s 3,..., s n ; w) problem. We prove this assertion by mathematical induction on n. The uniqueness for n = 0 is trivial. As an induction hypothesis, we suppose that the a; s, s 3,..., s n ; w) problem has a unique solution for every a [0, 1] and for every weight function w, which are integrable and nonnegative on the interval [, π), vanishing there only on a set of a measure zero. For ξ, η) [, π] we define the weight functions p ξ,η x) = sin x ξ as n 1 + sin x η as n + wx). According to the induction hypothesis, for every a [0, 1] the a; s, s 3,..., s n ; p ξ,η ) problem has a unique solution x ξ, η), x 3 ξ, η),..., x n ξ, η)), such that < x ξ, η) < < x n ξ, η) < π, and, according to Lemma 4.5, x ν ξ, η) depends continuously on ξ, η) [, π], for all ν =, 3,..., n. We are going to prove that the solution of the a; s,..., s n ; w) problem is unique for every a [0, 1]. Let us denote x n 1 ξ, η) = ξ, x n ξ, η) = η, Wxξ, η), a, x) = n ν= ξ, η) as ν +,

12 and φ k xξ, η), a) = T.V. Tomović, M.P. Stanić / Filomat 9:10 015), cos x ) as1 +1 Wxξ, η), a, x) sin x x wx) dx, k =, 3,..., n. kξ, η) Applying the induction hypothesis we obtain that φ k xξ, η), a) = 0, k =, 3,..., n, 0) for ξ, η) D, where D = {ξ, η) : x n ξ, η) < ξ < η < π}. Let us now consider the following problem in D with unknown t = ξ, η): ϕ 1 t, a) = ϕ t, a) = cos x ) as1 +1 Wxξ, η), a, x) sin x ξ cos x ) as1 +1 Wxξ, η), a, x) sin x η wx) dx = 0, 1) wx) dx = 0. If ξ, η) D is a solution of the problem 1), then, according to 0), xξ, η) is a solution of the a; s,..., s n ; w) problem in the simplex S n 1. To the contrary, if x = x, x 3,..., x n ) is a solution of the a; s,..., s n ; w) problem in the simplex S n 1, then x n 1, x n ) is a solution of the problem 1). Applying Lemma 4.3 we conclude that every solution of the problem 1) belongs to D ε = {ξ, η) : ε ξ x n ξ, η), ε η ξ, ε π η}, for some ε > 0. Let x be a solution of the problem a; s, s 3,..., s n ; w) in S ε. Then, differentiating ϕ 1 with respect to the x k, for all k =, 3,..., n, n, we have ϕ 1 x k = as k + 1) n ν= ν k, ν n 1 cos x ) as1 +1 n as ν +1 n sgn ν= cos x x k wx) dx. ν= Since n ν= ν k, ν n 1 cos x x k T n 1, applying Lemma 4.4, we conclude that ϕ 1 = 0, for all k =, 3,..., n, n. Further, by using the following x k simple equality for cos x y = cos x cos y + sin x y I = ϕ 1 = as n x n 1 we obtain cos x n 1 I = as n cos y sin y + cos x y cos x ) as1 +1 n ν= ν n 1 sin y, as ν + sin x x n 1 as n 1 x x cos n 1 wx) dx, I 1 + sin x ) n 1 I, )

13 where I 1 = cos x ) as1 +1 n ν= ν n 1 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), as ν + sin x x n 1 as n 1 x cos wx) dx and I = cos x ) as1 +1 n ν= ν n 1 as ν + sin x x n 1 as n 1 x x n 1 sin wx) dx. According to 1) we have that I = 0, while I 1 > 0 the integrand does not change its sign on [, π)). Since < x n 1 < π, it follows from ) that ) ϕ1 sgn = 1. x n 1 Analogously, ) ϕ ϕ = 0, k =, 3,..., n 1, sgn = 1. x k x n Thus, at any solution t = ξ, η) of the problem 1) from D ε we have ϕ 1 ξ = n k= k n 1 ϕ 1 x k x k ξ + ϕ 1 x n 1 = ϕ 1 x n 1, n 1 ϕ η = ϕ x k x k η + ϕ = ϕ, x n x n k= ϕ ξ = 0, ϕ 1 η = 0, so, the determinant of the corresponding Jacobian Jt) is positive. In order to finish the proof, we have to prove that the problem 1) has a unique solution in D ε/. We do it by using the topological degree of a mapping. Let us define the mapping ϕt, a) : D ε/ [0, 1] R, by ϕt, a) = ϕ 1 t, a), ϕ t, a)), t R, a [0, 1]. If xξ, η) is a solution of the a; s,..., s n ; w) problem in S ε, then the solutions of the problem ϕt, a) = 0, 0) in D ε/ belong to D ε. The problem ϕt, a) = 0, 0) has the unique solution in D ε/ for a = 0. It is obvious that ϕ, 0) is differentiable on D ε/, the mapping ϕt, a) is continuous in D ε/ [0, 1], and ϕt, a) 0, 0) for all t D ε/ and a [0, 1]. Then degϕ, a), D ε/, 0, 0)) = sgndetjt))) = 1, for all a [0, 1], i.e., degϕ, a), D ε/, 0, 0)) is a constant independent of a. Thus, the problem ϕt, a) = 0, 0) has the unique solution in D ε/ for all a [0, 1], which means that the a; s,..., s n ; w) problem has a unique solution in S n 1 which belongs to S ε. Theorem 4.7. The solution x = xa) of the problem 19) depends continuously on a [0, 1]. Proof. Let us suppose that {a m }, a m [0, 1], m N, is a convergent sequence, which converges to a [0, 1]. Then for every a m, m N, there exists the unique solution xa m ) = x a m ),..., x n a m )) of the system 19) with a = a m. Let xa ) = x a ),..., x n a )) be the unique solution of system 19) for a = a. Let x = x,..., x n ) be an arbitrary limit point of the sequence xa m ) when a m a. Then, according to Theorem 4.6, for each m N we obtain cos x ) am s 1 +1 Wxa m ), a m, x) sin x x wx) dx = 0, k =, 3,..., n. ka m )

14 When a m a the above equations lead to T.V. Tomović, M.P. Stanić / Filomat 9:10 015), cos x ) a s 1 +1 Wx, a, x) sin x wx) dx = 0, k =, 3,..., n. x k Hence, according to Theorem 4.6, we have that x = xa ), i.e., lim am a xa m) = xa ), which gives what is claimed. 5. Numerical construction of quadrature rules In this section we present one method for construction of quadrature rules of the form 1), based on the theoretical results given in the previous sections. So, we chose x 1 = and obtain the nodes x, x 3,, x n by solving problem 19) for a = 1. Since that problem is a system of nonlinear equations, Newton Kantorovič method can be applied. The theoretical results obtained in Section 4 suggest that for fixed n the system 19) can be solved progressively, for an increasing sequence of values for a, up to the value a = 1. We use the solution for some a i) as the initial iteration in Newton Kantorovič method for calculating the solution for a i+1), a i) < a i+1) 1. If for some chosen a i+1) Newton Kantorovič method does not converge, we decrease a i+1) such that it becomes convergent, which is always possible according to Theorem 4.7. Practically, we can in each step set a i+1) = 1, and, if that iterative process is not convergent, we set a i+1) := a i+1) + a i) )/ until it becomes convergent. As the initial iteration for the first iterative process, for some a 1) > 0, we choose the zeros of the corresponding orthogonal trigonometric polynomial of degree n, i.e., the solution of system 19) for a = 0. Let us introduce the following matrix notation x = [ ] T x x 3 x n, x m) = [ ] x m) x m) x m) T 3 n, m = 0, 1,..., By φx) = [ φ x) φ 3 x) φ n x) ] T. W = Wx) = [w i,j ] n 1) n 1) = [ φi+1 ] x j+1 n 1) n 1) we denote Jacobian of φx), with entries given by φ i x j = 1+as j ) for i j, i, j =, 3,..., n, and cos x ) as1 +1 n as ν +1 n sgn ν= ν= n ν= ν i,ν j cos x x j wx) dx, φ i x i = 1 + as i cos x ) as1 +1 n as ν + sin x x i as i x x i cos wx) dx, i =, 3,..., n. ν= ν i All of the above integrals can be computed by using a Gaussian type quadrature rule for trigonometric polynomials. The Newton Kantorovič method for calculating the zeros of the σ orthogonal trigonometric polynomial T σ,n is given as follows x m+1) = x m) W 1 x m) )φx m) ), m = 0, 1,...,

15 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), and for sufficiently good chosen initial approximation x 0), it has the quadratic convergence. When the nodes of quadrature rule 1) are known, it is possible to calculate the corresponding weights. They can be calculating by using the Hermite trigonometric interpolation polynomial see [4, 5]), but it is very difficult. Our method is based on the method given in [7] for construction of Gauss Turán quadrature rules for algebraic polynomials) and generalized for Chakalov Popoviciu s type quadrature rules in [11, 1] for algebraic polynomials, too). An adaptation of that method to the construction of quadrature rules with the maximal trigonometric degree of exactness with an odd number of multiple nodes was proposed in [9]. Here we use the similar adaptation for the construction of quadrature rule 1) with an even number of multiple nodes. We use the facts that quadrature rule 1) is of interpolatory type and that it has the maximal trigonometric degree of exactness, i.e., that it is exact for all trigonometric polynomials of degree less than or equal to n s ν + 1) 1 = n s ν + n 1. So, the weights can be calculated requiring that quadrature rule 1) integrates exactly all trigonometric polynomials of degree less than or equal to n s ν+n 1, and in addition cos n s ν + n ) x, when n s ν + 1)x ν = lπ for an odd l Z, or sin n s ν + n ) x when n s ν + 1)x ν = lπ for an even l Z. If n s ν + 1)x ν lπ, l Z, one can choose to require exactness for cos n s ν + n ) x or sin n s ν + n ) x arbitrary. In such a way we obtain a system of linear equations for the unknown weights. That system can be solved by decomposing into a set of n upper triangular systems. In what follows we present that method of decomposing in details in the case when n s ν + 1)x ν = lπ, l Z. Let us denote Ω ν x) = n i=1 i ν sin x x i ) si +1, ν = 1,,..., n, u k,ν x) = ) k Ω ν x), ν = 1,,..., n, k = 0, 1,..., s ν, and u k,ν x) cos x x ν, k even, t k,ν x) = u k,ν x), k odd, ν = 1,,..., n, k = 0, 1,..., s ν. 3) Since t k,ν is a trigonometric polynomial of degree 1 n i=1 i ν s i + 1) + [ ] k n i=1 i ν s i + 1) + s ν + 1 n = s ν + n, which has the leading term cos n s ν + n ) x or sin n s ν + n ) x but not the both of them), the quadrature rule is exact for t k,ν, i.e., R n t k,ν ) = 0 for all ν = 1,,..., n, k = 0, 1,..., s ν. Since for all i ν we have that t j) k,ν x i) = 0, j = 0, 1,..., s j, then µ k,ν = t k,ν x) wx) dx = s ν j=0 A j,ν t j) k,ν x ν), ν = 1,,..., n. 4) In such a way, we obtain n independent systems for calculating the weights A j,ν, j = 0, 1,..., s ν, ν = 1,,..., n. Here, we need to calculate the derivatives t j) k,ν x), k = 0, 1,..., s ν, j = 0, 1,..., s ν, and for that we use the following result the proof can be obtained by the similar arguments as in [9]). Lemma 5.1. For the trigonometric polynomials t k,ν, given by 3) we have t i) k,ν x ν) = 0, i < k; t k) k,ν x ν) = k! k Ω νx ν ),

16 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), and for i > k t i) k,ν x ν) = v i) k,ν x ν), u i) k,ν x ν), k-even, k-odd, where v i) k,ν x ν) = [i/] m=0 ) i 1) m ui m) x m m k,ν ν ), and the sequence u i) k,ν x ν), k N 0, i N 0, is the solution of the difference equation f i) k,ν = [i 1)/] m=[i k)/] with the initial conditions f i) 0,ν = Ωi) ν x ν ). ) i 1) m i m 1) f x m + 1 m+1 k 1,ν ν ), ν = 1,,..., n, k N, i N 0, The calculating of Ω i) ν x ν ), i N 0, ν = 1,,..., n, can be done as in [9, Lemma 3.3]. Finally, Lemma 5.1 implies that systems 4) are the following upper triangular systems t 0,ν x ν ) t 0,ν x ν) t s ν) 0,ν x ν) t 1,ν x ν) t s ν) 1,ν x ν).... s ν,ν x ν) t s ν) A 0,ν A 1,ν. A sν,ν = µ 0,ν µ 1,ν. µ sν,ν, ν = 1,..., n. Remark 5.. If n s ν + 1)x ν lπ, l Z, then in the case when k is even we choose t k,ν, k = 0, 1,..., s ν, ν = 1,,..., n, in one of the following ways t k,ν x) = u k,ν x) cos which provides that x + n s i + 1)x i + s ν x ν i=1 i ν or t k,ν x) = u k,ν x) sin x + n s i + 1)x i + s ν x ν i=1 i ν, t k,ν T n s ν+n span { cos n ) } s ν + n x k = 0, 1,..., s ν, ν = 1,,..., n, respectively. or t k,ν T n s ν+n span { sin n ) } s ν + n x, Now we present one example as an illustration of the obtained theoretical results. Example 5.3. We choose weight function wx) = 1 + cos x, x [, π), n = 3 and σ = 3, 3, 3, 4, 4, 4). For calculation of nodes we use described procedure, and we have two iterative processes: for a = 1/ with 9 iterations) and for a = 1 with 8 iterations). Obtained numerical values for the nodes x ν, ν = 1,,..., n, are the following: , , , , , The corresponding weight coefficients A j,ν, j = 0, 1,..., s ν, ν = 1,,..., 6, are given in Table 1 numbers in parentheses denote decimal exponents).

17 T.V. Tomović, M.P. Stanić / Filomat 9:10 015), j A j,1 A j, A j, ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) j A j,4 A j,5 A j, ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) Table 1: Weight coefficients A j,ν, j = 0, 1,..., s ν, ν = 1,,..., 6, for wx) = 1 + cos x, x [, π), n = 3 and σ = 3, 3, 3, 4, 4, 4). References [1] R. Cruz Barroso, L. Darius, P. Gonzáles Vera, and O. Njåstad, Quadrature rules for periodic integrands. Bi orthogonality and para orthogonality, Ann. Math. et Informancae ) [] R. Cruz Barroso, P. González Vera, and O. Njåstad, On bi orthogonal systems of trigonometric functions and quadrature formulas for periodic integrands, Numer. Algor. Vol. 444) 007) [3] R.A. DeVore, G.G. Lorentz, Constructive Approximation, Springer Verlag, Berlin Heildeberg, [4] D.P. Dryanov, Quadrature formulae with free nodes for periodic functions, Numer. Math ) [5] J. Du, H. Han, G. Jin, On trigonometric and paratrigonometric Hermite interpolation, J. Approx. Theory ) [6] H. Engels, Numerical Quadrature and Qubature, Academic Press, London, [7] W. Gautschi, G.V. Milovanović, S orthogonality and construction of Gauss Turán type quadrature formulae, J. Comput. Appl. Math ) [8] A. Ghizzeti, A. Ossicini, Quadrature Formulae, Acadenic Press, London, [9] G.V. Milovanović, A.S. Cvetković, M.P. Stanić, Quadrature formulae with multiple nodes and a maximal trigonometric degree of axactness, Numer. Math ) [10] G.V. Milovanović, D.S. Mitrinović, Th.M. Rassias, Topics in Polynomials: Extremal Problems, Inequalites, Zeros, World Scientific, Singapore, New Jersey, London, Hong Kong, [11] G.V. Milovanović, M.M. Spalević, Construction of Chakalov Popoviciu s type quadrature formulae, Rend. Circ. Mat. Palermo, Serie II, Suppl ) [1] G.V. Milovanović, M.M. Spalević, A.S. Cvetković, Calculation of Gaussian type quadratures with multiple nodes, Math. Comput. Modelling ) [13] B. Mirković, Theory of Measures and Integrals, Naučna knjiga, Beograd, 1990 in Serbian). [14] J.M. Ortega, W.C. Rheinboldt, Iterative solution of nonlinear equations in several variables, In: Classics in Applied Mathematics, vol. 30. SIAM, Philadelphia, 000. [15] A.H. Turetzkii, On quadrature rule that are exact for trigonometric polynomials, East J. Approx ) translation in English from Uchenye Zapiski, Vypusk 1149), Seria Math. Theory of Functions, Collection of papers, Izdatel stvo Belgosuniversiteta imeni V.I. Lenina, Minsk 1959) 31 54).