On an extension of symmetric coherent pairs

Transcripción

1 On an extension of symmetric coherent pairs Luis Alejandro Molano Molano Universidad Pedagógica y Tecnológica de Colombia. Duitama. Colombia. EIBPOA UNAM, Ciudad de México. México.

2 Introduction Linear Symmetric Functionals Let L be a quaside nite linear functional De nition L is called symmetric if L, x 2n+1 = 0. Theorem Let fp n (x)g n2n be the MOPS with respect to a quaside nite functional L. Then the following are equivalent: a). L is symmetric; b). For n 0, P n ( x) = ( 1) n P n (x), and c). P n (x) = xp n 1 (x) γ n P n 2 (x), n 2.

3 Introduction Linear Symmetric Functionals Let L be a quaside nite linear functional De nition L is called symmetric if L, x 2n+1 = 0. Theorem Let fp n (x)g n2n be the MOPS with respect to a quaside nite functional L. Then the following are equivalent: a). L is symmetric; b). For n 0, P n ( x) = ( 1) n P n (x), and c). P n (x) = xp n 1 (x) γ n P n 2 (x), n 2.

4 Introduction Symmetrically Coherent Pairs De nition Let u and v denote a quaside nite symmetric linear functionals and fp n g and f n g the corresponding MOPS. The pair (u, v) is called a symmetrically coherent pair ab if there exist non-zero constants σ n such that P n (x) = 0 n+1 (x) n σ 0 n 1 (x) n n 1, n 2. a A.Iserles, et al. On polynomials orthogonal with respect to certain Sobolev inner products, J.Approx.Theory 65 (1991) b Meijer, H. G. Determination of all coherent pairs. J. Approx. Theory 89 (1997),

5 Introduction Sobolev Inner Products Let u and v be two positive de nite symmetric linear functionals. There exist two Borel positive measures µ 1 and µ 2, associated to v and u respectively, and let Sn(x) λ be the MOPS with respect to inner product Z hp, qi S = with λ > 0. Z p(x)q(x)dµ 1 (x) + λ p 0 (x)q 0 (x)dµ 2 (x), (1)

6 Let u and v be two quaside nite symmetric linear functionals, and fp n g and f n g the corresponding MOPS. We suppose that there exist sequences fa n g and fb n g, with b n 6= 0, such that 1 P n+2 (x) + a n P n (x) = 0 n+3 (x) n b 0 n+1 (x) n n + 1, n 0. 1 A. M. Delgado and F. Marcellán, On an Extension of Symmetric Coherent Pairs of Orthogonal Polynomials, J. Comput. Appl. Math. 178 (2005)

7 Sobolev inner product and the algebraic Connection If u and v are positive de nite, we consider the Sobolev inner product Z hp, qi S = Z p(x)q(x)dµ 1 (x) + λ p 0 (x)q 0 (x)dµ 2 (x), where µ 1 and µ 2 are the positive Borel measures associated to v and u respectively, and λ > 0. Besides let Sn(x) λ be the MOPS associated. Then it have sense to de ne the sequence fw n (x)g n2n in the next way: W n (x) := lim S λ λ! n(x).

8 Sobolev inner product and the algebraic Connection Two important facts If deg(q) n Z W 0 n+1(x)q 0 (x)dµ 2 (x) = 0, and as consequence Wn+1(x) 0 = (n + 1)P n (x). Moreover, we can do the same calculus with q(x) = 1 and n 0 Z W n+1 (x)dµ 1 (x) = 0.

9 Sobolev inner product and the algebraic Connection Two important facts If deg(q) n Z W 0 n+1(x)q 0 (x)dµ 2 (x) = 0, and as consequence Wn+1(x) 0 = (n + 1)P n (x). Moreover, we can do the same calculus with q(x) = 1 and n 0 Z W n+1 (x)dµ 1 (x) = 0.

10 Sobolev inner product and the algebraic Connection Also, it is possible to nd the relation W n+3 (x) + ea n W n+1 (x) = n+3 (x) + e b n n+1 (x), n 0. where by notation ea n = a n n + 3 n + 1, e bn = b n n + 3 n + 1, n 0. Moreover it is consider the expansion of W n by using the basis S λ n (x) namely W n (x) = Sn(x) λ n 2 + j=0 σ n,j S λ j (x),

11 Sobolev inner product and the algebraic Connection where, is well known that σ n,j = D E W n,s λ j S S λ j 2 S = and S λ j 2 E DS := λ S j, Sλ j. S Then we get W n(x)s λ j (x)dµ 1 (x)+λ W0 n(x) S λ j 2 S W n+3 (x) + ea n W n+1 (x) = Sn+3(x) λ n+1 + j=0 S λ j 0(x)dµ2 (x) η n,j (λ)s λ j (x),, where η n,j (λ) = 0, for j n.

12 Sobolev inner product and the algebraic Connection Finally we get W n+3 (x) + ea n W n+1 (x) = S λ n+3(x) + η n,n+1 (λ)s λ n+1 (x), and as a consequence S λ n+3(x) + η n (λ)s λ n+1 (x) = n+3(x) + e b n n+1 (x), n 0. η n (λ) := η n,n+1 (λ)

13 Sobolev inner product and the algebraic Connection Finally we get W n+3 (x) + ea n W n+1 (x) = S λ n+3(x) + η n,n+1 (λ)s λ n+1 (x), and as a consequence S λ n+3(x) + η n (λ)s λ n+1 (x) = n+3(x) + e b n n+1 (x), n 0. η n (λ) := η n,n+1 (λ)

14 The Sobolev Coe cients It can be obtained the relation η n (λ) = W n+3(x)sn+1 λ (x)dµ 1 (x) + ea n. Sn+1 λ 2 S First Z W n+3 (x)sn+1 λ (x)dµ 1 (x) = ea n(n + 1) 2 kp n k 2 µ λ + e b n k n+1 2 k 2 µ, 1

15 The Sobolev Coe cients It can be obtained the relation η n (λ) = W n+3(x)sn+1 λ (x)dµ 1 (x) + ea n. Sn+1 λ 2 S First Z W n+3 (x)sn+1 λ (x)dµ 1 (x) = ea n(n + 1) 2 kp n k 2 µ λ + e b n k n+1 2 k 2 µ, 1

16 The Sobolev Coe cients Second, we get the initial conditions S λ 2 S λ 1 2 = λ + k 1k 2 µ, η S 1 0 (λ) = ea 0λ + e b 0 k 1 k 2 µ 1 λ + k 1 k 2, µ 1 2 = 4 kp 1k 2 µ λ + k 2 S 2 k 2 µ η 1 1 (λ) = 4ea 1 kp 1 k 2 µ λ + e b 1 k 2 2 k 2 µ 1 4 kp 1 k 2 µ λ + k 2 2 k 2. µ 1

17 The Sobolev Coe cients Second, we get the initial conditions S λ 2 S λ 1 2 = λ + k 1k 2 µ, η S 1 0 (λ) = ea 0λ + e b 0 k 1 k 2 µ 1 λ + k 1 k 2, µ 1 2 = 4 kp 1k 2 µ λ + k 2 S 2 k 2 µ η 1 1 (λ) = 4ea 1 kp 1 k 2 µ λ + e b 1 k 2 2 k 2 µ 1 4 kp 1 k 2 µ λ + k 2 2 k 2. µ 1

18 The Sobolev Coe cients Lemma For n 2 ea n e bn S λ n+1 2 = ea n (n + 1) 2 kp n k 2 µ 2 λ + ea n k n+1 k 2 µ 1 S + ea nη n 2 (λ) η n 2 (λ) ea n 2 S λ n 1 2 ea n e bn 2 ebn 2 ea n 2 k n 1 k 2 µ 1, with the above initial conditions. S

19 The Sobolev Coe cients If we de ne T n+1 (x) = W n+1 (x) + ea n 2 W n 1 (x), and then D E Sn+3(x), λ T n+1 (x) S = ht n+3 (x), T n+1 (x)i S η n (λ) ht n+1 (x), T n+1 (x)i S η n 2 (λ) ht n 1 (x), T n+1 (x)i S then = 0, ht n+3 (x), T n+1 (x)i η n (λ) = S ht n+1 (x), T n+1 (x)i S η n 2 (λ) ht n 1 (x), T n+1 (x)i S

20 The Sobolev Coe cients where 2 and η n (λ) = A n λ + B n (C n λ + D n ) η n 2 (λ) [A n 2 λ + B n 2 ], A n = ea n (n + 1) 2 kp n k 2 µ 2, B n = e b n k n+1 k 2 µ 1, C n = (n + 1) 2 kp n k 2 µ 2 + ea 2 n 2(n 1) 2 kp n 2 k 2 µ 2, D n = k n+1 k 2 µ 1 + e b 2 n 2 k n 1 k 2 µ 1 2 T.E. Perez, Polinomios ortogonales respecto a productos de Sobolev: el caso continuo, Ph.D. Thesis, Departamento de Matematica Aplicada, Universidad de Granada, 1994.

21 The Sobolev Coe cients Proposición n o There exist sequences of polynomials fq n (λ)g and Q en (λ) such that for n 0, deg (Q n ) = deg Q en = n, moreover η 2n (λ) = (ea 2n(2n + 1) 2 p 2n λ + A 2n ) σ 2n Q n (λ) Q n+1 (λ), and η 2n+1 (λ) = (A 2n+1 λ + B 2n+1 ) eq n (λ) eq n+1 (λ).

22 The Sobolev Coe cients Proposición n o There exist sequences of polynomials fq n (λ)g and Q en (λ) such that for n 0, deg (Q n ) = deg Q en = n, moreover η 2n (λ) = (ea 2n(2n + 1) 2 p 2n λ + A 2n ) σ 2n Q n (λ) Q n+1 (λ), and η 2n+1 (λ) = (A 2n+1 λ + B 2n+1 ) eq n (λ) eq n+1 (λ).

23 The Sobolev Coe cients Proposición and Q n+1 (λ) = (C 2n λ + D 2n ) Q n (λ) (A 2n 2 λ + B 2n 2 ) 2 Q n 1 (λ), and eq n+1 (λ) = (C 2n+1 λ + D 2n+1 ) eq n (λ) (A 2n 1 λ + B 2n 1 ) 2 eq n 1 (λ)

24 Orthogonality Conditions Classical case If v is classical the we get the relation P n+2 (x) + a n P n (x) = e n+2 (x) + b n e n (x), n 0, and if it is suppose that P n (x) = xp n 1 (x) γ n P n 2 (x), and e n (x) = x e n 1 (x) eγ n e n 2 (x), then γ n+1 = a n 1b n eγ a n b n+1, n 1. n 1

25 Orthogonality Conditions Is well known that kp n k 2 µ 2 = γ 1.γ 2... γ n+1, then if γ n+1 = a n 1b n a n b n and 1 eγ n+1, we get kp n k 2 µ 2 = e n 2 µ 1 = eγ 1.eγ 2... eγ n+1, e n 2 µ 1 γ 1 a 0 b n eγ 1 b 0 a n. (2)

26 Orthogonality Conditions General Situation With the starting point P n+2 (x) + a n P n (x) = 0 n+3 (x) n + 3 and with the initial condition + b 0 n+1 (x) n n + 1, we get eγ a 0 γ 2 = b 2 + eγ 3 0, 3 γ n+3 = n+3 (n + 2) (n + 4) eγ n (n + 3) (n + 4) + (b n a n ) (b n+1 a n+1 ). j=2 eγ j

27 Orthogonality Conditions General Situation With the starting point P n+2 (x) + a n P n (x) = 0 n+3 (x) n + 3 and with the initial condition + b 0 n+1 (x) n n + 1, we get eγ a 0 γ 2 = b 2 + eγ 3 0, 3 γ n+3 = n+3 (n + 2) (n + 4) eγ n (n + 3) (n + 4) + (b n a n ) (b n+1 a n+1 ). j=2 eγ j

28 Orthogonality Conditions and the other hand (n + 1)(n + 2) eγ 2 n+4 n+2 (n + 3) (n + 2) = γ 3 γ j + γ 2 n+3 j=2 (n + 2)(n + 3) (n + 1)(n + 4) + (a n b n ) + (b n+1 a n+1 ). 2 2 anyway we can to use kp n k 2 µ 2 = γ 1.γ 2... γ n+1, and e n 2 µ 1 = eγ 1.eγ 2... eγ n+1,

29 The end Thank you so much