Introduction to Orthogonal Polynomials: Definition and basic properties

Transcription

1 Introduction to Orthogonal Polynomials: Definition and basic properties Prof. Dr. Mama Foupouagnigni African Institute for Mathematical Sciences, Limbe, Cameroon and Department of Mathematics, Higher Teachers Training College University of Yaounde I, Cameroon AIMS-Volkswagen Stiftung Workshop on Introduction to Orthogonal Polynomials and Applications Hotel Prince de Galles, Douala, Cameroon, October 5-12, 2018

2 Table of Contents 1 Why should we study orthogonal polynomials? 2 An example of a system of orthogonal polynomials 3 Construction of a system of orthogonal polynomials 4 Definition of orthogonal polynomials 5 Basis properties of orthogonal polynomials 6 Tutorials: Solving assignments

3 Main objectives 1 Give an example of a system of orthogonal polynomials 2 Provide a method for constructing a system orthogonal Polynomials 3 Define the notion of orthogonal polynomials; 4 Provide (with some illustrations on the proof) some basic properties such as: the uniqueness of a family of orthogonal polynomials; the matrix representation; the three-term recurrence relation, the Christoffel-Darboux formula and some of its consequences such as the interlacing properties of the zeros. 5 Finally we discuss and solve, as a short tutorial, some assignments given within the first talk which are mainly proof of some results provided earlier.

4 Why should we study orthogonal polynomials? Orthogonal polynomials are to be seen as a sequence of polynomials (p n ) n with deg(p n ) = n with orthogonality property. They are very useful in practice in various domains of mathematics, physics, engineering and so on because of the many properties and relations they satisfy: 1 Orthogonality (all of them) 2 Three term recurrence relation (all of them) 3 Darboux-Christoffel formula (all of them) 4 Matrix representation (all of them); 5 Gauss quadrature (all of them): used for approximation of integrals; 6 second-order holonomic differential, difference or q-difference equation (classical ones); 7 Fourth-order holonomic differential, difference or q-difference equations (Laguerre-Hahn class); 8 Rodrigues formula (classical ones); 9 Partial differential, difference or q-difference equations (OP of several variables); 10 Expansion of continuous function with integrable square derivable in terms of Fourier series of OP (classical OP);

5 AMS subject classification for Orthogonal Polynomials

6 The Chebyshev polynomials of the first kind Chebyshev polynomials of the first kind are defined by T n (x) = cos(nθ), x = cos θ, 0 < θ < π (1) {w0} and fulfil the following properties: 1 T n is a polynomial of degree n in x: This can be seen from the recurrence relation T n+1 (x) + T n 1 (x) = 2xT n (x), n 1, T 0 (x) = 1, T 1 (x) = x; (2) {w1} 2 (T n ) n satisfies the orthogonality relation π 1 dx cos(nθ) cos(mθ)dθ = k n δ n,m = T n (x) T m (x) (3) x 2 (with k 0 = π, k n = π 2, n 1), obtained using the change of variable {w2} x = cos θ, 0 < θ < π and the linearization formula 2 cos nθ cos mθ = cos(n + m)θ + cos(n m)θ. 3 Monic Chebyshev polynomial of degree n is the polynomial deviating less { from zero on [ 1, 1] among monic polynomials } of degree n: min max q n(x), q n R[x], q n (x) = x n x 1 4 Second-order holonomic differential equation: = max 1 x 1 21 n T n (x) = 2 1 n. (4) {w3} (1 x 2 ) T n (x) x T n(x) + n 2 T n (x) = 0, n 0. (5) {w4}

7 First 10 Chebyshev I Polynomials From the three-term recurrence relation, one can generate any T n : T n+1 (x) = 2xT n (x) T n 1 (x), n 1, T 0 (x) = 1, T 1 (x) = x; T 2 (x) = 2 x 2 1, T 3 (x) = 4 x 3 3 x, T 4 (x) = 8 x 4 8 x 2 + 1, T 5 (x) = 16 x 5 20 x x, (6) {w5} T 6 (x) = 32 x 6 48 x x 2 1, T 7 (x) = 64 x x x 3 7 x, T 8 (x) = 128 x x x 4 32 x 2 + 1, T 9 (x) = 256 x x x x x. The zeros x n,k of T n ranked in increasing order are: ( ) (2(n k) + 1 x n,k = cos π, k = 1..n. 2n (7) {w6}

8 Graphic of the first 10 Chebyshev I Polynomials

9 Orthogonality relations for the Chebyshev I Polynomials Summing up, we have seen that the Chebyshev polynomials T n satisfy: and the orthogonality condition: deg(t n ) = n 0; 1 dx 1 T n (x) T m (x) = 0, n m, dx T n (x) T n (x) 1 x 2 1 x , n 0. The polynomial sequence (T n ) n is said to be orthogonal with respect to the weight function ρ(x) = 1 1 x 2 defined over the interval ] 1, 1[. It is an orthogonal polynomial sequence. Assignment 1: Establish relations (1)-(7).

10 Construction of a system of orthogonal polynomials:part 1 Let us consider a scalar product (, ) defined on R[x] R[x] where R[x] is the ring of polynomials with real variable. As scalar product, it fulfills the following properties: (p, p) 0, p R[x], and (p, p) = 0 = p = 0, (8) {w7} (p, q) = (q, p), p, q R[x], (9) (λ p, q) = λ (p, q), λ R, p, q R[x], (10) (p + q, r) = (p, r) + (q, r), p, q, r R[x]. (11) As examples of scalar products on R[x] with connections to known systems of orthogonal polynomials, we mention: (p, q) = 1 1 connected to Chebyshev polynomials; (p, q) = dx p(x) q(x), (12) {w8} 1 x 2 N w k p(k) q(k), N N { }, (13) {w9} k=0 leading to orthogonal polynomials of a discrete variable.

11 Construction of a system of orthogonal polynomials: Part 2 Theorem (Gram-Schmidt orthogonalisation process) The polynomial systems (q n ) n and (p n ) n defined recurrently by the relations n 1 q 0 = 1, q n = x n satisfy the relations k=0 (q k, x n ) (q k, q k ) q k, n 1, p k = deg(q n ) = deg(p n ) = n, n 0, (q n, q m ) = 0, n m, (q n, q n ), 0 n n, (p n, p m ) = 0, n m, (p n, p n ) = 1, n n. q k, k 0, (14) {w10 (qk, q k ) The proof is done by induction on n: Assignment 2. The polynomial systems (q n ) n and (p n ) n are said to be orthogonal with respect to the scalar product (, ). They represent the same orthogonal polynomial system with different normalisation: (q n ) n is monic (to say the coefficient of the leading monomial is equal to 1) while (p n ) n is orthonormal ( (p n, p n ) = 1).

12 Definition of orthogonal polynomials: Part 1 Orthogonality with respect to a scalar product A system (p n ) n of polynomials is said to be orthogonal with respect to the scalar product (, ) if it satisfies the following 2 conditions deg(p n ) = n, n 0, (15) {w11 (p n, p m ) = 0, n m, (p n, p n ) 0, n n. (16) {w12 When scalar product is defined by a Stieltjes integral When the scalar product (, ) is defined by a Stieltjes integral (p, q) = b a p(x) q(x) dα(x), (17) {w13 where α is an appropriate real-valued function, then (16) reads b p n (x) p m (x) dα(x) = 0, n m, b a a p n (x) p n (x) dα(x) 0, n 0. (18) {w14

13 Definition of the Stieltjes integral

14 Definition of orthogonal polynomials: Part 2 When scalar product is defined by the Riemann When the scalar product is defined by a Stieltjes integral (17) with dα(x) = w(x) dx where w is an appropriate function, then (16) reads b a p n (x) p m (x) w(x) dx = 0, n m, b a p n (x) p n (x) w(x) dx 0, n 0. (p n ) n is said to be orthogonal with respect to the weight function w. Because of the form of the orthogonality relation, the variable here is continuous. When scalar product is defined by a special Stieltjes function When the scalar product is defined by a Stieltjes integral (17) where w is an appropriate step function on N or on {0, 1,..., N}, then (14) reads N (p, q) = w(k) p(k) q(k), N N { }. (19) {w17 k=0 (p n ) n is said to be orthogonal with respect to the discrete weight function w. Because of the form of the orthogonality relation, the variable is discrete. Assignment 3: Find the first five monic polynomials orthogonal with respect to the weight w(x) = 1 defined on the interval [ 1, 1].

15 Basis properties of orthogonal polynomials: Part 1 In this section, we will assume that the polynomial sequence (p n ) n satisfies deg(p n ) = n, n 0 and the orthogonality relation (18) which we recall here: b a p n (x) p m (x) dα(x) = 0, n m, b a p n (x) p n (x) dα(x) 0, n 0. Then we have the following properties: Lemma (Equivalent orthogonality relation) The orthogonality relation (18) is equivalent to b a p n (x) x m dα(x) = 0, n 1, 0 m n 1, b a p n (x) x n dα(x) 0, n 0. The previous equation implies that p n is orthogonal to any polynomial of degree less than n. (20) {w18

16 Basis properties of orthogonal polynomials Theorem (Uniqueness of an orthogonal polynomial system) To a scalar product (, ) on R[x] is associated a unique up to a multiplicative factor system of orthogonal polynomials: If (p n ) n and (q n ) n are both orthogonal with respect to a scalar product (, ), then there exists a sequence (α n ) n with α n 0, n 0 such that p n = α n q n, n 0. Proof s Indication: For a fixed n 1, we expand p n in terms of the (q k ) k and obtain n p n = c k,n q k, with Hence k=0 c k,n = (p n, q k ) = 0, for 0 k n 1. (q k, q k ) p n = (p n, q n ) (q n, q n ) q n.

17 Basis properties of orthogonal polynomials Theorem (Matrix representation of a system of orthogonal polynomials) Denoting by µ n the moment with respect to the Stieltjes measure dα µ n = b and n the Hankel determinant defined by a x n dα(x), n 0, n = det(µ k+j ) 0 k,j n 0, n 0, then the monic polynomial sequence orthogonal with respect to the Stieltjes measure dα is given by p n = 1 n 1 µ 0 µ 1 µ n 1 µ n µ 1 µ 1 µ n 1 µ n (21) {w19 µ n 1 µ n µ 2n 2 µ 2n 1 1 x x n 1 x n. Assignment 4: Proof of the theorem. Proof s indication: Prove that (p n, x k ) = 0, 0 k n 1, (p n, x n ) 0.

18 Basis properties of orthogonal polynomials Theorem (Three-term recurrence relation) Any polynomial sequence (p n ), orthogonal with respect to a scalar product (, ) defined by the Stieltjes integral satisfies the following relation called three-term recurrence relation x p n (x) = with a n a n+1 p n+1 + ( bn a n b n+1 a n+1 ) p n + a n 1 a n d 2 n d 2 n 1 p n 1, p 1 = 0, p 0 = 1, (22) {w20 p n = a n x n + b n x n 1 + low factors, (23) {w21 and d 2 n = (p n, p n ).. Assignment 5: Prove this theorem. Remark When (p n ) is monic (ie. a n = 1) or orthonormal (ie. d n = 1), then Equation (refw20) can be written respectively in the following forms: p n+1 = (x β n ) p n γ n p n 1, p 1 = 0, p 0 = 1, x p n = α n+1 p n+1 + β n p n + α n p n 1, p 1 = 0, p 0 = 1.

19 Basis properties of orthogonal polynomials: TTRR Proof s indication For fixed n 0, we expand x p n in the basis {p 0, p 1,..., p n+1 } to obtain Hence n+1 x p n = c k,n p k, k=0 c k,n = (xp n, p k ) (p k, p k ) = (p n, xp k ) = 0, 0 k < n 1. (p k, p k ) x p n = c n+1,n p n+1 + c n,n p n + c n 1,n p n 1. (24) {w22 Inserting (23) and (24) into (22) and identifying the leading coefficients of the monomials x n+1 and x n yields c n+1,n = a n a n+1, c n,n = ( bn b ) n+1. (25) {w23 a n a n+1

20 Basis properties of orthogonal polynomials Three-term recurrence relation Using twice (24) combined with the scalar product gives c n,n 1 dn 2 = (x p n, p n 1 ) = (p n, x p n 1 ) = c n 1,n dn 1 2 from where we deduct using (25) that c n 1,n = a n 1 a n d 2 n d 2 n 1.

21 Basis properties of orthogonal polynomials Theorem (Christofell-Darboux formula) Any system of orthogonal polynomials satisfying the three-term recurrence relation xp n (x) = a ( n bn p n+1 + b ) n+1 p n + a n 1 dn 2 p n 1, p 1 = 0, p 0 = 1, a n+1 a n a n+1 a n d 2 n 1 satisfies a so-called Christofell-Darboux formula given respectively in its initial form and confluent form as n k=0 p k (x)p k (y) d 2 k n k=0 p k (x)p k (x) d 2 k = a n a n+1 1 d 2 n = a n a n+1 1 d 2 n p n+1 (x) p n (y) p n+1 (y) p n (x), x y (26) {w24 x y ( p n+1 (x) p n (x) p n+1 (x) p n(x) ). (27) {w25 Assignment 6: Prove the Christoffel-Darboux formula and its confluent form

22 Basis properties of orthogonal polynomials Proof of the Christoffel-Darboux formula For the proof of (26), we multiply by p k (y) Equation (24) in which n is replaced by k to obtain x p k (x) p k (y) = c k+1,k p k+1 (x) p k (y) + c k,k p k (x) p k (y) + c k 1,k p k 1 (x) p k (y). Interchanging the role of x and y in the previous equation, we obtain y p k (x) p k (y) = c k+1,k p k+1 (y) p k (x) + c k,k p k (x) p k (y) + c k 1,k p k 1 (y) p k (x). Subtracting the last equation from the last but one, we obtain that p k (x) p k (y) d 2 k A k+1 (x, y) = c k+1,k d 2 k = A k+1(x, y) A k (x, y), x y (p k+1 (x) p k (y) p k+1 (y) p k (x)) taking into account the relation c k+1,k = c k,k+1. dk 2 dk+1 2 Equation (27) is obtained by taking the limit of (26) when y goes to x. = a k a k+1 1 d 2 k

23 Basic properties of Orthogonal Polynomials Theorem (On the zeros of orthogonal polynomials) If (p n ) n is an orthogonal polynomial system, the we have: 1 p n and p n+1 have no common zero. The same applies for P n and P n; 2 p n has n real simple zeros x n,k satisfying a < x n,k < b, 1 k n. 3 if x n,1 < x n,2 < < x n,n are the n zeros of p n, then a < x n+1,k < x n,k < x n+1,k+1, 1 k n.

24 Graphic of the first 10 Chebyshev I Polynomials

25 Tutorials: Solving assignments As tutorial, please solve assignment 1 to 6.

26 Thanks for your kind attention