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 Email:mfoupouagnigni@aims-cameroon.org AIMS-Volkswagen Stiftung Workshop on Introduction to Orthogonal Polynomials and Applications Hotel Prince de Galles, Douala, Cameroon, October 5-12, 2018
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
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.
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); 11....
AMS subject classification for Orthogonal Polynomials
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) 0 1 1 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 +... 1 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}
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 3 + 5 x, (6) {w5} T 6 (x) = 32 x 6 48 x 4 + 18 x 2 1, T 7 (x) = 64 x 7 112 x 5 + 56 x 3 7 x, T 8 (x) = 128 x 8 256 x 6 + 160 x 4 32 x 2 + 1, T 9 (x) = 256 x 9 576 x 7 + 432 x 5 120 x 3 + 9 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}
Graphic of the first 10 Chebyshev I Polynomials
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 2 1 1 0, 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).
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.
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).
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
Definition of the Stieltjes integral
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].
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
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.
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+1...... (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.
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.
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
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.
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
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
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.
Graphic of the first 10 Chebyshev I Polynomials
Tutorials: Solving assignments As tutorial, please solve assignment 1 to 6.
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