Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments

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:[email protected]

AIMS-Volkswagen Stiftung Workshop on Introduction to Orthogonal Polynomials and Applications Hotel Prince de Galles, Douala, Cameroon, October 5-12, 2018 Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments 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 Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials 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 ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Why should we study orthogonal polynomials ?

Orthogonal polynomials are to be seen as a sequence of polynomials (pn)n with deg(pn) = 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 .... Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments AMS subject classification for Orthogonal Polynomials

Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments The Chebyshev polynomials of the first kind

Chebyshev polynomials of the first kind are defined by Tn(x) = cos(nθ), x = cos θ, 0 < θ < π (1) {w0} and fulfil the following properties:

1 Tn is a polynomial of degree n in x: This can be seen from the recurrence relation Tn+1(x) + Tn−1(x) = 2xTn(x), n ≥ 1, T0(x) = 1, T1(x) = x; (2) {w1} 2 (Tn)n satisfies the orthogonality relation Z π Z 1 dx cos(nθ) cos(mθ)dθ = knδn,m = Tn(x) Tm(x)√ (3) {w2} 2 0 π −1 1 − x (with k0 = π, kn = 2 , n ≥ 1), obtained using the change of variable 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:   n 1−n 1−n min max |qn(x)|, qn ∈ R[x], qn(x) = x + ... = max |2 Tn(x)| = 2 . −1≤x≤1 −1≤x≤1 (4) {w3} 4 Second-order holonomic differential equation: 2 00 0 2 (1 − x ) Tn (x) − x Tn(x) + n Tn(x) = 0, n ≥ 0. (5) {w4} Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments First 10 Chebyshev I Polynomials

From the three-term recurrence relation, one can generate any Tn:

Tn+1(x) = 2xTn(x) − Tn−1(x), n ≥ 1, T0(x) = 1, T1(x) = x; 2 T2(x) = 2 x − 1, 3 T3(x) = 4 x − 3 x, 4 2 T4(x) = 8 x − 8 x + 1, 5 3 T5(x) = 16 x − 20 x + 5 x, (6) {w5} 6 4 2 T6(x) = 32 x − 48 x + 18 x − 1, 7 5 3 T7(x) = 64 x − 112 x + 56 x − 7 x, 8 6 4 2 T8(x) = 128 x − 256 x + 160 x − 32 x + 1, 9 7 5 3 T9(x) = 256 x − 576 x + 432 x − 120 x + 9 x.

The zeros xn,k of Tn ranked in increasing order are:

(2(n − k) + 1  x = cos π , k = 1..n. (7) {w6} n,k 2n Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Graphic of the first 10 Chebyshev I Polynomials Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Orthogonality relations for the Chebyshev I Polynomials

Summing up, we have seen that the Chebyshev polynomials Tn satisfy:

deg(Tn) = n ≥ 0;

and the orthogonality condition:

Z 1 dx Z 1 dx Tn(x) Tm(x)√ = 0, n 6= m, Tn(x) Tn(x)√ 6= 0, n ≥ 0. 2 2 −1 1 − x −1 1 − x

The polynomial sequence (Tn)n is said to be orthogonal with respect to the weight function ρ(x) = √ 1 defined over the interval ] − 1, 1[. It is an 1−x2 orthogonal polynomial sequence.

Assignment 1: Establish relations (1)-(7). Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments 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: Z 1 dx (p, q) = p(x) q(x)√ , (12) {w8} 2 −1 1 − x connected to Chebyshev polynomials;

N X (p, q) = wk p(k) q(k), N ∈ N ∪ {∞}, (13) {w9} k=0 leading to orthogonal polynomials of a discrete variable. Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Construction of a system of orthogonal polynomials: Part 2

Theorem (Gram-Schmidt orthogonalisation process)

The polynomial systems (qn)n and (pn)n defined recurrently by the relations n−1 n n X (qk, x ) qk q0 = 1, qn = x − qk, n ≥ 1, pk = p , k ≥ 0, (14) {w10} (qk, qk) k=0 (qk, qk) satisfy the relations

deg(qn) = deg(pn) = n, ∀n ≥ 0,

(qn, qm) = 0, n 6= m, (qn, qn), 6= 0 ∀n ≥ n,

(pn, pm) = 0, n 6= m, (pn, pn) = 1, ∀n ≥ n.

The proof is done by induction on n: Assignment 2. The polynomial systems (qn)n and (pn)n are said to be orthogonal with respect to the scalar product ( , ). They represent the same orthogonal polynomial system with different normalisation: (qn)n is monic (to say the coefficient of the leading monomial is equal to 1) while (pn)n is orthonormal ( (pn, pn) = 1). Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Definition of orthogonal polynomials: Part 1

Orthogonality with respect to a scalar product

A system (pn)n of polynomials is said to be orthogonal with respect to the scalar product ( , ) if it satisfies the following 2 conditions

deg(pn) = n, ∀n ≥ 0, (15) {w11}

(pn, pm) = 0, n 6= m, (pn, pn) 6= 0, ∀n ≥ n. (16) {w12}

When scalar product is defined by a Stieltjes integral When the scalar product ( , ) is defined by a Stieltjes integral Z b (p, q) = p(x) q(x) dα(x), (17) {w13} a where α is an appropriate real-valued function, then (16) reads Z b Z b pn(x) pm(x) dα(x) = 0, n 6= m, pn(x) pn(x) dα(x) 6= 0, ∀n ≥ 0. a a (18) {w14} Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Definition of the Stieltjes integral

Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments 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 Z b Z b pn(x) pm(x) w(x) dx = 0, n 6= m, pn(x) pn(x) w(x) dx 6= 0, ∀n ≥ 0. a a (pn)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 X (p, q) = w(k) p(k) q(k), N ∈ N ∪ {∞}. (19) {w17} k=0 (pn)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]. Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Basis properties of orthogonal polynomials: Part 1

In this section, we will assume that the polynomial sequence (pn)n satisfies deg(pn) = n, n ≥ 0 and the orthogonality relation (18) which we recall here:

Z b Z b pn(x) pm(x) dα(x) = 0, n 6= m, pn(x) pn(x) dα(x) 6= 0, ∀n ≥ 0. a a Then we have the following properties: Lemma (Equivalent orthogonality relation) The orthogonality relation (18) is equivalent to

Z b Z b m n pn(x) x dα(x) = 0, ∀n ≥ 1, 0 ≤ m ≤ n − 1, pn(x) x dα(x) 6= 0, ∀n ≥ 0. a a (20) {w18}

The previous equation implies that pn is orthogonal to any polynomial of degree less than n. Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments 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 (pn)n and (qn)n are both orthogonal with respect to a scalar product ( , ), then there exists a sequence (αn)n with αn 6= 0, ∀n ≥ 0 such that

pn = αn qn, ∀n ≥ 0.

Proof’s Indication: For a fixed n ≥ 1, we expand pn in terms of the (qk)k and obtain n X pn = ck,n qk, k=0 with (pn, qk) ck,n = = 0, for 0 ≤ k ≤ n − 1. (qk, qk) Hence (pn, qn) pn = qn. (qn, qn) Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Basis properties of orthogonal polynomials

Theorem (Matrix representation of a system of orthogonal polynomials)

Denoting by µn the moment with respect to the Stieltjes measure dα

Z b n µn = x dα(x), n ≥ 0, ∗ a and ∆n the Hankel determinant defined by

∗ ∆n = det(µk+j)0≤k,j≤n 6= 0, ∀n ≥ 0, then the monic polynomial sequence orthogonal with respect to the Stieltjes measure dα is given by

µ0 µ1 ··· µn−1 µn µ1 µ1 ··· µn−1 µn+1 1 . . . . . pn = ∗ ...... (21) {w19} ∆n−1 µn−1 µn ··· µ2n−2 µ2n−1 1 x ··· xn−1 xn. Assignment 4: Proof of the theorem. k n Proof’s indication: Prove that (pn, x ) = 0, 0 ≤ k ≤ n − 1, (pn, x ) 6= 0. Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Basis properties of orthogonal polynomials Theorem (Three-term recurrence relation)

Any polynomial sequence (pn), orthogonal with respect to a scalar product ( , ) defined by the Stieltjes integral satisfies the following relation called three-term recurrence relation

  2 an bn bn+1 an−1 dn x pn(x) = pn+1 + − pn + 2 pn−1, p−1 = 0, p0 = 1, an+1 an an+1 an dn−1 (22) {w20} with n n−1 pn = an x + bn x + low factors, (23) {w21} 2 andd n = (pn, pn).. Assignment 5: Prove this theorem.

Remark When (pn) is monic (ie. an = 1) or orthonormal (ie. dn = 1), then Equation (refw20) can be written respectively in the following forms:

pn+1 = (x − βn) pn − γn pn−1, p−1 = 0, p0 = 1,

x pn = αn+1 pn+1 + βn pn + αn pn−1, p−1 = 0, p0 = 1. Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Basis properties of orthogonal polynomials: TTRR

Proof’s indication

For fixed n ≥ 0, we expand x pn in the basis {p0, p1,..., pn+1} to obtain

n+1 X x pn = ck,n pk, k=0

(xpn, pk) (pn, xpk) ck,n = = = 0, 0 ≤ k < n − 1. (pk, pk) (pk, pk) Hence x pn = cn+1,n pn+1 + cn,n pn + cn−1,n pn−1. (24) {w22} Inserting (23) and (24) into (22) and identifying the leading coefficients of the monomials xn+1 and xn yields   an bn bn+1 cn+1,n = , cn,n = − . (25) {w23} an+1 an an+1 Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Basis properties of orthogonal polynomials

Three-term recurrence relation Using twice (24) combined with the scalar product gives

2 2 cn,n−1 dn = (x pn, pn−1) = (pn, x pn−1) = cn−1,n dn−1

from where we deduct using (25) that

2 an−1 dn cn−1,n = 2 . an dn−1 Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Basis properties of orthogonal polynomials

Theorem (Christofell-Darboux formula) Any system of orthogonal polynomials satisfying the three-term recurrence relation

  2 an bn bn+1 an−1 dn xpn(x) = pn+1 + − pn + 2 pn−1, p−1 = 0, p0 = 1, an+1 an an+1 an dn−1 satisfies a so-called Christofell-Darboux formula given respectively in its initial form and confluent form as

n X pk(x)pk(y) an 1 pn+1(x) pn(y) − pn+1(y) pn(x) = , = { } 2 2 x 6 y (26) w24 d an+1 d x − y k=0 k n

n X pk(x)pk(x) an 1 = 0 ( ) ( ) ( ) 0 ( )
. { } 2 2 pn+1 x pn x − pn+1 x pn x (27) w25 d an+1 d k=0 k n

Assignment 6: Prove the Christoffel-Darboux formula and its confluent form Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Basis properties of orthogonal polynomials

Proof of the Christoffel-Darboux formula For the proof of (26), we multiply by pk(y) Equation (24) in which n is replaced by k to obtain

x pk(x) pk(y) = ck+1,k pk+1(x) pk(y) + ck,k pk(x) pk(y) + ck−1,k pk−1(x) pk(y). Interchanging the role of x and y in the previous equation, we obtain

y pk(x) pk(y) = ck+1,k pk+1(y) pk(x) + ck,k pk(x) pk(y) + ck−1,k pk−1(y) pk(x). Subtracting the last equation from the last but one, we obtain that

pk(x) pk(y) Ak+1(x, y) − Ak(x, y) 2 = , dk x − y

ck+1,k Ak+1(x, y) = 2 (pk+1(x) pk(y) − pk+1(y) pk(x)) dk

ck+1,k ak 1 ck,k+1 taking into account the relation 2 = a 2 = 2 . dk k+1 dk dk+1 Equation (27) is obtained by taking the limit of (26) when y goes to x. Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Basic properties of Orthogonal Polynomials

Theorem (On the zeros of orthogonal polynomials)

If (pn)n is an orthogonal polynomial system, the we have: 0 1 pn and pn+1 have no common zero. The same applies for Pn and Pn;

2 pn has n real simple zeros xn,k satisfying a < xn,k < b, 1 ≤ k ≤ n.

3 if xn,1 < xn,2 < ··· < xn,n are the n zeros of pn, then

a < xn+1,k < xn,k < xn+1,k+1, 1 ≤ k ≤ n. Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Graphic of the first 10 Chebyshev I Polynomials Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments Tutorials: Solving assignments

As tutorial, please solve assignment 1 to 6. Why should we study orthogonal polynomials ? An example of a system of orthogonal polynomials Construction of a system of orthogonal polynomials Definition of orthogonal polynomials Basis properties of orthogonal polynomials Tutorials: Solving assignments

Thanks for your kind attention