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Orthogonal Polynomials: an Illustrated Guide

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Orthogonal Polynomials: an Illustrated Guide Orthogonal Polynomials: An Illustrated Guide Avaneesh Narla December 10, 2018 Contents 1 Definitions 1 2 Example 1 2 3 Three-term Recurrence Relation 3 4 Christoffel-Darboux Formula 5 5 Zeros 6 6 Gauss Quadrature 8 6.1 Lagrange Interpolation . .8 6.2 Gauss quadrature formula . .8 7 Classical Orthogonal Polynomials 11 7.1 Hermite Polynomials . 11 7.2 Laguerre Polynomials . 12 7.3 Legendre Polynomials . 14 7.4 Jacobi Polynomials . 16 7.5 Chebyshev Polynomials of the First Kind . 17 7.6 Chebyshev Polynomials of the Second Kind . 19 7.7 Gegenbauer polynomials . 20 1 Definitions Orthogonal polynomials are orthogonal with respect to a certain function, known as the weight function w(x), and a defined interval. The weight function must be continuous and positive such that its moments (µn) exist. Z b n µn := w(x)x dx; n = 0; 1; 2; ::: a The interval may be infinite. We now define the inner product of two polynomials as follows Z 1 hf; giw(x) := w(x)f(x)g(x) dx −∞ 1 We will drop the subscript indicating the weight function in future cases. Thus, as always, a 1 sequence of polynomials fpn(x)gn=0 with deg(pn(x)) = n are called orthogonal polynomials for a weight function w if hpm; pni = hnδmn Above, the delta function is the Kronecker Delta Function There are two possible normalisations: If hn = 1 8n 2 f0; 1; 2:::g, the sequence is orthonormal. If the coefficient of highest degree term is 1 for all elements in the sequence, the sequence is monic. We note that orthogonal polynomials are thus linearly independent and generate bases for all poly- nomials of any arbitrary degree. Important Immediate Properties: 1. hQ; pniw(x) = ckpn and hQ; pniw(x) 6= 0 =) Q has a term of degree n. Question 1. Can you have two distinct sets of orthogonal polynomials corresponding to the same weight function but different intervals? What about the same interval? Is there any relation be- tween the two distinct sets? Also, do all weight functions have corresponding sequence of orthogonal polynomials? If not, what are the defining properties of the weight functions that do? What additional restrictions do you need on w(x) to make the inner product space complete? Answers: For two sets of orthogonal polynomials corresponding to the same weight function but different intervals, look at the example set and the Legendre Polynomials. For two sets of orthogonal polynomials corresponding to the same weight function but the same interval, look at the two kinds of Chebyshev Polynomials. 2 Example 1 We choose the weight function to be a constant (say c), and choose the interval (0,a). We can construct the sequence through the Gram-Schmidt process (by moving from the basis f1; x; x2; x3:::g 2 to the orthogonal basis). We start with the sequence f1; x; x ; :::g. We choose p0(x) = 1 for now, we can later normalise the sequence to be orthonormal. Thus by the Gram-Schmidt Process, Proof. hx; p0(x)i hx; 1i a p1(x) = x − projp0(x)x = x − p0(x) = x − = x − hp0(x); p0(x)i h1; 1i 2 R a ca2 R a Since, hx; 1i = 0 c · x dx = 2 and h1; 1i = 0 cdx = ca 2 2 2 2 2 2 hx ; p0(x)i hx ; p1(x)i =) p2(x) = x − projp0(x)x − projp1(x)x = x − p0(x) − p1(x) = hp0(x); p0(x)i hp1(x); p1(x)i 2 2 a 3 4 2 2 2 2 hx ; 1i hx ; x − 2 i a 2 ca 1 ca 12 a 2 a a 2 a x − − a a x − = x − · − · 3 x − = x − −ax+ = x −ax+ h1; 1i hx − 2 ; x − 2 i 2 3 ca 12 ca 2 3 2 6 as, Z a ca3 hx2; 1i = c · x2 dx = 0 3 2 Z a 4 3 a 4 4 4 2 2 cx cax ca ca ca hx ; (x − a=2)i = cx · (x − a=2) dx = − = − = 0 4 6 0 4 6 12 Z a 3 2 2 a 3 3 3 3 2 cx cax ca x ca ca ca ca h(x − a=2); (x − a=2)i = c(x − a=2) dx = − + = − + = 0 3 2 4 0 3 2 4 12 And similarly by continuing the Gram-Schmidt process, we get that 3a 3a2 a3 9a2 2a3 a4 p (x) = x3 − x2 + x − ; p (x) = x4 − 2ax3 + x2 − x + 3 2 5 20 4 7 7 70 5a 20a2 5a3 5a4 a5 p (x) = x5 − x4 + x3 − x2 + x − 5 2 9 6 42 252 p R a 2 The orthonormal polynomials would be qn(x) = pn(x)= hn, where hn = 0 cpn(x) dx Question 2. What is the general expression for pn(x) in the above example? What about hn(x)? Is this a complete inner product space? 3 Three-term Recurrence Relation 1 Theorem 1. A sequence of orthogonal polynomials fpn(x)gn=0 satisfies pn+1(x) = (Anx + Bn)pn(x) + Cnpn−1(x); n = 1; 2; 3::: 3 where, kn+1 An hn An = ; n = 0; 1; 2; 3::: and Cn = − · ; n = 1; 2; 3::: kn An−1 hn−1 where kn is the leading coefficient of pn(x) Proof. We first note that the degree of pn+1(x) is the same as the degree of xpn(x). Thus, by defining the coefficients of the leading terms of each polynomial, we note that the leading term of p (x) is k xn+1, which is equal to the leading term of kn+1 xp := A xp . n+1 n+1 kn n n n =) hpn+1; xpn(x)i = An (1) Now we consider the lower order terms. pn+1 − Anxpn is a polynomial of degree ≤ n and n X pn+1 − Anxpn = ckpk(x) (2) k=0 This is because orthogonal polynomials are the basis for all polynomials. The orthogonality prop- erty now gives us that, n X hpn+1(x) − Anxpn(x); pk(x)i = cmhpm(x); pk(x)i = ckhpk(x); pk(x)i = hkck (3) m=0 = hpn+1(x); pk(x)i − hAnxpn(x); pk(x)i = −Anhpn(x); xpk(x)i (4) −An =) ck = hpn(x); xpk(x)i hk But, hpn(x); xpk(x)i = 0 for k < n − 1. Thus, 8k < n − 1; ck = 0 Therefore, from (2) 1 1 =) pn+1 − Anxpn = −An hpn(x); xpn(x)i · pn(x) + hpn(x); xpn−1(x)i · pn−1(x) hn hn−1 Or, equivalently pn+1(x) − Anxpn(x) = cnpn(x) + cn−1pn−1(x); n = 1; 2; 3::: Further, by using the same argument as (1), we have that kn−1 An hn hn−1cn−1 = −Anhpn(x); xpn−1(x)i = −An hn =) cn−1 = − · kn An−1 hn−1 The term B in Th. 1 can be represented as −An hp (x); xp (x)i but the inner product cannot n hn n n be put in terms of coefficients as simply as the other terms. Also, we note that if we represent the polynomials as monic polynomials, we get the simpler representation, hn pn+1(x) = xpn(x) + Bnpn(x) + Cnpn−1(x) where Cn = − ; n = 1; 2; 3::: hn−1 If they are orthogonal polynomials, we get the representation, hn+1 An pn+1(x) = ((Anx + Bn)pn(x) + Cnpn−1(x)) where Cn = − ; n = 1; 2; 3::: hn An−1 Question 3. Which is more computationally viable, the Gram-Schmidt Process or the Three-Term Recurrence Relation? Is there any other algorithm to do it better (like FFT)? 4 4 Christoffel-Darboux Formula As a consequence of the Three-Term Recurrence Relation, we have that 1 Theorem 2. A sequence of orthogonal polynomials fpn(x)gn=0 satisfies n X pk(x)pk(y) kn pn+1(x)pn(y) − pn+1(y)pn(x) = · ; n = 0; 1; 2::: hk hnkn+1 x − y k=0 and as y ! x n 2 X pk(x) kn 0 0 = · (pn+1(x)pn(x) − pn+1(x)pn(x)); n = 0; 1; 2::: hk hnkn+1 k=0 Proof. The three term recurrence relation implies that pn+1(x)pn(y) = (Anx + Bn)pn(x)pn(y) + Cnpn−1(x)pn(y); n = 1; 2; 3::: and pn+1(y)pn(x) = (Anx + Bn)pn(y)pn(x) + Cnpn−1(y)pn(x); n = 1; 2; 3::: Subtracting the two equations above, we have pn+1(x)pn(y) − pn+1(y)pn(x) = An(x − y)pn(x)pn(y) + Cn[pn−1(x)pn(y) − pn−1(y)pn(x)] =) An(x − y)pn(x)pn(y) = Cnpn−1(x)pn(y) − Cnpn−1(y)pn(x) − pn+1(x)pn(y) + pn+1(y)pn(x) Dividing all through by Anhn, we get (x − y)p (x)p (y) −p (y)p (x) + p (x)p (y) −p (y)p (x) + p (x)p (y) =) n n = n−1 n n−1 n + n+1 n n+1 n hn An−1hn−1 Anhn n n n X (x − y)pk(x)pk(y) X −pk−1(y)pk(x) + pk−1(x)pk(y) X −pk+1(y)pk(x) + pk+1(x)pk(y) =) = + hk Ak−1hk−1 Akhk k=1 k=1 k=1 By reordering, n n n X pk(x)pk(y) X pk+1(x)pk(y) − pk(x)pk+1(y) X pk(x)pk−1(y) − pk−1(x)pk(y) =) (x − y) = − hk Akhk Ak−1hk−1 k=1 k=1 k=1 By eliminating all the common terms in the two sums on the RHS, n X pk(x)pk(y) pn+1(x)pn(y) − pn(x)pn+1(y) p1(x)p0(y) − p0(x)p1(y) =) (x − y) = − hk Anhn A0h0 k=1 p (x)p (y) − p (x)p (y) k (p (x)k − k p (y)) = n+1 n n n+1 − 0 1 0 0 1 Anhn k1h0 as A = k1 and p (x) = p (y) = k 0 k0 0 0 0 k2(x−y) Further, p (x) − p (y) = k (x − y) =) k0(p1(x)k0−k0p1(y)) = 0 = (x − y) p0(x)p0(y) 1 1 1 k1h0 h0 h0 n X pk(x)pk(y) pn+1(x)pn(y) − pn(x)pn+1(y) =) (x − y) = hk Anhn k=0 5 n X pk(x)pk(y) kn =) = (pn+1(x)pn(y) − pn(x)pn+1(y)) hk (x − y)kn+1hn k=0 Thus, we prove the Christoffel-Darboux formula.
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