Orthogonal polynomials We start with 1 De¯nition 1. A sequence of polynomials fpn(x)gn=0 with degree[pn(x)] = n for each n is called orthogonal with respect to the weight function w(x) on the interval (a; b) with a < b if ( Z b 0; m 6= n w(x)pm(x)pn(x) dx = hn ±mn with ±mn := a 1; m = n: The weight function w(x) should be continuous and positive on (a; b) such that the moments Z b n ¹n := w(x) x dx; n = 0; 1; 2;::: a exist. Then the integral Z b hf; gi := w(x)f(x)g(x) dx a denotes an inner product of the polynomials f and g. The interval (a; b) is called the interval of orthogonality. This interval needs not to be ¯nite. If hn = 1 for each n 2 f0; 1; 2;:::g the sequence of polynomials is called orthonormal, and if n pn(x) = knx + lower order terms with kn = 1 for each n 2 f0; 1; 2;:::g the polynomials are called monic. Example. As an example we take w(x) = 1 and (a; b) = (0; 1). Using the Gram-Schmidt process the orthogonal polynomials can be constructed as follows. Start with the sequence 2 f1; x; x ;:::g. Choose p0(x) = 1. Then we have hx; p0(x)i hx; 1i 1 p1(x) = x ¡ p0(x) = x ¡ = x ¡ ; hp0(x); p0(x)i h1; 1i 2 since Z Z 1 1 1 h1; 1i = dx = 1 and hx; 1i = x dx = : 0 0 2 Further we have 2 2 2 hx ; p0(x)i hx ; p1(x)i p2(x) = x ¡ p0(x) ¡ p1(x) hp0(x); p0(x)i hp1(x); p1(x)i 2 2 1 µ ¶ µ ¶ 2 hx ; 1i hx ; x ¡ 2 i 1 2 1 1 2 1 = x ¡ ¡ 1 1 x ¡ = x ¡ ¡ x ¡ = x ¡ x + ; h1; 1i hx ¡ 2 ; x ¡ 2 i 2 3 2 6 since Z 1 Z 1 µ ¶ 2 2 1 2 1 2 1 1 1 1 hx ; 1i = x dx = ; hx ; x ¡ 2 i = x x ¡ dx = ¡ = 0 3 0 2 4 6 12 and Z 1 µ ¶2 Z 1 µ ¶ 1 1 1 2 1 1 1 1 1 hx ¡ 2 ; x ¡ 2 i = x ¡ dx = x ¡ x + dx = ¡ + = : 0 2 0 4 3 2 4 12 1 2 1 The polynomials p0(x) = 1, p1(x) = x ¡ 2 and p2(x) = x ¡ x + 6 are the ¯rst three monic orthogonal polynomials on the interval (0; 1) with respect to the weight function w(x) = 1. 1 Repeating this process we obtain 3 3 1 9 2 1 p (x) = x3 ¡ x2 + x ¡ ; p (x) = x4 ¡ 2x3 + x2 ¡ x + ; 3 2 5 20 4 7 7 70 5 20 5 5 1 p (x) = x5 ¡ x4 + x3 ¡ x2 + x ¡ ; 5 2 9 6 42 252 and so on. p The orthonormal polynomials would be q0(x) = p0(x)= h0 = 1, µ ¶ p p p p2(x) 2 1 q1(x) = p1(x)= h1 = 2 3(x ¡ 1=2); q2(x) = p = 6 5 x ¡ x + ; h2 6 µ ¶ p p3(x) 3 3 2 3 1 q3(x) = p = 20 7 x ¡ x + x ¡ ; h3 2 5 20 etcetera. All sequences of orthogonal polynomials satisfy a three term recurrence relation: 1 Theorem 1. A sequence of orthogonal polynomials fpn(x)gn=0 satis¯es pn+1(x) = (Anx + Bn)pn(x) + Cnpn¡1(x); n = 1; 2; 3;:::; where kn+1 An hn An = ; n = 0; 1; 2;::: and Cn = ¡ ¢ ; n = 1; 2; 3;:::: kn An¡1 hn¡1 Proof. Since degree [pn(x)] = n for each n 2 f0; 1; 2;:::g the sequence of orthogonal polyno- 1 mials fpn(x)gn=0 is linearly independent. Let An = kn+1=kn. Then pn+1(x) ¡ Anxpn(x) is a polynomial of degree · n. Hence Xn pn+1(x) ¡ Anxpn(x) = ckpk(x): k=0 The orthogonality property now gives Xn hpn+1(x) ¡ Anxpn(x); pk(x)i = cmhpm(x); pk(x)i = ckhpk(x); pk(x)i = ck hk: m=0 This implies hk ck = hpn+1(x) ¡ Anxpn(x); pk(x)i = hpn+1(x); pk(x)i ¡ Anhxpn(x); pk(x)i = ¡Anhpn(x); xpk(x)i: For k < n ¡ 1 we have degree [xpk(x)] < n which implies that hpn(x); xpk(x)i = 0. Hence: ck = 0 for k < n ¡ 1. This proves that the polynomials satisfy the three term recurrence relation pn+1(x) ¡ Anxpn(x) = cnpn(x) + cn¡1pn¡1(x); n = 1; 2; 3;:::: Further we have kn¡1 An hn hn¡1 cn¡1 = ¡Anhpn(x); xpn¡1(x)i = ¡An hn =) cn¡1 = ¡ ¢ : kn An¡1 hn¡1 This proves the theorem. 2 Note that the three term recurrence relation for a sequence of monic (kn = 1) orthogonal 1 polynomials fpn(x)gn=0 has the form hn pn+1(x) = xpn(x) + Bnpn(x) + Cnpn¡1(x) with Cn = ¡ ; n = 1; 2; 3;:::: hn¡1 A consequence of the three term recurrence relation is 1 Theorem 2. A sequence of orthogonal polynomials fpn(x)gn=0 satis¯es Xn p (x)p (y) k p (x)p (y) ¡ p (y)p (x) k k = n ¢ n+1 n n+1 n ; n = 0; 1; 2;::: (1) hk hn kn+1 x ¡ y k=0 and Xn 2 fpk(x)g kn ¡ 0 0 ¢ = ¢ pn+1(x)pn(x) ¡ pn+1(x)pn(x) ; n = 0; 1; 2;:::: (2) hk hn kn+1 k=0 Formula (1) is called the Christo®el-Darboux formula and (2) its confluent form. Proof. The three term recurrence relation implies that pn+1(x)pn(y) = (Anx + Bn)pn(x)pn(y) + Cnpn¡1(x)pn(y) and pn+1(y)pn(x) = (Any + Bn)pn(y)pn(x) + Cnpn¡1(y)pn(x): Subtraction gives 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)] : This leads to the telescoping sum Xn p (x)p (y) Xn p (x)p (y) ¡ p (y)p (x) (x ¡ y) k k = k+1 k k+1 k hk Ak hk k=1 k=1 Xn p (x)p (y) ¡ p (y)p (x) ¡ k k¡1 k k¡1 Ak¡1 hk¡1 k=1 p (x)p (y) ¡ p (y)p (x) k2(x ¡ y) = n+1 n n+1 n ¡ 0 : An hn h0 This implies that Xn p (x)p (y) p (x)p (y) ¡ p (y)p (x) (x ¡ y) k k = n+1 n n+1 n hk An hn k=0 kn = (pn+1(x)pn(y) ¡ pn+1(y)pn(x)) ; hn kn+1 which proves (1). The confluent form (2) then follows by taking the limit y ! x: p (x)p (y) ¡ p (y)p (x) lim n+1 n n+1 n y!x x ¡ y p (x)(p (x) ¡ p (y)) ¡ p (x)(p (x) ¡ p (y)) = lim n n+1 n+1 n+1 n n y!x x ¡ y 0 0 = pn(x)pn+1(x) ¡ pn+1(x)pn(x): 3 Zeros 1 Theorem 3. If fpn(x)gn=0 is a sequence of orthogonal polynomials on the interval (a; b) with respect to the weight function w(x), then the polynomial pn(x) has exactly n real simple zeros in the interval (a; b). Proof. Since degree[pn(x)] = n the polynomial has at most n real zeros. Suppose that pn(x) has m · n distinct real zeros x1; x2; : : : ; xm in (a; b) of odd order (or multiplicity). Then the polynomial pn(x)(x ¡ x1)(x ¡ x2) ¢ ¢ ¢ (x ¡ xm) does not change sign on the interval (a; b). This implies that Z b w(x)pn(x)(x ¡ x1)(x ¡ x2) ¢ ¢ ¢ (x ¡ xm) dx 6= 0: a By orthogonality this integral equals zero if m < n. Hence: m = n, which implies that pn(x) has n distinct real zeros of odd order in (a; b). This proves that all n zeros are distinct and simple (have order or multiplicity equal to one). 1 Theorem 4. If fpn(x)gn=0 is a sequence of orthogonal polynomials on the interval (a; b) with respect to the weight function w(x), then the zeros of pn(x) and pn+1(x) separate each other. Proof. This follows from the confluent form (2) of the Christo®el-Darboux formula. Note that Z b 2 hn = w(x) fpn(x)g dx > 0; n = 0; 1; 2;:::: a This implies that Xn 2 kn ¡ 0 0 ¢ fpk(x)g ¢ pn+1(x)pn(x) ¡ pn+1(x)pn(x) = > 0: hn kn+1 hk k=0 Hence kn ¡ 0 0 ¢ ¢ pn+1(x)pn(x) ¡ pn+1(x)pn(x) > 0: kn+1 Now suppose that xn;k and xn;k+1 are two consecutive zeros of pn(x) with xn;k < xn;k+1. 0 0 Since all n zeros of pn(x) are real and simple pn(xn;k) and pn(xn;k+1) should have opposite signs. Hence we have 0 0 pn(xn;k) = 0 = pn(xn;k+1) and pn(xn;k) ¢ pn(xn;k+1) < 0: This implies that pn+1(xn;k) ¢ pn+1(xn;k+1) should be negative too. Then the continuity of pn+1(x) implies that there should be at least one zero of pn+1(x) between xn;k and xn;k+1.