Orthogonal polynomials

We start with ∞ Definition 1. A sequence of polynomials {pn(x)}n=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 δmn with δmn := a 1, m = n. The weight function w(x) should be continuous and positive on (a, b) such that the moments Z b n µn := 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 finite.

If hn = 1 for each n ∈ {0, 1, 2,...} the sequence of polynomials is called orthonormal, and if n pn(x) = knx + lower order terms with kn = 1 for each n ∈ {0, 1, 2,...} 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 {1, x, x ,...}. 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 first 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. √ The orthonormal polynomials would be q0(x) = p0(x)/ h0 = 1, µ ¶ p √ √ p2(x) 2 1 q1(x) = p1(x)/ h1 = 2 3(x − 1/2), q2(x) = √ = 6 5 x − x + , h2 6 µ ¶ √ p3(x) 3 3 2 3 1 q3(x) = √ = 20 7 x − x + x − , h3 2 5 20 etcetera.

All sequences of orthogonal polynomials satisfy a three term recurrence relation: ∞ Theorem 1. A sequence of orthogonal polynomials {pn(x)}n=0 satisfies

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 ∈ {0, 1, 2,...} the sequence of orthogonal polyno- ∞ mials {pn(x)}n=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 ∞ polynomials {pn(x)}n=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

∞ Theorem 2. A sequence of orthogonal polynomials {pn(x)}n=0 satisfies 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 {pk(x)} 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 Christoffel-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

∞ Theorem 3. If {pn(x)}n=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).

∞ Theorem 4. If {pn(x)}n=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 Christoffel-Darboux formula. Note that Z b 2 hn = w(x) {pn(x)} dx > 0, n = 0, 1, 2,.... a This implies that Xn 2 kn ¡ 0 0 ¢ {pk(x)} · 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. However, this holds for each two consecutive zeros of pn(x). This proves the theorem.

n n+1 Moreover, if {xn,k}k=1 and {xn+1,k}k=1 denote the consecutive zeros of pn(x) and pn+1(x) respectively, then we have

a < xn+1,1 < xn,1 < xn+1,2 < xn,2 < ··· < xn+1,n < xn,n < xn+1,n+1 < b.

4 Gauss quadrature

If f is a continuous function on (a, b) and x1 < x2 < ··· < xn are n distinct points in (a, b), then there exists exactly one polynomial P with degree ≤ n − 1 such that P (xi) = f(xi) for all i = 1, 2, . . . , n. This polynomial P can easily be found by using Lagrange interpolation as follows. Define p(x) = (x − x1)(x − x2) ··· (x − xn) and consider Xn p(x) Xn (x − x ) ··· (x − x )(x − x ) ··· (x − x ) P (x) = f(x ) = f(x ) 1 i−1 i+1 n . i (x − x )p0(x ) i (x − x ) ··· (x − x )(x − x ) ··· (x − x ) i=1 i i i=1 i 1 i i−1 i i+1 i n

∞ Let {pn(x)}n=0 be a sequence of orthogonal polynomials on the interval (a, b) with respect to the weight function w(x). Then for x1 < x2 < ··· < xn we take the n distinct real zeros of the polynomial pn(x). If f is a polynomial of degree ≤ 2n − 1, then f(x) − P (x) is a polynomial of degree ≤ 2n − 1 with at least the zeros x1 < x2 < ··· < xn. Now we define

f(x) = P (x) + r(x)pn(x), where r(x) is a polynomial of degree ≤ n − 1. This can also be written as

Xn p (x) f(x) = f(x ) n + r(x)p (x). i (x − x )p0 (x ) n i=1 i n i This implies that Z Z Z b Xn b w(x)p (x) b w(x)f(x) dx = f(x ) n dx + w(x)r(x)p (x) dx. i (x − x )p0 (x ) n a i=1 a i n i a Since degree[r(x)] ≤ n − 1 the orthogonality property implies that the latter integral equals zero. This implies that Z Z b Xn b w(x)p (x) w(x)f(x) dx = λ f(x ) with λ := n dx, i = 1, 2, . . . , n. n,i i n,i (x − x )p0 (x ) a i=1 a i n i This is the Gauss quadrature formula. This gives the value of the integral in the case that f is a polynomial of degree ≤ 2n−1 if the value of f(xi) is known for the n zeros x1 < x2 < ··· < xn of the polynomial pn(x). If f is not a polynomial of degree ≤ 2n − 1 this leads to an approximation of the integral: Z Z b Xn b w(x)p (x) w(x)f(x) dx ≈ λ f(x ) with λ := n dx, i = 1, 2, . . . , n. n,i i n,i (x − x )p0 (x ) a i=1 a i n i n The coefficients {λn,i}i=1 are called Christoffel numbers. Note that these numbers do not depend on the function f. These Christoffel numbers are all positive. This can be shown as follows. We have Z b pn(x) λn,i = w(x)`n,i(x) dx with `n,i(x) := 0 , i = 1, 2, . . . , n. a (x − xi)pn(xi)

5 2 n Then `n,i(x)−`n,i(x) is a polynomial of degree ≤ 2n−2 which vanishes at the zeros {xn,k}k=1 of pn(x). Hence 2 `n,i(x) − `n,i(x) = pn(x)q(x) for some polynomial q of degree ≤ n − 2. This implies that Z b Z b ¡ 2 ¢ w(x) `n,i(x) − `n,i(x) dx = w(x)pn(x)q(x) dx = 0 a a by orthogonality. Hence we have Z b Z b 2 λn,i = w(x)`n,i(x) dx = w(x) {`n,i(x)} dx > 0. a a

Now we can also prove ∞ Theorem 5. Let {pn(x)}n=0 be a sequence of orthogonal polynomials on the interval (a, b) with respect to the weight function w(x) and let m < n. Then we have: between any two zeros of pm(x) there is at least one zero of pn(x).

Proof. Suppose that xm,k and xm,k+1 are two consecutive zeros of pm(x) and that there is no zero of pn(x) in (xm,k, xm,k+1). Then consider the polynomial p (x) g(x) = m . (x − xm,k)(x − xm,k+1) Then we have g(x)pm(x) ≥ 0 for x∈ / (xm,k, xm,k+1). Now the Gauss quadrature formula gives Z b Xn w(x)g(x)pm(x) dx = λn,ig(xn,i)pm(xn,i), a i=1 n where {xn,i}i=1 are the zeros of pn(x). Since there are no zeros of pn(x) in (xm,k, xm,k+1) we conclude that g(xn,i)pm(xn,i) ≥ 0 for all i = 1, 2, . . . , n. Further we have λn,i > 0 for all i = 1, 2, . . . , n which implies that the sum at the right-hand side cannot vanish. However, the integral at the left-hand side is zero by orthogonality. This contradiction proves that there should be at least one zero of pn(x) between the two consecutive zeros of pm(x).

2

y 1

0 0.2 0.4 0.6 0.8 1 x

±1

±2

The polynomials q2(x), q3(x) and q4(x)

6 Classical orthogonal polynomials

The classical orthogonal polynomials are named after Hermite, Laguerre and Jacobi:

name pn(x) w(x) (a, b)

−x2 Hermite Hn(x) e (−∞, ∞)

(α) −x α Laguerre Ln (x) e x (0, ∞)

(α,β) α β Jacobi Pn (x) (1 − x) (1 + x) (−1, 1)

Legendre Pn(x) 1 (−1, 1)

The Hermite polynomials are orthogonal on the interval (−∞, ∞) with respect to the normal distribution w(x) = e−x2 , the Laguerre polynomials are orthogonal on the interval (0, ∞) with respect to the gamma distribution w(x) = e−xxα and the Jacobi polynomials are orthogonal on the interval (−1, 1) with respect to the beta distribution w(x) = (1 − x)α(1 + x)β.

The Legendre polynomials form a special case (α = β = 0) of the Jacobi polynomials.

These classical orthogonal polynomials satisfy an orthogonality relation, a three term recur- rence relation, a second order linear differential equation and a so-called Rodrigues formula. Moreover, for each family of classical orthogonal polynomials we have a generating function.

In the sequel we will often use the Kronecker delta which is defined by ( 0, m 6= n δmn := 1, m = n for m, n ∈ {0, 1, 2,...} and the notation

d D = (3) dx for the differentiation operator. Then we have Leibniz’ rule µ ¶ Xn n Dn [f(x)g(x)] = Dkf(x)Dn−kg(x), n = 0, 1, 2,... (4) k k=0 which is a generalization of the product rule. The proof is by mathematical induction and by use of Pascal’s triangle identity µ ¶ µ ¶ µ ¶ n n n + 1 + = , k = 1, 2, . . . , n. k k − 1 k

7 Hermite

The Hermite polynomials are orthogonal on the interval (−∞, ∞) with respect to the normal distribution w(x) = e−x2 . They can be defined by means of their Rodrigues formula:

n (−1) 2 2 H (x) = Dnw(x) = (−1)nex Dne−x , n = 0, 1, 2,..., (5) n w(x) where the differentiation operator D is defined by (3). Since Dn+1 = DDn, we obtain

n+1 n n n £ 0 0 ¤ D w(x) = D [D w(x)] = (−1) D [w(x)Hn(x)] = (−1) w (x)Hn(x) + w(x)Hn(x) n+1 £ 0 ¤ = (−1) w(x) 2xHn(x) − Hn(x) , n = 0, 1, 2,..., which implies that

0 Hn+1(x) = 2xHn(x) − Hn(x), n = 0, 1, 2,.... (6)

The definition (5) implies that H0(x) = 1. Then (6) implies by induction that Hn(x) is a polynomial of degree n. Further we have that H2n(x) is even and H2n+1(x) is odd and that n the leading coefficient of the polynomial Hn(x) equals kn = 2 . The Hermite polynomials satisfy the orthogonality relation Z ∞ 1 −x2 n √ e Hm(x)Hn(x) dx = 2 n! δmn, m, n ∈ {0, 1, 2,...}. (7) π −∞ To prove this we use the definition (5) to obtain

Z ∞ Z ∞ −x2 n n −x2 e Hm(x)Hn(x) dx = (−1) Hm(x)D e dx. −∞ −∞ Now we use integration by parts n times to conclude that the integral vanishes for m < n. For m = n we have using integration by parts Z ∞ Z ∞ Z ∞ −x2 n n −x2 n −x2 e Hn(x)Hn(x) dx = (−1) Hn(x)D e dx = D Hn(x) · e dx −∞ −∞ −∞ Z ∞ −x2 n √ = kn n! e dx = 2 n! π. −∞ This proves the orthogonality relation (7).

In order to find the three term recurrence relation we start with

2 w(x) = e−x =⇒ w0(x) = −2xw(x).

Then we have by using Leibniz’ rule (4)

Dn+1w(x) = Dnw0(x) = Dn [−2xw(x)] = −2xDnw(x) − 2nDn−1w(x), which implies that

Hn+1(x) = 2xHn(x) − 2nHn−1(x), n = 1, 2, 3,.... (8)

8 Combining (6) and (8) we find that

0 Hn(x) = 2nHn−1(x), n = 1, 2, 3,.... (9)

Differentiation of (6) gives

0 0 00 Hn+1(x) = 2xHn(x) + 2Hn(x) − Hn(x), n = 0, 1, 2,....

Now we use (9) to conclude that

0 00 2(n + 1)Hn(x) = 2xHn(x) + 2Hn(x) − Hn(x), n = 0, 1, 2,..., which implies that Hn(x) satisfies the second order linear differential equation

y00(x) − 2xy0(x) + 2ny(x) = 0, n ∈ {0, 1, 2,...}.

Finally we will prove the generating function

∞ 2 X H (x) e2xt−t = n tn. (10) n! n=0 We start with 2 2 2 f(t) = e−(x−t) = e−x · e2xt−t . The Taylor series for f(t) is X∞ f (n)(0) f(t) = tn n! n=0 with, by using the substitution x − t = u,

· n ¸ · n ¸ (n) d −(x−t)2 n d −u2 n n −x2 −x2 f (0) = n e = (−1) n e = (−1) D e = e Hn(x) dt t=0 du u=x for n = 0, 1, 2,.... Hence we have

∞ (n) ∞ 2 2 2 X f (0) 2 X H (x) e−x · e2xt−t = e−(x−t) = f(t) = tn = e−x n tn. n! n! n=0 n=0 This proves the generating function (10).

9 Laguerre

The Laguerre polynomials are orthogonal on the interval (0, ∞) with respect to the gamma distribution w(x) = e−xxα. They can be defined by means of their Rodrigues formula: 1 1 1 £ ¤ L(α)(x) = Dn [w(x) xn] = ex x−α Dn e−xxn+α , n = 0, 1, 2,.... (11) n n! w(x) n! By using Leibniz’ rule (4) we have µ ¶ £ ¤ Xn n Dn e−xxn+α = Dke−xDn−kxn+α k k=0 µ ¶ Xn n = (−1)ke−x(n + α)(n + α − 1) ··· (α + k + 1)xα+k k k=0 µ ¶ Xn n Γ(n + α + 1) = e−x xα (−1)k xk. k Γ(k + α + 1) k=0 Hence we have µ ¶ Xn n + α xk L(α)(x) = (−1)k , n = 0, 1, 2,..., (12) n n − k k! k=0 where µ ¶ n + α Γ(n + α + 1) (k + α + 1) = = n−k , k = 0, 1, 2, . . . , n. n − k (n − k)! Γ(k + α + 1) (n − k)!

(α) This proves that Ln (x) is a polynomial of degree n. Since µ ¶ n + α (−1)k (α + 1) (α + 1) (−n) (−1)k = n = n k , k = 0, 1, 2, . . . , n n − k (n − k)! (α + 1)k n! (α + 1)k we also have for n = 0, 1, 2,... µ ¶ µ ¶ Xn k (α) (α + 1)n (−n)k x n + α −n Ln (x) = = 1F1 ; x . (13) n! (α + 1)k k! n α + 1 k=0 Note that we have µ ¶ n + α (α + 1) L(α)(0) = = n , n = 0, 1, 2,... n n n!

(α) and that the leading coefficient of the polynomial Ln (x) equals (−1)n k = , n = 0, 1, 2,.... n n! Further we have µ ¶ µ ¶ d d Xn n + α xk Xn n + α xk−1 L(α)(x) = (−1)k = (−1)k dx n dx n − k k! n − k (k − 1)! k=0 k=1 µ ¶ nX−1 n + α xk = (−1)k−1 = −L(α+1)(x), n = 1, 2, 3,.... (14) n − k − 1 k! n−1 k=0

10 Now we can prove the orthogonality relation

Z ∞ −x α (α) (α) Γ(n + α + 1) e x Lm (x)Ln (x) dx = δmn, α > −1 (15) 0 n! for m, n ∈ {0, 1, 2,...}. First of all, the integral converges if the moments

Z ∞ −x n+α µn = e x dx 0 exists for all n ∈ {0, 1, 2,...}. This leads to the condition α > −1. Note that µn = Γ(n+α+1). Now we use (11) to obtain

Z ∞ Z ∞ −x α (α) (α) 1 (α) n £ −x n+α¤ e x Lm (x)Ln (x) dx = Lm (x)D e x dx. 0 n! 0 We apply integration by parts to obtain

Z ∞ Z ∞ (α) n £ −x n+α¤ n n (α) −x n+α Lm (x)D e x dx = (−1) D Lm (x)e x dx 0 0 which equals zero for m < n. For m = n we find

Z ∞ Z ∞ n (α) −x n+α −x n+α n D Ln (x)e x dx = kn n! e x dx = (−1) Γ(n + α + 1). 0 0 This proves the orthogonality relation (15).

The Laguerre polynomials can also be defined by their generating function µ ¶ xt X∞ (1 − t)−α−1 exp − = L(α)(x)tn. (16) 1 − t n n=0 The proof is straightforward. Start with (13) and change the order of summation to obtain

X∞ X∞ Xn k X∞ X∞ k k n (α) n (α + 1)n n (−n)k x (α + 1)n (−1) x t Ln (x)t = t = n! (α + 1)k k! (α + 1)k k!(n − k)! n=0 n=0 k=0 k=0 n=k X∞ X∞ (α + 1) (−1)kxktn+k X∞ (−xt)k X∞ (α + k + 1) = n+k = n tn (α + 1)k k! n! k! n! k=0 n=0 k=0 n=0 µ ¶ X∞ (−xt)k X∞ 1 xt k = (1 − t)−α−k−1 = (1 − t)−α−1 − . k! k! 1 − t k=0 k=0 This proves (16).

If we define µ ¶ xt F (x, t) := (1 − t)−α−1 exp − , 1 − t then we easily obtain µ ¶ ∂ xt ∂ F (x, t) = −t(1 − t)−α−2 exp − =⇒ (1 − t) F (x, t) + tF (x, t) = 0 ∂x 1 − t ∂x

11 and ½ ¾ µ ¶ ∂ −x(1 − t) − xt xt F (x, t) = (α + 1)(1 − t)−α−2 + (1 − t)−α−1 · exp − ∂t (1 − t)2 1 − t ½ ¾ µ ¶ x xt = α + 1 − (1 − t)−α−2 exp − , 1 − t 1 − t which implies that ∂ (1 − t)2 F (x, t) + [x − (α + 1)(1 − t)] F (x, t) = 0. ∂t The first relation leads to

X∞ d X∞ (1 − t) L(α)(x)tn + t L(α)(x)tn = 0 dx n n n=0 n=0 or equivalently

X∞ d X∞ d X∞ L(α)(x)tn − L(α)(x)tn+1 + L(α)(x)tn+1 = 0. dx n dx n n n=0 n=0 n=0 Hence we have d d L(α) (x) − L(α)(x) + L(α)(x) = 0, n = 0, 1, 2,... (17) dx n+1 dx n n or equivalently, by using (14),

d L(α)(x) = L(α)(x) − L(α+1)(x), n = 0, 1, 2,.... dx n n n The second relation leads to X∞ X∞ 2 (α) n−1 (α) n (1 − t) nLn (x)t + [x − (α + 1)(1 − t)] Ln (x)t = 0 n=1 n=0 or equivalently X∞ X∞ X∞ (α) n−1 (α) n (α) n+1 nLn (x)t − 2 nLn (x)t + nLn (x)t n=1 n=1 n=1 X∞ X∞ X∞ (α) n (α) n (α) n+1 + x Ln (x)t − (α + 1) Ln (x)t + (α + 1) Ln (x)t = 0. n=0 n=0 n=0 Equating the coefficients of equal powers of t we obtain the three term recurrence relation

(α) (α) (α) (n + 1)Ln+1(x) + (x − 2n − α − 1)Ln (x) + (n + α)Ln−1(x) = 0, n = 1, 2, 3,.... (18) Note that this can also be written as h i h i (α) (α) (α) (α) (α) xLn (x) + (n + 1) Ln+1(x) − Ln (x) − (n + α) Ln (x) − Ln−1(x) = 0, n = 1, 2, 3,....

12 Now we differentiate and use (17) to obtain

d x L(α)(x) + L(α)(x) − (n + 1)L(α)(x) + (n + α)L(α) (x) = 0, n = 1, 2, 3,.... dx n n n n−1 This implies that d x L(α)(x) = nL(α)(x) − (n + α)L(α) (x), n = 1, 2, 3,.... (19) dx n n n−1

Now we differentiate (19) and use (17) and (19) to find

2 d (α) d (α) d (α) d (α) x 2 Ln (x) + Ln (x) = n Ln (x) − (n + α) Ln−1(x) dx dx dx · dx ¸ d d d = (n + α) L(α)(x) − L(α) (x) − α L(α)(x) dx n dx n−1 dx n d = −(n + α)L(α) (x) − α L(α)(x) n−1 dx n d d = x L(α)(x) − nL(α)(x) − α L(α)(x). dx n n dx n

(α) This proves that the polynomial Ln (x) satisfies the second order linear differential equation

xy00(x) + (α + 1 − x)y0(x) + ny(x) = 0, n ∈ {0, 1, 2,...}.

Finally, we use the generating function (16) to prove the addition formula

Xn (α+β+1) (α) β) Ln (x + y) = Lk (x)Ln−k(y), n = 0, 1, 2,.... (20) k=0 The generating function (16) implies that µ ¶ X∞ (x + y)t L(α+β+1)(x + y)tn = (1 − t)−α−β−2 exp − n 1 − t n=0 µ ¶ µ ¶ xt yt = (1 − t)−α−1 exp − · (1 − t)−β−1 exp − 1 − t 1 − t à ! X∞ X∞ X∞ Xn (α) k (β) m (α) (β) n = Lk (x)t · Lm (y)t = Lk (x)Ln−k(y) t . k=0 m=0 n=0 k=0 This proves (20).

13 Jacobi

The Jacobi polynomials are orthogonal on the interval (−1, 1) with respect to the beta dis- tribution w(x) = (1 − x)α(1 + x)β. They can be defined by means of their Rodrigues formula: (−1)n 1 £ ¤ P (α,β)(x) = Dn w(x) (1 − x2)n n 2n n! w(x) (−1)n h i = (1 − x)−α(1 + x)−β Dn (1 − x)n+α(1 + x)n+β (21) 2n n! for n = 0, 1, 2,.... By using Leibniz’ rule (4) we have µ ¶ h i Xn n Dn (1 − x)n+α(1 + x)n+β = Dk(1 − x)n+αDn−k(1 + x)n+β k k=0 µ ¶ Xn n = (−1)k(n + α)(n + α − 1) ··· (n + α − k + 1)(1 − x)n+α−k k k=0 × (n + β)(n + β − 1) ··· (β + k + 1)(1 + x)β+k µ ¶µ ¶ Xn n + α n + β = n! (−1)k (1 − x)n+α−k(1 + x)β+k, n = 0, 1, 2,.... k n − k k=0 This implies that µ ¶µ ¶ (−1)n Xn n + α n + β P (α,β)(x) = (−1)k (1 − x)n−k(1 + x)k, n = 0, 1, 2,.... (22) n 2n k n − k k=0

(α,β) This shows that Pn (x) is a polynomial of degree n. Note that we have the symmetry (α,β) n (β,α) Pn (−x) = (−1) Pn (x), n = 0, 1, 2,... (23) and µ ¶ µ ¶ n + α n + β P (α,β)(1) = and P (α,β)(−1) = (−1)n , n = 0, 1, 2,.... n n n n In order to find a hypergeometric representation we write for x 6= 1 µ ¶ µ ¶µ ¶ µ ¶ x − 1 n Xn n + α n + β x + 1 k P (α,β)(x) = , n = 0, 1, 2,.... n 2 n n − k x − 1 k=0 Now we have for x 6= 1 µ ¶ µ ¶ µ ¶ µ ¶ x + 1 k 2 k Xk k 2 i = 1 + = , k = 0, 1, 2,.... x − 1 x − 1 i x − 1 i=0 Now we obtain by changing the order of summations for x 6= 1 and n = 0, 1, 2,... µ ¶ µ ¶µ ¶µ ¶ µ ¶ x − 1 n Xn Xn n + α n + β k 2 i P (α,β)(x) = n 2 k n − k i x − 1 i=0 k=i µ ¶ µ ¶µ ¶µ ¶ µ ¶ x − 1 n Xn Xn−i n + α n + β i + k 2 i = . 2 i + k n − i − k i x − 1 i=0 k=0

14 Now we reverse the order in the first sum to find for x 6= 1 and n = 0, 1, 2,... µ ¶ µ ¶µ ¶µ ¶ µ ¶ x − 1 n Xn Xn n + α n + β n − i + k 2 n−i P (α,β)(x) = n 2 n − i + k i − k n − i x − 1 i=0 k=0 µ ¶µ ¶µ ¶ µ ¶ Xn Xn n + α n + β n − i + k x − 1 i = n − i + k i − k n − i 2 i=0 k=0 Xn Xn Γ(n + α + 1) = (n − i + k)! Γ(i − k + α + 1) i=0 k=0 µ ¶ Γ(n + β + 1) (n − i + k)! x − 1 i × (i − k)! Γ(n − i + k + β + 1) (n − i)! k! 2 µ ¶ Γ(n + α + 1)Γ(n + β + 1) Xn (−n) 1 − x i = i n! Γ(i + α + 1) i! Γ(n − i + β + 1) 2 i=0 Xn (−i) (−i − α − 1) × k k . (n − i + β + 1)k k! k=0 Since i ∈ {0, 1, 2, . . . , n} we have by using the Chu-Vandermonde summation formula Xn µ ¶ (−i)k(−i − α − 1)k −i, −i − α − 1 (n + α + β + 1)i = 2F1 ; 1 = . (n − i + β + 1)k k! n − i + β + 1 (n − i + β + 1)i k=0

Hence we have by using Γ(n − i + β + 1) (n − i + β + 1)i = Γ(n + β + 1) µ ¶ Γ(n + α + 1) Xn (−n) (n + α + β + 1) 1 − x i P (α,β)(x) = i i n n! Γ(i + α + 1) i! 2 i=0 µ ¶ Γ(n + α + 1) Xn (−n) (n + α + β + 1) 1 − x i = i i , n = 0, 1, 2,.... n! Γ(α + 1) (α + 1) i! 2 i=0 i This proves the hypergeometric representation µ ¶ µ ¶ n + α −n, n + α + β + 1 1 − x P (α,β)(x) = F ; , n = 0, 1, 2,.... (24) n n 2 1 α + 1 2 Note that this result also holds for x = 1. In view of the symmetry (23) we also have µ ¶ µ ¶ n + β −n, n + α + β + 1 1 + x P (α,β)(x) = (−1)n F ; , n = 0, 1, 2,.... n n 2 1 β + 1 2 Note that the hypergeometric representation implies that µ ¶ µ ¶ d n + α (−n)(n + α + β + 1) 1 P (α,β)(x) = · − dx n n α + 1 2 µ ¶ −n + 1, n + α + β + 2 1 − x × F ; 2 1 α + 2 2 µ ¶ µ ¶ n + α + β + 1 n + α −n + 1, n + α + β + 2 1 − x = F ; 2 n − 1 2 1 α + 2 2 n + α + β + 1 = P (α+1,β+1)(x), n = 1, 2, 3,.... 2 n−1

15 Another consequence of the hypergeometric representation is that the leading coefficient of (α,β) the polynomial Pn (x) equals

µ ¶ n n + α (−n)n(n + α + β + 1)n (−1) (n + α + β + 1)n kn = n = n , n = 0, 1, 2,.... n (α + 1)n n! 2 2 n!

Now it can be shown that the Jacobi polynomials satisfy the orthogonality relation

Z 1 α+β+1 α β (α,β) (α,β) 2 Γ(n + α + 1) Γ(n + β + 1) (1 − x) (1 + x) Pm (x)Pn (x) dx = δmn −1 (2n + α + β + 1) Γ(n + α + β + 1) n! for α > −1, β > −1 and m, n ∈ {0, 1, 2,...}. This can be shown by using the definition (21) and integration by parts. The value of the integral in the case m = n can be computed by using the leading coefficient and then writing the integral in terms of a beta integral: Z 1 n o2 α β (α,β) (1 − x) (1 + x) Pn (x) dx −1 n Z 1 h i (−1) (α,β) n n+α n+β = n Pn (x)D (1 − x) (1 + x) dx 2 n! −1 Z 1 1 n (α,β) n+α n+β = n D Pn (x)(1 − x) (1 + x) dx 2 n! −1 Z 1 (n + α + β + 1)n n+α n+β = 2n (1 − x) (1 + x) dx 2 n! −1 Z 1 Γ(2n + α + β + 1) n+α n+β = 2n (1 − x) (1 + x) dx, n = 0, 1, 2,... Γ(n + α + β + 1) 2 n! −1 and by using the substitution 1 − x = 2t

Z 1 Z 1 (1 − x)n+α(1 + x)n+β dx = (2t)n+α(2 − 2t)n+β 2 dt −1 0 Z 1 = 22n+α+β+1 tn+α(1 − t)n+β dt = 22n+α+β+1 B(n + α + 1, n + β + 1) 0 Γ(n + α + 1)Γ(n + β + 1) = 22n+α+β+1 Γ(2n + α + β + 2) Γ(n + α + 1)Γ(n + β + 1) = 22n+α+β+1 , n = 0, 1, 2,.... (2n + α + β + 1) Γ(2n + α + β + 1)

(α,β) The Jacobi polynomials Pn (x) satisfy the second order linear differential equation

(1 − x2)y00(x) + [β − α − (α + β + 2)x] y0(x) + n(n + α + β + 1)y(x) = 0, n = 0, 1, 2,....

A generating function for the Jacobi polynomials is given by

2α+β X∞ p = P (α,β)(x)tn,R = 1 − 2xt + t2. R(1 + R − t)α(1 + R + t)β n n=0

16 Legendre

The Legendre polynomials are orthogonal on the interval (−1, 1) with respect to the weight function w(x) = 1. They can be defined by means of their Rodrigues formula:

(−1)n 1 £ ¤ (−1)n £ ¤ P (x) = Dn w(x) (1 − x2)n = Dn (1 − x2)n , n = 0, 1, 2,.... (25) n 2n n! w(x) 2n n!

This is the special case α = β = 0 of the Jacobi polynomials: µ ¶ −n, n + 1 1 − x P (x) = P (0,0)(x) = F ; , n = 0, 1, 2,.... (26) n n 2 1 1 2

Further we have

n n Pn(−x) = (−1) Pn(x),Pn(1) = 1 and Pn(−1) = (−1) , n = 0, 1, 2,....

The leading coefficient of the polynomial Pn(x) equals

n (−n)n (n + 1)n (−1) (2n)! kn = n = n 2 , n = 0, 1, 2,.... (1)n n! 2 2 (n!) The orthogonality relation is Z 1 2 Pm(x)Pn(x) dx = δmn, m, n ∈ {0, 1, 2,...}. (27) −1 2n + 1 This can be shown by using the Rodrigues formula (25) and integration by parts

Z 1 n Z 1 Z 1 (−1) n £ 2 n¤ 1 n 2 n Pm(x)Pn(x) dx = n Pm(x)D (1 − x ) dx = n D Pm(x) (1 − x ) dx, −1 2 n! −1 2 n! −1 which vanishes for m < n. For m = n we have

Z 1 Z 1 Z 1 n 2 n 2 n (2n)! 2 n D Pn(x) (1 − x ) dx = kn n! (1 − x ) dx = n (1 − x ) dx. −1 −1 2 n! −1 Finally we have by using the substitution 1 − x = 2t for n = 0, 1, 2,...

Z 1 Z 1 Z 1 (1 − x2)n dx = (1 − x)n(1 + x)n dx = (2t)n(2 − 2t)n 2 dx −1 −1 0 Γ(n + 1) Γ(n + 1) 22n+1 (n!)2 = 22n+1 B(n + 1, n + 1) = 22n+1 = . Γ(2n + 2) (2n + 1)!

Hence we have

Z 1 2n+1 2 2 (2n)! 2 (n!) 2 {Pn(x)} dx = 2n 2 = , n = 0, 1, 2,.... −1 2 (n!) (2n + 1)! 2n + 1 This proves the orthogonality relation (27).

17 In order to find a generating function for the Legendre polynomials we use the hypergeometric representation (26) to find X∞ X∞ Xn µ ¶k n (−n)k (n + 1)k 1 − x n Pn(x)t = t (1)k k! 2 n=0 n=0 k=0 µ ¶ X∞ X∞ (−n) (n + 1) 1 − x k = k k tn k! k! 2 k=0 n=k µ ¶ X∞ X∞ (−n − k) (n + k + 1) 1 − x k = k k tn+k k! k! 2 k=0 n=0 µ ¶ X∞ (2k)! x − 1 k X∞ (2k + 1) = tk n tn k! k! 2 n! k=0 n=0 µ ¶ X∞ (2k)! x − 1 k = tk(1 − t)−2k−1 k! k! 2 k=0 · ¸ X∞ (1/2) 2(x − 1)t −1/2 = k [2(x − 1)t]k (1 − t)−2k−1 = (1 − t)−1 1 − k! (1 − t)2 k=0 £ ¤−1/2 1 = (1 − t)2 − 2(x − 1)t = (1 − 2xt + t2)−1/2 = √ . 1 − 2xt + t2 This proves the generating function X∞ 1 n √ = Pn(x)t . (28) 2 1 − 2xt + t n=0 If we define F (x, t) = (1 − 2xt + t2)−1/2, then we have ∂ 1 x − t F (x, t) = − (1 − 2xt + t2)−3/2 (−2x + 2t) = . ∂t 2 (1 − 2xt + t2)3/2 This implies that ∂ (1 − 2xt + t2) F (x, t) = (x − t)F (x, t). ∂t Now we use (28) to obtain X∞ X∞ 2 n−1 n (1 − 2xt + t ) nPn(x)t = (x − t) Pn(x)t . n=1 n=0 This can also be written as X∞ X∞ X∞ X∞ X∞ n−1 n n+1 n n+1 nPn(x)t − 2x nPn(x)t + nPn(x)t = x Pn(x)t − Pn(x)t n=1 n=1 n=1 n=0 n=0 or equivalently X∞ X∞ X∞ n n n+1 (n + 1)Pn+1(x)t − x (2n + 1)Pn(x)t + (n + 1)Pn(x)t = 0. n=0 n=0 n=0

This leads to P1(x) = xP0(x) and the three term recurrence relation

(n + 1)Pn+1(x) − (2n + 1)xPn(x) + nPn−1(x) = 0, n = 1, 2, 3,.... (29)

18 Chebyshev

For x ∈ [−1, 1] the Chebyshev polynomials Tn(x) of the first kind and the Chebyshev poly- nomials Un(x) of the second kind can be defined by sin(n + 1)θ T (x) = cos(nθ) and U (x) = , x = cos θ, n = 0, 1, 2,.... (30) n n sin θ The orthogonality property is given by

Z 1 Z π 2 −1/2 (1 − x ) Tm(x)Tn(x) dx = cos(mθ) cos(nθ) dθ = 0, m 6= n −1 0 and Z 1 Z π 2 1/2 (1 − x ) Um(x)Un(x) dx = sin(m + 1)θ sin(n + 1)θ dθ = 0, m 6= n. −1 0 Both families of orthogonal polynomials satisfy the three term recurrence relation

Pn+1(x) = 2xPn(x) − Pn−1(x), n = 1, 2, 3,..., since we have

Tn+1(x) + Tn−1(x) = cos(n + 1)θ + cos(n − 1)θ = 2 cos θ cos(nθ) = 2xTn(x) and sin(n + 2)θ + sin(nθ) 2 cos θ sin(n + 1)θ U (x) + U (x) = = = 2xU (x). n+1 n−1 sin θ sin θ n Note that T0(x) = U0(x) = 1,T1(x) = x and U1(x) = 2x. We also have Tn(x) = Un(x) − xUn−1(x), n = 1, 2, 3,..., since sin(n + 1)θ − cos θ sin(nθ) sin θ cos(nθ) U (x) − xU (x) = = = cos(nθ) = T (x). n n−1 sin θ sin θ n

In order to find a generating function for the Chebyshev polynomials Tn(x) of the first kind, we multiply the recurrence relation by tn+1 and take the sum to obtain X∞ X∞ X∞ n+1 n+1 n+1 Tn+1(x)t = 2x Tn(x)t − Tn−1(x)t . n=1 n=1 n=1 If we define X∞ n F (x, t) = Tn(x)t , |t| < 1, n=0 then we have

2 F (x, t) − T1(x)t − T0(x) = 2xt [F (x, t) − T0(x)] − t F (x, t).

19 This implies that 2 (1 − 2xt + t )F (x, t) = T0(x) + T1(x)t − 2xtT0(x) = 1 + xt − 2xt = 1 − xt. Hence we have the generating function X∞ 1 − xt T (x)tn = F (x, t) = , |t| < 1. n 1 − 2xt + t2 n=0

In the same way we have for the Chebyshev polynomials Un(x) of the second kind: X∞ n G(x, t) = Un(x)t , |t| < 1 n=0 where 2 (1 − 2xt + t )G(x, t) = U0(x) + U1(x)t − 2xtU0(x) = 1 + 2xt − 2xt = 1. Hence we have X∞ 1 U (x)tn = G(x, t) = , |t| < 1. n 1 − 2xt + t2 n=0

This can be used, for instance, to prove that Xn n−k Tk(x)x = Un(x), n = 0, 1, 2,.... k=0 In fact, we have for |t| < 1 Ã ! X∞ Xn X∞ X∞ X∞ X∞ n−k n n−k n n n+k Tk(x)x t = Tk(x)x t = Tk(x)x t n=0 k=0 k=0 n=k k=0 n=0 X∞ X∞ 1 − xt 1 = T (x)tk · (xt)n = · k 1 − 2xt + t2 1 − xt k=0 n=0 1 X∞ = = U (x)tn. 1 − 2xt + t2 n n=0 In a similar way we can prove that Xn Pk(x)Pn−k(x) = Un(x), n = 0, 1, 2,..., k=0 where Pn(x) denotes the Legendre polynomial. In fact, we have for |t| < 1 Ã ! X∞ Xn X∞ X∞ X∞ X∞ n n n+k Pk(x)Pn−k(x) t = Pk(x)Pn−k(x)t = Pk(x)Pn(x)t n=0 k=0 k=0 n=k k=0 n=0 X∞ X∞ k n 1 1 = P (x)t · Pn(x)t = √ · √ k 2 2 k=0 n=0 1 − 2xt + t 1 − 2xt + t 1 X∞ = = U (x)tn. 1 − 2xt + t2 n n=0

20