Introduction to orthogonal polynomials Michael Anshelevich November 6, 2003
µ = probability measure on R with finite moments m n (µ) = R xn dµ(x) <. Induces a functional on polynomials C[x], ϕ [P (x)] = R P (x) dµ(x). On the polynomials C[x], define the sesquilinear inner product x n, x k µ = ϕ µ [ x n+k ] = m n+k (µ). The set {x n } n=0 is a basis for C[x]. Gram-Schmidt with respect to the inner product, µ, get a family of polynomials {P n } n=0. Note they have real coefficients. They are orthogonal with respect to µ: if n k. ϕ µ [P n P k ] = R P n(x)p k (x) dµ(x) = 0 1
Hermite polynomials: normal (Gaussian) distribution 1 2πt e x2 /2t dx. 0.4 0.3 0.2 0.1 4 2 0 2 4 x Laguerre polynomials: Gamma distribution 1 0.8 1 Γ(t) e x x t 1 1 [0, ) (x) dx. 0.35 0.3 0.25 0.6 0.2 0.4 0.15 0.1 0.2 0.05 0 1 2 3 4 5 0 1 2 3 4 5 x x Jacobi polynomials: 1 Z(α, β) (1 x)α (1 + x) β 1 [ 1,1] (x) dx. In particular: Ultraspherical (Gegenbauer), α = β = λ 1 2, 1 Z(λ) (1 x2 ) λ 1 2 1 [ 1,1] (x) dx, 2
Chebyshev of the 1st kind, α = β = 1 2, λ = 0, 1 π 1 1 1 x 2 [ 1,1](x) dx, 6 5 4 3 2 1 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 x Chebyshev of the 2nd kind, α = β = 1 2, λ = 1, 2 π 1 x 2 1 [ 1,1] (x) dx, 0.6 0.5 0.4 0.3 0.2 0.1 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 x Kravchuk: binomial distribution, for q = 1 p, N ( N k=0 k ) p k q N k δ 2k N (x). 0.3 0.25 0.2 0.15 0.1 0.05 4 2 0 2 4 k 3
Charlier: Poisson distribution k=0 e t tk k! δ k(x). 0.25 0.2 0.15 0.1 0.05 0 1 2 3 4 5 k Meixner: negative binomial (Pascal) distribution k t ( k 1) p t q k t δ k (x). t 1 0.26 0.24 0.22 0.2 0.18 0.16 0.14 0.12 0.1 0 1 2 3 4 5 6 7 8 k Can describe in various ways: Using the measure of orthogonality. 2nd order differential or difference equations. Explicit formulas using hypergeometric functions. Three-term recursion relations. 4
Favard s theorem. (Stone 1932) Let {P n } n=0 be a polynomial family, that is, P n has degree n. {P n } are orthogonal with respect to some measure µ with positive leading coefficients they satisfy a threeterm recursion relation P 0 (x) = 1, xp 0 (x) = α 0 P 1 (x) + β 0 P 0 (x), xp n (x) = α n P n+1 (x) + β n P n (x) + γ n P n 1 (x), where all β n are real, α n > 0, γ n 0. Moreover, {P n } are monic iff all α n = 1. {P n } are orthonormal iff γ n = α n 1. µ has finite support iff γ n = 0 for some n. 5
) Suppose the polynomials are orthogonal with respect to µ and have positive leading coefficients. Then So for n + 1 < k, and also P n span ( P 0, P 1,..., P n 1 ). xp n, P k µ = 0 xp n, P k µ = ϕ µ [xp n P k ] = P n, xp k = 0 for n > k + 1, n 1 > k. So xp n (x) = α n P n+1 (x) + β n P n (x) + γ n P n 1 (x) for some real α n, β n, γ n. Leading coefficients of P n, P n+1 are positive α n > 0. Also, (leading coefficients are 1) α n = 1. 6
From the recursion relation, xp n, P n 1 = γ n P n 1, P n 1. Also, from the recursion xp n 1 (x) = α n 1 P n (x) + β n 1 P n 1 (x) + γ n 1 P n 2 (x), it follows that xp n, P n 1 = P n, xp n 1 = α n 1 P n, P n. So γ n P n 1, P n 1 = α n 1 P n, P n. So γ n 0, and {P n } are orthonormal iff γ n = α n 1. Finally, γ n = 0 P n, P n = 0. 7
) Suppose the polynomials satisfy a three-term recursion relation xp n (x) = γ n+1 P n+1 (x) + β n P n (x) + γ n P n 1 (x). On C[x], define a functional ϕ by and extend linearly: for ϕ [P n P k ] = δ nk A(x) = n a n P n (x), B(x) = k b k P k (x), ϕ [AB] = n k a n b k ϕ [P n P k ] = n a n b n. Why consistent? Suppose A 1 B 1 = A 2 B 2, to show ϕ [A 1 B 1 ] = ϕ [A 2 B 2 ]. 8
Factoring over C, enough to show ϕ [(A(x)(x a))b(x)] = ϕ [A(x)((x a)b(x))] for a C. By linearity, enough to show ϕ [(P k (x)x)p n (x)] = ϕ [P k (x)(xp n (x))]. But from the recursion relation, these are γ k+1 δ k+1,n + β k δ k,n + γ k δ k 1,n = γ n+1 δ k,n+1 + β n δ k,n + γ n δ k,n 1. 9
Remains to show ϕ = ϕ µ for some measure µ. Note: µ not unique. n n n a i P i (x), a i P i (x) = a i 2 > 0 i=0 i=0 ϕ i=0 so the induced inner product is positive definite (assume non-degenerate) Let H be the Hilbert space completion of C[x] with respect to the inner product, ϕ, and X be the operator with dense domain C[x] defined by XA(x) = xa(x). Then X is an unbounded symmetric operator. Has a selfadjoint extension? von Neumann: look at the deficiency subspaces K ± = ker(x i). A self-adjoint extension exists iff the deficiency indices d ± = dim(k ± ) are equal. 10
Let Since ( n ) C a j x j = j=0 n j=0 ā j x j. C(P ), C(P ) ϕ = P, P ϕ = P, P ϕ, C can extend to an operator on H. It commutes with X, so Indeed, But C(K ± ) = K. K + = {η H : X η = iη} = {η H : P R[x], η, XP = iη, P }. C(η), X(P ) = η, CX(P ) = η, XC(P ) We conclude that = iη, C(P ) = iη, P. d + = d. 11
Let X be a self-adjoint extension of X. It has a spectral measure: for any subset S R, have a projection E(S). Define µ(s) = 1, E(S)1. This is a probability measure. In fact, m n (µ) = R xn dµ(x) = 1, R xn de(x) 1 = 1, X n 1 = 1, x n = ϕ [x n ]. X not unique µ not unique. 12
Hermite: xp n = P n+1 + tnp n 1. Laguerre: xl n = L n+1 + (t + 2n)L n + n(t + n 1)L n 1. Jacobi: β 2 α 2 xp n = P n+1 + (2n + α + β)(2n + α + β + 2) P n 4n(n + α)(n + β)(n + α + β) + (2n + α + β 1)(2n + α + β) 2 (2n + α + β + 1) P n 1. Ultraspherical: xp n = P n+1 + Chebyshev: n(n + 2λ 1) 4(n + λ 1)(n + λ) P n 1. xp n = P n+1 + 1 4 P n 1. 13
Kravchuk: for a = (1 2p), b = 4pq, a 2 b = 1, xk n = K n+1 + a(2n N)K n + bn(n n + 1)K n 1. Charlier: xp n = P n+1 + (t + n)p n + tnp n 1. Meixner: for a = p 1 1 2, b = qp 2, a 2 b = 1 4 > 0, xm n = M n+1 + a(t + 2n)M n + bn(t + n 1)M n 1. Meixner-Pollaczek: same for a 2 b < 0. Note for Laguerre, a 2 b = 0. 14
In- Can study zeros, asymptotics, continued fractions. stead: Denote {e n } the basis of H coming from {P n } Xe n = γ n+1 e n+1 + β n e n + γ n e n 1. X has a tri-diagonal matrix β 0 γ 1 γ 1 β 1 γ 2 γ 2 β 2 γ 3 γ 3.......... Thus X is a sum of three operators: creation annihilation and preservation A e n = γ n+1 e n+1, Ae n = γ n e n 1, Ne n = β n e n. 15
Since β n is real, N is symmetric, and A e n, e k = γ n+1 en+1, e k = γn+1 δ n+1,k = γ k en, e k 1 = en, Ae n 1. A, A are (formal) adjoints of each other. For the Hermite polynomials, N = 0, A, A are the usual annihilation and creation operators. For the Charlier polynomials, A, A are as before, and N is (almost) the number operator. 16
Another way to look at this. Heisenberg Lie algebra H t has the basis {A, A, t} subject to the relation [A, A ] = t, t scalar. Each H t has a representation on some Hilbert space H with basis {e n } n=0 : A e n = tn e n 1, A e n = t(n + 1) e n+1. (AA A A)e n = [t(n + 1) tn]e n = te n. 17
If we want A + A = X, where XP (x) = xp (x), can take H = L 2 (R, µ t ) for µ t = Gaussian measure, and e n = P n (x, t) = Hermite polynomials. In this representation, A = t x, A = x t x. 18
Oscillator Lie algebra Os t has the basis {A, A, N, t} subject to the relations [A, A ] = t, [A, N] = A, [N, A ] = A, t scalar. Each Os t has a representation on some H: A e n = tn e n 1, A e n = t(n + 1) e n+1, N e n = (t + n)e n. Note e 0 is a cyclic vector. In fact, the representation is irreducible. If we want can take H = L 2 (R, µ t ) for and A + A + N = X, µ t = Poisson measure, e n = P n (x, t) = Charlier polynomials. A = t, (P ) = P (x) P (x 1), A (P ) = xp (x 1) tp, N(P ) = x P + tp (x 1). 19
The Lie algebra sl(2, R) of real traceless 2 2 matrices has the basis { J, J +, J 0 } J = ( ) 0 0, J 1 0 + = ( ) 0 1, J 0 0 0 = ( ) 1 0, 0 1 subject to the relations [J, J + ] = J 0, [J, J 0 ] = 2J, [J 0, J + ] = 2J +. Abstract representations, parameterized by t > 0: Want J e n = n(t + n 1) e n 1, J + e n = (n + 1)(t + n) e n+1, J 0 e n = (t + 2n)e n. J + J + + cj 0 = X. 20
For each t, c, sl(2, R) has a representation on L 2 (R, µ t,c ), for µ t,c = negative binomial (Meixner), c > 1, Gamma (Laguerre), c = 1,... (Meixner-Pollaczek), 0 < c < 1. Where are the finite-dimensional representations? Take t = N, N > 0 integer. J e n = n(n n + 1) e n 1, J + e n = (n + 1)(N n) e n+1, J 0 e n = (2n N)e n. Finite-dimensional since J + e N = 0. Can take µ N = binomial measure, e n = P n (t, N) = Kravchuk polynomials. 21
For example, for N = 1, dµ(x) = qδ 1 (x) + pδ 1 (x) is the Bernoulli distribution, K 0 = 1, K 1 = x + (1 2p). In the basis (K 1, K 0 ) for L 2 (R, µ) = C 2, J, J +, J 0 have matrices as above, and ( q p 1 X = J + 4pqJ + (q p)j 0 = 4pq p q ) sends and X : (K 0 = 1) ( K 1 (1 2p)K 0 = x ) X : K 1 (q p)k 1 + 4pqK 0 = (q p)x + (1 2p) 2 + 4p(1 p) = (q p)x + 1. 22
Note that for x = 1, xk 1 (x) = 2p = (q p)x + 1, and for x = 1, xk 1 (x) = 2q = (q p)x + 1, so xk 1 (x) = (q p)x + 1 and J + 4pqJ + (q p)j 0 = multiplication by x. 23
Additional material: 1) The Hilbert space H (the closure of C[x] with respect to, µ ) is not always equal to L 2 (R, µ). In fact, C[x] is dense in L 2 (R, µ) if and only if µ comes from a self-adjoint extension of a symmetric operator as in slide 12; there may be measures that do not. 2) For the uses of representation theory to obtain results about orthogonal polynomials, see reference 4. 3) Different representations of the Lie algebra sl(2, R) on slide 21, distinguished by the parameter c, correspond to representations of different Lie groups all of which have the same Lie algebra. See reference 3. REFERENCES 1. N. I. Akhiezer, The classical moment problem and some related questions in analysis, Hafner Publishing Co., New York, 1965. 2. Barry Simon, The classical moment problem as a self-adjoint finite difference operator, Adv. Math. 137 (1998), no. 1, 82 203. 3. Philip Feinsilver and René Schott, Algebraic structures and operator calculus. Vol. I, Kluwer Academic Publishers Group, Dordrecht, 1993, Representations and probability theory. 4. H. T. Koelink and J. Van Der Jeugt, Convolutions for orthogonal polynomials from Lie and quantum algebra representations, SIAM J. Math. Anal. 29 (1998), no. 3, 794 822 (electronic). 5. Erik Koelink, Spectral theory and special functions, Lecture notes for the SIAM Activity Group OP-SF summer school 2000, Laredo, Spain. arxiv:math.ca/0107036. 24