On a structure formula for classical q-orthogonal polynomials

Abstract

The classical orthogonal polynomials are given as the polynomial solutions P n ( x) of the differential equation σ(x)y″(x)+τ(x)y′(x)+λ ny(x)=0, where σ( x) turns out to be a polynomial of at most second degree and τ( x) is a polynomial of first degree. In a similar way, the classical discrete orthogonal polynomials are the polynomial solutions of the difference equation σ(x) Δ∇y(x)+τ(x) Δy(x)+λ ny(x)=0, where Δ y( x)= y( x+1)− y( x) and ∇ y( x)= y( x)− y( x−1) denote the forward and backward difference operators, respectively. Finally, the classical q-orthogonal polynomials of the Hahn tableau are the polynomial solutions of the q-difference equation σ(x)D qD 1/qy(x)+τ(x)D qy(x)+λ q,ny(x)=0, where D qf(x)= f(qx)−f(x) (q−1)x , q≠1, denotes the q-difference operator. We show by a purely algebraic deduction — without using the orthogonality of the families considered — that a structure formula of the type σ(x)D 1/qP n(x)=α nP n+1(x)+β nP n(x)+γ nP n−1(x) (n∈ N≔{1,2,3,…}) is valid. Moreover, our approach does not only prove this assertion, but generates the form of this structure formula. A similar argument applies to the discrete and continuous cases and yields σ(x)∇P n(x)=α nP n+1(x)+β nP n(x)+γ nP n−1(x) (n∈ N) and σ(x)P n′(x)=α nP n+1(x)+β nP n(x)+γ nP n−1(x) (n∈ N). Whereas the latter formulas are well-known, their previous deduction used the orthogonality property. Hence our approach is also of interest in these cases.