By Rusev P.

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5) (α,β ) Pn 1 ζ 2 − 1 n 1 − ζ α 1 + ζ β dζ . (z) = 2πi γ 2(ζ − z) 1−z 1+z ζ −z Proof. Let a = ±1, r(a) := min{2−1 |1 − a|, 2−1 |1 + a|} and B(a) = C \ {l(1 : z) ∪ l(−1; z)}. |z−a|≤r(a) ˜n(α,β ) (z). 5) by P 43 Integral representations and generating functions := {ζ : |ζ − a| < r(a)} and n = 0, 1, 2, . . , 1 ζ 2 − 1 n 1 − ζ α 1 + ζ β dζ (α,β ) ˜ , Pn (z) = 2πi C (a;r(a)) 2(ζ − z) 1−z 1+z ζ −z where C(a; r(a)) := {ζ : |ζ − a| = r(a)}. 2) are holomorphic functions ˜n(α,β ) (z) is a holomorphic function of z in of z in the disk U (a; r(a)).

23), is a solution of the recurrence equation an yn+1 + (ζ − λn )yn + cn yn−1 = 0. 21), we find Ω(z, ζ) = − X0 (z) (ζ − z)B(X0 (w), (w − ζ)−1 ) A0 −k0 {X0 (z)B(P1 (w), (w − ζ −1 ) − X1 (z)B(X0 (w), (w − ζ)−1 )} =− X0 (z) (ζ − z)B(X0 (w), (w − ζ)−1 ) − k0 B(X0 (z), X1 (w) A0 (X0 (w))2 − 1 B(ζ − z, (w − ζ), (w − ζ)−1 ) −X1 (z)X0 (w), (w − ζ) = A0 −k0 B(p0,0 (p1,0 w + p1,1 ) − (p1,0 z + p1,1 )p0,0 , (w − ζ)−1 ) 29 Jacobi, Laguerre and Hermite polynomials and. . =− p0,0 (X0 (w))2 B(ζ − z, (w − ζ)−1 ) + B(p0,0 p1,0 (w − z), (w − ζ)−1 ) A0 p1,0 A0 (X0 (w))2 (X0 (w))2 1 B(w − ζ, (w − ζ)−1 ) = B(1, 1) = B(X0 (w), X0 (w)) = 1.

M = 0, 1, 2, . . m! 10)] Γ(1/2 − m)Γ(1/2 + m) = π = (−1)m π, m = 0, 1, 2, . . 22m , m = 0, 1, 2, . . 15) means that M2m = 2m , m = 0, 1, 2, . . 14) followed by setting ζ = 0 to find that M2m+1 = 2m+1/2 , m = 0, 1, 2, . . Hence, Mn = 2n/2 , n = 0, 1, 2, . . 16) Hn (z) = 2n/2 exp(z 2 /2)Dn (z 2), z ∈ C; n = 0, 1, 2, . . 5 Now we are going to show that the Jacobi, Laguerre and Hermite associated functions also can be expressed in terms of hypergeometric and Weber-Hermite functions. 17) Γ(2n + α + β + 2)(z − 1)n+1 Qn (z) 2(n + α + β + 1)Γ(n + α + 1)Γ(n + β + 1) = F (n + α + 1, n + 1, 2n + α + β + 1; 2/(1 − z)), n = 0, 1, 2, .

### Classical orthogonal polynomials and their associated functions in complex domain by Rusev P.

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