Pollaczek polynomials and hypergeometric representation
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Pollaczek polynomials and hypergeometric representation Jamel Benameur · Mongi Blel
Received: 10 October 2011 / Accepted: 13 May 2012 / Published online: 19 September 2012 © Springer Science+Business Media, LLC 2012
Abstract This paper gives a solution, without the use of the three-term recurrence relation, of the problem posed in Ismail (Classical and Quantum Orthogonal Polynomials in One Variable, Cambridge University Press, Cambridge, 2005) (Problem 24.8.2, p. 658): that the hypergeometric representation of the general Pollaczek polynomials is a polynomial in cos(θ ) of degree n. Chu solved in (Ramanujan J. 13(1–3): 221–225, 2007) the problem in a particular case. We use elementary properties of functions of complex variables and Pfaff’s transformation on hypergeometric 2 F1 -series. Keywords Orthogonal polynomials · Pollaczek polynomials · Hypergeometric functions · Pfaff–Euler transformation Mathematics Subject Classification 33C45 · 33C47 · 33C50 · 42C05 1 Introduction The general Pollaczek polynomials Pnλ (x, a, b), defined for λ + a not a non-negative integer, satisfy the three-term recurrence relation (Szegö [6], and Chihara [2]), λ (n + 1)Pn+1 (x, a, b) = 2 (n + λ + a)x + b Pnλ (x, a, b) λ − (n + 2λ − 1)Pn−1 (x, a, b),
n ≥ 1,
The research is supported by NPST Program of King Saud University; project number 10-MAT1293-02. J. Benameur () · M. Blel Department of Mathematics, College of Science, King Saud University, Riyadh 11451, Kingdom of Saudi Arabia e-mail: [email protected] M. Blel e-mail: [email protected]
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J. Benameur, M. Blel
with the initial conditions P0λ (x, a, b) = 1,
P1λ (x, a, b) = 2(λ + a)x + 2b.
Pollaczek [5] introduced these polynomials when λ = 12 and Szegö [6] generalized them by introducing the parameter λ. The monic polynomials associated with the sequence (Pnλ (x, a, b))n is the sequence (Qλn (x, a, b))n defined by λ Qn (x, a, b) n≥0 =
n! 2n (λ + a)n
λ Pn (x, a, b) n≥0 .
The generating function associated with the sequence (Pnλ (x, a, b))n (see [4]) is given by ∞
−λ+iΦ(θ) −λ−iΦ(θ) 1 − te−iθ Pnλ (x, a, b)t n = 1 − teiθ ,
n=0 θ+b where Φ(θ ) = a cos sin θ . A classical computation leads to ∞
−λ+iΦ(θ) −λ−iΦ(θ) 1 − te−iθ Pnλ (x, a, b)t n = 1 − teiθ
n=0
=
∞ (λ − iΦ(θ ))n n=0
=
n!
n n
t z
∞ (λ + iΦ(θ ))n n=0
n!
n ∞ (λ + iΦ(θ ))k (λ − iΦ(θ ))n−k n=0 k=0
k!(n − k)!
n −inθ
t e
e−2ikθ einθ t n .
Since (λ + x)n = (λ + x)n−k (λ + x + n − k)k and (λ + x + n − k)k = (−1)k (−λ − x − n + 1)k , we have n (λ + iΦ(θ ))k (λ − iΦ(θ ))n−k
e−i(n−2k)θ k!(n − k)! k=0 ⎛ ⎞ −n, λ + iΦ(θ ) (λ − iΦ(θ )) n = einθ ; e−2iθ ⎠ . 2 F1 ⎝ n! −λ + iΦ(θ ) − n + 1
Pnλ (x, a, b) =
(1)
In [4], Problem 24.8.2, Mourad Ismail posed the problem to prove that the righthand side of (1) is a polynomial in cos(θ ) of degree n without the use of the threeterm recurrence relation of the Pollaczek polynomials. In 2007, Chu [3] solved the problem when b = 0. In this paper, we solve the problem in the general case and we prove the following result.
Polla
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