Mathematics Subject Classification: 33C45.A more general form for a classical generating function for the Jacobi polynomials is given.* Partially supported by Project MM 1305 - National Science Fund, Bulgarian Ministry of Educ. Sci

Rusev, Peter

Fractional Calculus and Applied Analysis

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A NOTE ON A CLASSICAL GENERATING FUNCTIONFOR THE JACOBI POLYNOMIALSPeter Rusev 1Dedicated to Acad. Bogoljub Stankovic´on the occasion of his 80th birthdayAbstractA more general form for a classical generating function for the Jacobipolynomials is given.Mathematics Subject Classification: 33C45Key Words and Phrases: Jacobi polynomials, generating function*1. If z 6= ±1, then we define l(1; z) : ζ = 1 + t(1− z) and l(−1; z) : ζ =−1− t(1 + z) for 0 ≤ t <∞ as well as(1− ζ1− z)α:= exp{α log1− ζ1− z}for ζ ∈ S(1; z) := C \ l(1; z), and(1 + ζ1 + z)β:= exp{β log1 + ζ1 + z}for ζ ∈ S(−1; z) := C \ l(−1; z), provided α and β are arbitrary complexnumbers.It is clear that S(1;x) = C \ [1,∞), S(−1;x) = C \ (−∞,−1] for x ∈(−1, 1) and, moreover, that1 Partially supported by Project MM 1305 - National Science Fund, Bulgarian Ministryof Educ. Sci.64 P. Rusev(1− ζ1− x)α=(1− ζ)α(1− x)α ,(1 + ζ1 + x)β=(1 + ζ)α(1 + x)βfor ζ ∈ S(1;x) ∩ S(−1;x) = C \ {(−∞,−1)] ∪ [1,∞)} and x ∈ (−1, 1).Proposition 1. Let γ be a rectifiable Jordan curve such that γ∪intγ ⊂C \ l(1; z) ∪ l(−1; z), where z 6= −1, 1, and ind(γ; z) = 1. ThenP (α,β)n (z) =12pii∫γ{ζ2 − 12(ζ − z)}n(1− ζ1− z)α(1 + ζ1 + z)β dζζ − z . (1.1)The above integral representation is a corollary of the Rodrigues formulafor the Jacobi polynomials as well as of the Cauchy integral formulas forthe derivatives of a holomorphic function.2. A well-known fact is that there exists unique complex-valued functionh holomorphic in the region H = C \ [−1, 1], and such that h2(z) = z2 − 1for z ∈ H and h(x) > 0 when x > 1. Usually, the value of this functionat any point z ∈ H is denoted by √z2 − 1. The function ω, defined inH as ω(z) = z + h(z), is also holomorphic in H. Moreover, ω(z) 6= 0 and(ω(z)+(ω(z))−1)/2 = z when z ∈ H, i.e. ω(z) is an inverse of the Zhukovskifunction z = (ω + ω−1)/2. As it is well-known, the last one is univalent inthe domain D = {ω : |ω| > 1} and maps it onto H. Hence, the function ωmaps H onto D. Since limz→∞ ω(z) = ∞, ω is a meromorphic function inthe region C \ [−1, 1] with a (simple) pole at the point of infinity.If z ∈ H, then we define the function p(z.w) in the disk, U(0; |ω(z)|−1)by the requirements p2(z, w) = 1 − 2zw + w2 and p(z, 0) = 1. We denotethis function by√1− 2zw + w2. Its existence follows from the fact thatthe disk U(0; |ω(z)|−1) is a simply connected region and 1− 2zw + w2 6= 0whenever w is in this disk, and z ∈ H. Indeed, the equality 1−2zw+w2 = 0implies w = ω(z) or w = (ω(z))−1 which is impossible.Let us note that the function 1 + p(z, w) does not vanish in the diskU(0; |ω(z)|−1) when z ∈ H. Indeed, the equality p(z0, w0) = 0 yields thatw0(w0 − 2z0) = 0 which contradicts to p(z0, 0) = p(z0, 2z0) = 1. Hence,ζ(w) =2z − w1 + p(z, w)(2.1)is a holomorphic function in the disk U(0; |ω(z)|−1) for z ∈ H.If w 6= 0, then from (2.1) it follows that ζ(w) = (1 − p(z, w))w−1 and,hence, the equalities hold:(1−w+p(z, w))(1−ζ(w)) = 2(1−z), (1+w+p(z, w))(1+ζ(w)) = 2(1+z).A NOTE ON A CLASSICAL GENERATING FUNCTION . . . 65A direct verification shows that these equalities are still valid for w = 0.Moreover, as their implication we obtain that 1 − w + p(z, w) 6= 0 and1 + w + p(z, w) 6= 0 for z ∈ H and w ∈ U(0; |ω(z)|−1).We define the function P (α,β)(z, w) for z ∈ H and w ∈ U(0; |ω(z)|−1) \{0} byP (α,β)(z, w) =2α+βp(z, w)(1− w + p(z, w))α(1 + w + p(z, w))β=2α+β√1− 2zw + w2(1− w + p(z, w)(1 + w + p(z, w))β ,and assume that P (α,β)(z, 0) ≡ 1.Proposition 2. For z ∈ H and w ∈ U(0; |ω(z)|−1) it holds∞∑n=0P (α,β)n (z)wn = P (α)(z, w). (2.2)P r o o f. We note that P (α,β)(z, w), as a function of w, is holomorphicin the disk U(0; |ω(z)|−1) and, hence, by Taylor’s theorem it has a powerseries representation centered at the origin, i.e.P (α,β)(z, w) =∞∑n=0a(α,β)n (z)wn. (2.3)If 0 < r < |ω(z)|−1, then for the coefficient in the right-hand side of(2.3) we havea(α,β)n (z) =12pii∫C(0;r)P (α,β)(z, w)wn+1dw, n = 0, 1, 2, . . . ,where C(0; r) is the positively oriented circle centered at the origin andhaving radius r.From p2(z, w) = 1− 2zw + w2 it follows that p′w(z, 0) = −z, and using(2.1) we find that ζ ′(0) = −1 + z2 6= 0 for z ∈ H. Hence, there exists aneighbourhood U(0; δ) with 0 < δ < |ω(z)|−1, where the function ζ(w) isunivalent. Since ζ(0) = 0, it is clear that for arbitrary r ∈ (0, δ) the image ofthe circle C(0; r) by the map ζ(w) is a positively oriented rectifiable Jordancurve γ(z; r) such that ind(γ(z; r); z) = 1. Moreover, r can be chosen suchthat γ(z; r)∪intγ(z; r) ⊂ H ∩ S(1; z) ∩ S(−1; z).Using the representation (1.1) with γ = γ(z; r) and the equalitiesζ2(w)− 12(ζ(w)− z) =1w,ζ ′(w)ζ(w)− z =1wp(z, w), w ∈ U(0; |ω(z)|−1) \ {0},66 P. Rusevand denoting the variable ζ by w, we obtainP (α,β)n (z) =12pii∫C(0;r)P (α,β)(z, w)wn+1dw, n = 0, 1, 2, . . . .Hence, a(α,β)n (z) = P(α,β)n (z) for n = 0, 1, 2, . . .. Then from (2.3) it followsthat (2.2) holds in the disk U(0; |ω(z)|−1).3. Let g(z) be the unique complex-valued function which is holomorphicin the region G = C\{(−∞,−1]∪ [1,∞)}, and such that g2(z) = z2−1 andg(0) = 1. The function τ(z), defined as τ(z) = z+ ig(z), is holomorphic andnowhere vanishing in G and, moreover, (τ(z) + (τ(z))−1)/2 = z for z ∈ G.Hence, τ(z) is an inverse of the Zhukovski function z = (τ + τ−1)/2 and assuch it is univalent in the half-plane =τ > 0 and maps it onto the regionG. In particular, the image of the point i is the origin. More precisely, theimage of the half-plane =z > 0 is the region determined by the inequalities|τ | < 1 and =τ > 0, while the image of the interval (−1, 1) is the arc of theunit circle located in the half-plane =τ > 0. The image of the half-plane=z < 0 is the region determined by |τ | > 1 and =τ > 0.The proof of Proposition 2 leads to the following assertion:Proposition 3. The equality (2.2) holds for arbitrary z ∈ G and w ∈U(0; ρ(z)),where ρ(z) = min(|τ(z)|, |τ(z)|−1).A particular case of (2.2) is the representation∞∑n=0P (α,β)n (x)wn = P (α,β)(x,w),which holds for −1 < x < 1 and |w| < 1 [1, 10.8, (29)]. Indeed, in this casewe have that |τ(x)| = |x+ i√1− x2| = 1.References[1] H. Bateman, A. Erde´lyi, Higher Transcendental Functions, I,II,III. McGraw-Hill, New York (1953).Institute of Mathematics and Informatics Received: December 15, 2004Bulgarian Academy of Sciences, Block 8Sofia 1113, BULGARIAe-mail: pkrusev@math.bas.bg

A Note on a Classical Generating Function for the Jacobi Polynomials

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A Note on a Classical Generating Function for the Jacobi Polynomials

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