Rev. Real Academia de Ciencias. Zaragoza. 59: 71–77, (2004). Unification of a class of trilateral generating functions Manik Chandra Mukherjee Netaji Nagar Vidyamandir Calcutta 700 092. India Abstract In this paper we have proved a general theorem in connection with the unification of a class of trilateral generating functions of certain special functions. Some particular cases of interest are also pointed out. Keywords: Unification, Generating functions, Group-theoretic method. MSC: 33A65 1 Introduction Theories in connection with the unification of generating functions are of greater im- portance in the study of special functions. In this direction attempts have been made by some researchers [1–8]. The aim at presenting this article is to state and prove a theorem on the unification of a class of trilateral generating functions of certain special functions. In fact we have proved the following theorem. Theorem 1 For a set of functions {Sn (x)|n ≥ 0} generated by ∞ An (m)Sn+m (x)tn = n=0 f (x, t) Sm (h(x, t)) g(x, t)m (1) where m is a non-negative integer, An (m) are arbitrary constants and f, g, h are arbitrarily chosen functions of x and t, let F (x, y, t) = ∞ an (m)Sn+m (x)gn (t)tn , n=0 then the following trilateral generating relation holds: ∞ n=0 Sn+m (x)σn (y, z)tn = f (x, t) F (h(x, t), y, zt/g(x, t)) g(x, t)m 71 where σn (y, z) = n ak An−k (m + k)gk (y)z k . k=0 Proof. ∞ = = = n=0 ∞ Sn+m (x)σn (y, z)tn Sn+m (x) n=0 ∞ ∞ n ak An−k (m + k)gk (y)z k tn k=0 Sn+m+k (x)σk (x)An (m + k)gk (y)z k tn+k n=0 k=0 ∞ ak gk (y)(zt)k k=0 f (x, t) Sm+k (h(x, t)) g(x, t)m ∞ zt f (x, t) a S (h(x, t))g (y) = k m+k k g(x, t)m k=0 g(x, t) f (x, t) zt = F h(x, t), y, . m g(x, t) g(x, t) k The result given in [9] can be obtained directly from the theorem. Indeed, by putting m = 0 in the above theorem, we get F (x, y, t) = ∞ an Sn (x)gn (y)tn , n=0 then, ∞ zt Sn (x)σn (y, z)t = f (x, t)F h(x, t), y, , g(x, t) n=0 n where σn (y, z) = n ak An−k (k)gk (y)z k k=0 which is found derived in [9]. 2 Applications We now apply our theorem in the derivation of generating functions of various special functions 2.1 On Hermite polynomials We first consider the following generating relation involving Hermite polynomials [9] ∞ n=0 Hn+m tn = exp(2xt − t2 )Hm (x − t). n! (2) Now it is evident that the relation (2) being of the form (1), one can easily obtain the following result with the help of theorem 1. 72 Proposition 1 If there exists a generating relation of the form ∞ F (x, y, t) = an Hn+m (x)gn (y) n=0 then ∞ Hn+m (x)σn (y, z) n=0 tn = exp(2xt − t2 )F ((x − t), y, zt), n! where σn (y, z) = n k=0 2.2 tn , n! n ak gk (y)z k . k On Laguerre polynomials Next we consider the following generating relation involving Laguerre polynomials [10,11] ∞ n=0 m+n α −xt x Ln+m (x)tn = (1 − t)−1−α−m exp Lαn . n 1−t 1−t (3) The relation (3) being of the form (1), one can state the following result with the help of theorem 1. Proposition 2 If ∞ F (x, y, t) = an Lαn+m (x)gn (y)tn , n=0 then ∞ Lαn+m (x)σn (y, z)tn −1−α−m = (1 − t) n=0 where σn (y, z) = n −xt exp 1−t F zt x , , y, 1−t 1−t m+n ak gk (y)z k . m+k k=0 2.3 On modified Laguerre polynomials Next we consider the following generating relation involving modified Laguerre poly- nomials [10] exp(xt)(1 − β [x(1 t)−β−m fm − t)] = ∞ (m + 1)n β fm+n (x)tn . n! n=0 (4) The relation (4) being of the form (1), one can obtain the following result with the help of theorem 1. Proposition 3 If there exist a generating relation of the form F (x, y, t) = ∞ β an fm+n (x)gn (y)tn , n=0 73 then ∞ β −β−m fn+m (x)σn (y, z)t = exp(xt)(1 − t) n n=0 where σn (y, z) = 2.4 zt F x(1 − t), y, , 1−t n (m + k + 1)n−k ak gk (y)z k . (n − k)! k=0 On ultra spherical polynomials Next we consider the following generating relation involving ultra spherical polynomi- als [11] ∞ m+n λ x−t Pn+m (x)tn = ρ−m−2λ Pmλ , ρ n n=0 (5) where ρ = (1 − 2xt + t2 )1/2 . The relation (5) being of the form (1), one can state the following result with the help of theorem 1. Proposition 4 If F (x, y, t) = ∞ λ an Pn+m (x)gn (y)tn , n=0 then ∞ λ Pn+m (x)σn (y, z)tn =ρ −2λ−m n=0 where σn (y, z) = n k=0 2.5 F x−t zt , , y, ρ ρ m+n ak gk (y) z k . m+k On modified Jacobi polynomials Next we consider the following generating function involving modified Jacobi polyno- mials [12]. ∞ n=0 m + n (α−m−n,β−m−n) Pn+m (x)tn n α−m = 1 + 12 (x + 1) α−n 1 + 12 (x − 1) (6) Pm(α−m,β−n) x + 12 (x2 − 1)t . The relation (6) being of the form (1), one can easily state the following result with the help of theorem 1. Proposition 5 If F (x, y, t) = ∞ (α−m−n,β−m−n) an Pn+m n=0 74 gn (y) tn , then ∞ (α−m−n,β−m−n) Pn+m (x) σn (y, z) tn n=0 α−m = 1 + 12 (x + 1) α−n 1 + 12 (x − 1) ×F 1 + 12 (x2 − 1)t, y, where zt 1 + 12 (x + 1)t n σn (y, z) = , 1 + 12 (x − 1)t m+n ak gk (y) z k . m+k k=0 If instead of (6) we consider the following generating realtion [13] ∞ (r + 1)n (k−r−n,β) (x) tn Pn+r n! n=0 −1−β−k = (1 + t)k−r 1 + 12 (1 − x)t Pr(k−r,β) (7) x − 12 (1 − x)t , 1 + 12 (1 − x)t then we get the following result Proposition 6 If ∞ F (x, y, t) = (k−r−n,β) an Pn+r gn (y) tn , n=0 then ∞ (k−r−n,β) Pn+r (x) σn (y, z) tn n=0 k−r = (1 + t) 1+ 1 (1 2 − x)t where σn (y, z) = 2.6 −1−β−k ×F x − 12 (1 − x)t zt , y, , 1 1+t 1 + 2 (1 − x)t n (r + k + 1)n−k ak gk (y) z k . (n − k)! k=0 On Bessel functions Next we consider the following generating relation involving Bessel functions [14]. ∞ tn 2t Jn+m (x) = 1 − n! x n=0 −m/2 √ Jm ( x2 − 2xt). (8) The relation (8) being of the form (1), one can easily state the following result with the help of theorem 1. Proposition 7 If F (x, y, t) = ∞ an Jn+m (x) gn (y) n=0 75 tn , n! then ∞ tn 2t Jn+m (x) σn (y, z) = 1 − n! x n=0 where σn (y, z) = F n k=0 2.7 −m/2 √ xtz x2 − 2xt, y, √ 2 , x − 2xt n ak gk (y) z k . k On Bessel polynomials Next we consider the following generating relation involving Bessel polynomials [10]. √ ∞ 1 − 1 − 2xt x tn −(m+1)/2 Yn+m (x) = (1 − 2xt) exp Ym √ . (9) n! x 1 − 2xt n=0 The relation (9) being of the form (1), one can easily state the following result with the help of theorem 1. Proposition 8 If ∞ tn an Yn+m (x) gn (y) , F (x, y, t) = n! n=0 then ∞ Jn+m (x) σn (y, z) n=0 tn n! −(m+1)/2 = (1 − 2xt) exp 1− √ 1 − 2xt F x where σn (y, z) = n k=0 2.8 x zt √ , y, √ 1 − 2xt 1 − 2xt , n ak gk (y) z k . k On modified generalised Bessel polynomiasl Next we consider the following generating relation involving modified generalised Bessel polynomiasl [15,16]. ∞ β n α−m−n x . Yn+m (x)tn = (1 − xt)1−α exp(βt)Ymα−m 1 − xt n=0 n! (10) The relation (10) being of the form (1), one can easily state the following result with the help of theorem 1. Proposition 9 If ∞ F (x, y, t) = α−m−n an Yn+m (x) gn (y) tn , n=0 then ∞ α−m−n Yn+m (x) σn (y, z) tn = (1 − xt) 1−α exp(βt)F n=0 where σn (y, z) = n β n−k ak gk (y) z k . (n − k)! k=0 76 x , y, zt , 1 − xt References [1] Singhal, J. P. and Srivastava, H. M.: A class of bilateral generating functions for certain classical polynomials; Pacific J. Math. 42 (1972), 355-362. [2] Chatterjea, S. 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