The Generating Functions for Pochhammer Symbol { }, n N Aleksandar Petoević University of Novi Sad Teacher Training Faculty, Department of Mathematics Podgorička 4, 25000 Sombor SERBIA and MONTENEGRO Email address: apetoe@ptt.yu Abstract In this paper we give elementary proofs of the generating functions for the Pochhammer symbol { }, n N. Also, we define polynomial P k n (x) and we give connection between this polynomial and function e x x n. 1 Introduction For sequence {c n } the generating function, exponential generating function and the Direchlet series generating function, denoted respectively by g(x), G(x) and D(x), are defined as [8, p. 3, p. 21, p. 56] g(x) = c n x n, G(x) = x n c n, D(x) = c n n. (1) x Apart [8], the relevant theory on generating functions can be found in [2] and Chapter VII in [4]. The Pochhammer symbol (z) n is defined by n=1 (z) 0 = 1, (z) n = z(z + 1)...(z + n 1) = Γ(z + n) Γ(z), (2) where Γ(z) is the gamma function Γ(z) = + 0 t z 1 e t dt (Re (z) > 0). This work was supported in part by the Serbian Ministry of Science, Technology and Development under Grant # 2002: Applied Orthogonal Systems, Constructive Approximation and Numerical Methods. 1
For a fixed number b and sequence {a n }, the Pochhammer symbol (b) n obeys Euler s transform (b) n ( ) n a n z n = (1 z) b (b) n z n a 0, (3) 1 z where is the forward difference defined via a n = a n+1 a n. Higher order differences are obtained by repeated operations of the forward difference operator k a n = k 1 a n+1 k 1 a n, so that in general k a n = k ( ) k ( 1) m a n+k m. (4) m Applying relations (3) and (4) for a n = 1 to obtain the exponential generating function for the Pochhammer symbol (b) n as follows The exponential integral E n (x) is defined by and has the asymptotic series [3, p. 1] so that (b) n z n = (1 z) b. (5) E n (x) = 1 e xt t n (n 1)! E n (x) = ( x) n 1 E 1 (x) + e x 2(n k 2)! ( x) k, E n (x) = 1 xe x k=0 k=0 dt. ( 1) k (n) k x k. Hence, generating function is given as follows (b) n x n = E b( 1/x). (6) xe 1/x The generating functions are of particular importance in theory of integer sequences. In Table 1 you can see examples of integer sequences generated according Pochhammer symbol {(b) n }. The second possibility of generation of integer sequences by Pochhammer symbol is that for fixed n N, terms of the sequence are generated by index i = 0, 1, 2, 3, 4,..., i.e., { }. In this way, here we give for a fixed n N the generating functions for the Pochhammer symbol { }, denoted by g n (x) = x i x i, G n (x) = i!, D n(x) = i. ( (0) x n = 0 ) (7) 2
Table 1: Integer sequences (b) n for b = 1, 2, 3, 4, 5 and n N 0 n (b) n sequences in [7] name 1 1, 1, 2, 6, 24, 120,... A000142 2 1, 2, 6, 24, 120, 720,... (n + 1)! 3 1, 3, 12, 60, 360, 2520,... A001710 4 1, 4, 20, 120, 840, 6720,... A001715 5 1, 5, 30, 210, 1680, 15120,... A001720 24 2 6 2 Statement of results In what follows ζ(z), s(n, m) and Pk n (x) are respectively the Riemann zeta function, Stirling number of the first kind and the polynomials defined by 1 ζ(z) =, (Re (z) > 1), (8) nz n=1 x(x 1) (x n + 1) = k 1 Pk n (x) = k 1 s(n, m)x k, (9) ( ) k (n m) x, (n, k N). (10) For Re (z) 1, z 1, the function ζ(z) is defined as the analytic continuations of the foregoing series. Both are analytic over the whole complex plane, except at z = 1, where they have a simple pole. My main results are as follows. Theorem 1 Let be the Pochhammer symbol. Then for a fixed number n N we have x i x =. (11) (1 x) n+1 Theorem 2 Let Pk n(x) be the polinomials defined in (10) and let be the Pochhammer symbol. Then for a fixed number n N we have x i i! = [ xex x n 1 + Pn 1(x) ] n. (12) Theorem 3 Let ζ(z) be the Riemann zeta function, let s(n, m) be the Stirling number of the first kind and let be the Pochhammer symbol. Then for a fixed number n N we have = ( 1) +n s(n, )ζ(x ). (13) i x =1 3
Note 1. For 1 k 4 the polynomials Pk n (x) are listed below. P n 1 (x) = n P n 2 (x) = 2nx + n(n 1) P n 3 (x) = 3nx2 + 3n(n 1)x + n(n 1)(n 2) P n 4 (x) = 4nx3 + 6n(n 1)x 2 + 4n(n 1)(n 2)x + +n(n 1)(n 2)(n 3) Several well-known special cases of the polynomials Pk n (x) are presented in Table 2. Let be (x) (m) the falling factorial defined by (x) (m) = x(x 1) (x m + 1). Then: k 1 Pk n (x) = ( ) k (n) (k ) x. Note 2. Since (i + 1) n lim i = lim i (i + n)! (i 1)! (i + n 1)! i! the expansion (11) converges for x < 1 and (12) for each x R. Let [f(x)] (k) be the k th derivative of a function f(x). It is clear that the formula (12) could be rewritten in the representation of the e x x n function, since there exists the following relationship [e x x n ] (n 1) = xe x [ x n 1 + P n n 1(x) ] between e x x n and the polynomials P n k (x). Table 2: The special cases P n k (x) = 1 Pk n (x) sequences in [7] Pk 1 (2) 0, 1, 4, 12, 32, 80,... A001787 Pk 1 (3) 0, 1, 6, 27, 108, 405,... A027471 Pk 1 (4) 0, 1, 8, 48, 256, 1280,... A002697 Pk 2 (1) 0, 2, 6, 12, 20, 30,... A002378 Pk 3 (1) 0, 3, 12, 33, 72, 135,... A054602 P2 n (2) 0, 4, 10, 18, 28, 40,... A028552 P2 n (3) 0, 6, 14, 24, 36, 50,... A028557 3 Proof of results Proof of Theorem 1. Let x < 1 and g n be defined by (7). Then +1 x i = n x i + i x i. 4
Integrating this equation, we obtain +1 x i+1 = n i i x i+1 + x i+1 (i + 1) n x i+1 = n (i + 1) n 1 x i+1 + x i+1 x i = n 1 x i + x x i i.e., (1 x) x i = n 1 x i. g n (x) = n (1 x) g n 1(x) = n(n 1) (1 x) 2 g n 2(x) Now use g 1 (x) = x/(1 x) 2, to obtain = which completes the proof. n(n 1)(n 2) (1 x) 3 g n 3 (x) = = g n (x) = x (1 x) n+1, (1 x) n 1 g 1(x) Proof of Theorem 2. Let G n (x) = xe [ x x n 1 + Pn 1 n (x) ]. Since n 2 ( n 2 n 1 [G n (x)] (1) = x n e x + nx n 1 e x + e )x x +1 (n m) + induction on i N we have n 2 ( n 1 + e x ( + 1) [G n (x)] (i) = e x x n + e x i =1 n 2 ( n 1 + e x i 1 + e x s=0 ( ) i n 2 s + 1 =s )x n 2 (n m) ( 1 i )x n i (n m) + i )x +1 n 2 ( + 1)! ( s)! 5 (n m) + ( n 1 )x s n 2 (n m).
Hence [G n (0)] (i) = i 1 ( ) ( i n 1 (s + 1)! s + 1 s s=0 ) n 2 s (n m) i 1 1 = i! (n 1)! (i s 1)! (s + 1)! (n s 1)! s! s=0 = i! (n 1)! (n + i 1)! i! (i 1)! (n 1)! = (n + i 1)! (i 1)! =. Applying the standard formula for the Taylor series expansion about the point x = 0 we arrive at the formula in (12), which completes the proof. Proof of Theorem 3. Using +1 i x = n i x + i x 1 we have D n+1 (x) = nd n (x) + D n (x 1). (14) The recurrence relation for Stirling numbers of the first kind s(n + 1, ) = s(n, 1) ns(n, ) produces n+1 ( 1) +n+1 s(n + 1, )ζ(x ) = n ( 1) +n s(n, )ζ(x ) + (15) =1 =1 + ( 1) +n s(n, )ζ(x 1 ). =1 Induction on n and by combining (14) and (15) we obtain the result of the theorem. References [1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, National Bureau of Standards, Washington, 1970. [2] L. Comtet, Advanced Combinatorics; The art of finite and infinite expansions, D. Reidel Publ. Co., Boston, 1974. [3] A. Erdélyi, Asymptotic expansions, Dover, New York, 1987. 6
[4] R. Graham, D. Knuth and I. O. Patashnik, Concrete mathematics, Reading, MA: Addison Wesley, 1989. [5] N.E. Nørlund, Hypergeometric function, Acta Math. 94 (1955), 289 349. [6] A. Petoević, The function v M m (s; a, z) and some well-known sequences, Journal of Integer Sequences, Vol.5, (2002), Article 02.1.7 [7] N.J.A. Sloane, The On-Linea Encyclopedia of Integer Sequence, published elec. at http://www.research.att.com/~nas/sequences/ [8] H. S. Wilf, Generatingfunctionology, Academic Press, New York, 1990. 2000 Mathematics Subect Classification: Primary 11M41, 41A58; Secondary 11B83, 11B73. Keywords: Pochhammer symbol, generating function, Pk n (x) polynomial, Direchlet series, Riemann zeta function, Stirling number, falling factorial 7