On the Zeros of Some Generalized Hypergeometric Functions

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252 KI AND KIM where , c ,...,c are constants with , c 0. It is well known 1, p. 60 0 N N and easy to see that F Ž z. satisfies the differential equation p p d Ž b1 1. Ž bp 1. Ž a1. Ž ap. pFpŽ z. 0. dz Ž 3. Ž. Ž. From 2 and 3 , we obtain p1 p c z Np lower terms 0, Ž . N Ž. so that 0 or 1. Since 0, it follows that 1, and hence 2 becomes Ž a . Ž a . z z c cz c z . n n 1 n p n N Ý Ž 0 1 N . Ý n0 Ž b1. n Ž bp. n! n0 n! n Therefore we have Ž a1. n Ž ap . 1 c0 c1 c n N Ž b . b n! n! Ž n 1 . ! Ž n N . ! Ž . 1 n p n Ž. Let f t be the polynomial defined by Ž n N, N 1,... . . Ž 4. fŽ t. c0ct 1 c2tŽ t1. cNtŽ t1. Ž t N 1 . . Ž. Then 4 implies that and hence we have Ž a1. n Ž ap . n fŽ n. Ž nN, N 1,... . , b b Ž . Ž . 1 n p n Ž a n. Ž a n. Ž a n. Ž a n. Ž a1. Ž ap . fŽ n. Ž b n. b n Ž b n. b n Ž b . b 1 p 1 p n n Ž . Ž . Ž . 1 p 1 p 1 n p n Ž a1. n1 Ž ap . n1 fŽ n1. b b Ž . Ž . 1 n1 p n1 Ž n N, N 1,... . ,
so that ZEROS OF GENERALIZED HYPERGEOMETRIC FUNCTIONS 253 Ž . Ž a n. Ž a n. fŽ n. Ž b n. b n fŽ n1. 1 p 1 p Ž. Ž. Since f t is a polynomial, Eq. 5 implies that Ž . Ž . Ž n N, N 1, ... . . Ž 5. Ž a t. a t fŽ t. Ž b t. b t fŽ t1 . . Ž 6. 1 p 1 p Now, using Ž. 6 and an induction, we will show that we can re-index a 1,...,a p so that a1b 1,...,a pbp are non-negative integers. If p N p deg fŽ. t 1, then we must have p 1 and N 0, so that our assertion is trivial. Let p N 1 and assume the induction hypothesis. Since b1 t is a factor of the right-hand side of Ž. 6 , it is a factor of the left-hand side also. Therefore b t divides Ž a t. Ž a t . ,orb t divides fŽ t . 1 1 p 1 . If b t divides Ž a t. Ž a t . 1 1 p , then b1 ak for some k. Re-index a ,...,a so that a b . Then Ž. 6 implies that 1 p 1 1 Ž . Ž . Ž a t. a t fŽ t. Ž b t. b t fŽ t1 . . 2 p 2 p Hence, by the induction hypothesis, we can re-index a ,...,a so that 2 p a b ,...,a b are non-negative integers. 2 2 p p If b t divides fŽ. t , then there is a polynomial f Ž. 1 1 t of degree N 1 such that fŽ. t Ž b t. f Ž. t . From Ž. 6 , we obtain 1 1 Ž . Ž . Ž a t. a t f Ž t. Ž b t. b t fŽ t1. 1 p 1 2 p Ž . Ž b t. b t Ž b t 1. f Ž t1 . . 2 p 1 1 Again, the induction hypothesis implies that we can re-index a ,...,a 1 p so that a b 1, a b ,...,a b are non-negative integers. This 1 1 2 2 p p proves our assertion. 3. REALITY OF ZEROS We start this section with an asymptotic equality due to Barnes 1 . Let a 1,...,a p and b 1,...,bp be complex constants and assume that b 1,...,bp 0, 1, 2, . . . . In 1, pp. 8083 , it is shown that for each 0we
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On the Zeros of Some Generalized Hypergeometric Functions

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