tailieunhanh - Báo cáo toán học: "Counting distinct zeros of the Riemann zeta-function"

Tuyển tập các báo cáo nghiên cứu khoa học ngành toán học tạp chí toán học quốc tế đề tài: Counting distinct zeros of the Riemann zeta-function. | Counting distinct zeros of the Riemann zeta-function David W. Farmer Submitted December 1 1994 Accepted December 13 1994. Abstract. Bounds on the number of simple zeros of the derivatives of a function are used to give bounds on the number of distinct zeros of the function. The Riemann -function is dehned by s H s s where H s 1 s s 1 _2sr 1 s and s is the Riemann -function. The zeros of s and its derivatives are all located in the critical strip 0 ơ 1 where s ơ it. Since H s is regular and nonzero for ơ 0 the nontrivial zeros of s exactly correspond to those of s . Let pj 3 i denote a zero of the jth derivative j s and denote its multiplicity by m . Dehne the following counting functions N j T N T 1 p j fi i N 0 T zeros of j ơ it with 0 t T zeros of ơ it with 0 t T NT X 1 p jj P i m 1 simple zeros of j ơ it with 0 t T N j T s 2 X 1 p 11 i m 1 simple zeros of j 2 it with 0 t T Mr T X 1 p 0 f3 i m r zeros of ơ it of multiplicity r with 0 t T M r T X 1 p 0 i m r zeros of ơ it of multiplicity r with 0 t T where all sums are over 0 T and zeros are counted according to their multiplicity. It is well known that N j T ỳ-T log T. Let Nj T T c s 2V j lirnl N N IT 3j lirniisf NjT Thus 3j is the proportion of zeros of j s which are simple and j is the proportion which are simple and on the critical line. The best currently available bounds are 0 0A0219 0A9874 2 0 93469 3 0 9673 a4 0 98006 and 5 0 9863. These bounds were obtained by combining Theorem 2 of C2 with the methods of C1 . Trivially 3j j. 1991 Mathematical Subject Classihcation 05A20 11M26 THE ELECTRONIC JOURNAL OF COMBINATORICS 2 1995 R1 2 Let Nd T be the number of distinct zeros of s in the region 0 t T. That is Nd T X MT n 1 1 It is conjectured that all of the zeros of s are distinct Nd T N T or equivalently all of the zeros are simple N .11 T N T . From the bound on 0 we have N .11 T kN T with . We will use the bounds on 3j to obtain the following Theorem. For T sufficiently large Nd T kN T with k .