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Some Remarks on the Báez-Duarte Criterion for the Riemann Hypothesis

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Some Remarks on the Báez-Duarte Criterion for the Riemann Hypothesis CMST 20(2) 39-47 (2014) DOI:10.12921/cmst.2014.20.02.39-47 Some remarks on the Báez-Duarte criterion for the Riemann Hypothesis Marek Wolf Cardinal Stefan Wyszynski University Faculty of Mathematics and Natural Sciences. College of Sciences ul. Wóycickiego 1/3, Auditorium Maximum, (room 113) PL-01-938 Warsaw, Poland e-mail: [email protected] Received: 4 March 2014; revised: 9 April 2014; accepted: 11 April 2014; published online: 25 April 2014 Abstract: In this paper we are going to describe the results of the computer experiment, which in principle can rule out validity of the Riemann Hypothesis (RH). We use the sequence ck appearing in the Báez-Duarte criterion for the RH and compare two formulas for these numbers. We describe the mechanism of possible violation of the Riemann Hypothesis. Next we calculate c100000 with a thousand digits of accuracy using two different formulas for ck with the aim to disprove the Riemann Hypothesis in the case these two numbers will differ. We found the discrepancy only on the 996th decimal place (accuracy of 10−996). The computer experiment reported herein can be of interest for developers of Mathematica and PARI/GP. Key words: Riemann Hypothesis, zeta function, Baez-Duarte criterion I. INTRODUCTION to prove RH in Chapter 8. Presently the requirement that the 0 nontrivial zeros are simple ζ (ρl) 6= 0 is often added. With the advent of the computer era the computing ma- The first use of computers in connection with RH was by chines have been used to prove mathematical theorems. Allan Turing checking whether the nontrivial zeros of ζ(s) 1 The most spectacular examples of such a use of computers have indeed real part 2 [5]. Turing suspected that the RH is were proofs of the four color theorem [1, 2] and of the Ke- not true and the first counterexample is lying relatively low. pler conjecture about sphere packing in three-dimensional Let us quote the sentence from the first page of his paper: Euclidean space [3]. It seems to be not possible to use com- “The calculation were done in an optimistic hope that a zero puters for the proof of the Riemann Hypothesis (RH), but its would be found off the critical line”, but up to t = 1540 Tur- refutation by numerical calculations seems to be plausible. ing found that all zeros are on the critical line. The present The Riemann Hypothesis says that the series record belongs to Xavier Gourdon [6] who has checked that all 1013 first zeros of the Riemann ζ(s) lie on the critical 1 X 1 line. Andrew Odlyzko checked that RH is true in different ζ(s) = ; (s = σ + it; <(s) > 1) (1) 20 21 22 ns intervals around 10 [7], 10 [8], 10 [9], but his aim was n=1 not verifying the RH but rather provide evidence for con- analytically continued to the complex plane in addition to jectures that relate nontrivial zeros of ζ(s) to eigenvalues of trivial zeros ζ(−2n) = 0 has nontrivial zeros ζ(ρl) = 0 in random matrices. In fact, Odlyzko expressed the view that the critical strip 0 < <(s) < 1 only on the critical line: off critical line zeros could be encountered at least at t of the 1 1 1010000 <(ρl) = 2 i.e. ρl = 2 + iγl, see e.g. the modern guide to the order 10 , see [10]. Asked by Derbyshire “What do you RH [4]. In the same book there is a review of failed attempts think about this darn Hypothesis? Is it true, or not?” Odlyzko 40 M. Wolf replied: “Either it’s true, or else it isn’t”. Also other famous where mathematicians John E. Littlewood and Paul Turán did not 1 s ξ(s) = s(s − 1)Γ ζ(s) believe RH is true. Aleksandar Ivic´ gave a few arguments 2 2 against the truth of the RH, see [4, ] and on arxiv [11]. fulfills: There were several attempts to use computers to disprove some conjectures related to RH in the past. Sometimes it λn ≥ 0 for n = 1; 2;::: (5) was sufficient to find a counterexample to the given hypoth- The explicit expression has the form: esis, sometimes the disproof was not direct. For example, X n the Haselgrove [12] disproved the Pólya’s Conjecture stat- λn = (1 − (1 − 1/ρ) ): (6) ing that the function ρ X K. Maslanka´ [21-23] has performed extensive computer cal- L(x) := λ(n) (2) culations of these constants confirming (5) with large sur- n≤x plus. satisfies L(x) ≤ 0 for x ≥ 2, where λ(n) is Liouville’s Let us mention also the elementary Lagarias criterion function defined by [24]: to disprove the RH it suffices to find one n that has λ(n) = (−1)r(n) so many divisors, that: where r(n) is the number of, not necessarily distinct, prime X d > Hn + exp(Hn) ln(Hn); (7) r1 rn factors in n = p1 ··· pα(n), with multiple factors counted djn according to their multiplicities: r = r +:::+r . From the 1 n where H = Pn 1=k is the n-th harmonic number. truth of the Pólya Conjectures the RH follows, but the con- n k=1 The Lagarias criterion is not well suited for computer verifi- verse statement is not true. The Haselgrove proof was indi- cation and in [25] Keith Briggs has undertook instead the rect, and in 1960 Lehman [13] found on the computer ex- verification of the Robin [26] criterion for RH: plicit counter–example: L(906180359) = 1. X The next example is provided by the Mertens conjecture. RH , d < eγ n ln ln(n) for n > 5040 (8) Let M(x) denote the Mertens function defined by djn X M(x) = µ(n); (3) For appropriately chosen n Briggs obtained for the differ- n≤x ence between r.h.s. and l.h.s. of the above inequality value as −13 −6 where µ(n) is the Möbius function small as e ≈ 2:2 × 10 , hence again RH is very close to being violated. 8 1 n = 1 < In this paper we are going to propose a method which in µ(n) = 0 p2jn principle can provide a refutation of the RH. The idea is to : (−1)r n = p p : : : p ; p 1 2 r i distinct calculate a number with very high accuracy (one thousand From digits) in two ways: one without any knowledge on the zeros 1 jM(x)j < x 2 (4) of ζ(s) and second using the explicit formula involving all again the RH would follow. However, in 1985 A. Odlyzko ρl. Despite some estimation presented in Sect. 3 indicating and H. te Riele [14] disproved the Mertens conjecture, again that the discrepancy could be found merely at much higher not directly, but later it was shown by J. Pintz [15] that than a thousand decimal places we performed the calcula- the first counterexample appears below exp(3:21 × 1064). tions in an optimistic hope that we will find the discrepancy The upper bound has since been lowered to exp(1:59×1040) between these two numbers, paraphrasing the sentence of [16]. Turing. There is a lot of number theoretic functions often Especially interesting is the value of the de Bruijn- defined in an elementary way being expressed also by the Newman constant Λ, see e.g. §2.32 (pp. 203-205) in [17]. “explicit” formulas in terms of zeros of the ζ(s) function. Unconditionally Λ ≤ 1=2 and the Riemann Hypothesis is Let us mention here the Chebyshev function X equivalent to the inequality Λ ≤ 0. The fascinating run (x) = Λ(n) (= ln(lcm(2; 3; · · · bxc))); Λ Λ > for the best lower bound on ended with the value n≤x −2:7 × 10−9 obtained by Odlyzko [18]. Such a narrow gap where the von Mangoldt function Λ(n) is defined as for values of Λ being compatible with RH allowed Odlyzko to make the remark: “the Riemann Hypothesis, if true, is just ln p for n = pm Λ(n) = barely true”. However, Don Zagier [19] interpreted it as “the 0 in other cases RH, if false, is only slightly false”. The explicit formula reads, see eg. [27] (the term ln 1 − 1 In 1997 Xian-Jin Li proved [20], that Riemann Hypo- x2 comes from trivial zeros): thesis is true iff the sequence: ρl n 1 1 X x 1 d n−1 (x) = x−ln(2π)− ln 1 − 2 − (9) λn = n (s ln ξ(s))js=1 2 x ρl (n − 1)! ds nontr: zeros ρl Some remarks on the Báez-Duarte criterion for the Riemann Hypothesis 41 Also the Mertens’s function has the explicit representa- where tion (the last term is comprising contribution from all trivial k zeros) [27]: X k 1 c = (−1)j ≡ k j ζ(2j + 2) X X xρl j=0 µ(n) = 0 − 2 (14) ρlζ (ρl) k n≤x nontr: X k (4j + 4)! zeros ρl j+1 ≡ (−1) 2j+1 2j+2 : (10) j 2 B2j+2π 1 2n j=0 X 2π 1 − (−1)n x (2n)!nζ(2n + 1) n=1 The problem with these series is that they are extremely If additionally slow converging because the partial sums oscillate with −3=4 amplitudes diminishing at very slow rates. For example ck = O k (15) (1000001) = 999586:597 :::, while from (9) summing then all zeros of ζ(s) are simple. Báez-Duarte showed over 5,549,728 zeros gives 999587:15 :::, thus relative error unconditionally (regardless of validity of the RH) slower is 0.000055.
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