An Explicit Zero-Free Region for the Riemann Zeta-Function

Cheng, Yuanyou Furui

doi:10.1216/rmjm/1022008980

Cited by 11 publications

(12 citation statements)

References 12 publications

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“…It should be noted that using the method of Tyrina [26] when 4 9 k 2 < Dk; s < 1 2 k 2 gives slightly better values for Dk; s, but only enough to improve the constant B in Theorem 1 by 0.01 or less. The next de®nition is slightly different from that given in [30].…”

Section: Vinogradov's Integral: Complete Systemsmentioning

confidence: 99%

“…In this section, we derive bounds for J s; k P using the iterative methods of Wooley [30], modi®ed using an idea of Arkhipov and Karatsuba [1] (the introduction of the parameter r). It should be noted that using the method of Tyrina [26] when 4 9 k 2 < Dk; s < 1 2 k 2 gives slightly better values for Dk; s, but only enough to improve the constant B in Theorem 1 by 0.01 or less. The next de®nition is slightly different from that given in [30].…”

Section: Vinogradov's Integral: Complete Systemsmentioning

confidence: 99%

“…By comparison, Popov [20] showed that the above holds with c 0:00006888, and Cheng [4] proved a zero-free region with c 1=990.…”

Section: Introductionmentioning

confidence: 99%

“…Moreover, in [6], using the full strength of Theorem 1, the zero-free region t > 3; 1 À b < c log t 2 = 3 log log t 1 = 3 ; c 1 57:54 is proved. By comparison, Popov [20] showed that the above holds with c 0:00006888, and Cheng [4] proved a zero-free region with c 1=990.…”

Section: Introductionmentioning

confidence: 99%

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Vinogradov's Integral and Bounds for the Riemann Zeta Function

Ford

2002

Proc. Lond. Math. Soc.

127

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kevin ford 1 < x < P ea 1 x . . . a k x k 2 s eÀa 1 h 1 À . . . À a k h k da < J s; k P; 0; . . . ; 0 J s; k P:Hence, writing Q P, we obtainAlso, counting only the solutions of (1.4) with x i y i for each i gives J s; k P > Q s . Therefore J s; k P > max2s À k P 2 s À kk 1 = 2 ; P s : 1:5 kevin fordUpper bounds take the form of J s; k P < Ds; kP 2 s À kk 1 = 2 hs; k ; 1:6where hs; k > 0 and Ds; k is independent of P. Stechkin in 1975 [24] proved (1.6) with hrk; k 1 2 k 2 1 À 1=k r ; Drk; k expfC minr; kk 2 log kg for an absolute constant C. The constant factor was improved by Wooley [31]. Small improvements to the exponents of P were subsequently made by Arkhipov and Karatsuba [1] and Tyrina [26] (signi®cant for s p k 2 ). Also signi®cant is Wooley's result [32] when s p k 3 = 2 À « , which is very close to the`ideal' bounds Ck; sP s in that range of s. For our purposes, the most important improvement comes from Wooley [30], who improved the exponents substantially in a wide range of s, showing that (1.6) holds with hk; s < 1 2 k 2 e 1 = 2 À 2 s = k 2 valid for s p k 2 log k (see [5, Lemma 5.2]). In Theorem 3 below, we combine Wooley's method with the main idea from [1] to improve this to hk; s < 3 8 k 2 e 1 = 2 À 2 s = k 2 . In the application to bounding the Riemann zeta function, we will take s to be of order k 2 , so this small improvement is signi®cant.Theorem 3. Let k and s be integers with k > 1000 and 2k 2 < s < 1 2 k 2 1 2 log 3 8 k. Then J s; k P < k 2:055 k 3 À 5:91 k 2 3 s 1:06 sk 2 s 2 = k À 9:7278 k 3 P 2 s À kk 1 = 2 D s P > 1;where D s 3 8 k 2 e 1 = 2 À 2 s = k 2 1:7 = k : Further, if k > 129, there is an integer s < rk 2 such that for P > 1, J s; k P < k vk 3 P 2 s À kk 1 = 2 0:001 k 2

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Section: Vinogradov's Integral: Complete Systemsmentioning

confidence: 99%

Section: Vinogradov's Integral: Complete Systemsmentioning

confidence: 99%

“…By comparison, Popov [20] showed that the above holds with c 0:00006888, and Cheng [4] proved a zero-free region with c 1=990.…”

Section: Introductionmentioning

confidence: 99%

Section: Introductionmentioning

confidence: 99%

See 2 more Smart Citations

Vinogradov's Integral and Bounds for the Riemann Zeta Function

Ford

2002

Proc. Lond. Math. Soc.

127

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“…[12]) trouve R 1 = 14518, puis en 2000, Y. Cheng trouve R 1 = 990 (voir [1]), résultat dernièrement amélioré par K. Ford : R 1 = 57.54 (voir [4]). …”

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