1935
DOI: 10.2307/1989728
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Distribution Functions and the Riemann Zeta Function

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Cited by 132 publications
(208 citation statements)
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“…1 A proof of this result in the case that Φ(ω) takes only the two values ±1 with equal probabilities can be found in [4]. We can and shall restrict our study to the case E(Φ) = 0.…”
mentioning
confidence: 79%
“…1 A proof of this result in the case that Φ(ω) takes only the two values ±1 with equal probabilities can be found in [4]. We can and shall restrict our study to the case E(Φ) = 0.…”
mentioning
confidence: 79%
“…[4]) that F(x, r) is absolutely continuous. Since the smoothness of E(x, r) depends only on the tail of r, similar results hold when (1.17) is satisfied from some time on.…”
Section: -{)/*-mentioning
confidence: 99%
“…It is easy to show that F(x, r) is continuous. Since it can be shown (see [4]) that it is always "pure," i.e. either absolutely continuous or purely singular, the question arises for which sequences r, E(x, r) enjoys the former or the latter property.…”
mentioning
confidence: 99%
“…Например, для независимого равновероятного способа расстановки знаков в (1), т.е. когда P n (ti,... ,t") = 2~", в контексте бесконечных симметрических сверток Бер-нулли соответствующие вопросы изучались в [2], [3], в частности, имеем Обратимся сейчас к характеристической функции / п (г) частичной суммы 5" = ai<lH \-a n t n ряда(1). Имеем f n (z) = £{ <1= ± i,...,*"= ± 1} exp{t«S"}P"(*i,..., t n ), или, с учетом (7), /п(г)= 9nrh"-ir 53 expiizSn + J^lttt+lJ.…”
Section: S(w) = Yakx K (U)unclassified