1996
DOI: 10.1070/sm1996v187n09abeh000158
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Modular functions and transcendence questions

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Cited by 117 publications
(167 citation statements)
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“…, where B ∈ N is the least common denominator of the numbers b j,k , and therefore S − C 0 is either zero or transcendental in view of the algebraic independence of π and e π √ d [7].…”
Section: Resultsmentioning
confidence: 99%
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“…, where B ∈ N is the least common denominator of the numbers b j,k , and therefore S − C 0 is either zero or transcendental in view of the algebraic independence of π and e π √ d [7].…”
Section: Resultsmentioning
confidence: 99%
“…Obviously by (1) and (4), the numbers ψ (k) (1)/ζ(k + 1), g (k) (1)/ζ(k + 1), ψ (k) (1/2)/ζ(k + 1) are rational (here ζ(s) = ∞ n=1 1/n s is the Riemann zeta function) and therefore from (2), (5) we get the following inclusions: (7) ψ (2k−1) (m), g (2k−1) (m), ψ (2k−1) (m + 1/2) ∈ Q × · π 2k + Q, m ∈ N.…”
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confidence: 99%
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“…Il n'est pas a priori très naturel de passer par J pour étudier la fonction exponentielle ; mais c'est pourtant bien un tel détour qui a permis d'obtenir l'indépendance algébrique de 7r et e7r (travaux de Y.V. Nesterenko, voir [4] (1) Compte - (1) [3], [11]. Si exp(x2Yl) et exp(x2Y2) sont algébriques, alors -(~) [12].…”
Section: -Introductionunclassified
“…Note that a 1988 result by Bézivin [3] can be used to prove this irrationality. The irrationality of ζ q (2) was proven by Duverney [9] in 1995, the transcendence of ζ q (2) (and in fact of ζ q (2s), s ∈ N) is a consequence of a general result by Nesterenko [12], [13]. Moreover, the three values 1, ζ q (1), ζ q (2) have been shown to be Qlinearly independent by Bundschuh and Väänänen [8], by Zudilin [18] and by Postelmans and Van Assche [14].…”
mentioning
confidence: 99%