Download Advances in the theory of Riemann surfaces; proceedings of by Lars Valerian Ahlfors, Lipman Bers PDF

By Lars Valerian Ahlfors, Lipman Bers

The current quantity comprises all yet of the papers learn on the convention, in addition to a couple of papers and brief notes submitted afterwards. we are hoping that it displays faithfully the current nation of study within the fields lined, and that it could offer an entry to those fields for destiny investigations.

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60 Analyse en 30 fiches 1 1 Application Étudiez la convergence de la suite définie par u 0 = 0 et la relation de récurrence : 1 · ∀n ∈ N u n+1 = 1 + un Solution • Il s'agit d'une suite récurrente du type u n+1 = f (u n ) avec f (x) = À partir de u 0 = 0, on vérifie aisément par récurrence que : ∀n ∈ N 0 un 1 · 1+x 1 Il suffit donc d'étudier les variations de f sur [0,1]. −1 < 0. La fonction f est donc décroissante sur [0,1]. On a toujours f (x) = (1 + x)2 Dans ce cas, on considère les deux suites extraites (u 2n ) et (u 2n+1 ) qui sont monotones et de sens contraire car f décroissante entraîne f ◦ f croissante.

1, c'est la fonction réciproque de la fonction loga . Pour a = y = a x ⇐⇒ ln y = x ln a ⇐⇒ x = loga (y) . Sa dérivée est : (a x ) = ln a × a x . Remarquez bien qu'ici, la variable est en exposant. Ses propriétés algébriques sont les mêmes que celles de la fonction exp. IV Fonctions puissances et comparaisons • Fonctions puissances La fonction x → x r , pour x > 0 et r ∈ Q, est déjà connue. FICHE 7 – Logar ithmes et exponentielles 37 On la généralise, pour x > 0 et a ∈ R, en posant : x a = ea ln x .

La fonction arcsin, de [−1,1 ] dans − • L'équation (1) implique : 5 4 = sin (arcsin x) = x . + arcsin 5 13 sin arcsin On développe le premier membre, en sachant que : sin (arccos θ) = sin2 (arccos θ) = 1 − cos2 (arccos θ) = 1 − θ2 , et on obtient ainsi : x= 4 × 5 1− 5 13 2 + 1− 4 5 2 × 5 4 12 3 5 63 = × + × = · 13 5 13 5 13 65 © Dunod – La photocopie non autorisée est un délit. • L'équation (2) implique : cos arccos 1 1 = cos (arccos x) = x . + arccos 3 4 On développe le premier membre, ce qui donne : 1 1 x= × − 3 4 1− 1 3 2 × 1− 1 4 2 1 = − 12 √ 8 15 1 − 2 30 × = · 9 16 12 Maple trouve une solution dans les trois cas, alors que l'équation (3) n'a pas de solution !

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