By Évariste Galois

DÉMONSTRATION

D’UN

THÉORÈME SUR LES FRACTIONS maintains PÉRIODIQUES ([1]).

On sait que si, par los angeles méthode de Lagrange, on développe en fraction proceed une des racines d’une équation du moment degré, cette fraction proceed sera périodique, et qu’il en sera encore de même de l’une des racines d’une équation de degré quelconque, si cette racine est racine d’un facteur rationnel du moment degré du most efficient membre de l. a. proposée, auquel cas cette équation air of secrecy, tout au moins, une autre racine qui sera également périodique. Dans l’un et dans l’autre cas, los angeles fraction proceed pourra d’ailleurs être immédiatement périodique ou ne l’être pas immédiatement ; mais, lorsque celle dernière circonstance charisma lieu, il y air of mystery du moins une des transformées dont une des racines sera immédiatement périodique.

Or, lorsqu’une équation a deux racines périodiques répondant à un même facteur rationnel du moment degré, et que l’une d’elles est immédiatement périodique, il existe entre ces deux racines une relation assez singulière qui parait n’avoir pas encore été remarquée, et qui peut être exprimée par le théorème suivant :

Théorème. — Si une des racines d’une équation de degré quelconque est une fraction proceed immédiatement périodique, cette équation air of mystery nécessairement une autre racine également périodique que l’on obtiendra en divisant l’unité négative par cette même fraction proceed périodique, écrite dans un ordre inverse.

Démonstration. — Pour fixer les idées, ne prenons que des périodes de quatre termes ; vehicle los angeles marche uniforme du calcul prouve qu’il en serait de même si nous en admettions un plus grand nombre. Soit une des racines d’une équation de degré quelconque exprimée comme il swimsuit :

}}}}}}}}}}}}}}} }}}}}}}}}}}}}}}

l’équation du moment degré, à laquelle appartiendra cette racine, et qui contiendra conséquemment sa corrélative, sera..

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**Sample text**

Then μ is uniquely determined by the values μ(E), E ∈ E. If μ is a probability measure, the existence of the sequence (Ωn )n∈N is not needed. Proof Let ν be a (possibly different) σ -finite measure on (Ω, A) such that μ(E) = ν(E) for every E ∈ E. Let E ∈ E with μ(E) < ∞. Consider the class of sets DE = A ∈ A : μ(A ∩ E) = ν(A ∩ E) . 10: (i) Clearly, Ω ∈ DE . (ii) Let A, B ∈ DE with A ⊃ B. Then μ (A \ B) ∩ E = μ(A ∩ E) − μ(B ∩ E) = ν(A ∩ E) − ν(B ∩ E) = ν (A \ B) ∩ E . Hence A \ B ∈ DE . (iii) Let A1 , A2 , .

58(iv): A measure ∞ n=1 αn δxn is a Lebesgue–Stieltjes measure for a suitable function F if and only if n:|xn |≤K αn < ∞ for all K > 0. 2 Let Ω be an uncountably infinite set and let ω0 ∈ Ω be an arbitrary element. Let A = σ ({ω} : ω ∈ Ω \ {ω0 }). 4 (p. 11). (ii) Show that (Ω, A, δω0 ) is complete. 3 Let (μn )n∈N be a sequence of finite measures on the measurable space (Ω, A). Assume that for any A ∈ A there exists the limit μ(A) := limn→∞ μn (A). Show that μ is a measure on (Ω, A). Hint: In particular, one has to show that μ is ∅-continuous.

1 (Rolling a die twice) Consider the random experiment of rolling a die twice. Hence Ω = {1, . . 30(ii)). , if A depends only on the outcome of the first roll and B depends only on the outcome of the second roll. ˜ B˜ ⊂ {1, . . , 6} such that Formally, we assume that there are sets A, ˜ A = A˜ × {1, . . , 6} and B = {1, . . , 6} × B. A. 1). To this end, we compute #A˜ #B #B˜ P[A] = #A 36 = 6 and P[B] = 36 = 6 . Furthermore, P[A ∩ B] = ˜ #(A˜ × B) #A˜ #B˜ = · = P[A] · P[B]. 36 6 6 (ii) Stochastic independence can occur also in less obvious situations.