. La-valeur-e-i-sera-appelée-la-multiplicité-de-f-i, Si k ? 1, nous dirons que f 1 , . . . , f k décrivent les racines réelles de A sur S. Les racines de A sont descriptibles sur S s'il existe des fonctions f 1

I. Dense, Soit Soit S un sous-ensemble connexe de R comme un polynôme en Y ne s'annule pas sur S, Théorème 6.11 (Cas particulier du théorème 1 dans

. Ainsi and Y. Dans-la-suite, nous considérerons des sous-ensembles de R où le polynôme Res(A, A Y ) n'a pas de racines. En particulier, nous voulons que ce polynôme soit non nul. Nous montrons que c

. Démonstration and . Par-un-résultat-bien-connu, Supposons que R(X) = Res(A, A Y )(X) = 0. Cela implique que A et A Y ont un facteur commun, il existe C dans R(X) tel que A = CB. Nous avons donc deg Y (A) = deg Y (B) ? deg Y (A Y ). Ce qui est impossible car deg Y (A) ? 1

F. Si and Y. , alors comme G(X, 0) = 0 (sinon (x, 0) est une solution de (6.6) pour tout x dans R), par la règle de Descartes

F. Si, 0, alors F ne dépend pas de Y et il y a au plus d valeurs x 1 , . . . , x p de X telles que F (x l , Y ) = 0. Pour chacune de ces valeurs, G(x l , Y ) est un polynôme univarié t-creux donc a au plus 2t ? 1 racines réelles distinctes

. Si, I est dans I, les racines de F sont descriptibles sur I par le corollaire 6

P. Ainsi and I. <. Chaque-intervalle-i-dans, < ? I,m I : I ? R telles que F (x, y) = 0 sur I × R si et seulement s'il existe i ? m I tel que y = ? I,i (x) De plus, F Y (x, ? I,i (x)) = 0 car Res(F, F Y ) ne s'annule pas sur I (cf. Remarque 6.14). La version analytique du théorème des fonctions implicites montre que les fonctions ? I,i sont analytiques sur I. Nous noterons ? = I?I I. Bornons

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