Page:Annales de mathématiques pures et appliquées, 1819-1820, Tome 10.djvu/38

Le texte de cette page a été corrigé et est conforme au fac-similé.

34

DÉVELOPPEMENT EN SÉRIES

Soit $u$ une fonction de $y$ qu’il soit question de développer suivant les puissances de $x\ ;y$ étant donnée par l’équation

y=a+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\ldots

On aura d’abord cette suite d’équations

{\frac {\operatorname {d} u}{\operatorname {d} x}}={\frac {\operatorname {d} u}{\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} x}}={\frac {\operatorname {d} u}{\operatorname {d} y}}\left(a_{1}+2a_{2}x+3a_{3}x^{2}+4a_{4}x^{3}+\ldots \right)

\left.{\begin{array}{lll}{\frac {\operatorname {d} u}{\operatorname {d} a}}&={\frac {\operatorname {d} u}{\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} a}}&=\quad {\frac {\operatorname {d} u}{\operatorname {d} y}},\\{\frac {\operatorname {d} u}{\operatorname {d} a_{1}}}&={\frac {\operatorname {d} u}{\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} a_{1}}}&=x\ \ {\frac {\operatorname {d} u}{\operatorname {d} y}},\\{\frac {\operatorname {d} u}{\operatorname {d} a_{2}}}&={\frac {\operatorname {d} u}{\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} a_{2}}}&=x^{2}{\frac {\operatorname {d} u}{\operatorname {d} y}},\\\ldots &\ldots \ldots &\ldots \ldots \\{\frac {\operatorname {d} u}{\operatorname {d} a_{n}}}&={\frac {\operatorname {d} u}{\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} a_{n}}}&=x^{n}{\frac {\operatorname {d} u}{\operatorname {d} y}}\,;\end{array}}\right\}(1)

dans lesquelles $a,a_{1},a_{2},\ldots a_{n}$ sont traitées comme indépendantes.

Éliminant ${\frac {\operatorname {d} u}{\operatorname {d} y}}$ entre chacune des équations (1) et la dernière, on trouve

\left.{\begin{array}{rl}{\frac {\operatorname {d} u}{\operatorname {d} a_{n}}}&=x^{n}\quad {\frac {\operatorname {d} u}{\operatorname {d} a}},\\{\frac {\operatorname {d} u}{\operatorname {d} a_{n}}}&=x^{n-1}{\frac {\operatorname {d} u}{\operatorname {d} a_{1}}},\\\ldots &\ldots \ldots \ldots \ldots \\{\frac {\operatorname {d} u}{\operatorname {d} a_{n}}}&=x^{n-k}{\frac {\operatorname {d} u}{\operatorname {d} a_{k}}}.\end{array}}\right\}(2)