Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/24

Donc, en mettant $x$ à la place de $p,$

${\begin{aligned}x^{m}&={\frac {a^{m}}{b^{m}}}+m\left({\frac {ca^{m+n-1}}{b^{m+n}}}-{\frac {ca^{m+r-1}}{b^{m+r}}}\right)\\&+{\frac {m}{2}}\left[{\frac {(m+2n-1)c^{2}a^{m+2n-2}}{b^{m+2n}}}-2{\frac {(m+n+r-1)ca^{m+n+r-2}}{b^{m+n+r}}}\right.\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad +\left.{\frac {(m+2r-1)a^{m+2r-2}}{b^{m+2r}}}\right]\\\\&+{\frac {m}{2.3}}\left[{\frac {(m+3n-1)(m+3n-2)c^{3}a^{m+3n-3}}{b^{m+3n}}}\right.\\&\qquad -3{\frac {(m+2n+r-1)(m+2n+r-2)c^{2}a^{m+2n+r-3}}{b^{m+2n+r}}}\\&\qquad +3{\frac {(m+n+2r-1)(m+n+2r-2)ca^{m+n+2r-3}}{b^{m+n+2r}}}\\&\quad \qquad \qquad \qquad \qquad \qquad -\left.{\frac {(m+3r-1)(m+3r-2)a^{m+3r-3}}{b^{m+3r}}}\right]\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}$

14. Exemple IV. — Soit l’équation générale

$a-bx+cx^{2}-dx^{3}+ex^{4}-fx^{5}+\ldots =0\,;$

on aura

$bx\xi =cx^{2}-dx^{3}+ex^{4}-fx^{5}+\ldots =0,$

et, par conséquent,

${\begin{aligned}\xi \ \ &={\frac {cx-dx^{2}+ex^{3}-fx^{4}+\ldots }{b}},\\\xi ^{2}&={\frac {c^{2}x^{2}-2cdx^{3}+\left(d^{2}+2ce\right)x^{4}-\ldots }{b^{2}}},\\\xi ^{3}&={\frac {c^{3}x^{3}-3cdx^{4}+\ldots }{b^{3}}},\\\xi ^{4}&={\frac {c^{4}x^{4}-\ldots }{b^{4}}},\\\ldots &\ldots \ldots \ldots \ldots .\end{aligned}}$

Donc

${\begin{aligned}x^{m}&={\frac {a^{m}}{b^{m}}}\\&+m\left({\frac {a^{m+1}c}{b^{m+2}}}-{\frac {a^{m+2}d}{b^{m+3}}}+{\frac {a^{m+3}e}{b^{m+4}}}-{\frac {a^{m+5}f}{b^{m+5}}}+\ldots \right)\\&+{\frac {m}{2}}\left[{\frac {(m+3)a^{m+2}c^{2}}{b^{m+4}}}-{\frac {(m+4)a^{m+3}.2cd}{b^{m+5}}}+{\frac {(m+5)a^{m+4}\left(d^{2}+2ce\right)}{b^{m+6}}}-\ldots \right]\end{aligned}}$