Page:Joseph Louis de Lagrange - Œuvres, Tome 8.djvu/274

Or $\left(u^{r}\right)'=ru^{r-1}\,;$ donc

${\begin{aligned}\left(u^{r}\right)'f(u)\ \,=&r{\frac {cu^{r+1}-du^{r+2}+eu^{r+3}-fu^{r+4}+\ldots }{b}},\\\left(u^{r}\right)'f^{2}(u)=&r{\frac {c^{2}u^{r+3}-2cdu^{r+4}+\left(d^{2}+2ce\right)u^{r+5}+\ldots }{b^{2}}},\\\left(u^{r}\right)'f^{3}(u)=&r{\frac {c^{3}u^{r+5}-3cdu^{r+6}+\ldots }{b^{3}}},\\\left(u^{r}\right)'f^{4}(u)=&r{\frac {c^{4}u^{r+7}+\ldots }{b^{4}}}.\end{aligned}}$

Prenant les fonctions dérivées et substituant dans la formule dont il s’agit, on aura, après avoir fait $u={\frac {a}{b}}$ et changé $\alpha$ en $x,$

${\begin{aligned}x^{r}=&{\frac {a^{r}}{br^{2}}}+r\left({\frac {a^{r+1}c}{b^{r+2}}}-{\frac {a^{r+2}d}{b^{r+3}}}+{\frac {a^{r+3}e}{b^{r+4}}}-{\frac {a^{r+4}f}{b^{r+5}}}+\ldots \right)\\&+{\frac {r}{2}}\left[{\frac {(r+3)a^{r+2}c^{2}}{b^{r+4}}}-{\frac {(r+4)a^{r+3}.2cd}{b^{r+5}}}+{\frac {(r+5)a^{r+4}\left(d^{2}+2ce\right)}{b^{r+5}}}+\ldots \right]\\&+{\frac {r}{2.3}}\left[{\frac {(r+5)(r+4)a^{r+3}c^{3}}{b^{r+6}}}-{\frac {(r+6)(r+5)a^{r+4}.3cd}{b^{r+7}}}+\ldots \right]\\&+{\frac {r}{2.3.4}}\left[{\frac {(r+7)(r+6)(r+5)a^{r+4}c^{4}}{b^{r+8}}}+\ldots \right]\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}$

19 Si $r=1,$ on aura

${\begin{aligned}x=&{\frac {a}{b}}+{\frac {a^{2}c}{b^{3}}}-{\frac {a^{3}d}{b^{4}}}+{\frac {a^{4}e}{b^{5}}}-{\frac {a^{5}f}{b^{6}}}+\ldots \\&+{\frac {2a^{3}c^{2}}{b^{5}}}-{\frac {5a^{4}cd}{b^{6}}}+{\frac {3a^{5}\left(d^{2}+2ce\right)}{b^{7}}}+\ldots \\&+{\frac {5a^{4}c^{3}}{b^{7}}}-{\frac {21a^{5}cd}{b^{8}}}+\ldots \\&+{\frac {14a^{5}c^{4}}{b^{9}}}+\ldots \\&\ldots \ldots \ldots \ldots \ldots .\end{aligned}}$

C’est-la formule connue de Newton, pour le retour des suites, qu’on