Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/8

donc

${\begin{aligned}\log &\left(1-{\frac {x}{p}}\right)+\log \left(1-{\frac {x}{q}}\right)+\log \left(1-{\frac {x}{r}}\right)\ldots \\=&-\ x\ \ \left({\frac {1}{p}}\ \,+{\frac {1}{q}}\ \,+{\frac {1}{r}}\ \,+\ldots \right)\\&-{\frac {x^{2}}{2}}\left({\frac {1}{p^{2}}}+{\frac {1}{q^{2}}}+{\frac {1}{r^{2}}}+\ldots \right)\\&-{\frac {x^{3}}{3}}\left({\frac {1}{p^{3}}}+{\frac {1}{q^{3}}}+{\frac {1}{r^{3}}}+\ldots \right)\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}$

Donc, si l’on développe la quantité

$-\log \left(1-{\frac {bx-cx^{2}+dx^{3}-\ldots }{a}}\right)$

en une série de cette forme

$\mathrm {A} x+\mathrm {B} x^{2}+\mathrm {C} x^{3}+\ldots ,$

on aura

${\begin{aligned}\mathrm {A} &={\frac {1}{p}}\,\ +{\frac {1}{q}}\,\ +{\frac {1}{r}}\,\ +\ldots ,\\2\mathrm {B} &={\frac {1}{p^{2}}}+{\frac {1}{q^{2}}}+{\frac {1}{r^{2}}}+\ldots ,\\3\mathrm {C} &={\frac {1}{p^{3}}}+{\frac {1}{q^{3}}}+{\frac {1}{r^{3}}}+\ldots ,\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}$

2. Puisqu’on suppose

$-\log \left(1-{\frac {bx-cx^{2}+dx^{3}-\ldots }{a}}\right)=\mathrm {A} x+\mathrm {B} x^{2}+\mathrm {C} x^{3}+\ldots ,$

on aura, en prenant les différentielles de part et d’autre et divisant par $dx,$

${\frac {b-2cx+3dx^{2}-\ldots }{a-bx+cx^{2}-dx^{3}+\ldots }}=\mathrm {A} +2\mathrm {B} x+3\mathrm {C} x^{2}+\ldots ,$