Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/19

on aura

${\begin{aligned}{\frac {\mathrm {X} }{b}}+{\frac {\mathrm {X} ^{2}}{2b^{2}}}+{\frac {\mathrm {X} ^{3}}{3b^{3}}}+\ldots =&-\log \left(1-{\frac {a}{bx}}-\xi \right)\\&-\log \left(1-{\frac {a}{bx}}\right)-\log \left(1-{\frac {\xi }{1-{\cfrac {a}{bx}}}}\right)\,;\end{aligned}}$

de sorte qu’en réduisant ces deux logarithmes en série on aura

${\frac {\mathrm {X} }{b}}+{\frac {\mathrm {X} ^{2}}{2b^{2}}}+{\frac {\mathrm {X} ^{3}}{3b^{3}}}+\ldots ={\frac {a}{bx}}+{\frac {a^{2}}{2b^{2}x^{2}}}+{\frac {a^{3}}{3b^{3}x^{3}}}+\ldots$

$+{\frac {\xi }{1-{\cfrac {a}{bx}}}}+{\frac {\xi ^{2}}{2\left(1-{\cfrac {a}{bx}}\right)^{2}}}+{\frac {\xi ^{3}}{3\left(1-{\cfrac {a}{bx}}\right)^{3}}}+\ldots ,$

et, par conséquent (8),

(E)

$\left\{{\begin{aligned}{\frac {a}{bx}}+{\frac {a^{2}}{2b^{2}x^{2}}}&+\ldots +{\frac {\xi }{1-{\cfrac {a}{bx}}}}+\ldots \\=&\log {\frac {bp}{a}}+{\frac {p}{x}}++{\frac {p^{2}}{2x^{2}}}++{\frac {p^{3}}{3x^{3}}}+\ldots \\&+\ x\ \left({\frac {1}{q}}+{\frac {1}{r}}+\ldots \right)+{\frac {x^{2}}{2}}\left({\frac {1}{q^{2}}}+{\frac {1}{r^{2}}}+\ldots \right)\\&+{\frac {x^{3}}{3}}\left({\frac {1}{q^{3}}}+{\frac {1}{r^{3}}}+\ldots \right)+\ldots .\end{aligned}}\right.$

Or, nous avons déjà vu plus haut (3) qu’on a, en réduisant en série,

${\begin{aligned}&{\frac {1}{1-{\cfrac {a}{bx}}}}=1+{\frac {a}{bx}}+{\frac {a^{2}}{b^{2}x^{2}}}+{\frac {a^{3}}{b^{3}x^{3}}}+\ldots ,\\&{\frac {1}{\left(1-{\cfrac {a}{bx}}\right)^{2}}}=1+{\frac {2a}{bx}}+{\frac {3a^{2}}{b^{2}x^{2}}}+{\frac {4a^{3}}{b^{3}x^{3}}}+\ldots ,\\&{\frac {1}{\left(1-{\cfrac {a}{bx}}\right)^{3}}}={\frac {1}{2}}\left(1.2+{\frac {2.3.a}{bx}}+{\frac {3.4.a^{2}}{b^{2}x^{2}}}+\ldots \right),\\\end{aligned}}$

et ainsi de suite.

Donc, si l’on suppose, en général,

${\begin{aligned}\xi \,\ &=\varpi +\varpi _{1}x+\varpi _{2}x^{2}+\varpi _{3}x^{3}+\ldots ,\\\xi ^{2}&=\rho \ +\rho _{1}\,\ x+\rho _{2}\ x^{2}+\rho _{3}\ x^{3}+\ldots ,\\\xi ^{3}&=\sigma \ +\sigma _{1}\ x+\sigma _{2}\ x^{2}+\sigma _{3}\ x^{3}+\ldots ,\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$