Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/20

on trouvera, en faisant abstraction des termes qui renfermeraient des puissances positives de $x,$

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

${\begin{aligned}{\frac {\xi ^{2}}{2\left(1-{\cfrac {a}{bx}}\right)^{2}}}=&{\frac {1}{2}}\left(\rho +2\rho _{1}{\frac {a}{b}}+3\rho _{2}{\frac {a^{2}}{b^{2}}}+4\rho _{3}{\frac {a^{3}}{b^{3}}}+\ldots \right)\\&+\ {\frac {1}{2x}}\ \left(2\rho \,{\frac {a}{b}}\ +3\rho _{1}{\frac {a^{2}}{b^{2}}}+4\rho _{2}{\frac {a^{3}}{b^{3}}}+\ldots \right)\\&+{\frac {1}{2x^{2}}}\left(3\rho {\frac {a^{2}}{b^{2}}}+4\rho _{1}{\frac {a^{3}}{b^{3}}}+5\rho _{2}{\frac {a^{4}}{b^{4}}}+\ldots \right)\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\\\\{\frac {\xi ^{3}}{3\left(1-{\cfrac {a}{bx}}\right)^{3}}}=&{\frac {1}{2.3}}\left(1.2.\sigma +2.3.\sigma _{1}{\frac {a}{b}}+3.4.\sigma _{2}{\frac {a^{2}}{b^{2}}}+\ldots \right)\\&+\ {\frac {1}{2.3.x}}\ \left(2.3.\sigma \ {\frac {a}{b}}\,+3.4.\sigma _{1}{\frac {a^{2}}{b^{2}}}+4.5.\sigma _{2}{\frac {a^{3}}{b^{3}}}+\ldots \right)\\&+{\frac {1}{2.3.x^{2}}}\left(3.4.\sigma {\frac {a^{2}}{b^{2}}}+4.5.\sigma _{1}{\frac {a^{3}}{b^{3}}}+5.6.\sigma _{2}{\frac {a^{4}}{b^{4}}}+\ldots \right)\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

et ainsi de suite.

Donc enfin, substituant ces valeurs dans l’équation (E) et comparant les termes affectés de $x^{0},x^{-1},x^{-2},\ldots ,$ à cause de $\log {\frac {bp}{a}}=\log p-\log {\frac {a}{b}},$ on trouvera

${\begin{aligned}\log p=&\log {\frac {a}{b}}+\varpi +\varpi _{1}{\frac {a}{b}}+\varpi _{2}{\frac {a^{2}}{b^{2}}}+\varpi _{3}{\frac {a^{3}}{b^{3}}}+\ldots \\&{\frac {1}{2}}\left(\rho +2\rho _{1}{\frac {a}{b}}+3\rho _{2}{\frac {a^{2}}{b^{2}}}+4\rho _{3}{\frac {a^{3}}{b^{3}}}+\ldots \right)\end{aligned}}$