Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/17

donc

${\begin{aligned}\mathrm {X} ^{2}&={\frac {a^{2}}{x^{2}}}+2ac+c^{2}x^{2},\\\mathrm {X} ^{3}&={\frac {a^{3}}{x^{3}}}+{\frac {3a^{2}c}{x}}+3ac^{2}x+c^{3}x^{3},\\\mathrm {X} ^{4}&={\frac {a^{4}}{x^{4}}}+{\frac {4a^{3}c}{x^{2}}}+6a^{2}c^{2}+4ac^{3}x^{2}+c^{4}x^{4},\end{aligned}}$

et ainsi de suite : donc

${\begin{aligned}\alpha =&{\frac {2ac}{2b^{2}}}+{\frac {4.3a^{2}c^{2}}{2.4.b^{4}}}+{\frac {6.5.4.a^{3}c^{3}}{2.3.6.b^{6}}}+\ldots ,\\\beta =&{\frac {a}{b}}+{\frac {3a^{2}c}{3b^{3}}}+{\frac {5.4.a^{3}c^{2}}{2.5.b^{5}}}+{\frac {7.6.5.a^{4}c^{3}}{2.3.7.b^{7}}}+\ldots ,\\\gamma =&{\frac {a^{2}}{2b^{2}}}+{\frac {4a^{3}c}{4b^{4}}}+{\frac {6.5.a^{4}c^{2}}{2.6.b^{6}}}+{\frac {8.7.6.a^{6}c^{3}}{2.3.8.b^{8}}}+\ldots ,\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

et par conséquent, en mettant $x$ à la place de $p,$

${\begin{aligned}\log x&=\log {\frac {a}{b}}+{\frac {ac}{b^{2}}}+{\frac {3a^{2}c^{2}}{2.b^{4}}}+{\frac {5.4.a^{3}c^{3}}{2.3.b^{6}}}+\ldots ,\\x\ \ &={\frac {a}{b}}+{\frac {a^{2}c}{b^{3}}}+{\frac {4a^{3}c^{2}}{2b^{5}}}+{\frac {6.5.a^{4}c^{3}}{2.3.b^{7}}}+\ldots ,\\{\frac {x^{2}}{2}}&={\frac {a^{2}}{2b^{2}}}+{\frac {a^{3}c}{b^{4}}}+{\frac {5a^{4}c^{2}}{2b^{6}}}+{\frac {7.6.a^{6}c^{3}}{2.3.b^{8}}}+\ldots ,\\{\frac {x^{3}}{3}}&={\frac {a^{3}}{3b^{3}}}+{\frac {a^{4}c}{b^{5}}}+{\frac {6a^{5}c^{2}}{2b^{7}}}+{\frac {8.7.a^{6}c^{3}}{2.3.b^{9}}}+\ldots ,\end{aligned}}$

et ainsi de suite ; de sorte qu’on aura, en général,

${\frac {x^{m}}{m}}={\frac {a^{m}}{mb^{m}}}+{\frac {a^{m+1}c}{b^{m+2}}}+{\frac {(m+3)a^{m+2}c^{2}}{2b^{m+4}}}+{\frac {(m+5)(m+4)a^{m+3}c^{m}}{2.3.b^{m+6}}}+\ldots .$

En effet, cette équation, étant résolue, donne

$x={\frac {b}{2c}}\pm {\frac {\sqrt {b^{2}-4ac}}{2c}}\,;$