Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/95

des fonctions données de $a_{1},a_{2},$ dont on trouvera la valeur par les méthodes des Articles précédents.

Donc, si l’on fait ces substitutions dans la quantité ${\frac {1}{\Delta (r_{1},r_{2})^{3}}}$ (Article XVII), et que l’on développe les produits des sinus et des cosinus, on trouvera

${\begin{aligned}&{\frac {1}{\Delta (r_{1},r_{2})^{3}}}\\&=\Gamma (a_{1},a_{2})+\Gamma _{1}(a_{1},a_{2})\cos(\varphi _{2}-\varphi _{1})+\Gamma _{2}(a_{1},a_{2})\cos 2(\varphi _{2}-\varphi _{1})\\&\quad +\Gamma _{3}(a_{1},a_{2})\cos 3(\varphi _{2}-\varphi _{1})+\ldots \\&\quad -3nx_{1}[a_{1}^{2}\Lambda (a_{1},a_{2})-a_{1}a_{2}{\frac {\Lambda _{1}(a_{1},a_{2})}{2}}]\\&\quad -3nx_{1}[a_{1}^{2}\Lambda _{1}(a_{1},a_{2})-a_{1}a_{2}{\frac {2\Lambda (a_{1},a_{2})+\Lambda _{2}(a_{1},a_{2})}{2}}]\cos(\varphi _{2}-\varphi _{1})\\&\quad -3nx_{1}[a_{1}^{2}\Lambda _{2}(a_{1},a_{2})-a_{1}a_{2}{\frac {2\Lambda _{1}(a_{1},a_{2})+\Lambda _{3}(a_{1},a_{2})}{2}}]\cos 2(\varphi _{2}-\varphi _{1})\\&\quad -\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\&\quad -3nx_{2}[a_{2}^{2}\Lambda (a_{1},a_{2})-a_{1}a_{2}{\frac {\Lambda _{1}(a_{1},a_{2})}{2}}]\\&\quad -3nx_{2}[a_{2}^{2}\Lambda _{1}(a_{1},a_{2})-a_{1}a_{2}{\frac {2\Lambda (a_{1},a_{2})+\Lambda _{2}(a_{1},a_{2})}{2}}]\cos(\varphi _{2}-\varphi _{1})\\&\quad -3nx_{2}[a_{2}^{2}\Lambda _{2}(a_{1},a_{2})-a_{1}a_{2}{\frac {2\Lambda _{1}(a_{1},a_{2})+\Lambda _{3}(a_{1},a_{2})}{2}}]\cos 2(\varphi _{2}-\varphi _{1})\\&\quad -\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}$

XXII.

Soit fait, pour plus de simplicité,

${\begin{aligned}&\Pi \ (a_{1},a_{2})={\frac {a_{1}a_{2}\Lambda _{1}(a_{1},a_{2})-2a_{1}^{2}\Lambda (a_{1},a_{2})}{2}},\\&\Pi _{1}(a_{1},a_{2})={\frac {a_{1}a_{2}\Lambda _{2}(a_{1},a_{2})-2a_{1}^{2}\Lambda _{1}(a_{1},a_{2})+2a_{1}a_{2}\Lambda (a_{1},a_{2})}{2}},\\&\Pi _{2}(a_{1},a_{2})={\frac {a_{1}a_{2}\Lambda _{3}(a_{1},a_{2})-2a_{1}^{2}\Lambda _{2}(a_{1},a_{2})+a_{1}a_{2}\Lambda _{1}(a_{1},a_{2})}{2}},\\&\Pi _{3}(a_{1},a_{2})={\frac {a_{1}a_{2}\Lambda _{4}(a_{1},a_{2})-2a_{1}^{2}\Lambda _{3}(a_{1},a_{2})+a_{1}a_{2}\Lambda _{2}(a_{1},a_{2})}{2}},\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}$