Page:Joseph Louis de Lagrange - Œuvres, Tome 8.djvu/22

et l’on aura, comme l’on sait,

${\begin{aligned}&\mathrm {A_{1}=A} ,\\&\mathrm {A_{2}=AA_{1}-2B} ,\\&\mathrm {A_{3}=AA_{2}-BA_{1}+3C} ,\\&\mathrm {A_{4}=AA_{3}-BA_{2}+CA_{1}-4D} ,\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{aligned}}$

Supposons de plus

${\begin{aligned}&a_{1}=(\alpha -\beta )^{2}+(\alpha -\gamma )^{2}+(\beta -\gamma )^{2}+\ldots ,\\&a_{2}=(\alpha -\beta )^{4}+(\alpha -\gamma )^{4}+(\beta -\gamma )^{4}+\ldots ,\\&a_{3}=(\alpha -\beta )^{6}+(\alpha -\gamma )^{6}+(\beta -\gamma )^{6}+\ldots ,\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;\end{aligned}}$

il est facile de voir que l’on aura

${\begin{aligned}&a_{1}=(m-1)\mathrm {A_{2}-2\left({\frac {A_{1}^{2}-A_{2}}{2}}\right)} ,\\&a_{2}=(m-1)\mathrm {A_{4}-4(A_{1}A_{3}-A_{4})+6\left({\frac {A_{2}^{2}-A_{4}}{2}}\right)} ,\\&a_{3}=(m-1)\mathrm {A_{6}-6(A_{1}A_{5}-A_{6})+15(A_{2}A_{4}-A_{6})-20\left({\frac {A_{3}^{2}-A_{6}}{2}}\right)} ,\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

ou bien

${\begin{aligned}&a_{1}=m\mathrm {A_{2}-2{\frac {A_{1}^{2}}{2}}} ,\\&a_{2}=m\mathrm {A_{4}-4A_{1}A_{3}+6{\frac {A_{2}^{2}}{2}}} ,\\&a_{3}=m\mathrm {A_{6}-6A_{1}A_{5}+15A_{2}A_{4}-20{\frac {A_{3}^{2}}{2}}} ,\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$