Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/209

ce qui se réduit à

$(1,2)={\breve {\Psi }}_{1}(a_{1},a_{2})+2{\breve {\Gamma }}_{1}(a_{1},a_{2})-2{\widehat {\Psi }}_{1}(a_{1},a_{2})-4{\widehat {\Gamma }}_{1}(a_{1},a_{2}).$

On trouvera de même

${\begin{aligned}(1,3)=&{\breve {\Psi }}_{1}(a_{1},a_{3})+2{\widehat {\Gamma }}_{1}(a_{1},a_{3})-2{\widehat {\Psi }}_{1}(a_{1},a_{3})-4{\widehat {\Gamma }}_{1}(a_{1},a_{3}),\\(2,1)=&{\breve {\Psi }}_{1}(a_{2},a_{1})+2{\widehat {\Gamma }}_{1}(a_{2},a_{1})-2{\widehat {\Psi }}_{1}(a_{2},a_{1})-4{\widehat {\Gamma }}_{1}(a_{2},a_{1})\,;\end{aligned}}$

et ainsi des autres.

Or nous avons déjà donné les valeurs des quantités ${\breve {\Gamma }}_{1}$ et ${\widehat {\Gamma }}_{1}$ (Article XLVII) ; il ne reste plus qu’à chercher celles de ${\breve {\Psi }}_{1}$ et de ${\widehat {\Psi }}_{1}.$

Pour cela, il faut auparavant chercher les valeurs des quantités $\Psi ,\Psi _{1}$ et $\Psi _{2}.$ Or, en faisant $q={\frac {a_{1}}{a_{2}}},$ on trouve (Article XXII), à cause de $\Lambda (a_{1},a_{2})=a_{2}^{-5}(\mathrm {A} ),$ $\Lambda _{1}(a_{1},a_{2})=a_{2}^{-5}(\mathrm {B} ),\ldots$ (Article LXXX),

${\begin{aligned}\Psi \ (a_{1},a_{2})=&{\frac {q(\mathrm {B} )-2(\mathrm {A} )}{2a_{2}^{3}}},\\\Psi _{1}(a_{1},a_{2})=&{\frac {q(\mathrm {C} )-2(\mathrm {B} )+2q(\mathrm {A} )}{2a_{2}^{3}}},\\\Psi _{2}(a_{1},a_{2})=&{\frac {q(\mathrm {D} )-2(\mathrm {C} )+q(\mathrm {B} )}{2a_{2}^{3}}},\end{aligned}}$

et de même, à cause de $\Lambda (a_{2},a_{1})=\Lambda (a_{1},a_{2}),\ldots ,$ (Article LXXX),

${\begin{aligned}\Psi \ (a_{2},a_{1})=&{\frac {q(\mathrm {B} )-2q^{2}(\mathrm {A} )}{2a_{2}^{3}}},\\\Psi _{1}(a_{2},a_{1})=&{\frac {q(\mathrm {C} )-2q^{2}(\mathrm {B} )+2q(\mathrm {A} )}{2a_{2}^{3}}},\\\Psi _{2}(a_{2},a_{1})=&{\frac {q(\mathrm {D} )-2q^{2}(\mathrm {C} )+q(\mathrm {B} )}{2a_{2}^{3}}},\end{aligned}}$

Donc (Article LXXX)

${\begin{alignedat}{2}\Psi (a_{1},a_{2})=\Pi (a_{2},a_{1}),\qquad &\Psi _{1}(a_{1},a_{2})=\Pi _{1}(a_{2},a_{1}),\ldots ,\\\Psi (a_{2},a_{1})=\Pi (a_{1},a_{2}),\qquad &\Psi _{1}(a_{2},a_{1})=\Pi _{1}(a_{1},a_{2}),\ldots ,\end{alignedat}}$

c’est-à-dire, que les quantités $\Psi$ sont les réciproques des quantités $\Pi .$