Page:Lagrange - Œuvres (1867) vol. 1.djvu/716

${\begin{aligned}\mathrm {q} &=3\left(\mathrm {a} {\frac {2{\mathfrak {P+R}}}{2}}-\mathrm {a} ^{2}{\mathfrak {Q}}\right),\\\mathrm {r} &=3\left(\mathrm {a} {\frac {\mathfrak {Q+S}}{2}}-\mathrm {a} ^{2}{\mathfrak {R}}\right),\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;\\\mathrm {p} _{1}&=3\left(\mathrm {a} {\frac {\mathfrak {Q}}{2}}-{\mathfrak {P}}\right),\\\mathrm {q} _{1}&=3\left(\mathrm {a} {\frac {\mathfrak {2P+R}}{2}}-{\mathfrak {Q}}\right),\\\mathrm {r} _{1}&=3\left(\mathrm {a} {\frac {\mathfrak {Q+S}}{2}}-{\mathfrak {R}}\right),\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

on aura

${\begin{alignedat}{3}{\mathfrak {P}}_{2}=&{\frac {\mathrm {p} }{a'^{3}}},&{\mathfrak {Q}}_{2}=&{\frac {\mathrm {q} }{a'^{3}}},&{\mathfrak {R}}_{2}=&{\frac {\mathrm {r} }{a'^{3}}},\ldots ,\\{\mathfrak {P}}_{3}=&{\frac {\mathrm {p} }{a'^{3}}},&{\mathfrak {Q}}_{3}=&{\frac {\mathrm {q} }{a'^{3}}},&{\mathfrak {R}}_{3}=&{\frac {\mathrm {r} }{a'^{3}}},\ldots .\end{alignedat}}$

De là on trouvera, par les formules du n^o 75,

${\begin{aligned}{\mathfrak {P}}_{6}&=\mathrm {a} ^{3}({\mathfrak {A}}+\mathrm {p} )-\mathrm {a^{2}{\frac {q}{2}}} ,\\{\mathfrak {P}}_{7}&=\mathrm {a} ^{2}\left({\frac {2{\mathfrak {A-C}}}{2}}+{\frac {2\mathrm {p_{1}-r_{1}} }{2}}\right)+2\mathrm {a} ^{2}\\\end{aligned}}$

et

${\mathfrak {Q}}_{5}=\mathrm {a^{3}q_{1}-a^{2}} \left({\frac {2{\mathfrak {A+C}}}{2}}+{\frac {2\mathrm {p_{1}+r_{1}} }{2}}\right)-2\mathrm {a} ^{2}.$

On trouvera de même les autres quantités ${\mathfrak {A'_{2},B'_{3}}},\ldots \,;$ il n’y aura pour cela qu’à mettre, dans les formules des numéros cités, $a'$ au lieu de $a$ et $a$ au lieu de $a',$ et marquer ensuite toutes les autres lettres d’un trait, ce qui donnera, après les substitutions,

${\begin{aligned}{\mathfrak {A}}'_{2}&={\mathfrak {A}}-\mathrm {a} {\frac {\mathfrak {B}}{2}},\\{\mathfrak {A}}'_{3}&=\mathrm {a} {\frac {2{\mathfrak {A-C}}}{2}}-{\frac {1}{\mathrm {a} ^{2}}},\\{\mathfrak {A}}'_{4}&={\mathfrak {A-A'_{2}}}=\mathrm {a} {\frac {\mathfrak {B}}{2}},\\{\mathfrak {B}}'_{2}&={\mathfrak {B}}-\mathrm {a} {\frac {2{\mathfrak {A+C}}}{2}}+{\frac {1}{\mathrm {a} ^{2}}},\\{\mathfrak {B}}'_{5}&=-\mathrm {a} {\mathfrak {B}},\end{aligned}}$