Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 9.djvu/173

et, par l’article IX, on aura

${\frac {\text{ʎ}}{2g}}={\frac {3\Delta }{5\Delta ^{(1)}}}p+{\frac {8\Delta }{35\Delta ^{(1)}}}p^{(1)},\qquad {\frac {{\text{ʎ}}^{(1)}}{2g}}={\frac {\Delta }{3\Delta ^{(1)}}}\,;$

partant,

$p_{1}=p\left(1-{\frac {3\Delta }{5\Delta ^{(1)}}}\right)-{\frac {8\Delta }{35\Delta ^{(1)}}}p^{(1)},\quad {\text{et}}\quad p_{1}^{(1)}=p^{(1)}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)\,;$

les trois équations précédentes donneront ainsi les deux suivantes

${\frac {n-i}{i}}{\frac {g}{n^{2}}}p^{(1)}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)+{\frac {g}{i^{2}}}p\left(1-{\frac {3\Delta }{5\Delta ^{(1)}}}\right)-{\frac {g}{i^{2}}}{\frac {8\Delta }{35\Delta ^{(1)}}}p^{(1)}={\frac {\mathrm {K} }{2i^{2}}}\sin ^{2}\nu ,$

$p={\frac {g}{2n^{2}}}p^{(1)}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)\left({\frac {3i-2n}{i}}l+{\frac {2n-4i}{i}}q\right)$

$+{\frac {gl}{i^{2}}}p^{(1)}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)+{\frac {gp}{i^{2}}}p\left(1-{\frac {3\Delta }{5\Delta ^{(1)}}}\right)-{\frac {8\Delta }{35\Delta ^{(1)}}}{\frac {qg}{i^{2}}}p^{(1)}-{\frac {q\mathrm {K} }{2i^{2}}}\sin ^{2}\nu \,;$

et si, pour abréger, on fait

${\begin{aligned}\mathrm {P} '=&{\frac {i^{2}-ni}{n^{2}}}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)+{\frac {8\Delta }{35\Delta ^{(1)}}},\\\mathrm {Q} '=&\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)\left({\frac {3i^{2}-2ni+2n^{2}}{2n^{2}}}l+{\frac {ni-2i^{2}}{n^{2}}}q\right)-{\frac {8\Delta }{35\Delta ^{(1)}}},\end{aligned}}$

on aura

${\begin{aligned}p\quad =&{\frac {{\cfrac {\mathrm {K} }{2g}}\sin ^{2}\nu (\mathrm {Q} '+q\mathrm {P} ')}{\mathrm {Q} '\left(1-{\cfrac {3\Delta }{5\Delta ^{(1)}}}\right)-\mathrm {P} '\left[{\cfrac {i^{2}}{g}}-q\left(1-{\cfrac {3\Delta }{5\Delta ^{(1)}}}\right)\right]}},\\p^{(1)}=&{\frac {{\cfrac {\mathrm {K} }{2g}}{\frac {i^{2}}{g}}\sin ^{2}\nu }{\mathrm {Q} '\left(1-{\cfrac {3\Delta }{5\Delta ^{(1)}}}\right)-\mathrm {P} '\left[{\cfrac {i^{2}}{g}}-q\left(1-{\cfrac {3\Delta }{5\Delta ^{(1)}}}\right)\right]}}\,;\end{aligned}}$

on déterminera ϐ au moyen de l’équation

${\text{ϐ}}={\frac {g}{2n^{2}}}p^{(1)}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right).$

Si l’on suppose $n$ infiniment petit, on aura

$\mathrm {P} '={\frac {m^{2}}{n^{2}}}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right),\qquad \mathrm {Q} '={\frac {3m^{2}l}{2n^{2}}}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right),$