Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 9.djvu/174

d’où l’on tirera

${\begin{aligned}p\quad =&{\frac {3l\mathrm {K} \sin \nu }{6lg\left(1-{\cfrac {3\Delta }{5\Delta ^{(1)}}}\right)-4m^{2}}},\\p^{(1)}=&{\frac {2\mathrm {K} {\cfrac {n^{2}}{g}}\sin \nu }{\left(1-{\cfrac {\Delta }{3\Delta ^{(1)}}}\right)\left[6lg\left(1-{\cfrac {3\Delta }{5\Delta ^{(1)}}}\right)-4m^{2}\right]}}\,;\end{aligned}}$

partant,

${\text{ϐ}}={\frac {\mathrm {K} \sin \nu }{6lg\left(1-{\cfrac {3\Delta }{5\Delta ^{(1)}}}\right)-4m^{2}}},$

ce qui est conforme à ce que nous avons trouvé (art. XI). Si l’on fait $i=n,$ on aura

$\mathrm {P} '={\frac {8\Delta }{35\Delta ^{(1)}}},\qquad \mathrm {Q} '=\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)\left({\frac {3}{2}}l-q\right)-{\frac {8\Delta q}{35\Delta ^{(1)}}},$

d’où l’on tirera

${\begin{aligned}p\quad =&{\frac {\left(1-{\cfrac {\Delta }{3\Delta ^{(1)}}}\right)\left({\cfrac {3}{2}}l-q\right){\cfrac {\mathrm {K} }{2g}}\sin ^{2}\nu }{\left(1-{\cfrac {3\Delta }{5\Delta ^{(1)}}}\right)\left(1-{\cfrac {\Delta }{3\Delta ^{(1)}}}\right)\left({\cfrac {3}{2}}l-q\right)-{\cfrac {8\Delta }{35\Delta ^{(1)}}}{\cfrac {n^{2}}{g}}}},\\p^{(1)}=&{\frac {{\cfrac {\mathrm {K} }{2g}}\sin ^{2}\nu {\cfrac {n^{2}}{g}}}{\left(1-{\cfrac {3\Delta }{5\Delta ^{(1)}}}\right)\left(1-{\cfrac {\Delta }{3\Delta ^{(1)}}}\right)\left({\cfrac {3}{2}}l-q\right)-{\cfrac {8\Delta }{35\Delta ^{(1)}}}{\cfrac {n^{2}}{g}}}}\,;\end{aligned}}$

la valeur de $y$ sera donc, dans le cas où $i=n$ et $r=1,$

${\begin{aligned}y=&{\frac {\mathrm {K} }{6g\left(1-{\cfrac {3\Delta }{5\Delta ^{(1)}}}\right)}}\left(\cos ^{2}\nu -{\frac {1}{2}}\sin ^{2}\nu \right)(1+3\cos 2\theta )\\&-{\frac {{\cfrac {8\mathrm {K} }{g}}\sin \mu \cos \mu \sin \theta \cos \theta }{7-{\cfrac {19\Delta }{15\Delta ^{(1)}}}}}\cos(it+\varpi )\\&+{\frac {\mathrm {K} }{2g}}\sin ^{2}\nu {\frac {\left(1-{\cfrac {\Delta }{3\Delta ^{(1)}}}\right)\left({\cfrac {3}{2}}l-q\right)+{\cfrac {n^{2}}{g}}\sin \theta }{\left(1-{\cfrac {3\Delta }{5\Delta ^{(1)}}}\right)\left(1-{\cfrac {\Delta }{3\Delta ^{(1)}}}\right)\left({\cfrac {3}{2}}l-q\right)-{\cfrac {8\Delta }{35\Delta ^{(1)}}}{\cfrac {n^{2}}{g}}}}\end{aligned}}$