Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 9.djvu/170

ce qui donne

$f_{1}=f\left(1-{\frac {3\Delta }{5\Delta ^{(1)}}}\right)-{\frac {16\Delta }{105\Delta ^{(1)}}}f^{(1)},$

$f_{1}^{(1)}=f^{(1)}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)\,;$

partant, on aura les quatre équations

${\begin{aligned}0=&{\frac {2n-i}{i}}e+{\frac {gf}{i^{2}}}\left(1-{\frac {3\Delta }{5\Delta ^{(1)}}}\right)-{\frac {16\Delta }{105\Delta ^{(1)}}}{\frac {gf^{(1)}}{i^{2}}}-{\frac {2\mathrm {K} }{i^{2}}}\sin \nu \cos \nu ,\\\\f=&{\frac {2n-3i}{i}}\left(le^{(1)}+qe\right)+{\frac {gp}{i^{2}}}f\left(1-{\frac {3\Delta }{5\Delta ^{(1)}}}\right)-{\frac {gq}{i^{2}}}{\frac {16\Delta }{105\Delta ^{(1)}}}f^{(1)}\\\\&+{\frac {lg}{i^{2}}}f^{(1)}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)-{\frac {2q\mathrm {K} }{i^{2}}}\sin \nu \cos \nu ,\end{aligned}}$

${\begin{aligned}\left(4n^{2}-i^{2}\right)e^{(1)}-4n^{2}e=&2g{\frac {n+i}{i}}f\left(1-{\frac {3\Delta }{5\Delta ^{(1)}}}\right)-2g{\frac {n+i}{i}}{\frac {16\Delta }{105\Delta ^{(1)}}}f^{(1)}\\\\-{\frac {2n+3i}{i}}&gf^{(1)}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)-4\mathrm {K} {\frac {n+i}{i}}\sin \nu \cos \nu ,\\\\-2n^{2}e^{(1)}=&{\frac {n+2i}{i}}gf^{(1)}\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)\,;\end{aligned}}$

on tirera facilement de ces équations les valeurs de $e,e^{(1)},f$ et $f^{(1)}$ et si, pour abréger, l’on fait

${\begin{aligned}{\frac {n^{2}}{g}}=&\mu ,\\\\\mathrm {P} =&(2n-i)\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)\left[2n^{2}(2n+3i)-(n+2i)\left(4n^{2}-i^{2}\right)\right]\\&+{\frac {64\Delta n^{2}}{105\Delta ^{(1)}}}\left[(n+i)(2n-i)-2n^{2}\right],\\\\\mathrm {Q} =&4n^{2}\left(1-{\frac {3\Delta }{5\Delta ^{(1)}}}\right)\left[(n+i)(2n-i)-2n^{2}\right],\\\\\mathrm {R} =&\left(1-{\frac {\Delta }{3\Delta ^{(1)}}}\right)\left\{(2n-3i)q\left[2n^{2}(2n+3i)-(n+2i)\left(4n^{2}-i^{2}\right)\right]\right.\\&\quad \qquad \qquad \qquad \qquad \qquad \qquad \left.+4n^{2}l\left[2n^{2}-(2n-3i)(n+2i)\right]\right\}\\&+4n^{2}q\left[(2n-3i)(n+i)-2n^{2}\right]{\frac {16\Delta }{105\Delta ^{(1)}}},\\\\\mathrm {S} =&8n^{2}i^{2}\mu +4n^{2}q\left(1-{\frac {3\Delta }{5\Delta ^{(1)}}}\right)\left[(2n-3i)(n+i)-2n^{2}\right],\end{aligned}}$