Page:Joseph Louis de Lagrange - Œuvres, Tome 4.djvu/532

37. Cela posé, soit

${\begin{aligned}t=&1-{\frac {\theta ^{2}}{6r^{3}}}+{\frac {(n-1)\theta ^{3}}{4(n+1)r^{3}}}p+{\frac {\left(3n^{2}-4n+3\right)\theta ^{4}}{8(n+1)^{2}r^{3}}}\left({\frac {q}{5}}-p^{2}\right)+{\frac {\theta ^{4}}{3.5.8.r^{3}}}+\ldots ,\\s=&1-{\frac {n^{2}\theta ^{2}}{6(n+1)^{2}r^{3}}}+{\frac {n^{3}\theta ^{3}}{4(n+1)^{3}r^{3}}}p+{\frac {n^{4}\theta ^{4}}{8(n+1)^{4}r^{3}}}\left({\frac {3q}{5}}-3p^{2}+{\frac {1}{15r^{3}}}\right)-\ldots ,\\u=&1-{\frac {\theta ^{2}}{6(n+1)^{2}r^{3}}}-{\frac {\theta ^{3}}{4(n+1)^{3}r^{3}}}p+{\frac {\theta ^{4}}{8(n+1)^{4}r^{3}}}\left({\frac {3q}{5}}-3p^{2}+{\frac {1}{15r^{3}}}\right)-\ldots .\end{aligned}}$

Soit ensuite

${\begin{aligned}x=&\ \ n{\text{ϐ}}s\mathrm {\left({\frac {\sin CS}{\sin CD}}\right)} \mathrm {\left(-{\frac {\sin ECS}{\sin ECD}}\right)} -(n+1)\delta t\mathrm {\left({\frac {\sin CT}{\sin CD}}\right)} \mathrm {\left(-{\frac {\sin ECT}{\sin ECD}}\right)} \\&+\lambda u\mathrm {\left({\frac {\sin CV}{\sin CD}}\right)} \mathrm {\left(-{\frac {\sin ECV}{\sin ECD}}\right)} ,\\\\y=&\ \ n{\text{ϐ}}s\mathrm {\left({\frac {\sin DS}{\sin DC}}\right)} \mathrm {\left({\frac {\sin EDS}{\sin EDC}}\right)} -(n+1)\delta t\mathrm {\left({\frac {\sin DT}{\sin DC}}\right)} \mathrm {\left({\frac {\sin EDT}{\sin EDC}}\right)} \\&+\lambda u\mathrm {\left({\frac {\sin DV}{\sin DC}}\right)} \mathrm {\left({\frac {\sin EDV}{\sin EDC}}\right)} ,\\\\z=&\ \ \,n{\text{ϐ}}s\mathrm {\left({\frac {\sin DS}{\sin DE}}\right)} \mathrm {\left({\frac {\sin CDS}{\sin CDE}}\right)} -(n+1)\delta t\mathrm {\left({\frac {\sin DT}{\sin DE}}\right)} \mathrm {\left({\frac {\sin CDT}{\sin CDE}}\right)} \\&+\lambda u\mathrm {\left({\frac {\sin DV}{\sin DE}}\right)} \mathrm {\left({\frac {\sin CDV}{\sin CDE}}\right)} .\end{aligned}}$

Enfin soit

${\begin{aligned}\sigma ^{2}=r^{2}&-{\frac {2r^{2}\theta }{n+1}}p+{\frac {r^{2}\theta ^{2}}{(n+1)^{2}}}q+{\frac {\theta ^{3}}{3(n+1)^{3}r}}p-{\frac {\theta ^{4}}{4(n+1)^{4}r}}\left({\frac {q}{3}}-p^{2}\right)\\&-{\frac {\theta ^{5}}{4(n+1)^{5}r}}\left({\frac {3pq}{5}}+{\frac {p}{15r^{3}}}-p^{3}\right)+\ldots ,\\\\\upsilon ^{2}=r^{2}&+{\frac {2nr^{2}\theta }{n+1}}p+{\frac {n^{2}r^{2}\theta ^{2}}{(n+1)^{2}}}q-{\frac {n^{3}\theta ^{3}}{3(n+1)^{3}r}}p-{\frac {n^{4}\theta ^{4}}{4(n+1)^{4}r}}\left({\frac {q}{3}}-p^{2}\right)\\&+{\frac {n^{5}\theta ^{5}}{4(n+1)^{5}r}}\left({\frac {3pq}{5}}+{\frac {p}{15r^{3}}}-p^{3}\right)+\ldots .\end{aligned}}$

38. On aura ces trois équations

${\begin{aligned}&x^{2}+2(n+1)\delta tx\cos \mathrm {DT} +(n+1)^{2}\left(\delta ^{2}-r^{2}\right)t^{2}=0,\\&y^{2}-2n{\text{ϐ}}sy\cos \mathrm {CS} +n^{2}\left({\text{ϐ}}^{2}-\sigma ^{2}\right)s^{2}=0,\\&z^{2}-2\lambda uz\cos \mathrm {EV} +\left(\lambda ^{2}-\upsilon ^{2}\right)u^{2}=0,\end{aligned}}$

par lesquelles il faudra déterminer les trois inconnues $r,p,q.$