Page:Joseph Louis de Lagrange - Œuvres, Tome 8.djvu/262

Donc on aura

${\begin{aligned}\varphi (u)\ \,=&{\frac {\mathrm {P} }{2\cos \omega }}+{\frac {\mathrm {Q} }{2\cos \omega }}u,\\f(u)\ \,=&{\frac {u^{2}}{2\cos \omega }},\\f^{2}(u)=&{\frac {u^{4}}{(2\cos \omega )^{2}}},\\f^{3}(u)=&{\frac {u^{6}}{(2\cos \omega )^{3}}},\\..\ldots \ldots &\ldots \ldots \ldots .\end{aligned}}$

Donc

${\begin{aligned}{\frac {\varphi (u)}{u^{n+1}}}\quad =&{\frac {\mathrm {P} u^{-n-1}}{2\cos \omega }}\,\ \ +\ \ {\frac {\mathrm {Q} u^{-n}}{2\cos \omega }},\\{\frac {\varphi (u)f(u)\ \,}{u^{n+1}}}=&{\frac {\mathrm {P} u^{-n+1}}{(2\cos \omega )^{2}}}+{\frac {\mathrm {Q} u^{-n+2}}{(2\cos \omega )^{2}}},\\{\frac {\varphi (u)f^{2}(u)}{u^{n+1}}}=&{\frac {\mathrm {P} u^{-n+3}}{(2\cos \omega )^{3}}}+{\frac {\mathrm {Q} u^{-n+4}}{(2\cos \omega )^{3}}},\\.\ldots \ldots \ldots \ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}$

En prenant les fonctions dérivées par rapport à $u,$ on aura donc

${\begin{aligned}\left[{\frac {\varphi (u)f(u)\ \,}{u^{n+1}}}\right]'\ =&-{\frac {(n-1)\mathrm {P} u^{-n}}{(2\cos \omega )^{2}}}-{\frac {(n-2)\mathrm {Q} u^{-n+1}}{(2\cos \omega )^{2}}},\\\left[{\frac {\varphi (u)f^{2}(u)}{2u^{n+1}}}\right]''=&{\frac {(n-3)(n-2)\mathrm {P} u^{-n+1}}{2(2\cos \omega )^{3}}}+{\frac {(n-4)(n-3)\mathrm {Q} u^{-n+2}}{2(2\cos \omega )^{3}}},\\.\ldots \ldots \ldots \ldots \ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

et, par conséquent,

${\begin{aligned}(n)&=\mathrm {P} \left[{\frac {u^{-n-1}}{2\cos \omega }}-{\frac {(n-1)u^{-n}}{(2\cos \omega )^{2}}}\ \ \ +{\frac {(n-3)(n-2)u^{-n+1}}{2(2\cos \omega )^{3}}}-\ldots \right]\\&+\mathrm {Q} \left[{\frac {u^{-n}}{2\cos \omega }}-{\frac {(n-2)u^{-n+1}}{(2\cos \omega )^{2}}}+{\frac {(n-4)(n-3)u^{-n+2}}{2(2\cos \omega )^{3}}}-\ldots \right],\end{aligned}}$

où il n’y aura plus qu’a substituer au lieu de $u$ sa valeur ${\frac {1}{2\cos \omega }}.$