Page:Lagrange - Œuvres (1867) vol. 1.djvu/602

et prenant les différences logarithmiques,

${\frac {d\rho }{\rho }}={\frac {\sin \varphi }{1-\cos \varphi }}d\varphi \,;$

donc

${\frac {1}{2}}\rho {\frac {\mathrm {P} }{\rho }}={\frac {(1-\cos \varphi )\cos(n+1)\varphi }{2\sin ^{2}\varphi }}={\frac {\cos(n+1)\varphi }{2(1+\cos \varphi )}}.$

Maintenant on a $\varphi =0,$ lorsque $\rho =0,$ et, par conséquent

$\lambda ''=2\lambda ',\qquad \lambda '''=3\lambda ',\ldots \,;$

donc

${\begin{aligned}\chi =&\lambda '\left[\nu '+2\nu ''+3\nu '''+\ldots +n\nu ^{(n)}\right]\\=&{\frac {\nu '\lambda '}{\sin \varphi }}\left(\sin \varphi +2\sin 2\varphi +3\sin 3\varphi +\ldots +n\sin n\varphi \right)\\=&{\frac {\nu '\lambda '}{\sin \varphi }}{\frac {(n+1)\sin n\varphi -n\sin(n+1)\varphi }{2(1-\cos \varphi )}},\end{aligned}}$

et, à cause de $\sin(n+1)\varphi =0,$

$\chi =\nu '\lambda '{\frac {(n+1)\sin n\varphi }{2\sin \varphi (1-\cos \varphi )}}\,;$

donc

$\mathrm {Q} =-{\frac {1}{2}}\chi \rho {\frac {d\mathrm {P} }{d\rho }}=-{\frac {(n+1)\sin \varphi \cos(n+1)\varphi }{4\sin ^{2}\varphi }}\nu '\lambda ',$

et, à cause de $\sin(2n+1)\varphi =-\sin \varphi ,$

$\mathrm {Q} =-{\frac {(n+1)\left[\sin(2n+1)\varphi -\sin \varphi \right]}{4\sin ^{3}\varphi }}\nu '\lambda '={\frac {n+1}{2\sin ^{2}\varphi }}\nu '\lambda '.$

Donc, faisant ces substitutions dans l’expression de $y'$ (n^o 30), et supposant en général

$\mathrm {Y} _{m}=\mathrm {Y} '\sin {\frac {m\pi }{n+1}}+\mathrm {Y} ''\sin {\frac {2m\pi }{n+1}}+\mathrm {Y} '''\sin {\frac {3m\pi }{n+1}}+\ldots +\mathrm {Y} ^{(n)}\sin {\frac {nm\pi }{n+1}}$