Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/29

sances de $z,$ il est facile de voir qu’elles seront exprimées de cette manière

${\begin{aligned}\mathrm {P} &=\mathrm {A} z,\\\mathrm {Q} &=\mathrm {B} z,\\\mathrm {R} &=\mathrm {C} z+\mathrm {C} _{1}z^{2},\\\mathrm {S} &=\mathrm {D} z+\mathrm {D} _{1}z^{2},\\\mathrm {T} &=\mathrm {E} z+\mathrm {E} _{1}z^{2}+\mathrm {E} _{2}z^{2},\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

et l’on trouvera

${\begin{alignedat}{3}\mathrm {A} =&ac,\\\mathrm {B} =&a^{2}d,\\\mathrm {C} =&a^{3}e,&\mathrm {C} _{1}&=ac\mathrm {A} ,\\\mathrm {D} =&a^{4}f,&\mathrm {D} _{1}&=a^{2}d\mathrm {A} +ac\mathrm {B} ,\\\mathrm {E} =&a^{5}g,\qquad &\mathrm {E} _{1}&=a^{3}e\mathrm {A} +a^{2}d\mathrm {B} +ac\mathrm {C} ,\qquad &\mathrm {E} _{2}&=ac\mathrm {C} _{1},\\\ldots &\ldots ,&\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots ,&\ldots &\ldots \ldots .\end{alignedat}}$

Donc, on aura,

${\begin{aligned}{\frac {\psi '(y)}{z\left[1-{\cfrac {z\varphi (\alpha y)}{\alpha }}\right]}}=&{\frac {\psi '(y)}{z}}+{\frac {\mathrm {A} }{b^{2}}}y^{2}\psi '(y)-{\frac {\mathrm {B} }{b^{3}}}y^{3}\psi '(y)+{\frac {\mathrm {C} +\mathrm {C} _{1}z}{b^{4}}}y^{4}\psi '(y)\\&-{\frac {\mathrm {D} +\mathrm {D} _{1}z}{b^{5}}}y^{5}\psi '(y)+{\frac {\mathrm {E} +\mathrm {E} _{1}z+\mathrm {E} _{2}z^{2}}{b^{6}}}y^{6}\psi '(y)-\ldots .\end{aligned}}$

Donc, changeant ${\frac {1}{z}}$ en $\int dy,$ $z$ en ${\frac {1}{2}}{\frac {d}{dy}},\ldots$ (18), on aura

${\begin{aligned}\psi \left({\frac {bx}{a}}\right)=&\psi (y)\\&+{\frac {\mathrm {A} }{b^{2}}}y^{2}\psi '(y)\\&-{\frac {\mathrm {B} }{b^{3}}}y^{3}\psi '(y)\\&+{\frac {\mathrm {C} }{b^{4}}}y^{4}\psi '(y)+{\frac {\mathrm {C} _{1}}{b^{4}}}.{\frac {1}{2}}{\frac {d[y^{4}\psi '(y)]}{dy}}\\&+{\frac {\mathrm {D} }{b^{5}}}y^{5}\psi '(y)+{\frac {\mathrm {D} _{1}}{b^{5}}}.{\frac {1}{2}}{\frac {d[y^{5}\psi '(y)]}{dy}}\\&+{\frac {\mathrm {E} }{b^{6}}}y^{6}\psi '(y)+{\frac {\mathrm {E} _{1}}{b^{6}}}.{\frac {1}{2}}{\frac {d[y^{6}\psi '(y)]}{dy}}+{\frac {\mathrm {E} _{2}}{b^{6}}}.{\frac {1}{2.3}}{\frac {d^{2}[y^{6}\psi '(y)]}{dy^{2}}}\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

où il n’y aura plus qu’à faire $y=1,$ après avoir exécuté les différentiations qui ne sont qu’indiquées.