Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/38

tant $x^{r}$ à la place de $t$ et conservant les mêmes valeurs des coefficients $\mathrm {A,B,B'} ,\ldots ,$

${\begin{aligned}\psi ({\frac {x^{r}}{\rho ^{r}}})=&\psi (y)\\&+\mathrm {A} \rho ^{p-r}y^{\frac {p}{r}}\psi '(y)\\\\&+\mathrm {B} \rho ^{p+q-r}y^{\frac {p+q}{r}}\psi '(y)+\mathrm {B} _{1}\rho ^{2p-2r}.{\frac {1}{2}}{\frac {dy^{\frac {2p}{r}}\psi '(y)}{dy}}\\\\&+\mathrm {C} \rho ^{p+2q-r}y^{\frac {p+2q}{r}}\psi '(y)+\mathrm {C} _{1}\rho ^{2p+q-2r}.{\frac {1}{2}}{\frac {d\left[y^{\frac {2p+q}{r}}\psi '(y)\right]}{dy}}\\&\quad +\mathrm {C} _{2}\rho ^{3p-3r}.{\frac {1}{2.3}}{\frac {d^{2}\left[y^{\frac {3p}{r}}\psi '(y)\right]}{dy^{2}}}\\\\&+\mathrm {D} \rho ^{p+3q-r}y^{\frac {p+3q}{r}}\psi '(y)+\mathrm {D} _{1}\rho ^{2p+2q-2r}.{\frac {1}{2}}{\frac {d\left[y^{\frac {2p+2q}{r}}\psi '(y)\right]}{dy}}\\&\quad +\mathrm {D} _{2}\rho ^{3p+q-3r}.{\frac {1}{2.3}}{\frac {d^{2}\left[y^{\frac {3p+q}{r}}\psi '(y)\right]}{dy^{2}}}+\mathrm {D} _{3}\rho ^{4p-4r}.{\frac {1}{2.3.4}}{\frac {d^{3}\left[y^{\frac {4p}{r}}\psi '(y)\right]}{dy^{3}}}\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

où il faudra faire $y=1$ après les différentiations.

Si l’on suppose $\psi (y)=y^{\frac {m}{r}},$ on aura

$\psi \left({\frac {x^{r}}{\rho ^{r}}}\right)={\frac {x^{m}}{\rho ^{m}}}\,;$

donc on trouvera

${\begin{aligned}\left({\frac {x}{\rho }}\right)^{m}=&1\\&+{\frac {m}{r}}\mathrm {A} \rho ^{p-r}\\\\&+{\frac {m}{r}}\mathrm {B} \rho ^{p+q-r}+{\frac {m(m+2p-r)}{2r^{2}}}\mathrm {B} _{1}\rho ^{2p-2r}\\\\&+{\frac {m}{r}}\mathrm {C} \rho ^{p+2q-r}+{\frac {m(m+2p+q-r)}{2r^{2}}}\mathrm {C} _{1}\rho ^{2p+q-2r}\\&\quad +{\frac {m(m+3p-r)(m+3p-2r)}{2.3.r^{3}}}\mathrm {C} _{2}\rho ^{3p-3r}\end{aligned}}$