Page:Joseph Louis de Lagrange - Œuvres, Tome 5.djvu/423

Enfin on aura par les mêmes formules

${\begin{aligned}&2r{\frac {d[r,r']}{dr}}\ =-[r,r']-\left(r^{2}-r'^{2}\right)(r,r'),\\&2r{\frac {d[r,r']_{1}}{dr}}=-[r,r']_{1}-\left(r^{2}-r'^{2}\right)(r,r')_{1},\\&2r{\frac {d[r,r']_{2}}{dr}}=-[r,r']_{2}-\left(r^{2}-r'^{2}\right)(r,r')_{2},\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}$

4. Or, en faisant $z={\frac {r'}{r}}$ et prenant pour $\mathrm {M}$ et $\mathrm {N}$ les expressions en $z$ du n^o 3 de la seconde Partie de la Théorie citée, on a

$(r,r')={\frac {\mathrm {M} }{r^{3}\left(1-z^{2}\right)^{2}}},\quad (r,r')_{1}={\frac {6\mathrm {N} }{r^{3}\left(1-z^{2}\right)^{2}}}.$

Si donc on suppose

$(0)={\frac {\mathrm {M} }{r^{3}\left(1-z^{2}\right)^{2}}},\quad (1)={\frac {6\mathrm {N} }{r^{3}\left(1-z^{2}\right)^{2}}},$

et ensuite

${\begin{aligned}&(2)=2\left(z+{\frac {1}{z}}\right)(1)-6(0),\\&(3)={\frac {4}{3}}\left(z+{\frac {1}{z}}\right)(2)-{\frac {5}{3}}(1),\\&(4)={\frac {6}{5}}\left(z+{\frac {1}{z}}\right)(3)-{\frac {7}{5}}(2),\\&(5)={\frac {8}{7}}\left(z+{\frac {1}{z}}\right)(4)-{\frac {9}{7}}(3),\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$

on aura

${\begin{aligned}&r[r,r']\,\ =\left(1+z^{2}\right)(0)-z(1),\\&r[r,r']_{1}=4z(0)-\left(1+z^{2}\right)(1),\\&r[r,r']_{2}={\frac {2}{3}}z(1)-{\frac {1}{3}}\left(1+z^{2}\right)(2),\\&r[r,r']_{3}={\frac {2}{5}}z(2)-{\frac {1}{5}}\left(1+z^{2}\right)(3),\\&r[r,r']_{4}={\frac {2}{7}}z(2)-{\frac {1}{7}}\left(1+z^{2}\right)(4),\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$