Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 8.djvu/451

on aura, cela posé, dans le cas de $\mu ={\frac {3}{2}},$

$(b)={\frac {1}{\left(1+i^{2}\right)^{\frac {3}{2}}}}$

$\times \left\{{\begin{aligned}1&+\left(1-{\frac {1}{4^{2}}}\right)\left({\frac {2i}{1+i^{2}}}\right)^{2}\\&+\left(1-{\frac {1}{4^{2}}}\right)\left(1-{\frac {1}{8^{2}}}\right)\left({\frac {2i}{1+i^{2}}}\right)^{4}\\&+\left(1-{\frac {1}{4^{2}}}\right)\left(1-{\frac {1}{8^{2}}}\right)\left(1-{\frac {1}{12^{2}}}\right)\left({\frac {2i}{1+i^{2}}}\right)^{6}+\ldots \end{aligned}}\right\},$

$(b_{1})={\frac {3i}{\left(1+i^{2}\right)^{\frac {5}{2}}}}$

$\times \left\{{\begin{aligned}1&+\left[1+{\frac {3}{4(3^{2}-1)}}\right]\left({\frac {2i}{1+i^{2}}}\right)^{2}\\&+\left[1+{\frac {3}{4(3^{2}-1)}}\right]\left[1+{\frac {3}{4(5^{2}-1)}}\right]\left({\frac {2i}{1+i^{2}}}\right)^{4}\\&+\left[1+{\frac {3}{4(3^{2}-1)}}\right]\left[1+{\frac {3}{4(5^{2}-1)}}\right]\left[1+{\frac {3}{4(7^{2}-1)}}\right]\left({\frac {2i}{1+i^{2}}}\right)^{6}+\ldots \end{aligned}}\right\},$

${\begin{alignedat}{2}(b_{2})=&{\frac {2(b_{1})\left(1+i^{2}\right)-6(b)i}{i}},\qquad &(b_{3})=&{\frac {4(b_{2})\left(1+i^{2}\right)-5(b_{1})i}{3i}},\\(b_{4})=&{\frac {6(b_{3})\left(1+i^{2}\right)-7(b_{2})i}{5i}},\qquad &(b_{5})=&{\frac {8(b_{4})\left(1+i^{2}\right)-9(b_{3})i}{7i}},\\\ldots &\ldots \ldots \ldots \ldots \ldots ,&\ldots &\ldots \ldots \ldots \ldots \ldots .\end{alignedat}}$

La loi de ces termes est trop visible pour les continuer plus loin ; on aura ensuite

${\begin{aligned}\left({\frac {db}{dq}}\right)\ \,=&{\frac {(b_{1})\left(1+i^{2}\right)^{2}}{4\left(1-i^{2}\right)^{2}}}+{\frac {3(b)i\left(1+i^{2}\right)}{\left(1-i^{2}\right)^{2}}},\\\left({\frac {db_{1}}{dq}}\right)=&{\frac {3(b)\left(1+i^{2}\right)^{2}}{\left(1-i^{2}\right)^{2}}}-{\frac {(b_{1})\left(1+i^{2}\right)\left(1-4i^{2}+i^{4}\right)}{2i\left(1-i^{2}\right)^{2}}},\\\left({\frac {db_{2}}{dq}}\right)=&{\frac {6(b)\left(1+i^{2}\right)\left(1-i^{2}+i^{4}\right)}{i\left(1-i^{2}\right)^{2}}}-{\frac {(b_{1})\left(1+i^{2}\right)^{2}\left(4-9i^{2}+4i^{4}\right)}{2i^{2}\left(1-i^{2}\right)^{2}}}\,;\end{aligned}}$

et les équations (/) donnent, en les différenciant,

${\begin{aligned}\left({\frac {db_{3}}{dq}}\right)=&{\frac {4\left(1+i^{2}\right)\left({\cfrac {db_{2}}{dq}}\right)}{3i}}-{\frac {2(b_{2})\left(1+i^{2}\right)^{2}}{3i^{2}}}-{\frac {5}{3}}\left({\frac {db_{1}}{dq}}\right),\\\left({\frac {db_{4}}{dq}}\right)=&{\frac {6\left(1+i^{2}\right)\left({\cfrac {db_{3}}{dq}}\right)}{5i}}-{\frac {3(b_{3})\left(1+i^{2}\right)^{2}}{5i^{2}}}-{\frac {7}{5}}\left({\frac {db_{2}}{dq}}\right),\\\left({\frac {db_{5}}{dq}}\right)=&{\frac {8\left(1+i^{2}\right)\left({\cfrac {db_{4}}{dq}}\right)}{7i}}-{\frac {4(b_{4})\left(1+i^{2}\right)^{2}}{7i^{2}}}-{\frac {9}{7}}\left({\frac {db_{3}}{dq}}\right),\\\ldots \ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{aligned}}$