Page:Annales de mathématiques pures et appliquées, 1819-1820, Tome 10.djvu/42

38

DÉVELOPPEMENT EN SÉRIES

$u=A+{\frac {\operatorname {d} A}{\operatorname {d} a}}.tx+{\frac {\operatorname {d} ^{2}A}{\operatorname {d} a^{2}}}.{\frac {t^{2}x^{2}}{2!}}+{\frac {\operatorname {d} ^{3}A}{\operatorname {d} a^{3}}}.{\frac {t^{3}x^{3}}{3!}}+\ldots$

On a d’ailleurs

${\begin{aligned}{\frac {t^{n-1}}{(n-1)!}}&={\frac {1}{n!}}{\frac {\operatorname {d} t^{n}}{\operatorname {d} a_{1}}},\\{\frac {t^{n-2}}{(n-2)!}}&={\frac {1}{n!}}{\frac {\operatorname {d} ^{2}.t^{n}}{\operatorname {d} a_{1}^{2}}},\\\ldots \ldots &\ldots \ldots ,\\{\frac {t^{n-k}}{(n-k)!}}&={\frac {1}{n!}}{\frac {\operatorname {d} ^{k}.t^{n}}{\operatorname {d} a_{1}^{k}}}\,;\end{aligned}}$

ainsi

$u=A+{\frac {1}{n!}}\left(x{\frac {\operatorname {d} A}{\operatorname {d} a}}.{\frac {\operatorname {d} ^{n-1}.t^{n}}{\operatorname {d} a_{1}^{n-1}}}+x^{2}{\frac {\operatorname {d} ^{2}A}{\operatorname {d} a^{2}}}.{\frac {\operatorname {d} ^{n-2}.t^{n}}{\operatorname {d} a_{1}^{n-2}}}+\ldots +x^{n}{\frac {\operatorname {d} ^{n}A}{\operatorname {d} a^{n}}}.t^{n}\right)$

si donc l’on suppose

$t^{n}=B+B_{1}x+B_{2}x^{2}+B_{3}x^{3}+\ldots$

on aura

$A_{n}={\frac {1}{n!}}\left\{{\frac {\operatorname {d} A}{\operatorname {d} a}}.{\frac {\operatorname {d} ^{n-1}.B_{n-1}}{\operatorname {d} a_{1}^{n-1}}}+{\frac {\operatorname {d} ^{2}A}{\operatorname {d} a^{2}}}.{\frac {\operatorname {d} ^{n-2}.B_{n-2}}{\operatorname {d} a_{1}^{n-2}}}+\ldots +{\frac {\operatorname {d} ^{n}A}{\operatorname {d} a^{n}}}.B\right\}\,;$

ou, en renversant l’ordre des termes,

$A_{n}={\frac {1}{n!}}\left\{{\frac {\operatorname {d} ^{n}A}{\operatorname {d} a^{n}}}.B+{\frac {\operatorname {d} ^{n-1}A}{\operatorname {d} a^{n-1}}}.{\frac {\operatorname {d} B_{1}}{\operatorname {d} a_{1}}}+\ldots {\frac {\operatorname {d} A}{\operatorname {d} a}}.{\frac {\operatorname {d} ^{n-1}.B_{n-1}}{\operatorname {d} a_{1}^{n-1}}}\right\}.$

Mais, en suivant la marche qui nous a conduit aux formules (6) et (7), nous trouverions

$kB_{k}=a_{2}{\frac {\operatorname {d} B_{k-1}}{\operatorname {d} a_{1}}}+2a_{3}{\frac {\operatorname {d} B_{k-1}}{\operatorname {d} a_{2}}}+3a_{4}{\frac {\operatorname {d} B_{k-1}}{\operatorname {d} a_{3}}}+\ldots$

Partant, si l’on fait