Page:Annales de mathématiques pures et appliquées, 1819-1820, Tome 10.djvu/51

47

DES SUITES.

$\left.{\begin{array}{lll}&k{\frac {\operatorname {d} z}{\operatorname {d} a}}-1&=0,\\&k{\frac {\operatorname {d} z}{\operatorname {d} a_{1}}}+z&=0,\\&k{\frac {\operatorname {d} z}{\operatorname {d} a_{2}}}+z^{2}&=0,\\&\ldots \ldots \ldots &\ldots \\&k{\frac {\operatorname {d} z}{\operatorname {d} a_{n}}}+z^{n}&=0.\end{array}}\right\}$(2)

Éliminant $k$ entre la première des équations (2) et chacune des suivantes, on trouve

$\left.{\begin{aligned}{\frac {\operatorname {d} z}{\operatorname {d} a_{1}}}+z\ \ {\frac {\operatorname {d} z}{\operatorname {d} a}}&=0,\\{\frac {\operatorname {d} z}{\operatorname {d} a_{2}}}+z^{2}{\frac {\operatorname {d} z}{\operatorname {d} a}}&=0,\\\ldots \ldots \ldots &\ldots \\{\frac {\operatorname {d} z}{\operatorname {d} a_{n}}}+z^{n}{\frac {\operatorname {d} z}{\operatorname {d} a}}&=0.\end{aligned}}\right\}$(3)

D’ailleurs, en remettant pour $k$ sa valeur dans la première des équations (2), elle deviendra

$1=a_{1}{\frac {\operatorname {d} z}{\operatorname {d} a}}+2a_{2}z{\frac {\operatorname {d} z}{\operatorname {d} a}}+3a_{3}z^{2}{\frac {\operatorname {d} z}{\operatorname {d} a}}+4a_{4}z^{3}{\frac {\operatorname {d} z}{\operatorname {d} a}}+\ldots$

ou bien, en vertu des équations (3),

$a_{1}{\frac {\operatorname {d} z}{\operatorname {d} a}}=1+2a_{2}{\frac {\operatorname {d} z}{\operatorname {d} a_{1}}}+3a_{3}{\frac {\operatorname {d} z}{\operatorname {d} a_{2}}}+4a_{4}{\frac {\operatorname {d} z}{\operatorname {d} a_{3}}}+\ldots \qquad$(4)

Supposons