Page:Annales de mathématiques pures et appliquées, 1819-1820, Tome 10.djvu/53

49

DES SUITES.

${\frac {\operatorname {d} A_{n-1-k}}{\operatorname {d} a_{n-1}}}=0,$

tant que $k$ ne sera pas nul ; et

$z^{n}=-\left({\frac {\operatorname {d} A_{n-1}}{\operatorname {d} a_{n-1}}}a^{n}+{\frac {\operatorname {d} A_{n}}{\operatorname {d} a_{n-1}}}.{\frac {na^{n+1}}{n+1}}+{\frac {\operatorname {d} A_{n+1}}{\operatorname {d} a_{n-1}}}.{\frac {na^{n+2}}{n+2}}+\ldots \right)\,;$

qui donnera facilement la valeur de $z^{n},$ quand celle de $z$ sera connue.

Souvent on a

$a=b_{1}x+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4}+\ldots \,;$

et alors c’est suivant les puissances de $x$ qu’il faut ordonner le développement de $z.$ Dans ce cas on a

${\frac {\operatorname {d} z}{\operatorname {d} x}}={\frac {\operatorname {d} z}{\operatorname {d} a}}{\frac {\operatorname {d} a}{\operatorname {d} x}}\,;$

ou bien

${\frac {\operatorname {d} z}{\operatorname {d} x}}={\frac {\operatorname {d} z}{\operatorname {d} a}}\left(b_{1}+2b_{2}x+3b_{3}x^{2}+4b_{4}x^{3}+\ldots \right).$

On trouverait de même

$\left.{\begin{aligned}{\frac {\operatorname {d} z}{\operatorname {d} b_{1}}}&=x{\frac {\operatorname {d} z}{\operatorname {d} a}},\\{\frac {\operatorname {d} z}{\operatorname {d} b_{2}}}&=x^{2}{\frac {\operatorname {d} z}{\operatorname {d} a}},\\{\frac {\operatorname {d} z}{\operatorname {d} b_{3}}}&=x^{3}{\frac {\operatorname {d} z}{\operatorname {d} a}},\\\ldots \ldots &\ldots \ldots \ldots ,\\{\frac {\operatorname {d} z}{\operatorname {d} b_{n}}}&=x\ \ {\frac {\operatorname {d} z}{\operatorname {d} a}},\\\end{aligned}}\right\}$(60)

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