Page:Annales de mathématiques pures et appliquées, 1815-1816, Tome 6.djvu/105

99

DES DÉRIVATIONS.

$(58)\left\{{\begin{aligned}&\zeta ^{2}=a_{1}^{2}+\operatorname {D} .a_{1}^{2}.x+{\tfrac {1}{2}}\operatorname {D} ^{2}.a_{1}^{2}.x^{2}+{\tfrac {1}{6}}\operatorname {D} ^{3}.a_{1}^{2}.x^{3}+\ldots \\&\zeta ^{3}=a_{1}^{3}+\operatorname {D} .a_{1}^{3}.x+{\tfrac {1}{2}}\operatorname {D} ^{2}.a_{1}^{3}.x^{2}+{\tfrac {1}{6}}\operatorname {D} ^{3}.a_{1}^{3}.x^{3}+\ldots \\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\&\zeta ^{n}=a_{1}^{n}+\operatorname {D} .a_{1}^{n}.x+{\tfrac {1}{2}}\operatorname {D} ^{2}.a_{1}^{n}.x^{2}+{\tfrac {1}{6}}\operatorname {D} ^{3}.a_{1}^{n}.x^{3}+\ldots \\\end{aligned}}\right.$

où $a_{1}$ doit être considéré comme un premier terme de polynôme ; c’est-à-dire, qu’on a $\operatorname {D} a_{1}=a_{2},{\tfrac {1}{2}}\operatorname {D} ^{2}a_{1}=a_{3},{\tfrac {1}{6}}\operatorname {D} ^{2}a_{1}=a_{3},{\tfrac {1}{6}}\operatorname {D} ^{3}a_{1}=a_{4},\ldots$ ${\frac {1}{1.2\ldots (n-1)}}\operatorname {D} ^{n-1}a_{1}=a_{n}.$

Substituant ces valeurs dans l’équation. (57), on obtient

(59)$\qquad \phi \left(a+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\ldots \right)=$

$\qquad A+A_{1}x+A_{2}x^{2}\qquad +A_{3}x^{3}+\ldots \qquad \qquad +A_{n}x^{n}+\ldots$

${\begin{aligned}{\begin{aligned}&\left.{\begin{aligned}=\phi a+\operatorname {D} \phi a.a_{1}x+\operatorname {D} \phi a.\operatorname {D} a_{1}&\\+{\tfrac {1}{2}}\operatorname {D} ^{2}\phi a.a_{1}^{2}&\\\end{aligned}}\right\}x^{2}\\\\\\\\\\\\\\\\\\\end{aligned}}{\begin{aligned}\left.{\begin{aligned}&+\operatorname {D} \phi a.{\tfrac {1}{2}}\operatorname {D} ^{2}a_{1}\\&+{\tfrac {1}{2}}\operatorname {D} ^{2}\phi a.\operatorname {D} .a_{1}^{2}\\&+{\tfrac {1}{6}}\operatorname {D} ^{3}\phi a.\operatorname {D} .a_{1}^{3}\\\end{aligned}}\right\}\\\\\\\\\\\\\\\end{aligned}}{\begin{aligned}{\begin{aligned}&\ldots \\\\&x^{3}+\ldots \\\\&\ldots \\\end{aligned}}\\\\\\\\\\\\\\\end{aligned}}\left.{\begin{aligned}&+\operatorname {D} \phi a.{\tfrac {\operatorname {D} ^{n-1}a_{1}}{1.2\ldots (n-1)}}\\&+{\tfrac {1}{2}}\operatorname {D} ^{2}\phi a.{\tfrac {\operatorname {D} ^{n-2}a_{1}^{2}}{1.2\ldots (n-2)}}\\&+{\tfrac {1}{6}}\operatorname {D} ^{3}\phi a.{\tfrac {\operatorname {D} ^{n-3}a_{1}^{3}}{1.2\ldots (n-3)}}\\&+\ldots \ldots \ldots \\&+{\tfrac {\operatorname {D} ^{n-1}\phi a}{1.2\ldots (n-1)}}\operatorname {D} .a_{1}^{n-1}\\&+{\tfrac {\operatorname {D} ^{n}\phi a}{1.2\ldots n}}.a_{1}^{n}\\\end{aligned}}\right\}{\begin{aligned}&+\ldots \\\\&+\ldots \\\\&x^{n}\\\\&+\ldots \\\\&+\ldots \\\\&+\ldots \\\end{aligned}}\end{aligned}}$

d’où l’on tire

(60) $A_{n}={\frac {\operatorname {D} ^{n}\phi a}{1.2\ldots n}}=\operatorname {D} \phi a.{\frac {\operatorname {D} ^{n-1}a_{1}}{1.2\ldots (n-1)}}+{\tfrac {1}{2}}\operatorname {D} ^{2}\phi a.{\frac {\operatorname {D} ^{n-2}a_{1}^{2}}{1.2\ldots (n-2)}}$

$+{\tfrac {1}{6}}\operatorname {D} ^{3}\phi a.{\frac {\operatorname {D} ^{n-3}a_{1}^{3}}{1.2\ldots (n-3)}}+\ldots +{\frac {\operatorname {D} ^{n-1}\phi a}{1.2\ldots (n-1)}}\operatorname {D} .a_{1}^{n-1}+{\frac {\operatorname {D} ^{n}\phi a}{1.2\ldots n}}.a_{1}^{n}\,;$

ou, en écrivant cette formule à rebours,