Page:Annales de mathématiques pures et appliquées, 1821-1822, Tome 12.djvu/316

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DÉVELOPPEMENT

$\Gamma u=au+a_{1}{\frac {\operatorname {E} }{x}}u+a_{2}\left({\frac {\operatorname {E} }{x}}\right)^{2}u+\ldots +a_{m}\left({\frac {\operatorname {E} }{x}}\right)^{m}u\,;$

on aura

$\bigtriangledown u=\Gamma u\,;$

et, comme les caractéristiques $\bigtriangledown ,\Gamma$ sont alors de nature distributive, l’on en déduirait un développement de $u.$

L’on aurait encore

$\bigtriangledown u=\Gamma u,$

et l’on arriverait à un autre développement de la même fonction, en posant

$\bigtriangledown u=au+a_{1}{\frac {\operatorname {d} u}{\operatorname {d} y}}+a_{2}{\frac {\operatorname {d} ^{2}u}{\operatorname {d} y^{2}}}+\ldots +a_{m}{\frac {\operatorname {d} ^{m}u}{\operatorname {d} y^{m}}},$

$\Gamma u=au+a_{1}{\frac {\operatorname {d} u}{\operatorname {d} x}}+a_{2}{\frac {\operatorname {d} ^{2}u}{\operatorname {d} x^{2}}}+\ldots +a_{m}{\frac {\operatorname {d} ^{m}u}{\operatorname {d} x^{m}}}.$

L’on parviendrait encore à un nouveau développement, en posant

${\begin{aligned}&\bigtriangledown u=au+a_{1}{\frac {\operatorname {E} }{y}}u+a_{2}\left({\frac {\operatorname {E} }{y}}\right)^{2}u+\ldots +a_{m}\left({\frac {\operatorname {E} }{y}}\right)^{m}u\\+&b{\frac {\operatorname {d} u}{\operatorname {d} y}}+b_{1}{\frac {\operatorname {E} }{y}}{\frac {\operatorname {d} u}{\operatorname {d} y}}+b_{2}\left({\frac {\operatorname {E} }{y}}\right)^{2}{\frac {\operatorname {d} u}{\operatorname {d} y}}+\ldots +b_{m}\left({\frac {\operatorname {E} }{y}}\right)^{m}{\frac {\operatorname {d} u}{\operatorname {d} y}}\\+&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\+&p{\frac {\operatorname {d} ^{n}u}{\operatorname {d} y^{n}}}+p_{1}{\frac {\operatorname {E} }{y}}{\frac {\operatorname {d} ^{n}u}{\operatorname {d} y^{n}}}+p_{n}\left({\frac {\operatorname {E} }{y}}\right)^{2}{\frac {\operatorname {d} ^{n}u}{\operatorname {d} y^{n}}}+\ldots +p_{m}\left({\frac {\operatorname {E} }{y}}\right)^{m}{\frac {\operatorname {d} ^{n}u}{\operatorname {d} y^{n}}},\end{aligned}}$