Page:Annales de mathématiques pures et appliquées, 1829-1830, Tome 20.djvu/275

${\frac {\operatorname {d} S}{\operatorname {d} x}}=2(Ax+Cy+D),\qquad {\frac {\operatorname {d} S}{\operatorname {d} y}}=2(By+Cx+E),$

${\frac {\operatorname {d} ^{2}S}{\operatorname {d} x^{2}}}=2A,\qquad {\frac {\operatorname {d} ^{2}S}{\operatorname {d} x\operatorname {d} y}}=2C,\qquad {\frac {\operatorname {d} ^{2}S}{\operatorname {d} y^{2}}}=2B\,;$

et tous les autres coefficiens différentiels seront nuls. S’il s’agit de la fonction de trois variables

$S=Ax^{2}+By^{2}+Cz^{2}+2Dyz+2Ezx+2Fxy+2Gx+2Hy+2Kz+L,$

on trouvera

${\begin{alignedat}{3}{\frac {\operatorname {d} S}{\operatorname {d} x}}=2(Ax+Ez+Fy+G),&\qquad {\frac {\operatorname {d} ^{2}S}{\operatorname {d} x^{2}}}=2A,&\qquad {\frac {\operatorname {d} ^{2}S}{\operatorname {d} y\operatorname {d} z}}=2D,\\\\{\frac {\operatorname {d} S}{\operatorname {d} y}}=2(By+Fx+Dz+H),&\qquad {\frac {\operatorname {d} ^{2}S}{\operatorname {d} y^{2}}}=2B,&\qquad {\frac {\operatorname {d} ^{2}S}{\operatorname {d} z\operatorname {d} x}}=2E,\\\\{\frac {\operatorname {d} S}{\operatorname {d} z}}=2(Cz+Dy+Ex+K),&\qquad {\frac {\operatorname {d} ^{2}S}{\operatorname {d} z^{2}}}=2C,&\qquad {\frac {\operatorname {d} ^{2}S}{\operatorname {d} x\operatorname {d} y}}=2F\,;\end{alignedat}}$

et tous les autres coefficiens différentiels seront encore nuls.

Soit $P$ une fonction donnée quelconque de la seule variable $x$, et soit $S$ une fonction donnée quelconque de $P\,;$ si $x$ se change en$x+g,\ P$ deviendra (45)

$P+{\frac {\operatorname {d} P}{\operatorname {d} x}}{\frac {g}{1}}+{\frac {\operatorname {d} ^{2}P}{\operatorname {d} x^{2}}}{\frac {g^{2}}{1.2}}+{\frac {\operatorname {d} ^{3}P}{\operatorname {d} x^{3}}}{\frac {g^{3}}{1.2.3}}+\ldots \,;$

de sorte qu’en représentant par $G$ son accroissement, on aura

$G={\frac {\operatorname {d} P}{\operatorname {d} x}}{\frac {g}{1}}+{\frac {\operatorname {d} ^{2}P}{\operatorname {d} x^{2}}}{\frac {g^{2}}{1.2}}+{\frac {\operatorname {d} ^{3}P}{\operatorname {d} x^{3}}}{\frac {g^{3}}{1.2.3}}+\ldots \,;$

mais $P$ devenant ainsi $P+G,\ S$ deviendra (45)