![{\displaystyle {\frac {\operatorname {d} S}{\operatorname {d} x}}=2(Ax+Cy+D),\qquad {\frac {\operatorname {d} S}{\operatorname {d} y}}=2(By+Cx+E),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8d8356cb0f940f157274db5a3699a8d5a644148)
![{\displaystyle {\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\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe205b4745dacd71958532c72a368c8b4b4b6506)
et tous les autres coefficiens différentiels seront nuls. S’il s’agit de la fonction de trois variables
![{\displaystyle S=Ax^{2}+By^{2}+Cz^{2}+2Dyz+2Ezx+2Fxy+2Gx+2Hy+2Kz+L,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6af24624d04a4b4508bdd0c49ba587bae63b209c)
on trouvera
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b995bfd9063f9912b9c0b28ca64317217b99de8)
et tous les autres coefficiens différentiels seront encore nuls.
Soit
une fonction donnée quelconque de la seule variable
, et soit
une fonction donnée quelconque de
si
se change en
deviendra (45)
![{\displaystyle 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 \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/184f06e125d5049159a7a61c972b4171976d7ff8)
de sorte qu’en représentant par
son accroissement, on aura
![{\displaystyle 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 \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b9b542c96dc6ac47c5e283e85fa71590ce48fa0)
mais
devenant ainsi
deviendra (45)