Formule de Taylor

Bonjour,

Il y a une erreur dans les factorielles.
Tu as calculé les termes :
\[\frac{\partial_x^{\alpha}V(x_0)}{\vert\alpha\rvert!}(x-x_0)^{\alpha} = \frac{\partial_{x_1}^{\alpha_1}\partial_{x_2}^{\alpha_2}V(x_0)}{(\alpha_1+\alpha_2)!}(x-x_0)^{\alpha}\]
au lieu de :
\[\frac{\partial_x^{\alpha}V(x_0)}{\alpha!}(x-x_0)^{\alpha} = \frac{\partial_{x_1}^{\alpha_1}\partial_{x_2}^{\alpha_2}V(x_0)}{\alpha_1!\alpha_2!}(x-x_0)^{\alpha}\]

\begin{align}
V(x)&=V(x_{1,0},x_{2,0})+
\Bigl[\partial_{x_1}V(x_{1,0},x_{2,0})(x_1-x_{1,0})+
\partial_{x_2}V(x_{1,0},x_{0,2})(x_2- x_{2,0})\Bigr] \\
&\quad+\left[\frac{\partial_{x_1}^2V(x_{1,0},x_{2,0})}{2!}(x_1-x_{1,0})^2+
\frac{\partial_{x_2}^2V(x_{1,0},x_{2,0})}{2!}(x_2-x_{2,0})^2\right. \\
&\qquad+\left.\frac{\partial_{x_1}\partial_{x_2}V(x_{1,0},x_{2,0})}{1!1!}(x_1-x_{1,0})(x_2-x_{2,0})\right] \\
&\quad+\left[\frac{\partial_{x_1}^3V(x_{1,0},x_{2,0})}{3!0!}(x_1-x_{1,0})^3+
\frac{\partial_{x_2}^3V(x_{1,0},x_{2,0})}{0!3!}(x_2-x_{2,0})^3\right. \\
&\qquad+\left.\frac{\partial_{x_1}^2\partial_{x_2}V(x_{1,0},x_{2,0})}{2!1!}(x_1-x_{1,0})^2(x_2-x_{2,0})+
\frac{\partial_{x_1}\partial_{x_2}^2V(x_{1,0},x_{2,0})}{1!2!}(x_1-x_{1,0})(x_2-x_{2,0})^2\right] \\
&\quad+\left[\frac{\partial_{x_1}^4V(x_{1,0},x_{2,0})}{4!0!}(x_1-x_{1,0})^4+
\frac{\partial_{x_2}^4V(x_{1,0},x_{2,0})}{0!4!}(x_2-x_{2,0})^4\right. \\
&\qquad+\left.\frac{\partial_{x_1}^3\partial_{x_2}V(x_{1,0},x_{2,0})}{3!1!}(x_1-x_{1,0})^3(x_2-x_{2,0})+
\frac{\partial_{x_1}\partial_{x_2}^3V(x_{1,0},x_{2,0})}{1!3!}(x_1-x_{1,0})(x_2-x_{2,0})^3\right. \\
&\qquad+\left.\frac{\partial_{x_1}^2\partial_{x_2}^2V(x_{1,0},x_{2,0})}{2!2!}(x_1-x_{1,0})^2(x_2-x_{2,0})^2\right]
\end{align}