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Notice that by Taylor's theorem If the function $f : \mathbb{R}^{n} \rightarrow \mathbb{R}$ is $ k+1$ times continuously differentiable in the closed ball B, then one can derive an exact formula for the remainder in terms of (k+1)-th order partial derivatives of f in this neighborhood. Namely,

\begin{align}& f( \boldsymbol{x} ) = \sum_{|\alpha|\leq k} \frac{\mathrm D^\alpha f(\boldsymbol{a})}{\alpha!} (\boldsymbol{x}-\boldsymbol{a})^\alpha + \sum_{|\beta|=k+1} R_\beta(\boldsymbol{x})(\boldsymbol{x}-\boldsymbol{a})^\beta, \\& R_\beta( \boldsymbol{x} ) = \frac{|\beta|}{\beta!} \int_0^1 (1-t)^{|\beta|-1} \mathrm D^\beta f \big(\boldsymbol{a}+t( \boldsymbol{x}-\boldsymbol{a} )\big) \, dt. \end{align}

Until the degree two I Know that I can write the polynomial $ \sum_{|\alpha|\leq k} \frac{\mathrm D^\alpha f(\boldsymbol{a})}{\alpha!} (\boldsymbol{x}-\boldsymbol{a})^\alpha $ in this form \begin{equation} f(0) + \mathrm Df(0) \cdot X + \frac{1}{2} X^{t} \mathrm D^{2}f(0) X \end{equation} Can we write the polynomial above in a similar way for a degree grater than two?

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You should not write $\begin{equation} f(a) + Df(a) \cdot (x-a) + \frac{1}{2}D^{2}f(a)\cdot (x-a)^{2} \end{equation}$, in the penultimate line of your question? – Manoel Dec 19 '12 at 2:39

Dear @user29999 use the following equality, with a degree of times continuously differentiable less; 4, 5 ou 6, for example, will give a clear

$D^{k+1}f(a)\cdot (\boldsymbol{x}-\boldsymbol{a})^{k+1}=\sum_{i_{1}, i_{2},...,i_{k+1}} \frac{\partial^{k+1}f}{\partial x_{i_{1}} \partial x_{i_{2} \cdot \cdot \cdot} \partial x_{i_{k+1}}}(a)(x-a)_{i_{1}}(x-a)_{i_{2}}\cdot \cdot \cdot(x-a)_{i_{k+1}} $, where $(x-a)=((x-a)_{1},...,(x-a)_{n}).$

and I think this can help you ... with some calculations, but it will give ok.

A Hug, Manoel.

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@user29999, my answer did not help? You do not agree? – Manoel Feb 10 '13 at 1:20

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