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?