Taylor polynomials

For \(D \subseteq \mathbb{R}^n\) let \(f: D \to \mathbb{R}\) sufficiently often differentiable, we have \(\mathbf{a} \in D\). Find a polynomial \(T: \mathbb{R}^n \to \mathbb{R}\) such that

\[ \partial^\beta T(\mathbf{a}) = \partial^\beta f(\mathbf{a}). \]

Ansatz: let \(T(\mathbf{x}) = \sum_{|\alpha| \leq n} c_\alpha (\mathbf{x} - \mathbf{a})^\alpha\). Then

\[ \partial^\beta T(\mathbf{x}) = \sum_{|\alpha| \leq n,\; \alpha \geq \beta} c_\alpha \frac{\alpha!}{(\alpha - \beta)!} (\mathbf{x} - \mathbf{a})^{\alpha - \beta}. \]

Choose \(\mathbf{x} = \mathbf{a}\): \(\partial^\beta T(\mathbf{a}) = c_\beta \beta! = \partial^\beta f(\mathbf{a}) \implies c_\beta = \frac{\partial^\beta f(\mathbf{a})}{\beta!}\). Therefore we obtain

\[ T(\mathbf{x}) = \sum_{|\alpha| \leq n} \frac{\partial^\alpha f(\mathbf{a})}{\alpha!} (\mathbf{x} - \mathbf{a})^\alpha. \]

Theorem: suppose \(x \in D\) and the line segment \([\mathbf{a},\mathbf{x}]\) lies completely in \(D\). Set \(\mathbf{h} = \mathbf{x} - \mathbf{a}\). Then there is a \(\theta \in (0,1)\) such that

\[ f(\mathbf{x}) = T(\mathbf{x}) + \frac{1}{(n+1)!} \partial_\mathbf{h}^{n+1} f(\mathbf{a} + \theta \mathbf{h}). \]

Proof:

Apply Taylor’s theorem in 1D and the chain rule to the function \(\phi : [0, 1] \to \mathbb{R}\) given by

\[ \phi(\theta) := f(\mathbf{a} + \theta \mathbf{h}). \]

Other methods

Creating multivariable Taylor polynomials by using 1D Taylor polynomials of the different variables and composing them.

Example