1.5 KiB
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}).
??? note "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.