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Added Taylor polynomials.

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Luc Bijl 2023-10-30 18:12:29 +01:00
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*Definition*: let $D \subseteq \mathbb{R}^2$ and let $f: D \to \mathbb{R}$ then for $c \in \mathbb{R}$ we have $S_c := \big\{(x, y) \in D \;\big|\; f(x,y) = c \big\}$ is the level set of $f$. Observe that $S_c \subseteq \mathbb{R}^2$.
## Multi-index notation
*Definition*: an $n$-dimensional multi-index is an $n$-tuple of non-negative integers
$$
\alpha = (\alpha_1, \alpha_2, \dotsc, \alpha_n), \qquad \text{with } \alpha_i \in \mathbb{N}.
$$
### Properties
For the sum of components we have: $|\alpha| := \alpha_1 + \dotsc + \alpha_n$.
For $n$-dimensional multi-indeces $\alpha, \beta$ we have componentwise sum and difference
$$
\alpha \pm \beta := (\alpha_1 \pm \beta_1, \dotsc, \alpha_n \pm \beta_n).
$$
For the products of powers with $\mathbf{x} \in \mathbb{R}^n$ we have
$$
\mathbf{x}^\alpha := x_1^{\alpha_1} x_2^{\alpha_2} \dotsc x_n^{\alpha_n}.
$$
For factorials we have
$$
\alpha ! = \alpha_1 ! \cdot \alpha_2 ! \cdots \alpha_n !
$$
For the binomial coefficient we have
$$
\begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \alpha_1 \\ \beta_1 \end{pmatrix} \begin{pmatrix} \alpha_2 \\ \beta_2 \end{pmatrix} \cdots \begin{pmatrix} \alpha_n \\ \beta_n \end{pmatrix} = \frac{\alpha !}{\beta ! (\alpha - \beta)!}
$$
For polynomials of degree less or equal to $m$ we have
$$
p(\mathbf{x}) = \sum_{|\alpha| \leq m} c_\alpha \mathbf{x}^\alpha,
$$
as an example for $m=2$ and $n=2$ we have
$$
p(\mathbf{x}) = c_1 + c_2 x_1 + c_3 x_2 + c_4 x_1 x_2 + c_5 x_1 ^2 + c_6 x_2^2 \qquad c_{1,2,3,4,5,6} \in \mathbb{R}
$$
For partial derivatives of $f: \mathbb{R}^n \to \mathbb{R}$ we have
$$
\partial^\alpha f(\mathbf{x}) = \partial^{\alpha_1}_{x_1} \dotsc \partial^{\alpha_n}_{x_n} f(\mathbf{x}).
$$

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# Implicit equations
*Theorem*: for $D \subseteq \mathbb{R}^2$ (for simplicty), let $f: D \to \mathbb{R}$ be continuously differentiable and $\mathbf{a} \in D$. Assume
*Theorem*: for $D \subseteq \mathbb{R}^n$ ($n=2$ for simplicty), let $f: D \to \mathbb{R}$ be continuously differentiable and $\mathbf{a} \in D$. Assume
* $f(\mathbf{a}) = 0$,
* $\partial_2 f(\mathbf{a}) \neq 0$, nondegeneracy.

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# 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 Taylors 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