port from mathematics-physics notes

docs/mathematics/multivariable-calculus/differentation.md

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+# Differentation
+
+Generalization of derivatives to higher dimensions:
+
+* limit of difference quotient: partial derivatives,
+* linearization: total derivative.
+
+## Partial derivatives
+
+*Definition*: let $D \subseteq \mathbb{R}^n$ ($n=2$ for simplicity) and let $f: D \to \mathbb{R}$ and $\mathbf{a} \in D$, if the limit exists the partial derivates of $f$ are
+
+$$
+\begin{align*}
+    &\partial_1 f(\mathbf{a}) := \lim_{h \to 0} \frac{f(a_1 + h, a_2) - f(\mathbf{a})}{h}, \\
+    &\partial_2 f(\mathbf{a}) := \lim_{h \to 0} \frac{f(a_1, a_2 + h) - f(\mathbf{a})}{h}.
+\end{align*}
+$$
+
+*Theorem*: suppose that two mixed $n$th order partial derivatives of a function $f$ involve the same differentations but in different orders. If those partials are continuous at a point $\mathbf{a}$ and if $f$ and all partials of $f$ of order less than $n$ are continuous in a neighbourhood of $\mathbf{a}$, then the two mixed partials are equal at the point $\mathbf{a}$. We have for $n=2$ 
+
+$$
+    \partial_{12} f(P) = \partial_{21} f(P),
+$$
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+## Total derivatives
+
+*Definition*: let $D \subseteq \mathbb{R}^n$ ($n=2$ for simplicity) and let $f: D \to \mathbb{R}$, determining an affine linear approximation of $f$ around $\mathbf{a} \in D$
+
+$$
+    p(\mathbf{x}) = f(\mathbf{a}) + \big\langle L,\; \mathbf{x} - \mathbf{a} \big\rangle,
+$$
+
+with $f(\mathbf{x}) = p(\mathbf{x}) + r(\mathbf{x})$ demand $\frac{r(\mathbf{x})}{\|\mathbf{x} - \mathbf{a}\|} \to 0$ when $\mathbf{x} \to \mathbf{a}$. 
+
+if $L \in \mathbb{R}^2$ exists to satisfy this, then $f$ is called totally differentiable in $\mathbf{a}$. 
+
+*Theorem*: if $f$ is totally differentiable in $\mathbf{a}$, then $f$ is partially differentiable in $\mathbf{a}$ and the partial derivatives are
+
+$$
+    \partial_1 f(\mathbf{x}) = L_1, \qquad \partial_2 f(\mathbf{x}) = L_2,
+$$
+
+obtaining 
+
+$$
+    p(\mathbf{x}) = f(\mathbf{a}) + \big\langle \nabla f(\mathbf{a}),\; \mathbf{x} - \mathbf{a} \big\rangle.
+$$
+
+with $\nabla f(\mathbf{a})$ the gradient of $f$.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+## Chain rule
+
+*Definition*: let $D \subseteq \mathbb{R}^n$ ($n=2$ for simplicity) and let $f: D \to \mathbb{R}$, also let $g: \mathbb{R} \to \mathbb{R}$ given by
+
+$$
+    g(t) = f\big(\mathbf{x}(t)\big),
+$$
+
+if $f$ is continuously differentiable, then $g$ is differentiable with
+
+$$
+    g'(t) = \big\langle \nabla f\big(\mathbf{x}(t)\big),\; \mathbf{\dot x}(t) \big\rangle.
+$$
+
+## Gradients
+
+*Definition*: at any point $\mathbf{x} \in D$ where the first partial derivatives of $f$ exist, we define the gradient vector $\nabla$ by
+
+$$
+    \nabla f(\mathbf{x}) = \begin{pmatrix} \partial_1 f(\mathbf{x}) \\ \partial_2 f(\mathbf{x}) \end{pmatrix}.
+$$
+
+The direction of the gradient is the direction of steepest increase of $f$ at $\mathbf{x}$.
+
+<br>
+
+*Theorem*: gradients are orthogonal to level lines and level surfaces.
+
+??? note "*Proof*:"
+
+    let $\mathbf{r}(t) = \big(x(t),\; y(t) \big)^T$ be a parameterization of the level curve of $f$ such that $\mathbf{r}(0) = \mathbf{a}$. Then for all $t$ near $0$, $f(\mathbf{r}(t)) = f(\mathbf{a})$. Differentiating this equation with respect to $t$ using the chain rule, we obtain
+
+    $$
+        \partial_1 f(\mathbf{x}) \dot x(t) + \partial_2 f(\mathbf{x}) \dot y(t) = 0,
+    $$
+
+    at $t=0$, we can rewrite this to
+
+    $$
+        \big\langle \nabla f(\mathbf{a}),\; \mathbf{\dot r}(0) \big\rangle = 0,
+    $$
+
+    obtaining that $\nabla f$ is orthogonal to $\mathbf{\dot r}$. 
+
+## Directional derivatives
+
+*Definition*: let $D \subseteq \mathbb{R}^n$ and let $f: D \to \mathbb{R}$ with $\mathbf{v} \in D$ and $\|\mathbf{v}\| = 1$ a unit vector. The directional derivative is then the change of $f$ near a point $\mathbf{a} \in D$ in the direction of $\mathbf{v}$ 
+
+$$  
+    D_\mathbf{v} f(\mathbf{a}) = \big\langle \mathbf{v},\; \nabla f(\mathbf{a}) \big\rangle.
+$$
+
+## The general case
+
+*Definition*: let $D \subseteq \mathbb{R}^n$ and let $\mathbf{f}: D \to \mathbb{R}^m$, with $f_i: D \to \mathbb{R}$, with $i = 1, \dotsc, m$ being the components of $\mathbf{f}$. 
+
+* $\mathbf{f}$ is continuous at $\mathbf{a} \in D$ $\iff$ all $f_i$ continuous at $\mathbf{a}$,
+* $\mathbf{f}$ is partially/totally differentiable at $\mathbf{a}$ $\iff$ all $f_i$ are partially/totally differentiable at $\mathbf{a}$. 
+
+The linearization of every component $f_i$ we have
+
+$$
+    f_i(\mathbf{x}) = f_i(\mathbf{a}) + \big\langle \nabla f_i(\mathbf{a}),\; \mathbf{x} - \mathbf{a} \big\rangle + r_i(\mathbf{x}),
+$$
+
+so in total we have
+
+$$
+    \mathbf{f}(\mathbf{x}) = \mathbf{f}(\mathbf{a}) + D\mathbf{f}(\mathbf{a}) \big(\mathbf{x} - \mathbf{a}\big) + \mathbf{r}(\mathbf{x}),
+$$
+
+with $D\mathbf{f}(\mathbf{a})$ the Jacobian of $\mathbf{f}$.
+
+*Definition*: the Jacobian is given by $\big[D\mathbf{f}(\mathbf{a}) \big]_{i,\;j} = \partial_j f_i(\mathbf{a}).$
+
+### Chain rule
+
+Let $D \subseteq \mathbb{R}^n$ and let $E \subseteq \mathbb{R}^m$ be sets and let $\mathbf{f}: D \to \mathbb{R}^m$ and let $\mathbf{g}: E \to \mathbb{R}^k$ with $\mathbf{f}$ differentiable at $\mathbf{x}$ and $\mathbf{g}$ differentiable at $\mathbf{f}(\mathbf{x})$. Then $D\mathbf{f}(\mathbf{x}) \in \mathbb{R}^{m \times n}$ and $D\mathbf{g}\big(\mathbf{f}(\mathbf{x})\big) \in \mathbb{R}^{k \times m}$. 
+
+Then if we differentiate $\mathbf{g} \circ \mathbf{f}$ we obtain
+
+$$
+    D(\mathbf{g} \circ \mathbf{f})(\mathbf{x}) = D\mathbf{g}\big(\mathbf{f}(\mathbf{x})\big) D\mathbf{f}(\mathbf{x}).
+$$
+
+We have two interpretations:
+
+* the composition of linear maps,
+* the matrix multiplication of the Jacobian.
+
+??? note "*Proof*:"
+
+    Will be added later.

docs/mathematics/multivariable-calculus/extrema.md

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+# Extrema
+
+*Definition*: for $D \subseteq \mathbb{R}^n$ let $f: D \to \mathbb{R}$ be differentiable and $D$ contains no boundary points (open). A point $\mathbf{x^*} \in D$ is called a critical point for $f$ $\iff \nabla f(\mathbf{x^*}) = \mathbf{0}$.
+
+*Definition*: $f$ has (strict) global $\begin{matrix} \text{ maximum } \\ \text{ minimum } \end{matrix}$ in $\mathbf{x^*} \in D$ $\iff \forall \mathbf{x} \in D \backslash \{\mathbf{x^*}\} \Big[f(\mathbf{x^*}) \begin{matrix} (>) \\ \geq \\ \leq \\ (<) \end{matrix} f(\mathbf{x}) \Big]$.
+
+*Definition*: $f$ has (strict) local $\begin{matrix} \text{ maximum } \\ \text{ minimum } \end{matrix}$ in $\mathbf{x^*} \in D$ $\iff \exists r_{>0} \forall \mathbf{x} \in D \backslash \{\mathbf{x^*}\} \Big[f(\mathbf{x^*}) \begin{matrix} (>) \\ \geq \\ \leq \\ (<) \end{matrix} f(\mathbf{x}) \;\land\;  (0) < \|\mathbf{x} - \mathbf{x^*}\| < r \Big]$
+
+*Theorem*: if $f$ has local $\begin{matrix} \text{ maximum } \\ \text{ minimum } \end{matrix}$ at $\mathbf{x^*} \in D$ then $\mathbf{x^*}$ is a critical point of for $f$.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+## A second derivative test
+
+*Definition*: suppose $f: \mathbb{R}^n \to \mathbb{R}$ is differentiable with $\mathbf{x} \in \mathbb{R}^n$. The Hessian matrix of $f$ is defined as
+
+$$
+    H_f(\mathbf{x}) := \begin{pmatrix} \partial_{11} f(\mathbf{x}) & \dots & \partial_{1n} f(\mathbf{x}) \\ \vdots & \ddots & \vdots \\ \partial_{n1} f(\mathbf{x}) & \dots & \partial_{nn} f(\mathbf{x}) \end{pmatrix}.
+$$
+
+*Theorem*:
+
+* If $H_f(\mathbf{x^*})$ is positive definite (all eigenvalues are positive), then $f$ has a local minimum at $\mathbf{x^*}$.
+* If $H_f(\mathbf{x^*})$ is negative definite (all eigenvalues are negative), then $f$ has a local maximum at $\mathbf{x^*}$.
+* If $H_f(\mathbf{x^*})$ is indefinite (both positive and negative eigenvalues), then $f$ has a saddle point at $\mathbf{x^*}$.
+* If $H_f(\mathbf{x^*})$ is neither positive nor negative definite, nor indefinite, (eigenvalues equal to zero) this test gives no information.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+## Extrema on restricted domains
+
+*Theorem*: let $D \subseteq \mathbb{R}^n$ be bounded and closed ($D$ contains all boundary points). Let $f: D \to \mathbb{R}$ be continuous, then $f$ has a global maximum and minimum.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+**Procedure to find the global maximum and minimum**:
+
+* Find critical points in the interior.
+* Find global extrema on the boundary.
+* Find the largest/smallest among them.
+
+### Lagrange multipliers
+
+*Theorem*: let $f: M \to \mathbb{R}$ and $g: \mathbb{R}^n \to \mathbb{R}$ with $M$ the boundary of $D$ given by
+
+$$
+    M := \big\{\mathbf{x} \in \mathbb{R}^n \;\big|\; g(\mathbf{x}) = 0  \big\} \subseteq D,
+$$
+
+suppose that there is global maximum or minimum $\mathbf{x^*} \in M$ of $f$ that is not an endpoint of $M$ and $\nabla g(\mathbf{x^*}) \neq \mathbf{0}$. Then there exists a $\lambda^* \in \mathbb{R}$ such that $(\mathbf{x^*}, \lambda^*)$ is a critical point of the Lagrange function
+
+$$
+    L(\mathbf{x}, \lambda) := f(\mathbf{x}) - \lambda g(\mathbf{x}).
+$$
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+### The general case
+
+*Theorem*: Let $f: S \to \mathbb{R}$ and $\mathbf{g}: \mathbb{R}^m \to \mathbb{R}^n$ with $m \leq n -1$ restrictions given by
+
+$$
+    S := \big\{\mathbf{x} \in \mathbb{R}^n \;\big|\; \mathbf{g}(\mathbf{x}) = 0  \big\} \subseteq D,
+$$
+
+suppose that there is global maximum or minimum $\mathbf{x^*} \in S$ of $f$ that is not an endpoint of $S$ and $D \mathbf{g}(\mathbf{x^*}) \neq \mathbf{0}$. Then there exists a $\mathbf{\lambda^*} \in \mathbb{R^m}$ such that $(\mathbf{x^*}, \mathbf{\lambda^*})$ is a critical point of the Lagrange function
+
+$$
+    L(\mathbf{x}, \mathbf{\lambda}) := f(\mathbf{x}) - \big\langle \mathbf{\lambda},\; \mathbf{g}(\mathbf{x})  \big\rangle.
+$$
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+#### Example
+
+Let $f: M_1 \cap M_2 \to \mathbb{R}$ and $g_{1,2}: \mathbb{R}^n \to \mathbb{R}$ with the restrictions given by
+
+$$
+    M_{1,2} := \big\{\mathbf{x} \in \mathbb{R}^n \;\big|\; g_{1,2}(\mathbf{x}) = 0  \big\} \subseteq D,
+$$
+
+suppose that there is global maximum or minimum $\mathbf{x^*} \in M_1 \cap M_2$ of $f$ that is not an endpoint of $M_1 \cap M_2$ and $\nabla g_{1,2}(\mathbf{x^*}) \neq \mathbf{0}$. Then there exists a $\lambda_{1,2}^* \in \mathbb{R}$ such that $(\mathbf{x^*}, \lambda_{1,2}^*)$ is a critical point of the Lagrange function
+
+$$
+    L(\mathbf{x}, \lambda_1, \lambda_2) := f(\mathbf{x}) - \lambda_1 g_1(\mathbf{x}) - \lambda_2 g_2(\mathbf{x}).
+$$

docs/mathematics/multivariable-calculus/functions-of-several-variables.md

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+# Functions of several variables
+
+*Definition*: let $D \subseteq \mathbb{R}^m$ with $m>1$, and $f: D \to \mathbb{R}^n$ then $f$ is a function of several variables where:
+
+* for $n=1$, $f$ is a scalar function,
+* for $n>1$, $f$ is a vector valued function.
+
+<br>
+
+*Definition*: the domain convention specifies that the domain of a function of $m$ variables is the largest set of points for which the function makes sense as a real number, unless that domain is explicitly stated to be a smaller set. 
+
+## Graphical representations of scalar valued functions
+
+### Graphs
+
+*Definition*: let $D \subseteq \mathbb{R}^2$ and let $f: D \to \mathbb{R}$ then $G_f := \big\{\big(x, y, f(x,y)\big) \;\big|\; (x, y) \in D\big\}$ is the graph of $f$. Observe that $G_f \subseteq \mathbb{R}^3$. 
+
+### Level sets
+
+*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
+
+$$
+    \partial^\alpha f(\mathbf{x}) = \partial^{\alpha_1}_{x_1} \dotsc \partial^{\alpha_n}_{x_n} f(\mathbf{x}).
+$$

docs/mathematics/multivariable-calculus/implicit-equations.md

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+# Implicit equations
+
+*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. 
+
+then there exists an $I$ around $a_1$ and an $J$ around $a_2$ such that $\phi: I \to J$ is differentiable and 
+
+$$
+    \forall x \in I, y \in J: f(x,y) = 0 \iff y = \phi(x).
+$$
+
+Now calculating $\phi' (x)$ with the chain rule
+
+$$
+\begin{align*}
+    f\big(x,\phi(x)\big) &= 0, \\
+    \partial_1 f\big(x,\phi(x)\big) + \partial_2 f\big(x,\phi(x)\big) \phi' (x) &= 0,
+\end{align*}
+$$
+
+and we obtain
+
+$$
+    \phi' (x) = - \frac{\partial_1 f\big(x,\phi(x)\big)}{\partial_2 f\big(x,\phi(x)\big)}.
+$$
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+## General case
+
+*Theorem*: Let $\mathbf{F}: \mathbb{R}^{n+m} \to \mathbb{R}^m$ given by $F(\mathbf{x},\mathbf{y}) = \mathbf{0}$ with $\mathbf{x} \in \mathbb{R}^n$ and $\mathbf{y} \in \mathbb{R}^m$. Suppose $\mathbf{F}$ is continuously differentiable and assume $D_2 \mathbf{F}(\mathbf{x},\mathbf{y}) \in \mathbb{R}^{m \times m}$ is nonsingular. Then there exists in neighbourhoods $I$ of $\mathbf{x}$ and $J$ of $\mathbf{y}$ with $I \subseteq \mathbb{R}^n,\; J \subseteq \mathbb{R}^m$, such that $\mathbf{\phi}: I \to J$ is differentiable and 
+
+$$
+    \forall (\mathbf{x},\mathbf{y}) \in I \times J: \mathbf{F}(\mathbf{x},\mathbf{y}) = \mathbf{0} \iff \mathbf{y} = \mathbf{\phi}(\mathbf{x}).
+$$
+
+Now calculating $D \mathbf{\phi}(\mathbf{x})$ with the generalized chain rule
+
+$$
+\begin{align*}
+    \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big) &= \mathbf{0}, \\
+    D_1 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big) + D_2 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big) D \mathbf{\phi}(\mathbf{x}) &= \mathbf{0}, \\
+\end{align*}
+$$
+
+and we obtain
+
+$$
+    D \mathbf{\phi}(\mathbf{x}) = - \Big(D_2 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big) \Big)^{-1} D_1 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big).
+$$
+
+??? note "*Proof*:"
+
+    Will be added later.

docs/mathematics/multivariable-calculus/integration.md

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+# Integration
+
+*Theorem*: for $D \subseteq \mathbb{R}^n$ ($n=2$ for simplicity) with $D = X \times Y$, let $f: D \to \mathbb{R}$ then we have
+
+$$
+    \iint_D f = \int_X \Big(\int_Y f(x,y)dy \Big)dx = \int_Y \Big(\int_X f(x,y)dx \Big)dy
+$$
+
+implying that order can be interchanged, this is true for $n \in \mathbb{N}$. 
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+## Iteration of integrals
+
+*Theorem*: for $D \subseteq \mathbb{R}^n$ ($n=2$ for simplicity) bounded and piecewise smooth boundary, let $f: D \to \mathbb{R}$ be bounded and continuous. Let $R$ be a rectangle with $D \subseteq R$ then
+
+$$
+    \iint_D f dA = \iint_R F dA, \qquad \text{where } F(\mathbf{x}) = \begin{cases} F(\mathbf{x}) \quad &\mathbf{x} \in D, \\ 0 \quad &\mathbf{x} \notin D. \end{cases}
+$$
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+## Coordinate transformation for integrals
+
+*Theorem*: for $D \subseteq \mathbb{R}^n$ ($n=2$ for simplicity) bounded and piecewise smooth boundary, let $f: D \to \mathbb{R}$ be bounded and continuous and let $\phi: D \to \mathbb{R}^n$ be continuously differentiable and injective, define
+
+$$
+    E := \phi(D),
+$$
+
+then we have
+
+$$
+    \iint_D f = \iint_E f \circ \phi \;\Big|\mathrm{det} \big(D_\phi \big) \Big|,
+$$
+
+with $D_\phi$ the Jacobian of $\phi$.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+### Example
+
+Let $D = \big\{(x,y) \in \mathbb{R}^2 \;\big|\; x^2 + y^2 \leq 4 \land 0 \leq y \leq x \big\}$ and let $\phi: D \to \mathbb{R}^2$ be given by
+
+$$
+    \phi(r,\theta) = \begin{pmatrix} r\cos \theta \\ r\sin \theta \end{pmatrix},
+$$
+
+define $E := \phi(D) = [0,2] \times [0, \frac{\pi}{4}]$. Then $E$ is a rectangle which can be more easily integrated.

docs/mathematics/multivariable-calculus/limits-and-continuity.md

@@ -0,0 +1,21 @@
+# Limits and continuity
+
+## Limit
+
+*Definition*: let $D \subseteq \mathbb{R}^m$ and let $f: D \to \mathbb{R}^n$, with $m,n \in \mathbb{N}$. Let $\mathbf{a}$ be the point $\mathbf{x}$ approaches, then $f$ approaches the limit $L \in \mathbb{R}^n$
+
+$$
+\lim_{\mathbf{x} \to \mathbf{a}} f(\mathbf{x}) = L \iff \forall \varepsilon_{>0} \exists \delta_{>0} \Big[0 < \|\mathbf{x} - \mathbf{a}\|< \delta \implies \|f(\mathbf{x}) - L\| < \varepsilon \Big],
+$$
+
+with $\mathbf{a}, \mathbf{x} \in \mathbb{R}^m$.
+
+## Continuity
+
+*Definition*: let $D \subseteq \mathbb{R}^m$ and let $f: D \to \mathbb{R}^n$, with $m,n \in \mathbb{N}$. Then $f$ is called continuous at $\mathbf{a}$ if
+
+$$
+\lim_{\mathbf{x} \to \mathbf{a}} f(\mathbf{x}) = f(\mathbf{a}),
+$$
+
+with $\mathbf{a}, \mathbf{x} \in \mathbb{R}^m$.

docs/mathematics/multivariable-calculus/taylor-polynomials.md

@@ -0,0 +1,39 @@
+# 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. 
+
+### Example