Added extrema and integration sections to multivariable calculus.

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

@@ -172,7 +172,7 @@ $$
 \lim_{x \to 0} \frac{\frac{x^4}{6} + O(x^6)}{- \frac{x^4}{6} + O(x^6)} = -1
 $$
 
-## L'Hostpital's rule
+## L'Hôpital's rule
 
 Suppose the function $f$ and $g$ are differentiable on the interval $(a,b)$, and $g'(x) \neq 0$ there. Also suppose that $\lim_{x \downarrow a} f(x) = \lim_{x \downarrow a} g(x) = 0$ then
 

docs/en/mathematics/multivariable-calculus/extrema.md

@@ -0,0 +1,111 @@
+# 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 $f$.
+
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
+
+## 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.
+
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
+
+## 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.
+
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
+
+**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
+
+$$
+    L(\mathbf{x}, \lambda) := f(\mathbf{x}) - \lambda g(\mathbf{x}).
+$$
+
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
+
+### The general case
+
+*Theorem*: Let $f: S \to \mathbb{R}$ and $\mathbf{g}: \mathbb{R}^m \to \mathbb{R}^n$ with $m$ 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
+
+$$
+    L(\mathbf{x}, \mathbf{\lambda}) := f(\mathbf{x}) - \big\langle \mathbf{\lambda},\; \mathbf{g}(\mathbf{x})  \big\rangle.
+$$
+
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
+
+#### 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/en/mathematics/multivariable-calculus/integration.md

@@ -0,0 +1,64 @@
+# 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 $n \in \mathbb{N}$. 
+
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
+
+## Iteration of integralss
+
+*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
+
+$$
+    \iint_D f dA = \iint_R F dA, \qquad \text{where } F(\mathbf{x}) = \begin{cases} F(\mathbf{x}) \quad \mathbf{x} \in D, \\ 0 \quad \mathbf{x} \notin D. \end{cases}
+$$
+
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
+
+## 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 $\phi$.
+
+<details>
+<summary><em>Proof</em>:</summary>
+
+will be added later.
+</details>
+<br>
+
+### 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.