diff --git a/docs/en/mathematics/calculus/taylor-polynomials.md b/docs/en/mathematics/calculus/taylor-polynomials.md index d9481ab..da14b6b 100755 --- a/docs/en/mathematics/calculus/taylor-polynomials.md +++ b/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 diff --git a/docs/en/mathematics/multivariable-calculus/extrema.md b/docs/en/mathematics/multivariable-calculus/extrema.md index e69de29..d247012 100644 --- a/docs/en/mathematics/multivariable-calculus/extrema.md +++ b/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 $f$. + +
+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. + +
+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. + +
+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 + +$$ + L(\mathbf{x}, \lambda) := f(\mathbf{x}) - \lambda g(\mathbf{x}). +$$ + +
+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 $m$ 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 + +$$ + L(\mathbf{x}, \mathbf{\lambda}) := f(\mathbf{x}) - \big\langle \mathbf{\lambda},\; \mathbf{g}(\mathbf{x}) \big\rangle. +$$ + +
+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 + +$$ + L(\mathbf{x}, \lambda_1, \lambda_2) := f(\mathbf{x}) - \lambda_1 g_1(\mathbf{x}) - \lambda_2 g_2(\mathbf{x}). +$$ \ No newline at end of file diff --git a/docs/en/mathematics/multivariable-calculus/integration.md b/docs/en/mathematics/multivariable-calculus/integration.md index e69de29..71b0587 100644 --- a/docs/en/mathematics/multivariable-calculus/integration.md +++ b/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 $n \in \mathbb{N}$. + +
+Proof: + +will be added later. +
+
+ +## 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 + +$$ + \iint_D f dA = \iint_R F dA, \qquad \text{where } F(\mathbf{x}) = \begin{cases} F(\mathbf{x}) \quad \mathbf{x} \in D, \\ 0 \quad \mathbf{x} \notin D. \end{cases} +$$ + +
+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 $\phi$. + +
+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. \ No newline at end of file