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@@ -31,6 +31,7 @@ %% Layout \usepackage{enumitem} \usepackage{booktabs} +\usepackage{footmisc} %% Nice drwaings \usepackage{tikz} @@ -91,16 +92,17 @@ These are just my personal notes of the \themodule{} course, and definitively not a rigorously constructed mathematical text. The good looking \LaTeX{} typesetting may trick you into thinking it is rigorous, but really, it is not. -\section{Derivatives of vector valued functions} +\section{Derivatives of vector valued scalar functions} \begin{definition}[Partial derivative] - A vector values function \(f: \mathbb{R}^m\to\mathbb{R}\), with + A vector valued function \(f: \mathbb{R}^m\to\mathbb{R}\), with \(\vec{v}\in\mathbb{R}^m\), has a partial derivative with respect to \(v_i\) defined as \[ \partial_{v_i} f(\vec{v}) - = f_{v_i}(\vec{v}) - = \lim_{h\to 0} \frac{f(\vec{v} + h\vec{e}_j) - f(\vec{v})}{h} + % = f_{v_i}(\vec{v}) + = \frac{\partial f}{\partial v_i} + = \lim_{h\to 0} \frac{f(\vec{v} + h\vec{e}_i) - f(\vec{v})}{h} \] \end{definition} @@ -113,17 +115,41 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \] \end{theorem} -\begin{definition}[Linearization] +\begin{application}[Find the slope of an implicit curve] + Let \(f(x,y) = 0\) be an implicit curve. It's slope at any point where + \(\partial_y f \neq 0\) is \(m = - \partial_x f / \partial_y f\) +\end{application} + +\begin{definition}[Total differential] + The total differential \(df\) of \(f:\mathbb{R}^m\to\mathbb{R}\) is + \[ + df = \sum_{i=0}^m \partial_{x_i} f\cdot dx . + \] + That reads, the \emph{total} change is the sum of the change in each + direction. This implies + \[ + \frac{df}{dx_k} = \frac{\partial f}{\partial x_k} + + \sum_{i \in \{1 \leq i \leq m : i \neq k\}} + \frac{\partial f}{\partial x_i} \cdot \frac{dx_i}{dx_k} , + \] + i.e. the change in direction \(x_k\) is how \(f\) changes in \(x_k\) + (ignoring other directions) plus, how \(f\) changes with respect to each + other variable \(x_i\) times how it (\(x_i\)) changes with respect to \(x_k\). +\end{definition} + +\begin{application}[Linearization] A function \(f: \mathbb{R}^m\to\mathbb{R}\) has a linearization \(g\) at \(\vec{x}_0\) given by \[ g(\vec{x}) = f(\vec{x}_0) + \sum_{i=1}^m \partial_{x_i} f(\vec{x}_0)(x_i - x_{i,0}) , \] - if all partial derviatives are defined at \(\vec{x}_0\). -\end{definition} + if all partial derivatives are defined at \(\vec{x}_0\). With the gradient + (defined below) \(g(\vec{x}) = f(\vec{x}_0) + \grad f(\vec{x}_0) \dotp + (\vec{x} - \vec{x}_0)\). +\end{application} -\begin{theorem}[Propagation of uncertanty] +\begin{application}[Propagation of uncertanty] Given a measurement of \(m\) values in a vector \(\vec{x}\in\mathbb{R}^m\) with values given in the form \(x_i = \bar{x}_i \pm \sigma_{x_i}\), a linear approximation the error of a dependent variable \(y\) is computed with @@ -132,7 +158,7 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \pm \sqrt{\sum_{i=1}^m \left( \partial_{x_i} f(\bar{\vec{x}}) \sigma_{x_i}\right)^2} \] -\end{theorem} +\end{application} \begin{definition}[Gradient vector] The \emph{gradient} of a function \(f(\vec{x}), \vec{x}\in\mathbb{R}^m\) is a @@ -150,20 +176,20 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \] \end{definition} +\begin{theorem} + The gradient vector always points towards \emph{the direction of steepest + ascent}, and thus is always perpendicular to contour lines. +\end{theorem} + \begin{definition}[Directional derivative] A function \(f(\vec{x})\) has a directional derivative in direction - \(\vec{r}\) (with \(|\vec{r}| = 1\)) given by + \(\vec{r}\) (with \(|\vec{r}|=1\)) of \[ \frac{\partial f}{\partial\vec{r}} = \nabla_\vec{r} f = \vec{r} \dotp \grad f \] \end{definition} -\begin{theorem} - The gradient vector always points towards \emph{the direction of steepest - ascent}. -\end{theorem} - \begin{definition}[Jacobian Matrix] The \emph{Jacobian} \(\mx{J}_f\) (sometimes written as \(\frac{\partial(f_1,\ldots f_m)}{\partial(x_1,\ldots,x_n)}\)) of a function @@ -192,7 +218,7 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \begin{definition}[Hessian matrix] Given a function \(f: \mathbb{R}^m \to \mathbb{R}\), the square matrix whose entry at the \(i\)-th row and \(j\)-th column is the second derivative of - \(f\) first with respect to \(x_j\) and then to \(x_i\) is know as the + \(f\) first with respect to \(x_j\) and then to \(x_i\) is known as the \emph{Hessian} matrix. \( \left(\mx{H}_f\right)_{i,j} = \partial_{x_i}\partial_{x_j} f @@ -212,11 +238,14 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \section{Methods for maximization and minimization problems} +\subsection{Analytical methods} + \begin{method}[Find stationary points] Given a function \(f: D \subseteq \mathbb{R}^m \to \mathbb{R}\), to find its maxima and minima we shall consider the points \begin{itemize} - \item that are on the boundary of the domain \(\partial D\), + \item that are on the boundary\footnote{If it belongs to \(f\). + \label{ftn:boundary}} of the domain \(\partial D\), \item where the gradient \(\grad f\) is not defined, \item that are stationary, i.e. where \(\grad f = \vec{0}\). \end{itemize} @@ -243,14 +272,14 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \begin{remark} The previous method is obtained by studying the second directional derivative \(\nabla_\vec{r}\nabla_\vec{r} f\) at the stationary point in direction of a - vector \(\vec{r} = \vec{e}_1\cos(\alpha) + \vec{e}_2\sin(\alpha)\) + vector \(\vec{r} = \vec{e}_1\cos(\alpha) + \vec{e}_2\sin(\alpha)\). \end{remark} \begin{method}[Determine the type of stationary point in higher dimensions] - Given a scalar function of two variables \(f(x,y)\) and a stationary point - \(\vec{x}_s\) (where \(\grad f(\vec{x}_s) = \vec{0}\)), we compute the - Hessian matrix \(\mx{H}_f(\vec{x}_s)\). Then we compute its eigenvalues - \(\lambda_1, \ldots, \lambda_m\) and + Given a scalar function of multiple variables \(f(\vec{x})\) and a stationary + point \(\vec{x}_s\) (\(\grad f(\vec{x}_s) = \vec{0}\)), we compute the + Hessian matrix \(\mx{H}_f(\vec{x}_s)\) and its eigenvalues \(\lambda_1, + \ldots, \lambda_m\), then \begin{itemize} \item if all \(\lambda_i > 0\), the point is a minimum; \item if all \(\lambda_i < 0\), the point is a maximum; @@ -284,7 +313,8 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \(f: D \subseteq \mathbb{R}^2 \to \mathbb{R}\). To find the extrema we look for points \begin{itemize} - \item on the boundary \(\vec{u} \in \partial D\) where \(n(\vec{u}) = 0\); + \item on the boundary\footref{ftn:boundary} \(\vec{u} \in \partial D\) + where \(n(\vec{u}) = 0\); \item \(\vec{u}\) where the gradient either does not exist or is \(\vec{0}\), and satisfy \(n(\vec{u}) = 0\); @@ -316,7 +346,7 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. under \(k < m\) constraints \(n_1 = 0, \cdots, n_k = 0\). To find the extrema we consider the following points: \begin{itemize} - \item Points on the boundary \(\vec{u} \in \partial D\) that satisfy + \item Points on the boundary\footref{ftn:boundary} \(\vec{u} \in \partial D\) that satisfy \(n_i(\vec{u}) = 0\) for all \(1 \leq i \leq k\), \item Points \(\vec{u} \in D\) where either @@ -336,18 +366,56 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \] The \(\lambda\) values are known as \emph{Lagrange multipliers}. The same calculation can be written more compactly by defining the - \(m+k\) dimensional \emph{Lagrangian} + \emph{Lagrangian} \[ \mathcal{L}(\vec{u}, \vec{\lambda}) - = f(\vec{u}) - \sum_{i = 0}^k \lambda_i n_i(\vec{u}) + = f(\vec{u}) - \sum_{i = 0}^k \lambda_i n_i(\vec{u}), \] where \(\vec{\lambda} = \lambda_1, \ldots, \lambda_k\) and then solving - \(\grad \mathcal{L}(\vec{u}, \vec{\lambda}) = \vec{0}\). This is - generally used in numerical computations and not very useful by hand. + the \(m+k\) dimensional equation \(\grad \mathcal{L}(\vec{u}, + \vec{\lambda}) = \vec{0}\) (this is generally used in numerical + computations and not very useful by hand). \end{itemize} \end{method} -\section{Integration of vector values scalar functions} +\subsection{Numerical methods} + +\begin{method}[Newton's method] + For a function \(f:\mathbb{R}^m\to\mathbb{R}\) we wish to numerically find + its stationary points (where \(\grad f = \vec{0}\)). + \begin{enumerate} + \item Pick a starting point \(\vec{x}_0\) + \item Set the linearisation\footnote{The gradient becomes a hessian matrix.} + of \(\grad f\) at \(\vec{x}_k\) to zero and + solve for \(\vec{x}_{k+1}\) + \begin{gather*} + \grad f(\vec{x}_k) + \mx{H}_f (\vec{x}_k) + (\vec{x}_{k+1} - \vec{x}_k) = \vec{0} \\ + \vec{x}_{k+1} = \vec{x}_k - \mx{H}_f^{-1} (\vec{x}_k) \grad f(\vec{x}_k) + \end{gather*} + \item Repeat the last step until the magnitude of the error + \(|\vec{\epsilon}| = |\mx{H}_f^{-1} (\vec{x}_k) \grad f(\vec{x}_k)|\) is + sufficiently small. + \end{enumerate} +\end{method} + +\begin{method}[Gradient ascent / descent] + Given \(f:\mathbb{R}^m\to\mathbb{R}\) we wish to numerically find + the stationary points (where \(\grad f = \vec{0}\)). + \begin{enumerate} + \item Define an arbitrarily small length \(\eta\) and a starting point + \(\vec{x}_0\) + \item Compute \(\vec{v} = \pm\grad f(\vec{x}_k)\) (positive for ascent, + negative for descent), then \(\vec{x}_{k+1} = \vec{x}_k + \eta\vec{v}\) + if the rate of change \(\epsilon\) is acceptable (\(\epsilon = |\grad + f(\vec{x}_{k+1})| > 0\)) else recompute \(\vec{v} := \pm \grad + f(\vec{x}_{k+1})\). + \item Stop when the rate of change \(\epsilon\) stays small enough for many + iterations. + \end{enumerate} +\end{method} + +\section{Integration of vector valued scalar functions} \begin{figure} \centering @@ -385,7 +453,7 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \begin{theorem}[Transformation of coordinates] The generalization of theorem \ref{thm:transform-coords} is quite simple. - For an \(n\)-integral of a function \(f:\mathbb{R}^m\to\mathbb{R}\) over a + For an \(m\)-integral of a function \(f:\mathbb{R}^m\to\mathbb{R}\) over a region \(B\), we let \(\vec{x}(\vec{u})\) be ``nice'' functions that transform the coordinate system. Then as before \[ @@ -407,14 +475,153 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. Cylindrical & r\,dr\,d\phi\,dz & \uvec{z}r\,dr\,d\phi \\ & & \uvec{\phi}\,dr\,dz \\ & & \uvec{r}r\,d\phi\,dz \\ - Spherical & r^2\sin\theta\, dr\,d\theta\,d\phi & \uvec{r}r^2\sin\theta\,d\theta\,d\phi \\ + Spherical & r^2\sin\theta\, dr\,d\theta\,d\phi & + \uvec{r}r^2\sin\theta\,d\theta\,d\phi \\ Curvilinear & |\mx{J}_f|\,du\,dv\,dw & - \\ \bottomrule \end{tabular} \caption{Differential elements for integration.} \end{table} -\section{Derivatives of curves} +\begin{application}[Physics] + Given the mass \(m\) and density function \(\rho\) of an object, + its \emph{center of mass} is calculated with + \[ + \vec{x}_c = \frac{1}{m}\int_V \vec{x}\rho(\vec{x}) \,dv + \stackrel{\rho\text{ const.}}{=} \frac{1}{V} \int_V \vec{x}\,dv . + \] + The (scalar) \emph{moment of inertia} \(J\) of an object is given by + \[ + J = \int_V \rho(\vec{r}) r^2 \,dv . + \] + % and similarly the \emph{area moment of inertia} \(I\) +\end{application} + +\section{Parametric curves and line integrals} + +\begin{definition}[Parametric curve] + A parametric curve is a vector function \(\mathcal{C} : \mathbb{R} \to W + \subseteq \mathbb{R}^n, t \mapsto \vec{f}(t)\), that takes a parameter \(t\). +\end{definition} + +\begin{definition}[Multivariable chain rule] + Let \(\vec{x}: \mathbb{R} \to \mathbb{R}^m\) and \(f: \mathbb{R}^m \to + \mathbb{R}\), so that \(f\circ\vec{x}: \mathbb{R} \to \mathbb{R}\), then + the multivariable chain rule states: + \[ + \frac{d}{dt}f(\vec{x}(t)) = \grad f (\vec{x}(t)) \dotp \vec{x}'(t) + = \nabla_{\vec{x}'(t)} f(\vec{x}(t)) + \] +\end{definition} + +\begin{theorem}[Signed area enclosed by a planar parametric curve] + A planar (2D) parametric curve \((x(t), y(t))^t\) with \(t\in[r,s]\) that does + not intersect itself encloses a surface with area + \[ + A = \int_r^s x'(t)y(t) \,dt + = \int_r^s x(t)y'(t) \,dt + \] +\end{theorem} + +\begin{theorem}[Derivative of a curve] + The derivative of a curve is + \begin{align*} + \vec{f}'(t) &= \lim_{h\to 0} \frac{\vec{f}(t + h) - \vec{f}(t)}{h} \\ + &= \sum_{i=0}^n \left(\lim_{h\to 0} \frac{f_i(t+h) - f_i(t)}{h}\right) \vec{e}_i \\ + &= \sum_{i=0}^n \frac{df_i}{dt}\vec{e}_i + = \left(\frac{df_1}{dt}, \ldots, \frac{df_m}{dt}\right)^t + \end{align*} +\end{theorem} + +\begin{definition}[Line integral in a scalar field] + Let \(\mathcal{C}:[a,b]\to\mathbb{R}^n, t \mapsto \vec{x}(t)\) be a + parametric curve. The \emph{line integral} in a field \(f(\vec{x})\) is the + integral of the signed area under the curve traced in \(\mathbb{R}^n\), and + is computed with + \[ + \int_\mathcal{C} f(\vec{x}) \,d\ell + = \int_\mathcal{C} f(\vec{x}) \,|d\vec{x}| + = \int_a^b f(\vec{x}(t)) |\vec{x}'(t)| \, dt + \] +\end{definition} + +\begin{application}[Length of a parametric curve] + By computing the line integral of the function \(\vec{1}(t) = 1\) we get the + length of the parametric curve \(\mathcal{C}:[a,b]\to\mathbb{R}^n\). + \[ + \int_\mathcal{C}d\ell + = \int_\mathcal{C} |d\vec{x}| + = \int_a^b \sqrt{\sum_{i=1}^n x'_i(t)^2} \,dt + \] + In the special case with the scalar function \(f(x)\) results in + \(\int_a^b\sqrt{1+f'(x)^2}\,dx\) +\end{application} + +\begin{definition}[Line integral in a vector field] + The line integral in a vector field \(\vec{F}(\vec{x})\) is ``sum'' of the + projections of the field's vectors on the tangent of the parametric curve + \(\mathcal{C}\). + \[ + \int_\mathcal{C} \vec{F}(\vec{r})\dotp d\vec{r} + = \int_a^b \vec{F}(\vec{r}(t))\dotp \vec{r}'(t) \,dt + \] +\end{definition} + +\begin{theorem}[Line integral in the opposite direction] + By integrating while moving backwards (\(-t\)) on the parametric curve gives + \[ + \int_{-\mathcal{C}} \vec{F}(\vec{r})\dotp d\vec{r} + = -\int_{\mathcal{C}} \vec{F}(\vec{r})\dotp d\vec{r} + \] +\end{theorem} + +\begin{definition}[Conservative field] + A vector field is said to be \emph{conservative} the line integral over a + closed path is zero. + \[ + \oint_\mathcal{C} \vec{F}(\vec{r})\cdot d\vec{r} = 0 + \] +\end{definition} + +\begin{theorem} + For a twice partially differentiable vector field \(\vec{F}(\vec{x})\) in + \(n\) dimensions without ``holes'', i.e. in which each closed curve can be + contracted to a point (simply connected open set), the following statements + are equivalent: + \begin{itemize} + \item \(\vec{F}\) is conservative + \item \(\vec{F}\) is path-independent + \item \(\vec{F}\) is a \emph{gradient field}, i.e. there is a + function \(\phi\) called \emph{potential} such that \(\vec{F} = \grad + \phi\) + \item \(\vec{F}\) satisfies the condition \(\partial_{x_j} F_i = + \partial_{x_i} F_j\) for all \(i,j \in \{1,2,\ldots,n\}\). In the 2D case + \(\partial_x F_y = \partial_y F_x\), and in 3D + \[ + \begin{cases} + \partial_y F_x = \partial_x F_y \\ + \partial_z F_y = \partial_y F_z \\ + \partial_x F_z = \partial_z F_x \\ + \end{cases} + \] + \end{itemize} +\end{theorem} + +\begin{theorem} + In a conservative field \(\vec{F}\) with gradient \(\phi\), using the + multivariable the chain rule: + \begin{align*} + \int_\mathcal{C} \vec{F} \dotp d\vec{r} + &= \int_\mathcal{C} \vec{F}(\vec{r}(t)) \dotp \vec{r}'(t) \,dt \\ + &= \int_\mathcal{C} \grad \phi(\vec{r}(t)) \cdot \vec{r}'(t) \,dt \\ + &= \int_\mathcal{C} \frac{d\phi(\vec{r}(t))}{dt}\,dt + = \phi(\vec{r}(b)) - \phi(\vec{r}(a)) + \end{align*} +\end{theorem} + +\section{Surface integrals} + +\section{Vector analysis} \section*{License} \doclicenseText |