From 336d46fa6637fb38fef2527350cd72f926860ccb Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Fri, 23 Jul 2021 18:41:01 +0200 Subject: Rewrite everything from scratch --- FuVar.tex | 293 ++++++++++++++++++++++++++++++++++++++++++-------------------- 1 file changed, 200 insertions(+), 93 deletions(-) diff --git a/FuVar.tex b/FuVar.tex index aa4ba54..dba65b3 100644 --- a/FuVar.tex +++ b/FuVar.tex @@ -10,6 +10,7 @@ %% TODO: publish to CTAN \usepackage{tex/hsrstud} +\usepackage{mathtools} %% Language configuration \usepackage{polyglossia} @@ -46,14 +47,6 @@ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Macros and settings -%% number sets -\newcommand\Nset{\mathbb{N}} -\newcommand\Zset{\mathbb{Z}} -\newcommand\Qset{\mathbb{Q}} -\newcommand\Rset{\mathbb{R}} -\newcommand\Cset{\mathbb{C}} -\newcommand\T{\mathrm{T}} - %% Theorems \newtheoremstyle{fuvarzf} % name of the style to be used {\topsep} @@ -67,8 +60,14 @@ \theoremstyle{fuvarzf} \newtheorem{theorem}{Theorem} +\newtheorem{proposition}{Proposition} +\newtheorem{method}{Method} \newtheorem{definition}{Definition} \newtheorem{lemma}{Lemma} +\newtheorem{remark}{Remark} + +\DeclareMathOperator{\tr}{\mathrm{tr}} + \setlist[description]{ format = { \normalfont\itshape } @@ -80,115 +79,223 @@ \begin{document} \maketitle -\tableofcontents - -\section{Fields and vector spaces} - -\begin{definition}[Field] - A field is a set \(F\) with two binary operators \(+\) and \(\cdot\) that map - \(F\times F \to F\) and follow the \emph{field axioms} listed below. We let - \(a, b \in F\) and \(\star\) stands for \(\cdot\) or \(+\). - \begin{description} - \item[Associativity:] \((a \star b) \star c = a \star (b \star c)\) - \item[Commutativity:] \(a \star b = b \star a\) - \item[Identities:] \(0 + a = a\) and \(1\cdot a = a\) - \item[Inverses:] - \(a + (-a) = 0\) and - \(b \cdot b^{-1} = 1\) iff \(b \neq 0\) - \item[Distributivity:] \(a \cdot (b + c) = a\cdot b + a \cdot c\) - \end{description} -\end{definition} - -\begin{theorem} - \(\Rset\) is a field. -\end{theorem} +% \tableofcontents -\begin{definition}[Vector space] - A vector space \(U\) over a field \(F\) is a set of objects called - \emph{vectors} equipped with two operations: \emph{addition} - \(+: U \times U \to U\) and \emph{scalar multiplication} - \(\cdot: F\times U \to U\), that respect the following axioms. - Let \(\vec{u}, \vec{v}, \vec{w} \in U\) and \(a, b \in F\). - \begin{description} - \item[Additive associativity:] \((\vec{u} + \vec{v}) + \vec{w} - = \vec{u} + (\vec{v} + \vec{w})\) - \item[Additive commutativity:] \(\vec{u} + \vec{v} = \vec{v} + \vec{u}\) - \item[Identities:] There is an element - \(\vec{0} \in U : \vec{u} + \vec{0} = \vec{u}\) - and \(1 \in F : 1 \cdot \vec{u} = \vec{u}\) - \item[Additive inverse:] \(\vec{u} + (\vec{-u}) = 0\) - \item[Compatibility of multiplication] - \(a\cdot (b \cdot \vec{u}) = (a\cdot b) \cdot \vec{u}\) - \item[Distributivity:] - \((a + b) \cdot \vec{u} = a\cdot\vec{u} + b\cdot\vec{u}\) and conversely - \(a \cdot (\vec{u} + \vec{v}) = a\cdot\vec{u} + a\cdot\vec{v}\) - \end{description} - And of course elements in \(F\) follow the field axioms. -\end{definition} +\section{Preface} -\begin{theorem} - \(\Rset^n = \Rset\times\cdots\times\Rset\) is a vector space. -\end{theorem} +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. -\begin{definition}[Row and column vectors] - Although there is virtually no difference between the two, we need two type - of \(n\)-tuples that satisfy the vector space axioms. \emph{Row} vectors are - written horizontally and \emph{column} vectors vertically. -\end{definition} +\section{Derivatives of vector valued scalar functions} -\begin{definition}[Transposition] - Let \(\vec{u} \in \Rset^n\) be a row vector. The \emph{transpose} of - \(\vec{u}\) denoted with \(\vec{u}^\T\) is column vector with the same - components. Conversely if \(\vec{v}\) is a column vector then \(\vec{v}^\T\) - is a row vector. +\begin{definition}[Partial derivative] + A vector values 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 + \[ + \partial_{v_i} f(\vec{v}) + = f_{v_i}(\vec{v}) + = \lim_{h\to 0} \frac{f(\vec{v} + h\vec{e}_j) - f(\vec{v})}{h} + \] \end{definition} -\section{Scalar fields} +\begin{proposition} + Under some generally satisfied conditions (continuity of \(n\)-th order + partial derivatives) Schwarz's theorem states that it is possible to swap + the order of differentiation. + \[ + \partial_x \partial_y f(x,y) = \partial_y \partial_x f(x,y) + \] +\end{proposition} -\begin{definition}[Scalar field] - Confusingly we call a function \(f: \Rset^n \to \Rset\) a \emph{scalar - field}, but this is unrelated to the previously defined field. +\begin{definition}[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} -\begin{definition}[Partial derivative of a scalar field] - Let \(f: \Rset^n \to \Rset\), the \emph{partial} derivative of \(f\) with - respect to \(x_k\), (\(0 < k \leq n\)), is defined as +\begin{theorem}[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 + \[ + y = \bar{y} \pm \sigma_y \approx f(\bar{\vec{x}}) + \pm \sqrt{\sum_{i=1}^m \left( + \partial_{x_i} f(\bar{\vec{x}}) \sigma_{x_i}\right)^2} + \] +\end{theorem} + +\begin{definition}[Gradient vector] + The \emph{gradient} of a function \(f(\vec{x}), \vec{x}\in\mathbb{R}^m\) is a + vector containing the derivatives in each direction. \[ - \frac{\partial f}{\partial x_k} := - \lim_{h \to 0} \frac{f(x_1, \dots, x_k + h, \dots, x_n) - - f(x_1, \dots, x_k, \dots, x_n)}{h} - = \partial_{x_k} f(x, y) + \grad f (\vec{x}) = \sum_{i=1}^m \partial_{x_i} f(\vec{x}) \vec{e}_i + = \begin{pmatrix} + \partial_{x_1} f(\vec{x}) \\ + \vdots \\ + \partial_{x_m} f(\vec{x}) \\ + \end{pmatrix} \] - That is, we keep all variables of \(f\) fixed, except for \(x_k\). \end{definition} -\begin{definition}[Tangent plane] - For a scalar field \(f(x,y)\) we define the \emph{tangent plane} \(p(x,y)\) - at coordinates \((x_0, y_0)\) to be: +\begin{definition}[Directional derivative] + A function \(f(\vec{x})\) has a directional derivative in direction + \(\vec{r}\) (with \(|\vec{r}| = 1\)) given by \[ - p(x, y) = - f(x_0, y_0) - + \partial_x f(x_0, y_0) (x - x_0) - + \partial_y f(x_0, y_0) (y - y_0) + \frac{\partial f}{\partial\vec{r}} = \nabla_\vec{r} f = \vec{r} \dotp \grad f \] \end{definition} -The above can be used to calculate the one dimensional derivative of an implicit curve. +\begin{theorem} + The gradient vector always points towards \emph{the direction of steepest ascent}. +\end{theorem} -\begin{lemma}[Implicit derivative] - The slope \(m\) of an implicit curve \(f(x,y)\) at the point \((x_0, y_0)\) is given by +\subsection{Methods for maximization and minimization problems} + +\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 where the gradient \(\grad f\) is not defined, + \item that are stationary, i.e. where \(\grad f = \vec{0}\). + \end{itemize} +\end{method} + +\begin{method}[Determine the type of stationary point for 2 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 define the + \emph{discriminant} \[ - m = \partial_x f(x_0, y_0) / \partial_y f(x_0, y_0) + \Delta = \partial_x^2 f \partial_y^2 f - \partial_y \partial_x f \] - of course only if \(\partial_y f(x_0, y_0) \neq 0\). -\end{lemma} - -\begin{definition}[Total derivative] + \begin{itemize} + \item if \(\Delta > 0\) then \(\vec{x}_s\) is an extrema, if \(\partial_x^2 + f(\vec{x}_s) < 0\) it is a maximum, whereas if \(\partial_x^2 + f(\vec{x}_s) > 0\) it is a minimum; + + \item if \(\Delta < 0\) then \(\vec{x}_s\) is a saddle point; + + \item if \(\Delta = 0\) we need to analyze further. + \end{itemize} +\end{method} + +\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 \(\vec{r} = \vec{e}_1\cos(\alpha) + \vec{e}_2\sin(\alpha)\) +\end{remark} + +\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 + \emph{Hessian} matrix. + \( + \left(\mtx{H}_f\right)_{i,j} = \partial_{x_i}\partial_{x_j} f + \) + or \[ - \dd{f} + \mtx{H}_f = \begin{pmatrix} + \partial_{x_1}\partial_{x_1} f & \cdots & \partial_{x_1}\partial_{x_m} f \\ + \vdots & \ddots & \vdots \\ + \partial_{x_m}\partial_{x_1} f & \cdots & \partial_{x_m}\partial_{x_m} f \\ + \end{pmatrix} \] + Because (almost always) the order of differentiation + does not matter, it is a symmetric matrix. \end{definition} +\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 \(\mtx{H}_f(\vec{x}_s)\). Then we compute its eigenvalues + \(\lambda_1, \ldots, \lambda_m\) and + \begin{itemize} + \item if all \(\lambda_i > 0\), the point is a minimum; + \item if all \(\lambda_i < 0\), the point is a maximum; + \item if there are both positive and negative eigenvalues, + it is a saddle point. + \end{itemize} + In the other cases, when there are \(\lambda_i \leq 0\) and/or \(\lambda_i + \geq 0\) further analysis is required. +\end{method} + +\begin{remark} + Recall that to compute the eigenvalues of a matrix, one must solve the + equation \((\mtx{H} - \lambda\mtx{I})\vec{x} = \vec{0}\). Which can be done + by solving the characteristic polynomial \(\det\left(\mtx{H} - + \lambda\mtx{I}\right) = 0\) to obtain \(\dim(\mtx{H})\) \(\lambda_i\), which + when plugged back in result in a overdetermined system of equations. +\end{remark} + +\begin{method}[Quickly find the eigenvalues of a \(2\times 2\) matrix] + Let + \[ + m = \frac{1}{2}\tr \mtx{H} = \frac{a + d}{2} + \text{ and } + p = \det\mtx{H} = ad - bc , + \] + then + \[ + \lambda = m \pm \sqrt{m^2 - p} . + \] +\end{method} + +\begin{method}[Search for a constrained extremum in 2 dimensions] + Let \(n(x,y) = 0\) be a constraint in the search of the extrema of a function + \(f: D \subseteq \mathbb{R}^2 \to \mathbb{R}\). 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 \(\vec{u}\) where the gradient either does not exist or is + \(\vec{0}\), and satisfy \(n(\vec{u}) = 0\); + + \item that solve the system of equations + \[ + \begin{cases} + \partial_x f(\vec{u}) \cdot \partial_y n(\vec{u}) + = \partial_y f(\vec{u}) \cdot \partial_x n(\vec{u}) \\ + n(\vec{u}) = 0 + \end{cases} + \] + \end{itemize} +\end{method} + +\begin{method}[% + Search for a constrained extremum in higher dimensions, + method of Lagrange multipliers] + We wish to find the extrema of \(f: D \subseteq \mathbb{R}^m \to \mathbb{R}\) + under \(k < m\) constraints \(n_1 = 0, \cdots, n_k = 0\). For that we consider + the following points: + \begin{itemize} + \item Points on the 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 + \begin{itemize} + \item any of \(\grad f, \grad n_1, \ldots, \grad n_k\) do not exist, or + \item \(\grad n_1, \ldots, \grad n_k\) are linearly \emph{dependent}, + \end{itemize} + and that satisfy \(0 = n_1(\vec{u}) = \ldots = n_k(\vec{u})\). + + \item Points that solve the system of \(m+k\) equations + \[ + \begin{dcases} + \grad f(\vec{u}) = \sum_{i = 1}^k \lambda_k \grad n_i(\vec{u}) + & (m\text{-dimensional}) \\ + n_i(\vec{u}) = 0 & \text{ for } 1 \leq i \leq k + \end{dcases} + \] + The \(\lambda\) values are known as \emph{Lagrange multipliers}. + \end{itemize} +\end{method} \section*{License} \doclicenseText -- cgit v1.2.1