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@@ -30,6 +30,7 @@ %% Layout \usepackage{enumitem} +\usepackage{booktabs} %% Nice drwaings \usepackage{tikz} @@ -63,11 +64,11 @@ \theoremstyle{fuvarzf} \newtheorem{theorem}{Theorem} -\newtheorem{proposition}{Proposition} \newtheorem{method}{Method} +\newtheorem{application}{Application} \newtheorem{definition}{Definition} -\newtheorem{lemma}{Lemma} \newtheorem{remark}{Remark} +\newtheorem{note}{Note} \DeclareMathOperator{\tr}{\mathrm{tr}} @@ -90,7 +91,7 @@ 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 scalar functions} +\section{Derivatives of vector valued functions} \begin{definition}[Partial derivative] A vector values function \(f: \mathbb{R}^m\to\mathbb{R}\), with @@ -103,14 +104,14 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \] \end{definition} -\begin{proposition} +\begin{theorem}(Schwarz's theorem, symmetry of partial derivatives) 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} +\end{theorem} \begin{definition}[Linearization] A function \(f: \mathbb{R}^m\to\mathbb{R}\) has a linearization \(g\) at @@ -135,7 +136,10 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \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. + column vector\footnote{In matrix notation it is also often defined as row + vector to avoid having to do some transpositions in the Jacobian matrix and + dot products in directional derivatives} containing the derivatives in each + direction. \[ \grad f (\vec{x}) = \sum_{i=1}^m \partial_{x_i} f(\vec{x}) \vec{e}_i = \begin{pmatrix} @@ -151,7 +155,7 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \(\vec{r}\) (with \(|\vec{r}| = 1\)) given by \[ \frac{\partial f}{\partial\vec{r}} - = \nabla_\vec{r} f = \vec{r} \dotp \grad f + = \nabla\vec{r} f = \vec{r} \dotp \grad f \] \end{definition} @@ -160,6 +164,52 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. 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 + \(\vec{f}: \mathbb{R}^n \to \mathbb{R}^m\) is a matrix + \(\in\mathbb{R}^{n\times m}\) whose entry at the \(i\)-th row and \(j\)-th + column is given by \((\mx{J}_f)_{i,j} = \partial_{x_j} f_i\), so + \[ + \mx{J}_f = \begin{pmatrix} + \partial_{x_1} f_1 & \cdots & \partial_{x_n} f_1 \\ + \vdots & \ddots & \vdots \\ + \partial_{x_1} f_m & \cdots & \partial_{x_n} f_m \\ + \end{pmatrix} + = \begin{pmatrix} + (\grad f_1)^t \\ + \vdots \\ + (\grad f_m)^t \\ + \end{pmatrix} + \] +\end{definition} + +\begin{remark} + In the scalar case (\(m = 1\)) the Jacobian matrix is the transpose of the + gradient vector. +\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. + \( + \left(\mx{H}_f\right)_{i,j} = \partial_{x_i}\partial_{x_j} f + \) + or + \[ + \mx{H}_f = \begin{pmatrix} + \partial_{x_1}\partial_{x_1} f & \cdots & \partial_{x_1}\partial_{x_m} f \\ + \vdots & \ddots & \vdots \\ + \partial_{x_m}\partial_{x_1} f & \cdots & \partial_{x_m}\partial_{x_m} f \\ + \end{pmatrix} + \] + Because (almost always) the order of differentiation + does not matter, it is a symmetric matrix. +\end{definition} + + \section{Methods for maximization and minimization problems} \begin{method}[Find stationary points] @@ -196,30 +246,10 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. 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 - \[ - \mtx{H}_f = \begin{pmatrix} - \partial_{x_1}\partial_{x_1} f & \cdots & \partial_{x_1}\partial_{x_m} f \\ - \vdots & \ddots & \vdots \\ - \partial_{x_m}\partial_{x_1} f & \cdots & \partial_{x_m}\partial_{x_m} f \\ - \end{pmatrix} - \] - 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 + Hessian matrix \(\mx{H}_f(\vec{x}_s)\). Then we compute its eigenvalues \(\lambda_1, \ldots, \lambda_m\) and \begin{itemize} \item if all \(\lambda_i > 0\), the point is a minimum; @@ -233,23 +263,20 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \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 + equation \((\mx{H} - \lambda\mx{I})\vec{x} = \vec{0}\). Which can be done + by solving the characteristic polynomial \(\det\left(\mx{H} - + \lambda\mx{I}\right) = 0\) to obtain \(\dim(\mx{H})\) \(\lambda_i\), 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 - \[ - \lambda = m \pm \sqrt{m^2 - p} . + m = \frac{1}{2}\tr \mx{H} = \frac{a + d}{2} , + \qquad + p = \det\mx{H} = ad - bc , \] + then \(\lambda = m \pm \sqrt{m^2 - p}\). \end{method} \begin{method}[Search for a constrained extremum in 2 dimensions] @@ -273,6 +300,15 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \end{itemize} \end{method} +\begin{figure} + \centering + \includegraphics{img/lagrange-multipliers} + \caption{ + Intuition for the method of Lagrange multipliers. Extrema of a constrained + function are where \(\grad f\) is proportional to \(\grad n\). + } +\end{figure} + \begin{method}[% Search for a constrained extremum in higher dimensions, method of Lagrange multipliers] @@ -305,16 +341,80 @@ typesetting may trick you into thinking it is rigorous, but really, it is not. \mathcal{L}(\vec{u}, \vec{\lambda}) = f(\vec{u}) - \sum_{i = 0}^k \lambda_i n_i(\vec{u}) \] - where \(\vec{\lambda} = \lambda_1, \ldots, \lambda_k\) and then - evaluating \(\grad \mathcal{L}(\vec{u}, \vec{\lambda}) = \vec{0}\). + 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. \end{itemize} \end{method} -\section{Integration} -\begin{remark} - -\end{remark} +\section{Integration of vector values scalar functions} + +\begin{figure} + \centering + \includegraphics{img/double-integral} + \caption{ + Double integral. + \label{fig:double-integral} + } +\end{figure} + +\begin{theorem}[Change the order of integration for double integrals] For a + double integral over a region \(S\) (see Fig. \ref{fig:double-integral}) we + need to compute + \[ + \iint_S f(x,y) \,ds = + \int\limits_{x_1}^{x_2} \int\limits_{y_1(x)}^{y_2(x)} f(x,y) \,dydx . + \] + If \(y_1(x)\) and \(y_2(x)\) are bijective we can swap the order of + integration by finding the inverse functions \(x_1(y)\) and \(x_2(y)\). If + they are not bijective (like in Fig. \ref{fig:double-integral}), the region + must be split into smaller parts. If the region is a rectangle it is always + possible to change the order of integration. +\end{theorem} + +\begin{theorem}[Transformation of coordinates in 2 dimensions] + \label{thm:transform-coords} + Given two ``nice'' functions \(x(u,v)\) and \(y(u,v)\), that means are a + bijection from \(S\) to \(S'\) with continuous partial derivatives and + nonzero Jacobian determinant \(|\mx{J}_f| = \partial_u x \partial_v y - + \partial_v x \partial_u y\), which transform the coordinate system. Then + \[ + \iint_S f(x,y) \,ds = \iint_{S'} f(x(u,v), y(u,v)) |\mx{J}_f| \,ds + \] +\end{theorem} + +\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 + region \(B\), we let \(\vec{x}(\vec{u})\) be ``nice'' functions that + transform the coordinate system. Then as before + \[ + \int_B f(\vec{x}) \,ds = \int_{B'} f(\vec{x}(\vec{u})) |\mx{J}_f| \,ds + \] +\end{theorem} +\begin{table} + \centering + \begin{tabular}{l >{\(}l<{\)} >{\(}l<{\)}} + \toprule + & \text{Volume } dv & \text{Surface } d\vec{s}\\ + \midrule + Cartesian & - & dx\,dy \\ + Polar & - & rd\,rd\phi \\ + Curvilinear & - & |\mx{J}_f|\,du\,dv \\ + \midrule + Cartesian & dx\,dy\,dz & \uvec{z}\,dx\,dy \\ + 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 \\ + Curvilinear & |\mx{J}_f|\,du\,dv\,dw & - \\ + \bottomrule + \end{tabular} + \caption{Differential elements for integration.} +\end{table} + +\section{Derivatives of curves} \section*{License} \doclicenseText |