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command---while reading file lecture_12.aux +You've used 0 entries, + 0 wiz_defined-function locations, + 83 strings with 494 characters, +and the built_in function-call counts, 0 in all, are: += -- 0 +> -- 0 +< -- 0 ++ -- 0 +- -- 0 +* -- 0 +:= -- 0 +add.period$ -- 0 +call.type$ -- 0 +change.case$ -- 0 +chr.to.int$ -- 0 +cite$ -- 0 +duplicate$ -- 0 +empty$ -- 0 +format.name$ -- 0 +if$ -- 0 +int.to.chr$ -- 0 +int.to.str$ -- 0 +missing$ -- 0 +newline$ -- 0 +num.names$ -- 0 +pop$ -- 0 +preamble$ -- 0 +purify$ -- 0 +quote$ -- 0 +skip$ -- 0 +stack$ -- 0 +substring$ -- 0 +swap$ -- 0 +text.length$ -- 0 +text.prefix$ -- 0 +top$ -- 0 +type$ -- 0 +warning$ -- 0 +while$ -- 0 +width$ -- 0 +write$ -- 0 +(There were 3 error messages) diff --git a/lecture_12/lecture_12.ipynb b/lecture_12/lecture_12.ipynb index dc44a21..2937e47 100644 --- a/lecture_12/lecture_12.ipynb +++ b/lecture_12/lecture_12.ipynb @@ -2,7 +2,7 @@ "cells": [ { "cell_type": "code", - "execution_count": 27, + "execution_count": 2, "metadata": { "collapsed": true }, @@ -13,7 +13,7 @@ }, { "cell_type": "code", - "execution_count": 28, + "execution_count": 3, "metadata": { "collapsed": true }, @@ -22,6 +22,97 @@ "setdefaults" ] }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A =\n", + "\n", + " 0.447394 0.357071 0.720915 0.499926\n", + " 0.648313 0.323276 0.521677 0.288345\n", + " 0.084982 0.581513 0.466420 0.142342\n", + " 0.576580 0.658089 0.916987 0.923165\n", + "\n", + "L =\n", + "\n", + " 1.00000 0.00000 0.00000 0.00000\n", + " 0.13108 1.00000 0.00000 0.00000\n", + " 0.69009 0.24851 1.00000 0.00000\n", + " 0.88935 0.68736 0.68488 1.00000\n", + "\n", + "U =\n", + "\n", + " 0.64831 0.32328 0.52168 0.28834\n", + " 0.00000 0.53914 0.39804 0.10455\n", + " 0.00000 0.00000 0.26199 0.27496\n", + " 0.00000 0.00000 0.00000 0.40655\n", + "\n", + "P =\n", + "\n", + "Permutation Matrix\n", + "\n", + " 0 1 0 0\n", + " 0 0 1 0\n", + " 1 0 0 0\n", + " 0 0 0 1\n", + "\n", + "ans = 1\n" + ] + } + ], + "source": [ + "A=rand(4,4)\n", + "\n", + "[L,U,P]=lu(A)\n", + "\n", + "det(L)" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ans =\n", + "\n", + " 4 4\n", + "\n", + "ans = 23.586\n", + "ans = 35.826\n", + "ans = 14.869\n", + "C =\n", + "\n", + " 5.98549 4.28555 4.35707 4.31359\n", + " 0.00000 3.63950 1.35005 1.45342\n", + " 0.00000 0.00000 3.62851 1.50580\n", + " 0.00000 0.00000 0.00000 3.21911\n", + "\n" + ] + } + ], + "source": [ + "A=rand(4,100)';\n", + "A=A'*A;\n", + "size(A)\n", + "min(min(A))\n", + "max(max(A))\n", + "cond(A)\n", + "C=chol(A)" + ] + }, { "cell_type": "markdown", "metadata": {}, @@ -90,7 +181,7 @@ "1\\end{array}\\right]$\n", "\n", "$A^{-1}=\\frac{1}{2*3-1*1}\\left[ \\begin{array}{cc}\n", - "3 & 1 \\\\\n", + "3 & -1 \\\\\n", "-1 & 2 \\end{array} \\right]=\n", "\\left[ \\begin{array}{cc}\n", "3/5 & -1/5 \\\\\n", @@ -99,7 +190,7 @@ }, { "cell_type": "code", - "execution_count": 2, + "execution_count": 45, "metadata": { "collapsed": false }, @@ -203,7 +294,7 @@ "\n", "$A^{-1}=\\left[ \\begin{array}{cccc}\n", "| & | & & | \\\\\n", - "a_{1} & a_{2} & \\cdots & a_{3} \\\\\n", + "a_{1} & a_{2} & \\cdots & a_{n} \\\\\n", "| & | & & | \\end{array} \\right]$\n", "\n", "\n", @@ -236,9 +327,26 @@ "0 & -1 & 1 \\end{array} \\right]$\n" ] }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#### Note on solving for $A^{-1}$ column 1\n", + "\n", + "$Aa_1=I(:,1)$\n", + "\n", + "$LUa_1=I(:,1)$\n", + "\n", + "$(LUa_1-I(:,1))=0$\n", + "\n", + "$L(Ua_1-d_1)=0$\n", + "\n", + "$I(:,1)=Ld_1$" + ] + }, { "cell_type": "code", - "execution_count": 21, + "execution_count": 56, "metadata": { "collapsed": false }, @@ -278,7 +386,7 @@ }, { "cell_type": "code", - "execution_count": 22, + "execution_count": 57, "metadata": { "collapsed": false }, @@ -317,13 +425,23 @@ "1 \\\\ \n", "0 \\\\ \n", "\\vdots \\\\\n", - "0 \\end{array} \\right]$\n", - "$;~Ua_{1}=d_{1}$" + "0 \\end{array} \\right]=\n", + "\\left[\\begin{array}{ccc} \n", + "1 & 0 & 0 \\\\ \n", + "-1/2 & 1 & 0 \\\\\n", + "0 & -2/3 & 1 \\end{array} \\right]\\left[\\begin{array}{c} \n", + "d1(1) \\\\ \n", + "d1(2) \\\\ \n", + "d1(3)\\end{array} \\right]=\\left[\\begin{array}{c} \n", + "1 \\\\ \n", + "0 \\\\ \n", + "0 \\end{array} \\right]\n", + ";~Ua_{1}=d_{1}$" ] }, { "cell_type": "code", - "execution_count": 29, + "execution_count": 58, "metadata": { "collapsed": false }, @@ -350,7 +468,7 @@ }, { "cell_type": "code", - "execution_count": 30, + "execution_count": 59, "metadata": { "collapsed": false }, @@ -377,7 +495,7 @@ }, { "cell_type": "code", - "execution_count": 28, + "execution_count": 60, "metadata": { "collapsed": false }, @@ -404,7 +522,7 @@ }, { "cell_type": "code", - "execution_count": 31, + "execution_count": 61, "metadata": { "collapsed": false }, @@ -431,7 +549,7 @@ }, { "cell_type": "code", - "execution_count": 37, + "execution_count": 62, "metadata": { "collapsed": false }, @@ -458,7 +576,7 @@ }, { "cell_type": "code", - "execution_count": 38, + "execution_count": 63, "metadata": { "collapsed": false }, @@ -492,7 +610,7 @@ }, { "cell_type": "code", - "execution_count": 40, + "execution_count": 69, "metadata": { "collapsed": false }, @@ -507,18 +625,25 @@ " 1.00000 2.00000 2.00000\n", " 1.00000 2.00000 3.00000\n", "\n", - "ans =\n", + "I_app =\n", "\n", " 1.00000 0.00000 0.00000\n", " 0.00000 1.00000 -0.00000\n", " -0.00000 -0.00000 1.00000\n", - "\n" + "\n", + "ans = -4.4409e-16\n", + "ans = 2.2204e-16\n", + "ans = 0.0039062\n" ] } ], "source": [ "invA=[a1,a2,a3]\n", - "A*invA" + "I_app=A*invA\n", + "I_app(2,3)\n", + "eps\n", + "\n", + "2^-8" ] }, { @@ -530,7 +655,7 @@ }, { "cell_type": "code", - "execution_count": 44, + "execution_count": 70, "metadata": { "collapsed": false }, @@ -577,7 +702,7 @@ }, { "cell_type": "code", - "execution_count": 61, + "execution_count": 4, "metadata": { "collapsed": false }, @@ -703,7 +828,7 @@ "\t\n", "\n", "\n", - "\t\n", + "\t\n", "\t\n", "\tbackslash\n", "\n", @@ -712,7 +837,7 @@ "\t\n", "\n", "\n", - "\t\n", + "\t\n", "\t\n", "\tmultiplication\n", "\n", @@ -721,7 +846,7 @@ "\t\n", "\n", "\n", - "\t\n", + "\t\n", "\t\n", "\n", "\n", @@ -813,7 +938,7 @@ }, { "cell_type": "code", - "execution_count": 1, + "execution_count": 72, "metadata": { "collapsed": false }, @@ -864,7 +989,7 @@ }, { "cell_type": "code", - "execution_count": 12, + "execution_count": 75, "metadata": { "collapsed": false }, @@ -908,7 +1033,7 @@ }, { "cell_type": "code", - "execution_count": 11, + "execution_count": 74, "metadata": { "collapsed": false }, diff --git a/lecture_12/lecture_12.log b/lecture_12/lecture_12.log new file mode 100644 index 0000000..c4a5ed3 --- /dev/null +++ b/lecture_12/lecture_12.log @@ -0,0 +1,841 @@ +This is pdfTeX, Version 3.14159265-2.6-1.40.16 (TeX 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+{/usr/share/texlive/texmf-dist/fonts/enc/dvips/base/8r.enc}{/usr/share/texmf/ +fonts/enc/dvips/cm-super/cm-super-ts1.enc}{/usr/share/texmf/fonts/enc/dvips/cm- +super/cm-super-t1.enc} +Output written on lecture_12.pdf (14 pages, 232410 bytes). +PDF statistics: + 177 PDF objects out of 1000 (max. 8388607) + 140 compressed objects within 2 object streams + 34 named destinations out of 1000 (max. 500000) + 77 words of extra memory for PDF output out of 10000 (max. 10000000) + diff --git a/lecture_12/lecture_12.md b/lecture_12/lecture_12.md index 9befcbc..82f9bea 100644 --- a/lecture_12/lecture_12.md +++ b/lecture_12/lecture_12.md @@ -9,6 +9,75 @@ setdefaults ``` + +```octave +A=rand(4,4) + +[L,U,P]=lu(A) + +det(L) +``` + + A = + + 0.447394 0.357071 0.720915 0.499926 + 0.648313 0.323276 0.521677 0.288345 + 0.084982 0.581513 0.466420 0.142342 + 0.576580 0.658089 0.916987 0.923165 + + L = + + 1.00000 0.00000 0.00000 0.00000 + 0.13108 1.00000 0.00000 0.00000 + 0.69009 0.24851 1.00000 0.00000 + 0.88935 0.68736 0.68488 1.00000 + + U = + + 0.64831 0.32328 0.52168 0.28834 + 0.00000 0.53914 0.39804 0.10455 + 0.00000 0.00000 0.26199 0.27496 + 0.00000 0.00000 0.00000 0.40655 + + P = + + Permutation Matrix + + 0 1 0 0 + 0 0 1 0 + 1 0 0 0 + 0 0 0 1 + + ans = 1 + + + +```octave +A=rand(4,100)'; +A=A'*A; +size(A) +min(min(A)) +max(max(A)) +cond(A) +C=chol(A) +``` + + ans = + + 4 4 + + ans = 23.586 + ans = 35.826 + ans = 14.869 + C = + + 5.98549 4.28555 4.35707 4.31359 + 0.00000 3.63950 1.35005 1.45342 + 0.00000 0.00000 3.62851 1.50580 + 0.00000 0.00000 0.00000 3.21911 + + + ## My question from last class ![q1](det_L.png) @@ -68,7 +137,7 @@ x_{2} \end{array}\right]= 1\end{array}\right]$ $A^{-1}=\frac{1}{2*3-1*1}\left[ \begin{array}{cc} -3 & 1 \\ +3 & -1 \\ -1 & 2 \end{array} \right]= \left[ \begin{array}{cc} 3/5 & -1/5 \\ @@ -156,7 +225,7 @@ Which we can solve for each $a_{n}$ with LU-decomposition, knowing the lower and $A^{-1}=\left[ \begin{array}{cccc} | & | & & | \\ -a_{1} & a_{2} & \cdots & a_{3} \\ +a_{1} & a_{2} & \cdots & a_{n} \\ | & | & & | \end{array} \right]$ @@ -189,6 +258,18 @@ $A=\left[ \begin{array}{ccc} 0 & -1 & 1 \end{array} \right]$ +#### Note on solving for $A^{-1}$ column 1 + +$Aa_1=I(:,1)$ + +$LUa_1=I(:,1)$ + +$(LUa_1-I(:,1))=0$ + +$L(Ua_1-d_1)=0$ + +$I(:,1)=Ld_1$ + ```octave A=[2,-1,0;-1,2,-1;0,-1,1] @@ -245,8 +326,18 @@ $Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ -0 \end{array} \right]$ -$;~Ua_{1}=d_{1}$ +0 \end{array} \right]= +\left[\begin{array}{ccc} +1 & 0 & 0 \\ +-1/2 & 1 & 0 \\ +0 & -2/3 & 1 \end{array} \right]\left[\begin{array}{c} +d1(1) \\ +d1(2) \\ +d1(3)\end{array} \right]=\left[\begin{array}{c} +1 \\ +0 \\ +0 \end{array} \right] +;~Ua_{1}=d_{1}$ ```octave @@ -349,7 +440,11 @@ Final solution for $A^{-1}$ is $[a_{1}~a_{2}~a_{3}]$ ```octave invA=[a1,a2,a3] -A*invA +I_app=A*invA +I_app(2,3) +eps + +2^-8 ``` invA = @@ -358,12 +453,15 @@ A*invA 1.00000 2.00000 2.00000 1.00000 2.00000 3.00000 - ans = + I_app = 1.00000 0.00000 0.00000 0.00000 1.00000 -0.00000 -0.00000 -0.00000 1.00000 + ans = -4.4409e-16 + ans = 2.2204e-16 + ans = 0.0039062 Now the solution of $x$ to $Ax=y$ is $x=A^{-1}y$ @@ -430,7 +528,7 @@ legend('inversion','backslash','multiplication','Location','NorthWest') ``` -![svg](lecture_12_files/lecture_12_21_0.svg) +![svg](lecture_12_files/lecture_12_24_0.svg) ## Condition of a matrix diff --git a/lecture_12/lecture_12.out b/lecture_12/lecture_12.out new file mode 100644 index 0000000..e7b5af8 --- /dev/null +++ b/lecture_12/lecture_12.out @@ -0,0 +1,7 @@ +\BOOKMARK [2][-]{subsection.0.1}{My question from last class}{}% 1 +\BOOKMARK [2][-]{subsection.0.2}{Your questions from last class}{}% 2 +\BOOKMARK [1][-]{section.1}{Matrix Inverse and Condition}{}% 3 +\BOOKMARK [2][-]{subsection.1.1}{Condition of a matrix}{section.1}% 4 +\BOOKMARK [3][-]{subsubsection.1.1.1}{just checked in to see what condition my condition was in}{subsection.1.1}% 5 +\BOOKMARK [3][-]{subsubsection.1.1.2}{Matrix norms}{subsection.1.1}% 6 +\BOOKMARK [3][-]{subsubsection.1.1.3}{Condition of Matrix}{subsection.1.1}% 7 diff --git a/lecture_12/lecture_12.pdf b/lecture_12/lecture_12.pdf index ed0f569..e0b4e6c 100644 Binary files a/lecture_12/lecture_12.pdf and b/lecture_12/lecture_12.pdf differ diff --git a/lecture_12/lecture_12.tex b/lecture_12/lecture_12.tex new file mode 100644 index 0000000..1265a06 --- /dev/null +++ b/lecture_12/lecture_12.tex @@ -0,0 +1,1015 @@ + +% Default to the notebook output style + + + + +% Inherit from the specified cell style. + + + + + +\documentclass[11pt]{article} + + + + \usepackage[T1]{fontenc} + % Nicer default font (+ math font) than Computer Modern for most use cases + \usepackage{mathpazo} + + % Basic figure setup, for now with no caption control since it's done + % automatically by Pandoc (which extracts ![](path) syntax from Markdown). + \usepackage{graphicx} + % We will generate all images so they have a width \maxwidth. This means + % that they will get their normal width if they fit onto the page, but + % are scaled down if they would overflow the margins. + \makeatletter + \def\maxwidth{\ifdim\Gin@nat@width>\linewidth\linewidth + \else\Gin@nat@width\fi} + \makeatother + \let\Oldincludegraphics\includegraphics + % Set max figure width to be 80% of text width, for now hardcoded. + \renewcommand{\includegraphics}[1]{\Oldincludegraphics[width=.8\maxwidth]{#1}} + % Ensure that by default, figures have no caption (until we provide a + % proper Figure object with a Caption API and a way to capture that + % in the conversion process - todo). + \usepackage{caption} + \DeclareCaptionLabelFormat{nolabel}{} + \captionsetup{labelformat=nolabel} + + \usepackage{adjustbox} % Used to constrain images to a maximum size + \usepackage{xcolor} % Allow colors to be defined + \usepackage{enumerate} % Needed for markdown enumerations to work + \usepackage{geometry} % Used to adjust the document margins + \usepackage{amsmath} % Equations + \usepackage{amssymb} % Equations + \usepackage{textcomp} % defines textquotesingle + % Hack from http://tex.stackexchange.com/a/47451/13684: + \AtBeginDocument{% + \def\PYZsq{\textquotesingle}% Upright quotes in Pygmentized code + } + \usepackage{upquote} % Upright quotes for verbatim code + \usepackage{eurosym} % defines \euro + \usepackage[mathletters]{ucs} % Extended unicode (utf-8) support + \usepackage[utf8x]{inputenc} % Allow utf-8 characters in the tex document + \usepackage{fancyvrb} % verbatim replacement that allows latex + \usepackage{grffile} % extends the file name processing of package graphics + % to support a larger range + % The hyperref package gives us a pdf with properly built + % internal navigation ('pdf bookmarks' for the table of contents, + % internal cross-reference links, web links for URLs, etc.) + \usepackage{hyperref} + \usepackage{longtable} % longtable support required by pandoc >1.10 + \usepackage{booktabs} % table support for pandoc > 1.12.2 + \usepackage[inline]{enumitem} % IRkernel/repr support (it uses the enumerate* environment) + \usepackage[normalem]{ulem} % ulem is needed to support strikethroughs (\sout) + % normalem makes italics be italics, not underlines + + + + + % Colors for the hyperref package + \definecolor{urlcolor}{rgb}{0,.145,.698} + \definecolor{linkcolor}{rgb}{.71,0.21,0.01} + \definecolor{citecolor}{rgb}{.12,.54,.11} + + % ANSI colors + \definecolor{ansi-black}{HTML}{3E424D} + \definecolor{ansi-black-intense}{HTML}{282C36} + \definecolor{ansi-red}{HTML}{E75C58} + \definecolor{ansi-red-intense}{HTML}{B22B31} + \definecolor{ansi-green}{HTML}{00A250} + \definecolor{ansi-green-intense}{HTML}{007427} + \definecolor{ansi-yellow}{HTML}{DDB62B} + \definecolor{ansi-yellow-intense}{HTML}{B27D12} + \definecolor{ansi-blue}{HTML}{208FFB} + \definecolor{ansi-blue-intense}{HTML}{0065CA} + \definecolor{ansi-magenta}{HTML}{D160C4} + \definecolor{ansi-magenta-intense}{HTML}{A03196} + \definecolor{ansi-cyan}{HTML}{60C6C8} + \definecolor{ansi-cyan-intense}{HTML}{258F8F} + \definecolor{ansi-white}{HTML}{C5C1B4} + \definecolor{ansi-white-intense}{HTML}{A1A6B2} + + % commands and environments needed by pandoc snippets + % extracted from the output of `pandoc -s` + \providecommand{\tightlist}{% + \setlength{\itemsep}{0pt}\setlength{\parskip}{0pt}} + \DefineVerbatimEnvironment{Highlighting}{Verbatim}{commandchars=\\\{\}} + % Add ',fontsize=\small' for more characters per line + \newenvironment{Shaded}{}{} + \newcommand{\KeywordTok}[1]{\textcolor[rgb]{0.00,0.44,0.13}{\textbf{{#1}}}} + \newcommand{\DataTypeTok}[1]{\textcolor[rgb]{0.56,0.13,0.00}{{#1}}} + \newcommand{\DecValTok}[1]{\textcolor[rgb]{0.25,0.63,0.44}{{#1}}} + \newcommand{\BaseNTok}[1]{\textcolor[rgb]{0.25,0.63,0.44}{{#1}}} + \newcommand{\FloatTok}[1]{\textcolor[rgb]{0.25,0.63,0.44}{{#1}}} + \newcommand{\CharTok}[1]{\textcolor[rgb]{0.25,0.44,0.63}{{#1}}} + 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PY@tok@c\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}} +\expandafter\def\csname PY@tok@mf\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@err\endcsname{\def\PY@bc##1{\setlength{\fboxsep}{0pt}\fcolorbox[rgb]{1.00,0.00,0.00}{1,1,1}{\strut ##1}}} +\expandafter\def\csname PY@tok@mb\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@ss\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}} +\expandafter\def\csname PY@tok@sr\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.53}{##1}}} +\expandafter\def\csname PY@tok@mo\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@kd\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@mi\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@kn\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@cpf\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}} +\expandafter\def\csname PY@tok@kr\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@s\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@kp\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@w\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.73,0.73}{##1}}} +\expandafter\def\csname PY@tok@kt\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.69,0.00,0.25}{##1}}} +\expandafter\def\csname PY@tok@sc\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@sb\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@k\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@se\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.13}{##1}}} +\expandafter\def\csname PY@tok@sd\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} + +\def\PYZbs{\char`\\} +\def\PYZus{\char`\_} +\def\PYZob{\char`\{} +\def\PYZcb{\char`\}} +\def\PYZca{\char`\^} +\def\PYZam{\char`\&} +\def\PYZlt{\char`\<} +\def\PYZgt{\char`\>} +\def\PYZsh{\char`\#} +\def\PYZpc{\char`\%} +\def\PYZdl{\char`\$} +\def\PYZhy{\char`\-} +\def\PYZsq{\char`\'} +\def\PYZdq{\char`\"} +\def\PYZti{\char`\~} +% for compatibility with earlier versions +\def\PYZat{@} +\def\PYZlb{[} +\def\PYZrb{]} +\makeatother + + + % Exact colors from NB + \definecolor{incolor}{rgb}{0.0, 0.0, 0.5} + \definecolor{outcolor}{rgb}{0.545, 0.0, 0.0} + + + + + % Prevent overflowing lines due to hard-to-break entities + \sloppy + % Setup hyperref package + \hypersetup{ + breaklinks=true, % so long urls are correctly broken across lines + colorlinks=true, + urlcolor=urlcolor, + linkcolor=linkcolor, + citecolor=citecolor, + } + % Slightly bigger margins than the latex defaults + + \geometry{verbose,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in} + + + + \begin{document} + + + \maketitle + + + + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}27}]:} \PY{c}{\PYZpc{}plot \PYZhy{}\PYZhy{}format svg} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}28}]:} \PY{n}{setdefaults} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}29}]:} \PY{n}{A}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)} + + \PY{p}{[}\PY{n}{L}\PY{p}{,}\PY{n}{U}\PY{p}{,}\PY{n}{P}\PY{p}{]}\PY{p}{=}\PY{n+nb}{lu}\PY{p}{(}\PY{n}{A}\PY{p}{)} + + \PY{n+nb}{det}\PY{p}{(}\PY{n}{L}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 0.447394 0.357071 0.720915 0.499926 + 0.648313 0.323276 0.521677 0.288345 + 0.084982 0.581513 0.466420 0.142342 + 0.576580 0.658089 0.916987 0.923165 + +L = + + 1.00000 0.00000 0.00000 0.00000 + 0.13108 1.00000 0.00000 0.00000 + 0.69009 0.24851 1.00000 0.00000 + 0.88935 0.68736 0.68488 1.00000 + +U = + + 0.64831 0.32328 0.52168 0.28834 + 0.00000 0.53914 0.39804 0.10455 + 0.00000 0.00000 0.26199 0.27496 + 0.00000 0.00000 0.00000 0.40655 + +P = + +Permutation Matrix + + 0 1 0 0 + 0 0 1 0 + 1 0 0 0 + 0 0 0 1 + +ans = 1 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}44}]:} \PY{n}{A}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{100}\PY{p}{)}\PY{o}{\PYZsq{}}\PY{p}{;} + \PY{n}{A}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZsq{}}\PY{o}{*}\PY{n}{A}\PY{p}{;} + \PY{n+nb}{size}\PY{p}{(}\PY{n}{A}\PY{p}{)} + \PY{n+nb}{min}\PY{p}{(}\PY{n+nb}{min}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{)} + \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{max}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{)} + \PY{n+nb}{cond}\PY{p}{(}\PY{n}{A}\PY{p}{)} + \PY{n}{C}\PY{p}{=}\PY{n+nb}{chol}\PY{p}{(}\PY{n}{A}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ans = + + 4 4 + +ans = 23.586 +ans = 35.826 +ans = 14.869 +C = + + 5.98549 4.28555 4.35707 4.31359 + 0.00000 3.63950 1.35005 1.45342 + 0.00000 0.00000 3.62851 1.50580 + 0.00000 0.00000 0.00000 3.21911 + + + \end{Verbatim} + + \subsection{My question from last +class}\label{my-question-from-last-class} + +\begin{figure}[htbp] +\centering +\includegraphics{det_L.png} +\caption{q1} +\end{figure} + +\begin{figure}[htbp] +\centering +\includegraphics{chol_pre.png} +\caption{q2} +\end{figure} + +\subsection{Your questions from last +class}\label{your-questions-from-last-class} + +\begin{enumerate} +\def\labelenumi{\arabic{enumi}.} +\item + Will the exam be more theoretical or problem based? +\item + Writing code is difficult +\item + What format can we expect for the midterm? +\item + Could we go over some example questions for the exam? +\item + Will the use of GitHub be tested on the Midterm exam? Or is it more + focused on linear algebra techniques/what was covered in the lectures? +\item + This is not my strong suit, getting a bit overwhelmed with matrix + multiplication. +\item + I forgot how much I learned in linear algebra. +\item + What's the most exciting project you've ever worked on with + Matlab/Octave? +\end{enumerate} + + \section{Matrix Inverse and +Condition}\label{matrix-inverse-and-condition} + +Considering the same solution set: + +\(y=Ax\) + +If we know that \(A^{-1}A=I\), then + +\(A^{-1}y=A^{-1}Ax=x\) + +so + +\(x=A^{-1}y\) + +Where, \(A^{-1}\) is the inverse of matrix \(A\). + +\(2x_{1}+x_{2}=1\) + +\(x_{1}+3x_{2}=1\) + +\(Ax=y\) + +\(\left[ \begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array} \right] \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right]= \left[\begin{array}{c} 1 \\ 1\end{array}\right]\) + +\(A^{-1}=\frac{1}{2*3-1*1}\left[ \begin{array}{cc} 3 & -1 \\ -1 & 2 \end{array} \right]= \left[ \begin{array}{cc} 3/5 & -1/5 \\ -1/5 & 2/5 \end{array} \right]\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}45}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]} + \PY{n}{invA}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{5}\PY{o}{*}\PY{p}{[}\PY{l+m+mi}{3}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{]} + + \PY{n}{A}\PY{o}{*}\PY{n}{invA} + \PY{n}{invA}\PY{o}{*}\PY{n}{A} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 2 1 + 1 3 + +invA = + + 0.60000 -0.20000 + -0.20000 0.40000 + +ans = + + 1.00000 0.00000 + 0.00000 1.00000 + +ans = + + 1.00000 0.00000 + 0.00000 1.00000 + + + \end{Verbatim} + + How did we know the inverse of A? + +for 2$\times$2 matrices, it is always: + +$A=\left[ \begin{array}{cc} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array} \right]$ + +$A^{-1}=\frac{1}{det(A)}\left[ \begin{array}{cc} A_{22} & -A_{12} \\ -A_{21} & A_{11} \end{array} \right]$ + + $AA^{-1}=\frac{1}{A_{11}A_{22}-A_{21}A_{12}}\left[ \begin{array}{cc} A_{11}A_{22}-A_{21}A_{12} & -A_{11}A_{12}+A_{12}A_{11} \\ A_{21}A_{22}-A_{22}A_{21} & -A_{21}A_{12}+A_{22}A_{11} \end{array} \right] =\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$ + + What about bigger matrices? + +We can use the LU-decomposition + +\(A=LU\) + +\(A^{-1}=(LU)^{-1}=U^{-1}L^{-1}\) + +if we divide \(A^{-1}\) into n-column vectors, \(a_{n}\), then + +\(Aa_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right]\) +\(Aa_{2}=\left[\begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array} \right]\) +\(Aa_{n}=\left[\begin{array}{c} 0 \\ 0 \\ \vdots \\ 1 \end{array} \right]\) + +Which we can solve for each \(a_{n}\) with LU-decomposition, knowing the +lower and upper triangular decompositions, then + +\(A^{-1}=\left[ \begin{array}{cccc} | & | & & | \\ a_{1} & a_{2} & \cdots & a_{n} \\ | & | & & | \end{array} \right]\) + +\(Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right]\) +\(;~Ua_{1}=d_{1}\) + +\(Ld_{2}=\left[\begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array} \right]\) +\(;~Ua_{2}=d_{2}\) + +\(Ld_{n}=\left[\begin{array}{c} 0 \\ 1 \\ \vdots \\ n \end{array} \right]\) +\(;~Ua_{n}=d_{n}\) + +Consider the following matrix: + +\(A=\left[ \begin{array}{ccc} 2 & -1 & 0\\ -1 & 2 & -1\\ 0 & -1 & 1 \end{array} \right]\) + + \paragraph{\texorpdfstring{Note on solving for \(A^{-1}\) column +1}{Note on solving for A\^{}\{-1\} column 1}}\label{note-on-solving-for-a-1-column-1} + +\(Aa_1=I(:,1)\) + +\(LUa_1=I(:,1)\) + +\((LUa_1-I(:,1))=0\) + +\(L(Ua_1-d_1)=0\) + +\(I(:,1)=Ld_1\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}56}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{]} + \PY{n}{U}\PY{p}{=}\PY{n}{A}\PY{p}{;} + \PY{n}{L}\PY{p}{=}\PY{n+nb}{eye}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;} + \PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{p}{:}\PY{p}{)} + \PY{n}{L}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{/}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 2 -1 0 + -1 2 -1 + 0 -1 1 + +U = + + 2.00000 -1.00000 0.00000 + 0.00000 1.50000 -1.00000 + 0.00000 -1.00000 1.00000 + +L = + + 1.00000 0.00000 0.00000 + -0.50000 1.00000 0.00000 + 0.00000 0.00000 1.00000 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}57}]:} \PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)} + \PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +L = + + 1.00000 0.00000 0.00000 + -0.50000 1.00000 0.00000 + 0.00000 -0.66667 1.00000 + +U = + + 2.00000 -1.00000 0.00000 + 0.00000 1.50000 -1.00000 + 0.00000 0.00000 0.33333 + + + \end{Verbatim} + + Now solve for \(d_1\) then \(a_1\), \(d_2\) then \(a_2\), and \(d_3\) +then \(a_{3}\) + +\(Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right]= \left[\begin{array}{ccc} 1 & 0 & 0 \\ -1/2 & 1 & 0 \\ 0 & -2/3 & 1 \end{array} \right]\left[\begin{array}{c} d1(1) \\ d1(2) \\ d1(3)\end{array} \right]=\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] ;~Ua_{1}=d_{1}\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}58}]:} \PY{n}{d1}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{d1}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} + 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+\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +a1 = + + 1.00000 + 1.00000 + 1.00000 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}60}]:} \PY{n}{d2}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{p}{;} + \PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +d2 = + + 0.00000 + 1.00000 + 0.66667 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}61}]:} \PY{n}{a2}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;} + \PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}\PY{p}{;} + \PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +a2 = + + 1.0000 + 2.0000 + 2.0000 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}62}]:} \PY{n}{d3}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{p}{;} + \PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +d3 = + + 0 + 0 + 1 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}63}]:} \PY{n}{a3}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;} + \PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}\PY{p}{;} + \PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +a3 = + + 1.00000 + 2.00000 + 3.00000 + + + \end{Verbatim} + + Final solution for \(A^{-1}\) is \([a_{1}~a_{2}~a_{3}]\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}69}]:} \PY{n}{invA}\PY{p}{=}\PY{p}{[}\PY{n}{a1}\PY{p}{,}\PY{n}{a2}\PY{p}{,}\PY{n}{a3}\PY{p}{]} + \PY{n}{I\PYZus{}app}\PY{p}{=}\PY{n}{A}\PY{o}{*}\PY{n}{invA} + \PY{n}{I\PYZus{}app}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)} + \PY{n+nb}{eps} + + \PY{l+m+mi}{2}\PYZca{}\PY{o}{\PYZhy{}}\PY{l+m+mi}{8} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +invA = + + 1.00000 1.00000 1.00000 + 1.00000 2.00000 2.00000 + 1.00000 2.00000 3.00000 + +I\_app = + + 1.00000 0.00000 0.00000 + 0.00000 1.00000 -0.00000 + -0.00000 -0.00000 1.00000 + +ans = -4.4409e-16 +ans = 2.2204e-16 +ans = 0.0039062 + + \end{Verbatim} + + Now the solution of \(x\) to \(Ax=y\) is \(x=A^{-1}y\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}70}]:} \PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{2}\PY{p}{;}\PY{l+m+mi}{3}\PY{p}{]} + \PY{n}{x}\PY{p}{=}\PY{n}{invA}\PY{o}{*}\PY{n}{y} + \PY{n}{xbs}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{y} + \PY{n}{x}\PY{o}{\PYZhy{}}\PY{n}{xbs} + \PY{n+nb}{eps} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +y = + + 1 + 2 + 3 + +x = + + 6.0000 + 11.0000 + 14.0000 + +xbs = + + 6.0000 + 11.0000 + 14.0000 + +ans = + + -3.5527e-15 + -8.8818e-15 + -1.0658e-14 + +ans = 2.2204e-16 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}71}]:} \PY{n}{N}\PY{p}{=}\PY{l+m+mi}{100}\PY{p}{;} + \PY{n}{n}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{:}\PY{n}{N}\PY{p}{]}\PY{p}{;} + \PY{n}{t\PYZus{}inv}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{t\PYZus{}bs}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{t\PYZus{}mult}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{k}{for} \PY{n}{i}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{n}{N} + \PY{n}{A}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{n}{i}\PY{p}{,}\PY{n}{i}\PY{p}{)}\PY{p}{;} + \PY{n+nb}{tic} + \PY{n}{invA}\PY{p}{=}\PY{n+nb}{inv}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{;} + \PY{n}{t\PYZus{}inv}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;} + \PY{n}{b}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{n}{i}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n+nb}{tic}\PY{p}{;} + \PY{n}{x}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{b}\PY{p}{;} + \PY{n}{t\PYZus{}bs}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;} + \PY{n+nb}{tic}\PY{p}{;} + \PY{n}{x}\PY{p}{=}\PY{n}{invA}\PY{o}{*}\PY{n}{b}\PY{p}{;} + \PY{n}{t\PYZus{}mult}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;} + \PY{k}{end} + \PY{n+nb}{plot}\PY{p}{(}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}inv}\PY{p}{,}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}bs}\PY{p}{,}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}mult}\PY{p}{)} + \PY{n+nb}{axis}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{0} \PY{l+m+mi}{100} \PY{l+m+mi}{0} \PY{l+m+mf}{0.002}\PY{p}{]}\PY{p}{)} + \PY{n+nb}{legend}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{inversion\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{backslash\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{multiplication\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{Location\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{NorthWest\PYZsq{}}\PY{p}{)} +\end{Verbatim} + + \begin{center} + \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_12_files/lecture_12_24_0.pdf} + \end{center} + { \hspace*{\fill} \\} + + \subsection{Condition of a matrix}\label{condition-of-a-matrix} + +\subsubsection{\texorpdfstring{\emph{just checked in to see what +condition my condition was +in}}{just checked in to see what condition my condition was in}}\label{just-checked-in-to-see-what-condition-my-condition-was-in} + +\subsubsection{Matrix norms}\label{matrix-norms} + +The Euclidean norm of a vector is measure of the magnitude (in 3D this +would be: \(|x|=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}\)) in general the +equation is: + +\(||x||_{e}=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}\) + +For a matrix, A, the same norm is called the Frobenius norm: + +\(||A||_{f}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{2}}\) + +In general we can calculate any \(p\)-norm where + +\(||A||_{p}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{p}}\) + +so the p=1, 1-norm is + +\(||A||_{1}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{1}}=\sum_{i=1}^{n}\sum_{i=1}^{m}|A_{i,j}|\) + +\(||A||_{\infty}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{\infty}}=\max_{1\le i \le n}\sum_{j=1}^{m}|A_{i,j}|\) + +\subsubsection{Condition of Matrix}\label{condition-of-matrix} + +The matrix condition is the product of + +\(Cond(A) = ||A||\cdot||A^{-1}||\) + +So each norm will have a different condition number, but the limit is +\(Cond(A)\ge 1\) + +An estimate of the rounding error is based on the condition of A: + +\(\frac{||\Delta x||}{x} \le Cond(A) \frac{||\Delta A||}{||A||}\) + +So if the coefficients of A have accuracy to \$10\^{}\{-t\} + +and the condition of A, \(Cond(A)=10^{c}\) + +then the solution for x can have rounding errors up to \(10^{c-t}\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}72}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{4}\PY{p}{;}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{5}\PY{p}{]} + \PY{p}{[}\PY{n}{L}\PY{p}{,}\PY{n}{U}\PY{p}{]}\PY{p}{=}\PY{n}{LU\PYZus{}naive}\PY{p}{(}\PY{n}{A}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 1.00000 0.50000 0.33333 + 0.50000 0.33333 0.25000 + 0.33333 0.25000 0.20000 + +L = + + 1.00000 0.00000 0.00000 + 0.50000 1.00000 0.00000 + 0.33333 1.00000 1.00000 + +U = + + 1.00000 0.50000 0.33333 + 0.00000 0.08333 0.08333 + 0.00000 -0.00000 0.00556 + + + \end{Verbatim} + + Then, \(A^{-1}=(LU)^{-1}=U^{-1}L^{-1}\) + +\(Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]\), +\(Ux_{1}=d_{1}\) ... + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}75}]:} \PY{n}{invA}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;} + \PY{n}{d1}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;} + \PY{n}{d2}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;} + \PY{n}{d3}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{;} + \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d1}\PY{p}{;} + \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d2}\PY{p}{;} + \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d3} + \PY{n}{invA}\PY{o}{*}\PY{n}{A} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +invA = + + 9.0000 -36.0000 30.0000 + -36.0000 192.0000 -180.0000 + 30.0000 -180.0000 180.0000 + +ans = + + 1.0000e+00 3.5527e-15 2.9976e-15 + -1.3249e-14 1.0000e+00 -9.1038e-15 + 8.5117e-15 7.1054e-15 1.0000e+00 + + + \end{Verbatim} + + Find the condition of A, \(cond(A)\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}74}]:} \PY{c}{\PYZpc{} Frobenius norm} + \PY{n}{normf\PYZus{}A} \PY{p}{=} \PY{n+nb}{sqrt}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{o}{.\PYZca{}}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)} + \PY{n}{normf\PYZus{}invA} \PY{p}{=} \PY{n+nb}{sqrt}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{o}{.\PYZca{}}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)} + + \PY{n}{cond\PYZus{}f\PYZus{}A} \PY{p}{=} \PY{n}{normf\PYZus{}A}\PY{o}{*}\PY{n}{normf\PYZus{}invA} + + \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{fro\PYZsq{}}\PY{p}{)} + + \PY{c}{\PYZpc{} p=1, column sum norm} + \PY{n}{norm1\PYZus{}A} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)} + \PY{n}{norm1\PYZus{}invA} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)} + \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)} + + \PY{n}{cond\PYZus{}1\PYZus{}A}\PY{p}{=}\PY{n}{norm1\PYZus{}A}\PY{o}{*}\PY{n}{norm1\PYZus{}invA} + + \PY{c}{\PYZpc{} p=inf, row sum norm} + \PY{n}{norminf\PYZus{}A} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{)} + \PY{n}{norminf\PYZus{}invA} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{)} + \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n+nb}{inf}\PY{p}{)} + + \PY{n}{cond\PYZus{}inf\PYZus{}A}\PY{p}{=}\PY{n}{norminf\PYZus{}A}\PY{o}{*}\PY{n}{norminf\PYZus{}invA} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +normf\_A = 1.4136 +normf\_invA = 372.21 +cond\_f\_A = 526.16 +ans = 1.4136 +norm1\_A = 1.8333 +norm1\_invA = 30.000 +ans = 1.8333 +cond\_1\_A = 55.000 +norminf\_A = 1.8333 +norminf\_invA = 30.000 +ans = 1.8333 +cond\_inf\_A = 55.000 + + \end{Verbatim} + + Consider the problem again from the intro to Linear Algebra, 4 masses +are connected in series to 4 springs with spring constants \(K_{i}\). +What does a high condition number mean for this problem? + +\begin{figure}[htbp] +\centering +\includegraphics{../lecture_09/mass_springs.png} +\caption{Springs-masses} +\end{figure} + +The masses haves the following amounts, 1, 2, 3, and 4 kg for masses +1-4. Using a FBD for each mass: + +\(m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0\) + +\(m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0\) + +\(m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0\) + +\(m_{4}g-k_{4}(x_{4}-x_{3})=0\) + +in matrix form: + +\(\left[ \begin{array}{cccc} k_{1}+k_{2} & -k_{2} & 0 & 0 \\ -k_{2} & k_{2}+k_{3} & -k_{3} & 0 \\ 0 & -k_{3} & k_{3}+k_{4} & -k_{4} \\ 0 & 0 & -k_{4} & k_{4} \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]= \left[ \begin{array}{c} m_{1}g \\ m_{2}g \\ m_{3}g \\ m_{4}g \end{array} \right]\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}21}]:} \PY{n}{k1}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} \PY{c}{\PYZpc{} N/m} + \PY{n}{k2}\PY{p}{=}\PY{l+m+mi}{100000}\PY{p}{;} + \PY{n}{k3}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} + \PY{n}{k4}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} + \PY{n}{m1}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} kg} + \PY{n}{m2}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;} + \PY{n}{m3}\PY{p}{=}\PY{l+m+mi}{3}\PY{p}{;} + \PY{n}{m4}\PY{p}{=}\PY{l+m+mi}{4}\PY{p}{;} + \PY{n}{g}\PY{p}{=}\PY{l+m+mf}{9.81}\PY{p}{;} \PY{c}{\PYZpc{} m/s\PYZca{}2} + \PY{n}{K}\PY{p}{=}\PY{p}{[}\PY{n}{k1}\PY{o}{+}\PY{n}{k2} \PY{o}{\PYZhy{}}\PY{n}{k2} \PY{l+m+mi}{0} \PY{l+m+mi}{0}\PY{p}{;} \PY{o}{\PYZhy{}}\PY{n}{k2} \PY{n}{k2}\PY{o}{+}\PY{n}{k3} \PY{o}{\PYZhy{}}\PY{n}{k3} \PY{l+m+mi}{0}\PY{p}{;} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k3} \PY{n}{k3}\PY{o}{+}\PY{n}{k4} \PY{o}{\PYZhy{}}\PY{n}{k4}\PY{p}{;} \PY{l+m+mi}{0} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k4} \PY{n}{k4}\PY{p}{]} + \PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{n}{m1}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m2}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m3}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m4}\PY{o}{*}\PY{n}{g}\PY{p}{]} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +K = + + 100010 -100000 0 0 + -100000 100010 -10 0 + 0 -10 11 -1 + 0 0 -1 1 + +y = + + 9.8100 + 19.6200 + 29.4300 + 39.2400 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}25}]:} \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{n+nb}{inf}\PY{p}{)} + \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)} + \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{fro\PYZsq{}}\PY{p}{)} + \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ans = 3.2004e+05 +ans = 3.2004e+05 +ans = 2.5925e+05 +ans = 2.5293e+05 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}26}]:} \PY{n+nb}{e}\PY{p}{=}\PY{n+nb}{eig}\PY{p}{(}\PY{n}{K}\PY{p}{)} + \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{e}\PY{p}{)}\PY{o}{/}\PY{n+nb}{min}\PY{p}{(}\PY{n+nb}{e}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +e = + + 7.9078e-01 + 3.5881e+00 + 1.7621e+01 + 2.0001e+05 + +ans = 2.5293e+05 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor} }]:} +\end{Verbatim} + + + % Add a bibliography block to the postdoc + + + + \end{document} diff --git a/lecture_12/lecture_12_files/lecture_12_21_0.png b/lecture_12/lecture_12_files/lecture_12_21_0.png new file mode 100644 index 0000000..2633731 Binary files /dev/null and b/lecture_12/lecture_12_files/lecture_12_21_0.png differ diff --git a/lecture_12/lecture_12_files/lecture_12_24_0.pdf b/lecture_12/lecture_12_files/lecture_12_24_0.pdf new file mode 100644 index 0000000..e2d1d9b Binary files /dev/null and b/lecture_12/lecture_12_files/lecture_12_24_0.pdf differ diff --git a/lecture_12/lecture_12_files/lecture_12_24_0.png b/lecture_12/lecture_12_files/lecture_12_24_0.png new file mode 100644 index 0000000..c8959a7 Binary files /dev/null and b/lecture_12/lecture_12_files/lecture_12_24_0.png differ diff --git a/lecture_12/lecture_12_files/lecture_12_24_0.svg b/lecture_12/lecture_12_files/lecture_12_24_0.svg new file mode 100644 index 0000000..4705883 --- /dev/null +++ b/lecture_12/lecture_12_files/lecture_12_24_0.svg @@ -0,0 +1,148 @@ + + +Gnuplot +Produced by GNUPLOT 5.0 patchlevel 3 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + 0 + + + + + 0.0005 + + + + + 0.001 + + + + + 0.0015 + + + + + 0.002 + + + + + 0 + + + + + 20 + + + + + 40 + + + + + 60 + + + + + 80 + + + + + 100 + + + + + + + + + + + + + inversion + + + + + inversion + + + + + + backslash + + + backslash + + + + + + multiplication + + + multiplication + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/lecture_13/.GS_rel.m.swp b/lecture_13/.GS_rel.m.swp new file mode 100644 index 0000000..58db6b3 Binary files /dev/null and b/lecture_13/.GS_rel.m.swp differ diff --git a/lecture_13/.GaussSeidel.m.swp b/lecture_13/.GaussSeidel.m.swp new file mode 100644 index 0000000..43abe04 Binary files /dev/null and b/lecture_13/.GaussSeidel.m.swp differ diff --git a/lecture_13/.Jacobi_rel.m.swp b/lecture_13/.Jacobi_rel.m.swp new file mode 100644 index 0000000..8f17e8f Binary files /dev/null and b/lecture_13/.Jacobi_rel.m.swp differ diff --git a/lecture_13/.ipynb_checkpoints/lecture_12-checkpoint.ipynb b/lecture_13/.ipynb_checkpoints/lecture_13-checkpoint.ipynb similarity index 100% rename from lecture_13/.ipynb_checkpoints/lecture_12-checkpoint.ipynb rename to lecture_13/.ipynb_checkpoints/lecture_13-checkpoint.ipynb diff --git a/lecture_13/.lambda_fcn.m.swp b/lecture_13/.lambda_fcn.m.swp new file mode 100644 index 0000000..5031365 Binary files /dev/null and b/lecture_13/.lambda_fcn.m.swp differ diff --git a/lecture_13/GS_rel.m b/lecture_13/GS_rel.m new file mode 100644 index 0000000..4a6daf4 --- /dev/null +++ b/lecture_13/GS_rel.m @@ -0,0 +1,36 @@ +function [x,ea,iter] = GS_rel(A,b,lambda,es,maxit) +% GaussSeidel: Gauss Seidel method +% x = GaussSeidel(A,b): Gauss Seidel without relaxation +% input: +% A = coefficient matrix +% b = right hand side vector +% es = stop criterion (default = 0.00001%) +% maxit = max iterations (default = 50) +% output: +% x = solution vector +if nargin<3,error('at least 2 input arguments required'),end +if nargin<5|isempty(maxit),maxit=50;end +if nargin<4|isempty(es),es=0.00001;end +[m,n] = size(A); +if m~=n, error('Matrix A must be square'); end +C = A-diag(diag(A)); +x=zeros(n,1); +for i = 1:n + C(i,1:n) = C(i,1:n)/A(i,i); +end + +d = b./diag(A); + +iter = 0; +while (1) + xold = x; + for i = 1:n + x(i) = d(i)-C(i,:)*x; + x(i) = lambda*x(i)+(1-lambda)*xold(i); + if x(i) ~= 0 + ea(i) = abs((x(i) - xold(i))/x(i)) * 100; + end + end + iter = iter+1; + if max(ea)<=es | iter >= maxit, break, end +end diff --git a/lecture_13/GaussSeidel.m b/lecture_13/GaussSeidel.m new file mode 100644 index 0000000..2be52e1 --- /dev/null +++ b/lecture_13/GaussSeidel.m @@ -0,0 +1,35 @@ +function x = GaussSeidel(A,b,es,maxit) +% GaussSeidel: Gauss Seidel method +% x = GaussSeidel(A,b): Gauss Seidel without relaxation +% input: +% A = coefficient matrix +% b = right hand side vector +% es = stop criterion (default = 0.00001%) +% maxit = max iterations (default = 50) +% output: +% x = solution vector +if nargin<2,error('at least 2 input arguments required'),end +if nargin<4|isempty(maxit),maxit=50;end +if nargin<3|isempty(es),es=0.00001;end +[m,n] = size(A); +if m~=n, error('Matrix A must be square'); end +C = A-diag(diag(A)); +x=zeros(n,1); +for i = 1:n + C(i,1:n) = C(i,1:n)/A(i,i); +end + +d = b./diag(A); + +iter = 0; +while (1) + xold = x; + for i = 1:n + x(i) = d(i)-C(i,:)*x; + if x(i) ~= 0 + ea(i) = abs((x(i) - xold(i))/x(i)) * 100; + end + end + iter = iter+1; + if max(ea)<=es | iter >= maxit, break, end +end diff --git a/lecture_13/Jacobi.m b/lecture_13/Jacobi.m new file mode 100644 index 0000000..8a7b4ae --- /dev/null +++ b/lecture_13/Jacobi.m @@ -0,0 +1,39 @@ +function x = Jacobi(A,b,es,maxit) +% GaussSeidel: Gauss Seidel method +% x = GaussSeidel(A,b): Gauss Seidel without relaxation +% input: +% A = coefficient matrix +% b = right hand side vector +% es = stop criterion (default = 0.00001%) +% maxit = max iterations (default = 50) +% output: +% x = solution vector +if nargin<2,error('at least 2 input arguments required'),end +if nargin<4|isempty(maxit),maxit=50;end +if nargin<3|isempty(es),es=0.00001;end +[m,n] = size(A); +if m~=n, error('Matrix A must be square'); end +C = A-diag(diag(A)); +x=zeros(n,1); +for i = 1:n + C(i,1:n) = C(i,1:n)/A(i,i); +end + +d = b./diag(A); + +iter = 0; +while (1) + xold = x; + x = d-C*x; + % if any values of x are zero, we add 1 to denominator so error is well-behaved + i_zero=find(x==0); + i=find(x~=0); + if length(i_zero)>0 + ea(i_zero)=abs((x-xold)./(1+x)*100); + ea(i) = abs((x(i) - xold(i))./x(i)) * 100; + else + ea = abs((x - xold)./x) * 100; + end + iter = iter+1; + if max(ea)<=es | iter >= maxit, break, end +end diff --git a/lecture_13/Jacobi_rel.m b/lecture_13/Jacobi_rel.m new file mode 100644 index 0000000..5cdec33 --- /dev/null +++ b/lecture_13/Jacobi_rel.m @@ -0,0 +1,41 @@ +function [x,ea,iter]= Jacobi_rel(A,b,lambda,es,maxit) +% GaussSeidel: Gauss Seidel method +% x = GaussSeidel(A,b): Gauss Seidel without relaxation +% input: +% A = coefficient matrix +% b = right hand side vector +% es = stop criterion (default = 0.00001%) +% maxit = max iterations (default = 50) +% output: +% x = solution vector +if nargin<3,error('at least 2 input arguments required'),end +if nargin<5|isempty(maxit),maxit=50;end +if nargin<4|isempty(es),es=0.00001;end +[m,n] = size(A); +if m~=n, error('Matrix A must be square'); end +C = A-diag(diag(A)); +x=zeros(n,1); +for i = 1:n + C(i,1:n) = C(i,1:n)/A(i,i); +end + +d = b./diag(A); + +iter = 0; +while (1) + xold = x; + x = d-C*x; + % Add relaxation parameter lambda to current iteration + x = lambda*x+(1-lambda)*xold; + % if any values of x are zero, we add 1 to denominator so error is well-behaved + i_zero=find(x==0); + i=find(x~=0); + if length(i_zero)>0 + ea(i_zero)=abs((x-xold)./(1+x)*100); + ea(i) = abs((x(i) - xold(i))./x(i)) * 100; + else + ea = abs((x - xold)./x) * 100; + end + iter = iter+1; + if max(ea)<=es | iter >= maxit, break, end +end diff --git a/lecture_13/efficient_soln.png b/lecture_13/efficient_soln.png new file mode 100644 index 0000000..ef24ece Binary files /dev/null and b/lecture_13/efficient_soln.png differ diff --git a/lecture_13/gp_image_01.png b/lecture_13/gp_image_01.png new file mode 100644 index 0000000..ef291b5 Binary files /dev/null and b/lecture_13/gp_image_01.png differ diff --git a/lecture_13/lambda_fcn.m b/lecture_13/lambda_fcn.m new file mode 100644 index 0000000..435f9e9 --- /dev/null +++ b/lecture_13/lambda_fcn.m @@ -0,0 +1,8 @@ +function iters = lambda_fcn(A,b,lambda) + % function to minimize the number of iterations for a given Ax=b solution + % using default Jacobi_rel parameters of es=0.00001 and maxit=50 + + [x,ea,iters]= Jacobi_rel(A,b,lambda,1e-8); +end + + diff --git a/lecture_13/lecture_13.aux b/lecture_13/lecture_13.aux new file mode 100644 index 0000000..513ef62 --- /dev/null +++ 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0000000..fb66ca0 --- /dev/null +++ b/lecture_13/lecture_13.blg @@ -0,0 +1,48 @@ +This is BibTeX, Version 0.99d (TeX Live 2015/Debian) +Capacity: max_strings=35307, hash_size=35307, hash_prime=30011 +The top-level auxiliary file: lecture_13.aux +I found no \citation commands---while reading file lecture_13.aux +I found no \bibdata command---while reading file lecture_13.aux +I found no \bibstyle command---while reading file lecture_13.aux +You've used 0 entries, + 0 wiz_defined-function locations, + 83 strings with 494 characters, +and the built_in function-call counts, 0 in all, are: += -- 0 +> -- 0 +< -- 0 ++ -- 0 +- -- 0 +* -- 0 +:= -- 0 +add.period$ -- 0 +call.type$ -- 0 +change.case$ -- 0 +chr.to.int$ -- 0 +cite$ -- 0 +duplicate$ -- 0 +empty$ -- 0 +format.name$ -- 0 +if$ -- 0 +int.to.chr$ -- 0 +int.to.str$ -- 0 +missing$ -- 0 +newline$ -- 0 +num.names$ -- 0 +pop$ -- 0 +preamble$ -- 0 +purify$ -- 0 +quote$ -- 0 +skip$ -- 0 +stack$ -- 0 +substring$ -- 0 +swap$ -- 0 +text.length$ -- 0 +text.prefix$ -- 0 +top$ -- 0 +type$ -- 0 +warning$ -- 0 +while$ -- 0 +width$ -- 0 +write$ -- 0 +(There were 3 error messages) diff --git a/lecture_13/lecture_13.ipynb b/lecture_13/lecture_13.ipynb new file mode 100644 index 0000000..34a39e5 --- /dev/null +++ b/lecture_13/lecture_13.ipynb @@ -0,0 +1,5622 @@ +{ + "cells": [ + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "%plot --format svg" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "setdefaults" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## My question from last class \n", + "\n", + "![q1](efficient_soln.png)\n", + "\n", + "![A](https://lh4.googleusercontent.com/fmG7EnFxpvvjSgijOuwz8osuiH3cBDgOzTE64KnfQeeDDSG2oE86-BzcpYIQMVkkAgRRGEDEGi6-Nkr8qmEMeaAk-gcjEmXe42WFYUdOa5XoUaBkXRakkA77_XrkRjArCGZIFhjjDRoO7x0)\n", + "\n", + "![q2](norm_A.png)\n", + "\n", + "\n", + "## Your questions from last class\n", + "\n", + "1. Do we have to submit a link for HW #4 somewhere or is uploading to Github sufficient?\n", + " \n", + " -no, your submission from HW 3 is sufficient\n", + "\n", + "2. How do I get the formulas/formatting in markdown files to show up on github?\n", + " \n", + " -no luck for markdown equations in github, this is an ongoing request\n", + " \n", + "3. Confused about the p=1 norm part and ||A||_1\n", + "\n", + "4. When's the exam?\n", + " \n", + " -next week (3/9)\n", + "\n", + "5. What do you recommend doing to get better at figuring out the homeworks?\n", + "\n", + " -time and experimenting (try going through the lecture examples, verify my work)\n", + " \n", + "6. Could we have an hw or extra credit with a video lecture to learn some simple python?\n", + " \n", + " -Sounds great! how simple? \n", + " \n", + " -[Installing Python and Jupyter Notebook (via Anaconda) - https://www.continuum.io/downloads](https://www.continuum.io/downloads)\n", + " \n", + " -[Running Matlab kernel in Jupyter - https://anneurai.net/2015/11/12/matlab-based-ipython-notebooks/](https://anneurai.net/2015/11/12/matlab-based-ipython-notebooks/)\n", + " \n", + " -[Running Octave kernel in Jupyter - https://anaconda.org/pypi/octave_kernel](https://anaconda.org/pypi/octave_kernel)\n", + " \n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Condition of a matrix \n", + "### *just checked in to see what condition my condition was in*\n", + "### Matrix norms\n", + "\n", + "The Euclidean norm of a vector is measure of the magnitude (in 3D this would be: $|x|=\\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$) in general the equation is:\n", + "\n", + "$||x||_{e}=\\sqrt{\\sum_{i=1}^{n}x_{i}^{2}}$\n", + "\n", + "For a matrix, A, the same norm is called the Frobenius norm:\n", + "\n", + "$||A||_{f}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{2}}$\n", + "\n", + "In general we can calculate any $p$-norm where\n", + "\n", + "$||A||_{p}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{p}}$\n", + "\n", + "so the p=1, 1-norm is \n", + "\n", + "$||A||_{1}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{1}}=\\sum_{i=1}^{n}\\sum_{i=1}^{m}|A_{i,j}|$\n", + "\n", + "$||A||_{\\infty}=\\sqrt{\\sum_{i=1}^{n}\\sum_{i=1}^{m}A_{i,j}^{\\infty}}=\\max_{1\\le i \\le n}\\sum_{j=1}^{m}|A_{i,j}|$\n", + "\n", + "### Condition of Matrix\n", + "\n", + "The matrix condition is the product of \n", + "\n", + "$Cond(A) = ||A||\\cdot||A^{-1}||$ \n", + "\n", + "So each norm will have a different condition number, but the limit is $Cond(A)\\ge 1$\n", + "\n", + "An estimate of the rounding error is based on the condition of A:\n", + "\n", + "$\\frac{||\\Delta x||}{x} \\le Cond(A) \\frac{||\\Delta A||}{||A||}$\n", + "\n", + "So if the coefficients of A have accuracy to $10^{-t}\n", + "\n", + "and the condition of A, $Cond(A)=10^{c}$\n", + "\n", + "then the solution for x can have rounding errors up to $10^{c-t}$\n" + ] + }, + { + "cell_type": "code", + "execution_count": 72, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A =\n", + "\n", + " 1.00000 0.50000 0.33333\n", + " 0.50000 0.33333 0.25000\n", + " 0.33333 0.25000 0.20000\n", + "\n", + "L =\n", + "\n", + " 1.00000 0.00000 0.00000\n", + " 0.50000 1.00000 0.00000\n", + " 0.33333 1.00000 1.00000\n", + "\n", + "U =\n", + "\n", + " 1.00000 0.50000 0.33333\n", + " 0.00000 0.08333 0.08333\n", + " 0.00000 -0.00000 0.00556\n", + "\n" + ] + } + ], + "source": [ + "A=[1,1/2,1/3;1/2,1/3,1/4;1/3,1/4,1/5]\n", + "[L,U]=LU_naive(A)" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "Then, $A^{-1}=(LU)^{-1}=U^{-1}L^{-1}$\n", + "\n", + "$Ld_{1}=\\left[\\begin{array}{c}\n", + "1 \\\\\n", + "0 \\\\\n", + "0 \\end{array}\\right]$, $Ux_{1}=d_{1}$ ..." + ] + }, + { + "cell_type": "code", + "execution_count": 75, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "invA =\n", + "\n", + " 9.0000 -36.0000 30.0000\n", + " -36.0000 192.0000 -180.0000\n", + " 30.0000 -180.0000 180.0000\n", + "\n", + "ans =\n", + "\n", + " 1.0000e+00 3.5527e-15 2.9976e-15\n", + " -1.3249e-14 1.0000e+00 -9.1038e-15\n", + " 8.5117e-15 7.1054e-15 1.0000e+00\n", + "\n" + ] + } + ], + "source": [ + "invA=zeros(3,3);\n", + "d1=L\\[1;0;0];\n", + "d2=L\\[0;1;0];\n", + "d3=L\\[0;0;1];\n", + "invA(:,1)=U\\d1;\n", + "invA(:,2)=U\\d2;\n", + "invA(:,3)=U\\d3\n", + "invA*A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Find the condition of A, $cond(A)$" + ] + }, + { + "cell_type": "code", + "execution_count": 74, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "normf_A = 1.4136\n", + "normf_invA = 372.21\n", + "cond_f_A = 526.16\n", + "ans = 1.4136\n", + "norm1_A = 1.8333\n", + "norm1_invA = 30.000\n", + "ans = 1.8333\n", + "cond_1_A = 55.000\n", + "norminf_A = 1.8333\n", + "norminf_invA = 30.000\n", + "ans = 1.8333\n", + "cond_inf_A = 55.000\n" + ] + } + ], + "source": [ + "% Frobenius norm\n", + "normf_A = sqrt(sum(sum(A.^2)))\n", + "normf_invA = sqrt(sum(sum(invA.^2)))\n", + "\n", + "cond_f_A = normf_A*normf_invA\n", + "\n", + "norm(A,'fro')\n", + "\n", + "% p=1, column sum norm\n", + "norm1_A = max(sum(A,2))\n", + "norm1_invA = max(sum(invA,2))\n", + "norm(A,1)\n", + "\n", + "cond_1_A=norm1_A*norm1_invA\n", + "\n", + "% p=inf, row sum norm\n", + "norminf_A = max(sum(A,1))\n", + "norminf_invA = max(sum(invA,1))\n", + "norm(A,inf)\n", + "\n", + "cond_inf_A=norminf_A*norminf_invA\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with spring constants $K_{i}$. What does a high condition number mean for this problem? \n", + "\n", + "![Springs-masses](../lecture_09/mass_springs.png)\n", + "\n", + "The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:\n", + "\n", + "$m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0$\n", + "\n", + "$m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0$\n", + "\n", + "$m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0$\n", + "\n", + "$m_{4}g-k_{4}(x_{4}-x_{3})=0$\n", + "\n", + "in matrix form:\n", + "\n", + "$\\left[ \\begin{array}{cccc}\n", + "k_{1}+k_{2} & -k_{2} & 0 & 0 \\\\\n", + "-k_{2} & k_{2}+k_{3} & -k_{3} & 0 \\\\\n", + "0 & -k_{3} & k_{3}+k_{4} & -k_{4} \\\\\n", + "0 & 0 & -k_{4} & k_{4} \\end{array} \\right]\n", + "\\left[ \\begin{array}{c}\n", + "x_{1} \\\\\n", + "x_{2} \\\\\n", + "x_{3} \\\\\n", + "x_{4} \\end{array} \\right]=\n", + "\\left[ \\begin{array}{c}\n", + "m_{1}g \\\\\n", + "m_{2}g \\\\\n", + "m_{3}g \\\\\n", + "m_{4}g \\end{array} \\right]$" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "K =\n", + "\n", + " 100010 -100000 0 0\n", + " -100000 100010 -10 0\n", + " 0 -10 11 -1\n", + " 0 0 -1 1\n", + "\n", + "y =\n", + "\n", + " 9.8100\n", + " 19.6200\n", + " 29.4300\n", + " 39.2400\n", + "\n" + ] + } + ], + "source": [ + "k1=10; % N/m\n", + "k2=100000;\n", + "k3=10;\n", + "k4=1;\n", + "m1=1; % kg\n", + "m2=2;\n", + "m3=3;\n", + "m4=4;\n", + "g=9.81; % m/s^2\n", + "K=[k1+k2 -k2 0 0; -k2 k2+k3 -k3 0; 0 -k3 k3+k4 -k4; 0 0 -k4 k4]\n", + "y=[m1*g;m2*g;m3*g;m4*g]" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ans = 3.2004e+05\n", + "ans = 3.2004e+05\n", + "ans = 2.5925e+05\n", + "ans = 2.5293e+05\n" + ] + } + ], + "source": [ + "cond(K,inf)\n", + "cond(K,1)\n", + "cond(K,'fro')\n", + "cond(K,2)" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "e =\n", + "\n", + " 7.9078e-01\n", + " 3.5881e+00\n", + " 1.7621e+01\n", + " 2.0001e+05\n", + "\n", + "ans = 2.5293e+05\n" + ] + } + ], + "source": [ + "e=eig(K)\n", + "max(e)/min(e)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Iterative Methods\n", + "\n", + "## Gauss-Seidel method\n", + "\n", + "If we have an intial guess for each value of a vector $x$ that we are trying to solve, then it is easy enough to solve for one component given the others. \n", + "\n", + "Take a 3$\\times$3 matrix \n", + "\n", + "$Ax=b$\n", + "\n", + "$\\left[ \\begin{array}{ccc}\n", + "3 & -0.1 & -0.2 \\\\\n", + "0.1 & 7 & -0.3 \\\\\n", + "0.3 & -0.2 & 10 \\end{array} \\right]\n", + "\\left[ \\begin{array}{c}\n", + "x_{1} \\\\\n", + "x_{2} \\\\\n", + "x_{3} \\end{array} \\right]=\n", + "\\left[ \\begin{array}{c}\n", + "7.85 \\\\\n", + "-19.3 \\\\\n", + "71.4\\end{array} \\right]$\n", + "\n", + "$x_{1}=\\frac{7.85+0.1x_{2}+0.3x_{3}}{3}$\n", + "\n", + "$x_{2}=\\frac{-19.3-0.1x_{1}+0.3x_{3}}{7}$\n", + "\n", + "$x_{3}=\\frac{71.4+0.1x_{1}+0.2x_{2}}{10}$" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A =\n", + "\n", + " 3.00000 -0.10000 -0.20000\n", + " 0.10000 7.00000 -0.30000\n", + " 0.30000 -0.20000 10.00000\n", + "\n", + "b =\n", + "\n", + " 7.8500\n", + " -19.3000\n", + " 71.4000\n", + "\n", + "x =\n", + "\n", + " 3.0000\n", + " -2.5000\n", + " 7.0000\n", + "\n" + ] + } + ], + "source": [ + "A=[3 -0.1 -0.2;0.1 7 -0.3;0.3 -0.2 10]\n", + "b=[7.85;-19.3;71.4]\n", + "\n", + "x=A\\b" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "### Gauss-Seidel Iterative approach\n", + "\n", + "As a first guess, we can use $x_{1}=x_{2}=x_{3}=0$\n", + "\n", + "$x_{1}=\\frac{7.85+0.1(0)+0.3(0)}{3}=2.6167$\n", + "\n", + "$x_{2}=\\frac{-19.3-0.1(2.6167)+0.3(0)}{7}=-2.7945$\n", + "\n", + "$x_{3}=\\frac{71.4+0.1(2.6167)+0.2(-2.7945)}{10}=7.0056$\n", + "\n", + "Then, we update the guess:\n", + "\n", + "$x_{1}=\\frac{7.85+0.1(-2.7945)+0.3(7.0056)}{3}=2.9906$\n", + "\n", + "$x_{2}=\\frac{-19.3-0.1(2.9906)+0.3(7.0056)}{7}=-2.4996$\n", + "\n", + "$x_{3}=\\frac{71.4+0.1(2.9906)+0.2(-2.4966)}{10}=7.00029$\n", + "\n", + "The results are conveerging to the solution we found with `\\` of $x_{1}=3,~x_{2}=-2.5,~x_{3}=7$\n", + "\n", + "We could also use an iterative method that solves for all of the x-components in one step:\n", + "\n", + "### Jacobi method\n", + "\n", + "$x_{1}^{i}=\\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$\n", + "\n", + "$x_{2}^{i}=\\frac{-19.3-0.1x_{1}^{i-1}+0.3x_{3}^{i-1}}{7}$\n", + "\n", + "$x_{3}^{i}=\\frac{71.4+0.1x_{1}^{i-1}+0.2x_{2}^{i-1}}{10}$\n", + "\n", + "Here the solution is a matrix multiplication and vector addition\n", + "\n", + "$\\left[ \\begin{array}{c}\n", + "x_{1}^{i} \\\\\n", + "x_{2}^{i} \\\\\n", + "x_{3}^{i} \\end{array} \\right]=\n", + "\\left[ \\begin{array}{c}\n", + "7.85/3 \\\\\n", + "-19.3/7 \\\\\n", + "71.4/10\\end{array} \\right]-\n", + "\\left[ \\begin{array}{ccc}\n", + "0 & -0.1 & -0.2 \\\\\n", + "0.1 & 0 & -0.3 \\\\\n", + "0.3 & -0.2 & 0 \\end{array} \\right]\n", + "\\left[ \\begin{array}{c}\n", + "x_{1}^{i-1} \\\\\n", + "x_{2}^{i-1} \\\\\n", + "x_{3}^{i-1} \\end{array} \\right]$\n", + "\n", + "|x_{j}|Jacobi method |vs| Gauss-Seidel |\n", + "|--------|------------------------------|---|-------------------------------|\n", + "|$x_{1}^{i}=$ | $\\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$ | | $\\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$|\n", + "|$x_{2}^{i}=$ | $\\frac{-19.3-0.1x_{1}^{i-1}+0.3x_{3}^{i-1}}{7}$ | | $\\frac{-19.3-0.1x_{1}^{i}+0.3x_{3}^{i-1}}{7}$ |\n", + "|$x_{3}^{i}=$ | $\\frac{71.4+0.1x_{1}^{i-1}+0.2x_{2}^{i-1}}{10}$ | | $\\frac{71.4+0.1x_{1}^{i}+0.2x_{2}^{i}}{10}$|" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ba =\n", + "\n", + " 2.6167\n", + " -2.7571\n", + " 7.1400\n", + "\n", + "sA =\n", + "\n", + " 0.00000 -0.10000 -0.20000\n", + " 0.10000 0.00000 -0.30000\n", + " 0.30000 -0.20000 0.00000\n", + "\n", + "sA =\n", + "\n", + " 0.000000 -0.033333 -0.066667\n", + " 0.014286 0.000000 -0.042857\n", + " 0.030000 -0.020000 0.000000\n", + "\n", + "x1 =\n", + "\n", + " 2.6167\n", + " -2.7571\n", + " 7.1400\n", + "\n", + "x2 =\n", + "\n", + " 3.0008\n", + " -2.4885\n", + " 7.0064\n", + "\n", + "x3 =\n", + "\n", + " 3.0008\n", + " -2.4997\n", + " 7.0002\n", + "\n", + "solution is converging to [3,-2.5,7]]\n" + ] + } + ], + "source": [ + "ba=b./diag(A) % or ba=b./[A(1,1);A(2,2);A(3,3)]\n", + "sA=A-diag(diag(A)) % A with zeros on diagonal\n", + "sA(1,:)=sA(1,:)/A(1,1);\n", + "sA(2,:)=sA(2,:)/A(2,2);\n", + "sA(3,:)=sA(3,:)/A(3,3)\n", + "x0=[0;0;0];\n", + "x1=ba-sA*x0\n", + "x2=ba-sA*x1\n", + "x3=ba-sA*x2\n", + "fprintf('solution is converging to [3,-2.5,7]]\\n')" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ans =\n", + "\n", + " 3\n", + " 7\n", + " 10\n", + "\n", + "ans =\n", + "\n", + "Diagonal Matrix\n", + "\n", + " 3 0 0\n", + " 0 7 0\n", + " 0 0 10\n", + "\n" + ] + } + ], + "source": [ + "diag(A)\n", + "diag(diag(A))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "This method works if problem is diagonally dominant, \n", + "\n", + "$|a_{ii}|>\\sum_{j=1,j\\ne i}^{n}|a_{ij}|$\n", + "\n", + "If this condition is true, then Jacobi or Gauss-Seidel should converge\n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A =\n", + "\n", + " 0.10000 1.00000 3.00000\n", + " 1.00000 0.20000 3.00000\n", + " 5.00000 2.00000 0.30000\n", + "\n", + "b =\n", + "\n", + " 12\n", + " 2\n", + " 4\n", + "\n", + "ans =\n", + "\n", + " -2.9393\n", + " 9.1933\n", + " 1.0336\n", + "\n" + ] + } + ], + "source": [ + "A=[0.1,1,3;1,0.2,3;5,2,0.3]\n", + "b=[12;2;4]\n", + "A\\b" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ba =\n", + "\n", + " 120.000\n", + " 10.000\n", + " 13.333\n", + "\n", + "sA =\n", + "\n", + " 0 1 3\n", + " 1 0 3\n", + " 5 2 0\n", + "\n", + "sA =\n", + "\n", + " 0.00000 10.00000 30.00000\n", + " 5.00000 0.00000 15.00000\n", + " 16.66667 6.66667 0.00000\n", + "\n", + "x1 =\n", + "\n", + " 120.000\n", + " 10.000\n", + " 13.333\n", + "\n", + "x2 =\n", + "\n", + " -380.00\n", + " -790.00\n", + " -2053.33\n", + "\n", + "x3 =\n", + "\n", + " 6.9620e+04\n", + " 3.2710e+04\n", + " 1.1613e+04\n", + "\n", + "solution is not converging to [-2.93,9.19,1.03]\n" + ] + } + ], + "source": [ + "ba=b./diag(A) % or ba=b./[A(1,1);A(2,2);A(3,3)]\n", + "sA=A-diag(diag(A)) % A with zeros on diagonal\n", + "sA(1,:)=sA(1,:)/A(1,1);\n", + "sA(2,:)=sA(2,:)/A(2,2);\n", + "sA(3,:)=sA(3,:)/A(3,3)\n", + "x0=[0;0;0];\n", + "x1=ba-sA*x0\n", + "x2=ba-sA*x1\n", + "x3=ba-sA*x2\n", + "fprintf('solution is not converging to [-2.93,9.19,1.03]\\n')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Gauss-Seidel with Relaxation\n", + "\n", + "In order to force the solution to converge faster, we can introduce a relaxation term $\\lambda$. \n", + "\n", + "where the new x values are weighted between the old and new:\n", + "\n", + "$x^{i}=\\lambda x^{i}+(1-\\lambda)x^{i-1}$\n", + "\n", + "after solving for x, lambda weights the current approximation with the previous approximation for the updated x\n" + ] + }, + { + "cell_type": "code", + "execution_count": 105, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A =\n", + "\n", + " 3.00000 -0.10000 -0.20000\n", + " 0.10000 7.00000 -0.30000\n", + " 0.30000 -0.20000 10.00000\n", + "\n", + "b =\n", + "\n", + " 7.8500\n", + " -19.3000\n", + " 71.4000\n", + "\n" + ] + }, + { + "data": { + "image/svg+xml": [ + "\n", + "\n", + "Gnuplot\n", + "Produced by GNUPLOT 5.0 patchlevel 3 \n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t \n", + "\t \n", + "\t\n", + "\t\n", + "\t \n", + "\t \n", + "\t\n", + "\n", + "\n", + "\n", + "\n", + "\t\n", + "\t\t\n", + "\t\n", + "\n", + "\n", + "\n", + "\n", + "\t\t\n", + "\t\t0\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t10\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t20\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t30\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t40\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t50\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t0\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t0.5\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t1\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t1.5\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t2\n", + "\t\n", + "\n", + "\n", + "\n", + "\n", + "\t\n", + "\n", + "\n", + "\tgnuplot_plot_1a\n", + "\n", + "\n", + "\n", + "\t\n", + "\t\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "" + ], + "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "% rearrange A and b\n", + "A=[3 -0.1 -0.2;0.1 7 -0.3;0.3 -0.2 10]\n", + "b=[7.85;-19.3;71.4]\n", + "\n", + "iters=zeros(100,1);\n", + "for i=1:100\n", + " lambda=2/100*i;\n", + " [x,ea,iters(i)]=Jacobi_rel(A,b,lambda);\n", + "end\n", + "plot([1:100]*2/100,iters) " + ] + }, + { + "cell_type": "code", + "execution_count": 107, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "l = 0.99158\r\n" + ] + } + ], + "source": [ + "l=fminbnd(@(l) lambda_fcn(A,b,l),0.5,1.5)" + ] + }, + { + "cell_type": "code", + "execution_count": 108, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ans =\n", + "\n", + " 3.0000\n", + " -2.5000\n", + " 7.0000\n", + "\n" + ] + } + ], + "source": [ + "A\\b" + ] + }, + { + "cell_type": "code", + "execution_count": 109, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x =\n", + "\n", + " 3.0000\n", + " -2.5000\n", + " 7.0000\n", + "\n", + "ea =\n", + "\n", + " 1.8289e-07\n", + " 2.1984e-08\n", + " 2.3864e-08\n", + "\n", + "iter = 8\n", + "x =\n", + "\n", + " 3.0000\n", + " -2.5000\n", + " 7.0000\n", + "\n", + "ea =\n", + "\n", + " 1.9130e-08\n", + " 7.6449e-08\n", + " 3.3378e-08\n", + "\n", + "iter = 8\n" + ] + } + ], + "source": [ + "[x,ea,iter]=Jacobi_rel(A,b,l,0.000001)\n", + "[x,ea,iter]=Jacobi_rel(A,b,1,0.000001)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Nonlinear Systems\n", + "\n", + "Consider two simultaneous nonlinear equations with two unknowns:\n", + "\n", + "$x_{1}^{2}+x_{1}x_{2}=10$\n", + "\n", + "$x_{2}+3x_{1}x_{2}^{2}=57$\n", + "\n", + "Graphically, we are looking for the solution:\n" + ] + }, + { + "cell_type": "code", + "execution_count": 121, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/svg+xml": [ + "\n", + "\n", + "Gnuplot\n", + "Produced by GNUPLOT 5.0 patchlevel 3 \n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t \n", + "\t \n", + "\t\n", + "\t\n", + "\t \n", + "\t \n", + "\t\n", + "\n", + "\n", + "\n", + "\n", + "\t\n", + "\t\t\n", + "\t\n", + "\n", + "\n", + "\n", + "\n", + "\t\t\n", + "\t\t0\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t5\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t10\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t15\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t20\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t0\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t1\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t2\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t3\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t4\n", + "\t\n", + "\n", + "\n", + "\t\t\n", + "\t\t5\n", + "\t\n", + "\n", + "\n", + "\n", + "\n", + "\t\n", + "\n", + "\n", + "\tgnuplot_plot_1a\n", + "\n", + "\n", + "\n", + "\t\n", + "\t\n", + "\tgnuplot_plot_2a\n", + "\n", + "\t\n", + "\t\n", + "\tgnuplot_plot_3a\n", + "\n", + "\t \n", + "\t\n", + "\n", + "\t\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "" + ], + "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "x11=linspace(0.5,3);\n", + "x12=(10-x11.^2)./x11;\n", + "\n", + "x22=linspace(2,8);\n", + "x21=(57-x22).*x22.^-2/3;\n", + "\n", + "plot(x11,x12,x21,x22)\n", + "% Solution at x_1=2, x_2=3\n", + "hold on;\n", + "plot(2,3,'o')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Newton-Raphson part II\n", + "\n", + "Remember the first order approximation for the next point in a function is:\n", + "\n", + "$f(x_{i+1})=f(x_{i})+(x_{i+1}-x_{i})f'(x_{i})$\n", + "\n", + "then, $f(x_{i+1})=0$ so we are left with:\n", + "\n", + "$x_{i+1}=x_{i}-\\frac{f(x_{i})}{f'(x_{i})}$\n", + "\n", + "We can use the same formula, but now we have multiple dimensions so we need to determine the Jacobian\n", + "\n", + "$[J]=\\left[ \\begin{array}{cccc}\n", + "\\frac{\\partial f_{1,i}}{\\partial x_{1}} & \\frac{\\partial f_{1,i}}{\\partial x_{2}} & \n", + "\\cdots & \\frac{\\partial f_{1,i}}{\\partial x_{n}} \\\\\n", + "\\frac{\\partial f_{2,i}}{\\partial x_{1}} & \\frac{\\partial f_{2,i}}{\\partial x_{2}} & \n", + "\\cdots & \\frac{\\partial f_{2,i}}{\\partial x_{n}} \\\\\n", + "\\vdots & \\vdots & & \\vdots \\\\\n", + "\\frac{\\partial f_{n,i}}{\\partial x_{1}} & \\frac{\\partial f_{n,i}}{\\partial x_{2}} & \n", + "\\cdots & \\frac{\\partial f_{n,i}}{\\partial x_{n}} \\\\\n", + "\\end{array} \\right]$\n", + "\n", + "$\\left[ \\begin{array}{c}\n", + "f_{1,i+1} \\\\\n", + "f_{2,i+1} \\\\\n", + "\\vdots \\\\\n", + "f_{n,i+1}\\end{array} \\right]=\n", + "\\left[ \\begin{array}{c}\n", + "f_{1,i} \\\\\n", + "f_{2,i} \\\\\n", + "\\vdots \\\\\n", + "f_{n,i}\\end{array} \\right]+\n", + "\\left[ \\begin{array}{cccc}\n", + "\\frac{\\partial f_{1,i}}{\\partial x_{1}} & \\frac{\\partial f_{1,i}}{\\partial x_{2}} & \n", + "\\cdots & \\frac{\\partial f_{1,i}}{\\partial x_{n}} \\\\\n", + "\\frac{\\partial f_{2,i}}{\\partial x_{1}} & \\frac{\\partial f_{2,i}}{\\partial x_{2}} & \n", + "\\cdots & \\frac{\\partial f_{2,i}}{\\partial x_{n}} \\\\\n", + "\\vdots & \\vdots & & \\vdots \\\\\n", + "\\frac{\\partial f_{n,i}}{\\partial x_{1}} & \\frac{\\partial f_{n,i}}{\\partial x_{2}} & \n", + "\\cdots & \\frac{\\partial f_{n,i}}{\\partial x_{n}} \\\\\n", + "\\end{array} \\right]\n", + "\\left( \\left[ \\begin{array}{c}\n", + "x_{i+1} \\\\\n", + "x_{i+1} \\\\\n", + "\\vdots \\\\\n", + "x_{i+1}\\end{array} \\right]-\n", + "\\left[ \\begin{array}{c}\n", + "f_{1,i} \\\\\n", + "f_{2,i} \\\\\n", + "\\vdots \\\\\n", + "f_{n,i}\\end{array} \\right]\\right)$\n", + "\n", + "### Solution is again in the form Ax=b\n", + "\n", + "$[J]([x_{i+1}]-[x_{i}])=-[f]$\n", + "\n", + "so\n", + "\n", + "$[x_{i+1}]= [x_{i}]-[J]^{-1}[f]$\n", + "\n", + "## Example of Jacobian calculation\n", + "\n", + "### Nonlinear springs supporting two masses in series\n", + "\n", + "Two springs are connected to two masses, with $m_1$=1 kg and $m_{2}$=2 kg. The springs are identical, but they have nonlinear spring constants, of $k_1$=10 N/m and $k_2$=-4 N/m\n", + "\n", + "We want to solve for the final position of the masses ($x_1$ and $x_2$)\n", + "\n", + "$m_{1}g+k_{1}(x_{2}-x_{1})+k_{2}(x_{2}-x_{1})^{2}+k_{1}x_{1}+k_{2}x_{1}^{2}=0$\n", + "\n", + "$m_{2}g-k_{1}(x_{2}-x_{1})-k_{2}(x_2-x_1)^{2}=0$\n", + "\n", + "$J(1,1)=\\frac{\\partial f_{1}}{\\partial x_{1}}=-k_{1}-2k_{2}(x_{2}-x_{1})+k_{1}+2k_{2}x_{1}$\n", + "\n", + "$J(1,2)=\\frac{\\partial f_1}{\\partial x_{2}}=k_{1}+2k_{2}(x_{2}-x_{1})$\n", + "\n", + "$J(2,1)=\\frac{\\partial f_2}{\\partial x_{1}}=k_{1}+2k_{2}(x_{2}-x_{1})$\n", + "\n", + "$J(2,2)=\\frac{\\partial f_2}{\\partial x_{2}}=-k_{1}-2k_{2}(x_{2}-x_{1})$\n", + "\n", + "Use an initial guess of $x_1=x_2=0$\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "m1=1; % kg \n", + "m2=2; % kg\n", + "k1=10; % N/m\n", + "k2=-4; % N/m^2" + ] + }, + { + "cell_type": "code", + "execution_count": 214, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + "function [f,J]=mass_spring(x)\n", + " % Function to calculate function values f1 and f2 as well as Jacobian \n", + " % for 2 masses and 2 identical nonlinear springs\n", + " m1=1; % kg \n", + " m2=2; % kg\n", + " k1=100; % N/m\n", + " k2=-10; % N/m^2\n", + " g=9.81; % m/s^2\n", + " x1=x(1);\n", + " x2=x(2);\n", + " J=[-k1-2*k2*(x2-x1)-k1-2*k2*x1,k1+2*k2*(x2-x1);\n", + " k1+2*k2*(x2-x1),-k1-2*k2*(x2-x1)];\n", + " f=[m1*g+k1*(x2-x1)+k2*(x2-x1).^2-k1*x1-k2*x1^2;\n", + " m2*g-k1*(x2-x1)-k2*(x2-x1).^2];\n", + "end\n", + " " + ] + }, + { + "cell_type": "code", + "execution_count": 217, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "f =\n", + "\n", + " -190.19\n", + " 129.62\n", + "\n", + "J =\n", + "\n", + " -200 120\n", + " 120 -120\n", + "\n" + ] + } + ], + "source": [ + "[f,J]=mass_spring([1,0])" + ] + }, + { + "cell_type": "code", + "execution_count": 227, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x1 =\n", + "\n", + " -1.5142\n", + " -1.4341\n", + "\n", + "ea =\n", + "\n", + " 2.9812\n", + " 2.3946\n", + "\n", + "x2 =\n", + "\n", + " 0.049894\n", + " 0.248638\n", + "\n", + "ea =\n", + "\n", + " 31.3492\n", + " 6.7678\n", + "\n", + "x3 =\n", + "\n", + " 0.29701\n", + " 0.49722\n", + "\n", + "ea =\n", + "\n", + " 0.83201\n", + " 0.49995\n", + "\n", + "x =\n", + "\n", + " 0.29701\n", + " 0.49722\n", + "\n", + "ea =\n", + "\n", + " 0.021392\n", + " 0.012890\n", + "\n", + "ea =\n", + "\n", + " 1.4786e-05\n", + " 8.9091e-06\n", + "\n", + "ea =\n", + "\n", + " 7.0642e-12\n", + " 4.2565e-12\n", + "\n" + ] + } + ], + "source": [ + "x0=[3;2];\n", + "[f0,J0]=mass_spring(x0);\n", + "x1=x0-J0\\f0\n", + "ea=(x1-x0)./x1\n", + "[f1,J1]=mass_spring(x1);\n", + "x2=x1-J1\\f1\n", + "ea=(x2-x1)./x2\n", + "[f2,J2]=mass_spring(x2);\n", + "x3=x2-J2\\f2\n", + "ea=(x3-x2)./x3\n", + "x=x3\n", + "for i=1:3\n", + " xold=x;\n", + " [f,J]=mass_spring(x);\n", + " x=x-J\\f;\n", + " ea=(x-xold)./x\n", + "end" + ] + }, + { + "cell_type": "code", + "execution_count": 228, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x =\n", + "\n", + " 0.30351\n", + " 0.50372\n", + "\n", + "X0 =\n", + "\n", + " 0.30351\n", + " 0.50372\n", + "\n" + ] + } + ], + "source": [ + "x\n", + "X0=fsolve(@(x) mass_spring(x),[3;5])" + ] + }, + { + "cell_type": "code", + "execution_count": 236, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/svg+xml": [ + "\n", + "\n", + "Gnuplot\n", + "Produced by GNUPLOT 5.0 patchlevel 3 \n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + 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b/lecture_13/lecture_13.md @@ -0,0 +1,943 @@ + + +```octave +%plot --format svg +``` + + +```octave +setdefaults +``` + +## My question from last class + +![q1](efficient_soln.png) + +![A](https://lh4.googleusercontent.com/fmG7EnFxpvvjSgijOuwz8osuiH3cBDgOzTE64KnfQeeDDSG2oE86-BzcpYIQMVkkAgRRGEDEGi6-Nkr8qmEMeaAk-gcjEmXe42WFYUdOa5XoUaBkXRakkA77_XrkRjArCGZIFhjjDRoO7x0) + +![q2](norm_A.png) + + +## Your questions from last class + +1. Do we have to submit a link for HW #4 somewhere or is uploading to Github sufficient? + + -no, your submission from HW 3 is sufficient + +2. How do I get the formulas/formatting in markdown files to show up on github? + + -no luck for markdown equations in github, this is an ongoing request + +3. Confused about the p=1 norm part and ||A||_1 + +4. When's the exam? + + -next week (3/9) + +5. What do you recommend doing to get better at figuring out the homeworks? + + -time and experimenting (try going through the lecture examples, verify my work) + +6. Could we have an hw or extra credit with a video lecture to learn some simple python? + + -Sounds great! how simple? + + -[Installing Python and Jupyter Notebook (via Anaconda) - https://www.continuum.io/downloads](https://www.continuum.io/downloads) + + -[Running Matlab kernel in Jupyter - https://anneurai.net/2015/11/12/matlab-based-ipython-notebooks/](https://anneurai.net/2015/11/12/matlab-based-ipython-notebooks/) + + -[Running Octave kernel in Jupyter - https://anaconda.org/pypi/octave_kernel](https://anaconda.org/pypi/octave_kernel) + + + +## Condition of a matrix +### *just checked in to see what condition my condition was in* +### Matrix norms + +The Euclidean norm of a vector is measure of the magnitude (in 3D this would be: $|x|=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$) in general the equation is: + +$||x||_{e}=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}$ + +For a matrix, A, the same norm is called the Frobenius norm: + +$||A||_{f}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{2}}$ + +In general we can calculate any $p$-norm where + +$||A||_{p}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{p}}$ + +so the p=1, 1-norm is + +$||A||_{1}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{1}}=\sum_{i=1}^{n}\sum_{i=1}^{m}|A_{i,j}|$ + +$||A||_{\infty}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{\infty}}=\max_{1\le i \le n}\sum_{j=1}^{m}|A_{i,j}|$ + +### Condition of Matrix + +The matrix condition is the product of + +$Cond(A) = ||A||\cdot||A^{-1}||$ + +So each norm will have a different condition number, but the limit is $Cond(A)\ge 1$ + +An estimate of the rounding error is based on the condition of A: + +$\frac{||\Delta x||}{x} \le Cond(A) \frac{||\Delta A||}{||A||}$ + +So if the coefficients of A have accuracy to $10^{-t} + +and the condition of A, $Cond(A)=10^{c}$ + +then the solution for x can have rounding errors up to $10^{c-t}$ + + + +```octave +A=[1,1/2,1/3;1/2,1/3,1/4;1/3,1/4,1/5] +[L,U]=LU_naive(A) +``` + + A = + + 1.00000 0.50000 0.33333 + 0.50000 0.33333 0.25000 + 0.33333 0.25000 0.20000 + + L = + + 1.00000 0.00000 0.00000 + 0.50000 1.00000 0.00000 + 0.33333 1.00000 1.00000 + + U = + + 1.00000 0.50000 0.33333 + 0.00000 0.08333 0.08333 + 0.00000 -0.00000 0.00556 + + + +Then, $A^{-1}=(LU)^{-1}=U^{-1}L^{-1}$ + +$Ld_{1}=\left[\begin{array}{c} +1 \\ +0 \\ +0 \end{array}\right]$, $Ux_{1}=d_{1}$ ... + + +```octave +invA=zeros(3,3); +d1=L\[1;0;0]; +d2=L\[0;1;0]; +d3=L\[0;0;1]; +invA(:,1)=U\d1; +invA(:,2)=U\d2; +invA(:,3)=U\d3 +invA*A +``` + + invA = + + 9.0000 -36.0000 30.0000 + -36.0000 192.0000 -180.0000 + 30.0000 -180.0000 180.0000 + + ans = + + 1.0000e+00 3.5527e-15 2.9976e-15 + -1.3249e-14 1.0000e+00 -9.1038e-15 + 8.5117e-15 7.1054e-15 1.0000e+00 + + + +Find the condition of A, $cond(A)$ + + +```octave +% Frobenius norm +normf_A = sqrt(sum(sum(A.^2))) +normf_invA = sqrt(sum(sum(invA.^2))) + +cond_f_A = normf_A*normf_invA + +norm(A,'fro') + +% p=1, column sum norm +norm1_A = max(sum(A,2)) +norm1_invA = max(sum(invA,2)) +norm(A,1) + +cond_1_A=norm1_A*norm1_invA + +% p=inf, row sum norm +norminf_A = max(sum(A,1)) +norminf_invA = max(sum(invA,1)) +norm(A,inf) + +cond_inf_A=norminf_A*norminf_invA + +``` + + normf_A = 1.4136 + normf_invA = 372.21 + cond_f_A = 526.16 + ans = 1.4136 + norm1_A = 1.8333 + norm1_invA = 30.000 + ans = 1.8333 + cond_1_A = 55.000 + norminf_A = 1.8333 + norminf_invA = 30.000 + ans = 1.8333 + cond_inf_A = 55.000 + + +Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with spring constants $K_{i}$. What does a high condition number mean for this problem? + +![Springs-masses](../lecture_09/mass_springs.png) + +The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass: + +$m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0$ + +$m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0$ + +$m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0$ + +$m_{4}g-k_{4}(x_{4}-x_{3})=0$ + +in matrix form: + +$\left[ \begin{array}{cccc} +k_{1}+k_{2} & -k_{2} & 0 & 0 \\ +-k_{2} & k_{2}+k_{3} & -k_{3} & 0 \\ +0 & -k_{3} & k_{3}+k_{4} & -k_{4} \\ +0 & 0 & -k_{4} & k_{4} \end{array} \right] +\left[ \begin{array}{c} +x_{1} \\ +x_{2} \\ +x_{3} \\ +x_{4} \end{array} \right]= +\left[ \begin{array}{c} +m_{1}g \\ +m_{2}g \\ +m_{3}g \\ +m_{4}g \end{array} \right]$ + + +```octave +k1=10; % N/m +k2=100000; +k3=10; +k4=1; +m1=1; % kg +m2=2; +m3=3; +m4=4; +g=9.81; % m/s^2 +K=[k1+k2 -k2 0 0; -k2 k2+k3 -k3 0; 0 -k3 k3+k4 -k4; 0 0 -k4 k4] +y=[m1*g;m2*g;m3*g;m4*g] +``` + + K = + + 100010 -100000 0 0 + -100000 100010 -10 0 + 0 -10 11 -1 + 0 0 -1 1 + + y = + + 9.8100 + 19.6200 + 29.4300 + 39.2400 + + + + +```octave +cond(K,inf) +cond(K,1) +cond(K,'fro') +cond(K,2) +``` + + ans = 3.2004e+05 + ans = 3.2004e+05 + ans = 2.5925e+05 + ans = 2.5293e+05 + + + +```octave +e=eig(K) +max(e)/min(e) +``` + + e = + + 7.9078e-01 + 3.5881e+00 + 1.7621e+01 + 2.0001e+05 + + ans = 2.5293e+05 + + +# Iterative Methods + +## Gauss-Seidel method + +If we have an intial guess for each value of a vector $x$ that we are trying to solve, then it is easy enough to solve for one component given the others. + +Take a 3$\times$3 matrix + +$Ax=b$ + +$\left[ \begin{array}{ccc} +3 & -0.1 & -0.2 \\ +0.1 & 7 & -0.3 \\ +0.3 & -0.2 & 10 \end{array} \right] +\left[ \begin{array}{c} +x_{1} \\ +x_{2} \\ +x_{3} \end{array} \right]= +\left[ \begin{array}{c} +7.85 \\ +-19.3 \\ +71.4\end{array} \right]$ + +$x_{1}=\frac{7.85+0.1x_{2}+0.3x_{3}}{3}$ + +$x_{2}=\frac{-19.3-0.1x_{1}+0.3x_{3}}{7}$ + +$x_{3}=\frac{71.4+0.1x_{1}+0.2x_{2}}{10}$ + + +```octave +A=[3 -0.1 -0.2;0.1 7 -0.3;0.3 -0.2 10] +b=[7.85;-19.3;71.4] + +x=A\b +``` + + A = + + 3.00000 -0.10000 -0.20000 + 0.10000 7.00000 -0.30000 + 0.30000 -0.20000 10.00000 + + b = + + 7.8500 + -19.3000 + 71.4000 + + x = + + 3.0000 + -2.5000 + 7.0000 + + + +### Gauss-Seidel Iterative approach + +As a first guess, we can use $x_{1}=x_{2}=x_{3}=0$ + +$x_{1}=\frac{7.85+0.1(0)+0.3(0)}{3}=2.6167$ + +$x_{2}=\frac{-19.3-0.1(2.6167)+0.3(0)}{7}=-2.7945$ + +$x_{3}=\frac{71.4+0.1(2.6167)+0.2(-2.7945)}{10}=7.0056$ + +Then, we update the guess: + +$x_{1}=\frac{7.85+0.1(-2.7945)+0.3(7.0056)}{3}=2.9906$ + +$x_{2}=\frac{-19.3-0.1(2.9906)+0.3(7.0056)}{7}=-2.4996$ + +$x_{3}=\frac{71.4+0.1(2.9906)+0.2(-2.4966)}{10}=7.00029$ + +The results are conveerging to the solution we found with `\` of $x_{1}=3,~x_{2}=-2.5,~x_{3}=7$ + +We could also use an iterative method that solves for all of the x-components in one step: + +### Jacobi method + +$x_{1}^{i}=\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$ + +$x_{2}^{i}=\frac{-19.3-0.1x_{1}^{i-1}+0.3x_{3}^{i-1}}{7}$ + +$x_{3}^{i}=\frac{71.4+0.1x_{1}^{i-1}+0.2x_{2}^{i-1}}{10}$ + +Here the solution is a matrix multiplication and vector addition + +$\left[ \begin{array}{c} +x_{1}^{i} \\ +x_{2}^{i} \\ +x_{3}^{i} \end{array} \right]= +\left[ \begin{array}{c} +7.85/3 \\ +-19.3/7 \\ +71.4/10\end{array} \right]- +\left[ \begin{array}{ccc} +0 & -0.1 & -0.2 \\ +0.1 & 0 & -0.3 \\ +0.3 & -0.2 & 0 \end{array} \right] +\left[ \begin{array}{c} +x_{1}^{i-1} \\ +x_{2}^{i-1} \\ +x_{3}^{i-1} \end{array} \right]$ + +|x_{j}|Jacobi method |vs| Gauss-Seidel | +|--------|------------------------------|---|-------------------------------| +|$x_{1}^{i}=$ | $\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$ | | $\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$| +|$x_{2}^{i}=$ | $\frac{-19.3-0.1x_{1}^{i-1}+0.3x_{3}^{i-1}}{7}$ | | $\frac{-19.3-0.1x_{1}^{i}+0.3x_{3}^{i-1}}{7}$ | +|$x_{3}^{i}=$ | $\frac{71.4+0.1x_{1}^{i-1}+0.2x_{2}^{i-1}}{10}$ | | $\frac{71.4+0.1x_{1}^{i}+0.2x_{2}^{i}}{10}$| + + +```octave +ba=b./diag(A) % or ba=b./[A(1,1);A(2,2);A(3,3)] +sA=A-diag(diag(A)) % A with zeros on diagonal +sA(1,:)=sA(1,:)/A(1,1); +sA(2,:)=sA(2,:)/A(2,2); +sA(3,:)=sA(3,:)/A(3,3) +x0=[0;0;0]; +x1=ba-sA*x0 +x2=ba-sA*x1 +x3=ba-sA*x2 +fprintf('solution is converging to [3,-2.5,7]]\n') +``` + + ba = + + 2.6167 + -2.7571 + 7.1400 + + sA = + + 0.00000 -0.10000 -0.20000 + 0.10000 0.00000 -0.30000 + 0.30000 -0.20000 0.00000 + + sA = + + 0.000000 -0.033333 -0.066667 + 0.014286 0.000000 -0.042857 + 0.030000 -0.020000 0.000000 + + x1 = + + 2.6167 + -2.7571 + 7.1400 + + x2 = + + 3.0008 + -2.4885 + 7.0064 + + x3 = + + 3.0008 + -2.4997 + 7.0002 + + solution is converging to [3,-2.5,7]] + + + +```octave +diag(A) +diag(diag(A)) +``` + + ans = + + 3 + 7 + 10 + + ans = + + Diagonal Matrix + + 3 0 0 + 0 7 0 + 0 0 10 + + + +This method works if problem is diagonally dominant, + +$|a_{ii}|>\sum_{j=1,j\ne i}^{n}|a_{ij}|$ + +If this condition is true, then Jacobi or Gauss-Seidel should converge + + + + +```octave +A=[0.1,1,3;1,0.2,3;5,2,0.3] +b=[12;2;4] +A\b +``` + + A = + + 0.10000 1.00000 3.00000 + 1.00000 0.20000 3.00000 + 5.00000 2.00000 0.30000 + + b = + + 12 + 2 + 4 + + ans = + + -2.9393 + 9.1933 + 1.0336 + + + + +```octave +ba=b./diag(A) % or ba=b./[A(1,1);A(2,2);A(3,3)] +sA=A-diag(diag(A)) % A with zeros on diagonal +sA(1,:)=sA(1,:)/A(1,1); +sA(2,:)=sA(2,:)/A(2,2); +sA(3,:)=sA(3,:)/A(3,3) +x0=[0;0;0]; +x1=ba-sA*x0 +x2=ba-sA*x1 +x3=ba-sA*x2 +fprintf('solution is not converging to [-2.93,9.19,1.03]\n') +``` + + ba = + + 120.000 + 10.000 + 13.333 + + sA = + + 0 1 3 + 1 0 3 + 5 2 0 + + sA = + + 0.00000 10.00000 30.00000 + 5.00000 0.00000 15.00000 + 16.66667 6.66667 0.00000 + + x1 = + + 120.000 + 10.000 + 13.333 + + x2 = + + -380.00 + -790.00 + -2053.33 + + x3 = + + 6.9620e+04 + 3.2710e+04 + 1.1613e+04 + + solution is not converging to [-2.93,9.19,1.03] + + +## Gauss-Seidel with Relaxation + +In order to force the solution to converge faster, we can introduce a relaxation term $\lambda$. + +where the new x values are weighted between the old and new: + +$x^{i}=\lambda x^{i}+(1-\lambda)x^{i-1}$ + +after solving for x, lambda weights the current approximation with the previous approximation for the updated x + + + +```octave +% rearrange A and b +A=[3 -0.1 -0.2;0.1 7 -0.3;0.3 -0.2 10] +b=[7.85;-19.3;71.4] + +iters=zeros(100,1); +for i=1:100 + lambda=2/100*i; + [x,ea,iters(i)]=Jacobi_rel(A,b,lambda); +end +plot([1:100]*2/100,iters) +``` + + A = + + 3.00000 -0.10000 -0.20000 + 0.10000 7.00000 -0.30000 + 0.30000 -0.20000 10.00000 + + b = + + 7.8500 + -19.3000 + 71.4000 + + + + +![svg](lecture_13_files/lecture_13_22_1.svg) + + + +```octave +l=fminbnd(@(l) lambda_fcn(A,b,l),0.5,1.5) +``` + + l = 0.99158 + + + +```octave +A\b +``` + + ans = + + 3.0000 + -2.5000 + 7.0000 + + + + +```octave +[x,ea,iter]=Jacobi_rel(A,b,l,0.000001) +[x,ea,iter]=Jacobi_rel(A,b,1,0.000001) + +``` + + x = + + 3.0000 + -2.5000 + 7.0000 + + ea = + + 1.8289e-07 + 2.1984e-08 + 2.3864e-08 + + iter = 8 + x = + + 3.0000 + -2.5000 + 7.0000 + + ea = + + 1.9130e-08 + 7.6449e-08 + 3.3378e-08 + + iter = 8 + + +## Nonlinear Systems + +Consider two simultaneous nonlinear equations with two unknowns: + +$x_{1}^{2}+x_{1}x_{2}=10$ + +$x_{2}+3x_{1}x_{2}^{2}=57$ + +Graphically, we are looking for the solution: + + + +```octave +x11=linspace(0.5,3); +x12=(10-x11.^2)./x11; + +x22=linspace(2,8); +x21=(57-x22).*x22.^-2/3; + +plot(x11,x12,x21,x22) +% Solution at x_1=2, x_2=3 +hold on; +plot(2,3,'o') +``` + + +![svg](lecture_13_files/lecture_13_27_0.svg) + + +## Newton-Raphson part II + +Remember the first order approximation for the next point in a function is: + +$f(x_{i+1})=f(x_{i})+(x_{i+1}-x_{i})f'(x_{i})$ + +then, $f(x_{i+1})=0$ so we are left with: + +$x_{i+1}=x_{i}-\frac{f(x_{i})}{f'(x_{i})}$ + +We can use the same formula, but now we have multiple dimensions so we need to determine the Jacobian + +$[J]=\left[ \begin{array}{cccc} +\frac{\partial f_{1,i}}{\partial x_{1}} & \frac{\partial f_{1,i}}{\partial x_{2}} & +\cdots & \frac{\partial f_{1,i}}{\partial x_{n}} \\ +\frac{\partial f_{2,i}}{\partial x_{1}} & \frac{\partial f_{2,i}}{\partial x_{2}} & +\cdots & \frac{\partial f_{2,i}}{\partial x_{n}} \\ +\vdots & \vdots & & \vdots \\ +\frac{\partial f_{n,i}}{\partial x_{1}} & \frac{\partial f_{n,i}}{\partial x_{2}} & +\cdots & \frac{\partial f_{n,i}}{\partial x_{n}} \\ +\end{array} \right]$ + +$\left[ \begin{array}{c} +f_{1,i+1} \\ +f_{2,i+1} \\ +\vdots \\ +f_{n,i+1}\end{array} \right]= +\left[ \begin{array}{c} +f_{1,i} \\ +f_{2,i} \\ +\vdots \\ +f_{n,i}\end{array} \right]+ +\left[ \begin{array}{cccc} +\frac{\partial f_{1,i}}{\partial x_{1}} & \frac{\partial f_{1,i}}{\partial x_{2}} & +\cdots & \frac{\partial f_{1,i}}{\partial x_{n}} \\ +\frac{\partial f_{2,i}}{\partial x_{1}} & \frac{\partial f_{2,i}}{\partial x_{2}} & +\cdots & \frac{\partial f_{2,i}}{\partial x_{n}} \\ +\vdots & \vdots & & \vdots \\ +\frac{\partial f_{n,i}}{\partial x_{1}} & \frac{\partial f_{n,i}}{\partial x_{2}} & +\cdots & \frac{\partial f_{n,i}}{\partial x_{n}} \\ +\end{array} \right] +\left( \left[ \begin{array}{c} +x_{i+1} \\ +x_{i+1} \\ +\vdots \\ +x_{i+1}\end{array} \right]- +\left[ \begin{array}{c} +f_{1,i} \\ +f_{2,i} \\ +\vdots \\ +f_{n,i}\end{array} \right]\right)$ + +### Solution is again in the form Ax=b + +$[J]([x_{i+1}]-[x_{i}])=-[f]$ + +so + +$[x_{i+1}]= [x_{i}]-[J]^{-1}[f]$ + +## Example of Jacobian calculation + +### Nonlinear springs supporting two masses in series + +Two springs are connected to two masses, with $m_1$=1 kg and $m_{2}$=2 kg. The springs are identical, but they have nonlinear spring constants, of $k_1$=10 N/m and $k_2$=-4 N/m + +We want to solve for the final position of the masses ($x_1$ and $x_2$) + +$m_{1}g+k_{1}(x_{2}-x_{1})+k_{2}(x_{2}-x_{1})^{2}+k_{1}x_{1}+k_{2}x_{1}^{2}=0$ + +$m_{2}g-k_{1}(x_{2}-x_{1})-k_{2}(x_2-x_1)^{2}=0$ + +$J(1,1)=\frac{\partial f_{1}}{\partial x_{1}}=-k_{1}-2k_{2}(x_{2}-x_{1})+k_{1}+2k_{2}x_{1}$ + +$J(1,2)=\frac{\partial f_1}{\partial x_{2}}=k_{1}+2k_{2}(x_{2}-x_{1})$ + +$J(2,1)=\frac{\partial f_2}{\partial x_{1}}=k_{1}+2k_{2}(x_{2}-x_{1})$ + +$J(2,2)=\frac{\partial f_2}{\partial x_{2}}=-k_{1}-2k_{2}(x_{2}-x_{1})$ + +Use an initial guess of $x_1=x_2=0$ + + + +```octave +m1=1; % kg +m2=2; % kg +k1=10; % N/m +k2=-4; % N/m^2 +``` + + +```octave +function [f,J]=mass_spring(x) + % Function to calculate function values f1 and f2 as well as Jacobian + % for 2 masses and 2 identical nonlinear springs + m1=1; % kg + m2=2; % kg + k1=100; % N/m + k2=-10; % N/m^2 + g=9.81; % m/s^2 + x1=x(1); + x2=x(2); + J=[-k1-2*k2*(x2-x1)-k1-2*k2*x1,k1+2*k2*(x2-x1); + k1+2*k2*(x2-x1),-k1-2*k2*(x2-x1)]; + f=[m1*g+k1*(x2-x1)+k2*(x2-x1).^2-k1*x1-k2*x1^2; + m2*g-k1*(x2-x1)-k2*(x2-x1).^2]; +end + +``` + + +```octave +[f,J]=mass_spring([1,0]) +``` + + f = + + -190.19 + 129.62 + + J = + + -200 120 + 120 -120 + + + + +```octave +x0=[3;2]; +[f0,J0]=mass_spring(x0); +x1=x0-J0\f0 +ea=(x1-x0)./x1 +[f1,J1]=mass_spring(x1); +x2=x1-J1\f1 +ea=(x2-x1)./x2 +[f2,J2]=mass_spring(x2); +x3=x2-J2\f2 +ea=(x3-x2)./x3 +x=x3 +for i=1:3 + xold=x; + [f,J]=mass_spring(x); + x=x-J\f; + ea=(x-xold)./x +end +``` + + x1 = + + -1.5142 + -1.4341 + + ea = + + 2.9812 + 2.3946 + + x2 = + + 0.049894 + 0.248638 + + ea = + + 31.3492 + 6.7678 + + x3 = + + 0.29701 + 0.49722 + + ea = + + 0.83201 + 0.49995 + + x = + + 0.29701 + 0.49722 + + ea = + + 0.021392 + 0.012890 + + ea = + + 1.4786e-05 + 8.9091e-06 + + ea = + + 7.0642e-12 + 4.2565e-12 + + + + +```octave +x +X0=fsolve(@(x) mass_spring(x),[3;5]) +``` + + x = + + 0.30351 + 0.50372 + + X0 = + + 0.30351 + 0.50372 + + + + +```octave +[X,Y]=meshgrid(linspace(0,1,20),linspace(0,1,20)); +[N,M]=size(X); +F=zeros(size(X)); +for i=1:N + for j=1:M + [f,~]=mass_spring([X(i,j),Y(i,j)]); + F(i,j)=f(1); + end +end +pcolor(X,Y,F) +xlabel('x_1') +ylabel('x_2') +colorbar() +figure() +pcolor(X,Y,F) +xlabel('x_1') +ylabel('x_2') +colorbar() +``` + + +![svg](lecture_13_files/lecture_13_34_0.svg) + + + +![svg](lecture_13_files/lecture_13_34_1.svg) + + + +```octave + +``` diff --git a/lecture_13/lecture_13.out b/lecture_13/lecture_13.out new file mode 100644 index 0000000..edec246 --- /dev/null +++ b/lecture_13/lecture_13.out @@ -0,0 +1,16 @@ +\BOOKMARK [2][-]{subsection.0.1}{My question from last class}{}% 1 +\BOOKMARK [2][-]{subsection.0.2}{Your questions from last class}{}% 2 +\BOOKMARK [2][-]{subsection.0.3}{Condition of a matrix}{}% 3 +\BOOKMARK [3][-]{subsubsection.0.3.1}{just checked in to see what condition my condition was in}{subsection.0.3}% 4 +\BOOKMARK [3][-]{subsubsection.0.3.2}{Matrix norms}{subsection.0.3}% 5 +\BOOKMARK [3][-]{subsubsection.0.3.3}{Condition of Matrix}{subsection.0.3}% 6 +\BOOKMARK [1][-]{section.1}{Iterative Methods}{}% 7 +\BOOKMARK [2][-]{subsection.1.1}{Gauss-Seidel method}{section.1}% 8 +\BOOKMARK [3][-]{subsubsection.1.1.1}{Gauss-Seidel Iterative approach}{subsection.1.1}% 9 +\BOOKMARK [3][-]{subsubsection.1.1.2}{Jacobi method}{subsection.1.1}% 10 +\BOOKMARK [2][-]{subsection.1.2}{Gauss-Seidel with Relaxation}{section.1}% 11 +\BOOKMARK [2][-]{subsection.1.3}{Nonlinear Systems}{section.1}% 12 +\BOOKMARK [2][-]{subsection.1.4}{Newton-Raphson part II}{section.1}% 13 +\BOOKMARK [3][-]{subsubsection.1.4.1}{Solution is again in the form Ax=b}{subsection.1.4}% 14 +\BOOKMARK [2][-]{subsection.1.5}{Example of Jacobian calculation}{section.1}% 15 +\BOOKMARK [3][-]{subsubsection.1.5.1}{Nonlinear springs supporting two masses in series}{subsection.1.5}% 16 diff --git a/lecture_13/lecture_13.pdf b/lecture_13/lecture_13.pdf new file mode 100644 index 0000000..fad59c8 Binary files /dev/null and b/lecture_13/lecture_13.pdf differ diff --git a/lecture_13/lecture_13.tex b/lecture_13/lecture_13.tex new file mode 100644 index 0000000..87d8d25 --- /dev/null +++ b/lecture_13/lecture_13.tex @@ -0,0 +1,1255 @@ + +% Default to the notebook output style + + + + +% Inherit from the specified cell style. + + + + + +\documentclass[11pt]{article} + + + + \usepackage[T1]{fontenc} + % Nicer default font (+ math font) than Computer Modern for most use cases + \usepackage{mathpazo} + + % Basic figure setup, for now with no caption control since it's done + % automatically by Pandoc (which extracts ![](path) syntax from Markdown). + \usepackage{graphicx} + % We will generate all images so they have a width \maxwidth. This means + % that they will get their normal width if they fit onto the page, but + % are scaled down if they would overflow the margins. + \makeatletter + \def\maxwidth{\ifdim\Gin@nat@width>\linewidth\linewidth + \else\Gin@nat@width\fi} + \makeatother + \let\Oldincludegraphics\includegraphics + % Set max figure width to be 80% of text width, for now hardcoded. + \renewcommand{\includegraphics}[1]{\Oldincludegraphics[width=.8\maxwidth]{#1}} + % Ensure that by default, figures have no caption (until we provide a + % proper Figure object with a Caption API and a way to capture that + % in the conversion process - todo). + \usepackage{caption} + \DeclareCaptionLabelFormat{nolabel}{} + \captionsetup{labelformat=nolabel} + + \usepackage{adjustbox} % Used to constrain images to a maximum size + \usepackage{xcolor} % Allow colors to be defined + \usepackage{enumerate} % Needed for markdown enumerations to work + \usepackage{geometry} % Used to adjust the document margins + \usepackage{amsmath} % Equations + \usepackage{amssymb} % Equations + \usepackage{textcomp} % defines textquotesingle + % Hack from http://tex.stackexchange.com/a/47451/13684: + \AtBeginDocument{% + \def\PYZsq{\textquotesingle}% Upright quotes in Pygmentized code + } + \usepackage{upquote} % Upright quotes for verbatim code + \usepackage{eurosym} % defines \euro + \usepackage[mathletters]{ucs} % Extended unicode (utf-8) support + \usepackage[utf8x]{inputenc} % Allow utf-8 characters in the tex document + \usepackage{fancyvrb} % verbatim replacement that allows latex + \usepackage{grffile} % extends the file name processing of package graphics + % to support a larger range + % The hyperref package gives us a pdf with properly built + % internal navigation ('pdf bookmarks' for the table of contents, + % internal cross-reference links, web links for URLs, etc.) + \usepackage{hyperref} + \usepackage{longtable} % longtable support required by pandoc >1.10 + \usepackage{booktabs} % table support for pandoc > 1.12.2 + \usepackage[inline]{enumitem} % IRkernel/repr support (it uses the enumerate* environment) + \usepackage[normalem]{ulem} % ulem is needed to support strikethroughs (\sout) + % normalem makes italics be italics, not underlines + + + + + % Colors for the hyperref package + \definecolor{urlcolor}{rgb}{0,.145,.698} + \definecolor{linkcolor}{rgb}{.71,0.21,0.01} + \definecolor{citecolor}{rgb}{.12,.54,.11} + + % ANSI colors + \definecolor{ansi-black}{HTML}{3E424D} + \definecolor{ansi-black-intense}{HTML}{282C36} + \definecolor{ansi-red}{HTML}{E75C58} + \definecolor{ansi-red-intense}{HTML}{B22B31} + \definecolor{ansi-green}{HTML}{00A250} + \definecolor{ansi-green-intense}{HTML}{007427} + \definecolor{ansi-yellow}{HTML}{DDB62B} + \definecolor{ansi-yellow-intense}{HTML}{B27D12} + \definecolor{ansi-blue}{HTML}{208FFB} + \definecolor{ansi-blue-intense}{HTML}{0065CA} + \definecolor{ansi-magenta}{HTML}{D160C4} + \definecolor{ansi-magenta-intense}{HTML}{A03196} + 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PY@tok@nd\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.67,0.13,1.00}{##1}}} +\expandafter\def\csname PY@tok@ne\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.82,0.25,0.23}{##1}}} +\expandafter\def\csname PY@tok@nf\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.00,1.00}{##1}}} +\expandafter\def\csname PY@tok@si\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.53}{##1}}} +\expandafter\def\csname PY@tok@s2\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@nt\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@nv\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}} +\expandafter\def\csname PY@tok@s1\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@ch\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}} +\expandafter\def\csname PY@tok@m\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@gp\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.00,0.50}{##1}}} +\expandafter\def\csname PY@tok@sh\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@ow\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.67,0.13,1.00}{##1}}} +\expandafter\def\csname PY@tok@sx\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@bp\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@c1\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}} +\expandafter\def\csname PY@tok@o\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@kc\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@c\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}} +\expandafter\def\csname PY@tok@mf\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@err\endcsname{\def\PY@bc##1{\setlength{\fboxsep}{0pt}\fcolorbox[rgb]{1.00,0.00,0.00}{1,1,1}{\strut ##1}}} +\expandafter\def\csname PY@tok@mb\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@ss\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}} +\expandafter\def\csname PY@tok@sr\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.53}{##1}}} +\expandafter\def\csname PY@tok@mo\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@kd\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@mi\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@kn\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@cpf\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}} +\expandafter\def\csname PY@tok@kr\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@s\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@kp\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@w\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.73,0.73}{##1}}} +\expandafter\def\csname PY@tok@kt\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.69,0.00,0.25}{##1}}} +\expandafter\def\csname PY@tok@sc\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@sb\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@k\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@se\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.13}{##1}}} +\expandafter\def\csname PY@tok@sd\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} + +\def\PYZbs{\char`\\} +\def\PYZus{\char`\_} +\def\PYZob{\char`\{} +\def\PYZcb{\char`\}} +\def\PYZca{\char`\^} +\def\PYZam{\char`\&} +\def\PYZlt{\char`\<} +\def\PYZgt{\char`\>} +\def\PYZsh{\char`\#} +\def\PYZpc{\char`\%} +\def\PYZdl{\char`\$} +\def\PYZhy{\char`\-} +\def\PYZsq{\char`\'} +\def\PYZdq{\char`\"} +\def\PYZti{\char`\~} +% for compatibility with earlier versions +\def\PYZat{@} +\def\PYZlb{[} +\def\PYZrb{]} +\makeatother + + + % Exact colors from NB + \definecolor{incolor}{rgb}{0.0, 0.0, 0.5} + \definecolor{outcolor}{rgb}{0.545, 0.0, 0.0} + + + + + % Prevent overflowing lines due to hard-to-break entities + \sloppy + % Setup hyperref package + \hypersetup{ + breaklinks=true, % so long urls are correctly broken across lines + colorlinks=true, + urlcolor=urlcolor, + linkcolor=linkcolor, + citecolor=citecolor, + } + % Slightly bigger margins than the latex defaults + + \geometry{verbose,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in} + + + + \begin{document} + + + \maketitle + + + + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}1}]:} \PY{c}{\PYZpc{}plot \PYZhy{}\PYZhy{}format svg} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}2}]:} \PY{n}{setdefaults} +\end{Verbatim} + + \subsection{My question from last +class}\label{my-question-from-last-class} + +\begin{figure}[htbp] +\centering +\includegraphics{efficient_soln.png} +\caption{q1} +\end{figure} + +$A=\left[\begin{array}{ccc} + 2 & -2 & 0\\ + -1& 5 & 1 \\ +3 &4 & 5 \end{array}\right]$ + +\begin{figure}[htbp] +\centering +\includegraphics{norm_A.png} +\caption{q2} +\end{figure} + +\subsection{Your questions from last +class}\label{your-questions-from-last-class} + +\begin{enumerate} +\def\labelenumi{\arabic{enumi}.} +\item + Do we have to submit a link for HW \#4 somewhere or is uploading to + Github sufficient? + + -no, your submission from HW 3 is sufficient +\item + How do I get the formulas/formatting in markdown files to show up on + github? + + -no luck for markdown equations in github, this is an ongoing request +\item + Confused about the p=1 norm part and + \textbar{}\textbar{}A\textbar{}\textbar{}\_1 +\item + When's the exam? + + -next week (3/9) +\item + What do you recommend doing to get better at figuring out the + homeworks? + + -time and experimenting (try going through the lecture examples, + verify my work) +\item + Could we have an hw or extra credit with a video lecture to learn some + simple python? + + -Sounds great! how simple? + + -\href{https://www.continuum.io/downloads}{Installing Python and + Jupyter Notebook (via Anaconda) - https://www.continuum.io/downloads} + + -\href{https://anneurai.net/2015/11/12/matlab-based-ipython-notebooks/}{Running + Matlab kernel in Jupyter - + https://anneurai.net/2015/11/12/matlab-based-ipython-notebooks/} + + -\href{https://anaconda.org/pypi/octave_kernel}{Running Octave kernel + in Jupyter - https://anaconda.org/pypi/octave\_kernel} +\end{enumerate} + + \subsection{Condition of a matrix}\label{condition-of-a-matrix} + +\subsubsection{\texorpdfstring{\emph{just checked in to see what +condition my condition was +in}}{just checked in to see what condition my condition was in}}\label{just-checked-in-to-see-what-condition-my-condition-was-in} + +\subsubsection{Matrix norms}\label{matrix-norms} + +The Euclidean norm of a vector is measure of the magnitude (in 3D this +would be: $|x|=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$) in general the +equation is: + +$||x||_{e}=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}$ + +For a matrix, A, the same norm is called the Frobenius norm: + +$||A||_{f}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{2}}$ + +In general we can calculate any $p$-norm where + +$||A||_{p}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{p}}$ + +so the p=1, 1-norm is + +$||A||_{1}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{1}}=\sum_{i=1}^{n}\sum_{i=1}^{m}|A_{i,j}|$ + +$||A||_{\infty}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{\infty}}=\max_{1\le i \le n}\sum_{j=1}^{m}|A_{i,j}|$ + +\subsubsection{Condition of Matrix}\label{condition-of-matrix} + +The matrix condition is the product of + +$Cond(A) = ||A||\cdot||A^{-1}||$ + +So each norm will have a different condition number, but the limit is +$Cond(A)\ge 1$ + +An estimate of the rounding error is based on the condition of A: + +$\frac{||\Delta x||}{x} \le Cond(A) \frac{||\Delta A||}{||A||}$ + +So if the coefficients of A have accuracy to \$10\^{}\{-t\} + +and the condition of A, $Cond(A)=10^{c}$ + +then the solution for x can have rounding errors up to $10^{c-t}$ + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}72}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{4}\PY{p}{;}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{5}\PY{p}{]} + \PY{p}{[}\PY{n}{L}\PY{p}{,}\PY{n}{U}\PY{p}{]}\PY{p}{=}\PY{n}{LU\PYZus{}naive}\PY{p}{(}\PY{n}{A}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 1.00000 0.50000 0.33333 + 0.50000 0.33333 0.25000 + 0.33333 0.25000 0.20000 + +L = + + 1.00000 0.00000 0.00000 + 0.50000 1.00000 0.00000 + 0.33333 1.00000 1.00000 + +U = + + 1.00000 0.50000 0.33333 + 0.00000 0.08333 0.08333 + 0.00000 -0.00000 0.00556 + + + \end{Verbatim} + + Then, $A^{-1}=(LU)^{-1}=U^{-1}L^{-1}$ + +$Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]$, +$Ux_{1}=d_{1}$ ... + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}75}]:} \PY{n}{invA}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;} + \PY{n}{d1}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;} + \PY{n}{d2}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;} + \PY{n}{d3}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{;} + \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d1}\PY{p}{;} + \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d2}\PY{p}{;} + \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d3} + \PY{n}{invA}\PY{o}{*}\PY{n}{A} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +invA = + + 9.0000 -36.0000 30.0000 + -36.0000 192.0000 -180.0000 + 30.0000 -180.0000 180.0000 + +ans = + + 1.0000e+00 3.5527e-15 2.9976e-15 + -1.3249e-14 1.0000e+00 -9.1038e-15 + 8.5117e-15 7.1054e-15 1.0000e+00 + + + \end{Verbatim} + + Find the condition of A, $cond(A)$ + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}74}]:} \PY{c}{\PYZpc{} Frobenius norm} + \PY{n}{normf\PYZus{}A} \PY{p}{=} \PY{n+nb}{sqrt}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{o}{.\PYZca{}}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)} + \PY{n}{normf\PYZus{}invA} \PY{p}{=} \PY{n+nb}{sqrt}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{o}{.\PYZca{}}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)} + + \PY{n}{cond\PYZus{}f\PYZus{}A} \PY{p}{=} \PY{n}{normf\PYZus{}A}\PY{o}{*}\PY{n}{normf\PYZus{}invA} + + \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{fro\PYZsq{}}\PY{p}{)} + + \PY{c}{\PYZpc{} p=1, column sum norm} + \PY{n}{norm1\PYZus{}A} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)} + \PY{n}{norm1\PYZus{}invA} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)} + \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)} + + \PY{n}{cond\PYZus{}1\PYZus{}A}\PY{p}{=}\PY{n}{norm1\PYZus{}A}\PY{o}{*}\PY{n}{norm1\PYZus{}invA} + + \PY{c}{\PYZpc{} p=inf, row sum norm} + \PY{n}{norminf\PYZus{}A} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{)} + \PY{n}{norminf\PYZus{}invA} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{)} + \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n+nb}{inf}\PY{p}{)} + + \PY{n}{cond\PYZus{}inf\PYZus{}A}\PY{p}{=}\PY{n}{norminf\PYZus{}A}\PY{o}{*}\PY{n}{norminf\PYZus{}invA} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +normf\_A = 1.4136 +normf\_invA = 372.21 +cond\_f\_A = 526.16 +ans = 1.4136 +norm1\_A = 1.8333 +norm1\_invA = 30.000 +ans = 1.8333 +cond\_1\_A = 55.000 +norminf\_A = 1.8333 +norminf\_invA = 30.000 +ans = 1.8333 +cond\_inf\_A = 55.000 + + \end{Verbatim} + + Consider the problem again from the intro to Linear Algebra, 4 masses +are connected in series to 4 springs with spring constants $K_{i}$. +What does a high condition number mean for this problem? + +\begin{figure}[htbp] +\centering +\includegraphics{../lecture_09/mass_springs.png} +\caption{Springs-masses} +\end{figure} + +The masses haves the following amounts, 1, 2, 3, and 4 kg for masses +1-4. Using a FBD for each mass: + +$m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0$ + +$m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0$ + +$m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0$ + +$m_{4}g-k_{4}(x_{4}-x_{3})=0$ + +in matrix form: + +$\left[ \begin{array}{cccc} k_{1}+k_{2} & -k_{2} & 0 & 0 \\ -k_{2} & k_{2}+k_{3} & -k_{3} & 0 \\ 0 & -k_{3} & k_{3}+k_{4} & -k_{4} \\ 0 & 0 & -k_{4} & k_{4} \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]= \left[ \begin{array}{c} m_{1}g \\ m_{2}g \\ m_{3}g \\ m_{4}g \end{array} \right]$ + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}21}]:} \PY{n}{k1}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} \PY{c}{\PYZpc{} N/m} + \PY{n}{k2}\PY{p}{=}\PY{l+m+mi}{100000}\PY{p}{;} + \PY{n}{k3}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} + \PY{n}{k4}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} + \PY{n}{m1}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} kg} + \PY{n}{m2}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;} + \PY{n}{m3}\PY{p}{=}\PY{l+m+mi}{3}\PY{p}{;} + \PY{n}{m4}\PY{p}{=}\PY{l+m+mi}{4}\PY{p}{;} + \PY{n}{g}\PY{p}{=}\PY{l+m+mf}{9.81}\PY{p}{;} \PY{c}{\PYZpc{} m/s\PYZca{}2} + \PY{n}{K}\PY{p}{=}\PY{p}{[}\PY{n}{k1}\PY{o}{+}\PY{n}{k2} \PY{o}{\PYZhy{}}\PY{n}{k2} \PY{l+m+mi}{0} \PY{l+m+mi}{0}\PY{p}{;} \PY{o}{\PYZhy{}}\PY{n}{k2} \PY{n}{k2}\PY{o}{+}\PY{n}{k3} \PY{o}{\PYZhy{}}\PY{n}{k3} \PY{l+m+mi}{0}\PY{p}{;} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k3} \PY{n}{k3}\PY{o}{+}\PY{n}{k4} \PY{o}{\PYZhy{}}\PY{n}{k4}\PY{p}{;} \PY{l+m+mi}{0} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k4} \PY{n}{k4}\PY{p}{]} + \PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{n}{m1}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m2}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m3}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m4}\PY{o}{*}\PY{n}{g}\PY{p}{]} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +K = + + 100010 -100000 0 0 + -100000 100010 -10 0 + 0 -10 11 -1 + 0 0 -1 1 + +y = + + 9.8100 + 19.6200 + 29.4300 + 39.2400 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}25}]:} \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{n+nb}{inf}\PY{p}{)} + \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)} + \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{fro\PYZsq{}}\PY{p}{)} + \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ans = 3.2004e+05 +ans = 3.2004e+05 +ans = 2.5925e+05 +ans = 2.5293e+05 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}26}]:} \PY{n+nb}{e}\PY{p}{=}\PY{n+nb}{eig}\PY{p}{(}\PY{n}{K}\PY{p}{)} + \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{e}\PY{p}{)}\PY{o}{/}\PY{n+nb}{min}\PY{p}{(}\PY{n+nb}{e}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +e = + + 7.9078e-01 + 3.5881e+00 + 1.7621e+01 + 2.0001e+05 + +ans = 2.5293e+05 + + \end{Verbatim} + + \section{Iterative Methods}\label{iterative-methods} + +\subsection{Gauss-Seidel method}\label{gauss-seidel-method} + +If we have an intial guess for each value of a vector $x$ that we are +trying to solve, then it is easy enough to solve for one component given +the others. + +Take a 3$\times$3 matrix + +$Ax=b$ + +$\left[ \begin{array}{ccc} 3 & -0.1 & -0.2 \\ 0.1 & 7 & -0.3 \\ 0.3 & -0.2 & 10 \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right]= \left[ \begin{array}{c} 7.85 \\ -19.3 \\ 71.4\end{array} \right]$ + +$x_{1}=\frac{7.85+0.1x_{2}+0.3x_{3}}{3}$ + +$x_{2}=\frac{-19.3-0.1x_{1}+0.3x_{3}}{7}$ + +$x_{3}=\frac{71.4+0.1x_{1}+0.2x_{2}}{10}$ + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}9}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{3} \PY{o}{\PYZhy{}}\PY{l+m+mf}{0.1} \PY{o}{\PYZhy{}}\PY{l+m+mf}{0.2}\PY{p}{;}\PY{l+m+mf}{0.1} \PY{l+m+mi}{7} \PY{o}{\PYZhy{}}\PY{l+m+mf}{0.3}\PY{p}{;}\PY{l+m+mf}{0.3} \PY{o}{\PYZhy{}}\PY{l+m+mf}{0.2} \PY{l+m+mi}{10}\PY{p}{]} + \PY{n}{b}\PY{p}{=}\PY{p}{[}\PY{l+m+mf}{7.85}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mf}{19.3}\PY{p}{;}\PY{l+m+mf}{71.4}\PY{p}{]} + + \PY{n}{x}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{b} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 3.00000 -0.10000 -0.20000 + 0.10000 7.00000 -0.30000 + 0.30000 -0.20000 10.00000 + +b = + + 7.8500 + -19.3000 + 71.4000 + +x = + + 3.0000 + -2.5000 + 7.0000 + + + \end{Verbatim} + + \subsubsection{Gauss-Seidel Iterative +approach}\label{gauss-seidel-iterative-approach} + +As a first guess, we can use $x_{1}=x_{2}=x_{3}=0$ + +$x_{1}=\frac{7.85+0.1(0)+0.3(0)}{3}=2.6167$ + +$x_{2}=\frac{-19.3-0.1(2.6167)+0.3(0)}{7}=-2.7945$ + +$x_{3}=\frac{71.4+0.1(2.6167)+0.2(-2.7945)}{10}=7.0056$ + +Then, we update the guess: + +$x_{1}=\frac{7.85+0.1(-2.7945)+0.3(7.0056)}{3}=2.9906$ + +$x_{2}=\frac{-19.3-0.1(2.9906)+0.3(7.0056)}{7}=-2.4996$ + +$x_{3}=\frac{71.4+0.1(2.9906)+0.2(-2.4966)}{10}=7.00029$ + +The results are conveerging to the solution we found with +\texttt{\textbackslash{}} of $x_{1}=3,~x_{2}=-2.5,~x_{3}=7$ + +We could also use an iterative method that solves for all of the +x-components in one step: + +\subsubsection{Jacobi method}\label{jacobi-method} + +$x_{1}^{i}=\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$ + +$x_{2}^{i}=\frac{-19.3-0.1x_{1}^{i-1}+0.3x_{3}^{i-1}}{7}$ + +$x_{3}^{i}=\frac{71.4+0.1x_{1}^{i-1}+0.2x_{2}^{i-1}}{10}$ + +Here the solution is a matrix multiplication and vector addition + +$\left[ \begin{array}{c} x_{1}^{i} \\ x_{2}^{i} \\ x_{3}^{i} \end{array} \right]= \left[ \begin{array}{c} 7.85/3 \\ -19.3/7 \\ 71.4/10\end{array} \right]- \left[ \begin{array}{ccc} 0 & -0.1 & -0.2 \\ 0.1 & 0 & -0.3 \\ 0.3 & -0.2 & 0 \end{array} \right] \left[ \begin{array}{c} x_{1}^{i-1} \\ x_{2}^{i-1} \\ x_{3}^{i-1} \end{array} \right]$ + +\begin{longtable}[c]{@{}llll@{}} +\toprule +\begin{minipage}[b]{0.10\columnwidth}\raggedright\strut +x\_\{j\} +\strut\end{minipage} & +\begin{minipage}[b]{0.36\columnwidth}\raggedright\strut +Jacobi method +\strut\end{minipage} & +\begin{minipage}[b]{0.05\columnwidth}\raggedright\strut +vs +\strut\end{minipage} & +\begin{minipage}[b]{0.37\columnwidth}\raggedright\strut +Gauss-Seidel +\strut\end{minipage}\tabularnewline +\midrule +\endhead +\begin{minipage}[t]{0.10\columnwidth}\raggedright\strut +$x_{1}^{i}=$ +\strut\end{minipage} & +\begin{minipage}[t]{0.36\columnwidth}\raggedright\strut +$\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$ +\strut\end{minipage} & +\begin{minipage}[t]{0.05\columnwidth}\raggedright\strut +\strut\end{minipage} & +\begin{minipage}[t]{0.37\columnwidth}\raggedright\strut +$\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$ +\strut\end{minipage}\tabularnewline +\begin{minipage}[t]{0.10\columnwidth}\raggedright\strut +$x_{2}^{i}=$ +\strut\end{minipage} & +\begin{minipage}[t]{0.36\columnwidth}\raggedright\strut +$\frac{-19.3-0.1x_{1}^{i-1}+0.3x_{3}^{i-1}}{7}$ +\strut\end{minipage} & +\begin{minipage}[t]{0.05\columnwidth}\raggedright\strut +\strut\end{minipage} & +\begin{minipage}[t]{0.37\columnwidth}\raggedright\strut +$\frac{-19.3-0.1x_{1}^{i}+0.3x_{3}^{i-1}}{7}$ +\strut\end{minipage}\tabularnewline +\begin{minipage}[t]{0.10\columnwidth}\raggedright\strut +$x_{3}^{i}=$ +\strut\end{minipage} & +\begin{minipage}[t]{0.36\columnwidth}\raggedright\strut +$\frac{71.4+0.1x_{1}^{i-1}+0.2x_{2}^{i-1}}{10}$ +\strut\end{minipage} & +\begin{minipage}[t]{0.05\columnwidth}\raggedright\strut +\strut\end{minipage} & +\begin{minipage}[t]{0.37\columnwidth}\raggedright\strut +$\frac{71.4+0.1x_{1}^{i}+0.2x_{2}^{i}}{10}$ +\strut\end{minipage}\tabularnewline +\bottomrule +\end{longtable} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}15}]:} \PY{n}{ba}\PY{p}{=}\PY{n}{b}\PY{o}{./}\PY{n+nb}{diag}\PY{p}{(}\PY{n}{A}\PY{p}{)} \PY{c}{\PYZpc{} or ba=b./[A(1,1);A(2,2);A(3,3)]} + \PY{n}{sA}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZhy{}}\PY{n+nb}{diag}\PY{p}{(}\PY{n+nb}{diag}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{)} \PY{c}{\PYZpc{} A with zeros on diagonal} + \PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{/}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{/}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{;} + \PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{/}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)} + \PY{n}{x0}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;} + \PY{n}{x1}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x0} + \PY{n}{x2}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x1} + \PY{n}{x3}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x2} + \PY{n+nb}{fprintf}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{solution is converging to [3,\PYZhy{}2.5,7]]\PYZbs{}n\PYZsq{}}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ba = + + 2.6167 + -2.7571 + 7.1400 + +sA = + + 0.00000 -0.10000 -0.20000 + 0.10000 0.00000 -0.30000 + 0.30000 -0.20000 0.00000 + +sA = + + 0.000000 -0.033333 -0.066667 + 0.014286 0.000000 -0.042857 + 0.030000 -0.020000 0.000000 + +x1 = + + 2.6167 + -2.7571 + 7.1400 + +x2 = + + 3.0008 + -2.4885 + 7.0064 + +x3 = + + 3.0008 + -2.4997 + 7.0002 + +solution is converging to [3,-2.5,7]] + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}16}]:} \PY{n+nb}{diag}\PY{p}{(}\PY{n}{A}\PY{p}{)} + \PY{n+nb}{diag}\PY{p}{(}\PY{n+nb}{diag}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ans = + + 3 + 7 + 10 + +ans = + +Diagonal Matrix + + 3 0 0 + 0 7 0 + 0 0 10 + + + \end{Verbatim} + + This method works if problem is diagonally dominant, + +$|a_{ii}|>\sum_{j=1,j\ne i}^{n}|a_{ij}|$ + +If this condition is true, then Jacobi or Gauss-Seidel should converge + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}17}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mf}{0.1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mf}{0.2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{5}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mf}{0.3}\PY{p}{]} + \PY{n}{b}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{12}\PY{p}{;}\PY{l+m+mi}{2}\PY{p}{;}\PY{l+m+mi}{4}\PY{p}{]} + \PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{b} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 0.10000 1.00000 3.00000 + 1.00000 0.20000 3.00000 + 5.00000 2.00000 0.30000 + +b = + + 12 + 2 + 4 + +ans = + + -2.9393 + 9.1933 + 1.0336 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}20}]:} \PY{n}{ba}\PY{p}{=}\PY{n}{b}\PY{o}{./}\PY{n+nb}{diag}\PY{p}{(}\PY{n}{A}\PY{p}{)} \PY{c}{\PYZpc{} or ba=b./[A(1,1);A(2,2);A(3,3)]} + \PY{n}{sA}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZhy{}}\PY{n+nb}{diag}\PY{p}{(}\PY{n+nb}{diag}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{)} \PY{c}{\PYZpc{} A with zeros on diagonal} + \PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{/}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{/}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{;} + \PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{/}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)} + \PY{n}{x0}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;} + \PY{n}{x1}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x0} + \PY{n}{x2}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x1} + \PY{n}{x3}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x2} + \PY{n+nb}{fprintf}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{solution is not converging to [\PYZhy{}2.93,9.19,1.03]\PYZbs{}n\PYZsq{}}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ba = + + 120.000 + 10.000 + 13.333 + +sA = + + 0 1 3 + 1 0 3 + 5 2 0 + +sA = + + 0.00000 10.00000 30.00000 + 5.00000 0.00000 15.00000 + 16.66667 6.66667 0.00000 + +x1 = + + 120.000 + 10.000 + 13.333 + +x2 = + + -380.00 + -790.00 + -2053.33 + +x3 = + + 6.9620e+04 + 3.2710e+04 + 1.1613e+04 + +solution is not converging to [-2.93,9.19,1.03] + + \end{Verbatim} + + \subsection{Gauss-Seidel with +Relaxation}\label{gauss-seidel-with-relaxation} + +In order to force the solution to converge faster, we can introduce a +relaxation term $\lambda$. + +where the new x values are weighted between the old and new: + +$x^{i}=\lambda x^{i}+(1-\lambda)x^{i-1}$ + +after solving for x, lambda weights the current approximation with the +previous approximation for the updated x + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}105}]:} \PY{c}{\PYZpc{} rearrange A and b} + \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{3} \PY{o}{\PYZhy{}}\PY{l+m+mf}{0.1} \PY{o}{\PYZhy{}}\PY{l+m+mf}{0.2}\PY{p}{;}\PY{l+m+mf}{0.1} \PY{l+m+mi}{7} \PY{o}{\PYZhy{}}\PY{l+m+mf}{0.3}\PY{p}{;}\PY{l+m+mf}{0.3} \PY{o}{\PYZhy{}}\PY{l+m+mf}{0.2} \PY{l+m+mi}{10}\PY{p}{]} + \PY{n}{b}\PY{p}{=}\PY{p}{[}\PY{l+m+mf}{7.85}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mf}{19.3}\PY{p}{;}\PY{l+m+mf}{71.4}\PY{p}{]} + + \PY{n}{iters}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{100}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{k}{for} \PY{n}{i}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{100} + \PY{n}{lambda}\PY{p}{=}\PY{l+m+mi}{2}\PY{o}{/}\PY{l+m+mi}{100}\PY{o}{*}\PY{n}{i}\PY{p}{;} + \PY{p}{[}\PY{n}{x}\PY{p}{,}\PY{n}{ea}\PY{p}{,}\PY{n}{iters}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{]}\PY{p}{=}\PY{n}{Jacobi\PYZus{}rel}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n}{b}\PY{p}{,}\PY{n}{lambda}\PY{p}{)}\PY{p}{;} + \PY{k}{end} + \PY{n+nb}{plot}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{100}\PY{p}{]}\PY{o}{*}\PY{l+m+mi}{2}\PY{o}{/}\PY{l+m+mi}{100}\PY{p}{,}\PY{n}{iters}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 3.00000 -0.10000 -0.20000 + 0.10000 7.00000 -0.30000 + 0.30000 -0.20000 10.00000 + +b = + + 7.8500 + -19.3000 + 71.4000 + + + \end{Verbatim} + + \begin{center} + \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_13_files/lecture_13_22_1.pdf} + \end{center} + { \hspace*{\fill} \\} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}107}]:} \PY{n}{l}\PY{p}{=}\PY{n}{fminbnd}\PY{p}{(}\PY{p}{@}\PY{p}{(}\PY{n}{l}\PY{p}{)} \PY{n}{lambda\PYZus{}fcn}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n}{b}\PY{p}{,}\PY{n}{l}\PY{p}{)}\PY{p}{,}\PY{l+m+mf}{0.5}\PY{p}{,}\PY{l+m+mf}{1.5}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +l = 0.99158 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}108}]:} \PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{b} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ans = + + 3.0000 + -2.5000 + 7.0000 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}109}]:} \PY{p}{[}\PY{n}{x}\PY{p}{,}\PY{n}{ea}\PY{p}{,}\PY{n}{iter}\PY{p}{]}\PY{p}{=}\PY{n}{Jacobi\PYZus{}rel}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n}{b}\PY{p}{,}\PY{n}{l}\PY{p}{,}\PY{l+m+mf}{0.000001}\PY{p}{)} + \PY{p}{[}\PY{n}{x}\PY{p}{,}\PY{n}{ea}\PY{p}{,}\PY{n}{iter}\PY{p}{]}\PY{p}{=}\PY{n}{Jacobi\PYZus{}rel}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n}{b}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mf}{0.000001}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +x = + + 3.0000 + -2.5000 + 7.0000 + +ea = + + 1.8289e-07 + 2.1984e-08 + 2.3864e-08 + +iter = 8 +x = + + 3.0000 + -2.5000 + 7.0000 + +ea = + + 1.9130e-08 + 7.6449e-08 + 3.3378e-08 + +iter = 8 + + \end{Verbatim} + + \subsection{Nonlinear Systems}\label{nonlinear-systems} + +Consider two simultaneous nonlinear equations with two unknowns: + +$x_{1}^{2}+x_{1}x_{2}=10$ + +$x_{2}+3x_{1}x_{2}^{2}=57$ + +Graphically, we are looking for the solution: + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}121}]:} \PY{n}{x11}\PY{p}{=}\PY{n+nb}{linspace}\PY{p}{(}\PY{l+m+mf}{0.5}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;} + \PY{n}{x12}\PY{p}{=}\PY{p}{(}\PY{l+m+mi}{10}\PY{o}{\PYZhy{}}\PY{n}{x11}\PY{o}{.\PYZca{}}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{./}\PY{n}{x11}\PY{p}{;} + + \PY{n}{x22}\PY{p}{=}\PY{n+nb}{linspace}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{8}\PY{p}{)}\PY{p}{;} + \PY{n}{x21}\PY{p}{=}\PY{p}{(}\PY{l+m+mi}{57}\PY{o}{\PYZhy{}}\PY{n}{x22}\PY{p}{)}\PY{o}{.*}\PY{n}{x22}\PY{o}{.\PYZca{}}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{;} + + \PY{n+nb}{plot}\PY{p}{(}\PY{n}{x11}\PY{p}{,}\PY{n}{x12}\PY{p}{,}\PY{n}{x21}\PY{p}{,}\PY{n}{x22}\PY{p}{)} + \PY{c}{\PYZpc{} Solution at x\PYZus{}1=2, x\PYZus{}2=3} + \PY{n+nb}{hold} \PY{n}{on}\PY{p}{;} + \PY{n+nb}{plot}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{o\PYZsq{}}\PY{p}{)} +\end{Verbatim} + + \begin{center} + \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_13_files/lecture_13_27_0.pdf} + \end{center} + { \hspace*{\fill} \\} + + \subsection{Newton-Raphson part II}\label{newton-raphson-part-ii} + +Remember the first order approximation for the next point in a function +is: + +$f(x_{i+1})=f(x_{i})+(x_{i+1}-x_{i})f'(x_{i})$ + +then, $f(x_{i+1})=0$ so we are left with: + +$x_{i+1}=x_{i}-\frac{f(x_{i})}{f'(x_{i})}$ + +We can use the same formula, but now we have multiple dimensions so we +need to determine the Jacobian + +$[J]=\left[ \begin{array}{cccc} \frac{\partial f_{1,i}}{\partial x_{1}} & \frac{\partial f_{1,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{1,i}}{\partial x_{n}} \\ \frac{\partial f_{2,i}}{\partial x_{1}} & \frac{\partial f_{2,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{2,i}}{\partial x_{n}} \\ \vdots & \vdots & & \vdots \\ \frac{\partial f_{n,i}}{\partial x_{1}} & \frac{\partial f_{n,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{n,i}}{\partial x_{n}} \\ \end{array} \right]$ + +$\left[ \begin{array}{c} f_{1,i+1} \\ f_{2,i+1} \\ \vdots \\ f_{n,i+1}\end{array} \right]= \left[ \begin{array}{c} f_{1,i} \\ f_{2,i} \\ \vdots \\ f_{n,i}\end{array} \right]+ \left[ \begin{array}{cccc} \frac{\partial f_{1,i}}{\partial x_{1}} & \frac{\partial f_{1,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{1,i}}{\partial x_{n}} \\ \frac{\partial f_{2,i}}{\partial x_{1}} & \frac{\partial f_{2,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{2,i}}{\partial x_{n}} \\ \vdots & \vdots & & \vdots \\ \frac{\partial f_{n,i}}{\partial x_{1}} & \frac{\partial f_{n,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{n,i}}{\partial x_{n}} \\ \end{array} \right] \left( \left[ \begin{array}{c} x_{i+1} \\ x_{i+1} \\ \vdots \\ x_{i+1}\end{array} \right]- \left[ \begin{array}{c} f_{1,i} \\ f_{2,i} \\ \vdots \\ f_{n,i}\end{array} \right]\right)$ + +\subsubsection{Solution is again in the form +Ax=b}\label{solution-is-again-in-the-form-axb} + +$[J]([x_{i+1}]-[x_{i}])=-[f]$ + +so + +$[x_{i+1}]= [x_{i}]-[J]^{-1}[f]$ + +\subsection{Example of Jacobian +calculation}\label{example-of-jacobian-calculation} + +\subsubsection{Nonlinear springs supporting two masses in +series}\label{nonlinear-springs-supporting-two-masses-in-series} + +Two springs are connected to two masses, with $m_1$=1 kg and +$m_{2}$=2 kg. The springs are identical, but they have nonlinear +spring constants, of $k_1$=10 N/m and $k_2$=-4 N/m + +We want to solve for the final position of the masses ($x_1$ and +$x_2$) + +$m_{1}g+k_{1}(x_{2}-x_{1})+k_{2}(x_{2}-x_{1})^{2}+k_{1}x_{1}+k_{2}x_{1}^{2}=0$ + +$m_{2}g-k_{1}(x_{2}-x_{1})-k_{2}(x_2-x_1)^{2}=0$ + +$J(1,1)=\frac{\partial f_{1}}{\partial x_{1}}=-k_{1}-2k_{2}(x_{2}-x_{1})+k_{1}+2k_{2}x_{1}$ + +$J(1,2)=\frac{\partial f_1}{\partial x_{2}}=k_{1}+2k_{2}(x_{2}-x_{1})$ + +$J(2,1)=\frac{\partial f_2}{\partial x_{1}}=k_{1}+2k_{2}(x_{2}-x_{1})$ + +$J(2,2)=\frac{\partial f_2}{\partial x_{2}}=-k_{1}-2k_{2}(x_{2}-x_{1})$ + +Use an initial guess of $x_1=x_2=0$ + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor} }]:} \PY{n}{m1}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} kg } + \PY{n}{m2}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;} \PY{c}{\PYZpc{} kg} + \PY{n}{k1}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} \PY{c}{\PYZpc{} N/m} + \PY{n}{k2}\PY{p}{=}\PY{o}{\PYZhy{}}\PY{l+m+mi}{4}\PY{p}{;} \PY{c}{\PYZpc{} N/m\PYZca{}2} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}214}]:} \PY{k}{function}\PY{+w}{ }[f,J]\PY{p}{=}\PY{n+nf}{mass\PYZus{}spring}\PY{p}{(}x\PY{p}{)} + \PY{+w}{ }\PY{c}{\PYZpc{} Function to calculate function values f1 and f2 as well as Jacobian } + \PY{c}{\PYZpc{} for 2 masses and 2 identical nonlinear springs} + \PY{n}{m1}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} kg } + \PY{n}{m2}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;} \PY{c}{\PYZpc{} kg} + \PY{n}{k1}\PY{p}{=}\PY{l+m+mi}{100}\PY{p}{;} \PY{c}{\PYZpc{} N/m} + \PY{n}{k2}\PY{p}{=}\PY{o}{\PYZhy{}}\PY{l+m+mi}{10}\PY{p}{;} \PY{c}{\PYZpc{} N/m\PYZca{}2} + \PY{n}{g}\PY{p}{=}\PY{l+m+mf}{9.81}\PY{p}{;} \PY{c}{\PYZpc{} m/s\PYZca{}2} + \PY{n}{x1}\PY{p}{=}\PY{n}{x}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{x2}\PY{p}{=}\PY{n}{x}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{;} + 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120 + 120 -120 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}227}]:} \PY{n}{x0}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{2}\PY{p}{]}\PY{p}{;} + \PY{p}{[}\PY{n}{f0}\PY{p}{,}\PY{n}{J0}\PY{p}{]}\PY{p}{=}\PY{n}{mass\PYZus{}spring}\PY{p}{(}\PY{n}{x0}\PY{p}{)}\PY{p}{;} + \PY{n}{x1}\PY{p}{=}\PY{n}{x0}\PY{o}{\PYZhy{}}\PY{n}{J0}\PY{o}{\PYZbs{}}\PY{n}{f0} + \PY{n}{ea}\PY{p}{=}\PY{p}{(}\PY{n}{x1}\PY{o}{\PYZhy{}}\PY{n}{x0}\PY{p}{)}\PY{o}{./}\PY{n}{x1} + \PY{p}{[}\PY{n}{f1}\PY{p}{,}\PY{n}{J1}\PY{p}{]}\PY{p}{=}\PY{n}{mass\PYZus{}spring}\PY{p}{(}\PY{n}{x1}\PY{p}{)}\PY{p}{;} + \PY{n}{x2}\PY{p}{=}\PY{n}{x1}\PY{o}{\PYZhy{}}\PY{n}{J1}\PY{o}{\PYZbs{}}\PY{n}{f1} + \PY{n}{ea}\PY{p}{=}\PY{p}{(}\PY{n}{x2}\PY{o}{\PYZhy{}}\PY{n}{x1}\PY{p}{)}\PY{o}{./}\PY{n}{x2} + \PY{p}{[}\PY{n}{f2}\PY{p}{,}\PY{n}{J2}\PY{p}{]}\PY{p}{=}\PY{n}{mass\PYZus{}spring}\PY{p}{(}\PY{n}{x2}\PY{p}{)}\PY{p}{;} + 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4.2565e-12 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}228}]:} \PY{n}{x} + \PY{n}{X0}\PY{p}{=}\PY{n+nb}{fsolve}\PY{p}{(}\PY{p}{@}\PY{p}{(}\PY{n}{x}\PY{p}{)} \PY{n}{mass\PYZus{}spring}\PY{p}{(}\PY{n}{x}\PY{p}{)}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{5}\PY{p}{]}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +x = + + 0.30351 + 0.50372 + +X0 = + + 0.30351 + 0.50372 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}236}]:} \PY{p}{[}\PY{n}{X}\PY{p}{,}\PY{n}{Y}\PY{p}{]}\PY{p}{=}\PY{n+nb}{meshgrid}\PY{p}{(}\PY{n+nb}{linspace}\PY{p}{(}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{20}\PY{p}{)}\PY{p}{,}\PY{n+nb}{linspace}\PY{p}{(}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{20}\PY{p}{)}\PY{p}{)}\PY{p}{;} + \PY{p}{[}\PY{n}{N}\PY{p}{,}\PY{n}{M}\PY{p}{]}\PY{p}{=}\PY{n+nb}{size}\PY{p}{(}\PY{n}{X}\PY{p}{)}\PY{p}{;} + 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+ + + + + + + + + + \ No newline at end of file diff --git a/lecture_13/nohup.out b/lecture_13/nohup.out new file mode 100644 index 0000000..ee0677e --- /dev/null +++ b/lecture_13/nohup.out @@ -0,0 +1,2 @@ + +(evince:3288): Gtk-WARNING **: gtk_widget_size_allocate(): attempt to allocate widget with width -71 and height 20 diff --git a/lecture_13/norm_A.png b/lecture_13/norm_A.png new file mode 100644 index 0000000..5f2d273 Binary files /dev/null and b/lecture_13/norm_A.png differ diff --git a/lecture_13/octave-workspace b/lecture_13/octave-workspace new file mode 100644 index 0000000..8c437bb Binary files /dev/null and b/lecture_13/octave-workspace differ diff --git a/linear_algebra/LU_suggested.html b/linear_algebra/LU_suggested.html new file mode 100644 index 0000000..f1998e1 --- /dev/null +++ b/linear_algebra/LU_suggested.html @@ -0,0 +1,39 @@ + + + + + + + + + + + +

Linear Algebra Review

+

(Gauss Elimination) Suggested problems

+

No due date

+
    +

  1. Determine the lower (L) and upper (U) triangular matrices with LU-decomposition for the following matrices:

    +
      +

    1. \(A=\left[ \begin{array}{cc} 1 & 3 \\ 2 & 1 \end{array} \right]\)

      2. +

    2. \(A=\left[ \begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array} \right]\)

      3. +

    3. \(A=\left[ \begin{array}{cc} 1 & 1 \\ 2 & -2 \end{array} \right]\)

      4. +

    4. \(A=\left[ \begin{array}{ccc} 1 & 3 & 1 \\ -4 & -9 & 2 \\ 0 & 3 & 6\end{array} \right]\)

      5. +

    5. \(A=\left[ \begin{array}{ccc} 1 & 3 & 1 \\ -4 & -9 & 2 \\ 0 & 3 & 6\end{array} \right]\)

      6. +

    6. \(A=\left[ \begin{array}{ccc} 1 & 3 & -5 \\ 1 & 4 & -8 \\ -3 & -7 & 9\end{array} \right]\)

      7. +

    7. \(A=\left[ \begin{array}{ccc} 1 & 2 & -1 \\ 2 & 2 & 2 \\ 1 & -1 & 2\end{array} \right]\)

      8. +
    2. +

  2. Calculate the determinant of A from 1a-g.

    3. +

  3. Determine the Cholesky factorization, C, of the following matrices, where

    +

    \(C_{ii}=\sqrt{a_{ii}-\sum_{k=1}^{i-1}C_{ki}^{2}}\)

    +

    \(C_{ij}=\frac{a_{ij}-\sum_{k=1}^{i-1}C_{ki}C_{kj}}{C_{ii}}\).

    +
      +

    1. A=\(\left[ \begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array} \right]\)

      2. +

    2. A=\(\left[ \begin{array}{cc} 10 & 5 \\ 5 & 20 \end{array} \right]\)

      3. +

    3. A=\(\left[ \begin{array}{ccc} 10 & -10 & 20 \\ -10 & 20 & 10 \\ 20 & 10 & 30 \end{array} \right]\)

      4. +

    4. A=\(\left[ \begin{array}{cccc} 21 & -1 & 0 & 0 \\ -1 & 21 & -1 & 0 \\ 0 & -1 & 21 & -1 \\ 0 & 0 & -1 & 1 \end{array} \right]\)

      5. +
    4. +

  4. Verify that \(C^{T}C=A\) for 3a-d

    5. +
+ + diff --git a/linear_algebra/LU_suggested.md b/linear_algebra/LU_suggested.md new file mode 100644 index 0000000..f809813 --- /dev/null +++ b/linear_algebra/LU_suggested.md @@ -0,0 +1,68 @@ +# Linear Algebra Review +## (Gauss Elimination) Suggested problems +### No due date + +1. Determine the lower (L) and upper (U) triangular matrices with LU-decomposition for the +following matrices: + + a. $A=\left[ \begin{array}{cc} + 1 & 3 \\ + 2 & 1 \end{array} \right]$ + + a. $A=\left[ \begin{array}{cc} + 1 & 1 \\ + 2 & 3 \end{array} \right]$ + + a. $A=\left[ \begin{array}{cc} + 1 & 1 \\ + 2 & -2 \end{array} \right]$ + + b. $A=\left[ \begin{array}{ccc} + 1 & 3 & 1 \\ + -4 & -9 & 2 \\ + 0 & 3 & 6\end{array} \right]$ + + c. $A=\left[ \begin{array}{ccc} + 1 & 3 & 1 \\ + -4 & -9 & 2 \\ + 0 & 3 & 6\end{array} \right]$ + + d. $A=\left[ \begin{array}{ccc} + 1 & 3 & -5 \\ + 1 & 4 & -8 \\ + -3 & -7 & 9\end{array} \right]$ + + d. $A=\left[ \begin{array}{ccc} + 1 & 2 & -1 \\ + 2 & 2 & 2 \\ + 1 & -1 & 2\end{array} \right]$ + +2. Calculate the determinant of A from 1a-g. + +3. Determine the Cholesky factorization, C, of the following matrices, where + + $C_{ii}=\sqrt{a_{ii}-\sum_{k=1}^{i-1}C_{ki}^{2}}$ + + $C_{ij}=\frac{a_{ij}-\sum_{k=1}^{i-1}C_{ki}C_{kj}}{C_{ii}}$. + + a. A=$\left[ \begin{array}{cc} + 3 & 2 \\ + 2 & 1 \end{array} \right]$ + + a. A=$\left[ \begin{array}{cc} + 10 & 5 \\ + 5 & 20 \end{array} \right]$ + + a. A=$\left[ \begin{array}{ccc} + 10 & -10 & 20 \\ + -10 & 20 & 10 \\ + 20 & 10 & 30 \end{array} \right]$ + + a. A=$\left[ \begin{array}{cccc} + 21 & -1 & 0 & 0 \\ + -1 & 21 & -1 & 0 \\ + 0 & -1 & 21 & -1 \\ + 0 & 0 & -1 & 1 \end{array} \right]$ + +4. Verify that $C^{T}C=A$ for 3a-d +