diff --git a/lecture_09/octave-workspace b/lecture_09/octave-workspace new file mode 100644 index 0000000..f7e8b2b Binary files /dev/null and b/lecture_09/octave-workspace differ diff --git a/lecture_10/.ipynb_checkpoints/lecture_10-checkpoint.ipynb b/lecture_10/.ipynb_checkpoints/lecture_10-checkpoint.ipynb new file mode 100644 index 0000000..2fd6442 --- /dev/null +++ b/lecture_10/.ipynb_checkpoints/lecture_10-checkpoint.ipynb @@ -0,0 +1,6 @@ +{ + "cells": [], + "metadata": {}, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/lecture_10/GaussNaive.m b/lecture_10/GaussNaive.m new file mode 100644 index 0000000..d8c60d3 --- /dev/null +++ b/lecture_10/GaussNaive.m @@ -0,0 +1,25 @@ +function [x,Aug] = GaussNaive(A,y) +% GaussNaive: naive Gauss elimination +% x = GaussNaive(A,b): Gauss elimination without pivoting. +% input: +% A = coefficient matrix +% y = right hand side vector +% output: +% x = solution vector +[m,n] = size(A); +if m~=n, error('Matrix A must be square'); end +nb = n+1; +Aug = [A y]; +% forward elimination +for k = 1:n-1 + for i = k+1:n + factor = Aug(i,k)/Aug(k,k); + Aug(i,k:nb) = Aug(i,k:nb)-factor*Aug(k,k:nb); + end +end +% back substitution +x = zeros(n,1); +x(n) = Aug(n,nb)/Aug(n,n); +for i = n-1:-1:1 + x(i) = (Aug(i,nb)-Aug(i,i+1:n)*x(i+1:n))/Aug(i,i); +end diff --git a/lecture_10/GaussPivot.m b/lecture_10/GaussPivot.m new file mode 100644 index 0000000..20df9e8 --- /dev/null +++ b/lecture_10/GaussPivot.m @@ -0,0 +1,33 @@ +function [x,Aug,npivots] = GaussPivot(A,b) +% GaussPivot: Gauss elimination pivoting +% x = GaussPivot(A,b): Gauss elimination with pivoting. +% input: +% A = coefficient matrix +% b = right hand side vector +% output: +% x = solution vector +[m,n]=size(A); +if m~=n, error('Matrix A must be square'); end +nb=n+1; +Aug=[A b]; +npivots=0; % initially no pivots used +% forward elimination +for k = 1:n-1 + % partial pivoting + [big,i]=max(abs(Aug(k:n,k))); + ipr=i+k-1; + if ipr~=k + npivots=npivots+1; % if the max is not the current index ipr, pivot count + Aug([k,ipr],:)=Aug([ipr,k],:); + end + for i = k+1:n + factor=Aug(i,k)/Aug(k,k); + Aug(i,k:nb)=Aug(i,k:nb)-factor*Aug(k,k:nb); + end +end +% back substitution +x=zeros(n,1); +x(n)=Aug(n,nb)/Aug(n,n); +for i = n-1:-1:1 + x(i)=(Aug(i,nb)-Aug(i,i+1:n)*x(i+1:n))/Aug(i,i); +end diff --git a/lecture_10/Tridiag.m b/lecture_10/Tridiag.m new file mode 100644 index 0000000..ac4ec9b --- /dev/null +++ b/lecture_10/Tridiag.m @@ -0,0 +1,22 @@ +function x = Tridiag(e,f,g,r) +% Tridiag: Tridiagonal equation solver banded system +% x = Tridiag(e,f,g,r): Tridiagonal system solver. +% input: +% e = subdiagonal vector +% f = diagonal vector +% g = superdiagonal vector +% r = right hand side vector +% output: +% x = solution vector +n=length(f); +% forward elimination +for k = 2:n + factor = e(k)/f(k-1); + f(k) = f(k) - factor*g(k-1); + r(k) = r(k) - factor*r(k-1); +end +% back substitution +x(n) = r(n)/f(n); +for k = n-1:-1:1 + x(k) = (r(k)-g(k)*x(k+1))/f(k); +end diff 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b/lecture_10/lecture_10.bbl new file mode 100644 index 0000000..e69de29 diff --git a/lecture_10/lecture_10.blg b/lecture_10/lecture_10.blg new file mode 100644 index 0000000..36b5e7a --- /dev/null +++ b/lecture_10/lecture_10.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_10.aux +I found no \citation commands---while reading file lecture_10.aux +I found no \bibdata command---while reading file lecture_10.aux +I found no \bibstyle command---while reading file lecture_10.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 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matrix indicate the degrees of freedom of the system you are solving. \n", + "\n", + "If you have a set of known output, $y_{1},~y_{2},~...y_{N}$ and a set of equations that\n", + "relate unknown inputs, $x_{1},~x_{2},~...x_{N}$, then these can be written in a vector\n", + "matrix format as:\n", + "\n", + "$y=Ax$\n", + "\n", + "Consider a problem with 2 DOF:\n", + "\n", + "$x_{1}+3x_{2}=1$\n", + "\n", + "$2x_{1}+x_{2}=1$\n", + "\n", + "$\\left[ \\begin{array}{cc}\n", + "1 & 3 \\\\\n", + "2 & 1 \\end{array} \\right]\n", + "\\left[\\begin{array}{c} \n", + "x_{1} \\\\ \n", + "x_{2} \\end{array}\\right]=\n", + "\\left[\\begin{array}{c} \n", + "1 \\\\\n", + "1\\end{array}\\right]$\n", + "\n", + "The solution for $x_{1}$ and $x_{2}$ is the intersection of two lines:" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/svg+xml": [ + "\n", + "\n", + "Gnuplot\n", + "Produced by GNUPLOT 5.0 patchlevel 3 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\n", + "x_{1} \\\\ \n", + "x_{2} \\\\\n", + "x_{3} \\end{array}\\right]=\n", + "\\left[\\begin{array}{c} \n", + "1 \\\\\n", + "1 \\\\\n", + "1\\end{array}\\right]$" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "error: 'X22' undefined near line 1 column 16\n", + "error: 'X13' undefined near line 1 column 14\n", + "error: evaluating argument list element number 3\n" + ] + }, + { + "data": { + "image/svg+xml": [ + "\n", + "\n", + "Gnuplot\n", + "Produced by GNUPLOT 5.0 patchlevel 3 \n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t\n", + "\t \n", + "\t \n", + "\t\n", + "\t\n", + "\t \n", + "\t \n", + "\t\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + 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"[X11,X12]=meshgrid(x11,x12);\n", + "X13=1-10*X11-2*X22;\n", + "\n", + "x21=linspace(-2,2,5);\n", + "x22=linspace(-2,2,5);\n", + "[X21,X22]=meshgrid(x21,x22);\n", + "X23=1-2*X11-X22;\n", + "\n", + "x31=linspace(-2,2,5);\n", + "x32=linspace(-2,2,5);\n", + "[X31,X32]=meshgrid(x31,x32);\n", + "X33=1/10*(1-X31-2*X32);\n", + "\n", + "mesh(X11,X12,X13);\n", + "hold on;\n", + "mesh(X21,X22,X23)\n", + "mesh(X31,X32,X33)\n", + "x=[10,2, 1;2,1, 1; 1, 2, 10]\\[1;1;1];\n", + "plot3(x(1),x(2),x(3),'o')\n", + "view(45,45)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "After 3 DOF problems, the solutions are described as *hyperplane* intersections. Which are even harder to visualize" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Gauss elimination\n", + "### Solving sets of equations systematically\n", + "\n", + "$\\left[ \\begin{array}{ccc|c}\n", + " & A & & y \\\\\n", + "10 & 2 & 1 & 1\\\\\n", + "2 & 1 & 1 & 1 \\\\\n", + "1 & 2 & 10 & 1\\end{array} \n", + "\\right] $\n", + "\n", + "Ay(2,:)-Ay(1,:)/5 = ([2 1 1 1]-1/5[10 2 1 1])\n", + "\n", + "$\\left[ \\begin{array}{ccc|c}\n", + " & A & & y \\\\\n", + "10 & 2 & 1 & 1\\\\\n", + "0 & 3/5 & 4/5 & 4/5 \\\\\n", + "1 & 2 & 10 & 1\\end{array} \n", + "\\right] $\n", + "\n", + "Ay(3,:)-Ay(1,:)/10 = ([1 2 10 1]-1/10[10 2 1 1])\n", + "\n", + "$\\left[ \\begin{array}{ccc|c}\n", + " & A & & y \\\\\n", + "10 & 2 & 1 & 1\\\\\n", + "0 & 3/5 & 4/5 & 4/5 \\\\\n", + "0 & 1.8 & 9.9 & 0.9\\end{array} \n", + "\\right] $\n", + "\n", + "Ay(3,:)-1.8\\*5/3\\*Ay(2,:) = ([0 1.8 9.9 0.9]-3\\*[0 3/5 4/5 4/5])\n", + "\n", + "$\\left[ \\begin{array}{ccc|c}\n", + " & A & & y \\\\\n", + "10 & 2 & 1 & 1\\\\\n", + "0 & 3/5 & 4/5 & 4/5 \\\\\n", + "0 & 0 & 7.5 & -1.5\\end{array} \n", + "\\right] $\n", + "\n", + "now, $7.5x_{3}=-1.5$ so $x_{3}=-\\frac{1}{5}$\n", + "\n", + "then, $3/5x_{2}+4/5(-1/5)=1$ so $x_{2}=\\frac{8}{5}$\n", + "\n", + "finally, $10x_{1}+2(8/5)" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with K=10 N/m. What are the final positions of the masses? \n", + "\n", + "![Springs-masses](../lecture_09/mass_springs.svg)\n", + "\n", + "The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:\n", + "\n", + "$m_{1}g+k(x_{2}-x_{1})-kx_{1}=0$\n", + "\n", + "$m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0$\n", + "\n", + "$m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0$\n", + "\n", + "$m_{4}g-k(x_{4}-x_{3})=0$\n", + "\n", + "in matrix form:\n", + "\n", + "$\\left[ \\begin{array}{cccc}\n", + "2k & -k & 0 & 0 \\\\\n", + "-k & 2k & -k & 0 \\\\\n", + "0 & -k & 2k & -k \\\\\n", + "0 & 0 & -k & k \\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": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "K =\n", + "\n", + " 20 -10 0 0\n", + " -10 20 -10 0\n", + " 0 -10 20 -10\n", + " 0 0 -10 10\n", + "\n", + "y =\n", + "\n", + " 9.8100\n", + " 19.6200\n", + " 29.4300\n", + " 39.2400\n", + "\n" + ] + } + ], + "source": [ + "k=10; % N/m\n", + "m1=1; % kg\n", + "m2=2;\n", + "m3=3;\n", + "m4=4;\n", + "g=9.81; % m/s^2\n", + "K=[2*k -k 0 0; -k 2*k -k 0; 0 -k 2*k -k; 0 0 -k k]\n", + "y=[m1*g;m2*g;m3*g;m4*g]" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "K1 =\n", + "\n", + " 20.00000 -10.00000 0.00000 0.00000 9.81000\n", + " 0.00000 15.00000 -10.00000 0.00000 24.52500\n", + " 0.00000 -10.00000 20.00000 -10.00000 29.43000\n", + " 0.00000 0.00000 -10.00000 10.00000 39.24000\n", + "\n" + ] + } + ], + "source": [ + "K1=[K y];\n", + "K1(2,:)=K1(1,:)/2+K1(2,:)" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "K2 =\n", + "\n", + " 20.00000 -10.00000 0.00000 0.00000 9.81000\n", + " 0.00000 15.00000 -10.00000 0.00000 24.52500\n", + " 0.00000 0.00000 13.33333 -10.00000 45.78000\n", + " 0.00000 0.00000 -10.00000 10.00000 39.24000\n", + "\n" + ] + } + ], + "source": [ + "K2=K1;\n", + "K2(3,:)=K1(2,:)*2/3+K1(3,:)\n" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "K2 =\n", + "\n", + " 20.00000 -10.00000 0.00000 0.00000 9.81000\n", + " 0.00000 15.00000 -10.00000 0.00000 24.52500\n", + " 0.00000 0.00000 13.33333 -10.00000 45.78000\n", + " 0.00000 0.00000 0.00000 2.50000 73.57500\n", + "\n" + ] + } + ], + "source": [ + "K2(4,:)=-K2(3,:)*K2(4,3)/K2(3,3)+K2(4,:)" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x4 = 29.430\n", + "x3 = 25.506\n", + "x2 = 18.639\n", + "x1 = 9.8100\n" + ] + } + ], + "source": [ + "yp=K2(:,5);\n", + "x4=yp(4)/K2(4,4)\n", + "x3=(yp(3)+10*x4)/K2(3,3)\n", + "x2=(yp(2)+10*x3)/K2(2,2)\n", + "x1=(yp(1)+10*x2)/K2(1,1)" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ans =\n", + "\n", + " 9.8100\n", + " 18.6390\n", + " 25.5060\n", + " 29.4300\n", + "\n" + ] + } + ], + "source": [ + "K\\y" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Automate Gauss Elimination\n", + "\n", + "We can automate Gauss elimination with a function whose input is A and y:\n", + "\n", + "`x=GaussNaive(A,y)`" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x =\n", + "\n", + " 9.8100\n", + " 18.6390\n", + " 25.5060\n", + " 29.4300\n", + "\n" + ] + } + ], + "source": [ + "x=GaussNaive(K,y)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Problem (Diagonal element is zero)\n", + "\n", + "If a diagonal element is 0 or very small either:\n", + "\n", + "1. no solution found\n", + "2. errors are introduced \n", + "\n", + "Therefore, we would want to pivot before applying Gauss elimination\n", + "\n", + "Consider:\n", + "\n", + "(a) $\\left[ \\begin{array}{cccc}\n", + "0 & 2 & 3 \\\\\n", + "4 & 6 & 7 \\\\\n", + "2 & -3 & 6 \\end{array} \\right]\n", + "\\left[ \\begin{array}{c}\n", + "x_{1} \\\\\n", + "x_{2} \\\\\n", + "x_{3} \\end{array} \\right]=\n", + "\\left[ \\begin{array}{c}\n", + "8 \\\\\n", + "-3 \\\\\n", + "5\\end{array} \\right]$\n", + "\n", + "(b) $\\left[ \\begin{array}{cccc}\n", + "0.0003 & 3.0000 \\\\\n", + "1.0000 & 1.0000 \\end{array} \\right]\n", + "\\left[ \\begin{array}{c}\n", + "x_{1} \\\\\n", + "x_{2} \\end{array} \\right]=\n", + "\\left[ \\begin{array}{c}\n", + "2.0001 \\\\\n", + "1.0000 \\end{array} \\right]$" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "warning: division by zero\n", + "warning: called from\n", + " GaussNaive at line 16 column 12\n", + "warning: division by zero\n", + "warning: division by zero\n", + "ans =\n", + "\n", + " NaN\n", + " NaN\n", + " NaN\n", + "\n", + "ans =\n", + "\n", + " -5.423913\n", + " 0.021739\n", + " 2.652174\n", + "\n" + ] + } + ], + "source": [ + "format short\n", + "Aa=[0,2,3;4,6,7;2,-3,6]; ya=[8;-3;5];\n", + "GaussNaive(Aa,ya)\n", + "Aa\\ya" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x =\n", + "\n", + " -5.423913\n", + " 0.021739\n", + " 2.652174\n", + "\n", + "Aug =\n", + "\n", + " 4.00000 6.00000 7.00000 -3.00000\n", + " 0.00000 -6.00000 2.50000 6.50000\n", + " 0.00000 0.00000 3.83333 10.16667\n", + "\n", + "npivots = 2\n" + ] + } + ], + "source": [ + "[x,Aug,npivots]=GaussPivot(Aa,ya)" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ans =\n", + "\n", + " 0.325665420556713\n", + " 0.666666666666667\n", + "\n", + "ans =\n", + "\n", + " 0.333333333333333\n", + " 0.666666666666667\n", + "\n" + ] + } + ], + "source": [ + "format long\n", + "Ab=[0.3E-13,3.0000;1.0000,1.0000];yb=[2+0.1e-13;1.0000];\n", + "GaussNaive(Ab,yb)\n", + "Ab\\yb" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x =\n", + "\n", + " 0.333333333333333\n", + " 0.666666666666667\n", + "\n", + "Aug =\n", + "\n", + " 1.000000000000000 1.000000000000000 1.000000000000000\n", + " 0.000000000000000 2.999999999999970 1.999999999999980\n", + "\n", + "npivots = 1\n", + "ans =\n", + "\n", + " 0.333333333333333\n", + " 0.666666666666667\n", + "\n" + ] + } + ], + "source": [ + "[x,Aug,npivots]=GaussPivot(Ab,yb)\n", + "Ab\\yb\n", + "format short" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Spring-Mass System again\n", + "Now, 4 masses are connected in series to 4 springs with $K_{1}$=10 N/m, $K_{2}$=5 N/m, \n", + "$K_{3}$=2 N/m \n", + "and $K_{4}$=1 N/m. What are the final positions of the masses? \n", + "\n", + "![Springs-masses](../lecture_09/mass_springs.svg)\n", + "\n", + "The masses have 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": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "K =\n", + "\n", + " 15 -5 0 0\n", + " -5 7 -2 0\n", + " 0 -2 3 -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; k2=5;k3=2;k4=1; % N/m\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": "markdown", + "metadata": {}, + "source": [ + "## Tridiagonal matrix\n", + "\n", + "This matrix, K, could be rewritten as 3 vectors e, f and g\n", + "\n", + "$e=\\left[ \\begin{array}{c}\n", + "0 \\\\\n", + "-5 \\\\\n", + "-2 \\\\\n", + "-1 \\end{array} \\right]$\n", + "\n", + "$f=\\left[ \\begin{array}{c}\n", + "15 \\\\\n", + "7 \\\\\n", + "3 \\\\\n", + "1 \\end{array} \\right]$\n", + "\n", + "$g=\\left[ \\begin{array}{c}\n", + "-5 \\\\\n", + "-2 \\\\\n", + "-1 \\\\\n", + "0 \\end{array} \\right]$\n", + "\n", + "Where all other components are 0 and the length of the vectors are n and the first component of e and the last component of g are zero\n", + "\n", + "`e(1)=0` \n", + "\n", + "`g(end)=0`\n", + "\n", + "No need to pivot and number of calculations reduced enormously.\n", + "\n", + "|method |Number of Floating point operations for n$\\times$n-matrix|\n", + "|----------------|---------|\n", + "| Naive Gauss | n-cubed |\n", + "| Tridiagonal | n |" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ans =\n", + "\n", + " 9.8100 27.4680 61.8030 101.0430\n", + "\n" + ] + } + ], + "source": [ + "e=[0;-5;-2;-1];\n", + "g=[-5;-2;-1;0];\n", + "f=[15;7;3;1];\n", + "Tridiag(e,f,g,y)\n" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "% tic ... t=toc \n", + "% is Matlab timer used for debugging programs\n", + "t_GE = zeros(1,100);\n", + "t_GE_tridiag = zeros(1,100);\n", + "t_TD = zeros(1,100);\n", + "%for n = 1:200\n", + "for n=1:100\n", + " A = rand(n,n);\n", + " e = rand(n,1); e(1)=0;\n", + " f = rand(n,1);\n", + " g = rand(n,1); g(end)=0;\n", + " Atd=diag(f, 0) - diag(e(2:n), -1) - diag(g(1:n-1), 1);\n", + " b = rand(n,1);\n", + " tic;\n", + " x = GaussPivot(A,b);\n", + " t_GE(n) = toc;\n", + " tic;\n", + " x = GaussPivot(Atd,b);\n", + " t_GE_tridiag(n) = toc;\n", + " tic;\n", + " x = Tridiag(e,f,g,b);\n", + " t_TD(n) = toc;\n", + "end" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "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", 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Which are even harder to visualize + +## Gauss elimination +### Solving sets of equations systematically + +$\left[ \begin{array}{ccc|c} + & A & & y \\ +10 & 2 & 1 & 1\\ +2 & 1 & 1 & 1 \\ +1 & 2 & 10 & 1\end{array} +\right] $ + +Ay(2,:)-Ay(1,:)/5 = ([2 1 1 1]-1/5[10 2 1 1]) + +$\left[ \begin{array}{ccc|c} + & A & & y \\ +10 & 2 & 1 & 1\\ +0 & 3/5 & 4/5 & 4/5 \\ +1 & 2 & 10 & 1\end{array} +\right] $ + +Ay(3,:)-Ay(1,:)/10 = ([1 2 10 1]-1/10[10 2 1 1]) + +$\left[ \begin{array}{ccc|c} + & A & & y \\ +10 & 2 & 1 & 1\\ +0 & 3/5 & 4/5 & 4/5 \\ +0 & 1.8 & 9.9 & 0.9\end{array} +\right] $ + +Ay(3,:)-1.8\*5/3\*Ay(2,:) = ([0 1.8 9.9 0.9]-3\*[0 3/5 4/5 4/5]) + +$\left[ \begin{array}{ccc|c} + & A & & y \\ +10 & 2 & 1 & 1\\ +0 & 3/5 & 4/5 & 4/5 \\ +0 & 0 & 7.5 & -1.5\end{array} +\right] $ + +now, $7.5x_{3}=-1.5$ so $x_{3}=-\frac{1}{5}$ + +then, $3/5x_{2}+4/5(-1/5)=1$ so $x_{2}=\frac{8}{5}$ + +finally, $10x_{1}+2(8/5) + +Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with K=10 N/m. What are the final positions of the masses? + +![Springs-masses](../lecture_09/mass_springs.svg) + +The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass: + +$m_{1}g+k(x_{2}-x_{1})-kx_{1}=0$ + +$m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0$ + +$m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0$ + +$m_{4}g-k(x_{4}-x_{3})=0$ + +in matrix form: + +$\left[ \begin{array}{cccc} +2k & -k & 0 & 0 \\ +-k & 2k & -k & 0 \\ +0 & -k & 2k & -k \\ +0 & 0 & -k & k \end{array} \right] +\left[ \begin{array}{c} +x_{1} \\ +x_{2} \\ +x_{3} \\ +x_{4} \end{array} \right]= +\left[ \begin{array}{c} +m_{1}g \\ +m_{2}g \\ +m_{3}g \\ +m_{4}g \end{array} \right]$ + + +```octave +k=10; % N/m +m1=1; % kg +m2=2; +m3=3; +m4=4; +g=9.81; % m/s^2 +K=[2*k -k 0 0; -k 2*k -k 0; 0 -k 2*k -k; 0 0 -k k] +y=[m1*g;m2*g;m3*g;m4*g] +``` + + K = + + 20 -10 0 0 + -10 20 -10 0 + 0 -10 20 -10 + 0 0 -10 10 + + y = + + 9.8100 + 19.6200 + 29.4300 + 39.2400 + + + + +```octave +K1=[K y]; +K1(2,:)=K1(1,:)/2+K1(2,:) +``` + + K1 = + + 20.00000 -10.00000 0.00000 0.00000 9.81000 + 0.00000 15.00000 -10.00000 0.00000 24.52500 + 0.00000 -10.00000 20.00000 -10.00000 29.43000 + 0.00000 0.00000 -10.00000 10.00000 39.24000 + + + + +```octave +K2=K1; +K2(3,:)=K1(2,:)*2/3+K1(3,:) + +``` + + K2 = + + 20.00000 -10.00000 0.00000 0.00000 9.81000 + 0.00000 15.00000 -10.00000 0.00000 24.52500 + 0.00000 0.00000 13.33333 -10.00000 45.78000 + 0.00000 0.00000 -10.00000 10.00000 39.24000 + + + + +```octave +K2(4,:)=-K2(3,:)*K2(4,3)/K2(3,3)+K2(4,:) +``` + + K2 = + + 20.00000 -10.00000 0.00000 0.00000 9.81000 + 0.00000 15.00000 -10.00000 0.00000 24.52500 + 0.00000 0.00000 13.33333 -10.00000 45.78000 + 0.00000 0.00000 0.00000 2.50000 73.57500 + + + + +```octave +yp=K2(:,5); +x4=yp(4)/K2(4,4) +x3=(yp(3)+10*x4)/K2(3,3) +x2=(yp(2)+10*x3)/K2(2,2) +x1=(yp(1)+10*x2)/K2(1,1) +``` + + x4 = 29.430 + x3 = 25.506 + x2 = 18.639 + x1 = 9.8100 + + + +```octave +K\y +``` + + ans = + + 9.8100 + 18.6390 + 25.5060 + 29.4300 + + + +## Automate Gauss Elimination + +We can automate Gauss elimination with a function whose input is A and y: + +`x=GaussNaive(A,y)` + + +```octave +x=GaussNaive(K,y) +``` + + x = + + 9.8100 + 18.6390 + 25.5060 + 29.4300 + + + +## Problem (Diagonal element is zero) + +If a diagonal element is 0 or very small either: + +1. no solution found +2. errors are introduced + +Therefore, we would want to pivot before applying Gauss elimination + +Consider: + +(a) $\left[ \begin{array}{cccc} +0 & 2 & 3 \\ +4 & 6 & 7 \\ +2 & -3 & 6 \end{array} \right] +\left[ \begin{array}{c} +x_{1} \\ +x_{2} \\ +x_{3} \end{array} \right]= +\left[ \begin{array}{c} +8 \\ +-3 \\ +5\end{array} \right]$ + +(b) $\left[ \begin{array}{cccc} +0.0003 & 3.0000 \\ +1.0000 & 1.0000 \end{array} \right] +\left[ \begin{array}{c} +x_{1} \\ +x_{2} \end{array} \right]= +\left[ \begin{array}{c} +2.0001 \\ +1.0000 \end{array} \right]$ + + +```octave +format short +Aa=[0,2,3;4,6,7;2,-3,6]; ya=[8;-3;5]; +GaussNaive(Aa,ya) +Aa\ya +``` + + warning: division by zero + warning: called from + GaussNaive at line 16 column 12 + warning: division by zero + warning: division by zero + ans = + + NaN + NaN + NaN + + ans = + + -5.423913 + 0.021739 + 2.652174 + + + + +```octave +[x,Aug,npivots]=GaussPivot(Aa,ya) +``` + + x = + + -5.423913 + 0.021739 + 2.652174 + + Aug = + + 4.00000 6.00000 7.00000 -3.00000 + 0.00000 -6.00000 2.50000 6.50000 + 0.00000 0.00000 3.83333 10.16667 + + npivots = 2 + + + +```octave +format long +Ab=[0.3E-13,3.0000;1.0000,1.0000];yb=[2+0.1e-13;1.0000]; +GaussNaive(Ab,yb) +Ab\yb +``` + + ans = + + 0.325665420556713 + 0.666666666666667 + + ans = + + 0.333333333333333 + 0.666666666666667 + + + + +```octave +[x,Aug,npivots]=GaussPivot(Ab,yb) +Ab\yb +format short +``` + + x = + + 0.333333333333333 + 0.666666666666667 + + Aug = + + 1.000000000000000 1.000000000000000 1.000000000000000 + 0.000000000000000 2.999999999999970 1.999999999999980 + + npivots = 1 + ans = + + 0.333333333333333 + 0.666666666666667 + + + +### Spring-Mass System again +Now, 4 masses are connected in series to 4 springs with $K_{1}$=10 N/m, $K_{2}$=5 N/m, +$K_{3}$=2 N/m +and $K_{4}$=1 N/m. What are the final positions of the masses? + +![Springs-masses](../lecture_09/mass_springs.svg) + +The masses have 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; k2=5;k3=2;k4=1; % N/m +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 = + + 15 -5 0 0 + -5 7 -2 0 + 0 -2 3 -1 + 0 0 -1 1 + + y = + + 9.8100 + 19.6200 + 29.4300 + 39.2400 + + + +## Tridiagonal matrix + +This matrix, K, could be rewritten as 3 vectors e, f and g + +$e=\left[ \begin{array}{c} +0 \\ +-5 \\ +-2 \\ +-1 \end{array} \right]$ + +$f=\left[ \begin{array}{c} +15 \\ +7 \\ +3 \\ +1 \end{array} \right]$ + +$g=\left[ \begin{array}{c} +-5 \\ +-2 \\ +-1 \\ +0 \end{array} \right]$ + +Where all other components are 0 and the length of the vectors are n and the first component of e and the last component of g are zero + +`e(1)=0` + +`g(end)=0` + +No need to pivot and number of calculations reduced enormously. + +|method |Number of Floating point operations for n$\times$n-matrix| +|----------------|---------| +| Naive Gauss | n-cubed | +| Tridiagonal | n | + + +```octave +e=[0;-5;-2;-1]; +g=[-5;-2;-1;0]; +f=[15;7;3;1]; +Tridiag(e,f,g,y) + +``` + + ans = + + 9.8100 27.4680 61.8030 101.0430 + + + + +```octave +% tic ... t=toc +% is Matlab timer used for debugging programs +t_GE = zeros(1,100); +t_GE_tridiag = zeros(1,100); +t_TD = zeros(1,100); +for n = 1:200 + A = rand(n,n); + e = rand(n,1); e(1)=0; + f = rand(n,1); + g = rand(n,1); g(end)=0; + Atd=diag(f, 0) - diag(e(2:n), -1) - diag(g(1:n-1), 1); + b = rand(n,1); + tic; + x = GaussPivot(A,b); + t_GE(n) = toc; + tic; + x = GaussPivot(Atd,b); + t_GE_tridiag(n) = toc; + tic; + x = Tridiag(e,f,g,b); + t_TD(n) = toc; +end +``` + + +```octave +n=1:200; +loglog(n,t_GE,n,t_TD,n,t_GE_tridiag) +xlabel('number of elements') +ylabel('time (s)') +``` + + +![svg](lecture_10_files/lecture_10_27_0.svg) + + + +```octave +plot(t_TD) +``` + + +![svg](lecture_10_files/lecture_10_28_0.svg) + + + +```octave + +``` diff --git a/lecture_10/lecture_10.out b/lecture_10/lecture_10.out new file mode 100644 index 0000000..cb5b517 --- /dev/null +++ b/lecture_10/lecture_10.out @@ -0,0 +1,8 @@ +\BOOKMARK [1][-]{section.1}{Gauss Elimination}{}% 1 +\BOOKMARK [2][-]{subsubsection.1.0.1}{Solving sets of equations with matrix operations}{section.1}% 2 +\BOOKMARK [2][-]{subsection.1.1}{Gauss elimination}{section.1}% 3 +\BOOKMARK [3][-]{subsubsection.1.1.1}{Solving sets of equations systematically}{subsection.1.1}% 4 +\BOOKMARK [2][-]{subsection.1.2}{Automate Gauss Elimination}{section.1}% 5 +\BOOKMARK [2][-]{subsection.1.3}{Problem \(Diagonal element is zero\)}{section.1}% 6 +\BOOKMARK [3][-]{subsubsection.1.3.1}{Spring-Mass System again}{subsection.1.3}% 7 +\BOOKMARK [2][-]{subsection.1.4}{Tridiagonal matrix}{section.1}% 8 diff --git a/lecture_10/lecture_10.pdf b/lecture_10/lecture_10.pdf new file mode 100644 index 0000000..447bbe2 Binary files /dev/null and b/lecture_10/lecture_10.pdf differ diff --git a/lecture_10/lecture_10.tex b/lecture_10/lecture_10.tex new file mode 100644 index 0000000..1131881 --- /dev/null +++ b/lecture_10/lecture_10.tex @@ -0,0 +1,852 @@ + +% 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 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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}14}]:} \PY{c}{\PYZpc{}plot \PYZhy{}\PYZhy{}format svg} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}15}]:} \PY{n}{setdefaults} +\end{Verbatim} + + \section{Gauss Elimination}\label{gauss-elimination} + +\subsubsection{Solving sets of equations with matrix +operations}\label{solving-sets-of-equations-with-matrix-operations} + +The number of dimensions of a matrix indicate the degrees of freedom of +the system you are solving. + +If you have a set of known output, \(y_{1},~y_{2},~...y_{N}\) and a set +of equations that relate unknown inputs, \(x_{1},~x_{2},~...x_{N}\), +then these can be written in a vector matrix format as: + +\(y=Ax\) + +Consider a problem with 2 DOF: + +\(x_{1}+3x_{2}=1\) + +\(2x_{1}+x_{2}=1\) + +\(\left[ \begin{array}{cc} 1 & 3 \\ 2 & 1 \end{array} \right] \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right]= \left[\begin{array}{c} 1 \\ 1\end{array}\right]\) + +The solution for \(x_{1}\) and \(x_{2}\) is the intersection of two +lines: + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}16}]:} \PY{n}{x21}\PY{p}{=}\PY{p}{[}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{:}\PY{l+m+mi}{2}\PY{p}{]}\PY{p}{;} + \PY{n}{x11}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{l+m+mi}{3}\PY{o}{*}\PY{n}{x21}\PY{p}{;} + \PY{n}{x21}\PY{p}{=}\PY{p}{[}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{:}\PY{l+m+mi}{2}\PY{p}{]}\PY{p}{;} + \PY{n}{x22}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{x21}\PY{p}{;} + \PY{n+nb}{plot}\PY{p}{(}\PY{n}{x11}\PY{p}{,}\PY{n}{x21}\PY{p}{,}\PY{n}{x21}\PY{p}{,}\PY{n}{x22}\PY{p}{)} +\end{Verbatim} + + \begin{center} + \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_10_files/lecture_10_3_0.pdf} + \end{center} + { \hspace*{\fill} \\} + + For a $3\times3$ matrix, the solution is the intersection of the 3 +planes. + +\(10x_{1}+2x_{2}+x_{3}=1\) + +\(2x_{1}+x_{2}+x_{3}=1\) + +\(x_{1}+2x_{2}+10x_{3}=1\) + +$\left[ \begin{array}{ccc} 10 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 2 & 10\end{array} \right] + \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]= + \left[\begin{array}{c} 1 \\ 1 \\ 1\end{array}\right]$ + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}17}]:} \PY{n}{x11}\PY{p}{=}\PY{n+nb}{linspace}\PY{p}{(}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{5}\PY{p}{)}\PY{p}{;} + 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\emph{hyperplane} +intersections. Which are even harder to visualize + + \subsection{Gauss elimination}\label{gauss-elimination} + +\subsubsection{Solving sets of equations +systematically}\label{solving-sets-of-equations-systematically} + +$\left[ +\begin{array}{ccc|c} + & A & & y \\ +10 & 2 & 1 & 1\\ +2 & 1 & 1 & 1 \\ +1 & 2 & 10 & 1\end{array} +\right] $ + +Ay(2,:)-Ay(1,:)/5 = ({[}2 1 1 1{]}-1/5{[}10 2 1 1{]}) + +$\left[ +\begin{array}{ccc|c} + & A & & y \\ +10 & 2 & 1 & 1\\ +0 & 3/5 & 4/5 & 4/5 \\ +1 & 2 & 10 & 1\end{array} +\right] $ + +Ay(3,:)-Ay(1,:)/10 = ({[}1 2 10 1{]}-1/10{[}10 2 1 1{]}) + +$\left[ +\begin{array}{ccc|c} + & A & & y \\ +10 & 2 & 1 & 1\\ +0 & 3/5 & 4/5 & 4/5 \\ +0 & 1.8 & 9.9 & 0.9\end{array} +\right] $ + +Ay(3,:)-1.8*5/3*Ay(2,:) = ({[}0 1.8 9.9 0.9{]}-3*{[}0 3/5 4/5 4/5{]}) + +$\left[ +\begin{array}{ccc|c} + & A & & y \\ +10 & 2 & 1 & 1\\ +0 & 3/5 & 4/5 & 4/5 \\ +0 & 0 & 7.5 & -1.5\end{array} +\right] $ + +now, \(7.5x_{3}=-1.5\) so \(x_{3}=-\frac{1}{5}\) + +then, \(3/5x_{2}+4/5(-1/5)=1\) so \(x_{2}=\frac{8}{5}\) + +finally, \$10x\_\{1\}+2(8/5) + + Consider the problem again from the intro to Linear Algebra, 4 masses +are connected in series to 4 springs with K=10 N/m. What are the final +positions of the masses? + +\begin{figure}[htbp] +\centering +\includegraphics{../lecture_09/mass_springs.svg} +\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: + +\(m_{1}g+k(x_{2}-x_{1})-kx_{1}=0\) + +\(m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0\) + +\(m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0\) + +\(m_{4}g-k(x_{4}-x_{3})=0\) + +in matrix form: + +\(\left[ \begin{array}{cccc} 2k & -k & 0 & 0 \\ -k & 2k & -k & 0 \\ 0 & -k & 2k & -k \\ 0 & 0 & -k & k \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]= \left[ \begin{array}{c} m_{1}g \\ m_{2}g \\ m_{3}g \\ m_{4}g \end{array} \right]\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}18}]:} \PY{n}{k}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} \PY{c}{\PYZpc{} N/m} + \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{l+m+mi}{2}\PY{o}{*}\PY{n}{k} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{0} \PY{l+m+mi}{0}\PY{p}{;} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{0}\PY{p}{;} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k} \PY{o}{\PYZhy{}}\PY{n}{k}\PY{p}{;} \PY{l+m+mi}{0} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k} \PY{n}{k}\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 = + + 20 -10 0 0 + -10 20 -10 0 + 0 -10 20 -10 + 0 0 -10 10 + +y = + + 9.8100 + 19.6200 + 29.4300 + 39.2400 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}19}]:} \PY{n}{K1}\PY{p}{=}\PY{p}{[}\PY{n}{K} \PY{n}{y}\PY{p}{]}\PY{p}{;} + \PY{n}{K1}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{K1}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{/}\PY{l+m+mi}{2}\PY{o}{+}\PY{n}{K1}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +K1 = + + 20.00000 -10.00000 0.00000 0.00000 9.81000 + 0.00000 15.00000 -10.00000 0.00000 24.52500 + 0.00000 -10.00000 20.00000 -10.00000 29.43000 + 0.00000 0.00000 -10.00000 10.00000 39.24000 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}20}]:} \PY{n}{K2}\PY{p}{=}\PY{n}{K1}\PY{p}{;} + \PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{K1}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{*}\PY{l+m+mi}{2}\PY{o}{/}\PY{l+m+mi}{3}\PY{o}{+}\PY{n}{K1}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +K2 = + + 20.00000 -10.00000 0.00000 0.00000 9.81000 + 0.00000 15.00000 -10.00000 0.00000 24.52500 + 0.00000 0.00000 13.33333 -10.00000 45.78000 + 0.00000 0.00000 -10.00000 10.00000 39.24000 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}21}]:} \PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{o}{\PYZhy{}}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{*}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{/}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{+}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{p}{:}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +K2 = + + 20.00000 -10.00000 0.00000 0.00000 9.81000 + 0.00000 15.00000 -10.00000 0.00000 24.52500 + 0.00000 0.00000 13.33333 -10.00000 45.78000 + 0.00000 0.00000 0.00000 2.50000 73.57500 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}22}]:} \PY{n}{yp}\PY{p}{=}\PY{n}{K2}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{5}\PY{p}{)}\PY{p}{;} + \PY{n}{x4}\PY{p}{=}\PY{n}{yp}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{)}\PY{o}{/}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)} + \PY{n}{x3}\PY{p}{=}\PY{p}{(}\PY{n}{yp}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{+}\PY{l+m+mi}{10}\PY{o}{*}\PY{n}{x4}\PY{p}{)}\PY{o}{/}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)} + \PY{n}{x2}\PY{p}{=}\PY{p}{(}\PY{n}{yp}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{+}\PY{l+m+mi}{10}\PY{o}{*}\PY{n}{x3}\PY{p}{)}\PY{o}{/}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)} + \PY{n}{x1}\PY{p}{=}\PY{p}{(}\PY{n}{yp}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{+}\PY{l+m+mi}{10}\PY{o}{*}\PY{n}{x2}\PY{p}{)}\PY{o}{/}\PY{n}{K2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +x4 = 29.430 +x3 = 25.506 +x2 = 18.639 +x1 = 9.8100 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}23}]:} \PY{n}{K}\PY{o}{\PYZbs{}}\PY{n}{y} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ans = + + 9.8100 + 18.6390 + 25.5060 + 29.4300 + + + \end{Verbatim} + + \subsection{Automate Gauss +Elimination}\label{automate-gauss-elimination} + +We can automate Gauss elimination with a function whose input is A and +y: + +\texttt{x=GaussNaive(A,y)} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}24}]:} \PY{n}{x}\PY{p}{=}\PY{n}{GaussNaive}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{n}{y}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +x = + + 9.8100 + 18.6390 + 25.5060 + 29.4300 + + + \end{Verbatim} + + \subsection{Problem (Diagonal element is +zero)}\label{problem-diagonal-element-is-zero} + +If a diagonal element is 0 or very small either: + +\begin{enumerate} +\def\labelenumi{\arabic{enumi}.} +\tightlist +\item + no solution found +\item + errors are introduced +\end{enumerate} + +Therefore, we would want to pivot before applying Gauss elimination + +Consider: + +\begin{enumerate} +\def\labelenumi{(\alph{enumi})} +\item + \(\left[ \begin{array}{cccc} 0 & 2 & 3 \\ 4 & 6 & 7 \\ 2 & -3 & 6 \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right]= \left[ \begin{array}{c} 8 \\ -3 \\ 5\end{array} \right]\) +\item + \(\left[ \begin{array}{cccc} 0.0003 & 3.0000 \\ 1.0000 & 1.0000 \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right]= \left[ \begin{array}{c} 2.0001 \\ 1.0000 \end{array} \right]\) +\end{enumerate} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}25}]:} \PY{n}{format} \PY{n}{short} + \PY{n}{Aa}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{6}\PY{p}{,}\PY{l+m+mi}{7}\PY{p}{;}\PY{l+m+mi}{2}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{6}\PY{p}{]}\PY{p}{;} \PY{n}{ya}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{8}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{5}\PY{p}{]}\PY{p}{;} + \PY{n}{GaussNaive}\PY{p}{(}\PY{n}{Aa}\PY{p}{,}\PY{n}{ya}\PY{p}{)} + \PY{n}{Aa}\PY{o}{\PYZbs{}}\PY{n}{ya} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +warning: division by zero +warning: called from + GaussNaive at line 16 column 12 +warning: division by zero +warning: division by zero +ans = + + NaN + NaN + NaN + +ans = + + -5.423913 + 0.021739 + 2.652174 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}26}]:} \PY{p}{[}\PY{n}{x}\PY{p}{,}\PY{n}{Aug}\PY{p}{,}\PY{n}{npivots}\PY{p}{]}\PY{p}{=}\PY{n}{GaussPivot}\PY{p}{(}\PY{n}{Aa}\PY{p}{,}\PY{n}{ya}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +x = + + -5.423913 + 0.021739 + 2.652174 + +Aug = + + 4.00000 6.00000 7.00000 -3.00000 + 0.00000 -6.00000 2.50000 6.50000 + 0.00000 0.00000 3.83333 10.16667 + +npivots = 2 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}27}]:} \PY{n}{format} \PY{n}{long} + \PY{n}{Ab}\PY{p}{=}\PY{p}{[}\PY{l+m+mf}{0.3E\PYZhy{}13}\PY{p}{,}\PY{l+m+mf}{3.0000}\PY{p}{;}\PY{l+m+mf}{1.0000}\PY{p}{,}\PY{l+m+mf}{1.0000}\PY{p}{]}\PY{p}{;}\PY{n}{yb}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{o}{+}\PY{l+m+mf}{0.1e\PYZhy{}13}\PY{p}{;}\PY{l+m+mf}{1.0000}\PY{p}{]}\PY{p}{;} + \PY{n}{GaussNaive}\PY{p}{(}\PY{n}{Ab}\PY{p}{,}\PY{n}{yb}\PY{p}{)} + \PY{n}{Ab}\PY{o}{\PYZbs{}}\PY{n}{yb} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ans = + + 0.325665420556713 + 0.666666666666667 + +ans = + + 0.333333333333333 + 0.666666666666667 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}28}]:} \PY{p}{[}\PY{n}{x}\PY{p}{,}\PY{n}{Aug}\PY{p}{,}\PY{n}{npivots}\PY{p}{]}\PY{p}{=}\PY{n}{GaussPivot}\PY{p}{(}\PY{n}{Ab}\PY{p}{,}\PY{n}{yb}\PY{p}{)} + \PY{n}{Ab}\PY{o}{\PYZbs{}}\PY{n}{yb} + \PY{n}{format} \PY{n}{short} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +x = + + 0.333333333333333 + 0.666666666666667 + +Aug = + + 1.000000000000000 1.000000000000000 1.000000000000000 + 0.000000000000000 2.999999999999970 1.999999999999980 + +npivots = 1 +ans = + + 0.333333333333333 + 0.666666666666667 + + + \end{Verbatim} + + \subsubsection{Spring-Mass System again}\label{spring-mass-system-again} + +Now, 4 masses are connected in series to 4 springs with \(K_{1}\)=10 +N/m, \(K_{2}\)=5 N/m, \(K_{3}\)=2 N/m and \(K_{4}\)=1 N/m. What are the +final positions of the masses? + +\begin{figure}[htbp] +\centering +\includegraphics{../lecture_09/mass_springs.svg} +\caption{Springs-masses} +\end{figure} + +The masses have 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}29}]:} \PY{n}{k1}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} \PY{n}{k2}\PY{p}{=}\PY{l+m+mi}{5}\PY{p}{;}\PY{n}{k3}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;}\PY{n}{k4}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} N/m} + \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{p}{,} \PY{n}{k2}\PY{o}{+}\PY{n}{k3}\PY{p}{,} \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{p}{,} \PY{n}{k3}\PY{o}{+}\PY{n}{k4}\PY{p}{,} \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 = + + 15 -5 0 0 + -5 7 -2 0 + 0 -2 3 -1 + 0 0 -1 1 + +y = + + 9.8100 + 19.6200 + 29.4300 + 39.2400 + + + \end{Verbatim} + + \subsection{Tridiagonal matrix}\label{tridiagonal-matrix} + +This matrix, K, could be rewritten as 3 vectors e, f and g + +\(e=\left[ \begin{array}{c} 0 \\ -5 \\ -2 \\ -1 \end{array} \right]\) + +\(f=\left[ \begin{array}{c} 15 \\ 7 \\ 3 \\ 1 \end{array} \right]\) + +\(g=\left[ \begin{array}{c} -5 \\ -2 \\ -1 \\ 0 \end{array} \right]\) + +Where all other components are 0 and the length of the vectors are n and +the first component of e and the last component of g are zero + +\texttt{e(1)=0} + +\texttt{g(end)=0} + +No need to pivot and number of calculations reduced enormously. + +\begin{longtable}[c]{@{}ll@{}} +\toprule +method & Number of Floating point operations for +n\(\times\)n-matrix\tabularnewline +\midrule +\endhead +Naive Gauss & n-cubed\tabularnewline +Tridiagonal & n\tabularnewline +\bottomrule +\end{longtable} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}30}]:} \PY{n+nb}{e}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{5}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{;} + \PY{n}{g}\PY{p}{=}\PY{p}{[}\PY{o}{\PYZhy{}}\PY{l+m+mi}{5}\PY{p}{;}\PY{o}{\PYZhy{}}\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{p}{;} + \PY{n}{f}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{15}\PY{p}{;}\PY{l+m+mi}{7}\PY{p}{;}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{;} + \PY{n}{Tridiag}\PY{p}{(}\PY{n+nb}{e}\PY{p}{,}\PY{n}{f}\PY{p}{,}\PY{n}{g}\PY{p}{,}\PY{n}{y}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ans = + + 9.8100 27.4680 61.8030 101.0430 + + + \end{Verbatim} + + 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