diff --git a/lecture_13/LU_naive.m b/lecture_13/LU_naive.m new file mode 100644 index 0000000..92efde6 --- /dev/null +++ b/lecture_13/LU_naive.m @@ -0,0 +1,27 @@ +function [L, U] = LU_naive(A) +% 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; +L=diag(ones(n,1)); +U=A; +% forward elimination +for k = 1:n-1 + for i = k+1:n + fik = U(i,k)/U(k,k); + L(i,k)=fik; + U(i,k:nb) = U(i,k:nb)-fik*U(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_13/lecture_13.aux b/lecture_13/lecture_13.aux index 513ef62..30cf1b4 100644 --- a/lecture_13/lecture_13.aux +++ b/lecture_13/lecture_13.aux @@ -23,37 +23,41 @@ \@writefile{toc}{\contentsline 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{subsubsection}{\numberline {2.5.1}Nonlinear springs supporting two masses in series}{15}{subsubsection.2.5.1}} +\newlabel{nonlinear-springs-supporting-two-masses-in-series}{{2.5.1}{15}{Nonlinear springs supporting two masses in series}{subsubsection.2.5.1}{}} diff --git a/lecture_13/lecture_13.ipynb b/lecture_13/lecture_13.ipynb index 34a39e5..913f44f 100644 --- a/lecture_13/lecture_13.ipynb +++ b/lecture_13/lecture_13.ipynb @@ -68,6 +68,24 @@ " " ] }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Markdown examples\n", + "\n", + "` \" ' ` `\n", + "\n", + "```matlab\n", + "x=linspace(0,1);\n", + "y=x.^2;\n", + "plot(x,y)\n", + "for i = 1:10\n", + " fprintf('markdown is pretty')\n", + "end\n", + "```" + ] + }, { "cell_type": "markdown", "metadata": {}, @@ -82,7 +100,7 @@ "\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", + "$||A||_{f}=\\sqrt{\\sum_{i=1}^{n}\\sum_{j=1}^{m}A_{i,j}^{2}}$\n", "\n", "In general we can calculate any $p$-norm where\n", "\n", @@ -115,7 +133,7 @@ }, { "cell_type": "code", - "execution_count": 72, + "execution_count": 7, "metadata": { "collapsed": false }, @@ -166,7 +184,7 @@ }, { "cell_type": "code", - "execution_count": 75, + "execution_count": 8, "metadata": { "collapsed": false }, @@ -195,7 +213,7 @@ "d1=L\\[1;0;0];\n", "d2=L\\[0;1;0];\n", "d3=L\\[0;0;1];\n", - "invA(:,1)=U\\d1;\n", + "invA(:,1)=U\\d1; % shortcut invA(:,1)=A\\[1;0;0]\n", "invA(:,2)=U\\d2;\n", "invA(:,3)=U\\d3\n", "invA*A" @@ -210,7 +228,7 @@ }, { "cell_type": "code", - "execution_count": 74, + "execution_count": 9, "metadata": { "collapsed": false }, @@ -299,7 +317,7 @@ }, { "cell_type": "code", - "execution_count": 21, + "execution_count": 10, "metadata": { "collapsed": false }, @@ -341,7 +359,7 @@ }, { "cell_type": "code", - "execution_count": 25, + "execution_count": 11, "metadata": { "collapsed": false }, @@ -391,6 +409,15 @@ "max(e)/min(e)" ] }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## P=2 norm is ratio of biggest eigenvalue to smallest eigenvalue!\n", + "\n", + "no need to calculate the inv(K)" + ] + }, { "cell_type": "markdown", "metadata": {}, @@ -418,7 +445,7 @@ "-19.3 \\\\\n", "71.4\\end{array} \\right]$\n", "\n", - "$x_{1}=\\frac{7.85+0.1x_{2}+0.3x_{3}}{3}$\n", + "$x_{1}=\\frac{7.85+0.1x_{2}+0.2x_{3}}{3}$\n", "\n", "$x_{2}=\\frac{-19.3-0.1x_{1}+0.3x_{3}}{7}$\n", "\n", @@ -427,7 +454,7 @@ }, { "cell_type": "code", - "execution_count": 9, + "execution_count": 12, "metadata": { "collapsed": false }, @@ -511,9 +538,9 @@ "-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", + "0 & 0.1/3 & 0.2/3 \\\\\n", + "0.1/7 & 0 & -0.3/7 \\\\\n", + "0.3/10 & -0.2/10 & 0 \\end{array} \\right]\n", "\\left[ \\begin{array}{c}\n", "x_{1}^{i-1} \\\\\n", "x_{2}^{i-1} \\\\\n", @@ -528,7 +555,7 @@ }, { "cell_type": "code", - "execution_count": 15, + "execution_count": 14, "metadata": { "collapsed": false }, @@ -637,7 +664,7 @@ }, { "cell_type": "code", - "execution_count": 17, + "execution_count": 15, "metadata": { "collapsed": false }, @@ -675,7 +702,7 @@ }, { "cell_type": "code", - "execution_count": 20, + "execution_count": 16, "metadata": { "collapsed": false }, @@ -754,7 +781,7 @@ }, { "cell_type": "code", - "execution_count": 105, + "execution_count": 17, "metadata": { "collapsed": false }, @@ -1028,7 +1055,7 @@ }, { "cell_type": "code", - "execution_count": 121, + "execution_count": 19, "metadata": { "collapsed": false }, @@ -1075,86 +1102,96 @@ "\n", "\n", "\t\n", - "\t\t\n", + "\t\t\n", "\t\n", "\n", "\n", "\n", "\n", - "\t\t\n", + "\t\t\n", "\t\t0\n", "\t\n", "\n", "\n", - "\t\t\n", + "\t\t\n", "\t\t5\n", "\t\n", "\n", "\n", - "\t\t\n", + "\t\t\n", "\t\t10\n", "\t\n", "\n", "\n", - "\t\t\n", + "\t\t\n", "\t\t15\n", "\t\n", "\n", "\n", - "\t\t\n", + "\t\t\n", "\t\t20\n", "\t\n", "\n", "\n", - "\t\t\n", + "\t\t\n", "\t\t0\n", "\t\n", "\n", "\n", - "\t\t\n", + "\t\t\n", "\t\t1\n", "\t\n", "\n", "\n", - "\t\t\n", + "\t\t\n", "\t\t2\n", "\t\n", "\n", "\n", - "\t\t\n", + "\t\t\n", "\t\t3\n", "\t\n", "\n", "\n", - "\t\t\n", + "\t\t\n", "\t\t4\n", "\t\n", "\n", "\n", - "\t\t\n", + "\t\t\n", "\t\t5\n", "\t\n", "\n", "\n", "\n", "\n", - "\t\n", + "\t\n", + "\n", + "\t\n", + "\t\tx2\n", + "\t\n", + "\n", + "\n", + "\t\n", + "\t\tx1\n", + "\t\n", + "\n", "\n", "\n", "\tgnuplot_plot_1a\n", - "\n", + "\n", "\n", "\n", - "\t\n", + "\t\n", "\t\n", "\tgnuplot_plot_2a\n", "\n", - "\t\n", + "\t\n", "\t\n", "\tgnuplot_plot_3a\n", "\n", "\t \n", - "\t\n", + "\t\n", "\n", "\t\n", "\n", @@ -1186,7 +1223,9 @@ "plot(x11,x12,x21,x22)\n", "% Solution at x_1=2, x_2=3\n", "hold on;\n", - "plot(2,3,'o')" + "plot(2,3,'o')\n", + "xlabel('x_1')\n", + "ylabel('x_2')" ] }, { @@ -1240,10 +1279,10 @@ "\\vdots \\\\\n", "x_{i+1}\\end{array} \\right]-\n", "\\left[ \\begin{array}{c}\n", - "f_{1,i} \\\\\n", - "f_{2,i} \\\\\n", + "x_{1,i} \\\\\n", + "x_{2,i} \\\\\n", "\\vdots \\\\\n", - "f_{n,i}\\end{array} \\right]\\right)$\n", + "x_{n,i}\\end{array} \\right]\\right)$\n", "\n", "### Solution is again in the form Ax=b\n", "\n", @@ -1257,7 +1296,7 @@ "\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", + "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 $k_1$=100 N/m and $k_2$=-10 N/m\n", "\n", "We want to solve for the final position of the masses ($x_1$ and $x_2$)\n", "\n", @@ -1272,8 +1311,7 @@ "$J(2,1)=\\frac{\\partial f_2}{\\partial x_{1}}=k_{1}+2k_{2}(x_{2}-x_{1})$\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" + "\n" ] }, { @@ -1286,13 +1324,13 @@ "source": [ "m1=1; % kg \n", "m2=2; % kg\n", - "k1=10; % N/m\n", - "k2=-4; % N/m^2" + "k1=100; % N/m\n", + "k2=-10; % N/m^2" ] }, { "cell_type": "code", - "execution_count": 214, + "execution_count": 20, "metadata": { "collapsed": false }, @@ -1318,7 +1356,7 @@ }, { "cell_type": "code", - "execution_count": 217, + "execution_count": 21, "metadata": { "collapsed": false }, @@ -1346,7 +1384,7 @@ }, { "cell_type": "code", - "execution_count": 227, + "execution_count": 22, "metadata": { "collapsed": false }, @@ -1430,7 +1468,7 @@ }, { "cell_type": "code", - "execution_count": 228, + "execution_count": 23, "metadata": { "collapsed": false }, @@ -1459,7 +1497,7 @@ }, { "cell_type": "code", - "execution_count": 236, + "execution_count": 26, "metadata": { "collapsed": false }, @@ -1505,4046 +1543,3710 @@ "\n", "\n", "\n", - "\t\n", - "\t\t\n", - "\t\n", "\n", "\n", "\n", "\n", "\n", - "\n", - "\t\n", "\tgnuplot_plot_1a\n", - "\n", - "\n", - "\n", - "\n", - "\t\n", - "\t\n", - "\t\t\n", - "\t\n", - "\n", - "\t\t\n", - "\t\t\n", - "\t\n", - "\n", - "\n", - "\t\t\n", - "\t\t\n", - "\t\n", - "\n", - "\n", - "\t\t\n", - "\t\t\n", - "\t\n", - "\n", - "\n", - "\t\t\n", - "\t\t\n", - "\t\n", - "\n", - "\n", - "\t\t\n", - "\t\t\n", - "\t\n", - "\n", - "\n", - "\t\t\n", - "\t\t\n", - "\t\n", - "\n", - "\n", - "\t\t\n", - "\t\t\n", - "\t\n", - "\n", - "\n", - "\t\t\n", - "\t\t\n", - "\t\n", - "\n", - "\n", - "\t\t\n", - "\t\t\n", - "\t\n", - "\n", - "\n", - "\t\t\n", - "\t\t\n", - "\t\n", - "\n", - "\n", - "\t\t\n", - 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"\t\t100\n", - "\t\n", - "\n", - "\n", "\n", "\n", "\n", @@ -5553,7 +5255,7 @@ "\n", "\n", "\n", - "\t\n", + "\t\n", "\n", "\n", "\n", @@ -5568,21 +5270,22 @@ } ], "source": [ - "[X,Y]=meshgrid(linspace(0,1,20),linspace(0,1,20));\n", + "[X,Y]=meshgrid(linspace(0,10,20),linspace(0,10,20));\n", "[N,M]=size(X);\n", "F=zeros(size(X));\n", "for i=1:N\n", " for j=1:M\n", " [f,~]=mass_spring([X(i,j),Y(i,j)]);\n", - " F(i,j)=f(1);\n", + " F1(i,j)=f(1);\n", + " F2(i,j)=f(2);\n", " end\n", "end\n", - "pcolor(X,Y,F)\n", + "mesh(X,Y,F1)\n", "xlabel('x_1')\n", "ylabel('x_2')\n", "colorbar()\n", "figure()\n", - "pcolor(X,Y,F)\n", + "mesh(X,Y,F2)\n", "xlabel('x_1')\n", "ylabel('x_2')\n", "colorbar()" diff --git a/lecture_13/lecture_13.log b/lecture_13/lecture_13.log index bb369ae..a86b4c3 100644 --- a/lecture_13/lecture_13.log +++ b/lecture_13/lecture_13.log @@ -1,4 +1,4 @@ -This is pdfTeX, Version 3.14159265-2.6-1.40.16 (TeX Live 2015/Debian) (preloaded format=pdflatex 2017.1.11) 2 MAR 2017 09:22 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PDF statistics: - 281 PDF objects out of 1000 (max. 8388607) - 225 compressed objects within 3 object streams - 46 named destinations out of 1000 (max. 500000) - 164 words of extra memory for PDF output out of 10000 (max. 10000000) + 294 PDF objects out of 1000 (max. 8388607) + 237 compressed objects within 3 object streams + 49 named destinations out of 1000 (max. 500000) + 180 words of extra memory for PDF output out of 10000 (max. 10000000) diff --git a/lecture_13/lecture_13.md b/lecture_13/lecture_13.md index 914980b..88716ad 100644 --- a/lecture_13/lecture_13.md +++ b/lecture_13/lecture_13.md @@ -50,6 +50,19 @@ setdefaults +# Markdown examples + +` " ' ` ` + +```matlab +x=linspace(0,1); +y=x.^2; +plot(x,y) +for i = 1:10 + fprintf('markdown is pretty') +end +``` + ## Condition of a matrix ### *just checked in to see what condition my condition was in* ### Matrix norms @@ -60,7 +73,7 @@ $||x||_{e}=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}$ For a matrix, A, the same norm is called the Frobenius norm: -$||A||_{f}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{2}}$ +$||A||_{f}=\sqrt{\sum_{i=1}^{n}\sum_{j=1}^{m}A_{i,j}^{2}}$ In general we can calculate any $p$-norm where @@ -130,7 +143,7 @@ invA=zeros(3,3); d1=L\[1;0;0]; d2=L\[0;1;0]; d3=L\[0;0;1]; -invA(:,1)=U\d1; +invA(:,1)=U\d1; % shortcut invA(:,1)=A\[1;0;0] invA(:,2)=U\d2; invA(:,3)=U\d3 invA*A @@ -285,6 +298,10 @@ max(e)/min(e) ans = 2.5293e+05 +## P=2 norm is ratio of biggest eigenvalue to smallest eigenvalue! + +no need to calculate the inv(K) + # Iterative Methods ## Gauss-Seidel method @@ -308,7 +325,7 @@ x_{3} \end{array} \right]= -19.3 \\ 71.4\end{array} \right]$ -$x_{1}=\frac{7.85+0.1x_{2}+0.3x_{3}}{3}$ +$x_{1}=\frac{7.85+0.1x_{2}+0.2x_{3}}{3}$ $x_{2}=\frac{-19.3-0.1x_{1}+0.3x_{3}}{7}$ @@ -383,9 +400,9 @@ x_{3}^{i} \end{array} \right]= -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] +0 & 0.1/3 & 0.2/3 \\ +0.1/7 & 0 & -0.3/7 \\ +0.3/10 & -0.2/10 & 0 \end{array} \right] \left[ \begin{array}{c} x_{1}^{i-1} \\ x_{2}^{i-1} \\ @@ -600,7 +617,7 @@ plot([1:100]*2/100,iters) -![svg](lecture_13_files/lecture_13_22_1.svg) +![svg](lecture_13_files/lecture_13_24_1.svg) @@ -682,10 +699,12 @@ plot(x11,x12,x21,x22) % Solution at x_1=2, x_2=3 hold on; plot(2,3,'o') +xlabel('x_1') +ylabel('x_2') ``` -![svg](lecture_13_files/lecture_13_27_0.svg) +![svg](lecture_13_files/lecture_13_29_0.svg) ## Newton-Raphson part II @@ -735,10 +754,10 @@ x_{i+1} \\ \vdots \\ x_{i+1}\end{array} \right]- \left[ \begin{array}{c} -f_{1,i} \\ -f_{2,i} \\ +x_{1,i} \\ +x_{2,i} \\ \vdots \\ -f_{n,i}\end{array} \right]\right)$ +x_{n,i}\end{array} \right]\right)$ ### Solution is again in the form Ax=b @@ -752,7 +771,7 @@ $[x_{i+1}]= [x_{i}]-[J]^{-1}[f]$ ### 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 +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 $k_1$=100 N/m and $k_2$=-10 N/m We want to solve for the final position of the masses ($x_1$ and $x_2$) @@ -768,15 +787,14 @@ $J(2,1)=\frac{\partial f_2}{\partial x_{1}}=k_{1}+2k_{2}(x_{2}-x_{1})$ $J(2,2)=\frac{\partial f_2}{\partial x_{2}}=-k_{1}-2k_{2}(x_{2}-x_{1})$ -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 +k1=100; % N/m +k2=-10; % N/m^2 ``` @@ -909,32 +927,33 @@ X0=fsolve(@(x) mass_spring(x),[3;5]) ```octave -[X,Y]=meshgrid(linspace(0,1,20),linspace(0,1,20)); +[X,Y]=meshgrid(linspace(0,10,20),linspace(0,10,20)); [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); + F1(i,j)=f(1); + F2(i,j)=f(2); end end -pcolor(X,Y,F) +mesh(X,Y,F1) xlabel('x_1') ylabel('x_2') colorbar() figure() -pcolor(X,Y,F) +mesh(X,Y,F2) xlabel('x_1') ylabel('x_2') colorbar() ``` -![svg](lecture_13_files/lecture_13_34_0.svg) +![svg](lecture_13_files/lecture_13_36_0.svg) -![svg](lecture_13_files/lecture_13_34_1.svg) +![svg](lecture_13_files/lecture_13_36_1.svg) diff --git a/lecture_13/lecture_13.out b/lecture_13/lecture_13.out index edec246..f10cd22 100644 --- a/lecture_13/lecture_13.out +++ b/lecture_13/lecture_13.out @@ -1,16 +1,18 @@ \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 +\BOOKMARK [1][-]{section.1}{Markdown examples}{}% 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 +\BOOKMARK [2][-]{subsection.1.2}{P=2 norm is ratio of biggest eigenvalue to smallest eigenvalue!}{section.1}% 8 +\BOOKMARK [1][-]{section.2}{Iterative Methods}{}% 9 +\BOOKMARK [2][-]{subsection.2.1}{Gauss-Seidel method}{section.2}% 10 +\BOOKMARK [3][-]{subsubsection.2.1.1}{Gauss-Seidel Iterative approach}{subsection.2.1}% 11 +\BOOKMARK [3][-]{subsubsection.2.1.2}{Jacobi method}{subsection.2.1}% 12 +\BOOKMARK [2][-]{subsection.2.2}{Gauss-Seidel with Relaxation}{section.2}% 13 +\BOOKMARK [2][-]{subsection.2.3}{Nonlinear Systems}{section.2}% 14 +\BOOKMARK [2][-]{subsection.2.4}{Newton-Raphson part II}{section.2}% 15 +\BOOKMARK [3][-]{subsubsection.2.4.1}{Solution is again in the form Ax=b}{subsection.2.4}% 16 +\BOOKMARK [2][-]{subsection.2.5}{Example of Jacobian calculation}{section.2}% 17 +\BOOKMARK [3][-]{subsubsection.2.5.1}{Nonlinear springs supporting two masses in series}{subsection.2.5}% 18 diff --git a/lecture_13/lecture_13.pdf b/lecture_13/lecture_13.pdf index fad59c8..5a45770 100644 Binary files a/lecture_13/lecture_13.pdf and b/lecture_13/lecture_13.pdf differ diff --git a/lecture_13/lecture_13.tex b/lecture_13/lecture_13.tex index 87d8d25..72647e9 100644 --- a/lecture_13/lecture_13.tex +++ b/lecture_13/lecture_13.tex @@ -293,10 +293,12 @@ class}\label{my-question-from-last-class} \caption{q1} \end{figure} -$A=\left[\begin{array}{ccc} + + $A=\left[\begin{array}{ccc} 2 & -2 & 0\\ -1& 5 & 1 \\ -3 &4 & 5 \end{array}\right]$ + 3 &4 & 5 \end{array}\right]$ + \begin{figure}[htbp] \centering @@ -349,6 +351,21 @@ class}\label{your-questions-from-last-class} in Jupyter - https://anaconda.org/pypi/octave\_kernel} \end{enumerate} + \section{Markdown examples}\label{markdown-examples} + +\texttt{"\ \textquotesingle{}} ` + +\begin{Shaded} +\begin{Highlighting}[] +\NormalTok{x=linspace(}\FloatTok{0}\NormalTok{,}\FloatTok{1}\NormalTok{);} +\NormalTok{y=x.^}\FloatTok{2}\NormalTok{;} +\NormalTok{plot(x,y)} +\NormalTok{for i = }\FloatTok{1}\NormalTok{:}\FloatTok{10} + \NormalTok{fprintf(}\StringTok{'markdown is pretty'}\NormalTok{)} +\NormalTok{end} +\end{Highlighting} +\end{Shaded} + \subsection{Condition of a matrix}\label{condition-of-a-matrix} \subsubsection{\texorpdfstring{\emph{just checked in to see what @@ -365,7 +382,7 @@ $||x||_{e}=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}$ For a matrix, A, the same norm is called the Frobenius norm: -$||A||_{f}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{2}}$ +$||A||_{f}=\sqrt{\sum_{i=1}^{n}\sum_{j=1}^{m}A_{i,j}^{2}}$ In general we can calculate any $p$-norm where @@ -390,15 +407,15 @@ 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\} +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}{)} +{\color{incolor}In [{\color{incolor}7}]:} \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=\\\{\}] @@ -429,14 +446,14 @@ $Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]$, $Ux_{1}=d_{1}$ ... \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} +{\color{incolor}In [{\color{incolor}8}]:} \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{c}{\PYZpc{} shortcut invA(:,1)=A\PYZbs{}[1;0;0]} + \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=\\\{\}] @@ -458,27 +475,27 @@ ans = 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} +{\color{incolor}In [{\color{incolor}9}]:} \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=\\\{\}] @@ -523,7 +540,7 @@ 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} +{\color{incolor}In [{\color{incolor}10}]:} \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}{;} @@ -555,7 +572,7 @@ y = \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}{)} +{\color{incolor}In [{\color{incolor}11}]:} \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}{)} @@ -586,6 +603,11 @@ ans = 2.5293e+05 \end{Verbatim} + \subsection{P=2 norm is ratio of biggest eigenvalue to smallest +eigenvalue!}\label{p2-norm-is-ratio-of-biggest-eigenvalue-to-smallest-eigenvalue} + +no need to calculate the inv(K) + \section{Iterative Methods}\label{iterative-methods} \subsection{Gauss-Seidel method}\label{gauss-seidel-method} @@ -600,17 +622,17 @@ $Ax=b$ $\left[ \begin{array}{ccc} 3 & -0.1 & -0.2 \\ 0.1 & 7 & -0.3 \\ 0.3 & -0.2 & 10 \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right]= \left[ \begin{array}{c} 7.85 \\ -19.3 \\ 71.4\end{array} \right]$ -$x_{1}=\frac{7.85+0.1x_{2}+0.3x_{3}}{3}$ +$x_{1}=\frac{7.85+0.1x_{2}+0.2x_{3}}{3}$ $x_{2}=\frac{-19.3-0.1x_{1}+0.3x_{3}}{7}$ $x_{3}=\frac{71.4+0.1x_{1}+0.2x_{2}}{10}$ \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} +{\color{incolor}In [{\color{incolor}12}]:} \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=\\\{\}] @@ -670,7 +692,7 @@ $x_{3}^{i}=\frac{71.4+0.1x_{1}^{i-1}+0.2x_{2}^{i-1}}{10}$ Here the solution is a matrix multiplication and vector addition -$\left[ \begin{array}{c} x_{1}^{i} \\ x_{2}^{i} \\ x_{3}^{i} \end{array} \right]= \left[ \begin{array}{c} 7.85/3 \\ -19.3/7 \\ 71.4/10\end{array} \right]- \left[ \begin{array}{ccc} 0 & -0.1 & -0.2 \\ 0.1 & 0 & -0.3 \\ 0.3 & -0.2 & 0 \end{array} \right] \left[ \begin{array}{c} x_{1}^{i-1} \\ x_{2}^{i-1} \\ x_{3}^{i-1} \end{array} \right]$ +$\left[ \begin{array}{c} x_{1}^{i} \\ x_{2}^{i} \\ x_{3}^{i} \end{array} \right]= \left[ \begin{array}{c} 7.85/3 \\ -19.3/7 \\ 71.4/10\end{array} \right]- \left[ \begin{array}{ccc} 0 & 0.1/3 & 0.2/3 \\ 0.1/7 & 0 & -0.3/7 \\ 0.3/10 & -0.2/10 & 0 \end{array} \right] \left[ \begin{array}{c} x_{1}^{i-1} \\ x_{2}^{i-1} \\ x_{3}^{i-1} \end{array} \right]$ \begin{longtable}[c]{@{}llll@{}} \toprule @@ -725,7 +747,7 @@ $\frac{71.4+0.1x_{1}^{i}+0.2x_{2}^{i}}{10}$ \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)]} +{\color{incolor}In [{\color{incolor}14}]:} \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}{;} @@ -808,7 +830,7 @@ $|a_{ii}|>\sum_{j=1,j\ne i}^{n}|a_{ij}|$ 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}{]} +{\color{incolor}In [{\color{incolor}15}]:} \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} @@ -836,7 +858,7 @@ ans = \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)]} +{\color{incolor}In [{\color{incolor}16}]:} \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}{;} @@ -903,16 +925,16 @@ 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}{)} +{\color{incolor}In [{\color{incolor}17}]:} \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=\\\{\}] @@ -932,7 +954,7 @@ b = \end{Verbatim} \begin{center} - \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_13_files/lecture_13_22_1.pdf} + \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_13_files/lecture_13_24_1.pdf} \end{center} { \hspace*{\fill} \\} @@ -1005,20 +1027,22 @@ $x_{2}+3x_{1}x_{2}^{2}=57$ 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}{)} +{\color{incolor}In [{\color{incolor}19}]:} \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}{)} + \PY{n+nb}{xlabel}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{x\PYZus{}1\PYZsq{}}\PY{p}{)} + \PY{n+nb}{ylabel}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{x\PYZus{}2\PYZsq{}}\PY{p}{)} \end{Verbatim} \begin{center} - \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_13_files/lecture_13_27_0.pdf} + \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_13_files/lecture_13_29_0.pdf} \end{center} { \hspace*{\fill} \\} @@ -1038,7 +1062,7 @@ need to determine the Jacobian $[J]=\left[ \begin{array}{cccc} \frac{\partial f_{1,i}}{\partial x_{1}} & \frac{\partial f_{1,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{1,i}}{\partial x_{n}} \\ \frac{\partial f_{2,i}}{\partial x_{1}} & \frac{\partial f_{2,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{2,i}}{\partial x_{n}} \\ \vdots & \vdots & & \vdots \\ \frac{\partial f_{n,i}}{\partial x_{1}} & \frac{\partial f_{n,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{n,i}}{\partial x_{n}} \\ \end{array} \right]$ -$\left[ \begin{array}{c} f_{1,i+1} \\ f_{2,i+1} \\ \vdots \\ f_{n,i+1}\end{array} \right]= \left[ \begin{array}{c} f_{1,i} \\ f_{2,i} \\ \vdots \\ f_{n,i}\end{array} \right]+ \left[ \begin{array}{cccc} \frac{\partial f_{1,i}}{\partial x_{1}} & \frac{\partial f_{1,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{1,i}}{\partial x_{n}} \\ \frac{\partial f_{2,i}}{\partial x_{1}} & \frac{\partial f_{2,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{2,i}}{\partial x_{n}} \\ \vdots & \vdots & & \vdots \\ \frac{\partial f_{n,i}}{\partial x_{1}} & \frac{\partial f_{n,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{n,i}}{\partial x_{n}} \\ \end{array} \right] \left( \left[ \begin{array}{c} x_{i+1} \\ x_{i+1} \\ \vdots \\ x_{i+1}\end{array} \right]- \left[ \begin{array}{c} f_{1,i} \\ f_{2,i} \\ \vdots \\ f_{n,i}\end{array} \right]\right)$ +$\left[ \begin{array}{c} f_{1,i+1} \\ f_{2,i+1} \\ \vdots \\ f_{n,i+1}\end{array} \right]= \left[ \begin{array}{c} f_{1,i} \\ f_{2,i} \\ \vdots \\ f_{n,i}\end{array} \right]+ \left[ \begin{array}{cccc} \frac{\partial f_{1,i}}{\partial x_{1}} & \frac{\partial f_{1,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{1,i}}{\partial x_{n}} \\ \frac{\partial f_{2,i}}{\partial x_{1}} & \frac{\partial f_{2,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{2,i}}{\partial x_{n}} \\ \vdots & \vdots & & \vdots \\ \frac{\partial f_{n,i}}{\partial x_{1}} & \frac{\partial f_{n,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{n,i}}{\partial x_{n}} \\ \end{array} \right] \left( \left[ \begin{array}{c} x_{i+1} \\ x_{i+1} \\ \vdots \\ x_{i+1}\end{array} \right]- \left[ \begin{array}{c} x_{1,i} \\ x_{2,i} \\ \vdots \\ x_{n,i}\end{array} \right]\right)$ \subsubsection{Solution is again in the form Ax=b}\label{solution-is-again-in-the-form-axb} @@ -1057,7 +1081,7 @@ 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 +spring constants, of $k_1$=100 N/m and $k_2$=-10 N/m We want to solve for the final position of the masses ($x_1$ and $x_2$) @@ -1074,35 +1098,33 @@ $J(2,1)=\frac{\partial f_2}{\partial x_{1}}=k_{1}+2k_{2}(x_{2}-x_{1})$ $J(2,2)=\frac{\partial f_2}{\partial x_{2}}=-k_{1}-2k_{2}(x_{2}-x_{1})$ -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} + \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} \end{Verbatim} 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