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ME3255S2017/lecture_13/lecture_13.tex

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% 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
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% Set max figure width to be 80% of text width, for now hardcoded.
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% Ensure that by default, figures have no caption (until we provide a
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% in the conversion process - todo).
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\usepackage{enumerate} % Needed for markdown enumerations to work
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\usepackage{amsmath} % Equations
\usepackage{amssymb} % Equations
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\begin{document}
\maketitle
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{\color{incolor}In [{\color{incolor}1}]:} \PY{c}{\PYZpc{}plot \PYZhy{}\PYZhy{}format svg}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}2}]:} \PY{n}{setdefaults}
\end{Verbatim}
\subsection{My question from last
class}\label{my-question-from-last-class}
\begin{figure}[htbp]
\centering
\includegraphics{efficient_soln.png}
\caption{q1}
\end{figure}
$A=\left[\begin{array}{ccc}
2 & -2 & 0\\
-1& 5 & 1 \\
3 &4 & 5 \end{array}\right]$
\begin{figure}[htbp]
\centering
\includegraphics{norm_A.png}
\caption{q2}
\end{figure}
\subsection{Your questions from last
class}\label{your-questions-from-last-class}
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
Do we have to submit a link for HW \#4 somewhere or is uploading to
Github sufficient?
-no, your submission from HW 3 is sufficient
\item
How do I get the formulas/formatting in markdown files to show up on
github?
-no luck for markdown equations in github, this is an ongoing request
\item
Confused about the p=1 norm part and
\textbar{}\textbar{}A\textbar{}\textbar{}\_1
\item
When's the exam?
-next week (3/9)
\item
What do you recommend doing to get better at figuring out the
homeworks?
-time and experimenting (try going through the lecture examples,
verify my work)
\item
Could we have an hw or extra credit with a video lecture to learn some
simple python?
-Sounds great! how simple?
-\href{https://www.continuum.io/downloads}{Installing Python and
Jupyter Notebook (via Anaconda) - https://www.continuum.io/downloads}
-\href{https://anneurai.net/2015/11/12/matlab-based-ipython-notebooks/}{Running
Matlab kernel in Jupyter -
https://anneurai.net/2015/11/12/matlab-based-ipython-notebooks/}
-\href{https://anaconda.org/pypi/octave_kernel}{Running Octave kernel
in Jupyter - https://anaconda.org/pypi/octave\_kernel}
\end{enumerate}
\section{Markdown examples}\label{markdown-examples}
\texttt{"\ \textquotesingle{}} `
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\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
condition my condition was
in}}{just checked in to see what condition my condition was in}}\label{just-checked-in-to-see-what-condition-my-condition-was-in}
\subsubsection{Matrix norms}\label{matrix-norms}
The Euclidean norm of a vector is measure of the magnitude (in 3D this
would be: $|x|=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$) in general the
equation is:
$||x||_{e}=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}$
For a matrix, A, the same norm is called the Frobenius norm:
$||A||_{f}=\sqrt{\sum_{i=1}^{n}\sum_{j=1}^{m}A_{i,j}^{2}}$
In general we can calculate any $p$-norm where
$||A||_{p}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{p}}$
so the p=1, 1-norm is
$||A||_{1}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{1}}=\sum_{i=1}^{n}\sum_{i=1}^{m}|A_{i,j}|$
$||A||_{\infty}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{\infty}}=\max_{1\le i \le n}\sum_{j=1}^{m}|A_{i,j}|$
\subsubsection{Condition of Matrix}\label{condition-of-matrix}
The matrix condition is the product of
$Cond(A) = ||A||\cdot||A^{-1}||$
So each norm will have a different condition number, but the limit is
$Cond(A)\ge 1$
An estimate of the rounding error is based on the condition of A:
$\frac{||\Delta x||}{x} \le Cond(A) \frac{||\Delta A||}{||A||}$
So if the coefficients of A have accuracy to $10^{-t}$
and the condition of A, $Cond(A)=10^{c}$
then the solution for x can have rounding errors up to $10^{c-t}$
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}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=\\\{\}]
A =
1.00000 0.50000 0.33333
0.50000 0.33333 0.25000
0.33333 0.25000 0.20000
L =
1.00000 0.00000 0.00000
0.50000 1.00000 0.00000
0.33333 1.00000 1.00000
U =
1.00000 0.50000 0.33333
0.00000 0.08333 0.08333
0.00000 -0.00000 0.00556
\end{Verbatim}
Then, $A^{-1}=(LU)^{-1}=U^{-1}L^{-1}$
$Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]$,
$Ux_{1}=d_{1}$ ...
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}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=\\\{\}]
invA =
9.0000 -36.0000 30.0000
-36.0000 192.0000 -180.0000
30.0000 -180.0000 180.0000
ans =
1.0000e+00 3.5527e-15 2.9976e-15
-1.3249e-14 1.0000e+00 -9.1038e-15
8.5117e-15 7.1054e-15 1.0000e+00
\end{Verbatim}
Find the condition of A, $cond(A)$
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}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=\\\{\}]
normf\_A = 1.4136
normf\_invA = 372.21
cond\_f\_A = 526.16
ans = 1.4136
norm1\_A = 1.8333
norm1\_invA = 30.000
ans = 1.8333
cond\_1\_A = 55.000
norminf\_A = 1.8333
norminf\_invA = 30.000
ans = 1.8333
cond\_inf\_A = 55.000
\end{Verbatim}
Consider the problem again from the intro to Linear Algebra, 4 masses
are connected in series to 4 springs with spring constants $K_{i}$.
What does a high condition number mean for this problem?
\begin{figure}[htbp]
\centering
\includegraphics{../lecture_09/mass_springs.png}
\caption{Springs-masses}
\end{figure}
The masses haves the following amounts, 1, 2, 3, and 4 kg for masses
1-4. Using a FBD for each mass:
$m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0$
$m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0$
$m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0$
$m_{4}g-k_{4}(x_{4}-x_{3})=0$
in matrix form:
$\left[ \begin{array}{cccc} k_{1}+k_{2} & -k_{2} & 0 & 0 \\ -k_{2} & k_{2}+k_{3} & -k_{3} & 0 \\ 0 & -k_{3} & k_{3}+k_{4} & -k_{4} \\ 0 & 0 & -k_{4} & k_{4} \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]= \left[ \begin{array}{c} m_{1}g \\ m_{2}g \\ m_{3}g \\ m_{4}g \end{array} \right]$
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}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}{;}
\PY{n}{m1}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} kg}
\PY{n}{m2}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;}
\PY{n}{m3}\PY{p}{=}\PY{l+m+mi}{3}\PY{p}{;}
\PY{n}{m4}\PY{p}{=}\PY{l+m+mi}{4}\PY{p}{;}
\PY{n}{g}\PY{p}{=}\PY{l+m+mf}{9.81}\PY{p}{;} \PY{c}{\PYZpc{} m/s\PYZca{}2}
\PY{n}{K}\PY{p}{=}\PY{p}{[}\PY{n}{k1}\PY{o}{+}\PY{n}{k2} \PY{o}{\PYZhy{}}\PY{n}{k2} \PY{l+m+mi}{0} \PY{l+m+mi}{0}\PY{p}{;} \PY{o}{\PYZhy{}}\PY{n}{k2} \PY{n}{k2}\PY{o}{+}\PY{n}{k3} \PY{o}{\PYZhy{}}\PY{n}{k3} \PY{l+m+mi}{0}\PY{p}{;} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k3} \PY{n}{k3}\PY{o}{+}\PY{n}{k4} \PY{o}{\PYZhy{}}\PY{n}{k4}\PY{p}{;} \PY{l+m+mi}{0} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k4} \PY{n}{k4}\PY{p}{]}
\PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{n}{m1}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m2}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m3}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m4}\PY{o}{*}\PY{n}{g}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
K =
100010 -100000 0 0
-100000 100010 -10 0
0 -10 11 -1
0 0 -1 1
y =
9.8100
19.6200
29.4300
39.2400
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}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}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
ans = 3.2004e+05
ans = 3.2004e+05
ans = 2.5925e+05
ans = 2.5293e+05
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}26}]:} \PY{n+nb}{e}\PY{p}{=}\PY{n+nb}{eig}\PY{p}{(}\PY{n}{K}\PY{p}{)}
\PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{e}\PY{p}{)}\PY{o}{/}\PY{n+nb}{min}\PY{p}{(}\PY{n+nb}{e}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
e =
7.9078e-01
3.5881e+00
1.7621e+01
2.0001e+05
ans = 2.5293e+05
\end{Verbatim}
\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}
If we have an intial guess for each value of a vector $x$ that we are
trying to solve, then it is easy enough to solve for one component given
the others.
Take a 3$\times$3 matrix
$Ax=b$
$\left[ \begin{array}{ccc} 3 & -0.1 & -0.2 \\ 0.1 & 7 & -0.3 \\ 0.3 & -0.2 & 10 \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right]= \left[ \begin{array}{c} 7.85 \\ -19.3 \\ 71.4\end{array} \right]$
$x_{1}=\frac{7.85+0.1x_{2}+0.2x_{3}}{3}$
$x_{2}=\frac{-19.3-0.1x_{1}+0.3x_{3}}{7}$
$x_{3}=\frac{71.4+0.1x_{1}+0.2x_{2}}{10}$
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}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=\\\{\}]
A =
3.00000 -0.10000 -0.20000
0.10000 7.00000 -0.30000
0.30000 -0.20000 10.00000
b =
7.8500
-19.3000
71.4000
x =
3.0000
-2.5000
7.0000
\end{Verbatim}
\subsubsection{Gauss-Seidel Iterative
approach}\label{gauss-seidel-iterative-approach}
As a first guess, we can use $x_{1}=x_{2}=x_{3}=0$
$x_{1}=\frac{7.85+0.1(0)+0.3(0)}{3}=2.6167$
$x_{2}=\frac{-19.3-0.1(2.6167)+0.3(0)}{7}=-2.7945$
$x_{3}=\frac{71.4+0.1(2.6167)+0.2(-2.7945)}{10}=7.0056$
Then, we update the guess:
$x_{1}=\frac{7.85+0.1(-2.7945)+0.3(7.0056)}{3}=2.9906$
$x_{2}=\frac{-19.3-0.1(2.9906)+0.3(7.0056)}{7}=-2.4996$
$x_{3}=\frac{71.4+0.1(2.9906)+0.2(-2.4966)}{10}=7.00029$
The results are conveerging to the solution we found with
\texttt{\textbackslash{}} of $x_{1}=3,~x_{2}=-2.5,~x_{3}=7$
We could also use an iterative method that solves for all of the
x-components in one step:
\subsubsection{Jacobi method}\label{jacobi-method}
$x_{1}^{i}=\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$
$x_{2}^{i}=\frac{-19.3-0.1x_{1}^{i-1}+0.3x_{3}^{i-1}}{7}$
$x_{3}^{i}=\frac{71.4+0.1x_{1}^{i-1}+0.2x_{2}^{i-1}}{10}$
Here the solution is a matrix multiplication and vector addition
$\left[ \begin{array}{c} x_{1}^{i} \\ x_{2}^{i} \\ x_{3}^{i} \end{array} \right]= \left[ \begin{array}{c} 7.85/3 \\ -19.3/7 \\ 71.4/10\end{array} \right]- \left[ \begin{array}{ccc} 0 & 0.1/3 & 0.2/3 \\ 0.1/7 & 0 & -0.3/7 \\ 0.3/10 & -0.2/10 & 0 \end{array} \right] \left[ \begin{array}{c} x_{1}^{i-1} \\ x_{2}^{i-1} \\ x_{3}^{i-1} \end{array} \right]$
\begin{longtable}[c]{@{}llll@{}}
\toprule
\begin{minipage}[b]{0.10\columnwidth}\raggedright\strut
x\_\{j\}
\strut\end{minipage} &
\begin{minipage}[b]{0.36\columnwidth}\raggedright\strut
Jacobi method
\strut\end{minipage} &
\begin{minipage}[b]{0.05\columnwidth}\raggedright\strut
vs
\strut\end{minipage} &
\begin{minipage}[b]{0.37\columnwidth}\raggedright\strut
Gauss-Seidel
\strut\end{minipage}\tabularnewline
\midrule
\endhead
\begin{minipage}[t]{0.10\columnwidth}\raggedright\strut
$x_{1}^{i}=$
\strut\end{minipage} &
\begin{minipage}[t]{0.36\columnwidth}\raggedright\strut
$\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$
\strut\end{minipage} &
\begin{minipage}[t]{0.05\columnwidth}\raggedright\strut
\strut\end{minipage} &
\begin{minipage}[t]{0.37\columnwidth}\raggedright\strut
$\frac{7.85+0.1x_{2}^{i-1}+0.3x_{3}^{i-1}}{3}$
\strut\end{minipage}\tabularnewline
\begin{minipage}[t]{0.10\columnwidth}\raggedright\strut
$x_{2}^{i}=$
\strut\end{minipage} &
\begin{minipage}[t]{0.36\columnwidth}\raggedright\strut
$\frac{-19.3-0.1x_{1}^{i-1}+0.3x_{3}^{i-1}}{7}$
\strut\end{minipage} &
\begin{minipage}[t]{0.05\columnwidth}\raggedright\strut
\strut\end{minipage} &
\begin{minipage}[t]{0.37\columnwidth}\raggedright\strut
$\frac{-19.3-0.1x_{1}^{i}+0.3x_{3}^{i-1}}{7}$
\strut\end{minipage}\tabularnewline
\begin{minipage}[t]{0.10\columnwidth}\raggedright\strut
$x_{3}^{i}=$
\strut\end{minipage} &
\begin{minipage}[t]{0.36\columnwidth}\raggedright\strut
$\frac{71.4+0.1x_{1}^{i-1}+0.2x_{2}^{i-1}}{10}$
\strut\end{minipage} &
\begin{minipage}[t]{0.05\columnwidth}\raggedright\strut
\strut\end{minipage} &
\begin{minipage}[t]{0.37\columnwidth}\raggedright\strut
$\frac{71.4+0.1x_{1}^{i}+0.2x_{2}^{i}}{10}$
\strut\end{minipage}\tabularnewline
\bottomrule
\end{longtable}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}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}{;}
\PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{/}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}
\PY{n}{x0}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;}
\PY{n}{x1}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x0}
\PY{n}{x2}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x1}
\PY{n}{x3}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x2}
\PY{n+nb}{fprintf}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{solution is converging to [3,\PYZhy{}2.5,7]]\PYZbs{}n\PYZsq{}}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
ba =
2.6167
-2.7571
7.1400
sA =
0.00000 -0.10000 -0.20000
0.10000 0.00000 -0.30000
0.30000 -0.20000 0.00000
sA =
0.000000 -0.033333 -0.066667
0.014286 0.000000 -0.042857
0.030000 -0.020000 0.000000
x1 =
2.6167
-2.7571
7.1400
x2 =
3.0008
-2.4885
7.0064
x3 =
3.0008
-2.4997
7.0002
solution is converging to [3,-2.5,7]]
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}16}]:} \PY{n+nb}{diag}\PY{p}{(}\PY{n}{A}\PY{p}{)}
\PY{n+nb}{diag}\PY{p}{(}\PY{n+nb}{diag}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
ans =
3
7
10
ans =
Diagonal Matrix
3 0 0
0 7 0
0 0 10
\end{Verbatim}
This method works if problem is diagonally dominant,
$|a_{ii}|>\sum_{j=1,j\ne i}^{n}|a_{ij}|$
If this condition is true, then Jacobi or Gauss-Seidel should converge
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}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}
\begin{Verbatim}[commandchars=\\\{\}]
A =
0.10000 1.00000 3.00000
1.00000 0.20000 3.00000
5.00000 2.00000 0.30000
b =
12
2
4
ans =
-2.9393
9.1933
1.0336
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}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}{;}
\PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{sA}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{/}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}
\PY{n}{x0}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;}
\PY{n}{x1}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x0}
\PY{n}{x2}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x1}
\PY{n}{x3}\PY{p}{=}\PY{n}{ba}\PY{o}{\PYZhy{}}\PY{n}{sA}\PY{o}{*}\PY{n}{x2}
\PY{n+nb}{fprintf}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{solution is not converging to [\PYZhy{}2.93,9.19,1.03]\PYZbs{}n\PYZsq{}}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
ba =
120.000
10.000
13.333
sA =
0 1 3
1 0 3
5 2 0
sA =
0.00000 10.00000 30.00000
5.00000 0.00000 15.00000
16.66667 6.66667 0.00000
x1 =
120.000
10.000
13.333
x2 =
-380.00
-790.00
-2053.33
x3 =
6.9620e+04
3.2710e+04
1.1613e+04
solution is not converging to [-2.93,9.19,1.03]
\end{Verbatim}
\subsection{Gauss-Seidel with
Relaxation}\label{gauss-seidel-with-relaxation}
In order to force the solution to converge faster, we can introduce a
relaxation term $\lambda$.
where the new x values are weighted between the old and new:
$x^{i}=\lambda x^{i}+(1-\lambda)x^{i-1}$
after solving for x, lambda weights the current approximation with the
previous approximation for the updated x
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}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=\\\{\}]
A =
3.00000 -0.10000 -0.20000
0.10000 7.00000 -0.30000
0.30000 -0.20000 10.00000
b =
7.8500
-19.3000
71.4000
\end{Verbatim}
\begin{center}
\adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_13_files/lecture_13_24_1.pdf}
\end{center}
{ \hspace*{\fill} \\}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}107}]:} \PY{n}{l}\PY{p}{=}\PY{n}{fminbnd}\PY{p}{(}\PY{p}{@}\PY{p}{(}\PY{n}{l}\PY{p}{)} \PY{n}{lambda\PYZus{}fcn}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n}{b}\PY{p}{,}\PY{n}{l}\PY{p}{)}\PY{p}{,}\PY{l+m+mf}{0.5}\PY{p}{,}\PY{l+m+mf}{1.5}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
l = 0.99158
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}108}]:} \PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{b}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
ans =
3.0000
-2.5000
7.0000
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}109}]:} \PY{p}{[}\PY{n}{x}\PY{p}{,}\PY{n}{ea}\PY{p}{,}\PY{n}{iter}\PY{p}{]}\PY{p}{=}\PY{n}{Jacobi\PYZus{}rel}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n}{b}\PY{p}{,}\PY{n}{l}\PY{p}{,}\PY{l+m+mf}{0.000001}\PY{p}{)}
\PY{p}{[}\PY{n}{x}\PY{p}{,}\PY{n}{ea}\PY{p}{,}\PY{n}{iter}\PY{p}{]}\PY{p}{=}\PY{n}{Jacobi\PYZus{}rel}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n}{b}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mf}{0.000001}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
x =
3.0000
-2.5000
7.0000
ea =
1.8289e-07
2.1984e-08
2.3864e-08
iter = 8
x =
3.0000
-2.5000
7.0000
ea =
1.9130e-08
7.6449e-08
3.3378e-08
iter = 8
\end{Verbatim}
\subsection{Nonlinear Systems}\label{nonlinear-systems}
Consider two simultaneous nonlinear equations with two unknowns:
$x_{1}^{2}+x_{1}x_{2}=10$
$x_{2}+3x_{1}x_{2}^{2}=57$
Graphically, we are looking for the solution:
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}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_29_0.pdf}
\end{center}
{ \hspace*{\fill} \\}
\subsection{Newton-Raphson part II}\label{newton-raphson-part-ii}
Remember the first order approximation for the next point in a function
is:
$f(x_{i+1})=f(x_{i})+(x_{i+1}-x_{i})f'(x_{i})$
then, $f(x_{i+1})=0$ so we are left with:
$x_{i+1}=x_{i}-\frac{f(x_{i})}{f'(x_{i})}$
We can use the same formula, but now we have multiple dimensions so we
need to determine the Jacobian
$[J]=\left[ \begin{array}{cccc} \frac{\partial f_{1,i}}{\partial x_{1}} & \frac{\partial f_{1,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{1,i}}{\partial x_{n}} \\ \frac{\partial f_{2,i}}{\partial x_{1}} & \frac{\partial f_{2,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{2,i}}{\partial x_{n}} \\ \vdots & \vdots & & \vdots \\ \frac{\partial f_{n,i}}{\partial x_{1}} & \frac{\partial f_{n,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{n,i}}{\partial x_{n}} \\ \end{array} \right]$
$\left[ \begin{array}{c} f_{1,i+1} \\ f_{2,i+1} \\ \vdots \\ f_{n,i+1}\end{array} \right]= \left[ \begin{array}{c} f_{1,i} \\ f_{2,i} \\ \vdots \\ f_{n,i}\end{array} \right]+ \left[ \begin{array}{cccc} \frac{\partial f_{1,i}}{\partial x_{1}} & \frac{\partial f_{1,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{1,i}}{\partial x_{n}} \\ \frac{\partial f_{2,i}}{\partial x_{1}} & \frac{\partial f_{2,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{2,i}}{\partial x_{n}} \\ \vdots & \vdots & & \vdots \\ \frac{\partial f_{n,i}}{\partial x_{1}} & \frac{\partial f_{n,i}}{\partial x_{2}} & \cdots & \frac{\partial f_{n,i}}{\partial x_{n}} \\ \end{array} \right] \left( \left[ \begin{array}{c} x_{i+1} \\ x_{i+1} \\ \vdots \\ x_{i+1}\end{array} \right]- \left[ \begin{array}{c} x_{1,i} \\ x_{2,i} \\ \vdots \\ x_{n,i}\end{array} \right]\right)$
\subsubsection{Solution is again in the form
Ax=b}\label{solution-is-again-in-the-form-axb}
$[J]([x_{i+1}]-[x_{i}])=-[f]$
so
$[x_{i+1}]= [x_{i}]-[J]^{-1}[f]$
\subsection{Example of Jacobian
calculation}\label{example-of-jacobian-calculation}
\subsubsection{Nonlinear springs supporting two masses in
series}\label{nonlinear-springs-supporting-two-masses-in-series}
Two springs are connected to two masses, with $m_1$=1 kg and
$m_{2}$=2 kg. The springs are identical, but they have nonlinear
spring constants, of $k_1$=100 N/m and $k_2$=-10 N/m
We want to solve for the final position of the masses ($x_1$ and
$x_2$)
$m_{1}g+k_{1}(x_{2}-x_{1})+k_{2}(x_{2}-x_{1})^{2}+k_{1}x_{1}+k_{2}x_{1}^{2}=0$
$m_{2}g-k_{1}(x_{2}-x_{1})-k_{2}(x_2-x_1)^{2}=0$
$J(1,1)=\frac{\partial f_{1}}{\partial x_{1}}=-k_{1}-2k_{2}(x_{2}-x_{1})+k_{1}+2k_{2}x_{1}$
$J(1,2)=\frac{\partial f_1}{\partial x_{2}}=k_{1}+2k_{2}(x_{2}-x_{1})$
$J(2,1)=\frac{\partial f_2}{\partial x_{1}}=k_{1}+2k_{2}(x_{2}-x_{1})$
$J(2,2)=\frac{\partial f_2}{\partial x_{2}}=-k_{1}-2k_{2}(x_{2}-x_{1})$
\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}{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}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}20}]:} \PY{k}{function}\PY{+w}{ }[f,J]\PY{p}{=}\PY{n+nf}{mass\PYZus{}spring}\PY{p}{(}x\PY{p}{)}
\PY{+w}{ }\PY{c}{\PYZpc{} Function to calculate function values f1 and f2 as well as Jacobian }
\PY{c}{\PYZpc{} for 2 masses and 2 identical nonlinear springs}
\PY{n}{m1}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} kg }
\PY{n}{m2}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;} \PY{c}{\PYZpc{} kg}
\PY{n}{k1}\PY{p}{=}\PY{l+m+mi}{100}\PY{p}{;} \PY{c}{\PYZpc{} N/m}
\PY{n}{k2}\PY{p}{=}\PY{o}{\PYZhy{}}\PY{l+m+mi}{10}\PY{p}{;} \PY{c}{\PYZpc{} N/m\PYZca{}2}
\PY{n}{g}\PY{p}{=}\PY{l+m+mf}{9.81}\PY{p}{;} \PY{c}{\PYZpc{} m/s\PYZca{}2}
\PY{n}{x1}\PY{p}{=}\PY{n}{x}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
\PY{n}{x2}\PY{p}{=}\PY{n}{x}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{;}
\PY{n}{J}\PY{p}{=}\PY{p}{[}\PY{o}{\PYZhy{}}\PY{n}{k1}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k2}\PY{o}{*}\PY{p}{(}\PY{n}{x2}\PY{o}{\PYZhy{}}\PY{n}{x1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{k1}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k2}\PY{o}{*}\PY{n}{x1}\PY{p}{,}\PY{n}{k1}\PY{o}{+}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k2}\PY{o}{*}\PY{p}{(}\PY{n}{x2}\PY{o}{\PYZhy{}}\PY{n}{x1}\PY{p}{)}\PY{p}{;}
\PY{n}{k1}\PY{o}{+}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k2}\PY{o}{*}\PY{p}{(}\PY{n}{x2}\PY{o}{\PYZhy{}}\PY{n}{x1}\PY{p}{)}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{n}{k1}\PY{o}{\PYZhy{}}\PY{l+m+mi}{2}\PY{o}{*}\PY{n}{k2}\PY{o}{*}\PY{p}{(}\PY{n}{x2}\PY{o}{\PYZhy{}}\PY{n}{x1}\PY{p}{)}\PY{p}{]}\PY{p}{;}
\PY{n}{f}\PY{p}{=}\PY{p}{[}\PY{n}{m1}\PY{o}{*}\PY{n}{g}\PY{o}{+}\PY{n}{k1}\PY{o}{*}\PY{p}{(}\PY{n}{x2}\PY{o}{\PYZhy{}}\PY{n}{x1}\PY{p}{)}\PY{o}{+}\PY{n}{k2}\PY{o}{*}\PY{p}{(}\PY{n}{x2}\PY{o}{\PYZhy{}}\PY{n}{x1}\PY{p}{)}\PY{o}{.\PYZca{}}\PY{l+m+mi}{2}\PY{o}{\PYZhy{}}\PY{n}{k1}\PY{o}{*}\PY{n}{x1}\PY{o}{\PYZhy{}}\PY{n}{k2}\PY{o}{*}\PY{n}{x1}\PYZca{}\PY{l+m+mi}{2}\PY{p}{;}
\PY{n}{m2}\PY{o}{*}\PY{n}{g}\PY{o}{\PYZhy{}}\PY{n}{k1}\PY{o}{*}\PY{p}{(}\PY{n}{x2}\PY{o}{\PYZhy{}}\PY{n}{x1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{k2}\PY{o}{*}\PY{p}{(}\PY{n}{x2}\PY{o}{\PYZhy{}}\PY{n}{x1}\PY{p}{)}\PY{o}{.\PYZca{}}\PY{l+m+mi}{2}\PY{p}{]}\PY{p}{;}
\PY{k}{end}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}21}]:} \PY{p}{[}\PY{n}{f}\PY{p}{,}\PY{n}{J}\PY{p}{]}\PY{p}{=}\PY{n}{mass\PYZus{}spring}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
f =
-190.19
129.62
J =
-200 120
120 -120
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}22}]:} \PY{n}{x0}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{2}\PY{p}{]}\PY{p}{;}
\PY{p}{[}\PY{n}{f0}\PY{p}{,}\PY{n}{J0}\PY{p}{]}\PY{p}{=}\PY{n}{mass\PYZus{}spring}\PY{p}{(}\PY{n}{x0}\PY{p}{)}\PY{p}{;}
\PY{n}{x1}\PY{p}{=}\PY{n}{x0}\PY{o}{\PYZhy{}}\PY{n}{J0}\PY{o}{\PYZbs{}}\PY{n}{f0}
\PY{n}{ea}\PY{p}{=}\PY{p}{(}\PY{n}{x1}\PY{o}{\PYZhy{}}\PY{n}{x0}\PY{p}{)}\PY{o}{./}\PY{n}{x1}
\PY{p}{[}\PY{n}{f1}\PY{p}{,}\PY{n}{J1}\PY{p}{]}\PY{p}{=}\PY{n}{mass\PYZus{}spring}\PY{p}{(}\PY{n}{x1}\PY{p}{)}\PY{p}{;}
\PY{n}{x2}\PY{p}{=}\PY{n}{x1}\PY{o}{\PYZhy{}}\PY{n}{J1}\PY{o}{\PYZbs{}}\PY{n}{f1}
\PY{n}{ea}\PY{p}{=}\PY{p}{(}\PY{n}{x2}\PY{o}{\PYZhy{}}\PY{n}{x1}\PY{p}{)}\PY{o}{./}\PY{n}{x2}
\PY{p}{[}\PY{n}{f2}\PY{p}{,}\PY{n}{J2}\PY{p}{]}\PY{p}{=}\PY{n}{mass\PYZus{}spring}\PY{p}{(}\PY{n}{x2}\PY{p}{)}\PY{p}{;}
\PY{n}{x3}\PY{p}{=}\PY{n}{x2}\PY{o}{\PYZhy{}}\PY{n}{J2}\PY{o}{\PYZbs{}}\PY{n}{f2}
\PY{n}{ea}\PY{p}{=}\PY{p}{(}\PY{n}{x3}\PY{o}{\PYZhy{}}\PY{n}{x2}\PY{p}{)}\PY{o}{./}\PY{n}{x3}
\PY{n}{x}\PY{p}{=}\PY{n}{x3}
\PY{k}{for} \PY{n}{i}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{3}
\PY{n}{xold}\PY{p}{=}\PY{n}{x}\PY{p}{;}
\PY{p}{[}\PY{n}{f}\PY{p}{,}\PY{n}{J}\PY{p}{]}\PY{p}{=}\PY{n}{mass\PYZus{}spring}\PY{p}{(}\PY{n}{x}\PY{p}{)}\PY{p}{;}
\PY{n}{x}\PY{p}{=}\PY{n}{x}\PY{o}{\PYZhy{}}\PY{n}{J}\PY{o}{\PYZbs{}}\PY{n}{f}\PY{p}{;}
\PY{n}{ea}\PY{p}{=}\PY{p}{(}\PY{n}{x}\PY{o}{\PYZhy{}}\PY{n}{xold}\PY{p}{)}\PY{o}{./}\PY{n}{x}
\PY{k}{end}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
x1 =
-1.5142
-1.4341
ea =
2.9812
2.3946
x2 =
0.049894
0.248638
ea =
31.3492
6.7678
x3 =
0.29701
0.49722
ea =
0.83201
0.49995
x =
0.29701
0.49722
ea =
0.021392
0.012890
ea =
1.4786e-05
8.9091e-06
ea =
7.0642e-12
4.2565e-12
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{\color{incolor}In [{\color{incolor}23}]:} \PY{n}{x}
\PY{n}{X0}\PY{p}{=}\PY{n+nb}{fsolve}\PY{p}{(}\PY{p}{@}\PY{p}{(}\PY{n}{x}\PY{p}{)} \PY{n}{mass\PYZus{}spring}\PY{p}{(}\PY{n}{x}\PY{p}{)}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{5}\PY{p}{]}\PY{p}{)}
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\begin{Verbatim}[commandchars=\\\{\}]
x =
0.30351
0.50372
X0 =
0.30351
0.50372
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}26}]:} \PY{p}{[}\PY{n}{X}\PY{p}{,}\PY{n}{Y}\PY{p}{]}\PY{p}{=}\PY{n+nb}{meshgrid}\PY{p}{(}\PY{n+nb}{linspace}\PY{p}{(}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{10}\PY{p}{,}\PY{l+m+mi}{20}\PY{p}{)}\PY{p}{,}\PY{n+nb}{linspace}\PY{p}{(}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{10}\PY{p}{,}\PY{l+m+mi}{20}\PY{p}{)}\PY{p}{)}\PY{p}{;}
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\PY{n}{F}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{n+nb}{size}\PY{p}{(}\PY{n}{X}\PY{p}{)}\PY{p}{)}\PY{p}{;}
\PY{k}{for} \PY{n}{i}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{n}{N}
\PY{k}{for} \PY{n}{j}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{n}{M}
\PY{p}{[}\PY{n}{f}\PY{p}{,}\PY{o}{\PYZti{}}\PY{p}{]}\PY{p}{=}\PY{n}{mass\PYZus{}spring}\PY{p}{(}\PY{p}{[}\PY{n}{X}\PY{p}{(}\PY{n}{i}\PY{p}{,}\PY{n}{j}\PY{p}{)}\PY{p}{,}\PY{n}{Y}\PY{p}{(}\PY{n}{i}\PY{p}{,}\PY{n}{j}\PY{p}{)}\PY{p}{]}\PY{p}{)}\PY{p}{;}
\PY{n}{F1}\PY{p}{(}\PY{n}{i}\PY{p}{,}\PY{n}{j}\PY{p}{)}\PY{p}{=}\PY{n}{f}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;}
\PY{n}{F2}\PY{p}{(}\PY{n}{i}\PY{p}{,}\PY{n}{j}\PY{p}{)}\PY{p}{=}\PY{n}{f}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{;}
\PY{k}{end}
\PY{k}{end}
\PY{n+nb}{mesh}\PY{p}{(}\PY{n}{X}\PY{p}{,}\PY{n}{Y}\PY{p}{,}\PY{n}{F1}\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}{)}
\PY{n+nb}{colorbar}\PY{p}{(}\PY{p}{)}
\PY{n+nb}{figure}\PY{p}{(}\PY{p}{)}
\PY{n+nb}{mesh}\PY{p}{(}\PY{n}{X}\PY{p}{,}\PY{n}{Y}\PY{p}{,}\PY{n}{F2}\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}{)}
\PY{n+nb}{colorbar}\PY{p}{(}\PY{p}{)}
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\begin{center}
\adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_13_files/lecture_13_36_0.pdf}
\end{center}
{ \hspace*{\fill} \\}
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\end{center}
{ \hspace*{\fill} \\}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor} }]:}
\end{Verbatim}
% Add a bibliography block to the postdoc
\end{document}