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ME3255S2017/lecture_11/lecture_11.tex
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% Default to the notebook output style | |
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\maketitle | |
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{\color{incolor}In [{\color{incolor}1}]:} \PY{c}{\PYZpc{}plot \PYZhy{}\PYZhy{}format svg} | |
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\subsection{My question from last | |
class}\label{my-question-from-last-class} | |
\(A=\left[ \begin{array}{ccc} 10 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 10\end{array} \right]\) | |
\begin{figure}[htbp] | |
\centering | |
\includegraphics{det_A.png} | |
\caption{responses to determinant of A} | |
\end{figure} | |
\subsection{Your questions from last | |
class}\label{your-questions-from-last-class} | |
\begin{enumerate} | |
\def\labelenumi{\arabic{enumi}.} | |
\item | |
Need more linear algebra review | |
-We will keep doing Linear Algebra, try the practice problems in | |
'linear\_algebra' | |
\item | |
How do I do HW3? | |
-demo today | |
\item | |
For hw4 is the spring constant (K) suppose to be given? | |
-yes, its 30 N/m | |
\item | |
Deapool or Joker? | |
\end{enumerate} | |
\subsection{Midterm preference}\label{midterm-preference} | |
\begin{figure}[htbp] | |
\centering | |
\includegraphics{midterm_date.png} | |
\caption{responses to midterm date} | |
\end{figure} | |
\subsubsection{Midterm Questions}\label{midterm-questions} | |
\begin{enumerate} | |
\def\labelenumi{\arabic{enumi}.} | |
\item | |
Notes allowed | |
-no | |
\item | |
Will there be a review/study sheet | |
-yes | |
\end{enumerate} | |
\section{LU Decomposition}\label{lu-decomposition} | |
\subsubsection{efficient storage of matrices for | |
solutions}\label{efficient-storage-of-matrices-for-solutions} | |
Considering the same solution set: | |
\(y=Ax\) | |
Assume that we can perform Gauss elimination and achieve this formula: | |
\(Ux=d\) | |
Where, \(U\) is an upper triangular matrix that we derived from Gauss | |
elimination and \(d\) is the set of dependent variables after Gauss | |
elimination. | |
Assume there is a lower triangular matrix, \(L\), with ones on the | |
diagonal and same dimensions of \(U\) and the following is true: | |
\(L(Ux-d)=Ax-y=0\) | |
Now, \(Ax=LUx\), so \(A=LU\), and \(y=Ld\). | |
\(2x_{1}+x_{2}=1\) | |
\(x_{1}+3x_{2}=1\) | |
\(\left[ \begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array} \right] \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right]= \left[\begin{array}{c} 1 \\ 1\end{array}\right]\) | |
f21=0.5 | |
A(2,1)=1-1 = 0 | |
A(2,2)=3-0.5=2.5 | |
y(2)=1-0.5=0.5 | |
\(L(Ux-d)= \left[ \begin{array}{cc} 1 & 0 \\ 0.5 & 1 \end{array} \right] \left(\left[ \begin{array}{cc} 2 & 1 \\ 0 & 2.5 \end{array} \right] \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right]- \left[\begin{array}{c} 1 \\ 0.5\end{array}\right]\right)=0\) | |
\begin{Verbatim}[commandchars=\\\{\}] | |
{\color{incolor}In [{\color{incolor}3}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]} | |
\PY{n}{L}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mf}{0.5}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{]} | |
\PY{n}{U}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mf}{2.5}\PY{p}{]} | |
\PY{n}{L}\PY{o}{*}\PY{n}{U} | |
\PY{n}{d}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mf}{0.5}\PY{p}{]} | |
\PY{n}{y}\PY{p}{=}\PY{n}{L}\PY{o}{*}\PY{n}{d} | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
A = | |
2 1 | |
1 3 | |
L = | |
1.00000 0.00000 | |
0.50000 1.00000 | |
U = | |
2.00000 1.00000 | |
0.00000 2.50000 | |
ans = | |
2 1 | |
1 3 | |
d = | |
1.00000 | |
0.50000 | |
y = | |
1 | |
1 | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
{\color{incolor}In [{\color{incolor}5}]:} \PY{c}{\PYZpc{} what is the determinant of L, U and A?} | |
\PY{n+nb}{det}\PY{p}{(}\PY{n}{A}\PY{p}{)} | |
\PY{n+nb}{det}\PY{p}{(}\PY{n}{L}\PY{p}{)} | |
\PY{n+nb}{det}\PY{p}{(}\PY{n}{U}\PY{p}{)} | |
\PY{n+nb}{det}\PY{p}{(}\PY{n}{L}\PY{p}{)}\PY{o}{*}\PY{n+nb}{det}\PY{p}{(}\PY{n}{U}\PY{p}{)} | |
\PY{n+nb}{det}\PY{p}{(}\PY{n}{L}\PY{o}{*}\PY{n}{U}\PY{p}{)} | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
ans = 5.0000 | |
ans = 1 | |
ans = 5 | |
ans = 5 | |
ans = 5.0000 | |
\end{Verbatim} | |
\subsection{Pivoting for LU | |
factorization}\label{pivoting-for-lu-factorization} | |
LU factorization uses the same method as Gauss elimination so it is also | |
necessary to perform partial pivoting when creating the lower and upper | |
triangular matrices. | |
Matlab and Octave use pivoting in the command | |
\texttt{{[}L,U,P{]}=lu(A)} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
{\color{incolor}In [{\color{incolor}4}]:} \PY{n}{help} \PY{n+nb}{lu} | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
'lu' is a built-in function from the file libinterp/corefcn/lu.cc | |
-- Built-in Function: [L, U] = lu (A) | |
-- Built-in Function: [L, U, P] = lu (A) | |
-- Built-in Function: [L, U, P, Q] = lu (S) | |
-- Built-in Function: [L, U, P, Q, R] = lu (S) | |
-- Built-in Function: [{\ldots}] = lu (S, THRES) | |
-- Built-in Function: Y = lu ({\ldots}) | |
-- Built-in Function: [{\ldots}] = lu ({\ldots}, "vector") | |
Compute the LU decomposition of A. | |
If A is full subroutines from LAPACK are used and if A is sparse | |
then UMFPACK is used. | |
The result is returned in a permuted form, according to the | |
optional return value P. For example, given the matrix 'a = [1, 2; | |
3, 4]', | |
[l, u, p] = lu (A) | |
returns | |
l = | |
1.00000 0.00000 | |
0.33333 1.00000 | |
u = | |
3.00000 4.00000 | |
0.00000 0.66667 | |
p = | |
0 1 | |
1 0 | |
The matrix is not required to be square. | |
When called with two or three output arguments and a spare input | |
matrix, 'lu' does not attempt to perform sparsity preserving column | |
permutations. Called with a fourth output argument, the sparsity | |
preserving column transformation Q is returned, such that 'P * A * | |
Q = L * U'. | |
Called with a fifth output argument and a sparse input matrix, 'lu' | |
attempts to use a scaling factor R on the input matrix such that 'P | |
* (R \textbackslash{} A) * Q = L * U'. This typically leads to a sparser and more | |
stable factorization. | |
An additional input argument THRES, that defines the pivoting | |
threshold can be given. THRES can be a scalar, in which case it | |
defines the UMFPACK pivoting tolerance for both symmetric and | |
unsymmetric cases. If THRES is a 2-element vector, then the first | |
element defines the pivoting tolerance for the unsymmetric UMFPACK | |
pivoting strategy and the second for the symmetric strategy. By | |
default, the values defined by 'spparms' are used ([0.1, 0.001]). | |
Given the string argument "vector", 'lu' returns the values of P | |
and Q as vector values, such that for full matrix, 'A (P,:) = L * | |
U', and 'R(P,:) * A (:, Q) = L * U'. | |
With two output arguments, returns the permuted forms of the upper | |
and lower triangular matrices, such that 'A = L * U'. With one | |
output argument Y, then the matrix returned by the LAPACK routines | |
is returned. If the input matrix is sparse then the matrix L is | |
embedded into U to give a return value similar to the full case. | |
For both full and sparse matrices, 'lu' loses the permutation | |
information. | |
See also: luupdate, ilu, chol, hess, qr, qz, schur, svd. | |
Additional help for built-in functions and operators is | |
available in the online version of the manual. Use the command | |
'doc <topic>' to search the manual index. | |
Help and information about Octave is also available on the WWW | |
at http://www.octave.org and via the help@octave.org | |
mailing list. | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
{\color{incolor}In [{\color{incolor}9}]:} \PY{c}{\PYZpc{} time LU solution vs backslash} | |
\PY{n}{t\PYZus{}lu}\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{n}{t\PYZus{}bs}\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}{N}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{100} | |
\PY{n}{A}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{n}{N}\PY{p}{)}\PY{p}{;} | |
\PY{n}{y}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} | |
\PY{p}{[}\PY{n}{L}\PY{p}{,}\PY{n}{U}\PY{p}{,}\PY{n}{P}\PY{p}{]}\PY{p}{=}\PY{n+nb}{lu}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{;} | |
\PY{n+nb}{tic}\PY{p}{;} \PY{n}{d}\PY{p}{=}\PY{n+nb}{inv}\PY{p}{(}\PY{n}{L}\PY{p}{)}\PY{o}{*}\PY{n}{y}\PY{p}{;} \PY{n}{x}\PY{p}{=}\PY{n+nb}{inv}\PY{p}{(}\PY{n}{U}\PY{p}{)}\PY{o}{*}\PY{n}{d}\PY{p}{;} \PY{n}{t\PYZus{}lu}\PY{p}{(}\PY{n}{N}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;} | |
\PY{n+nb}{tic}\PY{p}{;} \PY{n}{x}\PY{p}{=}\PY{n+nb}{inv}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{o}{*}\PY{n}{y}\PY{p}{;} \PY{n}{t\PYZus{}bs}\PY{p}{(}\PY{n}{N}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\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{p}{,}\PY{n}{t\PYZus{}lu}\PY{p}{,}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{:}\PY{l+m+mi}{100}\PY{p}{]}\PY{p}{,}\PY{n}{t\PYZus{}bs}\PY{p}{)} | |
\PY{n+nb}{legend}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{LU decomp\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{Octave \PYZbs{}\PYZbs{}\PYZsq{}}\PY{p}{)} | |
\end{Verbatim} | |
\begin{center} | |
\adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_11_files/lecture_11_8_0.pdf} | |
\end{center} | |
{ \hspace*{\fill} \\} | |
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.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: | |
\(m_{1}g+k(x_{2}-x_{1})-kx_{1}=0\) | |
\(m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0\) | |
\(m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0\) | |
\(m_{4}g-k(x_{4}-x_{3})=0\) | |
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}10}]:} \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} | |
This matrix, K, is symmetric. | |
\texttt{K(i,j)==K(j,i)} | |
Now we can use, | |
\subsection{Cholesky Factorization}\label{cholesky-factorization} | |
We can decompose the matrix, K into two matrices, \(U\) and \(U^{T}\), | |
where | |
\(K=U^{T}U\) | |
each of the components of U can be calculated with the following | |
equations: | |
\(u_{ii}=\sqrt{a_{ii}-\sum_{k=1}^{i-1}u_{ki}^{2}}\) | |
\(u_{ij}=\frac{a_{ij}-\sum_{k=1}^{i-1}u_{ki}u_{kj}}{u_{ii}}\) | |
so for K | |
\begin{Verbatim}[commandchars=\\\{\}] | |
{\color{incolor}In [{\color{incolor}11}]:} \PY{n}{K} | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
K = | |
20 -10 0 0 | |
-10 20 -10 0 | |
0 -10 20 -10 | |
0 0 -10 10 | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
{\color{incolor}In [{\color{incolor}12}]:} \PY{n}{u11}\PY{p}{=}\PY{n+nb}{sqrt}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{)} | |
\PY{n}{u12}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{o}{/}\PY{n}{u11} | |
\PY{n}{u13}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}\PY{o}{/}\PY{n}{u11} | |
\PY{n}{u14}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)}\PY{p}{)}\PY{o}{/}\PY{n}{u11} | |
\PY{n}{u22}\PY{p}{=}\PY{n+nb}{sqrt}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u12}\PYZca{}\PY{l+m+mi}{2}\PY{p}{)} | |
\PY{n}{u23}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u12}\PY{o}{*}\PY{n}{u13}\PY{p}{)}\PY{o}{/}\PY{n}{u22} | |
\PY{n}{u24}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u12}\PY{o}{*}\PY{n}{u14}\PY{p}{)}\PY{o}{/}\PY{n}{u22} | |
\PY{n}{u33}\PY{p}{=}\PY{n+nb}{sqrt}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u13}\PYZca{}\PY{l+m+mi}{2}\PY{o}{\PYZhy{}}\PY{n}{u23}\PYZca{}\PY{l+m+mi}{2}\PY{p}{)} | |
\PY{n}{u34}\PY{p}{=}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u13}\PY{o}{*}\PY{n}{u14}\PY{o}{\PYZhy{}}\PY{n}{u23}\PY{o}{*}\PY{n}{u24}\PY{p}{)}\PY{o}{/}\PY{n}{u33} | |
\PY{n}{u44}\PY{p}{=}\PY{n+nb}{sqrt}\PY{p}{(}\PY{n}{K}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{u14}\PYZca{}\PY{l+m+mi}{2}\PY{o}{\PYZhy{}}\PY{n}{u24}\PYZca{}\PY{l+m+mi}{2}\PY{o}{\PYZhy{}}\PY{n}{u34}\PYZca{}\PY{l+m+mi}{2}\PY{p}{)} | |
\PY{n}{U}\PY{p}{=}\PY{p}{[}\PY{n}{u11}\PY{p}{,}\PY{n}{u12}\PY{p}{,}\PY{n}{u13}\PY{p}{,}\PY{n}{u14}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{,}\PY{n}{u22}\PY{p}{,}\PY{n}{u23}\PY{p}{,}\PY{n}{u24}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{,}\PY{n}{u33}\PY{p}{,}\PY{n}{u34}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{,}\PY{n}{u44}\PY{p}{]} | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
u11 = 4.4721 | |
u12 = -2.2361 | |
u13 = 0 | |
u14 = 0 | |
u22 = 3.8730 | |
u23 = -2.5820 | |
u24 = 0 | |
u33 = 3.6515 | |
u34 = -2.7386 | |
u44 = 1.5811 | |
U = | |
4.47214 -2.23607 0.00000 0.00000 | |
0.00000 3.87298 -2.58199 0.00000 | |
0.00000 0.00000 3.65148 -2.73861 | |
0.00000 0.00000 0.00000 1.58114 | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
{\color{incolor}In [{\color{incolor}17}]:} \PY{p}{(}\PY{n}{U}\PY{o}{\PYZsq{}}\PY{o}{*}\PY{n}{U}\PY{p}{)}\PY{o}{\PYZsq{}}\PY{o}{==}\PY{n}{U}\PY{o}{\PYZsq{}}\PY{o}{*}\PY{n}{U} | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
ans = | |
1 1 1 1 | |
1 1 1 1 | |
1 1 1 1 | |
1 1 1 1 | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
{\color{incolor}In [{\color{incolor}18}]:} \PY{c}{\PYZpc{} time solution for Cholesky vs backslash} | |
\PY{n}{t\PYZus{}chol}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{1000}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} | |
\PY{n}{t\PYZus{}bs}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{1000}\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}{1000} | |
\PY{n+nb}{tic}\PY{p}{;} \PY{n}{d}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZsq{}}\PY{o}{*}\PY{n}{y}\PY{p}{;} \PY{n}{x}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d}\PY{p}{;} \PY{n}{t\PYZus{}chol}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;} | |
\PY{n+nb}{tic}\PY{p}{;} \PY{n}{x}\PY{p}{=}\PY{n}{K}\PY{o}{\PYZbs{}}\PY{n}{y}\PY{p}{;} \PY{n}{t\PYZus{}bs}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;} | |
\PY{k}{end} | |
\PY{n+nb}{fprintf}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{average time spent for Cholesky factored solution = \PYZpc{}e+/\PYZhy{}\PYZpc{}e\PYZsq{}}\PY{p}{,}\PY{n+nb}{mean}\PY{p}{(}\PY{n}{t\PYZus{}chol}\PY{p}{)}\PY{p}{,}\PY{n+nb}{std}\PY{p}{(}\PY{n}{t\PYZus{}chol}\PY{p}{)}\PY{p}{)} | |
\PY{n+nb}{fprintf}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{average time spent for backslash solution = \PYZpc{}e+/\PYZhy{}\PYZpc{}e\PYZsq{}}\PY{p}{,}\PY{n+nb}{mean}\PY{p}{(}\PY{n}{t\PYZus{}bs}\PY{p}{)}\PY{p}{,}\PY{n+nb}{std}\PY{p}{(}\PY{n}{t\PYZus{}bs}\PY{p}{)}\PY{p}{)} | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
average time spent for Cholesky factored solution = 1.465964e-05+/-9.806001e-06 | |
average time spent for backslash solution = 1.555967e-05+/-1.048561e-05 | |
\end{Verbatim} | |
\begin{Verbatim}[commandchars=\\\{\}] | |
{\color{incolor}In [{\color{incolor} }]:} | |
\end{Verbatim} | |
% Add a bibliography block to the postdoc | |
\end{document} |