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\begin{document}
\maketitle
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}1}]:} \PY{c}{\PYZpc{}plot \PYZhy{}\PYZhy{}format svg}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}2}]:} \PY{n}{setdefaults}
\end{Verbatim}
\subsection{My question from last
class}\label{my-question-from-last-class}
\(A=\left[ \begin{array}{ccc} 10 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 10\end{array} \right]\)
\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}
You can’t perform that action at this time.