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diff --git a/Lectures_my/NumMet/Lecture6/mchrzasz.tex b/Lectures_my/NumMet/Lecture6/mchrzasz.tex
index 90d8dcd..4b958d9 100644
--- a/Lectures_my/NumMet/Lecture6/mchrzasz.tex
+++ b/Lectures_my/NumMet/Lecture6/mchrzasz.tex
@@ -206,7 +206,7 @@
 
 \author{ {\fontspec{Trebuchet MS}Marcin Chrz\k{a}szcz, Danny van Dyk} (UZH)}
 \institute{UZH}
-\title[Linear equation systems: exact methods]{Linear equation systems:exact methods}
+\title[Linear equation systems: iteration methods]{Linear equation systems:iteration methods}
 \date{21 September 2016}
 
 
@@ -220,7 +220,7 @@
 \begin{center}
 	\begin{columns}
 		\begin{column}{0.9\textwidth}
-			\flushright\fontspec{Trebuchet MS}\bfseries \Huge {Linear equation systems: exact methods}
+			\flushright\fontspec{Trebuchet MS}\bfseries \Huge {Linear equation systems: iteration methods}
 		\end{column}
 		\begin{column}{0.2\textwidth}
 		  %\includegraphics[width=\textwidth]{SHiP-2}
@@ -242,7 +242,7 @@
 \vspace{1em}
 %		\footnotesize\textcolor{gray}{With N. Serra, B. Storaci\\Thanks to the theory support from M. Shaposhnikov, D. Gorbunov}\normalsize\\
 \vspace{0.5em}
-	\textcolor{normal text.fg!50!Comment}{Numerical Methods, \\ 10 October, 2016}
+	\textcolor{normal text.fg!50!Comment}{Numerical Methods, \\ 17 October, 2016}
 \end{center}
 \end{frame}
 }
@@ -259,23 +259,17 @@
 \ARROW The problem: Find the $x$ vector.
 
 
-
 \end{frame}
 
-\begin{frame}\frametitle{Error digression}
+
+
+\begin{frame}\frametitle{LU method with main element selection}
 \begin{small}
-\ARROW There is enormous amount of ways to solve the linear equation system.\\
-\ARROW The choice of one over the other of them should be gathered by the {\it condition} of the matrix $A$ denoted at $cond(A)$.
-\ARROW If the $cond(A)$ is small we say that the problem is well conditioned, otherwise we say it's ill conditioned.\\
-\ARROW The {\it condition} relation is defined as:
-\begin{align*}
-cond(A) = \Vert A \Vert \cdot \Vert A^{-1} \Vert
-\end{align*}
-\ARROW Now there are many definitions of different norms... The most popular one (so-called ''column norm''):
-\begin{align*}
-\Vert A \vert_1 = \max_{1 \leq j \leq n} \sum_{i=1}^n \vert a_{i,j} \vert,
-\end{align*}
-where $n$ -is the dimension of $A$, $i,j$ are columns and rows numbers.
+\ARROW The algorithm for the LU matrix decomposition from previous lecture doesn't have  the main element selection. \\
+\ARROW This might be problematic. For example:
+
+\fixme
+
 
 
 
@@ -284,6 +278,297 @@
 
 
 
+\begin{frame}\frametitle{Chelosky decomposition}
+\begin{small}
+
+\ARROW If $A \in \textbf{R}^{N \times N}$ is a symmetric ($A^T=A$) and positively defined matrix, then there exits a special case of LU factorization such that:
+\begin{align*}
+A =C C^T
+\end{align*}
+where $C$ is a traingular matrixwith diagonal elements greater then zero. \\
+\ARROW Finding the Chelosky decomposition is two times faster the finding the LU decomposition. \\
+\ARROW The Chelosky decomposition has the same algorithm then the LU decomposition.
+
+
+
+\end{small}
+\end{frame}
+
+
+\begin{frame}\frametitle{LDL factorization}
+\begin{small}
+
+\ARROW If matrix can be Chelosky decomposed it can also be decomposed to:
+\begin{align*}
+A=LDL^T
+\end{align*}
+where $L$ is botton triangular matrix such that $\forall_i : l_{ii}=1$ and $D$ is diagonal matrix with positive elements. \\
+\ARROW The advantage of the LDL decomposition compared to Cehlosky decomposition is the fact that we don't need to square root in the calculations.
+
+
+
+\end{small}
+\end{frame}
+
+
+
+\begin{frame}\frametitle{Iterative methods}
+\begin{small}
+
+\ARROW The exact methods are the ones that require more computations to get the solutions.\\
+\ARROWR Because of this they are not really suited to solve big linear systems.\\
+\ARROW The iteration methods are simple and the main goal of them is to transform:
+\begin{align*}
+Ax=b
+\end{align*}
+to:
+\begin{align*}
+x= Mx + c
+\end{align*}
+where $A$,$M$ are matrices of $n \times n$ size. \\
+$b$ and $c$ are vectors of the size $n$.\\
+\ARROW Once we get the second system (btw. remember MC lectures?) we can use iterative methods to solve them.
+
+\end{small}
+\end{frame}
+
+
+\begin{frame}\frametitle{Expansion}
+\begin{small}
+
+\ARROW If $\overline{x}$ is the solution of the $A x = b$ system then also:
+\begin{align*}
+\overline{x} = M \overline{x} +c
+\end{align*}
+\ARROW We construct the a sequence that approximates the $\overline{x}$ in the following way:
+\begin{align*}
+x^{(k+1)} = M x^{(k)},~~~k=0,1,...
+\end{align*}
+\ARROW The necessary and sufficient requirement for the convergence of the set is:
+\begin{align*}
+\rho(M)<1
+\end{align*}
+
+
+
+\end{small}
+\end{frame}
+
+
+\begin{frame}\frametitle{Jakobi method}
+\begin{small}
+
+\ARROW The simplest method is the Jakobi method. \\
+\ARROW We start from decomposition of $A$ matrix:
+\begin{align*}
+A = L  + U +D
+\end{align*}
+where
+\begin{itemize}
+\item $L=(a_{ij})_{i>j}$ -  triangular lower matrix.
+\item $D=(a_{ik})_{i=j}$ - diagonal matrix.
+\item $U=(a_{ij})_{i<j}$ - triangular upper matrix.
+\end{itemize}
+\ARROW Now the new system:
+\begin{align*}
+(L+D+U)x =b
+\end{align*}
+\ARROW Translating them:
+\begin{align*}
+D x = - (L+U)x +b
+\end{align*}
+
+
+\end{small}
+\end{frame}
+
+\begin{frame}\frametitle{Jakobi method}
+\begin{small}
+
+\ARROW Now the $D$ matrix can be easily reverted (it is diagonal!):
+\begin{align*}
+x =  -D^{-1} (L+U) x  + D^{-1}b
+\end{align*}
+\ARROW So the iteration would have the form:
+\begin{align*}
+x^{(k+1)} =  -D^{-1} (L+U) x^{(k)} + D^{-1}b
+\end{align*}
+\ARROW The matrix:
+\begin{align*}
+M_J= - D^{-1}(L+U)
+\end{align*}
+is called the Jakobi matrix.
+
+
+\end{small}
+\end{frame}
+
+
+\begin{frame}\frametitle{Jakobi method}
+\begin{small}
+
+
+\ARROW The algorithm:
+\begin{itemize}
+\item Construct the matrix $A$ 
+\item Assume the precision of your calculations $\epsilon$.
+\item Decomposite the $A$ matrix into $L$, $D$, $U$.
+\item Calculate the Jocobi matrix $M_J$.
+\item Check the convergence of the method:
+\begin{itemize}
+\item Calculate the $\rho(M_J)$.
+\item Check the convergence conditions.
+\item If both are ok then continue, else stop and go home ;)
+\end{itemize}
+\item Choose the starting point $x^{(0)}$
+\item Calculate the $k+1$ point.
+\item Calculate the difference in each step:
+\begin{align*}
+\Vert x^{(k+1)} - x^{(k)} \Vert
+\end{align*}
+\item If above is smaller then $\epsilon$ the stop and assume $(k+1)$ is the solution.
+\end{itemize}
+
+
+
+\end{small}
+\end{frame}
+
+
+
+
+
+\begin{frame}\frametitle{Gauss-Seidle method}
+\begin{small}
+
+
+\ARROW A different iterative method is so-called Gauss-Seidle method.\\
+\ARROW We start from the previous: $(L+ D +U)x =b$.\\
+\ARROW Write the eq. in form:
+\begin{align*}
+(L+D)x = =Ux +b
+\end{align*}
+\ARROW The $(L+D)$ matrix is triangular and can easily be inverted.\\
+\ARROW From this one gets:
+\begin{align*}
+x= -(L+D)^{-1}  U x +(L+D)^{-1} b
+\end{align*}
+\ARROW So the iteration equation will take form:
+\begin{align*}
+x_i^{(k+1)} = -\frac{1}{a_{ii}} \left( \sum_{j<i} a_{ij} x_j^{(k+1)} + \sum _{i>i}a_{ij}x_{j}^{(k)} \right) + \frac{b_i}{a_{ii}}
+\end{align*}
+\ARROW The matrix:
+\begin{align*}
+M_{GS} = -(L+D)^{-1} U
+\end{align*}
+is so-called Gauss-Seidl matrix.
+
+\end{small}
+\end{frame}
+
+
+\begin{frame}\frametitle{Convergence of Gauss-Seidle and Jacobi methods}
+\begin{small}
+
+\begin{exampleblock}{Reminder:}
+The necessary and sufficient condition for the iteration to be convergence is that:
+\begin{align*}
+\rho(M)<1
+\end{align*}
+where $M$ is the matrix of a given method.
+\end{exampleblock}
+
+\ARROW Now calculating the $\rho(M)$ might be already a computationally expensive...\\
+\ARROW A very useful way of getting the $\rho(M_J)$ is:
+\begin{align*}
+r_J = \frac{\Vert x_{k+1} - x_k \Vert}{\Vert x_k - x_{k-1} \Vert} \approx \vert  \rho(M_J) -1 \vert
+\end{align*} 
+\ARROW Another useful equations:
+\begin{align*}
+\rho(M_{GS}) = \rho(M_J)^2
+\end{align*}
+
+
+
+\end{small}
+\end{frame}
+
+
+\begin{frame}\frametitle{Convergence of Gauss-Seidle and Jacobi methods}
+\begin{small}
+
+\begin{alertblock}{Theorem:}
+If matrix $(a_{ij})$ fulfils one of the conditions:\\
+\ARROW $\vert a_{ii} >  \sum_{j \neq i} \vert a_{ij} \vert~~ i=1,...,n$, so-called strong row criteria.\\
+\ARROW $\vert a_{jj} > \sum_{i \neq j} \vert a_{ij} \vert~~ j=1,...,n$, so-called strong column criteria.
+\end{alertblock}
+
+\begin{exampleblock}{Practical note:}
+\ARROWR One needs to note that calculating the $\rho(M)$ might be a as time consuming calculation as the solution is self...\\
+\ARROWR If that is the case usually one neglects this steps and computes the solution with extra care at each step checking the convergence.
+
+\end{exampleblock}
+
+
+\end{small}
+\end{frame}
+
+
+
+
+\begin{frame}\frametitle{SOR method}
+\begin{small}
+
+\ARROW The Successive Over-Relaxation method is an extension of the Gauss-Seidl methods. \\
+\ARROW The modification is that we can reuse the previous step to get a more precise approximation of the solution. \\
+\ARROW The ''hack'' happens when we calculate the $x^{(k+1)}_{GS}$ using the  Gauss-Seidl method. Instead assuming that the next step is the $x^{(k+1)}_{GS}$ we ''relax'' the condition using liner combination:
+\begin{align*}
+x^{(k+1)} =\omega x^{(k+1)}_{GS} + (1-\omega) x^{(k)}
+\end{align*}
+\ARROW From the above once can see that if $\omega=1$ then we end up with standard  Gauss-Seidl method. \\
+\ARROW The iteration method equation has the form:
+\begin{align*}
+x^{(k+1)} = (D+ \omega L)^{-1} ((1-\omega)D - \omega U)x^{(k)} + (D + \omega L)^{-1} \omega b
+\end{align*}
+
+
+
+
+\end{small}
+\end{frame}
+
+
+
+
+\begin{frame}\frametitle{SOR method, convergence}
+\begin{small}
+
+\ARROW The main difficulty in the SOR method is the fact we need to choose the $\omega$ parameter. \\
+\ARROW One can easily prove that if the method is converging the $\omega \in \left(0, 2 \right)$. \\
+\ARROW This is necessary condition but it's not enough! \\
+\ARROW If the $A$ is symmetric and positively defined  then the above condition is also sufficient!
+
+
+
+
+\end{small}
+\end{frame}
+
+
+
+
+
+\begin{frame}\frametitle{Conclusions}
+\begin{small}
+
+\ARROW We learned more exact method of solving the linear system.\\
+\ARROW For big matrix we need to use iterative method which are not exact.\\
+\ARROW My personal experience: Did not have big enough matrices to really need to apply the iterative methods but this is field dependent!
+
+
+\end{small}
+\end{frame}
+