--- a/day1/session4.tex	Thu Nov 05 13:51:00 2009 +0530
+++ b/day1/session4.tex	Thu Nov 05 14:07:26 2009 +0530
@@ -74,7 +74,7 @@
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 % Title page
-\title[Matrices \& Equations]{Python for Science and Engg: Matrices \& Solution of equations}
+\title[Matrices \& Equations]{Python for Science and Engg: Matrices, Least Square Fit, \& Solution of equations}
 
 \author[FOSSEE] {FOSSEE}
 
@@ -128,11 +128,9 @@
 
 \begin{frame}
 \frametitle{Matrices: Introduction}
-Let us now look at matrices in detail.\\
 \alert{All matrix operations are done using \kwrd{arrays}}
 \end{frame}
 
-\subsection{Initializing}
 \begin{frame}[fragile]
 \frametitle{Matrices: Initializing}
 \begin{lstlisting}
@@ -149,6 +147,104 @@
 \end{lstlisting}
 \end{frame}
 
+\begin{frame}[fragile]
+  \frametitle{Accessing elements}
+  \begin{lstlisting}
+In []: C = array([[1,1,2],
+                  [2,4,1],
+                  [-1,3,7]])
+
+In []: C[1][2]
+Out[]: 1
+
+In []: C[1,2]
+Out[]: 1
+
+In []: C[1]
+Out[]: array([2, 4, 1])
+  \end{lstlisting}
+\end{frame}
+
+\begin{frame}[fragile]
+  \frametitle{Changing elements}
+  \begin{small}
+  \begin{lstlisting}
+In []: C[1,1] = -2
+In []: C
+Out[]: 
+array([[ 1,  1,  2],
+       [ 2, -2,  1],
+       [-1,  3,  7]])
+
+In []: C[1] = [0,0,0]
+In []: C
+Out[]: 
+array([[ 1,  1,  2],
+       [ 0,  0,  0],
+       [-1,  3,  7]])
+  \end{lstlisting}
+  \end{small}
+How to change one \alert{column}?
+\end{frame}
+
+\begin{frame}[fragile]
+  \frametitle{Slicing}
+\begin{small}
+  \begin{lstlisting}
+In []: C[:,1]
+Out[]: array([1, 0, 3])
+
+In []: C[1,:]
+Out[]: array([0, 0, 0])
+
+In []: C[0:2,:]
+Out[]: 
+array([[1, 1, 2],
+       [0, 0, 0]])
+
+In []: C[1:3,:]
+Out[]: 
+array([[ 0,  0,  0],
+       [-1,  3,  7]])
+  \end{lstlisting}
+\end{small}
+\end{frame}
+
+\begin{frame}[fragile]
+  \frametitle{Slicing \ldots}
+\begin{small}
+  \begin{lstlisting}
+In []: C[:2,:]
+Out[]: 
+array([[1, 1, 2],
+       [0, 0, 0]])
+
+In []: C[1:,:]
+Out[]: 
+array([[ 0,  0,  0],
+       [-1,  3,  7]])
+
+In []: C[1:,:2]
+Out[]: 
+array([[ 0,  0],
+       [-1,  3]])
+  \end{lstlisting}
+
+\end{small}
+\end{frame}
+
+\begin{frame}[fragile]
+  \frametitle{Striding}
+  \begin{lstlisting}
+  \end{lstlisting}
+\end{frame}
+
+\begin{frame}[fragile]
+  \frametitle{Slicing \& Striding Exercises}
+  \begin{lstlisting}
+  \end{lstlisting}
+\end{frame}
+
 \subsection{Basic Operations}
 
 \begin{frame}[fragile]
@@ -236,6 +332,7 @@
 \end{lstlisting}
 \end{frame}
 
+%%use S=array(X,Y)
 \begin{frame}[fragile]
 \frametitle{Eigenvalues and Eigen Vectors}
 \begin{small}
@@ -255,185 +352,144 @@
 \end{small}
 \end{frame}
 
-\begin{frame}[fragile]
-\frametitle{Computing Norms}
-\begin{lstlisting}
-In []: norm(E)
-Out[]: 8.1240384046359608
-\end{lstlisting}
-\end{frame}
+%% \begin{frame}[fragile]
+%% \frametitle{Computing Norms}
+%% \begin{lstlisting}
+%% In []: norm(E)
+%% Out[]: 8.1240384046359608
+%% \end{lstlisting}
+%% \end{frame}
 
+%% \begin{frame}[fragile]
+%%   \frametitle{Singular Value Decomposition}
+%%   \begin{small}
+%%   \begin{lstlisting}
+%% In []: svd(E)
+%% Out[]: 
+%% (array(
+%% [[ -6.66666667e-01,  -1.23702565e-16,   7.45355992e-01],
+%%  [ -3.33333333e-01,  -8.94427191e-01,  -2.98142397e-01],
+%%  [ -6.66666667e-01,   4.47213595e-01,  -5.96284794e-01]]),
+%%  array([ 8.,  1.,  1.]),
+%%  array([[-0.66666667, -0.33333333, -0.66666667],
+%%         [-0.        ,  0.89442719, -0.4472136 ],
+%%         [-0.74535599,  0.2981424 ,  0.59628479]]))
+%%   \end{lstlisting}
+%%   \end{small}
+%% \inctime{15}
+%% \end{frame}
+
+\section{Least Squares Fit}
 \begin{frame}[fragile]
-  \frametitle{Singular Value Decomposition}
-  \begin{small}
-  \begin{lstlisting}
-In []: svd(E)
-Out[]: 
-(array(
-[[ -6.66666667e-01,  -1.23702565e-16,   7.45355992e-01],
- [ -3.33333333e-01,  -8.94427191e-01,  -2.98142397e-01],
- [ -6.66666667e-01,   4.47213595e-01,  -5.96284794e-01]]),
- array([ 8.,  1.,  1.]),
- array([[-0.66666667, -0.33333333, -0.66666667],
-        [-0.        ,  0.89442719, -0.4472136 ],
-        [-0.74535599,  0.2981424 ,  0.59628479]]))
-  \end{lstlisting}
-  \end{small}
-\inctime{15}
-\end{frame}
-
-\section{Solving linear equations}
-
-\begin{frame}[fragile]
-\frametitle{Solution of equations}
-Consider,
-  \begin{align*}
-    3x + 2y - z  & = 1 \\
-    2x - 2y + 4z  & = -2 \\
-    -x + \frac{1}{2}y -z & = 0
-  \end{align*}
-Solution:
-  \begin{align*}
-    x & = 1 \\
-    y & = -2 \\
-    z & = -2
-  \end{align*}
+\frametitle{$L$ vs. $T^2$}
+\vspace{-0.15in}
+\begin{figure}
+\includegraphics[width=4in]{data/L-Tsq-points.png}
+\end{figure}
 \end{frame}
 
 \begin{frame}[fragile]
-\frametitle{Solving using Matrices}
-Let us now look at how to solve this using \kwrd{matrices}
-  \begin{lstlisting}
-    In []: A = array([[3,2,-1],
-                      [2,-2,4],                   
-                      [-1, 0.5, -1]])
-    In []: b = array([[1], [-2], [0]])
-    In []: x = solve(A, b)
-    In []: Ax = dot(A,x)
-  \end{lstlisting}
-\end{frame}
-
-\begin{frame}[fragile]
-\frametitle{Solution:}
-\begin{lstlisting}
-In []: x
-Out[]: 
-array([[ 1.],
-       [-2.],
-       [-2.]])
-\end{lstlisting}
+\frametitle{$L$ vs. $T^2$}
+\vspace{-0.15in}
+\begin{figure}
+\includegraphics[width=4in]{data/L-Tsq-Line.png}
+\end{figure}
 \end{frame}
 
 \begin{frame}[fragile]
-\frametitle{Let's check!}
-\begin{lstlisting}
-In []: Ax
-Out[]: 
-array([[  1.00000000e+00],
-       [ -2.00000000e+00],
-       [  2.22044605e-16]])
-\end{lstlisting}
-\begin{block}{}
-The last term in the matrix is actually \alert{0}!\\
-We can use \kwrd{allclose()} to check.
-\end{block}
-\begin{lstlisting}
-In []: allclose(Ax, b)
-Out[]: True
-\end{lstlisting}
-\inctime{15}
+\frametitle{Least Squares Fit}
+\vspace{-0.15in}
+\begin{figure}
+\includegraphics[width=4in]{data/least-sq-fit.png}
+\end{figure}
 \end{frame}
 
-\subsection{Exercises}
-
-\begin{frame}[fragile]
-\frametitle{Problem 1}
-Given the matrix:\\
-\begin{center}
-\begin{bmatrix}
--2 & 2 & 3\\
- 2 & 1 & 6\\
--1 &-2 & 0\\
-\end{bmatrix}
-\end{center}
-Find:
+\begin{frame}
+\frametitle{Least Square Fit Curve}
 \begin{itemize}
-  \item[i] Transpose
-  \item[ii]Inverse
-  \item[iii]Determinant
-  \item[iv] Eigenvalues and Eigen vectors
-  \item[v] Singular Value decomposition
+\item $T^2$ and $L$ have a linear relationship
+\item Hence, Least Square Fit Curve is a line
+\item we shall use the \typ{lstsq} function
 \end{itemize}
 \end{frame}
 
 \begin{frame}[fragile]
-\frametitle{Problem 2}
-Given 
-\begin{center}
-A = 
-\begin{bmatrix}
--3 & 1 & 5 \\
-1 & 0 & -2 \\
-5 & -2 & 4 \\
-\end{bmatrix}
-, B = 
-\begin{bmatrix}
-0 & 9 & -12 \\
--9 & 0 & 20 \\
-12 & -20 & 0 \\
-\end{bmatrix}
-\end{center}
-Find:
+\frametitle{\typ{lstsq}}
 \begin{itemize}
-  \item[i] Sum of A and B
-  \item[ii]Elementwise Product of A and B
-  \item[iii] Matrix product of A and B
+\item We need to fit a line through points for the equation $T^2 = m \cdot L+c$
+\item The equation can be re-written as $T^2 = A \cdot p$
+\item where A is   
+  $\begin{bmatrix}
+  L_1 & 1 \\
+  L_2 & 1 \\
+  \vdots & \vdots\\
+  L_N & 1 \\
+  \end{bmatrix}$
+  and p is 
+  $\begin{bmatrix}
+  m\\
+  c\\
+  \end{bmatrix}$
+\item We need to find $p$ to plot the line
 \end{itemize}
 \end{frame}
 
 \begin{frame}[fragile]
-\frametitle{Solution}
-Sum: 
-\begin{bmatrix}
--3 & 10 & 7 \\
--8 & 0 & 18 \\
-17 & -22 & 4 \\
-\end{bmatrix}
-,\\ Elementwise Product:
-\begin{bmatrix}
-0 & 9 & -60 \\
--9 & 0 & -40 \\
-60 & 40 & 0 \\
-\end{bmatrix}
-,\\ Matrix product:
-\begin{bmatrix}
-51 & -127 & 56 \\
--24 & 49 & -12 \\
-66 & -35 & -100 \\
-\end{bmatrix}
+\frametitle{Generating $A$}
+\begin{lstlisting}
+In []: A = array([L, ones_like(L)])
+In []: A = A.T
+\end{lstlisting}
+\begin{small}
+\begin{block}{}
+  \begin{lstlisting}
+In []: ones((3,5))
+Out[]: 
+array([[ 1.,  1.,  1.,  1.,  1.],
+       [ 1.,  1.,  1.,  1.,  1.],
+       [ 1.,  1.,  1.,  1.,  1.]])
+
+In []: ones_like([1, 2, 3, 4, 5]) 
+Out[]: array([1, 1, 1, 1, 1])   
+  \end{lstlisting}
+Also available \alert{\typ{zeros, zeros_like, empty, empty_like}}
+\end{block}
+\end{small}
+
+%% \begin{itemize}
+%% \item A is also called a Van der Monde matrix
+%% \item It can also be generated using \typ{vander}
+%% \end{itemize}
+%% \begin{lstlisting}
+%% In []: A = vander(L, 2)
+%% \end{lstlisting}
 \end{frame}
 
 \begin{frame}[fragile]
-\frametitle{Problem 3}
-Solve the set of equations:
-\begin{align*}
-  x + y + 2z -w & = 3\\
-  2x + 5y - z - 9w & = -3\\
-  2x + y -z + 3w & = -11 \\
-  x - 3y + 2z + 7w & = -5\\
-\end{align*}
-\inctime{10}
+\frametitle{\typ{lstsq} \ldots}
+\begin{itemize}
+\item Now use the \typ{lstsq} function
+\item Along with a lot of things, it returns the least squares solution
+\end{itemize}
+\begin{lstlisting}
+In []: result = lstsq(A,TSq)
+In []: coef = result[0]
+\end{lstlisting}
 \end{frame}
 
+\subsection{Plotting}
 \begin{frame}[fragile]
-\frametitle{Solution}
-Use \kwrd{solve()}
-\begin{align*}
-  x & = -5\\
-  y & = 2\\
-  z & = 3\\
-  w & = 0\\
-\end{align*}
+\frametitle{Least Square Fit Line \ldots}
+We get the points of the line from \typ{coef}
+\begin{lstlisting}
+In []: Tline = coef[0]*L + coef[1]
+\end{lstlisting}
+\begin{itemize}
+\item Now plot Tline vs. L, to get the Least squares fit line. 
+\end{itemize}
+\begin{lstlisting}
+In []: plot(L, Tline)
+\end{lstlisting}
 \end{frame}
 
 \section{Summary}
@@ -442,16 +498,17 @@
   \begin{itemize}
   \item Matrices
     \begin{itemize}
+      \item Accessing elements
       \item Transpose
       \item Addition
       \item Multiplication
       \item Inverse of a matrix
       \item Determinant
       \item Eigenvalues and Eigen vector
-      \item Norms
-      \item Singular Value Decomposition
+      %% \item Norms
+      %% \item Singular Value Decomposition
     \end{itemize}
-  \item Solving linear equations
+  \item Least Square Curve fitting
   \end{itemize}
 \end{frame}