125 

   126 \section{Least Squares Fit}

   127 \begin{frame}[fragile]

   128 \frametitle{$L$ vs. $T^2$ - Scatter}

   129 Linear trend visible.

   130 \vspace{-0.1in}

   131 \begin{figure}

   132 \includegraphics[width=4in]{data/L-Tsq-points}

   133 \end{figure}

   134 \end{frame}

   135 

   136 \begin{frame}[fragile]

   137 \frametitle{$L$ vs. $T^2$ - Line}

   138 This line does not make any mathematical sense.

   139 \vspace{-0.1in}

   140 \begin{figure}

   141 \includegraphics[width=4in]{data/L-Tsq-Line}

   142 \end{figure}

   143 \end{frame}

   144 

   145 \begin{frame}[fragile]

   146 \frametitle{$L$ vs. $T^2$ - Least Square Fit}

   147 This is what our intention is.

   148 \vspace{-0.1in}

   149 \begin{figure}

   150 \includegraphics[width=4in]{data/least-sq-fit}

   151 \end{figure}

   152 \end{frame}

   153 

   154 \begin{frame}[fragile]

   155 \frametitle{Matrix Formulation}

   156 \begin{itemize}

   157 \item We need to fit a line through points for the equation $T^2 = m \cdot L+c$

   158 \item In matrix form, the equation can be represented as $T_{sq} = A \cdot p$, where $T_{sq}$ is

   159   $\begin{bmatrix}

   160   T^2_1 \\

   161   T^2_2 \\

   162   \vdots\\

   163   T^2_N \\

   164   \end{bmatrix}$

   165 , A is   

   166   $\begin{bmatrix}

   167   L_1 & 1 \\

   168   L_2 & 1 \\

   169   \vdots & \vdots\\

   170   L_N & 1 \\

   171   \end{bmatrix}$

   172   and p is 

   173   $\begin{bmatrix}

   174   m\\

   175   c\\

   176   \end{bmatrix}$

   177 \item We need to find $p$ to plot the line

   178 \end{itemize}

   179 \end{frame}

   180 

   181 \begin{frame}[fragile]

   182 \frametitle{Getting $L$ and $T^2$}

   183 %If you \alert{closed} IPython after session 2

   184 \begin{lstlisting}

   185 In []: L, T = loadtxt('pendulum.txt', 

   186                       unpack=True)

   187 In []: tsq = T*T

   188 \end{lstlisting}

   189 \end{frame}

   190 

   191 \begin{frame}[fragile]

   192 \frametitle{Generating $A$}

   193 \begin{lstlisting}

   194 In []: A = array([L, ones_like(L)])

   195 In []: A = A.T

   196 \end{lstlisting}

   197 \end{frame}

   198 

   199 \begin{frame}[fragile]

   200 \frametitle{\typ{lstsq} \ldots}

   201 \begin{itemize}

   202 \item Now use the \typ{lstsq} function

   203 \item Along with a lot of things, it returns the least squares solution

   204 \end{itemize}

   205 \begin{lstlisting}

   206 In []: result = lstsq(A,tsq)

   207 In []: coef = result[0]

   208 \end{lstlisting}

   209 \end{frame}

   210 

   211 \begin{frame}[fragile]

   212 \frametitle{Least Square Fit Line \ldots}

   213 We get the points of the line from \typ{coef}

   214 \begin{lstlisting}

   215 In []: Tline = coef[0]*L + coef[1]

   216 

   217 In []: Tline.shape

   218 \end{lstlisting}

   219 \begin{itemize}

   220 \item Now plot \typ{Tline} vs. \typ{L}, to get the Least squares fit line. 

   221 \end{itemize}

   222 \begin{lstlisting}

   223 In []: plot(L, Tline, 'r')

   224 

   225 In []: plot(L, tsq, 'o')

   226 \end{lstlisting}

   227 \end{frame}

   228 

   229 \begin{frame}[fragile]

   230 \frametitle{Least Squares Fit}

   231 \vspace{-0.15in}

   232 \begin{figure}

   233 \includegraphics[width=4in]{data/least-sq-fit}

   234 \end{figure}

   235 \end{frame}

   630   \item Least Square Fit