# HG changeset patch # User Puneeth Chaganti # Date 1256713501 -19800 # Node ID 9ed9539446bc35ea361b8f3f0fd9cca9da60916a # Parent 19592f802dde480b64a3fe4bf06ac2b3247752c4 Updated day1 session5 and session6. diff -r 19592f802dde -r 9ed9539446bc day1/session5.tex --- a/day1/session5.tex Wed Oct 28 11:15:49 2009 +0530 +++ b/day1/session5.tex Wed Oct 28 12:35:01 2009 +0530 @@ -124,9 +124,148 @@ % \pausesections \end{frame} -\section{Integration} +\section{Interpolation} + +\begin{frame}[fragile] +\frametitle{Interpolation} +\begin{itemize} +\item Let us begin with interpolation +\item Let's use the L and T arrays and interpolate this data to obtain data at new points +\end{itemize} +\begin{lstlisting} +In []: L = [] +In []: T = [] +In []: for line in open('pendulum.txt'): + l, t = line.split() + L.append(float(l)) + T.append(float(t)) +In []: L = array(L) +In []: T = array(T) +In []: Tsq = T*T +\end{lstlisting} +\end{frame} + +%% \begin{frame}[fragile] +%% \frametitle{Interpolation \ldots} +%% \begin{small} +%% \typ{In []: from scipy.interpolate import interp1d} +%% \end{small} +%% \begin{itemize} +%% \item The \typ{interp1d} function returns a function +%% \begin{lstlisting} +%% In []: f = interp1d(L, T) +%% \end{lstlisting} +%% \item Functions can be assigned to variables +%% \item This function interpolates between known data values to obtain unknown +%% \end{itemize} +%% \end{frame} + +%% \begin{frame}[fragile] +%% \frametitle{Interpolation \ldots} +%% \begin{lstlisting} +%% In []: Ln = arange(0.1,0.99,0.005) +%% # Interpolating! +%% # The new values in range of old data +%% In []: plot(L, T, 'o', Ln, f(Ln), '-') +%% In []: f = interp1d(L, T, kind='cubic') +%% # When kind not specified, it's linear +%% # Others are ... +%% # 'nearest', 'zero', +%% # 'slinear', 'quadratic' +%% \end{lstlisting} +%% \end{frame} + +\begin{frame}[fragile] +\frametitle{Spline Interpolation} +\begin{small} +\begin{lstlisting} +In []: from scipy.interpolate import splrep +In []: from scipy.interpolate import splev +\end{lstlisting} +\end{small} +\begin{itemize} +\item Involves two steps + \begin{enumerate} + \item Find out the spline curve, coefficients + \item Evaluate the spline at new points + \end{enumerate} +\end{itemize} +\end{frame} -\subsection{Quadrature} +\begin{frame}[fragile] +\frametitle{\typ{splrep}} +To find the B-spline representation +\begin{lstlisting} +In []: tck = splrep(L, T) +\end{lstlisting} +Returns a tuple containing +\begin{enumerate} +\item the vector of knots, +\item the B-spline coefficients +\item the degree of the spline (default=3) +\end{enumerate} +\end{frame} + +\begin{frame}[fragile] +\frametitle{\typ{splev}} +To Evaluate a B-spline and it's derivatives +\begin{lstlisting} +In []: Lnew = arange(0.1,1,0.005) +In []: Tnew = splev(Lnew, tck) + +#To obtain derivatives of the spline +#use der=1, 2,.. for 1st, 2nd,.. order +In []: Tnew = splev(Lnew, tck, der=1) +\end{lstlisting} +\end{frame} + +\begin{frame}[fragile] +\frametitle{} +\end{frame} + +\section{Differentiation} + +\begin{frame}[fragile] +\frametitle{Numerical Differentiation} +\begin{itemize} +\item Given function $f(x)$ or data points $y=f(x)$ +\item We wish to calculate $f^{'}(x)$ at points $x$ +\item Taylor series - finite difference approximations +\end{itemize} +\begin{center} +\begin{tabular}{l l} +$f(x+h)=f(x)+h.f^{'}(x)$ &Forward \\ +$f(x-h)=f(x)-h.f^{'}(x)$ &Backward +\end{tabular} +\end{center} +\end{frame} + +\begin{frame}[fragile] +\frametitle{Forward Difference} +\begin{lstlisting} +In []: x = linspace(0, 2*pi, 100) +In []: y = sin(x) +In []: deltax = x[1] - x[0] +\end{lstlisting} +Obtain the finite forward difference of y +\end{frame} + +\begin{frame}[fragile] +\frametitle{Forward Difference \ldots} +\begin{lstlisting} +In []: fD = (y[1:] - y[:-1]) / deltax +In []: plot(x, y, x[:-1], fD) +\end{lstlisting} +\begin{center} + \includegraphics[height=2in, interpolate=true]{data/fwdDiff} +\end{center} +\end{frame} + +\begin{frame}[fragile] +\frametitle{} +\end{frame} + +\section{Quadrature} \begin{frame}[fragile] \frametitle{Quadrature} @@ -135,7 +274,9 @@ \item Area under $(sin(x) + x^2)$ in $(0,1)$ \item scipy has functions to do that \end{itemize} -\small{\typ{In []: from scipy.integrate import quad}} +\begin{small} + \typ{In []: from scipy.integrate import quad} +\end{small} \begin{itemize} \item Inputs - function to integrate, limits \end{itemize} @@ -143,12 +284,15 @@ In []: x = 0 In []: quad(sin(x)+x**2, 0, 1) \end{lstlisting} +\begin{small} \alert{\typ{error:}} \typ{First argument must be a callable function.} +\end{small} \end{frame} \begin{frame}[fragile] \frametitle{Functions - Definition} +We have been using them all along. Now let's see how to define them. \begin{lstlisting} In []: def f(x): return sin(x)+x**2 @@ -233,69 +377,11 @@ \end{itemize} \end{frame} -\subsection{ODEs} - -\begin{frame}[fragile] -\frametitle{ODE Integration} -We shall use the simple ODE of a simple pendulum. -\begin{equation*} -\ddot{\theta} = -\frac{g}{L}sin(\theta) -\end{equation*} -\begin{itemize} -\item This equation can be written as a system of two first order ODEs -\end{itemize} -\begin{align} -\dot{\theta} &= \omega \\ -\dot{\omega} &= -\frac{g}{L}sin(\theta) \\ - \text{At}\ t &= 0 : \nonumber \\ - \theta = \theta_0\quad & \&\quad \omega = 0 \nonumber -\end{align} -\end{frame} - -\begin{frame}[fragile] -\frametitle{Solving ODEs using SciPy} -\begin{itemize} -\item We use the \typ{odeint} function from scipy to do the integration -\item Define a function as below -\end{itemize} -\begin{lstlisting} -In []: def pend_int(unknown, t, p): - .... theta, omega = unknown - .... g, L = p - .... f=[omega, -(g/L)*sin(theta)] - .... return f - .... -\end{lstlisting} -\end{frame} - -\begin{frame}[fragile] -\frametitle{Solving ODEs using SciPy \ldots} -\begin{itemize} -\item \typ{t} is the time variable \\ -\item \typ{p} has the constants \\ -\item \typ{initial} has the initial values -\end{itemize} -\begin{lstlisting} -In []: t = linspace(0, 10, 101) -In []: p=(-9.81, 0.2) -In []: initial = [10*2*pi/360, 0] -\end{lstlisting} -\end{frame} - -\begin{frame}[fragile] -\frametitle{Solving ODEs using SciPy \ldots} - -\small{\typ{In []: from scipy.integrate import odeint}} -\begin{lstlisting} -In []: pend_sol = odeint(pend_int, - initial,t, - args=(p,)) -\end{lstlisting} -\end{frame} - \begin{frame} \frametitle{Things we have learned} \begin{itemize} + \item Interpolation + \item Differentiation \item Functions \begin{itemize} \item Definition @@ -306,6 +392,5 @@ \item Quadrature \end{itemize} \end{frame} - \end{document} diff -r 19592f802dde -r 9ed9539446bc day1/session6.tex --- a/day1/session6.tex Wed Oct 28 11:15:49 2009 +0530 +++ b/day1/session6.tex Wed Oct 28 12:35:01 2009 +0530 @@ -73,7 +73,7 @@ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Title page -\title[]{Finding Roots} +\title[]{ODEs \& Finding Roots} \author[FOSSEE] {FOSSEE} @@ -123,6 +123,68 @@ %% % You might wish to add the option [pausesections] %% \end{frame} +\section{ODEs} + +\begin{frame}[fragile] +\frametitle{ODE Integration} +We shall use the simple ODE of a simple pendulum. +\begin{equation*} +\ddot{\theta} = -\frac{g}{L}sin(\theta) +\end{equation*} +\begin{itemize} +\item This equation can be written as a system of two first order ODEs +\end{itemize} +\begin{align} +\dot{\theta} &= \omega \\ +\dot{\omega} &= -\frac{g}{L}sin(\theta) \\ + \text{At}\ t &= 0 : \nonumber \\ + \theta = \theta_0\quad & \&\quad \omega = 0 \nonumber +\end{align} +\end{frame} + +\begin{frame}[fragile] +\frametitle{Solving ODEs using SciPy} +\begin{itemize} +\item We use the \typ{odeint} function from scipy to do the integration +\item Define a function as below +\end{itemize} +\begin{lstlisting} +In []: def pend_int(unknown, t, p): + .... theta, omega = unknown + .... g, L = p + .... f=[omega, -(g/L)*sin(theta)] + .... return f + .... +\end{lstlisting} +\end{frame} + +\begin{frame}[fragile] +\frametitle{Solving ODEs using SciPy \ldots} +\begin{itemize} +\item \typ{t} is the time variable \\ +\item \typ{p} has the constants \\ +\item \typ{initial} has the initial values +\end{itemize} +\begin{lstlisting} +In []: t = linspace(0, 10, 101) +In []: p=(-9.81, 0.2) +In []: initial = [10*2*pi/360, 0] +\end{lstlisting} +\end{frame} + +\begin{frame}[fragile] +\frametitle{Solving ODEs using SciPy \ldots} +\begin{small} + \typ{In []: from scipy.integrate import odeint} +\end{small} +\begin{lstlisting} +In []: pend_sol = odeint(pend_int, + initial,t, + args=(p,)) +\end{lstlisting} +\end{frame} + +\section{Finding Roots} \begin{frame}[fragile] \frametitle{Roots of $f(x)=0$}