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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Tutorial slides on Python.
%
% Author: FOSSEE
% Copyright (c) 2009, FOSSEE, IIT Bombay
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Taken from Fernando's slides.
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%%% This is from Fernando's setup.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Title page
\title[ODEs \& Root Finding]{Python for Science and Engg:\\ODEs \& Finding Roots}
\author[FOSSEE] {FOSSEE}
\institute[IIT Bombay] {Department of Aerospace Engineering\\IIT Bombay}
\date[] {31, October 2009\\Day 1, Session 6}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\pgfdeclareimage[height=0.75cm]{iitmlogo}{iitmlogo}
%\logo{\pgfuseimage{iitmlogo}}
%% Delete this, if you do not want the table of contents to pop up at
%% the beginning of each subsection:
\AtBeginSubsection[]
{
\begin{frame}<beamer>
\frametitle{Outline}
\tableofcontents[currentsection,currentsubsection]
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% DOCUMENT STARTS
\begin{document}
\begin{frame}
\maketitle
\end{frame}
%% \begin{frame}
%% \frametitle{Outline}
%% \tableofcontents
%% % 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 $f(x)=0$}
\begin{itemize}
\item Roots --- values of $x$ satisfying $f(x)=0$
\item $f(x)=0$ may have real or complex roots
\item Presently, let's look at real roots
\end{itemize}
\end{frame}
\begin{frame}[fragile]
\frametitle{Roots of $f(x)=0$}
\begin{itemize}
\item Given function $cosx-x^2$
\item First we find \alert{a} root in $(-\pi/2, \pi/2)$
\item Then we find \alert{all} roots in $(-\pi/2, \pi/2)$
\end{itemize}
\end{frame}
%% \begin{frame}[fragile]
%% \frametitle{Fixed Point Method}
%% \begin{enumerate}
%% \item Convert $f(x)=0$ to the form $x=g(x)$
%% \item Start with an initial value of $x$ and iterate successively
%% \item $x_{n+1}=g(x_n)$ and $x_0$ is our initial guess
%% \item Iterate until $x_{n+1}-x_n \le tolerance$
%% \end{enumerate}
%% \end{frame}
%% \begin{frame}[fragile]
%% \frametitle{Fixed Point \dots}
%% \begin{lstlisting}
%% In []: def our_g(x):
%% ....: return sqrt(cos(x))
%% ....:
%% In []: tolerance = 1e-5
%% In []: while abs(x1-x0)>tolerance:
%% ....: x0 = x1
%% ....: x1 = our_g(x1)
%% ....:
%% In []: x0
%% \end{lstlisting}
%% \end{frame}
\begin{frame}[fragile]
\frametitle{Bisection Method}
\begin{enumerate}
\item Start with the given interval $(-\pi/2, \pi/2)$ ($(a, b)$)
\item $f(a)\cdot f(b) < 0$
\item Bisect the interval. $c = \frac{a+b}{2}$
\item Change the interval to $(a, c)$ if $f(a)\cdot f(c) < 0$
\item Else, change the interval to $(c, b)$
\item Go back to 3 until $(b-a) \le$ tolerance
\end{enumerate}
\alert{\typ{tolerance = 1e-5}}
\end{frame}
%% \begin{frame}[fragile]
%% \frametitle{Bisection \dots}
%% \begin{lstlisting}
%% In []: tolerance = 1e-5
%% In []: a = -pi/2
%% In []: b = 0
%% In []: while b-a > tolerance:
%% ....: c = (a+b)/2
%% ....: if our_f(a)*our_f(c) < 0:
%% ....: b = c
%% ....: else:
%% ....: a = c
%% ....:
%% \end{lstlisting}
%% \end{frame}
\begin{frame}[fragile]
\frametitle{Newton-Raphson Method}
\begin{enumerate}
\item Start with an initial guess of x for the root
\item $\Delta x = -f(x)/f^{'}(x)$
\item $ x \leftarrow x + \Delta x$
\item Go back to 2 until $|\Delta x| \le$ tolerance
\end{enumerate}
\end{frame}
%% \begin{frame}[fragile]
%% \frametitle{Newton-Raphson \dots}
%% \begin{lstlisting}
%% In []: def our_df(x):
%% ....: return -sin(x)-2*x
%% ....:
%% In []: delx = 10*tolerance
%% In []: while delx > tolerance:
%% ....: delx = -our_f(x)/our_df(x)
%% ....: x = x + delx
%% ....:
%% ....:
%% \end{lstlisting}
%% \end{frame}
%% \begin{frame}[fragile]
%% \frametitle{Newton-Raphson \ldots}
%% \begin{itemize}
%% \item What if $f^{'}(x) = 0$?
%% \item Can you write a better version of the Newton-Raphson?
%% \item What if the differential is not easy to calculate?
%% \item Look at Secant Method
%% \end{itemize}
%% \end{frame}
\begin{frame}[fragile]
\frametitle{Initial Estimates}
\begin{itemize}
\item Given an interval
\item How to find \alert{all} the roots?
\end{itemize}
\begin{enumerate}
\item Check for change of signs of $f(x)$ in the given interval
\item A root lies in the interval where the sign change occurs
\end{enumerate}
\end{frame}
\begin{frame}[fragile]
\frametitle{Initial Estimates \ldots}
\begin{lstlisting}
In []: def our_f(x):
....: return cos(x)-x**2
....:
In []: x = linspace(-pi/2, pi/2, 11)
\end{lstlisting}
\begin{itemize}
\item Get the intervals of x, where sign changes occur
\end{itemize}
\end{frame}
\begin{frame}[fragile]
\frametitle{Initial Estimates \ldots}
\begin{lstlisting}
In []: pos = y[:-1]*y[1:] < 0
In []: rpos = zeros(x.shape, dtype=bool)
In []: rpos[:-1] = pos
In []: rpos[1:] += pos
In []: rts = x[rpos]
\end{lstlisting}
Now use Newton-Raphson?
\end{frame}
\begin{frame}[fragile]
\frametitle{Scipy Methods - \typ{roots}}
\begin{itemize}
\item Calculates the roots of polynomials
\item Array of coefficients is the only parameter
\end{itemize}
\begin{lstlisting}
In []: coeffs = [1, 6, 13]
In []: roots(coeffs)
\end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{Scipy Methods - \typ{fsolve}}
\begin{small}
\begin{lstlisting}
In []: from scipy.optimize import fsolve
\end{lstlisting}
\end{small}
\begin{itemize}
\item Finds the roots of a system of non-linear equations
\item Input arguments - Function and initial estimate
\item Returns the solution
\end{itemize}
\begin{lstlisting}
In []: fsolve(our_f, -pi/2)
\end{lstlisting}
\end{frame}
\begin{frame}[fragile]
\frametitle{Scipy Methods \dots}
\begin{small}
\begin{lstlisting}
In []: from scipy.optimize import fixed_point
In []: from scipy.optimize import bisect
In []: from scipy.optimize import newton
\end{lstlisting}
\end{small}
\end{frame}
\begin{frame}
\frametitle{Things we have learned}
\begin{itemize}
\item Solving ODEs
\item Finding Roots
\begin{itemize}
\item Estimating Interval
\item Newton-Raphson
\item Scipy methods
\end{itemize}
\end{itemize}
\end{frame}
\end{document}
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