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−%Tutorial slides on Python.
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−%
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−% Author: FOSSEE
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−% Copyright (c) 2009, FOSSEE, IIT Bombay
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−%\usepackage{times}
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−
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−% Taken from Fernando's slides.
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−
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−%%% This is from Fernando's setup.
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−% % Use and configure listings package for nicely formatted code
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−% }
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−
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−%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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−% Title page
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−\title[]{Finding Roots}
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−
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−\author[FOSSEE] {FOSSEE}
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−
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−\institute[IIT Bombay] {Department of Aerospace Engineering\\IIT Bombay}
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−\date[] {31, October 2009\\Day 1, Session 6}
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−%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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−
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−%\pgfdeclareimage[height=0.75cm]{iitmlogo}{iitmlogo}
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−%\logo{\pgfuseimage{iitmlogo}}
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−
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−
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−%% Delete this, if you do not want the table of contents to pop up at
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−%% the beginning of each subsection:
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−\AtBeginSubsection[]
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−{
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−  \begin{frame}<beamer>
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−    \frametitle{Outline}
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−    \tableofcontents[currentsection,currentsubsection]
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−}
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−{
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−  \begin{frame}<beamer>
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−    \frametitle{Outline}
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−    \tableofcontents[currentsection,currentsubsection]
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−}
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−
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−% If you wish to uncover everything in a step-wise fashion, uncomment
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−%\includeonlyframes{current,current1,current2,current3,current4,current5,current6}
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−
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−%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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−% DOCUMENT STARTS
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−\begin{document}
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−
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−\begin{frame}
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−  \maketitle
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−\end{frame}
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−
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−%% \begin{frame}
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−%%   \frametitle{Outline}
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−%%   \tableofcontents
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−%%   % You might wish to add the option [pausesections]
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−%% \end{frame}
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−
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−
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−\begin{frame}[fragile]
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−\frametitle{Roots of $f(x)=0$}
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−\begin{itemize}
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−\item Roots --- values of $x$ satisfying $f(x)=0$
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−\item $f(x)=0$ may have real or complex roots
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−\item Presently, let's look at real roots
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−\end{itemize}
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−\end{frame}
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−
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−\begin{frame}[fragile]
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−\frametitle{Initial Estimates}
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−\begin{itemize}
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−\item Find the roots of $cosx-x^2$ between $-\pi/2$ and $\pi/2$
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−\item We shall use a crude method to get an initial estimate first
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−\end{itemize}
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−\begin{enumerate}
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−\item Check for change of signs of $f(x)$ in the given interval
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−\item A root lies in the interval where the sign change occurs
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−\end{enumerate}
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−\end{frame}
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−
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−\begin{frame}[fragile]
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−\frametitle{Initial Estimates \ldots}
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−\begin{lstlisting}
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−  In []: def our_f(x):
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−   ....:     return cos(x)-x**2
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−   ....: 
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−  In []: x = linspace(-pi/2, pi/2, 11)
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−\end{lstlisting}
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−\begin{itemize}
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−\item Get the intervals of x, where sign changes occur
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−\end{itemize}
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−\end{frame}
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−
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−%% \begin{frame}[fragile]
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−%% \frametitle{Initial Estimates \ldots}
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−%% \begin{lstlisting}
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−%% In []: pos = y[:-1]*y[1:] < 0
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−%% In []: rpos = zeros(x.shape, dtype=bool)
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−%% In []: rpos[:-1] = pos
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−%% In []: rpos[1:] += pos
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−%% In []: rts = x[rpos]
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−%% \end{lstlisting}
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−%% \end{frame}
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−
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−\begin{frame}[fragile]
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−\frametitle{Fixed Point Method}
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−\begin{enumerate}
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−\item Convert $f(x)=0$ to the form $x=g(x)$
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−\item Start with an initial value of $x$ and iterate successively
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−\item $x_{n+1}=g(x_n)$ and $x_0$ is our initial guess
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−\item Iterate until $x_{n+1}-x_n \le tolerance$
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−\end{enumerate}
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−\end{frame}
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−
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−%% \begin{frame}[fragile]
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−%% \frametitle{Fixed Point \dots}
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−%% \begin{lstlisting}
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−%% In []: def our_g(x):
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−%%  ....:     return sqrt(cos(x))
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−%%  ....: 
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−%% In []: tolerance = 1e-5
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−%% In []: while abs(x1-x0)>tolerance:
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−%%  ....:     x0 = x1
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−%%  ....:     x1 = our_g(x1)
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−%%  ....:   
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−%% In []: x0
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−%% \end{lstlisting}
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−%% \end{frame}
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−
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−\begin{frame}[fragile]
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−\frametitle{Bisection Method}
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−\begin{enumerate}
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−\item Start with an interval $(a, b)$ within which a root exists
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−\item $f(a)\cdot f(b) < 0$
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−\item Bisect the interval. $c = \frac{a+b}{2}$
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−\item Change the interval to $(a, c)$ if $f(a)\cdot f(c) < 0$
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−\item Else, change the interval to $(c, b)$
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−\item Go back to 3 until $(b-a) \le$ tolerance
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−\end{enumerate}
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−\end{frame}
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−
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−%% \begin{frame}[fragile]
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−%% \frametitle{Bisection \dots}
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−%% \begin{lstlisting}
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−%% In []: tolerance = 1e-5
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−%% In []: a = -pi/2
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−%% In []: b = 0
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−%% In []: while b-a > tolerance:
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−%%  ....:     c = (a+b)/2
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−%%  ....:     if our_f(a)*our_f(c) < 0:
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−%%  ....:         b = c
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−%%  ....:     else:
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−%%  ....:         a = c
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−%%  ....:         
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−%% \end{lstlisting}
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−%% \end{frame}
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−
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−\begin{frame}[fragile]
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−\frametitle{Newton Raphson Method}
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−\begin{enumerate}
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−\item Start with an initial guess of x for the root
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−\item $\Delta x = -f(x)/f^{'}(x)$
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−\item $ x \leftarrow x + \Delta x$
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−\item Go back to 2 until $|\Delta x| \le$ tolerance
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−\end{enumerate}
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−\end{frame}
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−
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−%% \begin{frame}[fragile]
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−%% \frametitle{Newton Raphson \dots}
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−%% \begin{lstlisting}
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−%% In []: def our_df(x):
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−%%  ....:     return -sin(x)-2*x
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−%%  ....: 
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−%% In []: delx = 10*tolerance
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−%% In []: while delx > tolerance:
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−%%  ....:     delx = -our_f(x)/our_df(x)
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−%%  ....:     x = x + delx
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−%%  ....:     
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−%%  ....:     
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−%% \end{lstlisting}
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−%% \end{frame}
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−
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−\begin{frame}[fragile]
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−\frametitle{Newton Raphson \ldots}
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−\begin{itemize}
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−\item What if $f^{'}(x) = 0$?
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−\item Can you write a better version of the Newton Raphson?
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−\item What if the differential is not easy to calculate?
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−\item Look at Secant Method
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−\end{itemize}
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−\end{frame}
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−
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−\begin{frame}[fragile]
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−\frametitle{Scipy Methods - \typ{roots}}
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−\begin{itemize}
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−\item Calculates the roots of polynomials
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−\item Array of coefficients is the only parameter
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−\end{itemize}
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−\begin{lstlisting}
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−  In []: coeffs = [1, 6, 13]
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−  In []: roots(coeffs)
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−\end{lstlisting}
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−\end{frame}
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−
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−\begin{frame}[fragile]
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−\frametitle{Scipy Methods - \typ{fsolve}}
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−\begin{small}
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−\begin{lstlisting}
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−  In []: from scipy.optimize import fsolve
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−\end{lstlisting}
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−\end{small}
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−\begin{itemize}
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−\item Finds the roots of a system of non-linear equations
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−\item Input arguments - Function and initial estimate
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−\item Returns the solution
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−\end{itemize}
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−\begin{lstlisting}
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−  In []: fsolve(our_f, -pi/2)
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−\end{lstlisting}
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−\end{frame}
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−
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−\begin{frame}[fragile]
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−\frametitle{Scipy Methods \dots}
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−\small{
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−\begin{lstlisting}
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−In []: from scipy.optimize import fixed_point
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−
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−In []: from scipy.optimize import bisect
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−
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−In []: from scipy.optimize import newton
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−\end{lstlisting}}
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−\end{frame}
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−
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−
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−\end{document}