\documentclass[12pt]{article}
\title{Solving Equations \& ODEs}
\author{FOSSEE}
\usepackage{listings}
\lstset{language=Python,
    basicstyle=\ttfamily,
commentstyle=\itshape\bfseries,
showstringspaces=false,
}
\newcommand{\typ}[1]{\lstinline{#1}}
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\usepackage{times}
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\begin{document}
\date{}
\vspace{-1in}
\begin{center}
\LARGE{Solving Equations \& ODEs}\\
\large{FOSSEE}
\end{center}
\section{Solving linear equations}
Condier following sets of equations:\\
  \begin{align*}
    3x + 2y - z  & = 1 \\
    2x - 2y + 4z  & = -2 \\
    -x + $\frac{1}{2}$y -z & = 0
  \end{align*}\\
The matrix representation is:\\
\begin{center}
$A*x = B$
\end{center}
Where A is coefficient matrix(in this case 3x3)\\
B is constant matrix(1x3)\\
x is the required solution.\\
\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)
\end{lstlisting}
Solve the equation $A x = B$ for $x$.\\
To check whether solution is correct try this:
\begin{lstlisting}
In []: Ax = dot(A,x)  #Matrix multiplication of A and x(LHS)
In []: allclose(Ax, B)
Out[]: True
\end{lstlisting}
\typ{allclose} Returns \typ{True} if two arrays(in above case Ax and B) are element-wise equal within a tolerance. 
\newpage
\section{Finding roots}
\subsection{Polynomials}
\begin{center}
  $x^2+6x+13=0$
\end{center}
to find roots, pylab provides \typ{roots} function.
\begin{lstlisting}
In []: coeffs = [1, 6, 13] #list of all coefficients
In []: roots(coeffs)
\end{lstlisting}
\subsection{functions}
Functions can be defined and used by following syntax:
\begin{lstlisting}
def func_name(arg1, arg2):
    #function code
    return ret_value
\end{lstlisting}
A simple example can be
\begin{lstlisting}
def expression(x):
    y = x*sin(x)
    return y
\end{lstlisting}
Above function when called with a argument, will return $xsin(x)$ value for that argument.
\begin{lstlisting}
In [95]: expression(pi/2)
Out[95]: 1.5707963267948966

In [96]: expression(pi/3)
Out[96]: 0.90689968211710881
\end{lstlisting}
\subsection{Roots of non-linear eqations}
For Finding the roots of a non linear equation(defined as $f(x)=0$), around a starting estimate we use \typ{fsolve}:\\
\typ{In []: from scipy.optimize import fsolve}\\
\typ{fsolve} is not part of \typ{pylab}, instead it is part of \textbf{optimize} package of \textbf{scipy}, and hence we \textbf{import} it.\\
%\typ{fsolve} takes first argument as name of function, which evaluates $f(x)$, whose roots one wants to find. And second argument is starting estimate, around which roots are found.
For illustration, we want to find roots of equation:
\begin{center}
  $f(x)=sin(x)+cos(x)^2$
\end{center}
So just like we did above, we define a function:
\begin{lstlisting}
In []: def f(x):
   ....:        return sin(x)+cos(x)**2
   ....: 
\end{lstlisting}
Now to find roots of this non linear equation, around a initial estimate value, say 0, we use \typ{fsolve} in following way:
\begin{lstlisting}
In []: fsolve(f, 0) #arguments are function name and estimate
Out[]: -0.66623943249251527
\end{lstlisting}

\section{ODE}
\begin{lstlisting}
  In []: def epid(y, t):
  ....     k, L = 0.00003, 25000
  ....     return k*y*(L-y)
  ....
  
  In []: t = arange(0, 12, 0.2)

  In []: y = odeint(epid, 250, t)

  In []: plot(t, y)
\end{lstlisting}
\end{document}