\usepackage{listings}

\lstset{language=Python,

    basicstyle=\ttfamily,

commentstyle=\itshape\bfseries,

showstringspaces=false,

}

\newcommand{\typ}[1]{\lstinline{#1}}

\usepackage[english]{babel}

\usepackage[latin1]{inputenc}

\usepackage{times}

\usepackage[T1]{fontenc}

\usepackage{ae,aecompl}

\usepackage{mathpazo,courier,euler}

\usepackage[scaled=.95]{helvet}

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

\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}

\begin{lstlisting}

\end{lstlisting}