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%Slide 1
\title{{\LARGE\bf Newton's Method}\newline (and the square root)\newline \newline}
\author{
\begin{wrapfigure}{r}{0.35\textwidth}
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    {\fontsize{5}{5}\selectfont  
       http://csustan.csustan.edu/\~\ tom
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      Tom Carter \newline
      \newline
     Computer Science \newline
     CSU Stanislaus\newline
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{\hyperref{mailto:tom@csustan.csustan.edu}{}{}
     {tom@csustan.csustan.edu}}
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{\href{http://csustan.csustan.edu/\~tom/Lecture-Notes/Newtons-Method/Newtons-Method.pdf}{http://csustan.csustan.edu/\~.tom/Lecture-Notes/Newtons-Method/Newtons-Method.pdf}}
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\sectionhead{Newton's Method \ldots}

    Newton's method (also sometimes known as the Newton-Raphson method) is a way to approximate
    the root(s) of a differentiable real-valued function.
    
    The basic method is iterative.  We start with an initial guess of a root, say, for example,
         $$x_0 = 1$$
         
    We then iterate (until we get close enough to a nearby root, we hope \ldots :-)
         $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
   
   
   \pagedone
   
      An example \ldots finding the square root of a number $a$:
      $$f(x) = x^2 - a$$
      
      We then are using:
            \begin{eqnarray*}
               x_{n+1} & = & x_n - \frac{x_n^2 - a}{2xn} \\
                 & = & \frac{2x_n^2 - x_n^2 + a}{2x_n}\\
                 & = & \frac{x_n^2 + a}{2x_n} \\
                 & = & \frac{x_n + \frac{a}{x_n}}{2}\\
                 & = & \textrm{Avg}(x_n, \frac{a}{x_n}) \\
            \end{eqnarray*}
 \pagedone
 
   The ideas behind Newton's method:
   
   We assume that $f(x)$ is differentiable.  We are looking for a root of $f(x)$ (i.e., a place
   where $f(x) = 0$).  Suppose that our current estimate of a root is $x_n$.  We find the
   line tangent to $f(x)$ at $x_n$:
        $$y = f'(x_n)(x - x_n) + f(x_n)$$
   We find the place where this line crosses the $x$-axis, and let that value of $x$ be our
   next estimate of the root we are looking for:
        $$0 = f'(x_n)(x_{n+1} - x_n) + f(x_n)$$
   Solving this for $x_{n+1}$, we get:
        $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$

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