\documenttype{article}
\surtitle{Univ at Albany: W. F. Hammond: Honors Calculus}
\title{Notes on Newton's Method}
\subtitle{Supplementary Material for Honors Calculus\\
Originally prepared in the Fall Semester 1995}
\date{Revised for the Web: July 27, 2004}
\macro{\@.}{\aos;}
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\newcommand{\ritem}{\item[\label[:series="x" %
   ]{}(\series[:type="a"]{\evalref{\popkey}})]}
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\begin{document}

\bold{Proposition}: Let $f$ be a function that is differentiable on an
interval $I$ and assume further:
\begin{menu}
\ritem  The derivative $f'$ is either positive throughout $I$ or
negative throughout $I$\@.
\ritem  The derivative $f'$ is either steadily increasing in $I$ or steadily
decreasing in $I$\@.
\ritem  There are points $a, b$ in $I$ for which $f(a)$ and
$f(b)$ are both non-zero with opposite sign.
\end{menu}
Then:
\begin{enumerate}
\item  There is one and only one point $z$ in $I$ for
which $f(z) = 0$\@.

\item  If $x$ is on the \bold{convex} side of the graph of $f$ in $I$\@,
then so is $x' = x - f(x)/f'(x)$\@, and $x'$ lies
between $x$ and $z$\@.

\item  Successive iterations of Newton's method beginning with a point
$x$ on the \bold{convex} side of the graph of $f$ in $I$
will converge to $z$\@.

\item  \emph{Error control principle.} \ If $c$ is any point in $I$
on the \bold{concave} side of the graph of $f$ and $x$ is on
the \bold{convex} side, then the distance between
$x$ and $z$ is at most the absolute value of $f(x)/f'(c)$\@.
\end{enumerate}

\slnt{Proof:}
If $f'$ is positive in $I$ one has $f(x_1) < f(x_2)$ whenever
$x_1 < x_2$ in $I$.  If instead $f'$ is negative in $I$, then one
has $f(x_1) > f(x_2)$ for $x_1 < x_2$\@.
For this reason there is \emph{at most} one root
$z$ in $I$ with $f(z)=0$\@.  The \emph{Intermediate Value Theorem for
Continuous Functions} guarantees that there is at least one root between
$a$ and $b$\@.

We shall assume that $f'$ is positive and increasing.  One may
reduce each of the other three cases to this case by reflecting either
in the horizontal axis or in the vertical line $x=z$ or both.
Under this assumption the convex side of the graph of $f$ is the right
side.  Suppose that $z < x$: then $0=f(z) < f(x)$\@.
We apply the Mean Value Theorem to
$f$ on the interval $[z,x]$ to conclude that there is a number $u$
with $z < u < x$ for which
\[ f(x) - f(z) = f'(u)(x - z) \ \eos \]
Since $f(z)=0$ and $f'(u) > 0$\@, one obtains
\[ x - z = \frac{f(x)}{f'(u)} \ \eos \]
Since $f'$ is increasing, we find $f'(u) < f'(x)$\@, and, therefore,
$f(x)/f'(x) < f(x)/f'(u)$\@.  Consequently, $z < x' < x$\@.

In view of (2) one has
\[ z < \ldots < x_n < \ldots < x_2 < x_1 \ \eos \]
Letting
\[ x_* = \mbox{inf}_{(n \geq 1)} \ \balbr{x_n} \ \cma \]
one has
\[ x_* = \func{lim}_{n \rightarrow \infty} \ x_n \ \cma \]
and, therefore, taking the limit as \(n \rightarrow \infty\) on
both sides of the relation
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \ \cma \]
one finds that $f(x_*) = 0$\@.  Since by (1) there is only one root
of $f$ in $I$, it follows that $x_* = z$\@.

In the proof of (2) we saw that for $x$ in $I$ on the right side
of $z$ the distance from $x$ to $z$ is $f(x)/f'(u)$\@, where
$z<u<x$\@.  Since $c$ is on the concave side of the graph of $f$\@,
i.e., $c < z$\@, we find also $c < u$\@, hence, $f'(c) < f'(u)$\@.
Consequently,
\[ \mbox{the error} = x - z = \frac{f(x)}{f'(u)} \leq
   \frac{f(x)}{f'(c)} \ \eos \]
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