# Notes on Newton's Method

#### Revised for the Web: July 27, 2004

Proposition: Let $f$ be a function that is differentiable on an interval $I$ and assume further:

Then:

1. There is one and only one point $z$ in $I$ for which $f\left(z\right)=0$.

2. If $x$ is on the convex side of the graph of $f$ in $I$, then so is ${x}^{\prime }=x-f\left(x\right)⁄{f}^{\prime }\left(x\right)$, and ${x}^{\prime }$ lies between $x$ and $z$.

3. Successive iterations of Newton's method beginning with a point $x$ on the convex side of the graph of $f$ in $I$ will converge to $z$.

4. Error control principle. If $c$ is any point in $I$ on the concave side of the graph of $f$ and $x$ is on the convex side, then the distance between $x$ and $z$ is at most the absolute value of $f\left(x\right)⁄{f}^{\prime }\left(c\right)$.

Proof: If ${f}^{\prime }$ is positive in $I$ one has $f\left({x}_{1}\right) whenever ${x}_{1}<{x}_{2}$ in $I$. If instead ${f}^{\prime }$ is negative in $I$, then one has $f\left({x}_{1}\right)>f\left({x}_{2}\right)$ for ${x}_{1}<{x}_{2}$. For this reason there is at most one root $z$ in $I$ with $f\left(z\right)=0$. The Intermediate Value Theorem for Continuous Functions guarantees that there is at least one root between $a$ and $b$.

We shall assume that ${f}^{\prime }$ 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: then $0=f\left(z\right). We apply the Mean Value Theorem to $f$ on the interval $\left[z,x\right]$ to conclude that there is a number $u$ with $z for which $f\left(x\right)-f\left(z\right)={f}^{\prime }\left(u\right)\left(x-z\right)\phantom{\rule{0.5em}{0ex}}\text{.}$ Since $f\left(z\right)=0$ and ${f}^{\prime }\left(u\right)>0$, one obtains $x-z=\frac{f\left(x\right)}{{f}^{\prime }\left(u\right)}\phantom{\rule{0.5em}{0ex}}\text{.}$ Since ${f}^{\prime }$ is increasing, we find ${f}^{\prime }\left(u\right)<{f}^{\prime }\left(x\right)$, and, therefore, $f\left(x\right)⁄{f}^{\prime }\left(x\right). Consequently, $z<{x}^{\prime }.

In view of (2) one has $z<\dots <{x}_{n}<\dots <{x}_{2}<{x}_{1}\phantom{\rule{0.5em}{0ex}}\text{.}$ Letting ${x}_{*}={\mathrm{inf}}_{\left(n\ge 1\right)}\phantom{\rule{0.5em}{0ex}}\left\{{x}_{n}\right\}\phantom{\rule{0.5em}{0ex}}\text{,}$ one has ${x}_{*}={\phantom{\rule{0.1em}{0ex}}\mathrm{lim}}_{n\to \infty }\phantom{\rule{0.5em}{0ex}}{x}_{n}\phantom{\rule{0.5em}{0ex}}\text{,}$ and, therefore, taking the limit as $n\to \infty$ on both sides of the relation ${x}_{n+1}={x}_{n}-\frac{f\left({x}_{n}\right)}{{f}^{\prime }\left({x}_{n}\right)}\phantom{\rule{0.5em}{0ex}}\text{,}$ one finds that $f\left({x}_{*}\right)=0$. 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 $f\left(x\right)⁄{f}^{\prime }\left(u\right)$, where $z. Since $c$ is on the concave side of the graph of $f$, i.e., $c, we find also $c, hence, ${f}^{\prime }\left(c\right)<{f}^{\prime }\left(u\right)$. Consequently, $\mathrm{the error}=x-z=\frac{f\left(x\right)}{{f}^{\prime }\left(u\right)}\le \frac{f\left(x\right)}{{f}^{\prime }\left(c\right)}\phantom{\rule{0.5em}{0ex}}\text{.}$