Originally prepared in the Fall Semester 1995

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

Then:

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

If $x$ is on the

**convex**side of the graph of $f$ in $I$, then so is ${x}^{\prime}=x-f\left(x\right)\u2044{f}^{\prime}\left(x\right)$, and ${x}^{\prime}$ lies between $x$ and $z$.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 $z$.*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 $f\left(x\right)\u2044{f}^{\prime}\left(c\right)$.

*Proof:*
If ${f}^{\prime}$ is positive in $I$ one has $f\left({x}_{1}\right)<f\left({x}_{2}\right)$ 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 $z<x$: then $0=f\left(z\right)<f\left(x\right)$. We apply the Mean Value Theorem to $f$ on the interval $\left[z,x\right]$ to conclude that there is a number $u$ with $z<u<x$ 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, $f\left(x\right)\u2044{f}^{\prime}\left(x\right)<f\left(x\right)\u2044{f}^{\prime}\left(u\right)$. Consequently, $z<{x}^{\prime}<x$.

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 $f\left(x\right)\u2044{f}^{\prime}\left(u\right)$, 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, ${f}^{\prime}\left(c\right)<{f}^{\prime}\left(u\right)$. Consequently, $$\mathrm{the\; error}=x-z=\frac{f\left(x\right)}{{f}^{\prime}\left(u\right)}\le \frac{f\left(x\right)}{{f}^{\prime}\left(c\right)}\phantom{\rule{0.5em}{0ex}}\text{.}$$