# Ngô Quốc Anh

## August 1, 2009

### Solutions of Equations in One Variable

A root-finding algorithm is a numerical method, or algorithm, for finding a value $x$ such that $f(x) = 0$, for a given function $f$. Such an $x$ is called a root of the function $f$.

Suppose $f$ is a continuous function defined on the interval $[a, b]$, with $f(a)$ and $f(b)$ of opposite sign. By the Intermediate Value Theorem, there exists a number $p$ in $(a, b)$ with $f(p) = 0$. Although the procedure will work when there is more than one root in the interval $(a,b)$, we assume for simplicity that the root in this interval is unique. The method calls for a repeated halving of subintervals of $[a, b]$ and, at each step, locating the half containing $p$.

To begin, set $a_1=a$ and $b_1=b$, and let $p_1$ be the midpoint of $[a,b]$; that is,

$\displaystyle p_1 = a_1 + \frac{b_1-a_1}{2}=\frac{a_1+b_1}{2}$.

If $f(p_1)=0$, then $p=p_1$, and we are done. If $f(p_1)\ne 0$, then $f(p_1)$ has the same sign as either $f(a_1)$ or $f(b_1)$. When $f(p_1)$ and $f(a_1)$ have the same sign, $p \in (p_1, b_1)$, and we set $a_2=p_1$ and $b_2 = b_1$. When $f(p_1)$ and $f(a_1)$ have opposite signs, $p \in (a_1, p_1)$, and we set $a_2 = a_1$ and $b_2 = p_1$. We then reapply the process to the interval $[a_2, b_2]$.

A number p is a fixed point for a given function g if $g(p) = p$. In this section we consider the problem of finding solutions to fixed-point problems and the connection between the fixed-point problems and the root-finding problems we wish to solve. Root-finding problems and fixed-point problems are equivalent classes in the following sense:

Given a root-finding problem $f(p) = 0$, we can define functions $g$ with a fixed point at $p$ in a number of ways, for example, as $g(x) = x-f(x)$ or as $g(x) = x + 3f(x)$. Conversely, if the function $g$ has a fixed point at $p$, then the function defined by $f(x) = x-g(x)$ has a zero at $p$.

Although the problems we wish to solve are in the root-finding form, the fixed-point form is easier to analyze, and certain fixed-point choices lead to very powerful root-finding techniques.

• Newton’s method

Newton’s (or the Newton-Raphson) method is one of the most powerful and well-known numerical methods for solving aroot-finding problem. With an initial approximation $p_0$, the Newton’s method generates the sequence $\{p_n\}_{n=1}^\infty$ by

$\displaystyle {p_{n + 1}} = {p_n} - \frac{{f\left( {{p_n}} \right)}} {{f'\left( {{p_n}} \right)}}, \quad n \geqslant 0$.

To circumvent the problem of the derivative evaluation in Newton’s method, we  introduce a slight variation. By definition,

$\displaystyle f'(x) = \mathop {\lim }\limits_{x \to {p_n}} \frac{{f(x) - f({p_{n - 1}}){\text{ }}}}{{x - {p_{n - 1}}}}$.

Letting $x=p_{n-1}$, we have

$\displaystyle f'\left( {{p_n}} \right) \approx \frac{{f\left( {{p_{n - 1}}} \right) - f\left( {{p_n}} \right)}} {{{p_{n - 1}} - {p_n}}}$

which yields

$\displaystyle {p_{n + 1}} = {p_n} - {\text{ }}\frac{{f({p_n})\left( {{p_{n - 1}} - {p_n}} \right)}}{{f({p_{n - 1}}) - f({p_{n - 2}})}}$.

• The method of False Position

The method of False Position (also called Regula Falsi) generates approximations in the same manner as the Secant method, but it includes a test to ensure that the root is bracketed between successive iterations. Although it is not a method we generally recommend, it illustrates how bracketing can be incorporated.

First choose initial approximations $p_0$ and $p_1$ with $f(p_0) f(p_1) < 0$. The approximation $p_2$ is chosen in the same manner as in the Secant method, as the $x$-intercept of the line joining $(p_0, f(p_0))$ and $(p_1, f(p_1))$. To decide which secant line to use to compute $p_3$, we check $f(p_2) f(p_1)$. If this value is negative, then $p_1$ and $p_2$ bracket a root, and we choose $p_3$ as the $x$-intercept of the line joining $(p_1, f(p_1))$ and $(p_2, f(p_2))$. If not, we choose $p_3$ as the $x$-intercept of the line joining $(p_0, f(p_0))$ and $(p_2, f(p_2))$, and then interchange the indices on $p_0$ and $p_1$.

In a similar manner, once $p_3$ is found, the sign of $f(p_3)f(p_2)$ determines whether we use $p_2$ and $p_3$ or $p_3$ and $p_1$ to compute $p_4$. In the latter case a relabeling of $p_2$ and $p_1$ is performed.

Source: Richard L. Burden and J. Douglas Faires, Numerical Analysis, 8th edition, Thomson/Brooks/Cole, 2005.