Approximate Numerical Methods

Most numerical methods for solving equations in one variable are best illustrated graphically, indeed it is always a good idea to sketch the graph of the function before looking for the roots. Consider the roots of $F$ satisfying

$$F(x) = 0,$$ (1.8)

where $F(x)$ is a continuous function of $x$. Setting $y = F(x)$ we can sketch the graph of $F$. This is shown schematically in Figure 1.1.

Figure 1.1: A sketch of a continuous function, $F$, that has a root at $x = r$.

The object is to obtain the value of the root $r$ in the interval $a \le x \le b$, where $F(r) = 0$ or equivalently where the graph crosses the $x$-axis.

Before using the method described below, it is a good idea to identify the number of roots in the given interval. Consider Figure 1.2.

Figure: The graph of $\cos x$ illustrates a function that has an infinite number of roots where the graph crosses the $x$-axis.

$\cos x$ has an infinite number of roots. If we consider the interval $0 < x < \pi$, there is clearly only one root. Note that

$$\cos (0) = 1 > 0, \qquad \cos (\pi) = -1 < 0.$$

Thus, there is a change in the sign of the function values at the two ends of the interval. However, if we consider the interval $0 < x < 3\pi$, there is again a sign change since

$$\cos (0) = 1 > 0, \qquad \cos(3\pi) = -1 < 0,$$

but this time there are three roots in the interval. So a change in sign means that there are an odd number of roots. In the first interval the function does not have a turning point but in the second interval the function has two turning points, one lying above the $x$-axis and one lying below. These ideas are illustrated in the following example.

Example 1. .4

1. Show that $F(x) = x^{3} + x - 1$ has only one root in the interval $[0,1]$. (This is a short hand notation for $0 \le x \le 1$.) The function $F$ is continuous and
   
   $$F(0) = - 1, \qquad F(1) = 1.$$
   
   
   Hence, there is a sign change and there must at least one root in $[0,1]$. Now consider the derivative
   
   $$F^{\prime}(x) = 3x^{2} + 1 > 0.$$
   
   
   The derivative is positive and so there are no turning points at all. Hence, there can only be one root.

2. Show that $F(x) = x^{3} - x + 1$ has only one root in $[-2,-1]$. Note
   
   $$F(-2) = -8 + 2 + 1 = - 5, \qquad F(-1) = -1 + 1 + 1 = 1.$$
   
   
   Again there is a sign change and so there must be at least one root in the interval. To check the number of roots we look at the derivative
   
   $$F^{\prime}(x) = 3x^{2} - 1.$$
   
   
   There are two turning points where
   
   $$x^{2} = {1\over 3}, \qquad x = \pm {1\over \sqrt{3}} = \pm 0.57735$$
   
   
   In this example, the turning points are outside the interval and so there can only be one root in the interval $[-2,-1]$. In fact, it turns out for this function that there is only one root since at the turning points
   
   $$F\left ({1\over \sqrt{3}}\right ) = 1 - {2\over 3\sqrt{3}} = 0.615,$$
   
   
   and
   
   $$F\left (-{1\over \sqrt{3}}\right ) = 1 + {2\over 3\sqrt{3}} = 1.385.$$
   
   
   Both turning points lie above the $x$-axis and so there is only one root.