**Theorem**: *If f is continuous for x between a and b and if f (a) and f(b) have opposite signs, then there exists at least one real root of f (x) = 0 between a and b.*

- Procedure
*f ((a + b)*/2) = 0, in which case*(a + b)*/2 is the root;*f ((a + b)*/2) < 0, in which case the root lies between*(a + b)*/2 and*b*;*f ((a + b)*/2) > 0, in which case the root lies between*a*and*(a + b)*/2. The bisection method is almost certain to give a root. Provided the conditions of the above theorem hold; it can only fail if the accumulated error in the calculation of### Effectiveness

*f*at a bisection point gives it a small negative value when actually it should have a small positive value (or*vice versa*); the interval subsequently chosen would then be wrong. This can be overcome by working to sufficient accuracy, and this almost-assured convergence is not true of many other methods of finding roots.One drawback of the bisection method is that it applies only to roots of

*f*about which*f (x)*changes sign. In particular, double roots can be overlooked; one should be careful to examine*f(x)*in any range where it is small, so that repeated roots about which*f (x)*does not change sign are otherwise evaluated (for example, see Steps 9 and 10). Of course, such a close examination also avoids another nearby root being overlooked.Finally, note that bisection is rather slow; after

*n*iterations the interval containing the root is of length*$(b\; -\; a)/2n$*. However, provided values of*f*can be generated readily, as when a computer is used, the rather large number of iterations which can be involved in the application of bisection is of relatively little consequence. Let us solve 3$$### Example

*x*e^{x}= 1 to three decimal places by the bisection method.We can consider

*f(x) = 3x -*$e$*x*, which changes sign in the interval 0.25 <*x*< 0.27: one may tabulate (working to 4*D*) as follows:(The student should ascertain graphically that there is just one root.)

Let us denote the lower and upper endpoints of the interval bracketing the root at the

*n*-th iteration by*$a$*and_{n}*$b$*, respectively (with_{n}*$a$*= 0.25 and_{1}*$b$*= 0.27). Then the approximation to the root at the_{1}*n*-th iteration is given by*$x$*. Since the root is either in [_{n}= a_{n}+ b_{n})/2*$a$*] or [_{n}, b_{n}*$x$*] and both intervals are of length_{n}, b_{n}*$b$*/2, we see that_{n}- a_{n})*$x$*will be accurate to three decimal places when_{n}*$b$*/2 < 5*$10-4$. Proceeding to bisection:_{n}- a_{n})(Note that the values in the table are displayed to only 4

*D*.) Hence the root accurate to three decimal places is 0.258.### Checkpoint

- When may the bisection method be used to find a root of the equation
*f(x)*= 0? - What are the three possible choices after a bisection value is calcu lated?
- What is the maximum error after
*n*iterations of the bisection method`?

- Use the bisection method to find the root of the equation
*x*+cos*x*= 0.correct to two decimal places (2D ).

- Use the bisection method to find to 3
*D*the positive root of the equation .*x*- 0.2sin*x*- 0.5=0 - Each equation in Exercises
**2**(a)-2(c) of Step 6 has onlyone root. For each equation use the bisection method to find the root correct to 2*D*.