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160 5. Approximation and Location of Zeros

This quartic equation is easy to solve explicitly. However, if we look at

the possibility of a three-year cycle in which there are distinct population

sizes xl, 22 and xs for which

x2 = f(q) 23 = f(x2) x1 = f(x31,

then the xi will be roots of the polynomial equation

x = foYf(x)))

which is of the eighth degree. Two obvious roots are x = 0 and the positive
value of x for which x = f(x). But when the factors corresponding to these
roots are removed, this still leaves us with an equation of the sixth degree
to solve. Even if it were possible to solve this equation, it might be at the
expense of a fair bit of grind, and the answer might be in a form from which
it is hard to find out much about the roots, even whether they are relevant
to the problem at hand. What is really needed is a collection of techniques
which will yield useful information in an efficient way.
Many mathematical situations give rise to polynomial equations. If the
degree is higher than four, it is usually hopeless to think of solving them
exactly. However, this may not be necessary, and techniques which allow us
to locate the roots in a rough sense may be enough to support the analysis
we wish to make.
Information sought about roots falls into two main categories:

(i) numerical approximation

(ii) location.

Under the first heading are treated methods which yield roots to desired
numerical accuracy. Under the second are discussed methods which will
enable us to say whether there are real or nonreal roots, positive or negative
roots, or roots within a given region of the complex plane.
In this section, we will examine methods of numerical approximation.

Exercises

1. The method of bisection. Let p(t) b e any polynomial over R. Suppose
   further that p(a) is negative while p(b) is positive. The graph of p is
   a continuous curve which joins the point (a,p(a)) below the x-axis to
   the point (b,p(b)) a b ove the x-axis. Accordingly, it will cross the axis
   somewhere between a and b.

(a) Argue that p(t) h as at least one real zero between a and b.

(b) Show that, either (a + b)/2 is a zero of p(t), or else ~((a + b)/2)

differs in sign from either p(a) or p(b).