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5.1. Approximation of Roots 165

approximation ~1 to the solution, and let un = f(u,,-1) for

n 2 2. With suitable choice off and ~1, the sequence {u,,}

should converge to a solution r of t = f(t) and hence of

p(t) = 0.

For Newton’s Method, one makes the choice f(t) = t - p(t)/p’(t).

Other choices are possible. Consider the polynomial equation

(4

@I

(4

(4

(4

( f 1

t3 - 4t - 18 = 0.

Argue that there is a unique positive root r and that it lies

between 3 and 4.

The equation can be rewritten in the form t = f(t), where

4f(t) = t3 - 18. Begin with a value ~1 (say 3) and define suc-

cessively u, = f(~,-1) for n > 2. Does u,, approach a solution

of the given equation?

Sketch the graphs of the polynomial t and f(t). The solution of

the equation is determined by the point at which the graphs

cross. For the values of ‘~1 you used in (b), plot the points

(~1, ~4, (~2, 4, (w,w), (w,w), (~3, 4,.... How do YOU ac-

count for the behaviour observed in (b)?

Let g(t) = (4t + 18) ‘I3. Show that the given equation is equiv-

alent to the equation t = g(t). Let vr be equal to 3, 4 or some

other value of your choice, and define v, = g(vn-1) for n 2 2.

Does the sequence {vn} appear to approach a limit?

Sketch the graphs of the functions t and g(t), and plot the points

(vr, v2), (~2,212)~ (~2, us), (us, us),.... Indicate on the x-axis the

points 211, ~2, ~3,214,....

Let v > r where 3 < r < 4 and r3 = 4r + 18. Verify that

v - g(v) =

v3 - (4v + 18)

v2 + v(4v + 18)‘j3 + (4v + 18)2/3

and

4(v - r)

g(v)-r = (4v + 18)2/3 + (4v + 18)U3(4r + 18)li3 + (4r + 18)2/3’

Deduce that r < g(v) < v and that g(v) - r < (1/2)(v - r).

Argue that if vi = v, then the sequence {v,} should converge

towards the root r. Hence determine to five decimal places a

solution of the given polynomial equation.