James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 4.8 Newton’s Method 345
Suppose that a car dealer offers to sell you a car for $18,000 or for payments of $375 per
month for five years. You would like to know what monthly interest rate the dealer is, in
effect, charging you. To find the answer, you have to solve the equation
1
48xs1 1 xd 60 2 s1 1 xd 60 1 1 − 0
0.15
0 0.012
_0.05
FIGURE 1
Try to solve Equation 1 numerically
using your calculator or computer.
Some machines are not able to solve it.
Others are successful but require you to
specify a starting point for the search.
y
0 r x x¡ x
FIGURE 2
y=ƒ
{x¡, f(x¡)}
L
(The details are explained in Exercise 41.) How would you solve such an equation?
For a quadratic equation ax 2 1 bx 1 c − 0 there is a well-known formula for the
solutions. For third- and fourth-degree equations there are also formulas for the solutions,
but they are extremely complicated. If f is a polynomial of degree 5 or higher,
there is no such formula (see the note on page 211). Likewise, there is no formula that
will enable us to find the exact roots of a transcendental equation such as cos x − x.
We can find an approximate solution to Equation 1 by plotting the left side of the
equation. Using a graphing device, and after experimenting with viewing rectangles, we
pro duce the graph in Figure 1.
We see that in addition to the solution x − 0, which doesn’t interest us, there is a
solution between 0.007 and 0.008. Zooming in shows that the root is approximately
0.0076. If we need more accuracy we could zoom in repeatedly, but that becomes tiresome.
A faster alternative is to use a calculator or computer algebra system to solve the
equation numerically. If we do so, we find that the root, correct to nine decimal places,
is 0.007628603.
How do these devices solve equations? They use a variety of methods, but most of
them make some use of Newton’s method, also called the Newton-Raphson method.
We will explain how this method works, partly to show what happens inside a calculator
or computer, and partly as an application of the idea of linear approximation.
The geometry behind Newton’s method is shown in Figure 2. We wish to solve an
equation of the form f sxd − 0, so the roots of the equation correspond to the x-intercepts
of the graph of f. The root that we are trying to find is labeled r in the figure. We start
with a first approximation x 1 , which is obtained by guess ing, or from a rough sketch of
the graph of f , or from a computer-generated graph of f. Consider the tangent line L to
the curve y − f sxd at the point sx 1 , f sx 1 dd and look at the x-intercept of L, labeled x 2 .
The idea behind Newton’s method is that the tangent line is close to the curve and so
its x-intercept, x 2 , is close to the x-intercept of the curve (namely, the root r that we are
seeking). Because the tangent is a line, we can easily find its x-intercept.
To find a formula for x 2 in terms of x 1 we use the fact that the slope of L is f 9sx 1 d, so
its equation is
y 2 f sx 1 d − f 9sx 1 dsx 2 x 1 d
Since the x-intercept of L is x 2 , we know that the point sx 2 , 0d is on the line, and so
0 2 f sx 1 d − f 9sx 1 dsx 2 2 x 1 d
If f 9sx 1 d ± 0, we can solve this equation for x 2 :
We use x 2 as a second approximation to r.
x 2 − x 1 2 f sx 1d
f 9sx 1 d
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