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UAE_Math_Grade_12_Vol_1_SE_718383_ch3
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July 4, 2016
13:38
and so on. (See Figure 4.8.) In this way, we generate a sequence of successive approxi-
mations defined by
f(x )
x = x − n , for n = 0, 1, 2, 3, . . . . (1.8)
n
n+1
′
f (x )
n
This procedure is called the Newton-Raphson method, or simply Newton’s method. If
Figure 4.8 is any indication, x should get closer and closer to a zero as n increases.
n
Newton’s method is generally a very fast, accurate method for approximating the
zeros of a function, as we illustrate with example 1.5.

EXAMPLE 1.5 Using Newton’s Method to Approximate a Zero
5
Find an approximate zero of f(x) = x − x + 1.
y
Solution Figure 4.9 suggests that the only zero of f is located between x =−2 and
x =−1. Further, since f(−1) = 1 > 0, f(−2) =−29 < 0 and since f is continuous,
3 the Intermediate Value Theorem says that f must have a zero on the interval
(−2, −1). Because the zero appears to be closer to x =−1, we choose x =−1 as
0
′
4
our initial guess. Finally, f (x) = 5x − 1 and so, Newton’s method gives us
x
-2 -1 1
f(x )
x = x − n
′
-3 n+1 n f (x )
n
5
x − x + 1
= x − n n , n = 0, 1, 2,, . . . .
4
FIGURE 4.9 n 5x − 1
5
y = x − x + 1 n
Using the initial guess x =−1, we get
0
5
(−1) − (−1) + 1 1 5
x =−1 − =−1 − =− .
1
5(−1) − 1 4 4
4
5
Likewise, from x =− , we get the improved approximation
1
4
( 5 ) 5 ( 5 )
− − − + 1
5 4 4
x =− − ) 4 ≈−1.178459394
2
4
(
5 − 5 − 1
4
and so on. We find that x ≈−1.167537389,
3
x ≈−1.167304083
4
and x ≈−1.167303978 ≈ x .
6
5
Since x ≈ x , we will make no further progress by calculating additional steps. As
6
5
a final check on the accuracy of our approximation, we compute
f(x ) ≈ 1 × 10 −13 .
6
Since this is very close to zero, we say that x ≈−1.167303978 is an approximate
6
Copyright © McGraw-Hill Education As we illustrate in example 1.6, you may first need to rephrase the problem as a rootfind-
zero of f.
You can bring Newton’s method to bear on a variety of approximation problems.
ing problem.
233