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

Section 4.8 Newton’s Method 347
It turns out that this third approximation x 3 < 2.0946 is accurate to four decimal
places.
n
Suppose that we want to achieve a given accuracy, say to eight decimal places, using
Newton’s method. How do we know when to stop? The rule of thumb that is generally
used is that we can stop when successive approximations x n and x n11 agree to eight decimal
places. (A precise statement concerning accuracy in Newton’s method will be given
in Exercise 11.11.39.)
Notice that the procedure in going from n to n 1 1 is the same for all values of n. (It is
called an iterative process.) This means that Newton’s method is particularly convenient
for use with a programmable calculator or a computer.
Example 2 Use Newton’s method to find s 6 2 correct to eight decimal places.
SOLUtion First we observe that finding s 6 2 is equivalent to finding the positive root of
the equation
x 6 2 2 − 0
so we take f sxd − x 6 2 2. Then f 9sxd − 6x 5 and Formula 2 (Newton’s method)
becomes
x n11 − x n 2 fsx nd
f 9sx n d − x n 2 x n 6 2 2
5
6x n
If we choose x 1 − 1 as the initial approximation, then we obtain
x 2 < 1.16666667
x 3 < 1.12644368
x 4 < 1.12249707
x 5 < 1.12246205
x 6 < 1.12246205
Since x 5 and x 6 agree to eight decimal places, we conclude that
s 6 2 < 1.12246205
to eight decimal places.
n
Example 3 Find, correct to six decimal places, the root of the equation cos x − x.
SOLUtion We first rewrite the equation in standard form:
cos x 2 x − 0
y
y=x
Therefore we let f sxd − cos x 2 x. Then f 9sxd − 2sin x 2 1, so Formula 2 becomes
y=cos x
1
π
2
π
x
x n11 − x n 2 cos x n 2 x n
2sin x n 2 1 − x n 1 cos x n 2 x n
sin x n 1 1
FIGURE 6
In order to guess a suitable value for x 1 we sketch the graphs of y − cos x and y − x in
Figure 6. It appears that they intersect at a point whose x-coordinate is somewhat less
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