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

764 Chapter 11 Infinite Sequences and Series
Example 4 Find the Maclaurin series for sin x and prove that it represents sin x for
all x.
SOLUtion We arrange our computation in two columns as follows:
f sxd − sin x f s0d − 0
Figure 2 shows the graph of sin x
together with its Taylor (or Maclaurin)
polynomials
T 1sxd − x
T 3sxd − x 2 x 3
3!
T 5sxd − x 2 x 3
3! 1 x 5
5!
Notice that, as n increases, T nsxd
becomes a better approximation to
sin x.
FIGURE 2
y=sin x
y
1
T¡
0 1
x
T£
T∞
f 9sxd − cos x f 9s0d − 1
f 0sxd − 2sin x f 0s0d − 0
f -sxd − 2cos x f -s0d − 21
f s4d sxd − sin x f s4d s0d − 0
Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as
follows:
f s0d 1 f 9s0d
1!
x 1 f 0s0d
2!
x 2 1 f -s0d
3!
x 3 1 ∙ ∙ ∙
− x 2 x 3
3! 1 x 5
5! 2 x 7
7! 1 ∙ ∙ ∙ − òs21d n
Since f sn11d sxd is 6sin x or 6cos x, we know that | f sn11d sxd |
take M − 1 in Taylor’s Inequality:
14
| R nsxd | < M
sn 1 1d! | x n11 | −
n−0
| x | n11
sn 1 1d!
x 2n11
s2n 1 1d!
< 1 for all x. So we can
By Equation 10 the right side of this inequality approaches 0 as n l `, so
| R nsxd | l 0 by the Squeeze Theorem. It follows that R nsxd l 0 as n l `, so sin x is
equal to the sum of its Maclaurin series by Theorem 8.
n
We state the result of Example 4 for future reference.
15
sin x − x 2 x 3
3! 1 x 5
5! 2 x 7
7! 1 ∙ ∙ ∙
− òs21d n
n−0
x 2n11
s2n 1 1d!
for all x
Example 5 Find the Maclaurin series for cos x.
SOLUtion We could proceed directly as in Example 4, but it’s easier to differentiate
the Maclaurin series for sin x given by Equation 15:
cos x − d
d
ssin xd − Sx 2 D
x 3
dx dx 3! 1 x 5
5! 2 x 7
7! 1 ∙ ∙ ∙
− 1 2 3x 2
3!
1 5x 4
5!
2 7x 6
7!
1 ∙ ∙ ∙ − 1 2 x 2
2! 1 x 4
4! 2 x 6
6! 1 ∙ ∙ ∙
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