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

Section 11.11 Applications of Taylor Polynomials 775
There are three possible methods for estimating the size of the error:
1. If a graphing device is available, we can use it to graph | R nsxd | and thereby
estimate the error.
2. If the series happens to be an alternating series, we can use the Alternating
Series Estimation Theorem.
3. In all cases we can use Taylor’s Inequality (Theorem 11.10.9), which says that if
| f sn11d sxd | < M, then | R nsxd | < M
sn 1 1d! | x 2 a | n11
Example 1
(a) Approximate the function f sxd − s 3 x by a Taylor polynomial of degree 2 at a − 8.
(b) How accurate is this approximation when 7 < x < 9?
SOLUTION
(a) f sxd − s 3 x − x 1y3 f s8d − 2
f 9sxd − 1 3 x22y3 f 9s8d − 1
12
f 0sxd − 2 2 9 x25y3 f 0s8d − 2 1
144
f -sxd − 10
27 x28y3
Thus the second-degree Taylor polynomial is
The desired approximation is
T 2 sxd − f s8d 1 f 9s8d
1!
sx 2 8d 1 f 0s8d
2!
− 2 1 1
1
12 sx 2 8d 2 288 sx 2 8d2
sx 2 8d 2
s 3 x < T 2 sxd − 2 1 1
1
12 sx 2 8d 2 288 sx 2 8d2
(b) The Taylor series is not alternating when x , 8, so we can’t use the Alternating
Series Estimation Theorem in this example. But we can use Taylor’s Inequality with
n − 2 and a − 8:
| R 2sxd | < M 3! | x 2 8 | 3
where | f -sxd | < M. Because x > 7, we have x 8y3 > 7 8y3 and so
f -sxd − 10
27 ? 1 10
8y3
<
x 27 ? 1
8y3
, 0.0021
7
Therefore we can take M − 0.0021. Also 7 < x < 9, so 21 < x 2 8 < 1 and
| x 2 8 | < 1. Then Taylor’s Inequality gives
| R 2sxd | < 0.0021
3!
? 1 3 − 0.0021
6
, 0.0004
Thus, if 7 < x < 9, the approximation in part (a) is accurate to within 0.0004.
n
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