SECTION 9.7: Applications of Taylor and Maclaurin Series
Exercise 7
Using the Maclaurin series for cosine, we have:
cos 5◦= 1 − 1 2!
5π 180
2 + 1
4!
5π 180
4
− · · ·
Since the Maclaurin series of cosine can be viewed as a alternating series, the error E will be bounded by the first omitted term, that is, if the (n + 1)−th term is the first omitted term, we have:
|E| ≤ 1 (2n)!
5π 180
2n
Knowing that 1805π < 0.1, we observe that n = 2 is sufficient. Therefore,
cos 5◦≈ 1 − 1 2!
5π 180
2
≈ 0.99619.
Exercise 9
Using the Maclaurin series for ln(1 + x), we have:
ln 0.9 = ln(1 + (−0.1)) = (−0.1) −(−0.1)2
2 +(−0.1)3
3 −(−0.1)4 4 + · · · If the n−th term is the first omitted term, we can estimate the absolute value of the error as follows:
|E| =
∞
X
k=n
1
k(0.1)k< 1 n
∞
X
k=n
(0.1)k = 1 n
(0.1)n 1 − 0.1 = 10
9n(0.1)n.
When n = 4, |E| ≤ 1036(0.1)4 < 5 × 10−5, so we only need to compute the first three terms:
ln 0.9 ≈ (−0.1) −(−0.1)2
2 +(−0.1)3
3 ≈ −0.10533
Exercise 12
Using the Maclaurin series for tan−1x, we have:
tan−10.2 = 0.2 −(0.2)3
3 +(0.2)5 5 − · · · .
Since the Maclaurin series of tan−1x can be viewed as a alternating series when x = 0.2, the error E will be bounded by the first omitted term, that is, if the (n + 1)−th term is the first omitted term, we have:
|E| ≤ 1
2n + 1(0.2)2n+1
When n = 3, we have |E| ≤ 17(0.2)7< 0.0000128 < 5 × 10−5, hence comput- ing the first three nonzero terms is sufficient. Therefore,
tan−10.2 ≈ 0.2 −(0.2)3
3 +(0.2)5
5 ≈ 0.19740.
Exercise 14
Solution 1:
Using the Maclaurin series for ln(1 + x), we have:
ln3 2 = ln
1 + 1
2
= 1 2−1
2
1 2
2
+1 3
1 2
3
−1 4
1 2
4
+ · · · .
Since the Maclaurin series of ln(1 + x) can be viewed as a alternating series when x = 12, the error E will be bounded by the first omitted term, that is, if the n−th term is the first omitted term, we have:
|E| ≤ 1 n
1 2
n
When n = 11,
|E| ≤ 1 11
1 2
1
1 = 1
11 × 211 = 1
22528 < 1
20000 = 5 × 10−5. Therefore, it is sufficient to compute 10 nonzero terms:
ln3 2 ≈ ln
1 + 1
2
= 1 2−1
2
1 2
2
+ · · · − 1 10
1 2
10
≈ 0.40543.
Solution 2:
It will often be useful to use the Maclaurin series of ln1+x1−xinstead of ln(1+x), which converges faster and is capable to evaluate ln u for any positive u:
ln1 + x
1 − x = ln(1 + x) − ln(1 − x)
=
x −x2
2 +x3 3 −x4
4 + · · ·
−
−x −x2 2 −x3
3 −x4 4 − · · ·
=
∞
X
n=0
2
2n + 1x2n+1.
Substitute 0.2 for x in the Maclaurin series above, we get:
ln3
2 = ln1 + 0.2
1 − 0.2= 2 × 0.2 +2
3(0.2)3+2
5(0.2)5+2
7(0.2)7+ · · · . If the (n + 1)−th term is the first omitted term, we can estimate the absolute value of the error as follows:
|E| =
∞
X
k=n
2
2k + 1(0.2)2k+1< 1 n
∞
X
k=n
(0.2)2k+1= 1 n
(0.2)2n+1 1 − 0.04 = 25
24n(0.2)2n+1. If n = 3, |E| < 2572517 = 72×51 5 < 200001 = 5 × 10−5, hence it is accurate enough to compute three nonzero terms:
ln3
2 ≈ 2 × 0.2 +2
3(0.2)3+2
5(0.2)5≈ 0.40546.
Exercise 22
Solution 1:
lim
x→0
sin x2 sinh x= lim
x→0
x2−3!1(x2)3+ · · · x +3!1x3+ · · · = lim
x→0
x − 16x5+ · · · 1 + 16x2+ · · · =0
1 = 0.
Solution 2:
Using the l’Hˆopital’s rule, we can get:
lim
x→0
sin x2 sinh x = lim
x→0
2x cos x2 cosh x = 0
1 = 0.
Exercise 25
x→0lim
2 sin 3x − 3 sin 2x 5x − tan−15x = lim
x→0
2 3x −3!1(3x)3+ · · · − 3 2x −3!1(2x)3+ · · · 5x − 5x −13(5x)3+15(5x)5− · · ·
= lim
x→0
−54−246 x3+ Ax5+ · · ·
125
3 x3− Bx5+ · · ·
= lim
x→0
−5 + Ax2+ · · ·
125
3 − Bx2+ · · · = − 3 25.
(A and B are constants which are not necessary to compute explicitly here in order to evaluate the limit.)
Exercise 26
We first compute the first nonzero term of the Maclaurin series of the nom- inator and the denominator:
sin(sin x) − x = sin
x − x3
6 + · · ·
− x
=
" x − x3
6 + · · ·
−1 6
x − x3
6 + · · ·
3 + · · ·
#
− x
=
x − 1
6 +1 6
x3+ Ax5+ · · ·
− x
= −1
3x3+ Ax5+ · · · x(cos(sin x) − 1) = x
cos
x − x3
6 + · · ·
− 1
= x
1 − 1 2
x −x3
6 + · · ·
− 1
= x
−1
2x2+ Bx4+ · · ·
= −1
2x3+ Bx5+ · · ·
(A and B are constants which are not necessary to compute explicitly here in order to evaluate the limit.)
Therefore,
x→0lim
sin(sin x) − x
x(cos(sin x) − 1) = lim
x→0
−13x3+ Ax5+ · · ·
−12x3+ Bx5+ · · ·
= lim
x→0
−13+ Ax2+ · · ·
−12+ Bx2+ · · · = −13
−12 =2 3.
Exercise 27
Solution 1:
lim
x→0
sinh x − sin x cosh x − cos x
= lim
x→0
x + 3!1x3+5!1x5+7!1x7+ · · · − x − 3!1x3+5!1x5−7!1x7+ · · · 1 + 2!1x2+4!1x4+6!1x6+ · · · − 1 −2!1x2+4!1x4−6!1x6+ · · ·
= lim
x→0
2 3!1x3+7!1x7+ · · · 2 2!1x2+6!1x6+ · · · = lim
x→0 1
3!x + 7!1x5+ · · ·
1
2! +6!1x4+ · · · =0 1 = 0.
Solution 2:
Using the l’Hˆopital’s rule twice, we can get:
x→0lim
sinh x − sin x cosh x − cos x = lim
x→0
cosh x − cos x sinh x + sin x = lim
x→0
sinh x + sin x cosh x + cos x= 0
2 = 0.
