362 STEP 4. Review the Knowledge You Need to Score High
Step 2: For three decimal place accuracy, we want to find the value of nfor which
the remainder is less than or equal to 0.0005. Choose x 0 =1 in the interval
(0, 1). All derivatives ofex are equal toex, and therefore|f(n+1)(x)|≤e,so
M=e·
∣
∣∣
∣Rn
(
1
2
)∣∣
∣∣≤ e
(n+1)!
∣
∣∣
∣
1
2
− 1
∣
∣∣
∣
n+ 1
and
e
(n+1)!
∣
∣∣
∣
1
2
− 1
∣
∣∣
∣
n+ 1
=
e
2 n+^1 ·(n+1)!
≤
0 .0005 whenn≥4.
Step 3:
√
e=
∑^4
n= 0
1
2 n·n!
= 1 +
1
2
+
1
22 ·2!
+
1
23 ·3!
+
1
24 ·4!
= 1. 6484
Example 2
Estimate sin 4◦accurate to five decimal places.
Step 1: 4 ◦=
π
45
radians. Substitute
π
45
forxin the MacLaurin series that represents
sinx·sin
π
45
=
π
45
−
(π/45)^3
3!
+
(π/45)^5
5!
−
(π/45)^7
7!
+···
Step 2: For five decimal place accuracy, we must find the value ofnfor which the absolute
error is less than or equal to 5∣ × 10 −^6. For allx,|f(n+1)(x)|≤1. Choosex 0 =0.
∣
∣∣Rn
(
π
45
)∣∣
∣∣≤^1
(n+1)!
∣∣
∣∣π
45
− 0
∣∣
∣∣
n+ 1
=
(π/45)n+^1
(n+1)!
. The absolute error is less than or
equal to 5× 10 −^6 forn≥3.
Step 3: sin 4◦=sin
π
45
=
π
45
−
(π/45)
3!
3
+
(π/45)^5
5!
−
(π/45)^7
7!
= 0. 069756
14.8 Rapid Review
- Find the sum of the series 81+ 27 + 9 + 3 + 1 +
1
3
+···.
Answer:This is a geometric series with first term 81 and a ratio of
1
3
,
soS=
81
1 − 1 / 3
=
243
2
.
- Determine whether the series
∑∞
n= 1
5 n
n!
converges or diverges.
Answer:nlim→∞
5 n+^1
(n+1)!
·
n!
5 n
=nlim→∞
5
n+ 1
=0so
∑∞
n= 1
5 n
n!
converges by ratio test.
- Determine whether the series
∑∞
n= 1
lnn
√
n
converges or diverges.
Answer:
lnn
√
n
>
1
√
n
forn>e.
∑∞
n= 1
1
√
n
is ap-series withp<1, so it diverges.
∑∞
n= 1
lnn
√
n
=
ln 2
√
2
+
∑∞
n= 3
lnn
√
n
and
∑∞
n= 3
lnn
√
n
diverges by comparison to
∑∞
n= 3
1
√
n
, so the
series diverges.