ma112z02/HWSolutions/11.10 Solutions.pdf
Transcript of ma112z02/HWSolutions/11.10 Solutions.pdf
NOT FOR SALE
1040 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
41. By Example 7, tan−1 { =∞Sq=0
(−1)q{2q+1
2q+ 1for |{| ? 1. In particular, for { = 1√
3, we
have �
6= tan−1
�1√3
�=
∞Sq=0
(−1)q�1@√3�2q+1
2q+ 1=
∞Sq=0
(−1)q�1
3
�q1√3
1
2q+ 1, so
� =6√3
∞Sq=0
(−1)q
(2q+ 1)3q= 2√3∞Sq=0
(−1)q
(2q+ 1)3q.
42. (a)] 1@2
0
g{
{2 − {+ 1=
] 1@2
0
g{
({− 1@2)2 + 3@4
�{−
1
2=
√3
2x, x =
2√3
�{−
1
2
�, g{ =
√3
2gx
�
=
] 0
−1@√3
�√3@2�gx
(3@4)(x2 + 1)=2√3
3
ktan−1 x
l0−1@
√3
=2√3
k0−
�−�
6
�l=
�
3√3
(b) 1
{3 + 1=
1
({+ 1)({2 − {+ 1)⇒
1
{2 − {+ 1= ({+ 1)
�1
1 + {3
�= ({+ 1)
1
1− (−{3)= ({+ 1)
∞Sq=0
(−1)q{3q
=∞Sq=0
(−1)q{3q+1 +∞Sq=0
(−1)q{3q for |{| ? 1 ⇒
]g{
{2 − {+ 1= F +
∞Sq=0
(−1)q{3q+2
3q+ 2+
∞Sq=0
(−1)q{3q+1
3q+ 1for |{| ? 1 ⇒
] 1@2
0
g{
{2 − {+ 1=
∞Sq=0
(−1)q�
1
4 · 8q(3q+ 2)+
1
2 · 8q(3q+ 1)
�=1
4
∞Sq=0
(−1)q
8q
�2
3q+ 1+
1
3q+ 2
�.
By part (a), this equals �
3√3, so � = 3
√3
4
∞Sq=0
(−1)q
8q
�2
3q+ 1+
1
3q+ 2
�.
11.10 Taylor and Maclaurin Series
1. Using Theorem 5 with∞Sq=0
eq({− 5)q, eq =i (q)(d)
q!, so e8 =
i (8)(5)
8!.
2. (a) Using Equation 6, a power series expansion of i at 1 must have the form i(1) + i 0(1)({− 1) + · · · . Comparing to the
given series, 1=6− 0=8({− 1) + · · · , we must have i 0(1) = −0=8. But from the graph, i 0(1) is positive. Hence, the given
series is not the Taylor series of i centered at 1.
(b) A power series expansion of i at 2 must have the form i(2) + i 0(2)({− 2) + 12i 00(2)({− 2)2 + · · · . Comparing to the
given series, 2=8+ 0=5({− 2) + 1=5({− 2)2 − 0=1({− 2)3 + · · · , we must have 12i 00(2) = 1=5; that is, i 00(2) is positive.
But from the graph, i is concave downward near { = 2, so i 00(2) must be negative. Hence, the given series is not the
Taylor series of i centered at 2.
3. Since i (q)(0) = (q+ 1)!, Equation 7 gives the Maclaurin series
∞Sq=0
i (q)(0)
q!{q =
∞Sq=0
(q+ 1)!
q!{q =
∞Sq=0
(q+ 1){q. Applying the Ratio Test with dq = (q+ 1){q gives us
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 1041
limq→∞
����dq+1dq
���� = limq→∞
����(q+ 2){q+1
(q+ 1){q
���� = |{| limq→∞
q+ 2
q+ 1= |{| · 1 = |{|. For convergence, we must have |{| ? 1, so the
radius of convergence U = 1.
4. Since i (q)(4) = (−1)q q!3q(q+ 1)
, Equation 6 gives the Taylor series
∞Sq=0
i (q)(4)
q!({− 4)q =
∞Sq=0
(−1)q q!3q(q+ 1)q!
({− 4)q =∞Sq=0
(−1)q
3q(q+ 1)({− 4)q, which is the Taylor series for i
centered at 4. Apply the Ratio Test to find the radius of convergence U.
limq→∞
����dq+1dq
����= limq→∞
����(−1)q+1({− 4)q+1
3q+1(q+ 2)·
3q(q+ 1)
(−1)q({− 4)q
���� = limq→∞
����(−1)({− 4)(q+ 1)
3(q+ 2)
����
=1
3|{− 4| lim
q→∞
q+ 1
q+ 2=1
3|{− 4|
For convergence, 13|{− 4| ? 1 ⇔ |{− 4| ? 3, so U = 3.
5.q i (q)({) i (q)(0)
0 {h{ 0
1 ({+ 1)h{ 1
2 ({+ 2)h{ 2
3 ({+ 3)h{ 3
4 ({+ 4)h{ 4
Using Equation 6 with q = 0 to 4 and d = 0, we get
4Sq=0
i (q)(0)
q!({− 0)q =
0
0!{0 +
1
1!{1 +
2
2!{2 +
3
3!{3 +
4
4!{4
= {+ {2 + 12{3 + 1
6{4
6.q i (q)({) i (q)(2)
01
1 + {13
1 −1
(1 + {)2− 19
22
(1 + {)3227
3 −6
(1 + {)4− 681
3Sq=0
i (q)(2)
q!({− 2)q =
13
0!({− 2)0 −
19
1!({− 2)1
+227
2!({− 2)2 −
681
3!({− 2)3
= 13− 1
9({− 2) + 1
27({− 2)2 − 1
81({− 2)3
7.q i (q)({) i (q)(8)
0 3√{ 2
11
3{2@3112
2 −2
9{5@3− 2288
310
27{8@3106912
3Sq=0
i (q)(8)
q!({− 8)q =
2
0!({− 8)0 +
112
1!({− 8)1
−2288
2!({− 8)2 +
106912
3!({− 8)3
= 2 + 112({− 8)− 1
288({− 8)2 + 5
20,736 ({− 8)3
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
1042 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
8.q i (q)({) i (q)(1)
0 ln{ 0
1 1@{ 1
2 −1@{2 −1
3 2@{3 2
4 −6@{4 −6
4Sq=0
i (q)(1)
q!({− 1)q =
0
0!({− 1)0 +
1
1!({− 1)1 −
1
2!({− 1)2
+2
3!({− 1)3 −
6
4!({− 1)4
= ({− 1)− 12({− 1)2 + 1
3({− 1)3 − 1
4({− 1)4
9.q i (q)({) i (q)(�@6)
0 sin{ 1@2
1 cos{√3@2
2 − sin{ −1@2
3 − cos{ −√3@2
3Sq=0
i (q)(�@6)
q!
�{−
�
6
�q=1@2
0!
�{−
�
6
�0+
√3@2
1!
�{−
�
6
�1−1@2
2!
�{−
�
6
�2−√3@2
3!
�{−
�
6
�3
=1
2+
√3
2
�{−
�
6
�−1
4
�{−
�
6
�2−√3
12
�{−
�
6
�3
10.q i (q)({) i (q)(0)
0 cos2 { 1
1 −2 cos{ sin{ = − sin 2{ 0
2 −2 cos 2{ −2
3 4 sin 2{ 0
4 8 cos 2{ 8
5 −16 sin 2{ 0
6 −32 cos 2{ −32
6Sq=0
i (q)(0)
q!({− 0)q =
1
0!{0 −
2
2!{2 +
8
4!{4 −
32
6!{6
= 1− {2 + 13{4 − 2
45{6
11.q i (q)({) i (q)(0)
0 (1− {)−2 1
1 2(1− {)−3 2
2 6(1− {)−4 6
3 24(1− {)−5 24
4 120(1− {)−6 120
......
...
(1− {)−2 = i(0) + i 0(0){+i 00(0)
2!{2 +
i 000(0)
3!{3 +
i (4)(0)
4!{4 + · · ·
= 1 + 2{+ 62{2 + 24
6{3 + 120
24{4 + · · ·
= 1 + 2{+ 3{2 + 4{3 + 5{4 + · · · =∞Sq=0
(q+ 1){q
limq→∞
����dq+1dq
���� = limq→∞
����(q+ 2){q+1
(q+ 1){q
���� = |{| limq→∞
q+ 2
q+ 1= |{| (1) = |{| ? 1
for convergence, so U = 1.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 1043
12.q i (q)({) i (q)(0)
0 ln(1 + {) 0
1 (1 + {)−1 1
2 − (1 + {)−2 −1
3 2(1 + {)−3 2
4 −6(1 + {)−4 −6
5 24(1 + {)−5 24
......
...
ln(1 + {) = i(0) + i 0(0){+i 00(0)
2!{2
+i 000(0)
3!{3 +
i (4)(0)
4!{4 +
i (5)(0)
5!{5 + · · ·
= 0 + {− 12{2 + 2
6{3 − 6
24{4 + 24
120{5 − · · ·
= {−{2
2+
{3
3−
{4
4+
{5
5− · · · =
∞Sq=1
(−1)q−1
q{q
limq→∞
����dq+1dq
���� = limq→∞
����{q+1
q+ 1·q
{q
���� = limq→∞
|{|1 + 1@q
= |{| ? 1 for convergence,
so U = 1.
Notice that the answer agrees with the entry for ln(1 + {) in Table 1, but we obtained it by a different method. (Compare with
Example 11.9.6.)
13.q i (q)({) i (q)(0)
0 cos{ 1
1 − sin{ 0
2 − cos{ −1
3 sin{ 0
4 cos{ 1
......
...
cos{= i(0) + i 0(0){+i 00(0)
2!{2 +
i 000(0)
3!{3 +
i (4)(0)
4!{4 + · · ·
= 1−1
2!{2 +
1
4!{4 − · · ·
=∞Sq=0
(−1)q{2q
(2q)![Equal to (16).]
limq→∞
����dq+1dq
���� = limq→∞
����{2q+2
(2q+ 2)!·(2q)!
{2q
���� = limq→∞
{2
(2q+ 2)(2q+ 1)= 0 ? 1
for all {, so U =∞.
14.q i (q)({) i (q)(0)
0 h−2{ 1
1 −2h−2{ −2
2 4h−2{ 4
3 −8h−2{ −8
4 16h−2{ 16
......
...
h−2{ =∞Sq=0
i (q)(0)
q!{q =
∞Sq=0
(−2)q
q!{q.
limq→∞
����dq+1dq
����= limq→∞
����(−2)q+1{q+1
(q+ 1)!·
q!
(−2)q{q
���� = limq→∞
2 |{|q+ 1
= 0 ? 1 for all {, so U =∞=
15.q i (q)({) i (q)(0)
0 2{ 1
1 2{(ln 2) ln 2
2 2{(ln 2)2 (ln 2)2
3 2{(ln 2)3 (ln 2)3
4 2{(ln 2)4 (ln 2)4
......
...
2{ =∞Sq=0
i (q)(0)
q!{q =
∞Sq=0
(ln 2)q
q!{q.
limq→∞
����dq+1dq
����= limq→∞
����(ln 2)q+1{q+1
(q+ 1)!·
q!
(ln 2)q{q
����
= limq→∞
(ln 2) |{|q+ 1
= 0 ? 1 for all {, so U =∞.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
1044 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
16.q i (q)({) i (q)(0)
0 { cos{ 0
1 −{ sin{+ cos{ 1
2 −{ cos{− 2 sin{ 0
3 { sin{− 3 cos{ −3
4 { cos{+ 4 sin{ 0
5 −{ sin{+ 5 cos{ 5
6 −{ cos{− 6 sin{ 0
7 { sin{− 7 cos{ −7...
......
{ cos{ = i(0) + i 0(0){+i 00(0)
2!{2 +
i 000(0)
3!{3 +
i (4)(0)
4!{4 + · · ·
= 0 + 1{+ 0−3
3!{3 + 0 +
5
5!{5 + 0−
7
7!{7 + · · ·
= {−1
2!{3 +
1
4!{5 −
1
6!{7 + · · ·
=∞Sq=0
(−1)q1
(2q)!{2q+1
limq→∞
����dq+1dq
����= limq→∞
����(−1)q+1{2q+3
(2q+ 2)!·
(2q)!
(−1)q{2q+1
����
= limq→∞
{2
(2q+ 2)(2q+ 1)= 0 ? 1 for all {, so U =∞.
17.q i (q)({) i (q)(0)
0 sinh{ 0
1 cosh{ 1
2 sinh{ 0
3 cosh{ 1
4 sinh{ 0
......
...
i (q)(0) =
+0 if q is even
1 if q is oddso sinh{ =
∞Sq=0
{2q+1
(2q+ 1)!.
Use the Ratio Test to find U. If dq ={2q+1
(2q+ 1)!, then
limq→∞
����dq+1dq
����= limq→∞
����{2q+3
(2q+ 3)!·(2q+ 1)!
{2q+1
���� = {2 · limq→∞
1
(2q+ 3)(2q+ 2)
= 0 ? 1 for all {, so U =∞.
18.q i (q)({) i (q)(0)
0 cosh{ 1
1 sinh{ 0
2 cosh{ 1
3 sinh{ 0
......
...
i (q)(0) =
+1 if q is even
0 if q is oddso cosh{ =
∞Sq=0
{2q
(2q)!.
Use the Ratio Test to find U. If dq ={2q
(2q)!, then
limq→∞
����dq+1dq
����= limq→∞
����{2q+2
(2q+ 2)!·(2q)!
{2q
���� = {2 · limq→∞
1
(2q+ 2)(2q+ 1)
= 0 ? 1 for all {, so U =∞=
19.q i (q)({) i (q)(2)
0 {5 + 2{3 + { 50
1 5{4 + 6{2 + 1 105
2 20{3 + 12{ 184
3 60{2 + 12 252
4 120{ 240
5 120 120
6 0 0
7 0 0
......
...
i (q)({) = 0 for q ≥ 6, so i has a finite expansion about d = 2.
i({) = {5 + 2{3 + { =5S
q=0
i (q)(2)
q!({− 2)q
=50
0!({− 2)0 +
105
1!({− 2)1 +
184
2!({− 2)2 +
252
3!({− 2)3
+240
4!({− 2)4 +
120
5!({− 2)5
= 50 + 105({− 2) + 92({− 2)2 + 42({− 2)3
+ 10({− 2)4 + ({− 2)5
A finite series converges for all {> so U =∞=
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 1045
20.q i (q)({) i (q)(−2)
0 {6 − {4 + 2 50
1 6{5 − 4{3 −160
2 30{4 − 12{2 432
3 120{3 − 24{ −912
4 360{2 − 24 1416
5 720{ −1440
6 720 720
7 0 0
8 0 0
......
...
i (q)({) = 0 for q ≥ 7, so i has a finite expansion about d = −2.
i({) = {6 − {4 + 2 =6S
q=0
i (q)(−2)q!
({+ 2)q
=50
0!({+ 2)0 −
160
1!({+ 2)1 +
432
2!({+ 2)2 −
912
3!({+ 2)3
+1416
4!({+ 2)4 −
1440
5!({+ 2)5 +
720
6!({+ 2)6
= 50− 160({+ 2) + 216({+ 2)2 − 152({+ 2)3 + 59({+ 2)4 − 12({+ 2)5 + ({+ 2)6
A finite series converges for all {> so U =∞.
21.q i (q)({) i (q)(2)
0 ln{ ln 2
1 1@{ 1@2
2 −1@{2 −1@22
3 2@{3 2@23
4 −6@{4 −6@24
5 24@{5 24@25
......
...
i({) = ln{ =∞Sq=0
i (q)(2)
q!({− 2)q
=ln 2
0!({− 2)0 +
1
1! 21({− 2)1 +
−12! 22
({− 2)2 +2
3! 23({− 2)3
+−64! 24
({− 2)4 +24
5! 25({− 2)5 + · · ·
= ln2 +∞Sq=1
(−1)q+1(q− 1)!q! 2q
({− 2)q
= ln2 +∞Sq=1
(−1)q+11
q 2q({− 2)q
limq→∞
����dq+1dq
����= limq→∞
����(−1)q+2({− 2)q+1
(q+ 1) 2q+1·
q 2q
(−1)q+1({− 2)q
���� = limq→∞
����(−1)({− 2)q(q+ 1)2
���� = limq→∞
�q
q+ 1
�|{− 2|2
=|{− 2|2
? 1 for convergence, so |{− 2| ? 2 and U = 2.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
1046 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
22.q i (q)({) i (q)(−3)
0 1@{ −1@3
1 −1@{2 −1@32
2 2@{3 −2@33
3 −6@{4 −6@34
4 24@{5 −24@35
......
...
i({) =1
{=
∞Sq=0
i (q)(−3)q!
({+ 3)q
=−1@30!
({+ 3)0 +−1@32
1!({+ 3)1 +
−2@33
2!({+ 3)2
+−6@34
3!({+ 3)3 +
−24@35
4!({+ 3)4 + · · ·
=∞Sq=0
−q!@3q+1
q!({+ 3)q = −
∞Sq=0
({+ 3)q
3q+1
limq→∞
����dq+1dq
���� = limq→∞
����({+ 3)q+1
3q+2·3q+1
({+ 3)q
���� = limq→∞
|{+ 3|3
=|{+ 3|3
? 1 for convergence,
so |{+ 3| ? 3 and U = 3.
23.q i (q)({) i (q)(3)
0 h2{ h6
1 2h2{ 2h6
2 22h2{ 4h6
3 23h2{ 8h6
4 24h2{ 16h6
......
...
i({) = h2{ =∞Sq=0
i (q)(3)
q!({− 3)q
=h6
0!({− 3)0 +
2h6
1!({− 3)1 +
4h6
2!({− 3)2
+8h6
3!({− 3)3 +
16h6
4!({− 3)4 + · · ·
=∞Sq=0
2qh6
q!({− 3)q
limq→∞
����dq+1dq
���� = limq→∞
����2q+1h6({− 3)q+1
(q+ 1)!·
q!
2qh6({− 3)q
���� = limq→∞
2 |{− 3|q+ 1
= 0 ? 1 for all {, so U =∞.
24.q i (q)({) i (q)(�@2)
0 cos{ 0
1 − sin{ −1
2 − cos{ 0
3 sin{ 1
4 cos{ 0
5 − sin{ −1
6 − cos{ 0
7 sin{ 1
......
...
i({) = cos{ =∞Sq=0
i (q)(�@2)
q!
�{−
�
2
�q
=−11!
�{−
�
2
�1+1
3!
�{−
�
2
�3+−15!
�{−
�
2
�5+1
7!
�{−
�
2
�7+ · · ·
=∞Sq=0
(−1)q+1
(2q+ 1)!
�{−
�
2
�2q+1
limq→∞
����dq+1dq
����= limq→∞
�������
(−1)q+2�{−
�
2
�2q+3
(2q+ 3)!·
(2q+ 1)!
(−1)q+1�{−
�
2
�2q+1
�������
= limq→∞
�{−
�
2
�2
(2q+ 3)(2q+ 2)= 0 ? 1 for all {, so U =∞.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 1047
25.q i (q)({) i (q)(�)
0 sin{ 0
1 cos{ −1
2 − sin{ 0
3 − cos{ 1
4 sin{ 0
5 cos{ −1
6 − sin{ 0
7 − cos{ 1
......
...
i({) = sin{ =∞Sq=0
i (q)(�)
q!({− �)q
=−11!({− �)1 +
1
3!({− �)3 +
−15!({− �)5 +
1
7!({− �)7 + · · ·
=∞Sq=0
(−1)q+1
(2q+ 1)!({− �)2q+1
limq→∞
����dq+1dq
����= limq→∞
����(−1)q+2 ({− �)2q+3
(2q+ 3)!·
(2q+ 1)!
(−1)q+1 ({− �)2q+1
����
= limq→∞
({− �)2
(2q+ 3)(2q+ 2)= 0 ? 1 for all {, so U =∞.
26.q i (q)({) i (q)(16)
0√{ 4
1 12{−1@2
1
2·1
4
2 − 14{−3@2 −
1
4·1
43
3 38{−5@2
3
8·1
45
4 − 1516{−7@2 −
15
16·1
47...
......
i({) =√{ =
∞Sq=0
i (q)(16)
q!({− 16)q
=4
0!({− 16)0 +
1
2·1
4·1
1!({− 16)1 −
1
4·1
43·1
2!({− 16)2
+3
8·1
45·1
3!({− 16)3 −
15
16·1
47·1
4!({− 16)4 + · · ·
= 4 +1
8({− 16) +
∞Sq=2
(−1)q−11 · 3 · 5 · · · · · (2q− 3)
2q42q−1 q!({− 16)q
= 4 +1
8({− 16) +
∞Sq=2
(−1)q−11 · 3 · 5 · · · · · (2q− 3)
25q−2 q!({− 16)q
limq→∞
����dq+1dq
����= limq→∞
����(−1)q 1 · 3 · 5 · · · · · (2q− 1)({− 16)q+1
25q+3(q+ 1)!·
25q−2q!
(−1)q−1 1 · 3 · 5 · · · · · (2q− 3)({− 16)q
����
= limq→∞
(2q− 1) |{− 16|25(q+ 1)
=|{− 16|32
limq→∞
2− 1@q1 + 1@q
=|{− 16|32
· 2
=|{− 16|16
? 1 for convergence, so |{− 16| ? 16 and U = 16.
27. If i({) = cos{, then i (q+1)({) = ± sin{ or ± cos{. In each case,���i (q+1)({)
��� ≤ 1, so by Formula 9 with d = 0 and
P = 1, |Uq({)| ≤1
(q+ 1)!|{|q+1. Thus, |Uq({)|→ 0 as q→∞ by Equation 10. So lim
q→∞Uq({) = 0 and, by Theorem
8, the series in Exercise 13 represents cos{ for all {=
28. If i({) = sin{, then i (q+1)({) = ± sin{ or ± cos{. In each case,���i (q+1)({)
��� ≤ 1, so by Formula 9 with d = 0 and
P = 1, |Uq({)| ≤1
(q+ 1)!|{− �|q+1. Thus, |Uq({)|→ 0 as q→∞ by Equation 10. So lim
q→∞Uq({)→ 0 and, by
Theorem 8, the series in Exercise 25 represents sin{ for all {=
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
1048 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
29. If i({) = sinh{, then for all q, i (q+1)({) = cosh{ or sinh{. Since |sinh{| ? |cosh{| = cosh{ for all {, we have���i (q+1)({)
��� ≤ cosh{ for all q. If g is any positive number and |{| ≤ g, then���i (q+1)({)
��� ≤ cosh{ ≤ cosh g, so by
Formula 9 with d = 0 andP = cosh g, we have |Uq({)| ≤cosh g
(q+ 1)!|{|q+1. It follows that |Uq({)|→ 0 as q→∞ for
|{| ≤ g (by Equation 10). But g was an arbitrary positive number. So by Theorem 8, the series represents sinh{ for all {.
30. If i({) = cosh{, then for all q, i (q+1)({) = cosh{ or sinh{. Since |sinh{| ? |cosh{| = cosh{ for all {, we have���i (q+1)({)
��� ≤ cosh{ for all q. If g is any positive number and |{| ≤ g, then���i (q+1)({)
��� ≤ cosh{ ≤ cosh g, so by
Formula 9 with d = 0 andP = cosh g, we have |Uq({)| ≤cosh g
(q+ 1)!|{|q+1. It follows that |Uq({)|→ 0 as q→∞ for
|{| ≤ g (by Equation 10). But g was an arbitrary positive number. So by Theorem 8, the series represents cosh{ for all {.
31. 4√1− {= [1 + (−{)]1@4 =
∞Sq=0
#1@4
q
$(−{)q = 1 + 1
4(−{) +
14
�−34
�
2!(−{)2 +
14
�−34
� �− 74
�
3!(−{)3 + · · ·
= 1−1
4{+
∞Sq=2
(−1)q−1(−1)q · [3 · 7 · · · · · (4q− 5)]4q · q!
{q
= 1−1
4{−
∞Sq=2
3 · 7 · · · · · (4q− 5)4q · q!
{q
and |−{| ? 1 ⇔ |{| ? 1, so U = 1.
32. 3√8 + {= 3
u8�1 +
{
8
�= 2
�1 +
{
8
�1@3= 2
∞Sq=0
#1@3
q
$�{8
�q
= 2
%1 +
1
3
�{8
�+
13
�−23
�
2!
�{8
�2+
13
�− 23
��− 53
�
3!
�{8
�3+ · · ·
&
= 2
�1 +
1
24{+
∞Sq=2
(−1)q−1 · [2 · 5 · · · · · (3q− 4)]3q · 8q · q!
{q�
= 2 +1
12{+ 2
∞Sq=2
(−1)q−1[2 · 5 · · · · · (3q− 4)]24q · q!
{q
and���{
8
��� ? 1 ⇔ |{| ? 8, so U = 8.
33. 1
(2 + {)3=
1
[2(1 + {@2)]3=1
8
�1 +
{
2
�−3=1
8
∞Sq=0
#−3q
$�{2
�q. The binomial coefficient is
#−3q
$=(−3)(−4)(−5) · · · · · (−3− q+ 1)
q!=(−3)(−4)(−5) · · · · · [−(q+ 2)]
q!
=(−1)q · 2 · 3 · 4 · 5 · · · · · (q+ 1)(q+ 2)
2 · q!=(−1)q(q+ 1)(q+ 2)
2
Thus, 1
(2 + {)3=1
8
∞Sq=0
(−1)q(q+ 1)(q+ 2)2
{q
2q=
∞Sq=0
(−1)q(q+ 1)(q+ 2){q
2q+4for���{
2
��� ? 1 ⇔ |{| ? 2, so U = 2.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 1049
34. (1 + {)3@4 =∞Sq=0
#34
q
${q = 1 +
3
4{+
34
�− 14
�
2!{2 +
34
�− 14
��−54
�
3!{3 + · · ·
= 1 +3
4{+
∞Sq=2
(−1)q−1 · 3 · [1 · 5 · 9 · · · · · (4q− 7)]4q · q!
{q
for |{| ? 1, so U = 1.
35. arctan{ =∞Sq=0
(−1)q{2q+1
2q+ 1, so i({) = arctan({2) =
∞Sq=0
(−1)q�{2�2q+1
2q+ 1=
∞Sq=0
(−1)q1
2q+ 1{4q+2, U = 1.
36. sin{ =∞Sq=0
(−1)q{2q+1
(2q+ 1)!, so i({) = sin
��4{�=
∞Sq=0
(−1)q��4{�2q+1
(2q+ 1)!=
∞Sq=0
(−1)q�2q+1
42q+1(2q+ 1)!{2q+1, U =∞.
37. cos{ =∞Sq=0
(−1)q{2q
(2q)!⇒ cos 2{ =
∞Sq=0
(−1)q(2{)2q
(2q)!=
∞Sq=0
(−1)q22q{2q
(2q)!, so
i({) = { cos 2{ =∞Sq=0
(−1)q22q
(2q)!{2q+1, U =∞.
38. h{ =∞Sq=0
{q
q!, so i({) = h3{ − h2{ =
∞Sq=0
(3{)q
q!−
∞Sq=0
(2{)q
q!=
∞Sq=0
3q{q
q!−
∞Sq=0
2q{q
q!=
∞Sq=0
3q − 2q
q!{q, U =∞.
39. cos{ =∞Sq=0
(−1)q{2q
(2q)!⇒ cos
�12{
2�=
∞Sq=0
(−1)q�12{2�2q
(2q)!=
∞Sq=0
(−1)q{4q
22q (2q)!, so
i({) = { cos�12{2�=
∞Sq=0
(−1)q1
22q(2q)!{4q+1, U =∞.
40. ln(1 + {) =∞Sq=1
(−1)q−1{q
q⇒ ln(1 + {3) =
∞Sq=1
(−1)q−1{3q
q, so i({) = {2 ln(1 + {3) =
∞Sq=1
(−1)q−1{3q+2
q,
U = 1.
41. We must write the binomial in the form (1+ expression), so we’ll factor out a 4.
{√4 + {2
={s
4(1 + {2@4)=
{
2s1 + {2@4
={
2
�1 +
{2
4
�−1@2=
{
2
∞Sq=0
#− 12
q
$�{2
4
�q
={
2
%1 +
�− 12
�{2
4+
�− 12
��−32
�
2!
�{2
4
�2+
�− 12
��−32
��−52
�
3!
�{2
4
�3+ · · ·
&
={
2+
{
2
∞Sq=1
(−1)q1 · 3 · 5 · · · · · (2q− 1)
2q · 4q · q!{2q
={
2+
∞Sq=1
(−1)q1 · 3 · 5 · · · · · (2q− 1)
q! 23q+1{2q+1 and {2
4? 1 ⇔
|{|2
? 1 ⇔ |{| ? 2, so U = 2.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
1050 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
42. {2√2 + {
={2s
2 (1 + {@2)=
{2√2
�1 +
{
2
�−1@2=
{2√2
∞Sq=0
#− 12
q
$�{2
�q
={2√2
%1 +
�− 12
��{2
�+
�− 12
��−32
�
2!
�{2
�2+
�− 12
��−32
��−52
�
3!
�{2
�3+ · · ·
&
={2√2+
{2√2
∞Sq=1
(−1)q1 · 3 · 5 · · · · · (2q− 1)
q! 22q{q
={2√2+
∞Sq=1
(−1)q1 · 3 · 5 · · · · · (2q− 1)
q! 22q+1@2{q+2 and
���{
2
��� ? 1 ⇔ |{| ? 2, so U = 2.
43. sin2 { = 1
2(1− cos 2{) =
1
2
�1−
∞Sq=0
(−1)q(2{)2q
(2q)!
�=1
2
�1− 1−
∞Sq=1
(−1)q(2{)2q
(2q)!
�=
∞Sq=1
(−1)q+122q−1{2q
(2q)!,
U =∞
44. {− sin{{3
=1
{3
�{−
∞Sq=0
(−1)q{2q+1
(2q+ 1)!
�=1
{3
�{− {−
∞Sq=1
(−1)q{2q+1
(2q+ 1)!
�=1
{3
�−
∞Sq=0
(−1)q+1{2q+3
(2q+ 3)!
�
=1
{3
∞Sq=0
(−1)q{2q+3
(2q+ 3)!=
∞Sq=0
(−1)q{2q
(2q+ 3)!
and this series also gives the required value at { = 0 (namely 1@6); U =∞.
45. cos{ (16)=
∞Sq=0
(−1)q{2q
(2q)!⇒
i({) = cos({2) =∞Sq=0
(−1)q ({2)2q
(2q)!=
∞Sq=0
(−1)q{4q
(2q)!
= 1− 12{4 + 1
24{8 − 1
720{12 + · · ·
The series for cos{ converges for all {, so the same is true of the series for
i({), that is, U =∞. Notice that, as q increases, Wq({) becomes a better
approximation to i({).
46. ln(1 + {) =∞Sq=1
(−1)q−1{q
q⇒
i({) = ln(1 + {2) =∞Sq=1
(−1)q−1({2)q
q=
∞Sq=1
(−1)q−1{2q
q
= {2 − 12{4 + 1
3{6 − 1
4{8 + · · ·
The series for ln(1 + {) has U = 1 and��{2�� ? 1 ⇔ |{| ? 1,
so the series for i({) also has U = 1. From the graphs of i and
the first few Taylor polynomials, we see that Wq({) provides a
closer fit to i({) near 0 as q increases.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 1051
47. h{ (11)=
∞Sq=0
{q
q!, so h−{ =
∞Sq=0
(−{)q
q!=
∞Sq=0
(−1)q{q
q!, so
i({) = {h−{ =∞Sq=0
(−1)q1
q!{q+1
= {− {2 + 12{3 − 1
6{4 + 1
24{5 − 1
120{6 + · · ·
=∞Sq=1
(−1)q−1{q
(q− 1)!
The series for h{ converges for all {, so the same is true of the series
for i({); that is, U =∞. From the graphs of i and the first few Taylor
polynomials, we see that Wq({) provides a closer fit to i({) near 0 as q increases.
48. From Table 1, tan−1 { =∞Sq=0
(−1)q{2q+1
2q+ 1, so
i({) = tan−1({3) =∞Sq=0
(−1)q({3)2q+1
2q+ 1=
∞Sq=0
(−1)q{6q+3
2q+ 1
= {3 − 13{9 + 1
5{15 − 1
7{21 + · · ·
The series for tan−1 { has U = 1 and��{3�� ? 1 ⇔ |{| ? 1,
so the series for i({) also has U = 1. From the graphs of i and
the first few Taylor polynomials, we see that Wq({) provides a
closer fit to i({) near 0 as q increases.
49. 5◦ = 5◦� �
180◦
�=
�
36radians and cos{ =
∞Sq=0
(−1)q{2q
(2q)!= 1 −
{2
2!+
{4
4!−
{6
6!+ · · · , so
cos�
36= 1−
(�@36)2
2!+(�@36)4
4!−(�@36)6
6!+ · · · . Now 1− (�@36)2
2!≈ 0=99619 and adding (�@36)
4
4!≈ 2=4× 10−6
does not affect the fifth decimal place, so cos 5◦ ≈ 0=99619 by the Alternating Series Estimation Theorem.
50. 1@10√h = h−1@10 and h{ =
∞Sq=0
{q
q!= 1 + { +
{2
2!+
{3
3!+ · · · , so
h−1@10 = 1 −1
10+(1@10)2
2!−(1@10)3
3!+(1@10)4
4!−(1@10)5
5!+ · · · . Now
1−1
10+(1@10)2
2!−(1@10)3
3!+(1@10)4
4!≈ 0=90484 and subtracting (1@10)
5
5!≈ 8=3× 10−8 does not affect the fifth
decimal place, so h−1@10 ≈ 0=90484 by the Alternating Series Estimation Theorem.
51. (a) 1@√1− {2 =
�1 +
�−{2
��−1@2= 1 +
�− 12
��−{2
�+
�−12
��−32
�
2!
�−{2
�2+
�− 12
��− 32
��− 52
�
3!
�−{2
�3+ · · ·
= 1 +∞Sq=1
1 · 3 · 5 · · · · · (2q− 1)2q · q!
{2q
(b) sin−1 { =]
1√1− {2
g{ = F + {+∞Sq=1
1 · 3 · 5 · · · · · (2q− 1)(2q+ 1)2q · q!
{2q+1
= {+∞Sq=1
1 · 3 · 5 · · · · · (2q− 1)(2q+ 1)2q · q!
{2q+1 since 0 = sin−1 0 = F.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
1052 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
52. (a) 1@ 4√1 + { = (1 + {)−1@4 =
∞Sq=0
#− 14
q
${q = 1−
1
4{+
�− 14
��−54
�
2!{2 +
�− 14
��−54
� �− 94
�
3!{3 + · · ·
= 1−1
4{+
∞Sq=2
(−1)q1 · 5 · 9 · · · · · (4q− 3)
4q · q!{q
(b) 1@ 4√1 + { = 1− 1
4{+532{
2 − 15128{
3 + 1952048{
4 − · · · . 1@ 4√1=1 = 1@ 4
√1 + 0=1, so let { = 0=1. The sum of the first four
terms is then 1− 14(0=1) + 5
32(0=1)2 − 15
128(0=1)3 ≈ 0=976. The fifth term is 195
2048(0=1)4 ≈ 0=000 009 5, which does not
affect the third decimal place of the sum, so we have 1@ 4√1=1 ≈ 0=976. (Note that the third decimal place of the sum of the
first three terms is affected by the fourth term, so we need to use more than three terms for the sum.)
53.√1 + {3 = (1 + {3)1@2 =
∞Sq=0
#12
q
$({3)q =
∞Sq=0
#12
q
${3q ⇒
] s1 + {3 g{ = F +
∞Sq=0
#12
q
${3q+1
3q+ 1,
with U = 1.
54. sin{ =∞Sq=0
(−1)q{2q+1
(2q+ 1)!⇒ sin({2) =
∞Sq=0
(−1)q({2)2q+1
(2q+ 1)!=
∞Sq=0
(−1)q{4q+2
(2q+ 1)!⇒
{2 sin({2) =∞Sq=0
(−1)q{4q+4
(2q+ 1)!⇒
]{2 sin({2) g{ = F +
∞Sq=0
(−1)q{4q+5
(2q+ 1)!(4q+ 5), with U =∞.
55. cos{ (16)=
∞Sq=0
(−1)q{2q
(2q)!⇒ cos{− 1 =
∞Sq=1
(−1)q{2q
(2q)!⇒
cos{− 1{
=∞Sq=1
(−1)q{2q−1
(2q)!⇒
]cos{− 1
{g{ = F +
∞Sq=1
(−1)q{2q
2q · (2q)!, with U =∞.
56. arctan{ =∞Sq=0
(−1)q{2q+1
2q+ 1⇒ arctan({2) =
∞Sq=0
(−1)q({2)2q+1
2q+ 1=
∞Sq=0
(−1)q{4q+2
2q+ 1⇒
]arctan({2) g{ = F +
∞Sq=0
(−1)q{4q+3
(2q+ 1)(4q+ 3), with U = 1.
57. arctan{ =∞Sq=0
(−1)q{2q+1
2q+ 1for |{| ? 1, so {3 arctan{ =
∞Sq=0
(−1)q{2q+4
2q+ 1for |{| ? 1 and
]{3 arctan{g{ = F +
∞Sq=0
(−1)q{2q+5
(2q+ 1)(2q+ 5). Since 1
2? 1, we have
] 1@2
0
{3 arctan{g{ =∞Sq=0
(−1)q(1@2)2q+5
(2q+ 1)(2q+ 5)=(1@2)5
1 · 5−(1@2)7
3 · 7+(1@2)9
5 · 9−(1@2)11
7 · 11+ · · · . Now
(1@2)5
1 · 5−(1@2)7
3 · 7+(1@2)9
5 · 9≈ 0=0059 and subtracting (1@2)
11
7 · 11≈ 6=3× 10−6 does not affect the fourth decimal place,
soU 1@20
{3 arctan{g{ ≈ 0=0059 by the Alternating Series Estimation Theorem.
58. sin{ =∞Sq=0
(−1)q{2q+1
(2q+ 1)!for all {, so sin({4) =
∞Sq=0
(−1)q{8q+4
(2q+ 1)!for all { and
]sin({4) g{ = F +
∞Sq=0
(−1)q{8q+5
(2q+ 1)! (8q+ 5). Thus,
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 1053] 1
0
sin({4) g{ =∞Sq=0
(−1)q1
(2q+ 1)! (8q+ 5)=
1
1! · 5−
1
3! · 13+
1
5! · 21−
1
7! · 29+ · · · . Now
1
1! · 5−
1
3! · 13+
1
5! · 21≈ 0=1876 and subtracting 1
7! · 29≈ 6=84× 10−6 does not affect the fourth decimal place, so
U 10sin({4) g{ ≈ 0=1876 by the Alternating Series Estimation Theorem.
59.√1 + {4 = (1 + {4)1@2 =
∞Sq=0
#12
q
$({4)q, so
] s1 + {4 g{ = F +
∞Sq=0
#12
q
${4q+1
4q+ 1and hence, since 0=4 ? 1,
we have
L =
] 0=4
0
s1 + {4 g{ =
∞Sq=0
#12
q
$(0=4)4q+1
4q+ 1
= (1)(0=4)1
0!+
12
1!
(0=4)5
5+
12
�− 12
�
2!
(0=4)9
9+
12
�− 12
��−32
�
3!
(0=4)13
13+
12
�− 12
��− 32
��− 52
�
4!
(0=4)17
17+ · · ·
= 0=4 +(0=4)5
10−(0=4)9
72+(0=4)13
208−5(0=4)17
2176+ · · ·
Now (0=4)9
72≈ 3=6× 10−6 ? 5× 10−6, so by the Alternating Series Estimation Theorem, L ≈ 0=4 + (0=4)5
10≈ 0=40102
(correct to five decimal places).
60.] 0=5
0
{2h−{2
g{ =
] 0=5
0
∞Sq=0
(−1)q {2q+2
q!g{ =
∞Sq=0
�(−1)q {2q+3
q!(2q+ 3)
�1@2
0
=∞Sq=0
(−1)q
q!(2q+ 3)22q+3and since the term
with q = 2 is 1
1792? 0=001, we use
1Sq=0
(−1)q
q!(2q+ 3)22q+3=1
24−
1
160≈ 0=0354.
61. lim{→0
{− ln(1 + {)
{2= lim
{→0
{− ({− 12{
2 + 13{
3 − 14{
4 + 15{
5 − · · · ){2
= lim{→0
12{
2 − 13{
3 + 14{
4 − 15{
5 + · · ·{2
= lim{→0
( 12 −13{+
14{
2 − 15{
3 + · · · ) = 12
since power series are continuous functions.
62. lim{→0
1− cos{1 + {− h{
= lim{→0
1−�1− 1
2!{2 + 1
4!{4 − 1
6!{6 + · · ·
�
1 + {−�1 + {+ 1
2!{2 + 1
3!{3 + 1
4!{4 + 1
5!{5 + 1
6!{6 + · · ·
�
= lim{→0
12!{2 − 1
4!{4 + 1
6!{6 − · · ·
− 12!{2 − 1
3!{3 − 1
4!{4 − 1
5!{5 − 1
6!{6 − · · ·
= lim{→0
12! −
14!{
2 + 16!{
4 − · · ·− 12!− 1
3!{− 1
4!{2 − 1
5!{3 − 1
6!{4 − · · ·
=12 − 0− 12− 0
= −1
since power series are continuous functions.
63. lim{→0
sin{− {+ 16{3
{5= lim
{→0
�{− 1
3!{3 + 1
5!{5 − 1
7!{7 + · · ·
�− {+ 1
6{3
{5
= lim{→0
15!{5 − 1
7!{7 + · · ·
{5= lim
{→0
�1
5!−
{2
7!+
{4
9!− · · ·
�=1
5!=
1
120
since power series are continuous functions.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
1054 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
64. lim{→0
√1 + {− 1− 1
2{
{2= lim
{→0
�1 + 1
2{− 1
8{2 + 1
16{3 − · · ·
�− 1− 1
2{
{2= lim
{→0
− 18{2 + 1
16{3 − · · ·
{2
= lim{→0
�− 18+ 1
16{− · · ·
�= − 1
8since power series are continuous functions.
65. lim{→0
{3 − 3{+ 3 tan−1 {{5
= lim{→0
{3 − 3{+ 3�{− 1
3{3 + 1
5{5 − 1
7{7 + · · ·
�
{5
= lim{→0
{3 − 3{+ 3{− {3 + 35{5 − 3
7{7 + · · ·
{5= lim
{→0
35{5 − 3
7{7 + · · ·
{5
= lim{→0
�35− 3
7{2 + · · ·
�= 3
5since power series are continuous functions.
66. lim{→0
tan{− {
{3= lim
{→0
�{+ 1
3{3 + 2
15{5 + · · ·
�− {
{3= lim
{→0
13{3 + 2
15{5 + · · ·
{3= lim
{→0
�13+ 2
15{2 + · · ·
�= 1
3
since power series are continuous functions.
67. From Equation 11, we have h−{2
= 1−{2
1!+
{4
2!−
{6
3!+ · · · and we know that cos{ = 1− {2
2!+
{4
4!− · · · from
Equation 16. Therefore, h−{2cos{ =
�1− {2 + 1
2{4 − · · ·
��1− 1
2{2 + 1
24{4 − · · ·
�. Writing only the terms with
degree ≤ 4, we get h−{2cos{ = 1− 1
2{2 + 1
24{4 − {2 + 1
2{4 + 1
2{4 + · · · = 1− 3
2{2 + 25
24{4 + · · · .
68. sec{ = 1
cos{
(16)=
1
1− 12{2 + 1
24{4 − · · ·
.
1 + 12{2 + 5
24{4 + · · ·
1− 12{2 + 1
24{4 − · · · 1
1− 12{2 + 1
24{4 − · · ·
12{2 − 1
24{4 + · · ·
12{2 − 1
4{4 + · · ·
524{4 + · · ·
524{4 + · · ·
· · ·From the long division above, sec{ = 1 + 1
2{2 + 5
24{4 + · · · .
69. {
sin{
(15)=
{
{− 16{3 + 1
120{5 − · · ·
.1 + 1
6{2 + 7
360{4 + · · ·
{− 16{3 + 1
120{5 − · · · {
{− 16{
3 + 1120{
5 − · · ·
16{3 − 1
120{5 + · · ·
16{3 − 1
36{5 + · · ·
7360
{5 + · · ·7360
{5 + · · ·
· · ·From the long division above, {
sin{= 1+ 1
6{2 + 7
360{4 + · · · .
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 1055
70. From Table 1, we have h{ = 1 + {
1!+
{2
2!+
{3
3!+ · · · and that ln(1 + {) = {−
{2
2+
{3
3−
{4
4+ · · · . Therefore,
| = h{ ln(1 + {) =
�1 +
{
1!+
{2
2!+
{3
3!+ · · ·
��{−
{2
2+
{3
3−
{4
4+ · · ·
�. Writing only terms with degree ≤ 3,
we get h{ ln(1 + {) = {− 12{2 + 1
3{3 + {2 − 1
2{3 + 1
2{3 + · · · = {+ 1
2{2 + 1
3{3 + · · · .
71. | = (arctan{)2 =�{− 1
3{3 + 1
5{5 − 1
7{7 + · · ·
� �{− 1
3{3 + 1
5{5 − 1
7{7 + · · ·
�. Writing only the terms with
degree ≤ 6, we get (arctan{)2 = {2 − 13{
4 + 15{
6 − 13{
4 + 19{
6 + 15{
6 + · · · = {2 − 23{
4 + 2345{
6 + · · · .
72. | = h{ sin2 { = (h{ sin{) sin{ =�{+ {2 + 1
3{3 + · · ·
� �{− 1
6{3 + · · ·
�[from Example 13]. Writing only the terms
with degree ≤ 4, we get h{ sin2 { = {2 − 16{4 + {3 + 1
3{4 + · · · = {2 + {3 + 1
6{4 + · · · .
73.∞Sq=0
(−1)q{4q
q!=
∞Sq=0
�−{4
�q
q!= h−{
4
, by (11).
74.∞Sq=0
(−1)q �2q
62q(2q)!=
∞Sq=0
(−1)q��6
�2q
(2q)!= cos �
6 =√32 , by (16).
75.∞Sq=1
(−1)q−13q
q5q=
∞Sq=1
(−1)q−1(3@5)q
q= ln
�1 +
3
5
�[from Table 1] = ln 8
5
76.∞Sq=0
3q
5q q!=
∞Sq=0
(3@5)q
q!= h3@5, by (11).
77.∞Sq=0
(−1)q �2q+1
42q+1(2q+ 1)!=
∞Sq=0
(−1)q��4
�2q+1
(2q+ 1)!= sin �
4 =1√2, by (15).
78. 1− ln 2 + (ln 2)2
2!−(ln 2)3
3!+ · · · =
∞Sq=0
(− ln 2)q
q!= h− ln 2 =
�hln 2
�−1= 2−1 = 1
2 , by (11).
79. 3 + 9
2!+27
3!+81
4!+ · · · =
31
1!+32
2!+33
3!+34
4!+ · · · =
∞Sq=1
3q
q!=
∞Sq=0
3q
q!− 1 = h3 − 1, by (11).
80. 1
1 · 2−
1
3 · 23+
1
5 · 25−
1
7 · 27+ · · · =
∞Sq=0
(−1)q1
(2q+ 1)22q+1=
∞Sq=0
(−1)q(1@2)2q+1
2q+ 1= tan−1
�1
2
�[from Table 1]
81. If s is an qth-degree polynomial, then s(l)({) = 0 for l A q, so its Taylor series at d is s({) =qSl=0
s(l)(d)
l!({− d)l.
Put {− d = 1, so that { = d+ 1. Then s(d+ 1) =qSl=0
s(l)(d)
l!.
This is true for any d, so replace d by {: s({+ 1) =qSl=0
s(l)({)
l!
82. The coefficient of {58 in the Maclaurin series of i({) = (1 + {3)30 is i(58)(0)
58!. But the binomial series for i({) is
(1 + {3)30 =∞Sq=0
#30
q
${3q, so it involves only powers of { that are multiples of 3 and therefore the coefficient of {58 is 0.
So i (58)(0) = 0.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
1056 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
83. Assume that |i 000({)| ≤ P , so i 000({) ≤ P for d ≤ { ≤ d+ g. NowU {di 000(w) gw ≤
U {dP gw ⇒
i 00({)− i 00(d) ≤P({− d) ⇒ i 00({) ≤ i 00(d) +P({− d). Thus,U {di 00(w) gw ≤
U {d[i 00(d) +P(w− d)] gw ⇒
i 0({)− i 0(d) ≤ i 00(d)({− d) + 12P({− d)2 ⇒ i 0({) ≤ i 0(d) + i 00(d)({− d) + 1
2P({− d)2 ⇒
U {di 0(w) gw ≤
U {d
�i 0(d) + i 00(d)(w− d) + 1
2P(w− d)2
�gw ⇒
i({) − i(d) ≤ i 0(d)({ − d) + 12i 00(d)({ − d)2 + 1
6P({ − d)3. So
i({) − i(d) − i 0(d)({ − d) − 12i 00(d)({ − d)2 ≤ 1
6P({ − d)3. But
U2({) = i({)− W2({) = i({)− i(d)− i 0(d)({− d)− 12i 00(d)({− d)2, so U2({) ≤ 1
6P({− d)3.
A similar argument using i 000({) ≥ −P shows that U2({) ≥ − 16P({− d)3. So |U2({2)| ≤ 1
6P |{− d|3.
Although we have assumed that { A d, a similar calculation shows that this inequality is also true if { ? d.
84. (a) i({) =
+h−1@{
2if { 6= 0
0 if { = 0so i 0(0) = lim
{→0
i({)− i(0)
{− 0= lim
{→0
h−1@{2
{= lim
{→0
1@{
h1@{2= lim
{→0
{
2h1@{2= 0
(using l’Hospital’s Rule and simplifying in the penultimate step). Similarly, we can use the definition of the derivative and
l’Hospital’s Rule to show that i 00(0) = 0, i (3)(0) = 0, = = =, i (q)(0) = 0, so that the Maclaurin series for i consists
entirely of zero terms. But since i({) 6= 0 except for { = 0, we see that i cannot equal its Maclaurin series except
at { = 0.
(b) From the graph, it seems that the function is extremely flat at the origin.
In fact, it could be said to be “infinitely flat” at { = 0, since all of its
derivatives are 0 there.
85. (a) j({) =∞Sq=0
#n
q
${q ⇒ j0({) =
∞Sq=1
#n
q
$q{q−1, so
(1 + {)j0({) = (1 + {)∞Sq=1
#n
q
$q{q−1 =
∞Sq=1
#n
q
$q{q−1 +
∞Sq=1
#n
q
$q{q
=∞Sq=0
#n
q+ 1
$(q+ 1){q +
∞Sq=0
#n
q
$q{q
�Replace q with q+ 1
in the first series
�
=∞Sq=0
(q+ 1)n(n − 1)(n − 2) · · · (n − q+ 1)(n − q)
(q+ 1)!{q +
∞Sq=0
�(q)
n(n − 1)(n − 2) · · · (n − q+ 1)
q!
�{q
=∞Sq=0
(q+ 1)n(n − 1)(n − 2) · · · (n − q+ 1)
(q+ 1)![(n − q) + q]{q
= n∞Sq=0
n(n − 1)(n − 2) · · · (n − q+ 1)
q!{q = n
∞Sq=0
#n
q
${q = nj({)
Thus, j0({) = nj({)
1 + {.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY
NOT FOR SALE
LABORATORY PROJECT AN ELUSIVE LIMIT ¤ 1057
(b) k({) = (1 + {)−n j({) ⇒
k0({) = −n(1 + {)−n−1j({) + (1 + {)−n j0({) [Product Rule]
= −n(1 + {)−n−1j({) + (1 + {)−nnj({)
1 + {[from part (a)]
= −n(1 + {)−n−1j({) + n(1 + {)−n−1j({) = 0
(c) From part (b) we see that k({) must be constant for { ∈ (−1> 1), so k({) = k(0) = 1 for { ∈ (−1> 1).
Thus, k({) = 1 = (1 + {)−n j({) ⇔ j({) = (1 + {)n for { ∈ (−1> 1).
86. Using the binomial series to expand√1 + { as a power series as in Example 9, we get
√1 + { = (1 + {)1@2 = 1 +
{
2+
∞Sq=2
(−1)q−11 · 3 · 5 · · · · · (2q− 3){q
2q · q!, so
�1− {2
�1@2= 1−
1
2{2 −
∞Sq=2
1 · 3 · 5 · · · · · (2q− 3)2q · q!
{2q and
s1− h2 sin2 � = 1−
1
2h2 sin2 � −
∞Sq=2
1 · 3 · 5 · · · · · (2q− 3)2q · q!
h2q sin2q �. Thus,
O = 4d
] �@2
0
s1− h2 sin2 � g� = 4d
] �@2
0
�1−
1
2h2 sin2 � −
∞Sq=2
1 · 3 · 5 · · · · · (2q− 3)2q · q!
h2q sin2q �
�g�
= 4d
��
2−
h2
2V1 −
∞Sq=2
1 · 3 · 5 · · · · · (2q− 3)q!
�h2
2
�qVq
�
where Vq =] �@2
0
sin2q � g� =1 · 3 · 5 · · · · · (2q− 1)2 · 4 · 6 · · · · · 2q
�
2by Exercise 7.1.50.
O= 4d��2
��1−
h2
2·1
2−
∞Sq=2
1 · 3 · 5 · · · · · (2q− 3)q!
�h2
2
�q1 · 3 · 5 · · · · · (2q− 1)2 · 4 · 6 · · · · · 2q
�
= 2�d
�1−
h2
4−
∞Sq=2
h2q
2q·12 · 32 · 52 · · · · · (2q− 3)2(2q− 1)
q! · 2q · q!
�
= 2�d
%1−
h2
4−
∞Sq=2
h2q
4q
�1 · 3 · · · · · (2q− 3)
q!
�2(2q− 1)
&
= 2�d
�1−
h2
4−3h4
64−5h6
256− · · ·
�=
�d
128(256− 64h2 − 12h4 − 5h6 − · · · )
LABORATORY PROJECT An Elusive Limit
1. i({) = q({)
g({)=
sin(tan{)− tan(sin{)arcsin(arctan{)− arctan(arcsin{)
The table of function values were obtained using Maple with 10 digits of
precision. The results of this project will vary depending on the CAS and
precision level. It appears that as {→ 0+, i({)→ 103. Since i is an even
function, we have i({)→ 103as {→ 0.
{ i({)
1 1=1838
0=1 0=9821
0=01 2=0000
0=001 3=3333
0=0001 3=3333
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
FOR INSTRUCTOR USE ONLY