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

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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



NOT FOR SALE


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.



NOT FOR SALE


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 =∞.



NOT FOR SALE


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 =∞=



NOT FOR SALE


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.



NOT FOR SALE


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 =∞.



NOT FOR SALE


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 {=



NOT FOR SALE


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.



NOT FOR SALE


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.



NOT FOR SALE


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.



NOT FOR SALE


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.



NOT FOR SALE


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 =∞.


(−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.


(−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,



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


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




NOT FOR SALE


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


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 + · · · .



NOT FOR SALE


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.



NOT FOR SALE


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 + {.



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



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