IB Diploma Program Mathematics Course Companion Higher Level Option: Calculus Worked Solutions [1 ed.] 0198304846, 9780198304845

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O X

F O

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D

I B

D

I P L O

M

A

P

R

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G R

A

M

M

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WORKED SOLUTIONS M AT H E M AT I C S

HI GH E R

LE V E L

C A LC U LU S

Josip Harcet

Lorraine Heinrichs

Palmira Mariz Seiler

Marlene Torres-Skoumal

:

WORKED SOLUTIONS

1

Patterns to infinity

Skills check

3

3(4 n  1)  3(4 n + 1) 3 3 =  (4 n + 1)(4 n  1) 4n  1 4n  1

1 a

12n  3  12n  3 16n 2  1

=

6 16n 2  1

n 1 n = (n  1)(n  3)  n (n  2)  n2 n3 (n  2)(n  3)

b

⇒n>2∴m=3 4

n 2  n  3n  3  n 2  2n n 2  2n  3n  6 3 = 2 n  5n  6

=

Ÿ 0  3x  2 Ÿ 0 d x d b

2

2n 1  2n  1 5 1 Ÿn 8 1 No as this implies that n  . 8

Ÿ

Ÿ 2 x t 7 or 2 x d  1

1

(1 

n

n )(1 

n 1

b

7 1 or x d  2 2

2n (1  n )

2n

3 a

n 

n)

=

2n ( n  1) n 1

6

As n increases the terms get very close to 0.

7

1 § 1· §1· ¨  ¸  H Ÿ n  H Ÿ n ! log 3 ¨ ¸ 3 © 3¹ ©H ¹

n

(n  1)( n  n  1)

n 1

( n 

n  1)( n 

∴ ε = 0.01 ⇒ m = 6; ε = 0.001 ⇒ m = 7; ε = 0.0001 ⇒ m = 9

n  1) 8

(n  1)( n  n  1) n  n 1

(n  1)( n  1  n ) ( 1)n 1

4 a

b)

2n

d

( 1)n 1 n 1

c)

2n  1

∴ m = 1250 2 a

Investigation 1

1.1

u1(n) =

0.2 2

4

6

8

⇒ n > 1249.5

n+1 u1(n) = 3•n–1

.4

n+1 2•n+1

5 1  2n  1 500

Exercise 1A

1.1

.6

0.4

2

y 1 .8

0.6

0 –1

n 3 1 1 2n  6  2n  1 1 Ÿ    2n  1 2 1000 2(2n  1) 1000

Ÿ

Investigation 1 y 0.8

Exercise 1A 1

2n

|un – L| can take arbitrary small values as we consider just terms after an order m (as we increase this order the terms are closer to L).

2n  1 , n  ' 2n

(1)n 1

1

n 1 2 1 2n  2  4 n  2 1   Ÿ  2n  1 N 5 10 2(2n  1) 10 0 .4

3

|2 x  3| t 4 Ÿ 2 x  3 t 4 or 2 x  3 d  4

Ÿxt

n 1 1 2n  2  2n  1  H Ÿ H 2n  1 2 2(2n  1) 1 1§ 1 ·  1¸ Ÿ  2H Ÿ n ! ¨ 2n  1 2 © 2H ¹

∴ ε = 0.01 ⇒ m = 25; ε = 0.001 ⇒ m = 250; ε = 0.0001 ⇒ m = 2500 5

|3 x  1|  1 Ÿ  1  3 x  1  1

2 a

n 1 1 1 2n  2  2n  1 1   Ÿ  2n  1 2 10 2(2n  1) 10 1 1 Ÿ  2n  1 5

.2 0 –.2

10 12 14 x

3

6

9 12 15 18 21 x

1 2

As n increases the terms get very close to .

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 1

1

WORKED SOLUTIONS b

n 1 1 1 3n  3  3n  1 1 Ÿ    3n  1 3 1000 3(3n  1) 1000

5

n cos 2 (na ) n2  3

n 2 n  3

2 na ) as 0 d cos 2 (na ) d 1. ˜ cos  ( bounded

converges to zero

4 1  9n  3 1000 4003 Ÿn! 9

Ÿ

∴ m = 445 3 a y 2

n of

Exercise 1B n

1

b

n §3· · ©5¹ ¹

§ ©

3

6

9

2 a

12 15 18 21 x

Ÿ

4n  3 5 4n  3  4n 1  1  Ÿ  10000 2000 4n 4n 3 1 Ÿ n  2000 4

Convergent to zero as the absolute value of the terms get close to zero quickly.

c

Divergent as while the terms of even order get close to zero, the terms of odd order increase without bound.

3n  1 3n  sin(2n ) 3n  1 d d 4n  3 4n  3 4n  3

for all n  ] ; as lim n of

lim n of

c

1.1 1.2

1000 1 10000 2 3 100000 4 1000000 5 10000000 C6

B

C =√('n) =√('n+1) 31.6228 31.6228 100 100.005 316.228 316.229 1000. 1000 3162.28 3162.28

1 n lim n of 3 4 n 1 3 n lim n of 3 4 n

3n  1 4n  3

lim n of

3n  1 4n  3

lim n of

n of

b

1§ n2 · 1 3 ¨ 2© 2n  3 ¸¹

n2 approaches zero as n increases. 2n 3  3

As

3n  1 4n  3

3 by the squeeze 4 3 4

2n 2  Ÿ 6n  6n  2 Ÿ 0  2 3n  1 3

1 2n 2 d  for all n  ] . 2 3n  1 3

1 2n 2  d  for all n  ] and 2 3n  1 3 §1· ©2¹

n

lim ¨ ¸ n of

Diverges as the terms of even order are equal to 1 and the terms of odd order get close to zero.

and

4n  3

which is true for all n  ] .

Diverges as the terms follow the cyclic pattern 1, 0, −1, 0, 1, 0, −1, 0,….

1 n3  n  1 as 2 2n 3  3

3 4

3n  sin(2n ) 4n  3

for all n  ] ;

Diverges to positive infinity as the terms increase without bound.

Converges to

30 40

2n 1 t Ÿ 4n t 3n  1 Ÿ n t 1 which is true 3n  1 2

Therefore,

h

3 and 4

4n  3

theorem lim 3 a

30 40

As 3n  1 d 3n  sin(2n ) d 3n  1 and 4n  3

Exercise 1A

3n  1 4n  3

3

3n  1 b lim n of 4 n  3

Converges to zero as the square roots of consecutive integers are approximately equal for large values of n. An

g

n

as 4 n  3 ! 0

Divergent as the terms of even order approach

b

f

0, by the

1 d sin(2n ) d 1, for all n  ] 

3 3 and the terms of odd order approach  . 4 4

e

n

? 3n  1 d 3n  sin(2n ) d 3n  1

∴m=7

d

§3· ©5¹

lim 2 ˜ ¨ ¸ n of

§3· ©5¹

Ÿ n ! log 4 6000 4 a

n

squeeze theorem lim 2 ˜ (1)n 1 ˜ ¨ ¸ . n of

.5 0 –.5

n

lim ¨ 2 ˜ ¨ ¸ ¸ n of

n u1(n) = 4 –3 4n

1.5

0 , a ∈ R. QED

§3· §3· §3· 1 a 2 ˜ ¨ ¸ d 2 ˜ ( 1)n 1 ˜ ¨ ¸ d 2 ˜ ¨ ¸ ; as ©5¹ ©5¹ ©5¹

Exercise 1A

1.1

n cos 2 (na ) n2  3

Therefore lim

§ 2n · ¸ n of © 3n  1 ¹

lim ¨ 4

§3· ©4¹

n

lim ¨ ¸ n of

0 by the squeeze theorem,

n

0

Assumption: the sequence {an} converges to a real number a. As lim an 1 lim an a , then n of

lim an 1 n of

a 2 lim n Ÿa n of 3

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

n of

a2 Ÿa 3

1

Calculus: Chapter 1

2

WORKED SOLUTIONS

Investigation 2 1

1.1

y 20000

8

The absolute value of the terms of the sequence increase without upper bound.

Investigation 2

1.2

Investigation 3

16000

u1(n) = 3n–1

a

12000 8000

2

2

4

6

8

10 12

x

The terms of the sequence increase without upper bound. n 1

! 100 Ÿ n ! 5.19... Ÿ m

3

3

4

3n 1 t 1000 Ÿ n t 7.28... Ÿ m

b

6

8,

3n 1 t 10000 Ÿ n t 9.38... Ÿ m

1.1

1.2 1.3

log3(1000)+1

7.28771

log3(10000)+1

9.38361

log3(100000)+1

11.4795

10 and 12

2n d 1000 Ÿ n ! 9.96... Ÿ m

1.1

Investigation 3

1.1

Investigation 3

An

10 ,

n

2 d 10000 Ÿ n ! 13.2... Ÿ m

14 and 2 d 1000000 Ÿ n ! 16.6... Ÿ m 17 1.1

1.2 1.3

c

Investigation 2

9.96578

log2(10000)

13.2877

log2(100000)

16.6096

1.1 1.2 1.3

An

0

? lim c n

1

? lim dn

1

Investigation 3

B C =root(3,'n) 1.01105 1.0011 1.00011 1.00001

100 1 2 1000 3 10000 4 100000 5 B4 =1.0000109861832

3/99

n of

6

7

The terms of the sequence decrease without lower bound. 1.1

1.2 1.3

d

4.98289

log4(10000)

6.64386

log4(100000)

8.30482

1.1 1.2 1.3

An

Investigation 3

B C =root('n, 'n) 1.00693 100 1 1.00092 1000 2 1.00012 10000 3 1.00001 4 100000 1. 5 1000000 B5 =1.0000016118109

Investigation 2

log4(1000)

0

? lim bn n of

n

log2(1000)

n of

B C =('n)/3^('n) ='n/4^'n 0.33333333 1. 0.25 1 0.00016935 10. 0.00000954 2 3 100. 1.9403252E-46 6.2230153E-59 4 500. 1.3751265E-23... 4.6663181E-299 5 1000. 7.5638913E-47... 8.7098098E-600 C5 =8.7098098162032E-600

4/99

5

? lim an

B C ='n^2/3^('n) ='n^2/4^('n) 0.33333333 1. 0.25 1 0.00169351 2 10. 0.00009537 3 100. 1.9403252E-44 6.2230153E-57 4 500. 6.8756324E-23... 2.333159E-296 5 1000. 7.563891E-47... 8.7098098E-597 C5 =8.7098098162032E-597

Investigation 2

5.19181

C

Explore the limit for different values of k and b. For example: An

3n 1 t 100000 Ÿ n t 11.47... Ÿ m log3(100)+1

Investigation 3

B ='n/2^'n

0.5 1. 1 0.00976563 10. 2 7.8886091E-29 3 100. 1.5274682E-148 4 500. 5 1000. 9.3326362E-299 B5 =9.3326361850171E-299

4000 0

1.1 1.2 1.3

An

n of

3/99

e n

4 t 1000 Ÿ n ! 4.98... Ÿ m n

4 t 10000 Ÿ n ! 6.64... Ÿ m

4 n t 100000 Ÿ n ! 8.30... Ÿ m

lim en n of

f

5, 7 and

9

You need to use a powerful tool to calculate this one, e.g., Mathematica or its free version Wolfram Alpha.

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 1

3

WORKED SOLUTIONS n

Exercise 1C 5

5n 3  n 1 a lim 3 n of 2n  6n 2  1

lim n of

n  6n  1 n of n 1

50 200

6 1  3 n n

2

6 1  n n2 1 1  n n2

1

2

lim

1 n2

1 5  2 n n lim n of 2 3 1 2  3 n n

n 2  5n b lim 3 n of n  2n  3

c

k

lim n of

00 1 0  0

d

2

lim(n  n )

lim(n (n  1))

n of

2n 2  1

n of

1 n lim n of 1 2 2 n 1

n

0 l

¦ (2

lim k

k 1

0

n of

3

3

1 0 0 00

f

f

lim n n n of

ln( n n ) 1 Ÿ lim n of n of

n of

lim

n5

n of

n  2 

n5

n5 

n 2

3

lim n of

3 f

n5  n2

n  2

lim n of

n5 n2

ln 1

1 n

1 b

0 Ÿ lim n of

ln n n

lim n of

§5· © ¹

ln n  k n

ln n n

0

0, by the squeeze

ln(n  k ) n

0

n

lim n n ˜ lim

lim n n ˜ ¨ ¸ n of n

0

0

lim

n of

1 0 1 0

00 2

ln n ln(n  k ) ln n  k and d d n n n

theorem lim f

2n  1 n of 2 ˜ 3n

As

n of

0

5 1 n lim n of 2 1 n

1 2

lim

1 §2· ¨ ¸  n 3 3 lim © ¹ n of 2

lim  n  5  n  2



1 0 20

1 2n  1 lim 2 2n 1 n of 3

)

Ÿ lim ln n e

n2  n n of 2n 2  1

lim

n

( f) u ( f)

n of

k 0

5 2

2 a 3

lim

2n  2 ˜n lim 22 n of 2n  1

¦ (2k  1)

n of

n of

5 n

1u 0

0

n

n

g

§2· 1 ¨ ¸ lim ©n3 ¹ n of §4· ¨ ¸ 5 ©3¹

n

3 2 n of 4 n  5 ˜ 3n

lim

1 0 f  5

Exercise 1D 0 1 a b

n

( 3)n 1  7 n h lim n of 4 n 1  e n

i

lim n of

n3n n

lim( x 2  1)

§ 3· ¨  ¸ 1 lim © 7n¹ n n of 1 §4· §e· ˜¨ ¸  ¨ ¸ 4 ©7¹ ©7¹

0 1 00

1.1

j

§ n n  1·  lim ¨ ¸ n of n  1 n ¹ ©

4

4

{

–10

1.1

Exercise 1D

0 –4 –3 –2 –1 –1

1

( )

2

3

4

x

–2

lim sin(S x ) 

x o1

f

{

y sin(π•x), x < 1 2 f1(x) = π•x , x ≥ 1 cos 2 1

3

1 1  n 2 n3 1 1  n n2

1  0  0 00

2

2 f1(x) = x –1, x 1 the series does not

f

1 . n2  1

1 §§ 1 · §1 1 · § 1 1· ¨ ¨1  ¸  ¨  ¸  ¨  ¸ 2 ©© 3 ¹ ©2 4 ¹ © 3 5¹

n 1

n

3 6 9 12    ! 2 3 4 5

1 z 0, 2

1§ 1 1 ·  ¨ ¸ 2 © n 1 n 1¹

©

n 2

f

lim

g

1 2 n 2

n of

converge. f

1

lim

¦ 2 ¨ n  1  n  1¸

¦ ¨S n

1 1  2(n  1) 2(n  1)

1 § (n  1)  (n  1) · ¨ ¸ 2© n2  1 ¹

2 z 0 , hence the series 3

is a geometric series with r = 1 , hence 7 21 1 converges to its sum, Sf 3 ˜ 1 6 1 7 Hence S = 24.5 e

n of

2n 2 ˜ 2n  1

hence the series diverges by the nth term test.

diverges. d

lim

n of f

b

2n 2 1 n 1

Corollary 2: Let g ( x )

³

x

a

³

b

a

=− ³x f (t ) dt. f ( x ) dx

F (a )  F (b )

f (t ) dt , a d x d b . If furthermore F is

any anti-derivative of f, then by the mean value theorem, F (x) = g (x) + c, for some constant c. Evaluating therefore F (b) − F (a),

> g (b )  c @  > g (a )  c @

F (b )  F (a )

g (b )  g (a ) =

³

b

a

³

b

a

³

b

a

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

a

f (t )dt  ³ f (t ) dt a

f (t )dt  0 f (t )dt Calculus: Chapter 4

31

WORKED SOLUTIONS n

2

n

n

S

§ i 1·S ¸ n ¹n

i

i

i

i 1

i 1

i 1

³

c

¦ f (c 'x ) ¦ sin(c ) n ¦ sin ¨© S§

§S · § 2S · § 3S · ¨ sin ¨ ¸  sin ¨ ¸  sin ¨ ¸ n© n n © ¹ © ¹ © n ¹

§S · § 2S · ¨ sin ¨ ¸  sin ¨ ¸ n© ©n¹ © n ¹

b

F c( x )

c

F s( x )

S§

n of

§ 3S · § (n  1)S · ·  sin ¨ ¸  !  sin ¨ ¸¸ n n © ¹ © ¹¹

³

S

 cos x |S0

sin x dx

0

³

F (4)

Hence,

x 2  9 ; F c( 4 ) x

f ( xi ) n

f ( x i )'x

³

5

0

ii

2 x dx

1  x  x2

xi

1

5i ; n

n

50i

¦n

50 § n (n  1) · ¨ ¸ n2 © 2 ¹

n 1 25 ˜ n

0  ( 2) n

'x

(1  x  x 2 )2

2

i 1

50 n2

0

i 1

2

lim ³ e  x dx b of 0

0

sin x dx = lim

4 a b

³

4

2

x

1,

b

b

sin x dx = lim ( −cos x ⏐ ) b →∞

0

20i 12i 2  2 n n

³

f

0

ex 1 e

2x

dx

lim b of

³

ex

b

0

0

1 e

b 2x

dx

hence the integral converges to 4

³

∞

1

1 x

dx = lim ³ b →∞

1

b

1

x

x

lim ª¬arctan e º¼ b of 0

S . 4

S 4

,

b

dx = lim (2 x ⏐ ) = ∞, b →∞

1

hence the integral diverges. 5

³

f

0

b

cos S x dx

lim ³ cos S x dx b of 0

b

S lim sin x~ S lim(sin b  1). Since sinb b of

0

b of

oscillates between −1 and 1, the integral diverges.

(3x  2 x ) § 16 40 § n (n  1) · 24 § n (n  1)(2n  1) · ·  2¨ ¸ 3¨ ¸¸ n © 2 6 ¹ n © ¹¹ © n

4

6

³

∞

0

b

x 2 e − x dx = lim ³ x 2 e − x dx . Using integration by b →∞ 0

parts, lim

3  4x

2 tan t dt

0

b →∞

2

f c( x )

b of

between −1 and 1, the integral diverges.

2i · 2i · § § 3 ¨ 2  ¸  2 ¨ 2  ¸ n¹ n¹ © ©

lim ¨ n of

³

b →∞

16 40 § n (n  1) · 24 § n (n  1)(2n  1) ·  2¨ ¸ 3¨ ¸ 2 6 n3 n © ¹ n © ¹ 2

lim  e  x~

= lim ( − cos b + 1). Since cosb oscillates

§ 16 40i 24i 2 ·  2  3 ¸ f ( x i )'x ¦ ¨ n n ¹ i 1© n 16 40 n 24 n 2  2 ¦i  3 ¦i n n i 1 n i 1

³

2

b

b

e  x dx

25

3

8

³

∞

n

0

1

hence the integral converges to 1.

n

¦i

1· § 25 lim ¨ 1  ¸ n of n¹ ©

§ 24i 12i 2 · 4i  2 ¸4  ¨ 12  n n n ¹ ©

³

f

§ 1 · lim ¨  b  1¸ b of © e ¹

2 ; n

2  2  i ; f ( xi ) n

 (1  2 x )

Exercise 4C

5 0i n 10i n

n 1· § lim ¨ 25 ˜ ¸ n of n ¹ ©

5

y s !  Ÿ  (1  2 x ) ! 0, x < 

·· ¸¸ ¹¹

¦ ¨© i

4

; F s(4)

x2  9

5

1

y s x

§ 10i · § 5 · ¸¨ ¸ n ¹© n ¹ 1

25

The curve will be concave up at f ″ (x) > 0. yc

S

'x

0

7

§S · § 2S · lim ¨ sin ¨ ¸  sin ¨ ¸ n of n ©n¹ © n ¹ ©

3 i

t 2  9t dt

G′(x) = sin x

S§

5 ;x n i 5i 2˜ n

4

4

And therefore,

50 n

 x sec x

0

6

2

§ 3S · § (n  1)S  sin ¨ ¸  !  sin ¨ n © n ¹ © 2

t

 ³ t sec t dt

t sec t dt

5 a

§ (n  1)S · ·  !  sin ¨ ¸¸ n © ¹¹

S lim

0

t

b →∞

x

 ³ 2 tan t dt 4

³

b

0

x 2 e − x dx = lim ª¬ −e − x ( x 2 − 2 x − 2 ) º¼ = 2, b →∞ 0 b

hence the integral converges.  2 tan x

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 4

32

WORKED SOLUTIONS

Exercise 4D 1

3 is x 1

From the GDC, f ( x )

9

The integral test cannot be applied since the function is not decreasing for all x ≥ 1.

10

From the GDC, the function is continuous, positive and decreasing for all x ≥ 1, hence we can apply the integral test.

continuous, positive and decreasing for all x ≥ 1, hence we can apply the integral test.

³

f

1

3 dx x 1

3 dx x 1

b

lim ³1 b of

b

3 lim > ln( x  1)@1

³

b of

= 3 lim(ln(b + 1) − ln 2) = 3 lim ln n →∞

n →∞

1

³

1

x x2 +1

dx = lim b →∞

³

x

b

1

x2 +1

∞

1

1 dx = lim b →∞ x +1 2

dx = lim =

ln( x 2 + 1)

b

2

1

b →∞

11

⏐

1

5

12

0

ex dx 1  e 2x

b

lim ³0 b of

∞

1

ex dx 1  e 2x

³

lim

b

1

b of

ln x dx = lim b →∞ x2

S 4 13

b of

b

1

ln x § ln x + 1 · b dx = lim ¨ − ¸1 b →∞ x2 x © ¹

f

1

x2 e

x

b

x2

b of 1

x

lim ³

dx

e

b

lim( e  x ( x 2  2 x + 2))~

dx

b of

1

Using the p-series test, p series converges. Using the p-series test, p series diverges.

7

From the GDC, the function is continuous, positive and decreasing for all x ≥ 1, hence we can apply the integral test.

3

1

³

1

lim b of

³

1

§ 1 1· lim ¨  2  ¸ b of  2 3 5 b © ¹

1

The integral test cannot be applied since the function is not decreasing for all x ≥ 1.

2 2

1



3 3



S , S ! 1, hence the S S ,  1, hence the 4 4 f

1 4 4

¦n

 ...

1 n

n 1

∞

1

n =1

n n

∞

=

¦ n =1

1 n

3 3

3 2

, p = 2 , 2 > 1,

hence the series converges. 4

1

1 5

4



1 5

9



1 5

16

f

 ...

¦ n 1

f

1 5

n

2

¦ n 1

1 n

2 5

.

Using the p-series test, p = 2 , 2 < 1 hence the series 5 5 diverges.

Hence, the series converges. 8

1

Using the p-series test, ¦

b

ª º lim «  2 » b of ¬ 2 x  3 ¼1

1 5

5 . e

Exercise 4E

2

4x dx (2 x 2  3)2

1

5· § lim ¨ ( e b (b 2  2b  2))  ¸ b of © e¹ Hence the series converges.

The integral test cannot be applied since the function is not decreasing for all x ≥ 1.

b

ln 2 2

From the GDC, the function is continuous, positive and decreasing for all x ≥ 2, hence we can apply the integral test.

³

S 2

b

lim ª¬arctan e x º¼ 0

³

6

4x dx (2 x 2  3)2

3S 2 32

Hence the series converges.

Hence, the series converges.

f

b

ª (arctan x )2 º lim « » b of 2 ¬ ¼1

arctan x x2  1

§ ln b  1 ln 2 · lim ¨   ¸ b of 2 ¹ b ©

2

From the GDC, the function is continuous, positive and decreasing for all x ≥ 1, hence we can apply the integral test. f

arctan x x2  1

From the GDC, the function is continuous, positive and decreasing for all x ≥ 1, hence we can apply the integral test.

³

The integral test cannot be applied since the function is increasing for x ≥ 1. However, since §1· lim x sin ¨ ¸ 1 z 0, the series diverges by the nth b of ©x¹ test for divergence.

³

1

Hence the series converges.

hence the series converges. 4

f

b

1

b of

From the GDC, the function is continuous, positive and decreasing for all x ≥ 1, hence we can apply the integral test.

³

f,

b 1 dx = lim arctan x⏐ →∞ b 1 x +1 S S lim ª¬arctan b  arctan 1º¼  b of 2 4

³

x ln x

b →∞

From the GDC, the function is continuous, positive and decreasing for all x ≥ 1, hence we can apply the integral test.

³

b of 1

b

lim ln(ln( x ))~

dx

Hence the series diverges.

1 lim ª ln(b 2 + 1)  ln 2 º¼ 2 b of ¬ hence the series diverges. 3

1

b

= lim ( In(In(b ) − In(In(1) ) = ∞

From the GDC, the function is continuous, positive and decreasing for all x ≥ 1, hence we can apply the integral test. ∞

x ln x

lim ³

dx

b +1 =∞ 4

hence the series diverges. 2

1

f

5

1 1 < 1, hence the series diverges. 2

p = 2,

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 4

33

WORKED SOLUTIONS

Exercise 4F 1

2

2

1 1  3 , n t 1. Since 1 converges by n 1 n n3 3

1

1



n

n1

, n t 2. Since 1 diverges by

diverges. 3

n 1

n

3

4

5

3n2  2 n

diverges.

2 5

2 Since ¦ n

n

§

¦¨ 2 ©

n

n

2n n of 2 n  1 n2 n5

¦

5

1 1 ·  ¸  1 2n ¹

n

1, the series converges.

lim

.

1 4 n2  2 n 1 n of n2

n5 4  n5

· ¸ ¸ ¹

3.

1 , the series diverges. n

¦ 4n

2

1 1 . Compare with ¦ 2 . n  2n

p-series test, the original series converges.

converges by the p-series test, the series converges.

n1 2

n 1 n 1 1 1  5 for n ≥ 1. Since 5 ; 5 5 2 2 2 n n n n2

n

7

9 arctan x 

¦

S ; 2

S  3 2 n 1

arctan n

1 3

n 1

¦

1



1

1

. Since

converges by

¦ 8

1 n of

n 2 1 1 n2

lim n of

n of

n

series diverges since ¦

n of

1 n2

1 . Hence, since 3

2 n 

n2

diverges by the p-series test, the series diverges. n 2n n2

2n

by the ¦ n , which diverges n

n 2 n 1 3n2  2 n

n 2 2n  n n of 3n 2 2 n  2n 2 n

lim

n

2 n

1

lim n of

3

2 lim

1 1  n1

2

n2

z 0.

1 n 2n 2 n

1 . 3

Hence the series diverges.

Exercise 4H

1 . Hence, n2 lim

n

. n

Compare with ¦

lim

1

n2 n2  1

1

n

n of

Exercise 4G

lim

n of

1

1 1 . Hence, since ¦ 2 n 4

nth term test for divergence, since lim

converges by the p-series test, the original series converges.

Compare with ¦

lim

n

ln(n) ≥ 2 for n ≥ e2 hence 1ln n d 12 . Since 12 n

n  n2 1

n of

the p-series test, the original series also converges. n

n of

1

n3

n3

n2 4 n 2  2n

lim

Compare with ¦ lim 2

for n ≥ 1. Hence

n3

1.

1 diverges. n

lim

converges by the p-series test, the original series converges.

1

n

1 cos n 1 d 2 for n ≥ 1. Since 2 converges by the 2 n n n

n

10

n

n of

2

8

n

§ 3n 2  2n lim ¨ u 2 n of ¨ n ©

4  n5

lim

6

n of

2

Compare with ¦

n

diverges by the p-series test, the original series

lim

Compare with ¦ 1n , a convergent geometric series. Since lim

6 ln(n) > 1 for n ≥ 3, hence ln n t 1 for n ≥ 3. Since n n 1

7

n

1 2n 1 n of 1 2n

p-series test, the original series converges.

n

n n Compare with ¦ 1 . Hence lim 1 n of

The series diverges, since ¦

1 converges by the n3

; n t 1. Since

n n5

1

3

1 1 4 ln(n) ≤ n for n ≥ 2, hence t , n t 2. Since ln n n 1 diverges by the p-series test, the original series n 1

lim

2

converges.

n

p-series test, the original series converges.



2n 7  2n 2 n of 4 n 7  n 6  n 5

1 . The series 2 n2 1 converges since by the p-series test, ¦ 5 ¦ 3 n n

1 1 3  2 , n t 1. Since 12 converges by the (n + 1)! n n

1

2 n 2 1 4 n 5  n 1

n of

the p-series test, the original series also diverges.

5

n

lim

the p-series test, the original series also converges. 2

Compare with ¦ n 5 . Hence,

lim n of

2n 1 n 1 2n n

§ n 2 n 1 · ˜ n ¸ n of n  1 2 ¹ ©

lim ¨

n of

2.

Since L > 1 the series diverges.

1 1  n12

1

1, and the

1 ¦ n diverges.

2

lim ( n 11)! n of

n!

lim n of

n! (n  1)!

lim n of

1 n 1

0.

Since L < 1 the series converges. 3n 1

3

lim ( n31)! n of

n

n!

§ 3 n 1 n! · ˜ ¸ n n of  1)! ¹ ( n 3 ©

lim ¨

3 lim n of

1 n 1

0.

Since L < 1 the series converges. © Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 4

34

WORKED SOLUTIONS 4

lim

3 ( n 1)2n 1 3 n 2n

n of

n 2n (n  1)2n 1

lim n of

1 n lim n of 2 n 1

b

1 . 2

Since L < 1 the series converges. Sn

5

1

lim 3( n S1)  2

S lim n of

n

n of

3n  2

3n  2 3n  5

S.

5

6

lim ( n5 2 )! n of

(n  1)! (n  2)!

5 lim

n 2

n of

( n 1)!

5 lim n of

1 n2

n of

§ 2n n  1 · ˜ ¸ n of 2 n 1 n ¹ ©

n 1

n of

1 1· § lim ¨ 1  ¸ n of 2 n¹ ©

lim ¨

2n

1 . 2

sin n 2 ¦ 2 has terms of different signs, since n 1 n f 1 sin n 1 –1 ≤ sin n ≤ 1. Since 2 d 2 for all n, and ¦ 2 n n n n 1

converges by the p-series test, the series converges absolutely, therefore the series converges. 3 Using the ratio test, n of

2 lim

n

n of

n!

n! (n  1)!

2 lim n of

1 n 1

Using the comparison test, lim n of

The absolute value of the terms is decreasing, and lim un = 0, hence the series converges. For nof

the series of absolute values, using the ratio test,

absolutely.

lim n of

1 2 n! n

n of

n of

n! (n  1)!

lim n of

1 n 1

0.

The absolute value of the terms is decreasing, and lim un 0, hence the series converges. The n of series of absolute values converges because it is a geometric series, with |r| < 1. Hence the series converges absolutely.

f

¦

f

n

cos nS n 1

1 

1 1 1 1    ...  ( 1)n  ... 2 3 4 n

The absolute value of the terms is decreasing, and lim un = 0, hence the series converges. The nof

series of absolute values is the harmonic series, which diverges. Hence the series converges conditionally. g

The absolute value of the terms is decreasing, and lim un = 0, hence the series converges.

2 n! 2n 1 (n  1)!

1 1 lim 2 n of n  1

The integral test can be used on the series 1 of absolute values, since f ( x ) is

0.

x ln x

continuous, positive and decreasing for all x ≥ 1.

n! 6 Since lim n z 0, the series does not converge n of e

Hence,

³

f

1

1 x ln x

b

lim ³1 b of

1 x ln x

b

lim > ln(ln x )@1 n of

lim > ln(ln b )  ln(ln1)@

absolutely.

n of

f.

Since the series of absolute values diverges, the series converges conditionally.

Exercise 4J The absolute value of the terms is decreasing, and lim un 0, hence the series converges. n of

For the series of absolute values, ln(n) ≤ n for n ≥ 2, hence

n!

lim

e

Since L < 1, the series converges. Hence the series converges absolutely.

1 a

n of

nof

n

lim

lim ( n 11)!

Since L < 1, the series converges absolutely.

Using the ratio test, 1 2n 1 ( n 1)!

1

un 1 n of u n

lim

S arctan n d 23 3 n n

1 for all n ≥ 1. Since ¦ 3 converges by the p-series n test, so does S ¦ 13 . Hence the series converges 2 n 5

1 . 2

d

0.

Since L < 1 the series converges by the ratio test. Since the series converges absolutely, the series converges. 4

1 · 1º ¸˜ n ¹ 2 »¼

The absolute value of the terms is decreasing, and lim un 0, hence the series converges. The n of series of absolute values diverges by the p-series test, hence the series converges conditionally.

f

2n 1

ª§ ¬©

c

Since L < 1 the series converges by the ratio test. Since the series converges absolutely, the series converges.

lim ( n21)!

2n

Since L < 1, the series converges absolutely.

Using the ratio test, n 1

n of

lim «

lim Ǭ 1 

Exercise 4I lim 2 n

ª n  1 2n º ˜ » n of ¬ 2 n 1 n¼

n 1

lim 2 n

0.

Since L < 1 the series converges.

1

n 1

un 1 n of u n

lim

Since L > 1 the series diverges. n 3

The absolute value of the terms is decreasing, and lim un = 0, hence the series converges. nof Using the ratio test for the series of absolute values,

1 1 1 t , n t 2. Since diverges ln n n n

by the p-series test, the series of absolute values diverges by the comparison test. Hence the series converges conditionally.

h

The absolute value of the terms is decreasing, and lim un 0, hence the series converges. n of The series of absolute values converges by S comparison with ¦ 22 , hence the series n

converges absolutely.

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 4

35

WORKED SOLUTIONS 2

The conditions for Leibniz’ theorem are met since the absolute value of the terms is decreasing, and lim un 0. Hence the truncation error is

³

u11 3

The conditions for Leibniz’ theorem are met since the absolute value of the terms is decreasing, and lim un 0, hence the series converges. un 1

5

5. Hence,

The conditions for Leibniz’ theorem are met since the absolute value of the terms is decreasing, and lim un 0, hence the series converges. n of Since the series begins with n = 2, and we want the first three terms, i.e., n = 2, 3, 4, therefore 1 1 u5 . S S3  u5  0.012644. 3 5 5 u3

5

6

30.

1

7

6˜0

3n



1 3n  1

.

diverges by the p-series test, so does

n

. Hence the series does not converge

n 2

2

1

1

¦ ln n . Since n

¦ 2 ln n



1 for all ln n

n (n  1)

Since lim n of

n 3  5n 2

z 0, the series diverges by the

Since the absolute value of the terms of the series is decreasing, and lim un 0, the series converges. n of

0.

³

1

f

dx

x ln x

lim ³

1

b

b of 2

dx

x ln x b

§ 2 ˜ 2n (n  1) 3n · ˜ n¸ n n of n2 ¹ © 3˜3

lim ¨

lim ª¬2 ln x º¼ b of 2 2 n 1 lim n of 3 n

lim  2 ln b  2 ln 2 b of

f.

Hence, the series diverges absolutely, and converges conditionally.

2 . 3

Since L < 1, the series converges.

10 Using

This is a geometric series whose common ratio is 1 . Since |r| < 1, the series converges to its sum. e

4

1

3n  1, hence

Using the integral test, since the function representing the series is positive, decreasing and continuous for all x ≥ 2,

Using the ratio test,

2 1· § lim ¨ 1  ¸ 3 n of © n¹ 3

1

2

¦ ln n

2

(n  1)2n 1 n 1 lim 3 n n of n2 3n

1. Since L is a

n ≥ 2, the series diverges by comparison with the harmonic series.

9

Since L < 1, the series converges. 2

n of

2n 2  n 2n 2  3n  1

n 1 , or ¦ , 2 n n

Divergence Test.

§ 6 n 1 n! · lim ¨ n ˜ ¸ n of 6 (  1)! ¹ n © 1 n 1

 ln(cos(1)).

absolutely, and converges conditionally.

Using the ratio test,

n of

1

¦

3

Exercise 4K

6 lim

2

Since the absolute value of the terms of the series is decreasing, and lim un 0, the series converges.

Since ¦

8

6 n 1 (n  1)! lim of 6n n!

lim

For all n ≥ 1, 3n !

Hence, the truncated series has the required accuracy after the 29th term, since the first term corresponds to n = 2, and not n = 1.

1

§1· tan ¨ ¸ dx x ©x¹ 1

n of

The conditions for Leibniz’ theorem are met since the absolute value of the terms is decreasing, and lim un 0. n of un 1 d 10 5

1

positive real number, the series diverges since the harmonic series diverges.

30375

1 Ÿ  10 5 Ÿ n 3 (n  1) ln(n  1)

³

Using the limit comparison test with ¦ 2n  1 2n 2  3n  1 lim n of 1 n

S5 | 0.948. 4

b of

b

f

n of

1 d 0.001 Ÿ d 0.001 Ÿ n (n  1) 4

lim

§ 1 · lim ln ¨ ¸ n of © cos x ¹b Hence the series converges.

1 S10  | 0.902. 1331

1 .S 1331

§1· tan ¨ ¸ dx x ©x¹ 1

2

1

n of

1 113

f

Using the GDC, we see that the function that represents the series is continuous, positive and decreasing for all n ≥ 1. Hence, we can use 1 the integral test. Using the substitution u = ,

4

lim n of

the ratio test,

n 1

(n  1)2 (n  1)! 4n n2 n!

§ 4 n 1 (n  1)2 n! · u n 2¸ n of ( n )!  1 n ¹ 4 ©

lim ¨ lim n of

4 (n  1) n2

0.

Since L < 1, the series converges.

x

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 4

36

WORKED SOLUTIONS 11

Using the ratio test,

b

(n  1)!  1 , n Œ  . (n  1)! (n  1)!  1 For n = 1, show that (n  1)!

(n  1)! 2 ˜ 5 ˜ 8 ˜ ... ˜ (3(n  1)  2) lim n of n! 2 ˜ 5 ˜ 8 ˜ ... ˜ (3n  2)

Pn : Sn

§ (n  1)! · 2 ˜ 5 ˜ 8 ˜ ... ˜ (3n  2) ˜ ¸ n of n ! 2 ˜ 5 ˜ 8 ˜ ... ˜ ( 3 n  2 ) ˜ ( 3 ( n  1 )  2) ) © ¹

LHS:

lim ¨ lim n of

(n  1) 3n  5

2n

12

f

n

S k 1

(k  2)!  1 , k Œ  . (k  2)!

Sk 1

Sk  uk 1

Using the limit comparison test with ¦ §1· sin ¨ ¸ ©n¹ n lim n of 1 n n

§1· sin ¨ ¸ n lim © ¹ n of 1 n

1 n n

n2

Pk 1

1, and 1 > 0, hence the

2

(k  2)!  1 (k  2)!

,

Since Pn is true for n = 1, and proven true for n = k + 1 when n = k, k Œ , then by mathematical induction Pk 1 is true. ª (n  1)

b. Using the ratio test, lim « u n of ( n  2 )! ¬ Since L < 1, the series converges.

1

n2

f

n . Then n 1 ( n  1)! f f 1 1

n2

S e 2 S ¦ n 2

Review Exercise

b

The series does not have all non-negative terms.

1 a

The series does not have all non-negative terms. d i the sequence of absolute values is not decreasing. ii the limit of the nth term as n toes to infinity does not equal 0. c

§ n · ¸ n of n  4 © ¹

1 2 3   2! 3! 4!

( n  1)!

1

e  1.

n 1

n

§ · ¨ 1 ¸ lim ¨ ¸ n of ¨¨ 1  4 ¸¸ n¹ ©

n

1 z 0, hence the e4

series diverges.

sin

23 ; 24

1 n

Using the limit comparison test with ¦ ,

n 1

1 1 2 5 ; S2  ; S3 2! 2! 3! 6 1 2 3 4 119 S4    2 ! 3 ! 4 ! 5 ! 120 1 2 3 4 5 S5     2! 3! 4 ! 5! 6! 1 2 3 4 5 S6      2! 3! 4 ! 5! 6! (n  1)!  1 Conjecture: Sn (n  1)!

n 1

f

¦ n!

Using the nth term test for divergence, lim ¨

b

n

¦ (n  1)!.

n!

S ¦

Hence, S = 1.

The series is continuous and decreasing, but not positive for all n ≥ 1.

S1

0.

Let S denote ¦

15 a

16 a

(n  1)! º n »¼

*Finding the sum is best done after Chapter 5 using f(x) = ex and Taylor series.

series converges by the p-series test, the original series also converges.

f

(k  1)!  1 , k Œ . (k  1)!

(k  2)! (k  2)!  (k  2)  k  1 (k  2)!

series converges since ¦ converges by the n n p-series test. S S arctan n 1 S 2 .  14 0  3 3 ¦ 23 ¦ 3 . Since this n2

1 = RHS 2!

(k  1)!  1 k 1  by assumption (k  1)! (k  2)! (k  2) > (k  1)!  1@  k  1

Hence, the series diverges. (This one is best done with the Root Test, which is not on the syllabus.) 13

2!  1 2!

n . (n  1)!

We want to show that Pk 1 is true, i.e.,

( 2) 4 . Using the ratio test, ¦ n n n n 1 n 1 n 4 n 1 § nn · 4 ˜ 4n (n  1)n 1 lim lim u ¨ ¸ n of ( n  1)( n  1) n n of 4n 4n ¹ © nn 4n n lim 0. n of ( n  1)( n  1) n

¦

(1  1)!  1 (1  1)!

Assume Pk is true, i.e., Sk

1 3

Since L < 1 the series converges. f

Proof by Mathematical Induction:

lim n of

1 n

1 n

1, hence by comparison with the

harmonic series, the series diverges. 719 ; 720 6 5039 . 7 ! 5040

c

For n ≥ 2, and since 2 < e, ln(n) < n – 1. Hence, f f 1 1 1 1 ! Ÿ¦ !¦ ln n n  1 n 2 ln n n 2 n 1

f

1

¦ n . Since the n 1

harmonic series diverges, by the comparison test the given series will also diverge.

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 4

37

WORKED SOLUTIONS d

By the ratio test, n

§ n 1 e · u ¸ n 1 n¹ ©e

lim ¨ n of

n

§ n 1 e · u n 1 ¸ e ¹ © n

lim ¨ n of

1  1. Hence e

the series converges. (n  10 ) cos nS (n  10 ) e Since cos nS ( 1) , . (1)n n 1 .4 n1.4 (n  10 ) Since lim 1.4 0 and the sequence is n of n n

decreasing, the given series is convergent by the alternating series test.

2

§

k! ·

k

k

0. u By the ratio test, lim ¨ ¸ lim n of ( k  1)! (k  1) ¹ n of k 2  1 © f

(k  1) k! 1

Hence the series converges. ¦ k

Furthermore, S

f

7

Using the limit comparison test with bn § §1·· ¨ 1  cos ¨ n ¸ ¸ © ¹¸ lim ¨ n of ¨ 1 ¸ 2 ¨ ¸ n © ¹

1 , n2

0 , an indeterminate form which 0

allows us to apply L’Hopital’s rule. Taking the derivative of numerator and denominator, we 1 1 · § ¨ sin n u n 2 ¸ obtain lim ¨ ¸ n of 2 ¨¨ ¸¸ 3 n © ¹

1· § ¨ sin n ¸ 1 lim ¨ ¸ 2 n of ¨ 1 ¸ ¨ ¸ © n ¹

1 sin n lim 2 n o0 n

1 . 2

1

¦k.

Since ¦

k 1

1 1 2 1 3 1       ... 1! 1! 2 ! 2 ! 3 ! 3 ! §1 1· § 1 1· §1 1· = 1  ¨  ¸  ¨  ¸  ¨  ¸  ... © 1! 1! ¹ © 2 ! 2 ! ¹ © 3 ! 3 ! ¹

1 converges by the p-series test, the given n2

series also converges. 8 a y

= 1. 3

lim n of

1 (n  1)7

0, and for all n, an t an 1, hence the

series converges by the alternating series test 4. 8–7 = 0.000000476837. The sum of the series therefore, correct to 6d.p., is 1 1 1 1  7  7  ...  7 2 3 7 4

0

0.992594

f

nn lim n of 1  n 2

1. Hence the series

n

f

Since ¦ an converges, then lim an n of

n 1

0 Ÿ an  1

1 p

Since the area under the curve is between smaller areas and the larger f

converges. Similarly, bn  1 Ÿ bn2  bn , and

6 a

un

n

n

b

. Using the ratio test,

23 n § · 3  3 n 1 2 ¨ ¸ lim ¨ n 1 u n of 3n  2 ¸¸ ¨ 23 © ¹

1 2  1. 3

b

un

 2

n





1 2

lim n of

2n  1 2n  1

1 dx xp 1

¦n

p

n 1

f

1

. Then by the ratio test,

n § 2 ·¸ 2n  1 ¨ lim ¨ u n 1 n of 2n  1 ¸¸ ¨ 2 © ¹

1

³

Hence the series converges. 2n  1

³

f

f

1 § 3n  1 · 2 lim ¨ ¸ of n 3 © 3n  2 ¹

2

f

ª x  p 1 º « » ¬  p  1 ¼1 f

1 . p 1

f 1 1  1  ³1 p dx p x 2n

1 ¦ n

1 p  and n p p 1

Hence, 1

f f 1 1 1 .  dx  ¦ p p p ³ 1 x n 1 n 2 n

areas, ¦

hence ¦ bn2 converges because ¦ bn converges. 3n  2

.

n 1

By comparison, and since ¦ bn converges, then n n

x

5

1 . p n 2

¦n

for large n. Since bn ! 0 Ÿ an bn  bn .

¦a b

4

The sum of the areas of the larger rectangles is

f

b

3

is ¦

2

diverges by the limit comparison test. 5 a

2

The sum of the areas of the smaller rectangles

By comparison with the harmonic series, § 1 n n · lim ¨ u ¸ n of 1  n 2 1¹ ©

1

³

f

1

1

1 p 1

p . p 1

f 1 1 dx  ¦ p . p x n 1 n

f 1 1 p ¦ p  . p 1 n 1 n p 1

 1.

Hence the series converges. © Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 4

38

WORKED SOLUTIONS 9 a

c i

When n = 8, S < 1.08243… and S > 1.08219.., hence S = 1.082 to 3 d.p.

ii

Substituting this value for S, S4

N| 10 a S2 n

1.082

The total area of the upper rectangles is f

1 . ¦ 4 i n i

b

general, S2m n ! Sn 

1 . ¦ 4 i i n 1

S2m1 ! S2  c

infinity lies between the above two sums, hence 4

i n 1

b

³

f

n

f

 ³n

1 dx x4

1 dx  x4 f

ª 1 º « 3 » ¬ 3x ¼ n

f

1.

¦i

11 i

4

i n

1 . Hence, 3n 3

f

1

¦i

4



i n 1

Replacing n by 2n, S4 n ! S2 n 

1

! Sn  1.

2

3

f

The total area under the curve from x = n to 1

2

2

1 1 1 u1 u1 u 1  ... (n  1) 4 (n  2) 4 (n  3) 4

f

1 Sn  .

1 2n

Continuing this way, S8n ! Sn  . And in

The total area of the lower rectangles is

¦i

90 .

1 1 1 1 1   ...  ! Sn    ... 2n 2n 2n n 1 n  2

Sn 

 1 1 1 u1 u1 u 1  ... 4 4 n (n  1) (n  2) 4

| 90.0268... Ÿ N

1 . 3n 3

ii

m 2

m 2

. Then, letting n = 2,

.

m ! N , or if m ! 2( N  S2 ). 2 Hence the sequence is divergent. S2m1 ! N if S2 

The sequence of absolute values is decreasing and the nth term tends to 0 as n tends to infinity, hence by the alternating series test, the series converges. The partial sums are 0.333, 0.111, 0.269,

n

1 Then, adding ¦ 4 to both sides of the i 1 i

0.148, 0.246. Sf lies between any pair of successive partial sums, hence it lies between

n

1 1 inequality, S  ¦ 4  3 . In the same way 3n i 1 i n 1 1 1 1 ! 3 , and adding ¦ 4 to both sides, ¦ 4 3n i 1 i i n i f

0.148 and 0.246, therefore it is less than 0.25. 12 a

n 1

1 1  3 . Hence, the value of S lies 4 3n 1 i

S !¦ i

n 1 1 1 1 1 between ¦ 4  3 and ¦ 4  3 . 3n 3n i 1 i i 1 i n

b

The sequence of absolute values is decreasing and the nth term tends to 0 as n tends to infinity, hence by the alternating series test, the series converges. S4

1

1 3!



1 5!



1 7!

0.841468(6d.p.).

S4 error  a5, hence error 

1 9!

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

, or error  0.00000276.

Calculus: Chapter 4

39

WORKED SOLUTIONS

5

Everything polynomic Exercise 5B

Exercise 5A 1 a For f (x ) = r = 4x, 1  4x 1 1 1 º 1 1ª 4 x  1 Ÿ x  ,   x  , x  »  , «. 4 4 4 ¼ 4 4¬

a

f

R = 1. At x = 1, f

1  4 x  ( 4 x )2  !  ( 4 x )n  !

n

0

1 1  ( 5 x )

Hence, c

f (x )

( 1)n n

f

n 0

1 ,r 1  5x 1 1 º 1   x  , x  » , 5 5 ¼ 5

1ª . In this interval, 5 «¬

b

3

1

¦ (1)n n is the alternating harmonic f

¦

n 0

( 1)2 n n

f

1

¦ n is the harmonic

n 0

x n 1 ( n 1)!

x lim

xn n!

n! (n  1)!

x lim

1 n 1

0.

1  (5 x )  ( 5x )  (5 x )  !  ( 5x )  !

n of

1  5x  52 x 2  !  ( 1)n (5x )n  !

Hence, the interval of convergence is the set of Reals, and R = infinity.

f

1 1  5x

1 . 5

¦ (1)n (5x )n ; R

n

1 x 4

0

1 4  41 x  1

c

1 . 4 1  41 x

In this interval,

1 x 4



1 §1 · ¦ ¨ x¸ ; R 4 n 0© 4 ¹

n of

n of

Using the ratio test, lim

n of

2n 1 x n 1 ( n  2 )2 n

(n  1)2 n of ( n  2 ) 2

2 x lim

n

2 x ( n 1 ) 2

2 x . Hence, the series

converges in the interval 2|x|< 1, or  1 x=− , 2

n

f

4

f

¦

n 0

2n   12

n

f

( 1)n

¦ (n  1)

(n  1)2

2

1 1  x  . At 2 2

. Since the absolute

n 0

value of the series is decreasing, and lim un n of

f (x )

1 3x

· 1§ 1 ¨ ¸ 3 © 1  ( x  1) ¹

1§ 1 · ¨ ¸ 3© x ¹

x.

Using the ratio test, lim

n

1 x  1 Ÿ x  4,  4  x  4, x  @4, 4>. 4

d

n n 1

series, which diverges. Hence, the interval of convergence is ]−1, 1[, and R = 1.

1 5

 5x ; r  1 Ÿ x  ,

2

n of

series, which converges conditionally. At x = −1,

¦ (1)n

For f ( x )

x lim

n xn n

n 0

¦ ( 4 x )n ,

1 and R = . 4 b

( 1)

n of

The series converges in the interval |x| < 1, hence

In this interval, 1 1  4x

x n 1 n 1

( 1)n  1

Using the ratio test, lim

0,

1 2

the series converges conditionally. At x = , f

¦

· 1§ 1 ¨ ¸ ; x  1  1 Ÿ 0  x  2. 3 © 1  [ ( x  1)] ¹ 2 Hence, the interval is º» 0, ª« ¼ 3¬ f 1 In this interval, f ( x ) ¦ (1)n ( x  1)n ; R . 3 n 0

n 0

2n ( 12 )n (n  1)2

n

1

¦ (n  1)

2

. This series converges by

n 0

comparison with ¦

1 , which converges by the n2

p-series test. Hence, the interval of convergence is

[− 12 , 12 ], and R = 12 . n 1

e

(2 x  3) x ; x  1 Ÿ  1  x  1 , x  @1, 1> . In this d Using the ratio test, lim ( 1) ( n  1) ln( n n 1) 1 x n of ( 1)n (2nxln 3)n x 2 n x (1  ( x )  ( x )  !  (  x )  !) interval, n ln n 1 x 2 x  3 lim 2 x  3 . The interval f f n o f ( n  1) ln( n  1) n n n n 1 x (1) x (1) x ; R 1

f (x )

¦

n

f

f (x )

¦

0

n

0

2 1 1 º 1 1ª ; 4 x  1 Ÿ   x  , x  »  , «. 1  4x 4 4 ¼ 4 4¬

In this interval, 2 1  4x

2

2 ¦ (1)n (4 x )n n

0

f

¦ 2 ˜ (1)n (4 x )n ; R

n

0

At x = −2, ¦

1 is divergence by comparison with n ln n

the harmonic series, and when x = −1, the series converges conditionally. Hence, the interval of

n

2(1  (4 x )  (4 x )  !  ( 4 x )  !) f

of convergence is |2x + 3|< 1, or −2 < x < −1.

1 . 4

1 2

convergence is ]−2, −1], and R = .

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 5

40

WORKED SOLUTIONS e

( 1)n 2 ( x 1)n 2 n2

Using the ratio test, lim n of

( 1)n 1 ( x 1)n 1 n 1

n 1 n2

1( x  1) lim n of

j

x  1 . The interval of

convergence is |x – 1| < 1, or 0 < x < 2. At x = 0,

¦ (1) with

n 1

1 n

( 1)n 1 n 1

x 2 n 3 ( 2 n  3 )!

x 2 lim

x 2n+1 ( 2 n 1)!

n of

x n 1 ( n 1) n 1˜3n

n of

x2 ˜ 0

(2n + 3)! (2n  1)!

2 a

( kx )n 1

lim

0. Hence, the

n of

f

n3

n of

x

(n  1)

n

f

1 n n ˜3

¦n

n

n 1

f

p-series test. At x = −3, ¦ n 1

Using the ratio test, lim n of

x 3

1

x

n

3

¦

n

n 1

( 3)

1 n

converges by

3 2

n

f

¦

n n ˜ 3n

n 1

( 1) n

3 2

3n 1 n ˜ xn

3

Applying the ratio test, lim n of

n

x h k

n of

sin x n2

x n 1 lim 3 n of n

. The interval of convergence

1

n also diverges. Hence, the interval

n of

lim n of

x h

n n 1

k

. Since

x h k

1

Exercise 5C 2 1 x2

I

n 1

sin x lim

( x  h )n

Ÿ~x  h~ k Ÿ  k  x  h  k Ÿ  k  h  x  k  h Since the interval of convergence is ]  1, 7[Ÿ  k  h 1 and k  h 7. Solving simultaneously, h = 4 and k = 3.

2

2 (1  x 2  x 4  ...  x 2n  ...)

1 ( x  1)2

f

2¦ x 2 n . n 0

@1, 1> , R

1.

1 1 ˜ 1 x 1 x

f

f

n 0

n 0

¦ (1)n x n ˜ ¦ (1)n x n

f

¦ (1)n (n  1) x n . I @1, 1> ; R

Using the ratio test, lim

( n 1) k n 1 nk n

, the

n 1

sin n 1 x ( n 1 ) 2

( x  k )2 . The interval

( x  h )n 1

of convergence is ]−3, 3[, and R = 3. i

( n 1)4 ( x  k )2n

of convergence is ( x  k )2  1 Ÿ  1  x  k  1 Ÿ  1  k  x  1  k . Since the interval of convergence is given as [2, 4] Ÿ  1  k 2 and k  1 4. Hence, k = 3.

f

x = −3 ¦ (1)

2.

n4

n4 ( x  k ) lim n of ( n  1) 4

 1 Ÿ x  3. At x = 3, ¦ n diverges, and at n

kx. ª 1 1º

2

 1 Ÿ x  3.

f

1

3n

f

n of

1

.

3

3

n 1 ˜ x n 1

n of

( kx )n

n2 (n  1)2

k x  1 Ÿ x  . Since x  «  , » Ÿ k k ¬ 2 2¼

series converges conditionally by the alternating series test. Hence the interval of convergence is [−3,3], and R = 3.

lim 1 

x lim k ˜

n2

n n

lim

3n

At x = 3, ¦

( n 1)2

b Applying the ratio test, lim n of

The interval of convergence is

3

x 1 . 3

Applying the ratio test,

(n  1) n  1 ˜ 3

n of

3

x

x 1 3

( x  k )2 (n 1)

x lim

xn n n ˜3n

x

is

n of

interval of convergence is ]−4, 2[, and R = 3.

Using the ratio test,

3

ex

3 f ( 3)n At x = −4, ¦ (1)n, which diverges. At ¦ 3 n 1 n 1 f f 3n x = 2, ¦ n ¦1 which also diverges. Hence, the n 13 n 1

interval of convergence is the set of Reals, and R = infinity.

x

n n 1

f

1 x 2 lim n o f (2n  3)(2n  2)

h

n of

This converges for x  1  1 Ÿ  4  x  2 .

, which diverges by the p-series test. At

Using the ratio test, lim

n of

lim

e nx n

n of

n 1

lim

e ( n 1 ) x n 1

lim

1 ¦ n  1 diverges by comparison

the alternating series test. Hence, the interval of convergence is ]0, 2], and R = 1.

g

e x lim

enx n

n of

Hence, the interval of convergence is ex < 1, or x < 0, and R = infinity. k Using the ratio test,

x = 2, ¦ (1)n 1 1 converges conditionally by

f

e( n 1) x n 1

Using the ratio test, lim

1.

n 0

n2 (n  1)2

sin x

The interval of convergence is |sin(x)| < 1, and this is true for all x. Hence, the interval of convergence is the set of Reals, and R = infinity.

3

1 4x 2 1

1  4 x 2  (4 x 2 )2  (4 x 2 )3  ... f

f

¦ (1)n (4 x 2 )n ¦ (1)n (2 x )2n . n 0

I

º 1 1ª » 2 , 2 «; R ¼ ¬

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

n 0

1 . 2 Calculus: Chapter 5

41

WORKED SOLUTIONS 4

This example should be done in 5F using the Binomial Series. The solution is shown here. 1

1 x

(1  x ) 2 1

2 § 1·x

1

1

3 § 1· § 3·x

1  x  u ¨  ¸  u ¨  ¸ u ¨  ¸  ... 2 2 © 2 ¹ 2! 2 © 2 ¹ © 2 ¹ 3!

(1  x )

1 x



1 2

2 § 1 ·§ 3 · x

§ 1·

1 3 5 1  x  x 2  x 3  ..., −1 < x < 1. 2 8 16

(1  x ) (1  x )



1 1 2 1 3 1 3 5 § ·§ · x  ... ¸ ¨ 1  x  x 2  x 3  ... ¸ ¨1  x  x  2 8 16 2 8 12 © ¹© ¹

5

1 1 ³x;x

ln x

2( x  1)  ( x  2) ( x  2)( x  1)

∞ 2 =¦ x + 2 n =0

1

x 2 x 3 3x 4 + + ...; º¼ −1, 1ª¬ ; R = 1 2 2 8

1 1  ( x  1)

f

1 x 1

³

1

1 dx 1  (t  1)

³

x

3x = 2 x + x −2

ª

= «t − ¬

§

= ¨x − ©

f

n 0

+

n=0

∞

∞

n =0

n =0

2, f (3 ) ( x )

24 Ÿ f (1  x )5

(4)

(0 )

1  x  x2  x3  x4

1

sin x  x cos x Ÿ f c(0)

hence R = 1.

f cc( x )

cos x  cos x  x sin x

x n 1 x , lim × x = which is 2 n 2 n → ∞ (n + 1) 2 2 n =1 n 2

For ¦

f

(3)

(x )

convergent for |x|< 2, so R = 2. Testing the end1 points, for x = 2, ¦ 2 is convergent by the p-series n n 1 f 1 test. For x = −2, ¦ (1)n 2 converges absolutely. n n 1

f

(4)

(x )

2

T4 ( x )

2,

(3)

(0 )

0,

 3 cos x  cos x  x sin x  4 cos x  x sin x Ÿ f

Hence, the interval of convergence is [−2, 2]

convergent for |x|< 2. Testing the end points, at 1 f 1 1 f ( 1)n 1 x = 2, ¦ diverges, and at x = −2, ¦ 2n 1n 2n 1 n converges conditionally.

0,

 2 sin x  sin x  x cos x  3 sin x  x cos x Ÿ f

f

n

0,

2 cos x  x sin x Ÿ f cc(0 )

2

x , which is also 2

6 (1  x ) 4

1 x 2 x 2 6 x 3 24 x 4     0! 1! 2! 3! 4!

f c( x )

nx n 1 n 1 , lim u x 2 n n of n 1 2 1 n 2

¼

24

x sin x Ÿ f (0)

f

− 1» x n .

6,

f (x )

For ¦

¬

º

n

1,

2 Ÿ f "(0) (1  x )3

T4 ( x )

b

n

ª 1

1.

¦ (an x n  bn x n ) is convergent when |x| > 1, ∞

;

¦ − x n = ¦ « (−2)

1 Ÿ f '(0) 1, (1  x )2

f (4) (x )

n 0

7

n

f '( x )

3

f

6

§1 ·

Ÿ f ( 3 ) (0 )

( x  1)n 1 n 1

@0, 2> ; R

x 1  1 Ÿ I

I

n

∞

¦ (−1)n ¨© 2 x ¸¹

1 Ÿ f (0) 1 x

f "( x )

· ( x − 1) ( x − 1) + + ... ¸ − 1 2 3 ¹

¦ (1)

n

n 0

f (x )

3

º (t − 1) (t − 1) + + ...» 2 3 ¼1 2

¦ x

Exercise 5D

x

3

n

@1, 1> ; R

Hence, I

(1  (t  1)  (t  1)  (t  1)  ...)dt 2

f

1 1  x 0

1 ¦

n 0

3

2

1

f

1

¦ 1 x

1  ( x  1)  ( x  1)  ( x  1)  ... x

n=0

R = 2.

1 a

2

n

∞

§1 · ¦ (−1)n ¨© 2 x ¸¹ . I @2, 2> ;

=

§ 1 · 1− ¨ − x ¸ © 2 ¹

3x . x  x 2 2

I = ]−1, 1[ ; R = 1.

1 2

=1+ x +

2 1  x  2 x 1

3 § 1 ·§ 3 ·§ 5 · x

1  ¨  ¸ x  ¨  ¸ ¨  ¸  ¨  ¸ ¨  ¸ ¨  ¸  ... © 2¹ © 2 ¹ © 2 ¹ 2! © 2 ¹ © 2 ¹ © 2 ¹ 3!

1 2

convergence would be the same, except for possibly the end points. Hence, the interval of convergence is [−2, 2[. 8

1 1 1 1  x  x 2  x 3  ...;  1  x  1 2 8 16 1

xn nx n 1 Since 2 n is the derivative of 2 n , the interval of n 2 n 2

2x 4 x  2! 4!

4

(4)

(0 )

4

1 6

x2  x4

1

c

f (x )

(1  x ) 3 Ÿ f (0 ) 2  3

1,

1 (1  x ) Ÿ f c(0) 3 5 − 2 f ′′( x ) = − (1 + x ) 3 Ÿ f ′′(0) 9

f c( x )

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

1 , 3 2 9

=− , Calculus: Chapter 5

42

WORKED SOLUTIONS 8  10 (1  x ) 3 Ÿ f 27

f (3 ) ( x )

(3 )

10 , 27

(0 )

c

x 2 cos x Ÿ f (S )

f (x )

f ′( x ) = 2 x cos x − x 2 sin x Ÿ f ′(π ) = − 2π ,

8  80 80 f ( x )  (1  x ) 3 Ÿ f ( 4 ) (0)  81 81 80 2 2 10 3 1 x  x x  x4 1 3 T4 ( x )   9  27  81 4! 3! 2! 0 ! 1! 1 1 2 5 3 10 4 = 1+ x − x + x − x 3 9 81 243

(4)

f ′′( x ) = 2 cos x − 2 x sin x − 2 x sin x − x 2 cos x = (2 − x 2 )cos x − 4 x sin x

f

(3)

 2 x cos x  (2  x 2 )sin x  4 sin x  4 x cos x

(x )

 6 x cos x  ( x 2  6)sin x

Ÿ f '(0) = 1,

f "( x )

2e x  xe x Ÿ f "(0 )

f

(3)

(x )

3e x  xe x Ÿ f

(3)

f

(4)

(x )

4 e x  xe x Ÿ f

(4)

3,

(0 )

(π ) = −

=

f (x )

3

1

+−

π

a

 5, f c(0 )

f (0 )

Ÿ T3 ( x ) =

π2

(x − π ) +

2

π3

x 1 sin x Ÿ f (S )

( x − π )2 −

3

b

1

π4

f cc(1)

( x − π )3

6, f 3 (1)

1

π

,

4

2

T1 ( x ) =

f (a ) f ′(a )( x − a ) + = f (a ) + f ′(a )( x − a ) 0! 1!

= f ′(a )( x − a ) Ÿ y = f (a ) + f ′(a )( x − a )

−1

+ 2 x sin x + x sin x − x cos x  6 x 4 sin x  6 x 3 cos x  3x 2 sin x  x 1 cos x S3



1

1

(x  S )

S 2

2

5

f (1)

4, f c(1)

 1, f cc(1)

3, f 3 (1)

2Ÿ

4 1( x  1) 3( x  1)2 2( x  1)3    0! 1! 2! 3! 3 1 2 4  ( x  1)  ( x  1)  ( x  1)3 2 4 3 1 f (1.2) | T3 (1.2) 4  0.2  u 0.22  u 0.23 2 4

T3 ( x )

§ 6 1· 3 ¨  3  ¸ (x  S ) S S¹ © S S   1! 2! 3! 1 1 1 1 § ·  ( x  S )  2 ( x  S )2  ¨  3 ¸ ( x  S )3 S S © 6S S ¹ 

T3 ( x )

6

12 Ÿ

Equation: y − y1 = m ( x − x1 ) Ÿ y − f (a )

π2 f (3) ( x ) = − 6 x −4 sin x + 2 x −3 cos x + 4 x −3 cos x



4,

Equation of a tangent to the curve y f ( x ) at the point where x a is obtained in the following way: Point (a, f (a )), slope m = f ′(a )

2 x 3 sin x  2 x 2 cos x  x 1 sin x

Ÿ f (3) (S )

664

 2  4 ( x  1)  3( x  1)2  2( x  1)3

2

−2

12  6

 2, f c(1)

2 4 ( x  1) 6( x  1)2 12( x  1)3    0! 1! 2! 3!

T3 ( x )

2 x sin x  x cos x  x cos x  x 1 sin x

−2

 6, f (3) (0) = 12

−5 4 x −6 x 2 12 x 3 + + + 0! 1! 2! 3!

23 4 5

f (1)

0,

2

4, f cc(0 )

 5  4 x  3x 2  2 x 3

3

1

Ÿ f "(π ) = −

2 x 3  3 x 2  4 x  5 Ÿ f ′( x ) = 6 x 2 − 6 x + 4

f (x )

Ÿ f ′′( x ) = 12 x − 6 Ÿ f (3) ( x ) = 12

f ′( x ) = − x −2 sin x + x −1 cos x Ÿ f ′(π ) = f "( x )

§S2 ·  1¸ ( x  S )2  S ( x  S )3 2 © ¹

 S 2  2S ( x  S )  ¨

1 1 x  x  x3  x4 2 6 2

1 2 1 1  (x  S ) ( x  S ) 2  2 ( x  S )3 3 S  S2 S S   0! 1! 2! 3!

T3 ( x )

b

π

 6S

S 2 2S ( x  S ) (S 2  2)( x  S )2 6S ( x  S )3    0! 1! 2! 3!

T3 ( x )

1 1 1 Ÿ f (π ) = , f ′( x ) = − 2 2 a f (x ) = x π x 1 2 Ÿ f ′(π ) = − 2 , f ′′( x ) = 3 π x 2 6 Ÿ f ′′(π ) = 3 , f (3) ( x ) = − 4 π x 6 (3)

Ÿ f

(S )

4

x 2 x 2 3x 3 4 x 4    1! 2! 3! 4!

T4 ( x )

(3)

Ÿ f

2,

(0 )

S2 2

Ÿ f cc(S )

f ( x ) = xe x Ÿ f (0) = 0, f '( x ) = e x + xe x

d

 S 2,

( x  S )2

3.86267

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 5

43

WORKED SOLUTIONS

Exercise 5E

f

1

f (x ) =

f ( x ) cos x Ÿ f c( x )  sin x Ÿ f cc( x )  cos x Ÿ f (3 ) ( x ) sin x Ÿ f ( 4 ) ( x ) cos x We can summarize this into the following form (similar to the powers of the imaginary unit i):

f

(n )

­  sin x , n 4 k  3 ° cos x , n 4 k  2 ° , k  ] ® ° sin x , n 4 k  1 °¯ cos x , n 4 k

(x )

f ( n 1) (t ) n 1 x Ÿ f (n  1)!

Rn ( x ) 0

( n 1)

(n )

 (n  1)!, n  ] 

(0 ) ∞

n =1

x 

(3)

Ÿ f

(4)

Ÿ f

0

n of

 6(cos x ) 4 u (  sin x ) u sin x

(x )

 8(cos x ) 5 u (  sin x )(1  2(sin x )2 )

(x )

8 sin x (cos x ) 5 (2  (sin x )2 )

x 2n

8(cos x ) 5 (2 sin x  (sin x )3 )

¦ (1)n (2n )! for all x  \. sin 2 x

(5)

 8(cos x )5 (2 cos x  3(sin x )2 cos x ) 8(cos x )6 (10(sin x )2  5(sin x )4  2(cos x )2  3(sin x )2 (cos x )2 )

§2 2x 4 1 · 6 §2 2 · 8 x  x ¨  ¨  ¸ x  ... 2 ¸ u 5! ¹ 3! 5! (3!) 7! 3! © ¹ © 2

 3(sin x )2 (1  (sin x )2 ))

2x 2 8x 4 32 6 128 8 22 n 1 x 2 n x  x  ...  ( 1)n    ... 2! 4! 6! 8! (2n )!

16(sin x ) 4  88(sin x )2  16 (cos x )6

4

¦ (1)n n 1

22 n 1 x 2 n (2n )!

We use the expension of e of x we input 3x 0

1  3x 

Ÿ f cc( x )

f (0)



3

3

n

n

9x 27 x 3 x   ...   ... 2! 3! n!

ln(1  x ) Ÿ f c( x )

f (n ) ( x )

2

(3 x ) (3 x ) (3 x ) (3 x )     ... 0! 1! 2! 3!

e3 x

f (x )

x where instead ¦ n 0 n!

x



1  Ÿ f cc( x ) (1  x )2

Ÿ f cc(0)

0 Ÿ f c(0)

0 Ÿ f c(0 )

(3)

2Ÿ f

(0 )

1.1

d dx

(

2  (1  x )3

0Ÿ f

(0 )

0Ÿ (5)

(0 )

16 Ÿ

Exercise 5E

(cos(x))6

(

)

0

)

272

|x = 0

d2 2∙(sin(x))4+88∙(sin(x))2+16 |x = 0 dx2 (cos(x))6

2/99

x 

f (x )

=x +  1 Ÿ f cc(0)

(4)

1 Ÿ f cc(0 )

16∙(sin(x))4+88∙(sin(x))2+16

1 1 x

(n  1)! , n  ] (1  x )n

ln(1)

f (0 ) f

n

f

1

2

c

8(cos x )6 (10(sin x )2  5(sin x ) 4  2(1  (sin x )2 )

2x 8x 12 + 20 6 16 + 112 8 x − x + ... − + 2! 4! 6! 8! 2

f

f (x )

 40(cos x )6 u (  sin x )(2sin x  (sin x )3 )

(x )

sin x u sin x

§ · § · x3 x5 x7 x3 x5 x7  ... ¸ u ¨ x   ... ¸     ¨x  3! 5! 7! 3! 5! 7! © ¹ © ¹

b

x x xn   ...   ... 2 3 n

2(cos x ) 4 (1  2(sin x )2 )

Ÿ f

=

3

 2(cos x )3 u (  cos x )

n 0

f (x )

xn n

Ÿ f ′′( x ) = − 2(cos x ) −3 × ( − sin x ) = 2(cos x )−3 sin x

(t ) d 1

0 Ÿ lim Rn

f

2 a

n =1

−

 2(cos x )4 4 sin x cos x

x (n  1)!

Hence cos x

∞

¦

f ( x ) = tan x Ÿ f ′( x ) = (cos x )−2

n 1

n of

n!

xn =

2

d

f  n 1 (t ) n 1 x n 1 x d (n  1)! (n  1)!

Ÿ lim

( n − 1) !

−

¦

2 x 3 16 x 5 272 x 7    ... 3! 5! 7! x3 2 x 5 17 x 7 + + + ... 3 15 315

1

2

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 5

44

WORKED SOLUTIONS e

The derivatives are best found using a computer algebra system. arctan( x 2 ); f (0 )

f (x )

2x ; f c(0) x 1

f cc( x )

2(3 x 4  1) ; f cc(0) ( x 4  1)2

f ccc( x )

8 x 3 (3 x 4  5) ; f ccc(0) ( x 4  1)3

0

12

f

(0)

0

8

1  ix  (4)

(0 )

2 eix

4

0 240(21x 16  336 x 12  546 x 8  120 x 4  1) ; ( x 4  1)6

f ( 6 ) (0 )

 240

a

x6 x 10 x 14 x 4n+2 + ... + (−1)n + ... + − 3 5 7 2n + 1

§ ©

§ ©

= 2ix − § ©

b

· (2 x )2 (2 x ) 4 (2 x )6 1 1§  ... ¸  ¨1    2 2© 2! 4! 6! ¹

sin x x

1

n 1

§ · x2 ix 3 x4 ix 5 x6 ix 7       ... ¸ ¨ 1  ix  2! 3! 4! 5! 6! 7! © ¹ §

+ ¨ 1 − ix − ©

22 n 1 x 2 n (2n )!

=2−

§ · 1 x3 x5 x7    ... ¸ u ¨x  3! 5! 7! © ¹ x

§ ©

§ · sin x x2 x4 x6 = lim ¨ 1 − + ... ¸ + − x →0 x → 0 x 3! 5! 7! © ¹

1

· x2 x4 x6 − + + ... ¸ 2! 4! 6! ¹

2 cos x Ÿ cos x

lim

02 04 06   ... 1  3! 5! 7!

· x 2 ix 3 x 4 ix 5 x 6 ix 7 − + + − + + ... ¸ 2! 3! 4 ! 5! 6! 7! ¹

2x 2 2x 4 2x 6 + − + ... 2! 4! 6!

= 2 ¨1 −

x 2n x2 x4 x6 + ...   ... + ( −1)n  (2n + 1)! 3! 5! 7!

6

eix − e −ix 2i

If we add equations (1) and (2) we obtain the following:

eix  eix

1 1 2x 2 23 x 4    2 2 2! 4! f

· x3 x5 x7 + ... ¸ + − 3! 5! 7! ¹

= 2i sin x Ÿ sin x =

1 1  cos(2 x ) 2 2

¦ (1)n

· x2 ix 3 x 4 ix 5 x 6 ix 7 + − + + ... ¸ + − 2! 3! 4 ! 5! 6! 7! ¹

2ix 3 2ix 5 2ix 7 + ... + − 3! 5! 7!

= 2i ¨ x −

1 1 5 x  x2  x3  x  ... 3 30

f (x )

· x2 ix 3 x4 ix 5 x6 ix 7 − + ... ¸ − + + − 2! 3! 4! 5! 6! 7! ¹

− ¨1 − ix −

1· 1· 1 1· §1 §1 §1  ¸ x3  ¨  ¸ x4  ¨   ¸ x 5  ... © 2! 3! ¹ © 3! 3! ¹ © 5! 2! u 3! 4 ! ¹

25 x 6  ... 6!

x2 ix 3 x4 ix 5 x6 ix 7       ... 2! 3! 2! 5! 6! 7!

eix − e−ix = ¨ 1 + ix −

x  x2  ¨

4

( ix )6 ( ix )7  ...  6! 7!

If we subtract equation (2) from equation (1) we obtain the following:

§ · § · x2 x3 x3 x5 x7  ... ¸ u ¨ x   ... ¸    ¨1  x  2! 3! 3! 5! 7! © ¹ © ¹



( ix )0 ( ix )1 ( ix )2 ( ix )3 ( ix ) 4 ( ix )5      0! 1! 2! 3! 4! 5!

1  ix 

f ( x ) e x u sin x

sin 2 x

x2 ix 3 x4 ix 5 x6 ix 7   ...     2! 3! 2! 5! 6! 7!

0

f (6) ( x )

f (x )

(ix )6 (ix )7   ... 6! 7!



f (x ) = x 2 −

3

(ix )0 (ix )1 (ix )2 (ix )3 (ix ) 4 (ix )5      0! 1! 2! 3! 4! 5!



Hence,

f

eix

2

48 x (15 x  135 x  101x  5) ; ( x 4  1)5

f (5) ( x ) (5)

1

24 x 2 (5 x 8  22 x 4  5) ; f ( x 4  1) 4

f (4) (x )

xn where instead ¦ n 0 n! f

We use the expansion of e x

of x we input ix and  ix respectively.

0

f c( x )

4

5

eix  e ix 2

We can solve this question by using the results from the previous question directly or using algebra.

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 5

45

WORKED SOLUTIONS Method I

e

T

2

3

4

5

6

c

7

iT T iT T iT       ... 1  iT  2! 3! 4! 5! 6! 7!

iT

T2

T4

1

f (x )

1  4 x

§ · T3 T5 T7    ...  i ¨ T     ... ¸ 1 2! 4! 6! 3! 5! 7! ©  ¹  

cos T

§ 2 · ¨ ¸ ©n ¹

sin T

( 1)n

2 u 3 u ... u (n  1) n!

( 1)n

(n  1)! n!

Method II

iT

e e 2

 iT

iT

e e 2



eiT  e iT Ÿ cos T  i sin T 2i  iT

iT

2e 2

f

1 a

f (x )

f

n

(1  x )

4

6

§1· ¨ 2 ¸ (  x )n, we are ¦ ¨ ¸¸ n 0¨ ©n¹

1 º 1 1ª Ÿ x  » , « 4 ¼ 2 2¬

f

d

1

f (x )

4

1  2x 3

1  2x 3

( 1)n f

1  ¦ ( 1)n f

1  ¦ ( 1)n n 1

f

(2n )! 1  ¦ (1)n 1 n (1)n x n (2 u n !)2 (2n  1) n 0

n 0

n

(2n )! xn u n !)2 (2n  1)

2x 3  1 Ÿ x 3 

2

f (x )

b

f (x )

1 (1  x )3

(1  x )3

§ 3 · ©

( 3) u ( 4) u ... u ( 3  n  1) n!

(1)n

3 u 4 u ... u (n  2) n!

( 1)n

(n  1)(n  2) 2

f

§ 1· ¨ 2 ¸ ¨¨ ¸¸ © n ¹

1 1 x2

§ 1· ¨  2 ¸ (  x 2 )n ¦ ¨ ¸¸ n 0¨ © n ¹

1 2

f

§ 1· § 3· § 1 · ¨  ¸ u ¨  ¸ u ... u ¨   n  1¸ © 2¹ © 2¹ © 2 ¹ n!

1 u 3 u 5 u ... u (2n  1) 2n u n !

§ n  2 ·, so ¸ © 2 ¹

( 1)n

(2n )! 2 u n ! u 2n u n !

( 1)n ¨

n

f

§n  2· n ¸x © 2 ¹

x  1 Ÿ x  @1, 1>

1 1 1 1 ª º Ÿ x  3 x »  3 , 3 « 2 2 2 2¬ ¼

( 1)n

1  3x  6 x 2  10 x 3  21x 4  28 x 5  ..



1 u 5 u 9 u ... u (4 n  3) 3n x 2n u n !

(n  2)! 2un!

(1)n

¦ (1)n

¦ (1)n ¨ n 0



¹

going to simplify the binomial coefficient § 3 · ¨ ¸ ©n¹

(1  x 2 )

¦ ¨ n ¸ x n, we are n 0



1 u 5 u 9 u ... u (4 n  3) (2 x 3 ) n 4n u n !

arcsin x Ÿ f c( x )

 x  1 Ÿ x  @1,1> f

f

1 u 5 u 9 u ... u (4 n  3) , so 4n u n !

n 1

f

§ 1· ¨  4 ¸ 2x 3 n , ¦ ¨ ¸¸ n 0¨ © n ¹

1 4

§ 1· § 5· § 9· § 1 · ¨  ¸ u ¨  ¸ u ¨  ¸ u ... u ¨   n  1¸ © 4¹ © 4¹ © 4¹ © 4 ¹ n!

§ 1· ¨ 4 ¸ ¨¨ ¸¸ © n ¹

(2n )! , so (2n u n !)2 (2n  1)

¦ (2



we are going to the binomial coefficient

1 u 3 u 5 u ... u (2n  3) 2n u n ! (2n )! (1)n 1 n 2 u n ! u 2n u n ! u (2n  1)

1

(n  1) x 2 n

1  8 x  48 x  256 x  ..

(1)n 1

( 1)

n

n 0

2

1 § 1· § 3· §1 · u ¨  ¸ u ¨  ¸ u ... u ¨  n  1 ¸ 2 © 2¹ © 2¹ ©2 ¹ n!

n 1

¦4

(n  1)(  4 )n x 2 n

 4x 2  1 Ÿ x 2 

1 x

),

¹

( 1)n (n  1) ,

n 0

eiT

going to simplify the binomial coefficient §1· ¨2¸ ¨¨ ¸¸ ©n¹

¦ (1)

so

Exercise 5F 1 2

©

( 2) u ( 3) u ... u ( 2  n  1) n!

cos T  i sin T

cos T

n 0

2 n

we can simplify this binomial coefficient

T6

eiT  e iT , sin T 2

§ 2 ·

f

¦ ¨ n ¸ ( 4 x

(1  4 x 2 )2

2 2

n 0

f

f (x )

³ ¦ (2

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

n 0

( 1)n

(2n )! ( 1)n x 2 n 2 (2 u n !) n

(2n )! , so (2 u n !)2 n

f

¦

n 0

(2n )! x 2n 2 (2 u n !) n

(2n )! x 2n dx u n !)2

n

Calculus: Chapter 5

46

WORKED SOLUTIONS (2n )! x 2 n 1 ¦ n 2 n 0 ( 2 u n !) 2n  1 f

1 6

x  x3 

3 5 x  ... 40 §1·

1

3 a

x 3 1 9

9x

§

1 2

3 ¨1  u ©

1 6

1 x 

f

§x· 0¨ ¸© ¹ ©n¹

1 (1  x ) 4

n

3 u ¦¨ 2 ¸¨ ¸ ¨ ¸ 9 n

· x 1 x2 1 x3  u  u  ... ¸ 9 8 81 16 729 ¹

 sin c u x 6 º 1 1ª ; c  »  , «. 6! ¼ 3 3¬ 1 f (6)(c) is maximized when c = . 3

(1  x )4

f

3.017

(c ) x 6 6!

6

§1· ¨ ¸ © 3 ¹ u sin § 1 · | 6.2337 u 10 7. ¨ ¸ 6! ©3¹

§ 4 · ©

(6)

Hence, the maximum error is

¦¨ n ¸ xn n 0

f

R5 ( x )

1 2 1 x  x 3  ... 216 3888

1 1 9  0.1 | 3  u 0.1  u 0.12 6 216

9.1 b

x ·2 § 3 ¨1  ¸ 9¹ ©

3  0.00005 Ÿ n t 8. (n  1)! 1 1 1 Hence, e | 1  1    ...  2.7183. 2 3 8 x3 x5 4 sin x x    R5 ( x ) ; 3! 5!

ec  e  1 Ÿ Rn (1) 

¹

2

1  4 x  10 x  20 x 3  35 x 4  ... 1 1.034

5

1 (1  0.03) 4 2

3

| 1  4 u 0.03  10 u 0.03  20 u 0.03

0.8885

ex

1 x 

x2 x3 xn   ...   ... 2! 3! n!

1 §1· Rn ¨ ¸ 3 ©3¹

For x

Exercises 5G

1

2

1

ex

When x = –1, 11

1 1 1 1 1 1       .... 2 6 24 120 720 5040

§1·

1

³ 1  x dx ³ ¦ (1)

n

xn

x

2

¦ (1)n n  1  c .

n 0

0 .5 Ÿ  0.0001 Ÿ n 9 , and n 1

9

2

3

ex e1

1 x  11

1

0.405435

Since Rn (1)

1

3. Hence, e < 3.

4

2 9



5

2; f c(8)

10  38 x . 27

1 ; f cc(8) 12



1 . 144

f (8)  f c(8)( x  8) 

1

ec , and ex is a (n  1)!

strictly increasing function,

f cc(8) ( x  8)2 2!

and x 3 | T2 ( x ) . R2 ( x )

f ccc(c ) ( x  8)3 3!

10  38 ( x  8)3 c 27 3!

2

f ( n 1) (c ) (n  1)!

5

1  32 x ; 3

f cc( x )  x 3 ; f ccc( x )

Hence, T2 ( x )

1 1 1 1   .. d 1  1  2  3  ... 2 2 2! 3!

1

x 3 ; f c( x )

f (x )

3

§1· §1· §1· §1· ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ 1 3 3 3 3 1   © ¹  © ¹  © ¹  © ¹ | 1.3956. 4! 3! 2! 3 5! 1

6

f (8)

3

x x   ... 2! 3!

1

§1· © ¹

T5 ¨ ¸ 3

n 1

n 1

( 0 .5 ) n  1 (1)n ¦ n 1 n 0

1 | 0.000006  0.00005 . (5  1)! u 35

Hence,

1 1 1 1 1 | 11     | . 2 6 24 120 720 f

1

Therefore, n = 5 since

hence

Rn  un 1

1

n 1

Rn ¨ ¸  0.00005 Ÿ  0.00005. (n  1)! u 3n ©3¹

1 , which is approximately 0.0002, Error < 5040

2 ln(1  x )

1 ,0c  . 3

and we need

The series alternates and the last term is less than 0.001

e1

n 1

1 e3 §1· Hence, Rn §¨ ·¸  u ¨ ¸ . Since e < 27, e 3  3, (n  1)! © 3 ¹ ©3¹

3

x x 1 x    ... 2! 3!

e1

§1· (c ) ¨ ¸ ©3¹ (n  1)!

( n 1)

f

5( x  8)3 , 8

81c 3

where 8 < c < x. Since 7 ≤ x ≤ 9, −1 ≤ (x − 8), or |x − 8| ≤ 1, and therefore 3

8

8

x  8 d 1 . x ! 7 Ÿ c 3 ! 7 3 ! 179.

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Calculus: Chapter 5

47

WORKED SOLUTIONS 5 x 8

Hence, R2 ( x )

3



8

81c 3

5 u1  0.0004 , i.e., 81 u 179

lim x o0

the approximation is accurate to within 0.0004. 7

x

sin x

³

6 0

sin x  x x3

x3 x5   ... Ÿ sin 2 x 3! 5!

(sin x )(sin x ) S

x2 



1 4 2 x  x 6  ... 3 45

2x 

ª x3 1 2 7º  x5  x » « 15 315 ¼ 0 ¬3

³

6 0

lim

0.045294...

S

The exact integral is,

S

sin 2 x dx

³

6 0

x o0

1  2 cos x dx 2

c

lim x o0

8 lim

³

9

S 6

ª x º « » ¬ 2835 ¼ 0

x dx 315

8 8 x , 315

§1· © ¹

R5 ¨ ¸ 2 R5 ( x ) d

9

R4 (0.2)

f

(c ) § 1 · 1 u ¨ ¸ , 0  c  , and f 6! 2 2 © ¹

(6)

8 lim x o0

d

(c ) u (0.2)5 ; 0  c  0.2. Since 5!

e0.2 u (0.2)5 | 0.0000326. 5!

sin ( n ) (1) . Since all derivates of sin(x) are (n  1)!

between −1 and 1, Rn (1) d 1 7!

sin 2 x

Hence, lim

1 . By trial and (n  1)!

x o0

e

lim x o0

x3 x5 x    ... 3! 5!

1 8

8 lim x 6 x o0

1

x4 2  x 6  ... 3 45

x2 

x 2  sin 2 x x 2 sin 2 x

1

lim 3

x 4  452 x 6  ...

x o0

6

x 4  x3  ...

1 . 3

(sin 3 x  3 x )2 (cos 5 x  1)3

lim x o0

sin 2 3 x  3 x sin 3 x  9 x 2 cos 5 x  3 cos 2 5 x  3 cos 5 x  1 3

ª ¬

Numerator: «(3 x )2 

1  0.001 5050

Exercise 5H sin x

3 x 2 7 x 4 61x 6    ... 2 8 240

1

1  3 cos x  3 cos 2 x  cos3 x x6

ª ¬



162 15625

º (3 x ) 4 2(3 x )6   ...» 3 45 ¼

 3 x «3 x 

The degree of the polynomial is therefore 5.

1 a

(1  cos x )3 x6

x 4 2x 6   ... 3 45

(cos x )(cos2 x )

cos x

(5)

error, n = 6,

2

x2 x4 x6    ... 2! 4 ! 6!

1 x 2 

(c )  1.

1 | 0.0000217. 6 ! u 26

hence R4 (0.2) d Rn (1) d

x o0

x o0

2x 2x 3   ... 2! 4!

(cos x )(cos x )

0.000001. The actual

f (5)(x) = ex, the maximum that f (5)(c) can be in this interval is e0.2,

10

8 lim

2 lim

6

(6)

f

1

cos2 x

error is 0.0452940 − 0.045293 = 0.000001 and |R4(x)| ≤ actual error. 8

1 6



1  3 cos x  3 cos 2 x  cos3 x x6

cos x

The truncated term R4 ( x ) 8

x o0

0.045293. . .

2x 2 2x 4  ... 2! 4! x

(2  2 cos x )3 x6

S

S º6 ª sin » «x 3 «  » 2 4 « » ¬ ¼0

1 x2 x4 lim   ... 3! x o0 5! 7!

§ · § · x2 x2  .... ¸  ¨ 1  x   ... ¸ ¨1  x  2! 2! © ¹ © ¹ b lim x o0 x

§ 2 1 4 2 6· x ¸ dx ¨x  x  3 45 ¹ © S 6

S 6 0

§ · x3 x5   ... ¸  x ¨x  3 5 ! ! © ¹ lim 3 x o0 x

º (3 x )3 (3 x )5   ...»  (3 x )2 6 120 ¼

Denominator: ª º 3(5 x )2 7(5 x ) 4   ...»  3 «1  2 8 ¬ ¼ ª

 3 «1  ¬

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

ª º (5 x ) 4 2  ...» «1  (5x )  3 ¬ ¼

º (5 x )2 (5 x )4   ...»  1 ¼ 2 24

Calculus: Chapter 5

48

WORKED SOLUTIONS 2

y (0)  y c(0) 

For c = 0, y

y cc(0 ) y ccc(0 )   .... 2! 3!

Review Exercise 1

Therefore, y(0) = 0;

§ (2n )! x n 1

yc

y 2  x ; y c(0)

y cc

2 yy c  1; y cc(0 )

y ccc

2 yy cc  2( y c)2 ; y ccc(0 )

1

lim ¨ n of ¨

1 4

n of

y (5)

2 yy ( 4 )  8 y cy ccc  6( y cc)2 ; y ( 5 ) (0 ) 1 2

§ 2n (2n  1) · ¸x © n (n  1) ¹

14

1 1

if |4x|< 1, or x  º»  , ª« , hence R ¼ 4 4¬

66

4 3 14 4 66 5 x  x  x  ... 3! 4! 5!

2 a i

y(0.2) = 1.2264 (4d.p.) 3

3 x 2 y;

y cc

3x 2 y c  6 xy

y ccc

6 y  6 xy c  36 x 3 y  9 x 4 y c

For (1,1), y c

3; y cc

1  ( x  1)(3) 

Hence, f cc( x )  2 f c( x )  5 f ( x )

b

( x  1)2 ( x  1)3 (15)  (87 ) 2! 3!

x yc

x

¦ nc

n

x

n 1

f

¦ nc

n

3 a

x n 1

f

b

 1  x  ¦ c n x n . Moving the

summation index on the left by −1, i.e., replace n = 1 with n = 2 and n with (n − 1) we obtain

¦ (n  1) c n 1x n n 2

c0

1 u c1

c4

3 u c3

1! ; c 3

f

x e

x

2

b

f

b of

x 

Hence, y ( x )

1 x 

Hence, e x

2!;

1 x 

c

(n  1)!, n t 2 . f

2

e x sin x

¦ (n  1)!x . n

2

x  2

The radius of convergence is lim n of

n!

lim n of

2

º

b

1

1 . 2e

x2 x3   ... 2! 3!

1  x2 

1 n

Therefore the series converges only for x = 0. Hence, the differential equation does not have a convergent power series solution.

0.

d

x4 x6   .. 2! 3!

3 5 7 § · x4 x6 2   .. ¸ §¨ x  x  x  x  ... ·¸ ¨1  x  2! 3! 3! 5! 7! © ¹© ¹

n 2

(n  1)!

ª 1

lim «  e  x » b of ¬ 2 ¼

x3 x5 x7    ... and 3! 5! 7!

n 2

3!; ... ; c n

2

lim ³1 xe  x dx

dx

sin x ex

 1  x  c 0  c1 x  ¦ c n x n . 2 u c2

3 2 11 3 7 4 x  x  x  .... 2 6 24

Since the integral converges, the series converges.

n

n 1

f

0.

(0 )

2x 4  ... 3

1x

n 0

n 1

(n )

f is a continuous and positive decreasing function, hence we can use the integral test. 1

 1  x  ¦ c n x , so that

n 1

(0 )  5 f

x2 x3 x4    ... and 2 6 24

1  2x 2 

e x cos 2 x

³ f

1 x 

( n 1)

Multiplying these,

y  x  1,

f

e x

cos 2 x

¦ c n x n ; y c ¦ nc n x n 1.

Since x y c

(0 )  2 f

87

In this example we will purposely take a different approach to show you another way of solving this type of problem. 2

(n 2)

f

15; y ccc

0.

Differentiating n times and putting x = 0,

ii

1  3( x  1)  7.5( x  1)2  14.5( x  1)3

2

4e x sin 2 x  3e  x cos 2 x

 4e  x sin 2 x  5e  x cos 2 x .

Therefore, using the Taylor series expansion,

2

f cc( x )

4e  x sin 2 x  3e  x cos 2 x  2e  x cos 2 x

6 y  54 x 3 y  27 x 6 y

Let y

 e x cos 2 x  2e x sin 2 x

6 xy  9 x 4 y

3x 2 (3x 2 y )  6 xy

6 y  6 x (3x 2 y )  36 x 3 y  9 x 4 (3x 2 y )

4

f c( x )

1 . 4

f cc( x )  2 f c( x )  5( x )

yc

y

· ¸ ¸ ¹

4 x . The series is convergent

lim ¨

2 yy ccc  6 y cy cc; y ( 4 ) (0 )

1  x  x2 

n !(n  1)! (2n  2)! x n

u

n !(n  1)!

©

y (4)

Hence, y

Using the ratio test,

e x sin x  x x3

5 3 41 5 x  x  ... 6 120 5 41 2  x. 6 120 2

§ e x sin x  x · Hence, lim ¨¨ ¸¸ x o0 x3 © ¹

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5 . 6

Calculus: Chapter 5

49

WORKED SOLUTIONS an

4 a i

f

(0 ) or f n!

(n )

(0 )

7

n ! an.

When x = 0, (n  2)! an  2  n 2 u n ! an

a3

12 ; a5 2u3

0.

x

x2 x3 x4    ...  c 2 3 4

Since ln(1) = 0, c = 0. ln(1  x )

12 32 . u 2u3 4u5

hence

12 × 32 ... (n − 2)2 Hence, an = , n ∈ Odds, n ≥ 3. n! b

1  x  x 2  x 3  x 4  ....

1 x

ln(1  x )

n 2 an , n t 1.

Hence, (n  1)(n  2)an  2 ii

1

(n )

x 

x2 x3 x4    ..., 2 3 4

1 §1  x · ln ¨ ¸ 2 ©1 x ¹

Using the ratio test, for odd n,

· 1 §§ x2 x3 x4    ... ¸ ¨¨ x  ¨ 2 ©© 2 3 4 ¹

n2 x2 n of ( n  1)( n  2 )

 ¨ x 

lim

§ ©

x 2. Hence, the series is

convergent for |x|< 1, and R = 1. c

1  ln(1  x )  ln(1  x 2

6

| 1u

· 1§ 2x 3 2x 5   ... ¸ ¨ 2x  2© 3 5 ¹

=

3

S

§1· © ¹

arcsin ¨ ¸ 2

5

1 1 §1· 3 §1·  u¨ ¸  ¨ ¸. 2 6 ©2¹ 40 © 2 ¹

x 

| 3.139.

x3 x5   ... 3 5

1

5

lim n of

( x  3 ) n 1 2 n 1

2n n n n  1 ( x  3)

converge for

u

8

(1  x ) 3 Ÿ f c( x )

f (x )

| x  3| . The series will 2

| x  3|  1 Ÿ| x  3| 2 Ÿ 1  x  5. 2

Checking the endpoints, x = 1, ¦ n 1

( 2)

n

f

¦

2n n

n 1

( 1) n

n

.

(2)n

¦2 n 1

n

n

f

¦ n 1

1 n

f

i 0

, which diverges by the p-series test. 9 3D

S u3 180

(0 ) i x i!

6

Maclaurin series for e x

x

¦ n ! , Rn ( x ) n 0

d

M x n 1. (n  1)!

Ÿ sin x

0 .2

e Rn (0.2) d 0.2n 1  0.0005. (n  1)!

f

e  3 Ÿ e0.2  e0.5  2

Then, Rn (0.2) d

2 0.2n 1  0.0005 Ÿ n (n  1)!

Hence, e0.2 | 1  0.2 

0.22 0.23  2! 3!

1

8

(1  x ) 3 . 1 1 5 x  x 2  x3; 3 9 81

S . 60

f (0)  xf c(0) 

f (x ) n

27

(i )

Hence, the interval of convergence is [1, 5] f

10

9

Hence, f ( x ) | ¦

2

f (0.2) | 1.06272.

This converges by the alternating series test, since the terms decrease to zero. For x = 5, f

3



(1  x ) 3 ;

 (1  x ) 3 ; f ccc( x ) 3

f

1

5

2

f cc( x )

·· x2 x3 x4    ... ¸ ¸¸ 2 3 4 ¹¹

( n 1)

x3

0x

3!

x2 2!

f cc(0)  ...  Rn

 ... 

x n 1 f (n  1)!

r sin x or r cos x , then f

(x )

( n 1)

( n 1)

(c )

(c )

1.

n 1

3

S 60 Rn d d 0.000005 Ÿ n (n  1)!

3.

1.221 3

§S · ¨ ¸ S 60 sin 3D |  © ¹ | 0.05234. 60 3!

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 5

50

WORKED SOLUTIONS 10 a

e x  (  1)n e  x 2

f (n ) ( x )

b Coefficient of x

f

n

12 y(0)

(n )

1  ( 1) 2n !

(0 ) n!

n

x2 x4 Ÿ f (x ) 1    .... 2 24 c

1 2

1

1 1  8 16 u 24

error term

f ( n 1) (c ) n 1 x (n  1)!

f( )

433 . The Lagrange 384

11 a

f (x )

y ccc

2 ª¬( y c)2  yy ccº¼ ; y ccc(0)

 sin x  1 (1  sin x )2

(4)

cos x ; 1  sin x

f ccc(0)

(4)

1; f

c

ln(1  sin x )

d

y tan x  cos x , f c(0 )

y cc

y sec 2 x  y c tan x  sin x



Ÿy

 1; b

 2.



 x2 

e³

 tan xdx

Ÿ y cos x

Ÿ y cos x x4  ... 6

1

S 2

integrating factor

x2 x3 x4    .... 2 6 12

x2 x3 x4    ... 2 6 12

ln cos2 x

x 3 2x 5   ..., 3 15

S S x2 x  2 4

Adding, ln(1  sin 2 x )

e

yc

2

x 

x 

which is the Maclaurin expansion of

13 a

ln(1  sin( x ))

x 

2 x 3 16 x 5   ... 3! 5!

x 

y

Ÿ f cc(0)

Therefore,ln(1  sin x )



2

(x )

1; f cc(0)

(0)



0

2 ª¬3  y cc  y cy ccc  y cy ccc  yy ( 4 ) º¼

y (5)

 sin x (1  sin x )  2(1  sin x ) cos x . (1  sin x )4

0; f c(0)

cos x

³ cos

2

1 (1  cos 2 x )dx 2³

xdx

x sin 2 x  c 2 4

when

S, y

0Ÿ c



S 2

x

ln cos x

x2 x4    ... 2 12

ln sec x

x2 x4   ... 2 12

Ÿ y cos x

x sin 2 x S   2 4 2

Ÿy

sin 2 x S · §x   ¸ 4 2¹ ©2

ln sec x x x

x x2 x   ... 2 12

Hence, lim x o0

ln sec x x x

16

y = tan(x).

2

f (0)

2; y (5) (0)

2 ª¬3( y cc)2  4 y cy cc  yy ( 4 ) º¼

1 . 1  sin x

cos x ;f (1  sin x )2

0

2 > 2 y cy cc  y cy cc  yy ccc@

y (4)

0.000136 .

 sin x (1  sin x )  cos x (1  sin x )2

f ccc( x )

2 yy c; y cc(0)

1

2 >3 y cy cc  yy ccc@ ; y ( 4 ) (0)

2

b

y cc

5

ln(1  sin x ). Hence, f c( x )

f cc( x )

y 2  1; y c(0)

(c ) § 1 · u¨ ¸. 120 ©2¹

From the GDC, f (5) (c ) is an increasing function, hence the expression is maximized when x = 0.5. Therefore the upper bound (e 0.5  e 0.5 ) § 1 · u¨ ¸ 2 u 120 ©2¹

yc

5

(5)

f

=0

sec x ¨

0.

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 5

51

WORKED SOLUTIONS 1

1 x 

14

16

(1  x ) 2 2 §1· § 1·x

§1·

1

S

1

4

1

1  2 x

1  2x

17 e x

3 § 1 · § 3 · § 5 · ( 2 x )  ... 3!

 ¨  ¸˜¨  ¸¨  ¸ © 2 ¹ © 2 ¹© 2 ¹

(1  x ) (1  2 x )

2

e

2

f

¦ (1)

e

2

3

§ x2 x3 ·  ¨x  ¸ 2 6 ¹  ... © 6 x2 x3 x2 x3     ... 2 6 2 6 5 6

1  x  x 2  x 3  ...

lim x o0

e

=lim x o0

x

 1)

(e 1)

lim x o0

 11 e(e

e

1

x

 x 1)

(e x  x 1)

1 x

u (e  1)

e(e e

x

 1)

x

(e  x 1)

1

1

√2•π

•

∫

0.341345

1

0

–x2 e 2 dx

1 1 |n = {0,1,2,3,4,5} • √2 • π (2 • n+3) • 2n+1) • (n+1)! {0.06649,0.009974,0.001187,0.000115,0.000009,6.65969E–7}

4

(

1

√2 • π

• (–1)π •

1

(2n+1) • 2n–n!

)

n=0

When n = 4, we have the appropriate approximation, i.e., 0.341354.

Using l’hopital’s rule, x

ª ( 1)n x 2 n 1 º ¦ « ». n n 0 ¬ ( 2n  1)2 n ! ¼ 0 f

Exercise 5E

1.1 3

§ x x · x   ¸ § x2 x 3 · ¨© 2 6 ¹ 1  ¨x   ¸ 2 6 ¹ © 2

e(e

1 2S

n =0

1

From the GDC:

x2 x3   ... 2 6

1 x 

( −1)n x 2 n dx 0 2n n !

¦³

1  0.0001. (2n  3)2n 1 (n  1)!

un 1

2

e x 1

∞

1 2π

Since this is an alternating series,

º 1 1ª » 2 , 2 « ¼ ¬

x 

x 2n 2n n !

3 2 5 3 § · ¨ 1  x  x  x  ... ¸ 2 2 © ¹

3 15 51 x  x 2  x 3  ... 2 8 16

ex  1

n

Hence, P(0 < x < 1) =

1  2

n

§ x2 · § x2 ·  ¸ ¨ ¨ ¸ § x · 1  ¨  ¸  © 2 ¹  ...  © 2 ¹  ... © 2 ¹ 2! n!

x2  2

3 2 5 3 x  x  ... 2 2

1 1 2 1 3 § · x  ... ¸ ¨1  x  x  2 8 16 © ¹

I

x2 xn  ...   ... n! 2!

0

1 2

1

4 . 2n  1

n 1

1 x 

º 1 1ª » 2 , 2 « ¼ ¬

I

15

¦ (1)

4  5 u 10 5 Ÿ n t 4, 001 2n  1

1  2

1 3 ( 2 x )2 § 1· 1  ¨  ¸ ( 2 x )  §¨  ·¸ ˜ §¨  ·¸ © 2¹ © 2 ¹ © 2 ¹ 2!

1 x 

f

n 1

1 1 1 1  x  x 2  x 3  ... 2 8 16

@1,1>

1 1 1 § · 4 ˜ ¨ 1     ... ¸ 3 5 7 © ¹

3 x3

1

§ · § · § · 1  ¨ ¸ x  ¨ ¸˜¨  ¸  ¨ ¸ ˜ ¨  ¸ ˜ ¨  ¸  ... ©2¹ © 2 ¹ © 2 ¹ 2! © 2 ¹ © 2 ¹ © 2 ¹ 3 !

I

1 1 1    ... Ÿ S 3 5 7

1 1

1 2

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Calculus: Chapter 5

52