Since the length of the graph is greater than ∑ a n , the length of the graph is infinite.

Since the length of the graph is greater than ∑ a n , the length of the graph is infinite.

This is a Riemann sum for the function fx () = from x = 1 to 2 where Δ=. x

Instructor’s Resource Manual

Section 9.5 547

9.6 Concepts Review

1. power series

2. where it converges

3. interval; endpoints

Problem Set 9.6

n 1 − 1)! x n +

; lim ρ =

= lim

= 0 . Series converges for all x .

2. ∑ n ; lim ρ = n + 1 n = lim = ; convergence on ( 3,3) − . n = 1 3 n →∞ 3 x

n →∞ 3 3

For x = 3, a n = 1 and the series diverges. For x =− 3, a =− ( 1) n n and the series diverges.

Series converges on ( 3,3) −

∞ n 2 x 1 nx n +

3. ∑ 2 ; lim ρ =

2 n = lim

= x ; convergence on ( 1,1) − .

n →∞ ( n + 1) x

n →∞ (1 + 2 + 1

For x = 1, a n = 2 (p-series, p=2) and the series converges.

n ( 1) − n

For x =− 1, a n = 2 ( alternating p-series, p=2) and the series converges. by the Absolute Convergence Test.

Series converges on [ 1,1] −

( n + 1) x n + n 1

4. ∑ nx ; lim ρ =

= lim ⎜ 1 + ⎟ x = x ; convergence on ( 1,1) − .

For x = 1, a n = n and the series diverges. For x =− 1, a =− ( 1) n n n and the series diverges ( lim ( 1) − n n ≠ 0 )

n →∞

Series converges on ( 1,1) −

5. This is the alternating series for problem 3; thus it converges on [ 1,1] − by the Absolute Convergence Test.

= lim ⎛ n ∑ ⎞ ρ

⎟ = x ; convergence on ( 1,1) − .

(Alternating Harmonic Series) and the series converges.

1 For x =− 1, a n = (Harmonic Series) and the series diverges.

Series converges on ( 1,1] −

7. Let u =− x 2 ; then, from problem 6, the series converges when u ∈− ( 1,1] ; that is when x ∈ (1,3] .

Section 9.6 Instructor’s Resource Manual

∞ ( x + 1) n

nx !( + 1) n + 1 x + 1

; lim ρ =

= lim

= 0 . Series converges for all x.

n →∞ ( n + 1)!( x + 1) n

n →∞ n + 1

∞ ( 1) − n + 1 x n

When x = 1, the series is

which converges absolutely by comparison with the series

n 1 nn ( 1)

(–1) 2 n + 1 ∑ 1

(–1) n + 1 (–1)

When x = –1, the series is

which converges since

n = 1 nn ( + 1)

converges.

The series converges on –1 ≤ x ≤ 1.

10. ∑ ; lim ρ =

The series converges for all x. ∞ ( 1) − n + 12 x n − 1 x 2 n + 1 x 2 n − 1 1

The series converges for all x.

∞ ( 1) − n x 2 n

When x = 1, the series is

12. 2 ∑ ; ρ = lim ÷ n which clearly

n = 0 (2 )! n

n →∞ (2 n + 2)! (2 )!

2 1 = diverges. lim

n →∞

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

n ∑ 2 (–1) ; n n = 1

When x = –1, the series is

The series converges for all x.

a n = n 2 ; lim a n ≠ 0, thus the series diverges.

n →∞

∑ The series converges on –1 < x < 1.

When x = 1, the series is ∑ n which clearly

When x = 1, the series is 1 +

(–1) n , which is

When x =− 1 , the series is ∑ n (–1) ; ; a n = n 1 added to the alternating harmonic series n = 1 multiplied by –1, which converges.

lim a n ≠ 0, thus the series diverges.

n →∞

When x = –1, the series is

The series converges on –1 < x < 1.

(–1) ∑ n

=+ 1 , which diverges.

( n + 1) ∑ x

14. nx 2 n

; ρ = lim

The series converges on –1 < x ≤ 1.

Instructor’s Resource Manual

Section 9.6 549

16. 1 + ∑ ; ρ = lim

converges by comparison with ∑ 2 .

n →∞ n + 1 n

(–1) n n

= lim x

= x When x = –1, the series is ∑

∞ 1 converges absolutely by comparison with

When x = 1, the series is 1 + ∑ which

1 n = 1 n diverges since < 1. The series converges on –1 ≤ x ≤ 1.

n 1 When x = –1, the series is n 1 + n ∞ + ∑

1 nn

19. ∑ n ; lim ρ =

;1 < when –2 < x < 2.

1 lim

= 0, so the series converges.

(–1) n n ∑ n ∑ The series converges on –1 ≤ x < 1.

n →∞

When x = 2, the series is

(–1) n

which diverges.

∞ ( 1) − nn x

When x = –2, the series is

n = 1 nn ( + 2)

(–1) n (–2) =

(–1) (–1) n ∑ n n ∑ = ∑ 1 which

ρ = lim

diverges. The series converges on –2 < x < 2. n →∞ ( n + 1)( n + 3) nn ( + 2)

20. ∑ 2 nn x ; ρ = lim nn = lim 2 x = 2; x

When x = 1 the series is 1 + (–1) n

2 x < 1 when – << x . ∑

n = 1 nn ( + 2)

which converges absolutely by comparison with

When x = , the series is

n = 0 ⎝⎠ 2

the series ∑ 2 . which diverges.

When x = –1, the series is

When x = –, the series is

∑ n (–1)

(–1) n

∑ n which ∞

2 n ⎛ ∑ 1 ⎜ – ⎟ ⎞= ∑ (–1) n which diverges.

converges by comparison with ∑ 2 . 1 1

The series converges on – << x .

The series converges on –1 ≤ x ≤ 1.

n →∞ ( n + 2) 2 − 1 ( n + 1) 2 − 1 The series converges for all x.

= lim x

n →∞

When x = 1, the series is

Section 9.6 Instructor’s Resource Manual

22. ∑ ; lim ρ =

n + 1 n →∞ n + 2 n + 1 ∑ n 2 ; lim ρ =

n = 1 n →∞ ( n + 1) 2 n 2

= lim x

2 = x = lim x − 2 2 =− x 2 ; x −< 2 1 n when →∞ n + 2 n n →∞ ( n + 1)

When x = 1, the series is ∑

n = 1 n + 1 When x = 1, the series is ∑ 2 which

diverges since lim

n →∞ n + 1

converges absolutely since ∑ 2 converges.

n (–1)

When x = –1, the series is

which

When x = 3, the series is ∑ 2 which

diverges since lim

=≠ 1 0. n = 1 n

n →∞ n + 1 converges. The series converges on 1 ≤ x ≤ 3.

The series converges on –1 < x < 1.

; lim ρ =

n →∞ ( n + 1)( n + 2) nn ( + ∑ 1)

lim + 5 =+ 5 ; +< 5 1 when

= lim x − 1 =− x 1 ; x –1 < 1 when

When x = –4, the series is ∑

nn ( + ∑ 1) which n = 1

∞ (–1) n

which

When x = 0, the series is

converges by comparison with ∑ 2 .

converges.

When x = 2, the series is ∑ which diverges.

∞ (–1) n

When x = –6, the series is ∑

which

The series converges on 0 ≤ x < 2.

n = 1 nn ( + 1)

n = 1 nn ( + ∑ 1)

( x + 2)

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

converges absolutely since

1 The series converges on –6 ≤ x ≤ –4.

The series converges for all x.

28. ∑ ( 1) − nx ( + 3) ; lim ρ =

; lim ρ =

x + 1) n

= lim x + 3 =+ x 3 ; x +< 3 1 when

n →∞ 2 2 2 When x = –2, the series is

(–1) n + ∑ 1 n which

–3 < x < 1.

∞ (–2) n

When x = –3, the series is ∑ n = ∑ (–1) n →∞

diverges since lim n ≠ 0.

n = 0 2 n = 0 When x = –4, the series is

which diverges.

+ ∑ 1 (–1) n n (–1) n = ∑ – n , which diverges.

When x = 1, the series is ∑ n = ∑ 1 which

n = 0 2 n = 0 The series converges on –4 < x < –2.

diverges. The series converges on –3 < x < 1.

29. If for some x , lim 0 ≠

0, then

∑ could n !

n →∞ n !

not converge.

Instructor’s Resource Manual

Section 9.6 551

30. For any number k, since

32. Using the Absolute Ratio Test,

k – n < k – n + 1 < … < k – 2 < k – 1 < k,

( pn + p )! n + 1 ( pn )!

( – 1)( – 2) k k … (–) k n < k n , thus

lim

n →∞ (( n + 1)!) p

x n . Since –1 < x < 1, lim x n = 0,

and by Problem 21, lim

= hence 0,

n →∞ n !

The radius of convergence is p − p .

33. This is a geometric series, so it converges for x − < , 2 < x < 4. For these values of x, the 3 1

31. The Absolute Ratio Test gives

( 1)! 2 n + 3 ! 2 n + 1 series converges to

2 2 34. ∑ a n n ( – 3) x converges on an interval of the

lim

< when 1 n = 0

n →∞

2 n + 1 2 2 form (3 – a, 3 + a), where a ≥ 0. If the series converges at x = –1, then 3 – a ≤ –1, or a ≥ 4,

x < 2. since x = –1 could be an endpoint where the The radius of convergence is 2 .

series converges. If a ≥ 4, then 3 + a ≥ 7 so the series will converge at x = 6. The series may not converge at x = 7, since x = 7 may be an endpoint of the convergence intervals, where the series might or might not converge.

(3 x + 1) n + 1 (3 x + 1) n

35. a. ρ = lim

= lim 3 x + 1 = 3 x + 1; 3 x + < when 1 1 –1 << x .

When x = –1, the series is ∑ n = ∑ (–1) n , which converges.

When x = , the series is ∑ n = ∑ , which diverges. The series converges on –1 ≤< x .

(–1) n + 1 (2 – 3) x n + 1 (–1) (2 – 3) n x n

1 1 7 2–3 x < when 1 – << x .

When x = –, the series is ∑ (–1) n n =

(–4) n

which diverges since < 1.

When x = , the series is ∑ (–1) n n = ∑ (–1) n ; a n = ;, > so a n > a n + 1 ;

lim

= so 0, ∑ (–1) n converges. The series converges on – <≤ x .

n →∞ n

Section 9.6 Instructor’s Resource Manual

36. From Problem 52 of Section 9.1,

ρ = lim

n →∞ 2 n + 1 ⎢ (

x ( 1 + 5 )( –1–5

n + 1 n + 1 1– 5 n

= lim

n →∞ 2 )( n )

n →∞ 2 n

x ;1 x < when –

<< x

Note that lim ⎜ ⎜

37. If a n + 3 = a n , then a 0 = a 3 = a 6 = a 3 n , a 1 = a 4 = a 7 = a 3 n + 1 , and a 2 = a 5 = a 8 = a 3 n + 2 . Thus,

ax ∑ n n = a 0 + ax 1 + ax 2 + 3 4 5 2 3 2 6 ax 0 + ax 1 + ax 2 + = ( a 0 + ax 1 + ax 2 )(1 + x + x + )

0 + ax

1 + ax 2 ) ∑ x = ( a 0 + ax 1 + ax 2 ) ∑ ( x ) .

a 0 + ax 1 + ax 2 2 is a polynomial, which will converge for all x.

x 3 is a geometric series which, converges for x ∑ 3 ( ) n < 1 , or, equivalently, x < 1 .

( x 3 1 ) a n = for x < 1 , Sx () = 0 + ax 1 + ax ∑ 2 3 3 for x < 1 .

Since

38. If a n = a np + , then a 0 = a p = a 2 p = a np ,, a 1 = a p + 1 = a 2 p + 1 = a np + 1 etc. Thus,

∑ n ax

1 ++ a p 1 p np − 1 x ) ∑ x

a + ax ++ a x p − 0 1 1 p − 1 is a polynomial, which will converge for all x.

x ∑ np = ∑ ( x pn ) is a geometric series which converges for x p < 1 , or, equivalently, x < 1 .

Since ∑ ( x ) = p for x < 1 , Sx () = ( a 0 + ax ++ a x p − 1 1 p − 1 ) ⎜ p ⎟ for x < 1 .

Instructor’s Resource Manual

Section 9.6 553

9.7 Concepts Review

5. From the geometric series for

1x −

with x

1. integrated; interior

replaced by x , we get 2 x 2 x 3 x 4 x 5

radius of convergence .

2 3 Problem Set 9.7 4 1 ⎛ 1 ⎞ 1 2 x 4 x 8 x 16 x

1. From the geometric series for

1x −

with x

radius of convergence .

replaced by –x, we get

1 =−+ 1 2 − 3 + 4 − x 5 x x x x +

7. From the geometric series for

1x −

with x

radius of convergence 1.

replaced by 4 x , we get

radius of convergence 1.

1 1 9. From the geometric series for ln(1+ x) with x so

of the second derivative of

replaced by t, we get

(1 − x ) 3 2

∫ 0 ln(1 + t dt ) = – + – +… . Thus, ; 2 6 12 20

1x −

radius of convergence 1.

3 =+ 13 x + 6 x 3 + 10 x + (1 ;

10. ∫ tan t dt = – + − +… ;

radius of convergence 1.

4. Using the result of Problem 2, x

=− x 2 x 2 + 3 x 3 − 4 x 4 + 5 x 5

(1 x )

radius of convergence 1.

Section 9.7 Instructor’s Resource Manual

11. ln(1 + x ) = x – +

+… ,–1 <≤ x 1

ln(1 – ) x = –– x

+… ,–1 ≤< x 1

= ln(1 + x ) – ln(1 – ) x

1– x 2 x 3 2 x 5

+ ⋅⋅⋅ ; radius of convergence 1.

12. If M =

, then M – Mx = 1 + x;

1– x

M – 1 = (M + 1)x; x =

M –1

M –1 <

1 is equivalent to –M – 1 < M – 1 < M + 1 or 0 < 2M < 2M + 2 which is true for M > 0. Thus, the natural

7 logarithm of any positive number can be found by using the series from Problem 11. For M = 8, x = , so 9

13. Substitute –x for x in the series for x e to get:

15. Add the result of Problem 13 to the series for e x to get:

1 ln(1 + x ) 1 +− x x + x

Instructor’s Resource Manual

Section 9.7 555

2 4 2 x 3 13 x 5 29 x 21. 7 (tan x )(1 + x + x ) = ⎜ x −

23. The series representation of

is 1 +

+ , so

2 3 8 30 ∫ dt =+ x x – x + x – … . . 0 1 + t 6 12 40

tan − 1 x

x 2 x 4 x 6 x tan − 1 t

24. The series representation of

c. − ln(1 − x ) =+ x

+ , so − ln(1 2 ) − x = 2 x +

26. a. Since

=++ 1 2 3 4 2 4 6 x 8 x + x + x + , =+ 1 x + x + x + x + .

b. Again using

=++ 1 2 3 4 2 x 3 x + x + x + , −= 1 cos x + cos x + cos x + .

− 1 cos x

c. ln(1 − x ) =−− x −

− , so ln(1 − 2 ) =− x 2 x −

− , and

− ln(1 − 2 x ) = ln

27. Differentiating the series for

multiplying this series by x gives

(1 − x )

=+ x + 3

3 x + 4 x 4 + , hence

nx ∑ n =

28. Differentiating the series for

twice yields

=+ 26 2 x 3 + 12 x + 20 x + ⋅⋅⋅ . Multiplying this series by x

3 2 x +⋅ 32 x +⋅ 43 x +⋅ 54 x + ⋅⋅⋅ hence , ∑ nn ( + 1) x (1 – ) = x

Section 9.7 Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this

+ =− y , so y ′′ += y 0 . It is clear that y(0) = 0 and y′ (0) = 1 . Both the sine and cosine

functions satisfy y ′′ + = y 0 , however, only the sine function satisfies the given initial conditions. Thus,

n –1 f n –2 ) x =++ 0 x ∑ ( f n + 2 – f n + 1 – f n ) x

Since

n + 2 = f n + 1 + f n ,– f n + 2 f n + 1 – f n = 0 . Thus Fx ()– xF x ()– xFx () = x .

(Recall that 0! = 1.)

yx ′′ ()– yx ′ ()–() yx = 2 3 ⎜ 4 + x + x 2 +

= ∑ ( f n + 2 – f n + 1 – f n ) x n = 0 since f n + 2 = f n + 1 + f n for all n ≥ 0.

Instructor’s Resource Manual

Section 9.7 557

36. For any positive integer k ≤ n, both

np !

and

are positive integers. Thus, since q < n, ne ! =

is a positive

integer and M = ne ! – !– !– n n – – –

is also an integer. M is positive since

contradicts that M is a positive integer since for n ≥ 1, ≤ 1 and there are no positive integers

less than 1.

9.8 Concepts Review

Problem Set 9.8

sinh x

2. tanh x =

cosh x 1 + x + x +…

5. cos ln(1 x + x ) = ⎜ 1–

Section 9.7 Instructor’s Resource Manual

x 2 7 x 3 x 4 19 x 5

6. (sin ) 1 x += x

x sin x =+++ x ⎜ 1 x

– ln(1 + x ) – ln(1 + x ) ⎛

( 1 − x )( =− 1 x ) ∑ x n =−+ 1 x x − 4 x + , x < 1

1 ++ x x

1 ++ x x 1 − x 1 − x 3 n = 0

12. =+ 1 sin x + (sin ) 2 x + (sin ) 3 x +…

14. (sin 2 + sin 3 ) =

15. sec( 2 x

2 + sin x =

cos( x )

16. = (cos )(1 x + x )

Instructor’s Resource Manual

Section 9.8 559

⎜⎟ = 1; 2; 4; 16 f ′ ⎜⎟ = f ′′ ⎜⎟ = f ′′′ ⎜⎟ = ∑ n

be an odd function

⎝⎠ 4 ⎝⎠ 4 ⎝⎠ 4 ⎝⎠ 4 (f(–x) = –f(x)) for x in (–R, R). Then a n = 0 if n

tan x ≈+ 12 ⎜ x − ⎟ + 2 ⎜ x − ⎟ + ⎜ x − ⎟

is even.

The derivative of an even function is an odd function and the derivative of an odd function is

23. f(1) = 3; f′ (1) =+= 23 5; an even function (Problem 50 of Section 3.2).

f ′′ (1) =+= 26 8; (1) f ′′′ = 6 Since fx () = ∑ ax n is an odd function, fx ′ ()

1 + x 2 + x 3 =+ 3 5( – 1) 4( – 1) x + x 2 + ( – 1) x 3 is an even function, so f ′′ () x is an odd function,

hence f ′′′ () x is an even function, etc. Thus,

() n

This is exact since f () x = 0 for n ≥ 4.

f () n () x is an even function when n is odd and an

24. f ( 1) −=+++= 2131 7; odd function when n is even.

f′ n ( 1) −= –1 6 – 3 − =− 10; By the Uniqueness Theorem, if fx () = ∑

ax n ,

f ′′ ( 1) −=+= 6 6 12; (1) f ′′′ = –6 f () n (0)

then a n =

. If g(x) is an odd function,

2– x + 3 x – x =− 7 10( x ++ 1) 6( x + 1) 2 −+ ( x 1) 3 n ! (0) = 0, hence

This is exact since () () x = 0 for n ≥ 4. g a n () f n () x is an odd function for all even n.

for all even n since

25. The derivative of an even function is an odd function and the derivative of an odd function is

27. =+ [1 (– )] 2 an even function. (Problem 50 of Section 3.2). –1/ 2 t

1– t 2

Since ( ) fx =

∑ n ax

is an even function, fx ′ ()

is an odd function, so f ′′ () x is an even function,

= 1– (– ) 2 + 22 t 23 (– ) – t (– ) t +

hence f ′′′ () x is an odd function, etc.

Thus () n () is an even function when n is even

and an odd function when n is odd.

Thus, sin –1

By the Uniqueness Theorem, if fx () = ∑ ax n n , ∫ 0 1– t 2

() f n (0)

t then 2 3 a 4 n = . If g(x) is an odd function, t 5 t 6 ⎞

⎟ dt

g (0) = 0, thence a n = 0 for all odd n since

() n

f () x is an odd function for odd n.

Section 9.8 Instructor’s Resource Manual

x 8 x t 12 x 16

29. cos( x ) = 1–

2 8 16 ∫ 0 x dx ∫ 0 ⎜ ⎜

Thus, sinh () x = ∫

= ∫ ⎜ 1– + – + ⎟ dt

∫ 0 sin xdx = ∫ 0 ⎜ x –

=−−+− 1( x 1) ( x 1) 2 −− ( x 1) 3 +

4 64 n = 0 n ! Note that () n f (0) = 0 when n is odd.

and (51) f (0) = 0.

Instructor’s Resource Manual

Section 9.8 561

2 6 24 n ∑ = 0 n !

fx () = ∫ ⎜ 1 + + + +… ⎟ dt =+ ⎢ t + +… ⎥ =+ x + +… =

4! (cos – 1) x 2 (cos x − 1) 3

d. e cos –1 x =+ 1 (cos – 1) x +

Hence fx () = e – x + x – …= ∑

Section 9.8 Instructor's Resource Manual

2 sin 4 x sin 2 6 ln(1 sin x −

2 Hence 1 4 2 6 ∞ f (0)

Thus a 0 = a 1 1, 0, – = a 2 0 = a 3 1 0, – = a 0, – 2 + 4 0 = 0, so

and therefore sec x =+ 1 2 +

35. tanh =

sinh x

0 ax 1 + ax 2 +…

cosh x

so sinh x = cosh ( xa +

0 ax 1 + ax 2 +… )

Thus a 0 = a 1 0, 1, = a 2 + 0 = 0, a 3 1 + = ,

and therefore

tanh x = – 3 x x +

Instructor’s Resource Manual

Section 9.8 563

36. sech x =

= a ax +

0 + 1 ax 2 +…

cosh x so 1 = cosh (

xa 2

0 + ax 1 + ax 2 +… )

( a 0 ax 1 ax 2 +… ) ⎟ ⎟

Thus, a 0 = a 1 1, 0, = ⎜ a 2 + ⎟ = 0, ⎛ + 1 3 ⎞ = 0, ⎛ 4 + ⎜ 2 a ⎟ ⎜ a + 0 ⎞ ⎟ = 0, ⎛ +

, 0 a 5 = and therefore

37. a. First define Rx 3 () by

Rx 3 () = fx () − fa () − fax '( )( −−

f ''( ) a f '''( ) a

For any t in the interval [,] ax we define

f ''( ) t

f '''( ) t

( xt − ) 4

gt () = fx () − ft () − ftxt '( )( −− )

Next we differentiate with respect to t using the Product and Power Rules:

gt '( ) =− 0 ft '( ) −− ft '( ) f txt )

[ 2 + ''( )( − ] − ⎡ − f 2 ''( )( txt −+ ) f '''( )( txt − ) ⎤

− f 3 '''( )( txt − ) 2 + f (4) ( )( txt − ) 3 ⎤ + Rx 3 ()

Since gx () = 0 , ga () = Rx 3 () − Rx 3 () = 0 , and gt () is continuous on [,] ax , we can apply the Mean Value Theorem for Derivatives. There exists, therefore, a number c between a and x such that gc '( ) = 0 . Thus,

which leads to:

b. Like the previous part, first define R n () x by

For any t in the interval [,] ax we define

f () 2 n () t

( xt − ) n + 1

gt () = fx () − ft () − ftxt '( )( −− )

Next we differentiate with respect to t using the Product and Power Rules:

Section 9.8 Instructor's Resource Manual

Since gx () = 0 , ga () = R n () x − R n () x = 0 , and gt () is continuous on [,] ax , we can apply the Mean Value Theorem for Derivatives. There exists, therefore, a number c between a and x such that gc '( ) = 0 . Thus,

which leads to:

38. a. For ∑ ⎜⎟ x , lim ρ = ⎜ ⎟ x ÷ ⎜⎟ x = lim x

Thus fx () =+ 1 x ∑ n ⎜⎟ converges for x < 1.

n = 1 ⎝⎠ n

b. It is clear that f(0) = 1.

∞ ⎛⎞ p

n fx –1 ∑ ⎜⎟

Since () =+ 1 n

∑ n ⎜⎟ x and

( x + 1) fx ′ () = ∑ nx ( + 1) ⎜⎟ x =

n –1

∑ ⎢ nx ⎜⎟ n ⎜⎟ x ⎥ =⋅ 1 ⎜⎟ x + ∑ ⎢ n ⎜⎟ ++ ( n 1) ⎜ ⎟ ⎥ x

and since ⎜⎟ = p , (1 + ) ′ ()

xfx n =+ p ∑

p ⎜⎟ x = pf x () .

⎝⎠ 1 n = 1 ⎝⎠ n

c. Let y = f(x), then the differential equation is (1 + xy ) ′ = py or

Since f (0) = C (1) p = C and f(0) = 1, C = 1 and fx () =+ (1 p x ).

Instructor’s Resource Manual

Section 9.8 565

2 5 x 43. 3 3sin – 2 exp

3sin x = 3– x

lim + f () t = 24 while lim f () t = 0, thus

f (4) (0) does not exist, and f(t) cannot be

represented by a Maclaurin series.

Thus, 3sin – 2 exp x x = –2 + – 2 x x –

Suppose that g(t) as described in the text is

represented by a Maclaurin series, so

gt () = a 0 + at + at 2 2 +… = ∑

g () n (0)

t 1 n for all

44. exp( x 2 ) =+ 1 x 2 +

t in (–R, R) for some R > 0. It is clear that, for

2 2 ( x 22 ) ( x 23 )

t ≤ 0 , g(t) is represented by

exp( x ) =+ 1 x +

gt () =++ 00 t 0 t 2 +… . However, this will not

=+ 1 2 x

represent g(t) for any t > 0 since the car is

moving for t > 0. Similarly, any series that represents g(t) for t > 0 cannot be 0 everywhere,

so it will not represent g(t) for t < 0. Thus, g(t) cannot be represented by a Maclaurin series.

= 0 (by l’Hôpital’s Rule)

= (by using l’Hôpital’s Rule 0

h → 0 1/ h 2

e twice)

c. If f () n (0) = 0 for all n, then the Maclaurin

series for f(x) is 0.

d. No, f (x) ≠ 0 for x ≠ 0. It only represents

f (x) at x = 0.

e. Note that for any n and x ≠ 0,

1 Rx 2 () = e − x .

Section 9.8 Instructor's Resource Manual

45. sin(exp – 1) x =+ x

sin(exp – 1) x = ⎜ x +

46. exp(sin ) x =++ 1 x

sin x = x –

exp(sin ) x =+ 1 ⎜ x –

47. (sin )(exp ) x x =+ 2 x x +

x 3 x 5 ⎞⎛

(sin )(exp ) x x = ⎜ x –

Instructor’s Resource Manual

Section 9.8 567

9.9 Concepts Review

1. f(1); f ′ (1); (1) f ′′

f (6) (0)

3. error of the method; error of calculation

4. increase; decrease

Problem Set 9.9

f (3) () x = 2sec 4 x + 4 sec 2 x tan 2 x

fx () ≈ 2– x x 3 = 2– x x 3 f (4) () x = 16 sec 4 x tan x + 8sec 2 x tan 3 x

3 f (0) = 0 f (0.12) ≈ 2(0.12) − (0.12) ≈ 0.2377

Section 9.9 Instructor Solutions Manual

5. f(x) = ln(1 + x)

f (0) = 0

8. fx () = sinh x f (0) = 0

fx ′ () =

f′ (0) = 1 fx ′ () = cosh x f′ (0) = 1

f ′′ () x = sinh x (0) f ′′ = 0

f ′′ () x = –

f ′′ (0) =

f ′′′ () = 2 cosh –1 x x f ′′′ (0) = 1

(1 ) 4 f (0.12) ≈ 0.12 + (0.12) 3 ≈ 0.1203 + x 6

6. fx () = 1 + x f(0) = 1

fx ′ () = (1 + x ) –1/ 2

2 2 10. f(x) = sin x f ⎜⎟ ⎛⎞=

11. f(x) = tan x;

7. fx () = tan –1 x f(0) = 0

Instructor's Resource Manual

Section 9.9 569

12. f(x) = sec x ;

3 x 3 = 5sec x tan x + sec tan x x ;

2 Using n = 4, () ≈++ 1 2 3 + fx 4 x x + x x

f′ (2) =

1 2 a. f(0.1) ≈ 1.1111

4 16 b. f(0.5) ≈ 1.9375

8 64 c. f(0.9) ≈ 4.0951

d. f(2) ≈ 31

2 18. f(x) = sin x ; f(0) = 0 3

–2 fx 2 = 3 x x + 3 x + 5 ; f(1) = 7

When n is odd,

Using n = 5, sin x ≈− x

a. sin(0.1) ≈ 0.0998

Section 9.9 Instructor's Resource Manual

c. sin(1) ≈ 0.8417

d. sin(10) ≈ 676.67

Instructor's Resource Manual

Section 9.9 571

29. 2 c –2 c 2 1

≤ 6 e + 1 37. fx () = ln(2 + x ); ( ) fx ′ =

e 4 R (0.5) ≤

if is even n

5040 e c

( 0.5) − R 7 6 (0.5) ≤

0.5 ≈ 9.402 10 × 10 ln 2 7 5040 e

2 2 1 Rx 6 () =

= (Note that x − x is maximum at

⎦ .) cos 0.5 0.5 ( − )

Section 9.9 Instructor's Resource Manual

41. If fx () = , it is easily verified that

45. This is a Binomial Series ( p = ), so the third-order 1 2

Maclaurin polynomial is (see section 9.8, Thm D

( 1) − n n ! 2 f 3 n () x = ( n 1) . Thus for a = 1 x x

and example 6) Px 3 () =+− 1 + ; further,

, where c is between x and 1 .

Rx 3 () =

7 . Now if x ∈− [ 0.5, 0.5] and c

128(1 + c ) 2 (0.5) 7 is between 0 and x , then

Thus, R 6 (0.5) = 8 , where c ∈ (0.5,1) .

c 1 +>

0.5 and 4 1 c 1 x ≤ ⎛⎞ ⎜⎟ = so that, for all x,

Therefore, R 6 (0.5) ≤

, it is easily verified that

() =+ 46. 3/2 fx (1 x ) f(0) = 1

. Thus for = 3

, where c is

between x and 1 . Thus,

, where c ∈ (0.5,1) .

Therefore, R 6 (0.5) ≤

Note that e 1 <. 3

47. () =+ (1 ) fx –1/ 2 x f(0) = 1

( n + 1)!

0.000005 or 600000 < (n + 1)! when

44. To find a formula for () n

f () x (and thus for

is difficult, but we can use another approach: From

section 9.8 we know that

4(arctan ) = 4 , which is an

16 16 alternating series.(because of the ( 1) k + 1 1 3 − 5 and the (1 + x ) –1/ 2 ≈ 1– + 2 – x 3 x x 2 8 16

fact that all powers are odd) for all x ∈− [ 1,1] . Thus,

(1 + ) by the Alternating Series Test, –9 / 2 4 Rx 3 c x

we want R n (1) ≤ 0.000005 , we set

0.000005, which yields n > 399,999 .

Instructor’s Resource Manual

Section 9.9 573

51. Assume n is odd; that is n = 2 m + for 1 m ≥. 0

Then, R n + 1 () x = R

Note that, for all m,

23 therefore, R

n + 1 () x =

(2 m + 3)!

where c is

(1 – x )

x 48 (1 + x )

between 0 and x . For x ∈ [0, π ], c ∈ (0, ) x so

that cos c < 1 and x ≤ ; hence

48(1 10 + x 2 + 5 x 4 )

48(1 10 + c 2 + 5 c 4 )

(2 m + 3)!

. Now, for

k = 2,3, ,2 m + 3, 2

⎟ ⎞≈+ 2 x x k 4

so that

2 m +≥ 2 42.8666 ⇒= n 2 m +> 1 42

(0.5) R 4 () x ≤

52. Assume n is even; that is n = 2 m for m ≥. 0

f (2 m + 2) () c

∫ sin x dx ≈ ∫ ⎜ x – x ⎟ dx Then, R n + 1 () x = R 2 m + 1 () =

(2 m + 2)!

Note that, for all m,

x ⎢ 2 – x 4 ⎥ ≈ 0.1224 (4 ) m

therefore, R () x =

cos c

x 2 m + 2 where c is

between 0 and x . For x ∈ [0, π 2 ], c ∈ (0, ) x so

Rx 5 () ≤ ≈ 0.001389 6!

π that cos c < 1 and x ≤ ; hence

0 cos x dx ≈ ∫ 0 1– + dx ⎜ m ⎜ ⎟ ⎟ ⎛⎞ π ⎝

1 n + 1 . Now, for

k = 2,3, … ,2 m + 2, 2 ≤

so that

Section 9.9 Instructor's Resource Manual

2 m + ππ 1 ⎛⎞

b. f(x) = ln(1 + x); f(0) = 0

ln(1 + x ) ≈− x

53. The area of the sector with angle t is

area of the triangle is

ln 2 ln 2 ≈ 0.693 + ≈ +

24 r

Using n = 3, sin t ≈ t – t .

6 We let 24 – r ≈ 24 since the interest rate r is going to be close to 0.

A ≈ tr – r 2 ⎜ t – t 3 ⎟ 23 = rt

2 2 ⎝ 6 ⎠ 12 c. r n (exact) n (approx.) n (rule 72)

56. fx () = 1– e –(1 + kx ) ; f(0) = 0

The Maclaurin polynomial of order 2 is:

For x = 2k, the polynomial is

2 k − 4 k 3 − 2 k 4 ≈ 2 k when k is very small.

57. fx () = x 4 –3 x 3 + 2 x 2 + x –2 ; f(1) = –1

f (5) () x = 0 Since f (5) () x = 0, Rx 5 () = 0. x 4 –3 x 3 + 2 x 2 + x –2

= –1 – ( – 1) x 2 + ( – 1) x 3 + ( – 1) x 4

Instructor’s Resource Manual

Section 9.9 575

f ''( ) a f '''( ) a f () n () a

f '''( ) a f () n

Pa ' n () = fa '( ) 0 0 ++++= 0 fa '( )

() n

P n ''

f '''( ) a f () =+ a

= f ''( ) a + f '''( )( ax −++ a )

( n − 2)! P n '' = f ''( ) 0 0 a ++++= 0 f ''( ) a

59. f(x) = sin x ;

2 Since f () n () x is ±sin x or ±cos x,

≤ 0.0005 when n ≥ 2.

Section 9.9 Instructor's Resource Manual

5 sin c 6

x (i) Since ( n + 1) f () x is continuous near c, then

62. a. sin

f n () a < when a is near c. 3 0

sin – x x + x 6 ⎛ 1 sin c 1 Thus

() < when x is near c, so 0

f (x) < f(c) when x is near c. f(c) is a local maximum.

1 2 1 4 1 6 sin c (ii) Since 7 ( f n + 1) () is continuous near c, then

a > when a is near c. 0

2 cos – 1 4

x + x – x 2 24 Thus R n () x > when x is near c, so 0

lim

f (x) > f(c) when x is near c.

lim – ⎛ 1 sin c ⎞

f (c) is a local minimum.

Suppose fx () = 4 x . f(x) > 0 when x > 0 and

63. The kth derivative of h(x)f(x) is

f (x) < 0 when x > 0. Thus x = 0 is a local

() i () (–) ∑ ki ()

9.10 Chapter Review

Thus for i

≤ n + 1, () h i (0) = Let 0.

() = n qx 1 x + fx ( ). Then

Concepts Test

If b = 100 and a = + 50 (–1) n ∑ then () i h (0) f (0) = 0 n n

i = 0 ⎧ 51 if n is even

for k ≤ n + 1.

since a n =⎨

g () k (0) = p () k (0) + q () k (0) = p () k (0) lim b n = 100 while lim a n does not

The Maclaurin polynomial of order n for g is

x n which

2. True:

It is clear that ! n ≤ n n . The inequality

n ! ≤ n n ≤ (2 – 1)! n is equivalent to

is the Maclaurin polynomial of order n for p(x).

Since p(x) is a polynomial of degree at most n,

n n (2 – 1)! n

the remainder R n () x of Maclaurin’s Formula for

p (x) is 0, so the Maclaurin polynomial of order n

Expanding the terms gives

for g(x) is p(x).

n n nnn

n =⋅⋅⋅⋅ n ⋅

64. Using Taylor’s formula,

≤ ( n + 1)( n +⋅⋅+ 2) ( n n – 1) or

Since fc ′ () = f ′′ () c = f ′′′ () =…= () c f n () c = 0,

The left-hand side consists of n – 1 terms, each of which is less than or

fx () = fc () + R n () x equal to n, while the right-hand side

f ( n + 1) () a n + 1 consists of n – 1 terms, each of which

R n () x =

( n + 1)!

( )( – ) ax c where a is

is greater than n. Thus, the inequality

is true so ! n ≤ n ≤ (2 – 1)! n

between x and c.

Instructor’s Resource Manual

Section 9.10 577

3. True:

If lim a n = then for any L ε>0

n 1 →∞ 8. False: {}{ ( 1) − n and ( 1) − n + } both diverge

there is a number M > 0 such that

but

a n – L < ε for all n ≥ M. Thus, for

( 1) − n +− ( 1) n + { 1 }{ =− ( 1) (1 1) n − } = {} 0

the same ε, a 3 n + 4 – L < ε for converges.

If {} a n converges, then for some N, was arbitrary, lim a 3 n + 4 = L . there are numbers m and M with

n →∞

m ≤ a n ≤ M for all n ≥ N. Thus

4. False: Suppose a n = if n = 2k or n = 3k 1 m a ≤ M n ≤ for all n ≥ N. Since where k is any positive integer and

a n = if n is not a multiple of 2 or 3. 0 ⎧⎫ m

⎧ M ⎫ ⎨⎬ and ⎨ ⎬ both converge to 0,

Then lim a 2 n = lim a 3

n does not exist.

⎨ ⎬ must also converge to 0.

be given by

⎧ 1 if is prime n

⎩ 0 if n is composite

a = (–1) n 1 2 1 Then 1 a mn = for all mn since 0 m ≥, 2 n so a n = and

n 1 n hence lim

lim a n does not exist since for any

n →∞

11. True:

The series converges by the M > 0 there will be a n 's with a n = 1 Alternating Series Test.

= ,– = ,– = + , since there are infinitely many prime

S 1 aS 1 2 a 1 a 2 S 3 a 1 a 2 a 3 numbers.

S 4 = a 1 – a 2 + a 3 – a 4 , etc.

0 < a 2 < a 1 ⇒< 0 a 1 – a 2 = S 2 < a 1 ;

6. True:

Given ε > 0 there are numbers M 1

0 < a 3 < a 2 ⇒ – a 2 < – a 2 + a 3 < so 0

and M 2 such that a 2 n – L < when ε

0 < a 4 < a 3 ⇒< 0 a 3 – a 4 < a 3 , so n ≥ M 2 . Let M = max{ M 1 , }, M 2 – a 2 < – a 2 + a 3 – a 4 < – a 2 + a 3 < 0, then when n ≥ M we have

n ≥ M 1 and a 2 n + 1 – L < ε when

< a 1 – a 2 + a 3 < a 1 ; etc.

For each even n, 0 < S n –1 – a n while is either even (k = 2n) or odd

n ≥ 2M + 1 since every k ≥ 2M + 1

for each odd n, n > 1, S n –1 + a n < a 1 . (k = 2n + 1).

Let a n =+++ 1 . Then

⎛⎞ 1 n ∞ 1 ⎛⎞ n

converges since it is a geometric

n →∞

series with =

not finite since lim a

∑ ⎜⎟ ⎛ ⎞ = + + +…> 1 1 since

which diverges.

n = 1 ⎝⎠ n

all terms are positive.

Section 9.10 Instructor’s Resource Manual

⎛⎞ n 1 ∞

⎛⎞ n 1 ∞ ⎛⎞ 1 n 4 + 1

18. False:

Since lim

= there is some 0,

number M such that n > 1 4 1 e n + for 1

all n ≥ M, thus n > ln( 4 n + and 1)

n ⎛⎞ –1 1 1 1 1 1

for n ≥ M. Hence,

⎛⎞= S with

Thus,

n = 1 ⎝⎠ n

diverges by the

n = 1 ln( n 4 + 1)

1 <≤< S 2.

Comparison Test.

19. True:

n = 2 ( ln ) 2 n ∑ n

13. False:

(–1) diverges but the partial

sums are bounded ( S n = –1 for odd n

2 ⎥ n = 2 ⎢ ( ln ) n ( ln ) n n and S n = for even n.) 0 ⎣

is continuous, positive, and

x (ln ) 2

converges while ∑ diverges.

nonincreasing on [2,

= Ratio Test is 1,

n →∞ a n

2 dx = ∫

inconclusive. (See the discussion

2 x (ln ) x

ln 2 u 2

before Example 5 in Section 9.4.)

n = 2 n (ln ) ∑ n

converges, but

For n ≥ 3, ln n > 1, so (ln ) 2 n > and 1

2 = 1. 2 2 < 2 . n Thus →∞ a n n →∞ ( n + 1) n (ln ) n n

lim n + 1 = lim

1 ∑ 2 2 < ∑ 2 so

17. False:

lim 1 ⎜ − ⎟ =≠ 0 so the series

n = 3 n (ln ) n

n →∞ ⎝ n ⎠

cannot converge.

2 2 converges by the

n = 3 n (ln ) n Comparison Test. Since both series converge, so does their sum.

20. False:

This series is

1 which n = 1 2 n − 1 3 5

diverges.

Instructor’s Resource Manual

Section 9.10 579

21. True:

If 0 ≤ a n + 100 ≤ b n for all n in N, then

26. True

Since a n ≥ for all n, 0

n so ∑ a n also

∑ ( 1) − a n = ∑ a n so the series

n = 101

converges, since adding a finite

number of terms does not affect the n ∑ ( 1) − a

n converges absolutely. convergence or divergence of a series.

22. True:

If ca ≥ for all n in N with c > 0,

27. True:

∑ ( 1) − n + 1 −

n ∑ + ( 1) − 1

then a n ≥

1 1 ∞ 1 ∑ a n ≥ ∑ = cn c ∑ which

28. True:

Suppose ∑ a n converges. Thus,

a + a diverges. Thus,

2 a converges, so

a diverges by the

n = 1 converges since 0 ≤ a n + a n ≤ 2 a n . Comparison Test.

But by the linearity of convergent

series ∑ a n = ∑ ( a n + a n ) − ∑ a n

23. True:

3 3 ⎜⎟ ⎝⎠ converges, which is a contradiction. ⎝⎠ 3 n ∑ = 1 ⎝⎠ 3

n –1

1 1 29. True:

−− 3 ( 1.1) = 4.1 , so the radius of

convergence of the series is at least n = 1 3 3 2 4.1.

sum of the first thousand terms is less

37 −=< 4 4.1 so x = 7 is within the

than . interval of convergence.

30. False:

If the radius of convergence is 2, then

24. False: Consider the series with

the convergence at x = 2 is

( 1) − n + 1

independent of the convergence at

. Then

x = –2.

( 1) − 2 n + 1 − 1 31. True:

The radius of convergence is at least

( 1) − a n =

1.5, so 1 is within the interval of

convergence.

∑ 1 ( 1) − n

a n =−−−− 1 which

1 ⎡ ∞ ax n + 1 n ⎤

⎥ diverges.

n = 1 2 3 Thus ∫ () 0 f x dx =⎢ ∑

If b n ≤ a n ≤ for all n in N then 0 = ∑ .

25. True:

n ≤ b n for all n in N.

∑ The convergence set of a power series

– b n = (–1) ∑ b n which converges

32. False:

may consist of a single point.

33. False:

Consider the function

since ∑ b n converges.

Thus, by the Comparison Test,

a n converges, hence

The Maclaurin series for this function

represents the function only at x = 0.

a n =− ( 1) ∑ ( – a n ) also

34. True:

On (–1, 1), fx () =

Section 9.10 Instructor’s Resource Manual

∞ ( 1) − nn x

35. True:

e − ∑ x + e − x = 0 Sample Test Problems

If p(x) and q(x) are polynomials of

The sequence converges to 3.

degree less than or equal to n, satisfying p(a) = q(a) = f(a) and

2. Using l’Hôpital’s Rule,

k ≤ n, then p(x) = q(x).

n →∞ n n →∞ 1 n →∞ n

38. True:

f (0) = f ′ (0) = f ′′ (0) = its second 0,

lim

order Maclaurin polynomial is 0.

n →∞ n

39. True:

After simplifying, Px 3 () = fx () .

Any Maclaurin polynomial for cos x

n →∞

→∞ ⎝ ⎝ n ⎠

involves only even powers of x.

The sequence converges to e 4 .

41. True:

The Maclaurin polynomial of an even function involves only even powers of

x , so f′ (0) = if f(x) is an even 0 4. a n + 1 =

a n thus for n > 3, since > 1,

function.

a n + 1 > a n and the sequence diverges.

42. True:

Taylor’s Formula with Remainder for

n = 0 is ( ) fx = fa () + fcx ′ ( )( – ) a 5. Let y = n n = n 1/ n

1 then ln y = ln . n n

which is equivalent to the Mean Value

Theorem. 1 1 ln n

using l’Hôpital’s Rule. Thus,

lim n

n ln = lim e y = The sequence 1.

n →∞

converges to 1.

1/ 1 n 1 ⎛⎞ 1

6. lim 3 = while 0 n =⎜⎟ . As n →∞ ,

The sequence converges to 1.

The sequence converges to 0.

8. The sequence does not converge, since whenever n is an even multiple of 6, a n = , while 1 whenever n is an odd multiple of 6, a n =−. 1

Instructor’s Resource Manual

Section 9.10 581

⎟ = 1 →∞ . The series converges to 1. →∞

⎞ = .The series converges to .

n →∞ ⎜ ⎝ 2 n + 1 n + 2 ⎟ ⎠ 2 2

n →∞

11. ln + ln + ln +… = ∑ ln

= ∑ [ln – ln( n n + 1)]

S n = (ln1 – ln 2) (ln 2 – ln 3) + +…+ (ln( – 1) – ln ) (ln – ln( n n + n n + 1)) = ln1 – ln( n += 1) ln

The series diverges.

π=⎨ x

⎧ 1 if is even k

(–1) ∑ k which diverges since

converges to

if is even

so { S n } does not

1 e 2 1 −+ −+ = e ≈ 0.3679 .

13. –2 ∑ k e = ∑ ⎜ 2 ⎟ =

14. ∑ k = 3 ∑ ⎜⎟ = = 6 n 2

k = 0 2 k = 0 ⎝⎠ 2 1– 1 2 0<1< ∞.

By the Limit Comparison Test, since

k = 4 ∑ ⎛⎞ ⎜⎟ = 1 = 6 ∞

k = 0 3 k = 0 ⎝⎠ 3 1– 3 ∑ b n = ∑ diverges, ∑ a n = ∑

also

Since both series converge, their sum converges

diverges.

to 6 + 6 = 12.

The series converges since

By the Limit Comparison Test, since

⎜ ⎟ diverges since

b n = ∑ 2 converges, ∑ a n = ∑ 2

also converges.

Section 9.10 Instructor’s Resource Manual

( n + 1)! 2 n !

21. Since the series alternates,

= , the series converges by the 0 2( n + 1)

Alternating Series Test.

The series diverges.

22. The series diverges since

= lim 3 − 1/ n =. 1 30. lim 1 −

=≠ 0 , so the series does not

= ∑ ⎜ ⎛⎞ ⎛⎞ ⎜⎟ + ⎜⎟ ⎟ ( n + 1) 22 n + 1 2

= < , so the series converges. 1

The series converges to 4. The 1’s must be subtracted since the index starts with n = 1.

32. Since the series alternates,

= 0, the series converges by the

n →∞ + 1 ln n

⎛ 1 + 1 n ⎞ = Alternating Series Test.

lim ⎜

⎟ =< 0 1, so the series converges. n →∞ ⎜ e 2 n + 1 ⎟

33. a n =

so a n > a n + 1 ;

3–13–1 n n 3 n + 2

25. lim

≠ 0 , so the series diverges.

n →∞ 10 n + 12 10 lim a n = lim

= so the series 0,

n →∞

n →∞ 3–1 n

26. Let a n = 2 and b n = 3/2 . ∑ (–1)

converges by the Alternating

3–1 n

a 2 n Series Test. n 1

1 + 7 n →∞ 1 + 7 n 2 Let b n = , then

n →∞ b n n →∞ n 2

By the Limit Comparison Test, since

b n = ∑ 3/2 converges ⎜ > 1, ⎟ 1

0 < < ∞ By the Limit Comparison Test, since .

also converges.

b n = ∑ diverges, ∑ a n = ∑

The series is conditionally convergent.

n →∞ ( n + 1)! n ! n →∞ n 2 the series converges.

() 1 + n 1

n + 1) 3 n () 1 +

=< 01 The series is absolutely convergent.

n →∞ 3 nn ( + 2) n →∞ 1 + 2

The series converges.

Instructor’s Resource Manual

Section 9.10 583

1 n ⎛⎞ 3 1 3 3 3 lim 8 ⎜⎟ = 8 lim ⎛⎞ ⎜⎟ =∞ since > 1. n + 1

The series is divergent.

When x = 1, the series is

≤+ 1 ∑ , which

When x = –1, the series is ∑ 3 which

1/ x

2 [ln – (ln ) – ], x x 2 x for x ≥ 3, ln x > 1

converges absolutely since ∑ 3 converges.

( ln ) x x

so (ln ) x 2

> ln x hence f(x) is decreasing on

The series converges on –1 ≤ x ≤ 1.

[3, ∞ ). Thus, if a n =

= 2 x ; 2 x < 1 when

Using l’Hôpital’s Rule,

lim = thus 0,

e 0 = e = Hence, lim 1. When x = , the series is

is of the form

so lim

n = 0 2 n + Thus, by the Alternating Series Test, 3 ∞ (–1) nn

n →∞ ln n

, ∑ so

n →∞ 2 n + 3 ∑ 2 n + 3

converges by the Alternating Series Test.

Thus if ∑ diverges, ∑ also diverges.

When x = –, the series is

ln n

Let a n =

and b n = . Then

1 ∞ (–2) n + 1 – 1 n

= lim n = as shown above; 1 n →∞ b

n n →∞

, let b

n = then

1 0<1< ∞ . Since ∑ b n = ∑ diverges,

also diverges by the Limit Comparison

01 < < ∞ hence since ∑ b n = ∑ diverges,

Test, hence ∑ also diverges.

n = 2 ln n

and also – ∑

diverges.

The series is conditionally convergent.

The series converges on – <≤ x .

Section 9.10 Instructor’s Resource Manual

( x − 4) n + 1 ( x − 4) n

When x = 1, the series is

39. ρ = lim

n →∞

n + 2 n + 1 (–2)

= lim x − 4 =− 4 ; –4 < when

n →∞

n + 2 so the series diverges.

3 < x < 5.

The series converges on 1 < x < 5.

When x = 5, the series is

(–1) (1) n n

∞ (–1) n

( n + 1)!( x + 1) n + nx !( + 1) n

n + = ∞ unless x = –1. 1

n →∞

1 ∞ (–1) n

The series converges only for x = –1.

lim

= 0 so

converges by the

Alternating Series Test.

When x = 3, the series is

∑ . Thus,

1 1 If fx () =

, then fx ′ () =−

= ∑ . a = , let

(1 + x ) 2

differentiating the series for

and

b n = then lim

n →∞ b n n →∞ n + 1 multiplying by –1 yields

hence since ∑ b n = ∑ diverges, ∑

2 =− 12 x + 3 x − 4 x + . The series

n = 0 n + 1 (1 + x )

also diverges.

converges on –1 < x < 1.

The series converges on 3 <≤. x 5

, then f ′′ () x =

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

Differentiating the series for

1x +

twice and

The series converges for all x.

dividing by 2 gives

< 1 The series converges on –1 < x < 1.

when 1 < x < 5.

45. sin 2 x = ⎜ x − +

When x = 5, the series is

n = 0 n = 0 1 + () 2 1 + () 2 3 45 315

so the series diverges.

Since the series for sin x converges for all x, so does the series for sin x . 2

2 e 3 e 46. If ( ) 4

2 2 e () 2 n

fx 4 = e , then f () x = e . Thus, e = e + e ( x −+ 2) ( x − 2) + ( x − 2) + ( x − 2) + .

2! 3! 4! 5! Since the series for sin x and cos x converge for all x, so does the series for sin x + cos x.

Instructor’s Resource Manual

Section 9.10 585

48. Let a k =

2 and define fx () =

2 ; then ( ) fk = a k 9 and f is positive, continuous and non-increasing + k 9 + x

(since fx ′ () =

22 < 0 ) on [1, ) ∞ . Thus, by the Integral Test, E n < ∫

2 dx = lim ⎡ tan − 1 ⎤ = + x n A ⎣ (9 ⎢ ) 9 + x →∞ 3 3 ⎥ ⎦

. Now − tan − 1 n 0.00005 ⇒≥ n ⎢ ⎡ ⎥ ≤ 3 tan 3 ⎛ ⎜ π − 0.00005 ⎞ ⎤ ≈

and define fx () =

; then ( ) fk = a and f is positive, continuous and non-increasing (since

12 − x

fx ′ () =

< ) on [1, ) 0 ∞ . Thus, by the Integral Test, E <

≤ 0.000005 ⇒ n e ≥

50, 000 ⇒ 2 n ≥ ln(50, 000) ≈ 10.82 ⇒ n >. 3

2 ⎣ ⎢ A →∞ 2 e A 2 e n ⎥ ⎦ 4 e n 2 4 2 e n

50. One million terms are needed to approximate the sum to within 0.001 since

< 0.001 is equivalent to

999,999 < n.

51. a. From the Maclaurin series for

, we have

3 =+ 1 x 3 + x 6 + 1x . − 1 −

b. In Example 6 of Section 9.8 it is shown that 1 +=+ x 1 x − x 2 + x 3 −

d. Using division with the Maclaurin series for cos x, we get sec x =+ 1 +

Thus, x sec x =+ x +

+ ; Using division, we get

f (1) () x =− sin x f (1) (0) = 0 f '( ) x = cos 2 x − 2 x 2 sin 2 x

f (2) () x =− cos x f (2) (0) =− 1 f '(0) = 1

2 px () = xp ; (0.2) = 0.2; (0.2) f = 0.1998 x

∴ Px 2 () =− 1

Thus, cos(0.1) ≈− 1 =− 1 0.005 = 0.995 2

Section 9.10 Instructor’s Resource Manual

54. a.

fx () = xe x f(0) = 0

f ′′ () x = 2 e x + xe x f ′′ (0) = 2 (0.2)

b. f(x) = cosh x

f(0) = 1

fx ′ () = sinh x f′ (0) = 0 (4) () = –8 cos 2

f (4) x x f (0) = –8

f ′′ () x = cosh x

f ′′ (0) = 1 (5)

f (5) () x = 16 sin 2 x f (0) = 0

f (3) () x = sinh x f (3) (0) = 0 (6)

f (4) () x = cosh x f (4) (0) = 1 2 2 2 8 4

(cos 2 ) cx ≤ (0.2)

(3) () = 6 g (3) x g (2) = 6 (–1) n

Since g () x = Rx 3 0, ( ) = 0, so the Taylor

( n + 1) c n + 1

polynomial of order 3 based at 2 is an exact

0.2 n + 1 (0.25) n + 1

gx () = Px 4 () =+ 3 9( – 2) 4( – 2) x + x + ( – 2) x (0.25) n + 1

< 0.00005 when n ≥ 5 .

56. (2.1) =+ 3 9(0.1) 4(0.1) + 2 g 3 + (0.1) = 3.941 ( n + 1)

61. From Problem 60,

8 ∫ 0.8 ln x dx ≈

Instructor’s Resource Manual

Section 9.10 587

Review and Preview Problems

5. Since we are given a point, all we need is the slope to determine the equation of our tangent

line.

1. fx () =

so that fx ′ () = and f′ (2) = 1 ⎛

() 1 a. The tangent line will be the line through the

dx ⎝

point (2,1) having slope m = 1. Using the point

y dy

slope formula: ( y −= 1) 1( x − 2) or y =− x 1 .

2 dx

b. The normal line will be the line through the y dy =− 2 x

point (2,1) having slope m =−=− 1 . Using

1 2 dx

1 dy − 4 = x

the point slope formula: ( y −=− 1) 1( x − 2) or

y =−+ x 3 or x += y 3 .

At the point ⎜

,1 ⎟ ⎟ , we get

2. fx () == y , fx ′ () =

4 2 dy −− 4 ( 3/2 )

a. The line y = x has slope = 1, so we seek x

= 23 = m tan

such that fx ′ () =

1 or

x = 2 . The point is (2,1) .

Therefore, the equation of the tangent line to the

b. Since fx ′ () = is the slope of the tangent

curve at ⎜ ⎜ −

,1 ⎟ is given by

x 2 line at the point 2 (, x

4 ) , − will be the slope

y −= 1 23 ⎜ −− ⎜ x ⎜ ⎜ ⎟ ⎟ ⎟ ⎟

of the normal line at the same point. Since y = x

has slope 1, we seek x such that

y −= 1 23 x + 3

1 or x = − The point is 2. ( 2,1) − .

y = 23 x + 4

3. Solving equation 1 for y 2 : 9 y 2 =− 9 x 2 and

6. Since we are given a point, all we need is the

slope to determine the equation of our tangent

16 line.

putting this result into equation 2:

175 x = 1008 x = 5.76 x =± 2.4 −

9 8 dx

Putting these values into equation 1 we get

y dy

y =− 9 (5.76) =− 9 3.24 = 5.76 so y =± 2.4 8 dx

dy 16 x

also. Thus the points of intersection are

dx 9 y

Since the length of the graph is greater than ∑ a n , the length of the graph is infinite.

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