Transcript
Page 1: Link full download:  ... · Karl Byleen Link full download:

Karl Byleen

Link full download: https://getbooksolutions.com/download/solutions-manual-for-

additional-calculus-topics-11th-edition-by-raymond-barnett-karl-byleen-michael-

ziegler/

1.f(x) = 1 = x

-1 x

f’(x) = -x-2

(Using Power Rule)

f”(x) = 2x-3

6

f(3)

(x) = -6x-4

= -

x 4

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

f'(x) =

1 = (1 + x)

-1

1 x

f"(x) = (-1)(1 + x)-2

(Using Power Rule)

f(3)

(x) = (-1)(-2)(1 + x)-3

(Using Power Rule)

=

2

(1 x)3

3. f(x) = e-x

f’(x) = -e-x

f”(x) = e-x

f(3)

(x) = -e-x

4. f(x) = ln(1 + 3x) 3

f’(x) =

1 3x

f”(x) = 3(-1)(3)(1 + 3x)-2

= -9(1 + 3x)-2

(Using Power

Rule) f(3)

(x) = (-9)(-2)(3)(1 + 3x)-3

= 54(1 + 3x)-3

f(4)

(x) = (54)(-3)(3)(1 + 3x)-4

= -486(1 + 3x)-4

= -

486

(1 3x)4

5. f(x) = e5x

f’(x) = 5e5x

f”(x) = 5(5)e

5x = 5

2e5x

f(3)

(x) = 52(5)e

5x = 5

3e5x

f(4)

(x) = 53(5)e

5x = 5

4e5x

= 625e5x

96 TAYLOR POLYNOMIALS AND INFINITE SERIES

= 3(1 + 3x)-1

Page 2: Link full download:  ... · Karl Byleen Link full download:

6. f(x) =

1 = (2 + x)

-1

2 x

f’(x) = (-1)(2 + x)-2

f”(x) = (-1)(-2)(2 + x)-3

= 2(2 + x)-3

f(3)

(x) = (2)(-3)(2 + x)-4

= -6(2 + x)-4

f(4)

(x) = (-6)(-4)(2 + x)-5

= 24(2 + x)-5

7. f(x) = e-x

f'(x) = -e

-x

f"(x) = e-x

f(3)

(x) = -e-x

f(4)

(x) = e-x

Using 2,

f(0) = e-0

= 1

f'(0) = -e-0

= -1

f"(0) = e-0

= 1

f(3)

(0) = -e-0

= -1

f(4)

(0) = e-0

= 1

f"(0) (3) (4)

p (x) = f(0) + f'(0)x + x2 + f (0) x

3 + f (0) x

4

4

3!

Thus,

2! 4!

1

1

1

1

1

1

p (x) = 1 - x + x2

- x3 + x

4 = 1 - x + x

2 - x

3 + x

4

4

24 2! 3! 4! 2 6

8. f(x) = e4x

f(0) = 1

f'(x) = 4e4x

f'(0) = 4

f"(x) = 16e4x

f"(0) = 16

f(3)

(x) = 64e4x

f(3)

(0) = 64

f"(0) (3)

Thus, p (x) = f(0) + f'(0)x +

x2 +

f (0) x3

3 2! 3!

= 1 + 4x + 16 x2 + 64 x3 = 1 + 4x + 8x

2 + 32 x3

2!

3

3!

9. f(x) = (x + 1)3, f(0) = 1

f'(x) = 3(x + 1)2, f'(0) = 3

f"(x) = 6(x + 1), f"(0) = 6

f(3)

(x) = 6, f(3)

(0) = 6

f(4)

(x) = 0 f(4)

(0) = 0

p (x) = 1 + 3x + 6 x2 + 6 x3 = 1 + 3x + 3x

2 + x

3

4

2! 3!

EXERCISE 2-1 97

Page 3: Link full download:  ... · Karl Byleen Link full download:

10. f(x) = (1 - x)4, f(0) = 1

f'(x) = -4(1 - x)3, f'(0) = -4

f"(x) = 12(1 - x)2, f"(0) = 12

f(3)

(x) = -24(1 - x), f(3)

(0) = -24

Thus, 12

24

p

(x) = 1 - 4x +

x2 -

x3 = 1 - 11.

3 2! 3!

f(x) = ln(1 + 2x) 1 2

f'(x) =

(2) =

1 2x 1 2x

f"(x) = -2(1 + 2x)-2

(2) = 4

(1 2x)2

f(3)

(x) = 8(1 + 2x)-3

(2) = 16

(1 2x)3

4x + 6x2 - 4x

3

f(0) = ln(1) = 0

2

f'(0) = 1 2 0 = 2 f"(0) = -4

f(3)

(0) = 16

f"(0) (3)

Using 2, p (x) = f(0) + f'(0)x + x2 + f (0) x3

3 2!

4

16

3!

8

Thus, p (x) = 0 + 2x - x2 + x

3 = 2x - 2x

2 + x3

3

3!

2! 3

12. f(x) = 3

= (x + 1)1/3

f(0) = 3

= 1

x 1 1

f'(x) =

1

f'(0) =

1

3(x 1)2/3

3 2 2

f"(x) =

f"(0) = -

9(x 1)5/3

9

f(3)

(x) = 10

f(3)

(0) = 10

27(x 1)8/3

27

f"(0) (3)

p

(x) = f(0) + f'(0)x +

x2 + f (0)

x3

3 2! 3!

1

2

10

1

1

5

Thus, p (x) = 1 + x - x2

+ x3

= 1 + x - x2 + x

3

3

3 9 2! 27 3! 3 9 81

Page 4: Link full download:  ... · Karl Byleen Link full download:

13. f(x) = 4

= (x + 16)1/4

f(0) = 2

x 16

f'(x) =

1 (x + 16)

-3/4 f'(0) =

1

4 32

3

3

f"(x) = - (x + 16)-7/4

f"(0) = -

16 1

3

2048 p (x) = 2 + x - x

2

2

4096

32

14. (A) f(x) = x4 - 1 f(0) = -1

f'(x) = 4x3

f'(0) = 0

f"(x) = 12x2

f"(0) = 0

f(3)

(x) = 24x f(3)

(0) = 0

f"(0) (3)

Using 2, p (x) = f(0) + f'(0)x + x2 + f (0) x3

3 2!

3!

Thus, p3(x) = -1

Now |p 3 (x) - f(x)| = |-1 - (x4

- 1)| = |x4| = |x|

4

and |x|4 < 0.1 implies |x| < (0.1)

1/4 ≈ 0.562

Therefore, -0.562 < x < 0.562

(B) From part (A),

f(4)

(x) = 24 f(4)

(0) = 24 (3)

f"(0)

Using 2, p (x) = f(0) + f'(0)x + x2

+ f (0) x3

4

24

2!

3!

Thus, p

(x) = -1 + x4

= -1 + x4 = x

4 - 1 = f(x) 4 4!

and |p4(x) - f(x)| = 0 < 0.1 for all x.

15. (A) f(x) = x5

f(0) = 0

f'(x) = 5x4

f'(0) = 0

f"(x) = 20x3

f"(0) = 0

f(3)

(x) = 60x2

f(3)

(0) = 0

f(4)

(x) = 120x f(4)

(0) = 0

p4(x) = 0

|p4(x) - f(x)| = |0 - x5| = |x|

5 = <

0.01 or |x| < (0.01)1/5

= 0.398

Therefore, -0.398 < x < 0.398

(B) f(0) = f'(0) = f"(0) = f(3)

(0) = f(4)

(0) =

0 f(5)

(x) = 120 and hence f(5)

(0) = 120.

p5(x) = 5! x5 = x

5

|p5(x) - f(x)| = |x5 - x

5| = 0 < 0.01

Thus for all x, |p5(x) - f(x)| < 0.01.

(4)

+ f(0)

x4 4!

EXERCISE 2-1 99

Page 5: Link full download:  ... · Karl Byleen Link full download:

16. f(x) = x3

f(1) = 1

f'(x) = 3x2

f'(1) = 3 f"(x) = 6x f"(1) = 6

f(3)

(x) = 6 f(3)

(1) = 6

p (x) = 1 + 3(x - 1) + 6 (x - 1)2 + 6 (x - 1)

3

3

2! 3!

= 1 + 3(x - 1) + 3(x - 1)2 + (x - 1)

3

17. f(x) = x2 - 6x + 10 f(3) = 1

f'(x) = 2x - 6 f'(3) = 0

f"(x) = 2 2

f"(4) = 2

p (x) = 1 + (x - 3)2 = 1 + (x - 3)

2

2

2!

Page 6: Link full download:  ... · Karl Byleen Link full download:

18. f(x) = ln(2 - x) f(1) = 0

f'(x) = -(2 - x)-1

f'(1) = -1

f"(x) = -(2 - x)-2

f"(1) = -1

f(3)

(x) = -2(2 - x)-3

f(3)

(1) = -2

f(4)

(x) = -6(2 - x)-4

f(4)

(1) = -6

p (x) = -(x - 1) - 1 (x - 1)2

- 2 (x - 1)3 - 6 (x - 1)

4

4

4! 2! 3!

= -(x - 1) -

1 (x - 1)

2 -

1 (x - 1)

3 -

1 (x - 1)

4

2 3 4

19. f(x) = e-2x

f'(x) = -2e-2x

f"(x) = 4e-2x

f(3)

(x) = -8e-2x

f(0) = 1 f'(0) = -2

f"(0) = 4

f(3)

(0) = -8

f"(0) (3)

Using 2, p (x) = f(0) + f'(0)x + x2 + f (0) x

3

3

2!

4

8

3!

4

Thus, p (x) = 1 - 2x + x2 - x

3 = 1 - 2x + 2x

2 - x

3.

3

2! 3! 3

Now, e-0.5

= e-2(0.25)

= f(0.25) ≈ p 3 (0.25)

4

= 1 - 2(0.25) + 2(0.25)

2 - (0.25)

3 = 0.60416667. 3

Page 7: Link full download:  ... · Karl Byleen Link full download:

20.

f(x) =

= x1/2

f(1) = 1

x

f'(x) =

1

x-1/2

f'(1) =

1

2 2

f"(x) = - 1 x-3/2 f"(1) = - 1

4

4

f(3)

(x) = 3

x-5/2

f(3)

(1) = 3

8 8

f(4)

(x) = - 15

x-7/2

f(4)

(1) = - 15

16 16

Thus,

(3)

(4)

f"(1)

p

(x) = f(1) + f'(1)(x - 1) + (x - 1)2 +

f (1)

(x - 1)3 +

f (1)

(x - 1)4

4 2! 3! 4!

= 1 + 1 (x - 1) - 1 (x - 1)2

+ 3 (x - 1)3 - 15 (x - 1)

4

2

4 2!

8

3! 16

4!

= 1 +

1 (x - 1) -

1

(x - 1)2 +

1 (x - 1)

3 -

5 (x - 1)

4.

2 8 16 128

Now,

1.2 = f(1.2) ≈ p4(1.2) = 1 +

1 (1.2 - 1) - 1 (1.2 - 1)2 + 1 (1.2 - 1)

3 - 5 (1.2 - 1)

4

16

2 8 128

= 1 + 1 (0.2) - 1 (0.2)2 + 1 (0.2)

3 - 5 (0.2)

4

2

16

128

8

= 1.0954375.

Page 8: Link full download:  ... · Karl Byleen Link full download:

21. f(x) = 1 = (4 - x)-1

x 4

f'(x) = -1(4 - x)-2

(-1) = (4 - x)-2

f"(x) = -2(4 - x)-3

(-1) = 1·2(4 - x)-3

f(3)

(x) = 2(-3)(4 - x)-4

(-1) = 1·2·3(1 - x)-4

f(4)

(x) = 2·3(-4)(4 - x)-5

(-1) = 1·2·3·4(1 - x)-5

f(n)

(x) = n!(4 - x)-(n+1)

22. f(x) = 4 = 4(1 +x)-x

x 1

f'(x) = -4(1 + x)-2

f"(x) = (-1)2(4)(2)(1 + x)

-3 = 4(2!)(-1)

2(1 + x)

-3

f(3)

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

-4 = 4(3!)(-1)

3(1 + x)

-4

M

f(n)

(x) = 4(n!)(-1)n(1 + x)

-(n+1)

23. f(x) = e3x

f'(x) = e3x

(3) = 3e3x

f"(x) = 3e3x

(3) = 9e3x

= 32e3x

f(3)

(x) = 9e3x

(3) = 27e3x

= 33e3x

f(4)

(x) = 27e3x

(3) = 81e3x

= 34e3x

f(n)

(x) = 3ne3x

24. f(x) = ln(2x + 1)

f'(x) = 2(2x + 1)-1

= (-1)2(2x + 1)

-1

f"(x) = -(2)2(2x + 1)

-2 = (-1)

3 22(2x +

1)-2

f(3)

(x) = (-1)4 23(2!)(2x + 1)

-3

f(n)

(x) = (-1)n+1

2n((n - 1)!)(2x + 1)

-n

25. f(x) = ln(6 - x)

f'(x) = 1 (-1) = - 1 = -(6 - x)-1

6 x

x 6

f"(x) = (6 - x)-2

(-1) = -(6 - x)-2

f(3)

(x) = 2(6 - x)-3

(-1) = -1·2(6 - x)-3

f(4)

(x) = 1·2·3(6 - x)-4

(-1) = -1·2·3(6 - x)-4

f(n)

(x) = -(n - 1)!(6 - x)-n

Page 9: Link full download:  ... · Karl Byleen Link full download:

26. f(x) = ex/2

f'(x) = 12 e

x/2

1

2 1

f"(x) =

ex/2 =

ex/2

2 22

f(3)

(x) = 1

ex/2

3

2

1

f(n)

(x) =

ex/2

n

2

27. From Problem 31,

f(0) = 1 , f'(0) =

1 , f"(0) =

2! , f

(3)(0) =

3! , … , f

(n)(0) =

n!

4 2 3 4 n 1

4 4 4 4

Thus, 1

1

1

1

1

p (x) = + x + x2 + x

3 + … + x

n.

n

42

43

44

4n 1

4

28. f(x) =

4 = 4(1 + x)

-1

1 x

From problem 32, f(n)

(x) = 4(n!)(-1)n(1 + x)

-(n+1)

and hence f(n)

(0) = 4(n!)(-1)n. (n) n

The coefficient of xn is

f (0)

4(n!)(1)

= 4(-1)n.

n! n!

29.From Problem 33,

f(0) = e0 = 1, f'(0) = 3e

0 = 3, f"(0) = 3

2e0 =

32, f

(3)(0) = 3

3e0 = 3

3, …, f

(n)(0) = 3

ne0 = 3

n

Thus,

p (x) = 1 + 3x + 32 x2 + 3

3 x3 + … + 3

n xn.

n

2! 3! n!

30.f(x) = ln(2x + 1)

From problem 34, (n) n

f(n)

(x) = (-1)n+1

2n((n - 1)!)(2x + 1)

-n and

f (0)

= (-1)n+1 2

n! n

22

23

24

2n

Thus, p (x) = 2x - x2

+ x3 - x

4 + … + (-1)

n+1 xn

n 2

3

4

n

31. From Problem 35, 1

1

2!

(n 1)! f(0) = ln 6, f'(0) = - , f"(0) = -

, f

(3)(0) = - , …, f

(n)(0) = -

2

6 3 n

6 6 6

Thus, 1

1

1

1

p (x) = ln 6 - x - x2 -

x3 - … - xn.

n

2

3

n

6

2

3

n

6 6 6

EXERCISE 2-1 105

32. f(x) = ex/2

1 (n)

1

From problem 36, f(n)

(x) = ex/2 and f (0) = . Thus, n

n! n

Page 10: Link full download:  ... · Karl Byleen Link full download:

2 n!2

p (x) = 1 +

1 x +

1 x2 +

1 x3 + … +

1 x2

+xn

n 2 2!2

2 3!2

3 n!2

n

33. f(x) = 2 = 2x-1

f(1) = 2

x

f'(x) = -2x-2

f"(x) = (-1)2 2(2!)x

-3

f(3)

(x) = (-1)3 2(3!)x

-4

f(n)

(x) = (-1)n 2(n!)x

-(n+1)

(n)

Therefore f

(1)

= 2(-1)n and

n!

pn(x) = 2 - 2(x - 1) + 2(x - 1)2 - 2(x - 1)

3 + … + 2(-1)

n(x - 1)

n

34. Step 1. Step 2. Step 3.

f(x) = ln x f(1) = 0 a0 = f (1) = 0

1

f'(x) =

f'(1) = 1 a1 = f'(1) = 1

x 1

f"(1) 1 1

f"(x) = -

f"(1) = -1 a2 =

= -

= -

x2

2! 2! 2

2 (3)

2

1

f(3)

(x) =

f(3)

(1) = 2 a

= f (1)

=

=

3

x3

3!

3!

3

Page 11: Link full download:  ... · Karl Byleen Link full download:

f(4)

(x) = - 2 3

f(4)

(1) = -3! x4

f(n)

(x) = f(n)

(1) =

(1)n 1(n 1)!

(-1)n+1

(n - 1)!

xn

(4) 3! 1 f (1)

a4 =

= -

= -

4! 4! 4

an =

(n) n 1 1)!

n 1

f (1) = (1) (n = (1)

n!

n! n Step 4. The nth degree Taylor polynomial is:

1 (x - 1)2 +

1 (x - 1)3 - 1 (x - 1)4 + … +

(1)n 1

(x - 1)n. p (x) = (x - 1) -

n

2

3

4 n

35. f(x) = ex

(n)

1

f(n)

(x) = ex

and f (2)

=

n!

n!e

2

Thus, p

(x) =

1 +

1

(x + 2) +

1 (x + 2)

2 + … +

1 (x + 2)

n

n e2 e2

2!e

2 n!e

2

Page 12: Link full download:  ... · Karl Byleen Link full download:

36.f(x) = x5 + 2x

3 + 8x

2 + 1

(A) Fourth-degree Taylor polynomial p4(x) for f at 0 is:

p (x) = f(0) + F’(0) x + F’’(0) x2 + f

(3)(0) x

3 + f

(4)(0) x4

4 1! 2! 3! 4!

f(0) = 1

f’(x) = 5x4 + 6x

2 + 16x ; f’(0) = 0

f”(x) = 20x3 + 12x + 16 ; f”(0) = 16

f(3)

(x) = 60x2 + 12 ; f

(3)(0) = 12

f(4)

(x) = 120x ; f(4)

(0) = 0 Thus,

p 4 (x) = 1 + 8x

2 + 2x

3

= 2x3 + 8x

2 + 1

(B) The degree of the polynomial is 3.

37.f(x) = x6 + 2x

3 + 1

f(0) = 1

f’(x) = 6x5 + 6x

2 ; f’(0) = 0

f”(x) = 30x4 + 12x ; f”(0) = 0

f(3)

(x) = 120x3 + 12 ; f

(3)(0) = 12

f(4)

(x) = 360x2

; f(4)

(0) = 0

f(5)

(x) = 720x ; f(5)

(0) = 0

f(6)

(x) = 720 ; f(6)

(0) = 720

f(n)

(x) = 0 for n ≥ 7. Thus, for n = 0, 3 and 6, the nth-degree Taylor polynomial for f at 0

has degree n.

38.f(x) = x4 – 1

f(0) = -1

f’(x) = 4x3

; f’(0) = 0

f”(x) = 12x2

; f”(0) = 0

f(3)

(x) = 24x ; f(3)

(0) = 0

f(4)

(x) = 24 ; f(4)

(0) = 24

f(n)

(x) = 0 for n ≥ 5.

Thus, for n = 0 and n = 4; the nth degree Taylor polynomial for f at

0 has degree n.

EXERCISE 2-1 10

9

Page 13: Link full download:  ... · Karl Byleen Link full download:

39. f(x) = ln(1 + x) f(0) = 0

f'(x) =

1

f'(0) = 1

1 x

f"(x) =

1

f"(0) = -1

(1 x)2

f

(3)(x) = 2 f

(3)(0) = 2

(1 x)3

Thus, x2

x2

x3

p1(x) = x, p2(x) = x -

, p3(x) = x -

+

.

2 2 3

x p1(x) p2(x) p3(x) f(x)

-0.2 -0.2 -0.22 -0.222667 -0.223144

-0.1 -0.1 -0.105 -0.105333 -0.105361

0 0 0 0 0

0.1 0.1 0.095 0.095333 0.09531

0.2 0.2 0.18 0.182667 0.182322

x

p1(x) - f(x)

p2(x) - f(x)

p3(x) - f(x)

-0.2 0.023144 0.003144 0.000477

-0.1 0.005361 0.000361 0.000028

0 0 0 0

0.1 0.00469 0.00031 0.000023

0.2 0.017678 0.002322 0.000345

Page 14: Link full download:  ... · Karl Byleen Link full download:

40. f(x) = ln(1 + x) f(0) = 0

f'(x) = (1 + x)-1

f'(0) = 1

f"(x) = -(1 + x)-2

f"(0) = -1

f(3)

(x) = 2(1 + x)-3

f(3)

(0) = 2

Thus, 1

1

p (x) = x - x2 + x

3.

3

2 3

Using a graphing utility, we find that

|p3(x) - ln(1 + x)| < 0.1 for -0.654 < x < 0.910.

Page 15: Link full download:  ... · Karl Byleen Link full download:

41. f(x) = ex

f(a) = ea

f'(x) = ex

f'(a) = ea

f"(x) = ex

f"(a) = ea

f(3)

(x) = ex

f(3)

(a) = ea

M M

f(n)

(x) = ex

f(n)

(a) = ea

Thus, (3)

f"(a)

(x - a)2

p n

(x) = f(a) + f'(a)(x - a) + + f (a)

(x - a)3

2! 3! (n)

+ … +

f (a)

(x - a)n n!

= ea + e

a(x - a) + e

a (x - a)

2 + e

a (x - a)

3 +

… + e

a (x - a)

n

2! 3! n!

= ea 1 (x a) 1 (x a)

2 1 (x a)

3 1 (x a)

n

2! 3! n!

= ea n k

1!(x - a)

kk0

42. f(x) = ln x f(a) = ln a

f'(x) = x-1

f'(a) =

1

a

f"(x) = -x

-2 f"(a) = -

1

2

a 1

f(3)

(x) = (-1)4(2!)x

-3 f(3)

(a) = (-1)4(2!)

3

a

f(4)

(x) = (-1)5(3!)x

-4 f(4)

(a) = (-1)5(3!)

1 4

a

1 f(n)

(x) = (-1)n+1

((n - 1)!)x-n

f(n)

(a) = (-1)n+1

((n - 1)!) n

a

Thus,

1

1

p (x) = ln a + 1 (x - a) - (x - a)2 + (x - a)

3

n

2a2

a 3a3

- … + (-1)

n+1 1 (x - a)

n n na

n (1)k 1

= ln a +

(x - a)k

k

k 1 ka

Page 16: Link full download:  ... · Karl Byleen Link full download:

43. f(x) = (x + c)n

f(0) = cn

f'(x) = n(x + c)n-1

f'(0) = ncn-1

f"(x) = n(n - 1)(x + c)n-2

f"(0) = n(n - 1)cn-2

M

f(n-1)

(x) = n!(x + c) f(n-1)

(0) = n!c

f(n)

(x) = n! f(n)

(0) = n!

Thus, n(n 1)

n!

n!xn p (x) = c

n + nc

n-1x + c

n-2x2 + … + cx

n-1 +

n

2! (n 1)! n!

n!

n!

= cn + x +

cn-2x2 + … + xn

(n 1)! 2!(n

n

2)!

= n!

cn-kxk.

k 0 k!(n k)!

Page 17: Link full download:  ... · Karl Byleen Link full download:

44.Let f(x) be a polynomial of degree k, k ≥ 0. Then

f(x) = a 0 + a 1 x + a 2 x2 + a 3 x3 + … + a k xk f(0) = a 0

f'(x) = a 1 + 2a

2 x + 3a

3 x2 + … + ka k xk-1 f'(0) = a 1

xk-2

f"(x) = 2a 2 + 6a 3 x + … + k(k - 1)a k f"(0) = 2a 2 + 2!a 2

f(3)

(x) = 6a 3 + … + k(k - 1)(k - 2)a

k xk-3 f(3)

(0) = 6a 3 = 3!a

3

In general, m!am

for m 0,1,2,K k

f (m) (0) =

0

for m k

f"(0) (3) (n)

Since p (x) = f(0) + f'(0)x + x2 + f (0) x3 + … + f (0) xn

n 2!

3! n!

it follows that pn(x) = f(x) for all n ≥ k.

45.f(x) = ex

f(0) = 1

f’(x) = ex ; f’(0) = 1

f”(x) = ex

; f”(0) = 1

f(k)

(x) = ex ; f

(k)(0) = 1

Therefore,

p (x) = 1 + 1 x + 1 x2 + … + 1 x10 10 1! 2! 10!

p (x) = 1 + 1 x + 1 x2 + … + 1 x

10 + 1 x11

11 1! 2! 10! 11!

Page 18: Link full download:  ... · Karl Byleen Link full download:

For x > 0, ex

> p 11 (x) and hence

1

ex – p (x) > p (x) – p (x) = x11

10 11 10

11!

Take x = 2(11!)1/11

, then

ex – p (x) > 1 (2(11!)

1/11)11

= 211

= 2048. 10

11!

So, there exist values of x for which

|p 10 (x) – e

x| = |e

x – p

10 (x)| ≥ 100.

46.f(x) = 1 = x

-1

f(1) = 1

f (1)

f’(x) = -x-2

; f’(1) = -1 = -1! or = -1

1!

f”(x) = (-1)(-2)x-3

; f”(x) = 2 = 2! or f (1)

= 1 2!

M f(k)

(1)

f(k)

(1) = (-1)kk! or = (-1)

k

k!

Therefore,

p 12 (x) = 1 – (x – 1) + (x – 1)2 -

… - (x – 1)

11 + (x – 1)

12,

and

|p12(x) – f(x)| = 1 (x 1) (x 1)2 L (x 1)

12

1

x

=

1 |x – x(x – 1) + x(x – 1)

2 -

… + x(x – 1)

12 – 1|

x

If we take x = 0.001, then 1

= 1000 and every term involving x on

x

the right-hand side of the above equation is positive and

smaller than x.

Thus,

|p12(x) – f(x)| ≥ 1000(1 – 13x) = 1000(1 – 0.013) = 987.

So there exist values of x ≠ 0 for

which |p12(x) – f(x)| ≥ 100.

EXERCISE 2-1 115

Page 19: Link full download:  ... · Karl Byleen Link full download:

47.ln 1.1

Let f(x) = ln(1 + x)

f’(x) =

1 = (1 + x)

-1

1 x

f”(x) = -(1 + x)-2

f(3)

(x) = 2(1 + x)-3

f(n)

(x) = (n – 1)!(-1)n-1

(1 + x)-n

Rn(x) = f(n 1)(t)xn 1

for some t between 0 and x. (n 1)!

Note that f(n+1)

(t) = n!(-1)n(1 + t)

-(n+1) and hence

|f(n+1)

(t)| = |n!(-1)n(1 + t)

-(n+1)| = n!(1 + t)

-(n+1) < n! for t >

0. Therefore,

|Rn(x)| = f(n 1)(t)xn 1

< n!

x

n 1 =

x

n 1

(n 1)!

n 1

(n 1)!

and

Rn(0.1) < (0.1)

n 1

(n 1)

For n = 4, |R4(0.1)| < (0.1)5

= 0.000 002 < 0.000 005, and hence the

5

polynomial with the lowest degree is p (x) = x - 1 x2 + 1 x

3 - 1 x

4 4

2 3 4

which has degree 4.

1

1

1

ln(1.1) ≈ p (0.1) = 0.1 - (0.1)2 + (0.1)

3 - (0.1)

4

4

2 3 4

≈ 0.095308

CHAPTER 2 REVIEW 209


Top Related