Karl

Byleen

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) 1

= (1 + x)-1

f'(x) =

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

f”(x) = e

(3) -x

f (x) = -e

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

= 3(1 + 3x)

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)e5x

= 52e5x

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

1

= (2 + x)-1

6. f(x) =

2 x f’(x) = (-1)(2 + x)

f”(x) = (-1)(-2)(2 + x) = 2(2 + x)

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

f(4)

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

= 24(2 + x)-5

7. f(x) = e-x

f(0) = e-0

= 1 -x -0

= -1

f'(x) = -e f'(0) = -e

f"(x) = e-x

f"(0) = e-0

= 1

f(3)

(x) = -e-x

f(3)

(0) = -e-0

= -1

f(4)

(x) = e-x

f(4)

(0) = e-0

= 1 Using 2,

p (x) = f(0) + f'(0)x + f"(0) (3)

x + f (0) x +

4 2! 3!

Thus, 1

1

1

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

4 2! 3! 4!

8. f(x) = e4x

f(0) = 1 f'(x) = 4e f'(0) = 4

f"(x) = 16e f"(0) = 16

f(3)

(x) = 64e4x

f(3)

(0) = 64

f"(0) 2 (3)

3 f (0)

Thus, p 3 (x) = f(0) + f'(0)x +

x

+

x

2! 3!

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

2 +

2!

f(x) = (x + 1)

3,

3!

9. f(0) = 1

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

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

f (x) = 6, f (0) = 6

f (x) = 0 f (0) = 0

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

2 + x

3

4 2! 3!

(4)

f (0) x

4!

1 x - 1 x + 1 x

2 6 24

323 x

3

EXERCISE 2-1 97

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

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

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

f(3)

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

(0) = -24 Thus,

12

24

4x + 6x2 - 4x

3

x3 = 1 -

p 3 (x) = 1 - 4x + 2! x2 - 3! 11. f(x) = ln(1 + 2x) f(0) = ln(1) = 0

1 2 2

f'(x) =

(2) =

1 2x 1 2x 4

f'(0) = 1 2 0 -2

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

= 2 f"(0) = -4

(1 2x)

(3) -3 16 (3)

f

(x) = 8(1 + 2x)

(2) = (1 2x) f (0) = 16

(3)

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

2 + f (0) 3

3

2!

x

3! x

Thus, p (x) = 0 + 2x - 4 x2 + 16 x

3 = 2x - 2x

2 + 8 x3

3!

3 2! 3

12. f(x) = 3

= (x + 1)1/3

f(0) = 3

= 1

x 1 1

1 1

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

f"(x) = 9(x 1)

(3) 10

f (x) = 27(x 1)8/3

f"(0) = -9

(3) 10

f (0) = 27

p 3(x) = f(0) + f'(0)x +

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

9

3 3 2!

f"(0)

(3) f (0)

x2 + x3

2! 3!

x + 10 x = 1 + 1 x -

1 x + 5 x

27 3! 3 9 81

13. f(x) = 4

= (x + 16)

1/4 f(0) = 2 x 16

1 1

f'(x) = -3/4

f'(0) =

4(x + 16) 32

f"(x) = - 3 (x + 16) f"(0) = - 3 16

1

3 2048

p (x) = 2 + x - x

2 32 4096

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

f'(x) = 4x f'(0) = 0

f"(x) = 12x f"(0) = 0

f(3)

(x) = 24x f(3)

(0) = 0

f"(0) (3)

Using 2, p (x) = f(0) + f'(0)x + 2 + f (0) 3

3

2! x

3! x

Thus, p3(x) = -1

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

4| = |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 + x + f (0) x

4

24

2!

3!

Thus, p 4(x) = -1 + x

4

= -1 + x

4

= x

4

- 1 = f(x)

4!

and |p4(x) - f(x)| = 0 < 0.1 for all x. 15. (A) f(x) = x f(0) = 0

f'(x) = 5x f'(0) = 0

f"(x) = 20x f"(0) = 0

f (x) = 60x f (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

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 f"(4) = 2

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

2 2!

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

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

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

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

f(4)

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

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

4 2!

1

(x - 1)2

= -(x - 1) -

2

f(1) = 0

f'(1) = -1

f"(1) = -1

f (1) = -2

f(4)

(1) = -6

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

4

3! 4!

1 1

-

3

-

4

3 (x - 1) 4(x - 1)

19. f(x) = e-2x

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

f"(x) = 4e-2x

(3) - 2x

f (x) = -8e

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 + x + f (0) x

3

4 x2 -

2!

3!

- 4 x3.

Thus, p (x) = 1 - 2x + 8 x3 = 1 - 2x + 2x

2

3 3 2! 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

20.

= x1/2

f(x) = x f(1) = 1

1

x-1/2 1

f'(x) =

f'(1) =

2 2

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

4 4

3 3

f(3)

(x) =

x-5/2

f(3)

(1) =

8 8 15 15

f(4)

(x) = -

f(4)

(1) = -

16 x-7/2

16 Thus,

(3)

(4)

f"(1)

f (1) f (1)

p4(x) = f(1) + f'(1)(x - 1) +

(x - 1)2

+

(x - 1)3

+

(x - 1)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 1 5

= 1 +

(x - 1) -

(x - 1)2 +

3

-

(x - 1)4.

2 8 16 (x - 1) 128 Now,

1.2 = f(1.2) ≈ p4(1.2) (1.2 - 1)

2 +

(1.2 - 1)

3 (1.2 - 1)

4 = 1

+ 1 (1.2 - 1) - 1 1 - 5

2 8 16 128

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

3- 5 (0.2)

4

2 8 16 128 = 1.0954375.

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

4 x -2 -2

f'(x) = -1(4 - x)-3(-1) = (4 - x) -3 f"(x) = -2(4 - x) (-1) = 1·2(4 - x)

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

1 x

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

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

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) 2 -1 -1

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

6 x 6 x -2 -2

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

= -(6 - x)

f(4)

(x) = 1·2·3(6 - x)-4

(-1) = -1·2·3(6 -

x)-4

f(n)

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

26. f(x) = ex/2

1 x/2

f'(x) = 2 e

1

2 1 f"(x) =

ex/2 = 2 ex/2

2 2

f(3)

(x) = 1 ex/2 3

2 (n )

1

f (x) = n ex/2

27. From Problem 31, 1 1 2! (3) 3! (n) n! f(0) = , f'(0) =

, f"(0) = , f (0) = , … , f (0) =

4 2 3 4 n 1

4 4 4 4

Thus,

1

1

1 x .

p (x) = 1

+ x + x + 1 x + … +

n 4 42

43

44

4n 1

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

From problem 32, f (x) = 4(n!)(-1) (1 + x) and

hence f(n)

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

(n) n

The coefficient of xn f (0)

4(n!)(1)

= 4(-1)n. is

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 +

n

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

1) From problem 34,

32 x2 + 3

3 x3 + … + 3

n xn.

2! 3! n!

(n )

n+1

n

-n

(n) n+1

n

f (0) 2

f (x) = (-1) 2 ((n - 1)!)(2x + 1) and n! = (-1) n

Thus, p (x) = 2x - 22

x2 + 23 x3 - 24 x4 + … + (-1)

n+1 2n xn

n 2 3 4 n

31. From Problem 35, 1

1

(3)

2!

(n)

(n 1)!

f(0) = ln 6, f'(0) = -

, f"(0) = -

, f

(0) = -

, …, f

(0) = -

6 2 3 n

6 6 6

Thus, 1

1

1

1

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

x3 - … - x .

2

3

n

6

2

3

n n

6 6 6

EXERCISE 2-1 105

32. f(x) = ex/2

From problem 36, f(n)

(x) =

1 (n) 1

. Thus,

e x/2 and f (0) =

2 n!2

1 1 x2 +

1 x3 1

x2 +x

n p (x) = 1 +

x +

+ … +

2!2 2

3!2 3

n!2 n

n 2

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

f(1) = 2

x f'(x) = -2x

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)

Therefore f(n) (1) n

and

= 2(-1)

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.

a

f(x) = ln x

f(1) = 0 = f(1) = 0

0

1

a1 = f'(1) = 1

f'(x) = x 1

f'(1) = 1 f"(1) 1 1

x2

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

f(3)

(x) =

2

f(3)

(1) = 2

(3)

2

1

f (1)

x 3 a = = =

3 3! 3! 3

(4)

2

3

(4)

(4) 3!

1

4

a f (1)

f (x) = -

f (1)

= -3!

x

4 = 4! = -4! = - 4

f (n)

(x) =

f (n)

(1) = a

n =

(1)n 1

(n 1)! (-1)

n+1(n - 1)!

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

= (1) (n

= (1)

n! n!

xn n

Step 4. The nth degree Taylor polynomial is: (1)n 1

1 (x - 1) 2

+ 1 (x - 1) 3 -

1

(x - 1) n .

p (x) = (x - 1) - (x - 1)4 + … +

n 2 34 n

35. f(x) = ex

(n)

f(n)

(x) = ex

1

f (2)

and

=

n! n!e

Thus, p (x) = 1

+ 1

(x + 2) + 1

(x + 2)2

+ … + 1

(x + 2)n

e2 e2 2 2

n 2!e n!e

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) = 5x + 6x + 16x ; f’(0) = 0

f”(x) = 20x + 12x + 16 ; f”(0) = 16

f (x) = 60x + 12 ; f (0) = 12

f(4)

(x) = 120x ; f(4)

(0) = 0 Thus,

p 4(x) = 1 + 8x2 + 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) = 6x + 6x ; f’(0) = 0

f”(x) = 30x + 12x ; f”(0) = 0

f (x) = 120x + 12 ; f (0) = 12

f (x) = 360x ; f (0) = 0

f (x) = 720x ; f (0) = 0

f (x) = 720 ; f (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) = 4x ; f’(0) = 0

f”(x) = 12x ; f”(0) = 0

f (x) = 24x ; f (0) = 0

f (x) = 24 ; f (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

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

f'(x) =

f'(0) = 1

1 x

1

f"(x) = (1 x) f"(0) = -1 (3) 2 (3)

f (x) = f (0) = 2

(1 x)3

Thus,

x2

x2

x3

p1(x) = x, p 2(x) = x - 2 , p3(x) = x - 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

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

f'(x) = (1 + x) f'(0) = 1

f"(x) = -(1 + x) f"(0) = -1

f(3)

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

f(3)

(0) = 2

Thus,

1

2

1

3

p (x) = x - x + x .

3 2 3 Using a graphing utility, we find that

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

41. f(x) = ex

f(a) = ea

f'(x) = e f'(a) = e

f"(x) = e f"(a) = e

f(3)

(x) = ex

f(3)

(a) = ea

M M

f(n)

(x) = ex

f(n)

(a) = ea

Thus, (3)

(a)

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

(x - a)2 + f

2! 3! (x - a)

3

n (n)

(a)

+ +

f

(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) 1 (x a) 1 (x a)

n

1

2! 3! n!

= e

a n k (x - a) k 0

k !

42. f(x) = ln x -1

f'(x) = x

f(3)

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

-3

f(4)

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

-4

f( n)

(x) = (-1)n+1

((n - 1)!)x-n

Thus,

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

n a 2a2

n (1)k 1

f(a) = ln a

1

f'(a) = a

f"(a) = -

1

2

a

f(3)

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

1 3

a

f(4)

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

1 4

a

f (n)

(a) = (-1) n+1

((n - 1)!) 1

n

a

(x - a)2 +

1 (x - a)

3

3a

n+1

n - … + (-1)

1 (x - a) n

na

= ln a + k (x - a) k 1 ka

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

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

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

f(0) = cn

f'(0) = nc

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

M

f (x) = n!(x + c) f (0) = n!c

f(n)

(x) = n! f(n)

(0) = n!

Thus, x + n(n 1) c

n! cx

n! p (x) = c + nc x + … + +x

n

n n!

2! n!

(n 1)! n!

= c +

x +

cn-2x2 + … + x

(n 1)! 2!(n

2)!

=

n n!

cn-kxk.

k 0 k!(n k)!

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

f(x) = a 0 + a 1x + a 2 x2 + a3x3

+ … + a k xk f(0) = a 0

f'(x) = a 1 + 2a 2x + 3a 3x2 + … + ka kxk-1 f'(0) = a 1 f"(x) = 2a 2 + 6a3x + … + k(k - 1)a kxk-2 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) =

for m k

0 (3)

(n)

f"(0)

Since p (x) = f(0) + f'(0)x + 2+ f (0) 3 + … + f (0) n x x x

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 x + … + 1 x + 1x11

11

1! 2! 10! 11!

For x > 0, ex > p 11 (x) and hence

ex – p (x) > p (x) – p (x) = 1 x 11

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) – ex| = |e

x – p 10 (x)| ≥ 100.

46.f(x) =

1 = x

-1

f(1) = 1

f’(x) = -x-2

; f’(1) = -1 = -1! or f (1) = -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

1

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

12

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

47.ln 1.1

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

= (1 + x)-1

f’(x) =

1 x f”(x) = -(1 + x)

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,

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

<

n!

x

n 1

=

x

n 1

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

and

(0.1)n 1

R (0.1) < n (n 1)

(0.1)5

For n = 4, |R (0.1)| < 4 = 0.000 002 < 0.000 005, and hence the

5 x2 +

x3 -

x4

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

4 2 3 4 which has degree 4.

ln(1.1) ≈ p (0.1) = 0.1 - 1 (0.1) + 1 (0.1) - 1 (0.1)

4 2 3 4 ≈ 0.095308 CHAPTER 2 REVIEW 209