Page 1
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
Page 2
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
Page 3
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
Page 4
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
Page 5
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!
Page 6
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
Page 7
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.
Page 8
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
Page 9
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) =
Page 10
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
Page 11
(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
Page 12
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
Page 13
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
Page 14
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.
Page 15
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
Page 16
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)!
Page 17
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!
Page 18
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
Page 19
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