link full download: …€¦ · 13. f(x) = 4 = (x + 16)1/4 (0) = 2 x 16 f'(x) = 1 (x + 16)-3/4...
TRANSCRIPT
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
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
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
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
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!
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
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.
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
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
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
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
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
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
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.
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
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)!
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!
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
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