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numerical analysis 2nd edition solution.pdf
CHAPTER 0
Fundamentals
EXERCISES 0.1 Evaluating a Polynomial
3(1 + 1
3(5 + 1
3(1 + 1
3(6)))) = 2.
1 (a) P (x) = 1 + x(1 + x(5 + x(1 + x(6)))).
P ( 1
P ( 1
P ( 1
3)2 + 1
3)4 + ( 1
3)4 + ( 1
3) = 6( 1
3)3 + 5( 1
3)3 + 5( 1
3)4 + 4( 1
3)3 − ( 1
3) + 1 = 1 + 1
3) = −3( 1
3) = 2( 1
3 + 1 = 1 + 1
1 (b) P (x) = 1 + x(−5 + x(5 + x(4 + x(−3))))
1 (c) P (x) = 1 + x(0 + x(−1 + x(1 + x(2))))
3)2 − 5( 1
3(−5 + 1
3)2 + 1 = 1 + 1
3(0 + 1
3(−1 + 1
3(1 + 1
2 (a) P (x) = 7+x(−3+x(−2+x(6))); P (− 1
2) = 7+(− 1
2)(−3+(− 1
2 (b) P (x) = 1 + x(−3 + x(1 + x(−3 + x(−1 + x(8)))));
2)(−1 + (− 1
2)(−3 + (− 1
2)(1 + (− 1
2 (c) P (x) = 4 + x(−2 + x(0 + x(0 + x(−2 + x(0 + x(4))))));
2)(0 + (− 1
2)(0 + (− 1
2)(−2 + (− 1
2)2(1))) = 81/64.
P (− 1
P (− 1
2) = 1 + ( 1
2) = 1 + (− 1
2) = 4 + (− 1
2)(−3 + (− 1
2)(−2 + (− 1
2)2(2 + ( 1
3 P ( 1
2)2(−4 + ( 1
3(5 + 1
3(4 + 1
3(−3)))) = 0
3(2)))) = 77/81.
2)(−2+(− 1
2)(8))))) = 45/16.
2)(6))) = 29/4.
2)(0 + (− 1
2)(4)))))) = 79/16.
2 + (5 − 2)( 1
2 + (5 − 3)(− 1
2) = 4 + 1
2) = 4 − 1
2 + (−1 − 2)( 1
2 − 1)(1 + ( 1
2))) = 8
2 − 3)(2)))) = 5
2))) = −4
2 + (−1 − 3)(− 1
2 − 2)(3 + ( 1
4 (a) P (5) = 1 + 5( 1
4 (b) P (−1) = 1 + (−1)( 1
5 (a) P ( 1
2(4 + ( 1
5 (b) P (− 1
2(4 + (− 1
6 (a) P (x) = a0 + x5(a5 + x5(a10 + x5a15)). The three multiplications x2 = x · x, x4 =
x2 · x2, x5 = x4 · x are needed, together with 3 multiplications and 3 additions from the nested
multiplication. Total of 6 multiplications and 3 additions.
6 (b) P (x) = x7(a7 + x5(a12 + x5(a17 + x5(a22 + x5a27)))). The four multiplications x2 =
x · x, x4 = x2 · x2, x5 = x4 · x, x7 = x5 · x2 are needed, together with 5 multiplications and 4
additions from the nested multiplication. Total of 9 multiplications and 4 additions.
2 − 3)(2)))) = 41/4
2 − 1)(1 + (− 1
2 − 2)(3 + (− 1
7 The degree n polynomial with base points is P (x) = c1 + (x − r1)(c2 + (x − r2)(c3 + (x −
r3)(c4 + . . . + (x− rn)cn+1))). The operations needed are n multiplications and 2n additions.
COMPUTER PROBLEMS 0.1
1 The MATLAB command nest(50,ones(51,1),1.00001) gives 51.01275208274999,
differing from (x51 − 1)/(x − 1) with x = 1.00001 by 4.76 × 10−12.
c#2012 Pearson Education, Inc.
2
CHAPTER 0 FUNDAMENTALS
2 The command nest(99,(-1).ˆ(0:99),1.00001) gives −0.00050024507964763. The
equivalent expression (1 − x100)/(1 + x) for x = 1.00001 differs by 1.713 × 10−16.
EXERCISES 0.2 Binary Numbers
1 (a) (64)10 = (26)10 = (1000000)2
1 (b) (17)10 = (16 + 1)10 = (10001)2
1 (c)
Therefore (79)10 = (1001111)2.
1 (d)
79 ÷ 2 = 39 R 1
39 ÷ 2 = 19 R 1
19 ÷ 2 = 9 R 1
9 ÷ 2 = 4 R 1
4 ÷ 2 = 2 R 0
2 ÷ 2 = 1 R 0
1 ÷ 2 = 0 R 1
227 ÷ 2 = 113 R 1
113 ÷ 2 = 56 R 1
56 ÷ 2 = 28 R 0
28 ÷ 2 = 14 R 0
7 R 0
14 ÷ 2 =
3 R 1
7 ÷ 2 =
1 R 1
3 ÷ 2 =
0 R 1
1 ÷ 2 =
Therefore (227)10 = (11100011)2.
2 (a) (1/8)10 = (2−3)10 = (0.001)2
2 (b) (7/8)10 = (2−1 + 2−2 + 2−3)10 = (0.111)2
2 (c) (35/16)10 = (2 + 3/16)10 = (2 + 1/8 + 1/16)10 = (10.0011)2
c#2012 Pearson Education, Inc.
SECTION 0.2 BINARY NUMBERS
3
2 (d)
31/64 × 2 = 31/32 + 0
31/32 × 2 = 15/16 + 1
15/16 × 2 = 7/8 + 1
7/8 × 2 = 3/4 + 1
3/4 × 2 = 1/2 + 1
1/2 × 2 = 0 + 1
Therefore (31/64)10 = (0.011111)2.
3 (a) 10.5 = 10 + 0.5. Integer part: (10)10 = (8 + 2)10 = (1010)2. Fractional part: (0.5)10 =
(0.1)2, so (10.5)10 = (1010.1)2.
3 (b)
Therefore ( 1
3)10 = (0.01)2.
3 (c)
2
3
1
3
2
3
3
7
6
7
5
7
3
7
6
7
+ 0
+ 1
+ 0
+ 1
+ 0
+ 1
+ 1
+ 0
1
3 × 2 =
2
3 × 2 =
1
3 × 2 =
...
5
7 × 2 =
3
7 × 2 =
6
7 × 2 =
5
7 × 2 =
3
7 × 2 =
...
Therefore ( 5
7)10 = (0.101)2.
c#2012 Pearson Education, Inc.
4
CHAPTER 0 FUNDAMENTALS
3 (d) (12.8)10 = (12)10 + (0.8)10; (12)10 = (1100)2.
0.8 × 2 = 0.6 + 1
0.6 × 2 = 0.2 + 1
0.2 × 2 = 0.4 + 0
0.4 × 2 = 0.8 + 0
0.8 × 2 = 0.6 + 1
...
Therefore (12.8)10 = (1100.1100)2.
3 (e) (55.4)10 = (55)10 + (0.4)10; (55)10 = (32 + 16 + 4 + 2 + 1)10 = (110111)2.
0.4 × 2 = 0.8 + 0
0.8 × 2 = 0.6 + 1
0.6 × 2 = 0.2 + 1
0.2 × 2 = 0.4 + 0
0.4 × 2 = 0.8 + 0
...
Therefore (55.4)10 = (110111.0110)2.
3 (f)
0.1 × 2 = 0.2 + 0
0.2 × 2 = 0.4 + 0
0.4 × 2 = 0.8 + 0
0.8 × 2 = 0.6 + 1
0.6 × 2 = 0.2 + 1
0.2 × 2 = 0.4 + 0
...
Therefore (0.1)10 = (0.00011)2.
4 (a)
11.25 = 11 + 0.25. Integer part: (11)10 = (8 + 2 + 1)10 = (1011)2. Fractional part:
(0.25)10 = (0.01)2, so (10.25)10 = (1011.01)2.
c#2012 Pearson Education, Inc.
SECTION 0.2 BINARY NUMBERS
5
4 (b)
Therefore ( 2
3)10 = (0.10)2.
4 (c)
1
3
2
3
1
3
1
5
2
5
4
5
3
5
1
5
+ 1
+ 0
+ 1
+ 1
+ 0
+ 0
+ 1
+ 1
2
3 × 2 =
1
3 × 2 =
2
3 × 2 =
...
3
5 × 2 =
1
5 × 2 =
2
5 × 2 =
4
5 × 2 =
3
5 × 2 =
...
Therefore ( 3
5)10 = (0.1001)2.
4 (d) (3.2)10 = (3)10 + (0.2)10; (3)10 = (11)2.
0.2 × 2 = 0.4 + 0
0.4 × 2 = 0.8 + 0
0.8 × 2 = 0.6 + 1
0.6 × 2 = 0.2 + 1
0.2 × 2 = 0.4 + 0
...
Therefore (3.2)10 = (11.0011)2.
c#2012 Pearson Education, Inc.
6
CHAPTER 0 FUNDAMENTALS
4 (e) (30.6)10 = (30)10 + (0.6)10; (30)10 = (16 + 8 + 4 + 2)10 = (11110)2.
0.6 × 2 = 0.2 + 1
0.2 × 2 = 0.4 + 0
0.4 × 2 = 0.8 + 0
0.8 × 2 = 0.6 + 1
0.6 × 2 = 0.2 + 1
...
Therefore (30.6)10 = (11110.1001)2.
4 (f) (99.9)10 = (99)10 + (0.9)10; (99)10 = (64 + 32 + 2 + 1)10 = (1100011)2.
0.9 × 2 = 0.8 + 1
0.8 × 2 = 0.6 + 1
0.6 × 2 = 0.2 + 1
0.2 × 2 = 0.4 + 0
0.4 × 2 = 0.8 + 0
0.8 × 2 = 0.6 + 1
...
Therefore (99.9)10 = (1100011.11100)2.
5 (π)10 = (3)10 + (π − 3)10
0.14159265 × 2 = 0.28318531 + 0
0.28318531 × 2 = 0.56637061 + 0
0.56637061 × 2 = 0.13274123 + 1
0.13274123 × 2 = 0.26548246 + 0
0.26548246 × 2 = 0.53096491 + 0
0.53096491 × 2 = 0.06192983 + 1
0.06192983 × 2 = 0.12385966 + 0
0.12385966 × 2 = 0.24771932 + 0
0.24771932 × 2 = 0.49543864 + 0
0.49543864 × 2 = 0.99087728 + 0
0.99087728 × 2 = 0.98175455 + 1
0.98175455 × 2 = 0.96350910 + 1
0.96350910 × 2 = 0.92701821 + 1
...
c#2012 Pearson Education, Inc.
SECTION 0.2 BINARY NUMBERS
7
Therefore (π)10 = (11.0010010000111 . . .)2.
6 (e)10 = (2)10 + (e − 2)10
0.71828183 × 2 = 0.43656366 + 1
0.43656366 × 2 = 0.87312731 + 0
0.87312731 × 2 = 0.74625463 + 1
0.74625463 × 2 = 0.49250926 + 1
0.49250926 × 2 = 0.98501851 + 0
0.98501851 × 2 = 0.97003702 + 1
0.97003702 × 2 = 0.94007404 + 1
0.94007404 × 2 = 0.88014809 + 1
0.88014809 × 2 = 0.76029617 + 1
0.76029617 × 2 = 0.52059234 + 1
0.52059234 × 2 = 0.04118468 + 1
0.04118468 × 2 = 0.08236937 + 0
0.08236937 × 2 = 0.16473874 + 0
...
Therefore (e)10 = (10.1011011111100 . . .)2.
implies x = 1
7 (f)
7 (h)
2 + 1
8)10 = (93/8)10.
Therefore (110.10)2 = (6 + 2
3. Therefore (10111.01)2 = (23 + 1
3)10 = (70/3)10.
7 (a) (1010101)2 = (20 + 22 + 24 + 26)10 = (1 + 4 + 16 + 64)10 = (85)10
7 (b) (1011.101)2 = (23 + 21 + 20 + 2−1 + 2−3)10 = (11 + 1
7 (c) (10111.01)2 = (24 + 22 + 21 + 20)10 + (0.01)2. Set x = (0.01)2. Then 22x− x = (01)2 = 1
7 (d) (110.10)2 = (22 + 21)10 + (0.10)2. Set x = (0.10)2. Then 22x − x = (10)2 implies x = 2
3.
7 (e)
(10.110)2 = (2)10 + (0.110)2. Set x = (0.110)2. Then 23x − x = (110)2 = 6 implies
2 )10, where x = (0.101)2. Since
(110.1101)2 = (6)10 + ( 1
8(0.1101)2. Set x = (0.1101)2. Then 24x−x = (1101)2 =
(111.1)2 = (7)10 + (0.1)2 = (7)10 + x, where x = (0.1)2. Since 21x − x = (1)2, x = 1,
23x − x = (101)2 = 5, x = 5/7. Therefore (110.1101)2 = ( 13
13, implying that x = 13
4 + 1
7)10 = (20/7)10.
2)10 + (0.0101)2 = ( 13
x = 6/7. Therefore (10.110)2 = (2 + 6
15. Therefore (10.0101101)2 = ( 9
7 (g) (10.0101101)2 = (2)10 +( 1
13
15)10 = (283/120)10.
2 + 5
7
1
2)10 = (48/7)10.
2 + x
8
3)10 = (20/3)10.
4)10 + 1
and (111.1)2 = (7 + 1)10 = (8)10.
8 (a) (11011)2 = (20 + 21 + 23 + 24)10 = (1 + 2 + 8 + 16)10 = (27)10
8 (b) (110111.001)2 = (25 + 24 + 22 + 21 + 20 + 2−3)10 = (55 + 1
8)10.
8 (c)
(111.001)2 = (22 + 21 + 20)10 + (0.001)2. Set x = (0.001)2. Then 23x − x = (001)2 = 1
implies x = 1/7. Therefore (111.001)2 = (7 + 1/7)10.
c#2012 Pearson Education, Inc.
8
CHAPTER 0 FUNDAMENTALS
8 (d) (1010.01)2 = (23 + 21)10 + (0.01)2. Set x = (0.01)2. Then 22x− x = (01)2 implies x = 1
3.
8 (e)
(10111.10101)2 = (10111.10)2 = (24 + 22 + 21 + 20)10 + (0.10)2. Set x = (0.10)2. Then
Therefore (1010.01)2 = (10 + 1
3)10 = (10 + 1/3)10.
8 (f) (1111.010001)2 = (15)10 + (1/4)10 + 1
22x − x = (10)2 = 2 implies x = 2/3. Therefore (10111.10101)2 = (23 + 2
Since 23x − x = (001)2 = 5, x = 1/7. Therefore (1111.010001)2 = (15 + 1/4 + 1
(15 + 15/56)10.
8(0.001)2 = (15 + 1/4 + x
8 )10, where x = (0.001)2.
1
7)10 =
3)10.
8
EXERCISES 0.3 Floating Point Representation of Real Numbers
1 (a) ( 1
1 (b) ( 1
4)10 = (0.01)2; fl( 1
3)10 = (0.01)2 =
4) = +1.0 × 2−2.
+1. 0101010101010101010101010101010101010101010101010101 0101 . . . × 2−2.
The Rounding to Nearest Rule says to round down when the 53rd bit is 0.
3) = +1. 0101010101010101010101010101010101010101010101010101 × 2−2.
fl( 1
1 (c) ( 2
3)10 = (0.10)2 =
+1. 0101010101010101010101010101010101010101010101010101 0101 . . . × 2−1.
3) = +1. 0101010101010101010101010101010101010101010101010101 × 2−1.
fl( 2
+1. 1100110011001100110011001100110011001100110011001100 1100 . . . × 2−1.
The Rounding to Nearest Rule says to round up since the 53rd bit is nonzero, and further bits
are nonzero.
fl(0.9) = +1. 1100110011001100110011001100110011001100110011001101 × 2−1.
1 (d) (0.9)10 = (0.11100)2 =
2 (a) (9.5)10 = (1001.1)2; fl(9.5) = 1.0011 × 23.
2 (b) (9.6)10 = (1001.1001)2 = 1.0011001 × 23 =
+1. 0011001100110011001100110011001100110011001100110011 0011 . . . × 23.
fl(9.6) = +1. 0011001100110011001100110011001100110011001100110011 × 23.
2 (c) (100.2)10 = (1100100.0011)2 = 1.1001000011 × 26 =
+1. 1001000011001100110011001100110011001100110011001100 1100 . . . × 26.
fl(100.2) = +1. 1001000011001100110011001100110011001100110011001101 × 26.
2 (d) ! 44
7"10 = (6 + 2
+1. 1001001001001001001001001001001001001001001001001001 0010 . . . × 22.
fl! 44
7" = +1. 1001001001001001001001001001001001001001001001001001 × 22.
3 Note that fl(5) = 1.01× 22. Adding 1 as bit 3, 4, . . . , 52 of the mantissa will not incur rounding
7)10 = (110.010)2 =
error. These correspond to 2−k for k = 1, 2, . . . , 50.
4 Note that fl(19) = 1.0011× 24. Adding 1 to bit 52 of the mantissa, corresponding to 19 + 2−48,
will not be rounded away, and so 48 is the largest such k.
c#2012 Pearson Education, Inc.
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