Solution manual for numerical analysis 2nd edition by sauer

× 2 −1.The Rounding to Nearest Rule says to round up since the 53rd bit is nonzero, and further bitsare nonzero.. , 52 of the mantissa will not incur rounding error... After subtracting

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 4additions from the nested multiplication Total of 9 multiplications and 4 additions

7 The degree n polynomial with base points is P (x) = c1+ (x − r1)(c2+ (x − r2)(c3 + (x −

r3)(c4+ + (x − r n )c n+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

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

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

SECTION 0.2 BINARYNUMBERS 3

3 × 2 = 1

3+ 11

3 × 2 = 2

3+ 0

7 × 2 = 6

7+ 06

7 × 2 = 57+ 15

7 × 2 = 37+ 13

7 × 2 = 67+ 0

Therefore (5

7)10= (0.101)2

SECTION 0.2 BINARYNUMBERS 5

4 (b)

2

3 × 2 = 1

3+ 11

3 × 2 = 2

3+ 02

3 × 2 = 1

3+ 1

5 × 2 = 2

5+ 02

5 × 2 = 4

5+ 04

5 × 2 = 3

5+ 13

5 × 2 = 1

5+ 1

SECTION 0.2 BINARYNUMBERS 7

8 (d) (1010.01)2 = (23+ 21)10+ (0.01)2 Set x = (0.01)2 Then 22x − x = (01)2 implies x = 1

(15 + 15/56)10.EXERCISES 0.3 Floating Point Representation of Real Numbers

3) = +1 0101010101010101010101010101010101010101010101010101 × 2 −1

1 (d) (0.9)10 = (0.11100)2 =

+1 1100110011001100110011001100110011001100110011001100 1100 × 2 −1.The Rounding to Nearest Rule says to round up since the 53rd bit is nonzero, and further bitsare nonzero

"

= +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

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.

SECTION 0.3 FLOATINGPOINT REPRESENTATION OFREALNUMBERS 9

5 (a) 1 + (2−51+ 2−53) =

+1 0000000000000000000000000000000000000000000000000010 1 × 20.fl(1 + (2−51+ 2−53)) =

+1 0000000000000000000000000000000000000000000000000010 ×20,using the ing to Nearest Rule Therefore fl((1 + (2−51+ 2−53))− 1) =

Round-. 0000000000000000000000000000000000000000000000000010

= 1 0000000000000000000000000000000000000000000000000000 × 2 −51= 2−51

5 (b) 1 + (2−51+ 2−52+ 2−53) =

+1 0000000000000000000000000000000000000000000000000011 1 × 20.fl(1 + (2−51+ 2−52+ 2−53)) =

+1 0000000000000000000000000000000000000000000000000100 ×20,using the ing to Nearest Rule Therefore fl((1 + (2−51+ 2−52+ 2−53))− 1) =

Round-. 0000000000000000000000000000000000000000000000000100

= 1 0000000000000000000000000000000000000000000000000000 × 2 −50= 2−50

6 (a) 1 + (2−51+ 2−52+ 2−54)

= +1 0000000000000000000000000000000000000000000000000011 01 × 20.fl(1 + (2−51+ 2−52+ 2−54)) =

+1 0000000000000000000000000000000000000000000000000011 ×20,using the ing to Nearest Rule Therefore fl((1 + (2−51+ 2−52+ 2−54))− 1) =

+1 0000000000000000000000000000000000000000000000000011 ×20,using the ing to Nearest Rule Therefore fl((1 + (2−51+ 2−52+ 2−60))− 1) =

Round-. 0000000000000000000000000000000000000000000000000011 =

1 1000000000000000000000000000000000000000000000000000 × 2 −51

= 2−51+ 2−52 = 3"mach

7 (a) (8)10 = (1000.)2 = 1.0 ×23 The biased exponent is 3+1023 = 1026, which is 210+2 The

sign is 0 (positive), so the sign/exponent is represented by the binary string 0100 0000 0010

The mantissa is 52 zeros, so the machine representation is the 64 bits

0100 0000 0010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

or 4020000000000000 in hex format

7 (b) (21)10 = (10101.)2 = 1.0101 × 24 The biased exponent is 4 + 1023 = 1027 = 210+ 3,

represented by 100 0000 0011 The machine representation is

0100 0000 0011 0101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

or 4035000000000000 in hex format.

7 (c) (1/8)10 = 1.0 × 2 −3 The biased exponent is −3 + 1023 = 1020 = 210− 4, represented by

011 1111 1100 The machine representation is

0011 1111 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

or 3fc0000000000000 in hex format

7 (d) (1/3)10 = 1.01 × 2 −2 , and after rounding down, fl(1/3) = 1.0101 0101 × 2 −2 The

biased exponent is −2 + 1023 = 1021 = 210− 3, represented by 011 1111 1101 The machine

representation is

0011 1111 1101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101

or 3fd5555555555555 in hex format

7 (e) (2/3)10 = 1.01 × 2 −1 , and after rounding down, fl(1/3) = 1.0101 0101 × 2 −1 The

biased exponent is −1 + 1023 = 1022 = 210− 2, represented by 011 1111 1110 The machine

representation is

0011 1111 1110 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101

or 3fe5555555555555 in hex format

7 (f) (0.1)10 = 1.1001 × 2 −4 , and after rounding up, fl(0.1) = 1.1001 1001 1010 × 2 −4 The

biased exponent is −4 + 1023 = 1019 = 210− 5, represented by 011 1111 1011 The machine

representation is

0011 1111 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1010

or 3fb999999999999a in hex format

7 (g) (−0.1)10 =−1.1001 × 2 −4 , and after rounding, fl(−0.1) = −1.1001 1001 1010 × 2 −4

The biased exponent is −4 + 1023 = 1019 = 210 − 5, represented by 011 1111 1011 The

machine representation is

1011 1111 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1010

or bfb999999999999a in hex format

7 (h) (−0.2)10 =−1.1001 × 2 −3 , and after rounding, fl(−0.2) = −1.1001 1001 1010 × 2 −3

The biased exponent is −3 + 1023 = 1020 = 210 − 4, represented by 011 1111 1100 The

machine representation is

1011 1111 1100 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1010

or bfc999999999999a in hex format

8 Yes Yes No, under chopping, 1/3 + 2/3 = 1 − "mach

9 (a) (7/3)10 = 1.0010 × 21, and after rounding, fl(7/3) = 1.0010 1010 1011 × 21 (4/3)10=

1.01 × 20, and after rounding, fl(4/3) = 1.01 0101 0101 × 20 Subtracting gives

1 0010101010101010101010101010101010101010101010101011 0 × 21

− 0 1010101010101010101010101010101010101010101010101010 1 × 21

= 0 1000000000000000000000000000000000000000000000000000 1 × 21

SECTION 0.3 FLOATINGPOINT REPRESENTATION OFREALNUMBERS 11that is normalized to

= 1 0000000000000000000000000000000000000000000000000001 × 20,

which is 1 + "mach After subtracting 1, the result is that the double precision floating point

version of (7/3 − 4/3) − 1 is "mach

9 (b) (4/3)10 = 1.01 × 20, and after rounding, fl(4/3) = 1.01 0101 0101 × 20 (1/3)10 =

1.01 × 2 −2 , and after rounding, fl(1/3) = 1.01 0101 0101 × 2 −2 Subtracting gives

11 The associative law of addition fails for floating point addition with the Rounding to Nearest

Rule, for example, because 1 + ("mach/2 + "mach/2) = 1 + "mach > 1 , while (1 + "mach/2) +

"mach/2 = 1 , because 1 + "mach/2 = 1

12 (a) fl (1/3) = 1.0101 01 × 2 −2, with relative rounding error of 2−54 < "mach/2 = 2 −53

12 (b) fl (3.3) = 1.101001100110 0110×21, 3.3 − fl (3.3) = 0.4×2 −51with relative rounding

error of 8"mach/33

12 (c) fl (9/7) = 1.010010 0100101 × 20, fl(9/7) − 9/7 = 3"mach/7, with relative rounding

error of "mach/3

13 (a) 2, represented by 010 0 (b) 2 −511 , represented by 0010 0 (c) 0, represented by 10 0.

When bit 4 through 12 is the nonzero bit, the floating point number is positive but less than

2−511 When bit 13 through 64 is the nonzero bit, the number is positive and subnormal, soless than 2−511

14 (a) 0 (b) 2−51(c) 2−51

15(a) (8.3)10 = 1.00001001 × 23, and rounded, fl(8.3) = 1.0000 1001 1001 1001 1010 × 23.

(7.3)10 = 1.1101001 × 22, and rounded, fl(7.3) = 1.1101 0011 0011 0011 0011 × 22.Subtracting gives

SECTION 0.4 LOSS OFSIGNIFICANCE 13

16 (a) fl (11/4) = 1.011 × 21, with rounding error of 0

16 (b) fl (2.7) = 1.010110011001 100110010 × 21, fl (2.7) − 2.7 = 4"mach/5with relative

rounding error of 8"mach/27

16 (c) fl (10/3) = 1.1010 1011 × 21, fl(10/3) − 10/3 = 2"mach/3, with relative rounding error

of "mach/5.EXERCISES 0.4 Loss of Significance

1 (a) For x near 2πn for integer n, sec x ≈ 1, and the numerator exhibits subtraction of nearly

equal numbers An algebraically equivalent expression avoids the difficulty:

eliminates the loss of significance

1 (c) For x near 0, there is subtraction of nearly equal numbers Using common denominators

eliminates the problem:

2 (a) p = 8

2 (b) p = 5

SECTION 0.5 REVIEW OFCALCULUS 15

3 Since a is large and negative, the expression represents subtraction of nearly equal numbers.

Multiply numerator and denominator by the conjugate:

5 Set x = 3344556600 and y = 1.2222222 The difference between the lengths of the hypotenuse

and the longer leg is

where we have rewritten the expression to eliminate the subtraction of nearly equal numbers

Although calculating the leftmost expression in double precision yields no correct significant

digits, the rightmost expression gives the correct answer 2.23322 × 10 −10.EXERCISES 0.5 Review of Calculus

1 (a) Since f(0)f(1) = (1)(−2) < 0, there exists c between 0 and 1 such that f(c) = 0 by the

Intermediate Value Theorem

1 (b) Since f(0)f(1) = (1)(−9) < 0, f(c) = 0 for some c between 0 and 1 as in (a).

1 (c) Since f(0)f(1/2) = (1)(−1/2) < 0, f(c) = 0 for some c between 0 and 1/2 by the

Intermediate Value Theorem, thus 0 ≤ c ≤ 1.

1

2.

Since f(x) = x2, this implies c2 = 1/2, or c = 1/ √2

3 (c) According to the Mean Value Theorem for Integrals, there exists c between 0 and 1

4 (a) P (x) = 1 + x2

4 (b) P (x) = 1 −25

2x2

4 (c) P (x) = 1 − x + x2

5 (a) The derivatives evaluated at x = 0 are f(0) = 1, f " (0) = 0, f "" (0) = 2, f """(0) =

0, f (iv)(0) = 12, and f(v)(0) = 0 Then the degree 5 Taylor polynomial is P (x) = 1+x2+1

2x4

5 (b) The derivatives evaluated at x = 0 are f(0) = 1, f " (0) = 0, f ""(0) = −4, f """(0) =

0, f (iv)(0) = 16, and f(v)(0) = 0 The degree 5 Taylor polynomial is P (x) = 1 − 2x2+ 2

3x4

5 (c) The derivatives at x = 0 are f(0) = 0, f " (0) = 1, f ""(0) =−1, f """ (0) = 2, f (iv)(0) =−6,

and f (v)(0) = 24 The degree 5 Taylor polynomial is P (x) = x − 1

2x2+ 13x3− 1

4x4+15x5

5 (d) The derivatives at x = 0 are f(0) = 0, f " (0) = 0, f "" (0) = 2, f """ (0) = 0, f (iv)(0) = −8,

and f (v)(0) = 0 The degree 5 Taylor polynomial is P (x) = x2 −1

7 (a) The derivatives at x = 1 are f(1) = 0, f " (1) = 1, f ""(1) = −1, f """(1) = 2, and f(iv)(1) =

−6 The degree 4 Taylor polynomial is P (x) = x − 1 −1

2(x − 1)2+13(x − 1)3− 1

4(x − 1)4

7 (b) f(0.9) can be approximated by P (0.9) = −0.1053583 Likewise, f(1.1) ≈ P (1.1) =

0.0953083.

7 (c) The remainder term is (x − 1)5/(5c5), where c lies between x and 1 At x = 0.9, the error

is (0.1)5/(5c5)≤ (0.1)5/(5(0.9)5)≈ 0.000003387, where the upper bound results from

eval-uating c at the worst case c = 0.9 At x = 1.1, the error is (0.1)5/(5c5)≤ (0.1)5/(5(1.0)5)≈

0.000002 On the basis of the remainder, we predict smaller error at x = 1.1.

7 (d) The error at x = 0.9 is |f(0.9) − P (0.9)| = 0.00000218, and the error at x = 1.1 is

|f(1.1) − P (1.1)| = 0.00000185.

8 (a) P (x) = 1 − x2/2 + x4/24

8 (b) 0.000326

9 The degree one Taylor polynomial is P (x) = 1 +1

2x , with Taylor remainder E = x2/(8(1 + c) 3/2)for c between x and 0 Setting x = 0.02, E ≤ (0.02)2/(8(1) 3/2 ) = 0.00005 The actual

values are√ 1.02 ≈ 1.0099505 and 1 + 1

2(0.02) = 1.01, which is a difference of 0.0000495, slightly less than the upper bound E.

CHAPTER 1

Solving Equations

EXERCISES 1.1 The Bisection Method

1 (a) Check that f(x) = x3−9 satisfies f(2) = −1 and f(3) = 27−9 = 18 By the Intermediate

Value Theorem, f(2)f(3) < 0 implies the existence of a root between x = 2 and x = 3.

1 (b) Define f(x) = 3x3+ x2− x − 5 Check that f(1) = −2 and f(2) = 21, so there is a root

2) = 538 > 0 , which implies the new interval is [2,5

2] The second step is to

evaluate f(9

4) = 729

64 − 9 > 0, giving the interval [2,9

4] The best estimate is the midpoint

x c = 178

3 (b) Start with f(x) = 3x3+x2−x−5 on [1, 2], where f(1) > 0 and f(2) < 0 Since f(3

2) > 0, the second interval is [1,3

2] Since f(5

4) > 0, the third interval is [1,5

4] The best estimate is

the endpoint x c = 98

3 (c) Start with f(x) = cos2x + 6 − x on [6, 7], where f(6) > 0 and f(7) < 0 Since f(6.5) > 0,

the second interval is [6.5, 7] Since f(6.75) > 0, the third interval is [6.75, 7] The best estimate is the midpoint x c = 6.875.

5 (b) According to (1.1), the error after n steps is less than (3−2)/2 n+1 Ensuring that the error is

less than 10−10requires!1

2

"n+1

< 10 −10, or 2n+1 > 1010, which yields n > 10/ log10(2)−1 ≈

32.2 Therefore 33 steps are required.

6 Bisection Method converges to 0, but 0 is not a root

COMPUTER PROBLEMS 1.1

1 (a) There is a root in [2, 3] (see Exercise 1.1.1) In MATLAB, use the textbook’s Program 1.1,

bisect.m Six correct decimal places corresponds to error tolerances 5 × 10 −7, according

to Def 1.3 The calling sequence

>> f=@(x) xˆ3-9;

>> xc=bisect(f,2,3,5e-7)

returns the approximate root 2.080083.

1 (b) Similar to (a), on interval [1, 2] The command

>> xc=bisect(@(x) 3*xˆ3+xˆ2-x-5,1,2,5e-7)

returns the approximate root 1.169726.

1 (c) Similar to (a), on interval [6, 7] The command

(b) There are roots in [−2, −1], [−0.5, 0.5], and [0.5, 1.5] Using bisect as in part (a) yields

the approximate roots −1.023482, 0.163823, and 0.788942.

(c) There are roots in [−1.7, −0.7], [−0.7, 0.3], and [0.3, 1.3] Using bisect as in part (a) yields

the approximate roots −0.818094, 0, and 0.506308.

SECTION 1.2 FIXED-POINT ITERATION 19

yields the approximate cube root 1.25992105 in 27 steps.

5 (b) There is a root in the interval [1, 2] Using bisect as in (a) gives the approximate cube

root 1.44224957 in 27 steps.

5 (c) There is a root in the interval [1, 2] Using bisect as in (a) gives the approximate cube

root 1.70997595 in 27 steps.

6 0.785398

7 Trial and error, or a plot of f(x) = det(A) − 1000, shows that f(−18)f(−17) < 0 and

f (9)f (10) < 0 Applying bisect to f(x) yields the roots −17.188498 and 9.708299 The backward errors of the roots are |f(−17.188498)| = 0.0018 and |f(9.708299)| = 0.00014.

8 2.948011

9 The desired height is the root of the function f(H) = πH2(1− 1

3H) − 1 Using

>> bisect(@(H) pi*Hˆ2*(1-H/3)-1,0,1,0.001)

gives the solution 636 mm

EXERCISES 1.2 Fixed-Point Iteration