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The BabyEonian Algorithm (so called because it was used by the ancient Babylonians) computes -\/2 by repeated replacing one estimate x with the closer estimate (X + 2/x)/2. Note that this is simply the average of x and 2/x. #include <iostream.h> #include <math.h> // needed for the fabs() function main0 1 const double tolerance = 5e-8; double x = 2.0; while (fabs(x*x - 2.0) > tolerance) { tout << x << endl; x = (x + 2.0/x)/2.0; // average of x and 2/x tout << 'lx = l1 << x << II, x*x = 'I << x*x << endl; 2 345

17.1. Find, to the nearest integer (using p (a) A cube with an edge of 7 yd (b) A rectangular solid with dimensions of 8 ft, 62ft, and 14 ft (c) A sphere with radius of 30 m (d) A cylinder of revolution with a radius of 10 yd and a height of 42 yd. [Hint: Use T

function (for floating-point absolute value ), defined in the <math. h> header file, returns the absolute value of the exnression nassed to it. So the loon continues until x*x is within the given tolerance

(a) AC BD AD 3 5 3 5 6 5 8 11 6 14 (b) mj AEC mj BED mj AED 60 40 60 40 30 40 100 70 30 130

3.24 Convert the following for loop into a while loop: for (int i = 20; i > 10; i--) tout << i*i; 3.25

Applying postulates 4, 5, and 6 In each part, state a conclusion that follows when Postulates 4, 5, and 6 are applied to the given data. (a) Given: a (b) Given: a e (Fig. 2-16) c, b d (Fig. 2-16) mj DAE (Fig. 2-17) mj BCA, mj 1 mj 3 (Fig. 2-17)

Run the program in Example 3.13 to find the sizes of the 12 fundamental C++ types on your system. Run the program in Example 3.14 to find the precision and magnitude range for f 1 oats on your system. Describe the output from the following fragment:

Fig. 2-16

int f0 = fl = f2 = 1; for (int i = 0; i < 10; i++) { f0 = fl;

Most of the functions listed in Table 2.1 have a straightforward interpretation. A few, however, require some additional explanation. The next several examples should clarify any confusion. EXAMPLE 2.21

fl = f2; f2 = f0 + fl;

Fig. 2-17

tout << f2 << endl;

(a) a b a b CD a b a b CD e b b e EF c d c d AB Given Identity Add. Post . Subst. Given Given Add. Post. Subst. m/DAE m/1 m/DAE m/ 5 m/BCA m/3 m/BCA m/4 Given Given Subt. Post. Subst. Given Given Subt. Post. Subst.

Describe the output from the following fragment:

m/BAC m/ 1 m/BAC m/1 m/2 m/BAC m/ 1 m/BAC m/1 m/2

for (int i = 0; i < 8; i++) if (i%2 == 0) tout -CC i + 3 << endl; else if (i%3 == 0) tout -CC 2*i - 1 << endl; else if (i%5 == 0) tout << i*i << endl; else tout CC i << endl;

m/ 3

Describe the output from the following fragment:

Applying postulates 7 and 8 State the conclusions that follow when the multiplication and division axioms are applied to the data in (a) Fig. 2-18 and (b) Fig. 2-19.

int i = 0; while (++i <= 9) { if (i == 5) continue; tout << i C-C endl;

The Int function can be confusing, particularly with negative arguments. The values resulting from several typical function calls are shown below.

Describe the output from the following fragment:

Fig. 2-18

int i = 0; while (i < 5) { if (i < 2) { i += 2; continue; else tout CC ++i << endl; tout << "Bottom of loop.\n";

Fig. 2-19

In a 64-bit double type, 52 bits are used to store the mantissa and 11 bits are used to store the exponent.

(a) If a b, then 2a 2b since doubles of equals are equal. Hence, AF using the Multiplication Postulate. Hence, AB AC. (b) If mj A mj C, then 1m/A 2