CSCI 125 & 161 / ENGR 144 Lecture 13

CSCI 125 & 161 / ENGR 144 Lecture 13 Martin van Bommel

Floating Point Data • Floating point numbers are not exact • Value 0.1 is very close to 1/10, but not precisely equal to it • 0.110 = 0.000110011001100110011... 2 • Cannot rely on accuracy of floating point values

for With Floating Point Data • Following for loop is syntactically correct for (x = 1.0; x <= 2.0; x += 0.1) ... • On some machines, x will never take on value 2.0, but will become 2.00000000000001, which fails the test

for With Floating Point Values • Better to use the for loop for (i = 10; i <= 20; i++) ... • Then use the calculation x = i / 10.0; to give the floating point values for x

Equality With Floating Points • Testing floating point data for equality leads to problems with precision • Want to test if equal within some epsilon • In reality, test if absolute value of difference between numbers, divided by the smaller of their absolute values, is less than 

Approximately Equal #define Epsilon 0.000001 bool ApproximatelyEqual(double x, double y) { double diff = abs(x - y); y = abs(y); x = abs(x); if (y < x) x = y; return ((diff / x) < Epsilon); }

Numerical Algorithms • Techniques used by computers to implement mathematical functions like sqrt • sqrt function exists in math library • Most people just use it • How is it implemented? • Successive approximation? • Series expansion?

Successive Approximation • Steps: 1. Make a guess at the answer 2. Use guess to generate a better answer 3. If getting closer to actual answer, repeat until guess is close enough • How to generate next guess? • need a converging sequence of guesses

Newton’s Method • e.g. • Suppose want square root of 16 • Use 8 as first guess - too large, 8 x 8 = 64 • Derive next guess by dividing value by guess • 16 / 8 = 2, answer must be between 2 and 8 • Use (2 + 8) / 2 = 5 as next guess • 16 / 5 = 3.2, (3.2 + 5) / 2 = 4.1 as next guess • Next guesses: 4.001219512 and 4.00000018584

Problem with Successive Approx • Process will never guarantee an exact answer • Can continue until close enough (?) • Use ApproximatelyEqual function • Continue until square of guess approx. equal to original number

Newton’s Algorithm double Sqrt(double x) { double g = x; if (x == 0) return 0; if (x < 0) { cout << ”Sqrt on negative number\n”); return 0; } while (!ApproximatelyEqual(x, g*g)) g = (g + x / g) / 2; return g; }

Series Expansion • Value of function approximated by adding terms in a mathematical series • If addition of new term brings total closer to desired value, series converges and can be used to approximate result • Taylor Series Approximation for 0 < x < 2

Implement Series Expansion • For a series such as:Think of each inside term as • Initially sign = odd = 1; • Then, for the next term sign = -sign;odd = odd + 2;

Implementing Taylor Series • Think of each term as • Then, for next term • xpower *= (x-1); • factorial *= (i+1); • coeff *= (0.5-i);

Taylor Algorithm double TaylorSqrt(double x) { double sum, factorial, coeff, term, xpower; int i; factorial = coeff = xpower = term = 1.0; sum = 0.0; for (i = 0; sum != sum + term; i++) { sum += term; coeff *= (0.5 - i); xpower *= (x - 1); factorial *= (i + 1); term = coeff * xpower / factorial; } return sum; }

Fixing Limitation • Taylor Series Approx works for 0 < x < 2 • How can we use it for larger x? • Recall • Then divide by 4 until within range, then calcuate sqrt, and multiply by 2’s to recover • e.g. sqrt(24) = sqrt(4*4*1.5) = 2*2*sqrt(1.5)

Taylor Fix double TSqrt(double x) { int mult = 1; if (x == 0) return (0); if (x < 0) { cout << "TSqrt of negative value " << x << endl; return 0; } while (x >= 2){ x /= 4; mult *= 2; } return mult * TaylorSqrt(x); }

CSCI 125 & 161 / ENGR 144 Lecture 13