Question

Newtons method for approximating square roots.

The next iteration is the average of the previous iteration and the ratio of the number in question to the previous iteration.x_i = ( x_(i-1) + number/x_(i-1) ) / 2

if i is 0 the approximation is half the number.

Pre: number & interations are non-negative integers

Post: return an approximation of sqrt(number) , the return value is double

double newton(size_t number, size_t iterations){}

Answer 1

In this program, we have to approximate square root of a number using newtons method.

Initially set squareRoot = 0.5*number

then create a loop that repeats iterations times. Inside the loop set sqaureRoot = (squareRoot + (double)number/squareRoot)/2

After the loop terminates, return squareRoot

function:

double newton(size_t number, size_t iterations){
double squareRoot = (double)number/2;
for(int i = 1; i <= iterations; i++)
squareRoot = (squareRoot + (double)number/squareRoot)/2;
return squareRoot;
}

sample program to test this function:

#include <stdio.h>
#include <math.h>

double newton(size_t number, size_t iterations){
double squareRoot = (double)number/2;
for(int i = 1; i <= iterations; i++)
squareRoot = (squareRoot + (double)number/squareRoot)/2;
return squareRoot;
}

int main(){
double number; int iterations;
printf("Enter a number: ");
scanf("%lf", &number);
printf("Number of iterations: ");
scanf("%d",&iterations);
printf("\n\n Real square root: %lf\n",sqrt(number));
printf("Approximated square root: %lf\n",newton(number,iterations));
}

sample inputs and outputs: