Precise Square Root using Newton Raphson Method.
We want to calculate square root n.
n^(1/2) = x Lets say x is the square root of n.
n = x^2 squared both side.
x^2 – n = 0 moved n to the other side.
f(x) = x^2 – n.
Newton raphson iterative method formula.
Lets find derivative of our equation.
f ‘(x) = 2x
Simplify the Equation:
Xn+1 = Xn – (Xn^2 – n) / 2Xn
Xn+1 = (2Xn^2 – Xn^2 + n) / 2Xn
Xn+1 = (Xn^2 + n) / 2Xn
Xn+1 = (Xn + n/Xn)/2
take guess value of Xn lets say 1,
and keep iterating to calculate Xn+1
and stop when (Xn+1 – Xn) <= required precision.
Pseudo Code:
float square_root(int n)
{
float x = 1; //guess value.
while (true)
{
float lastX = x;
x = (x + n/x) / 2;
if (abs(x - lastX) <= 0.00001)
{
break;
}
}
return x;
}
public class Solution {
public static double squareRoot(int n) {
final double precison = 0.0000001;
final double guess = 1;
double x = guess;
double lastX;
do {
System.out.println(x);
lastX = x;
x = (x + n/x)/2;
} while ((Math.abs(x - lastX)) > precison);
return x;
}
public static void main(string[] args) {
System.out.println(squareRoot(2));
}
}
Output:
1.0
1.5
1.4166666666666665
1.4142156862745097
1.4142135623746899
1.414213562373095
buddhaTech 