A version of Newton Raphson (Newton's method)

Question

public class Sqrt
  {
     public static void main(String[] args)
       {
           double EPS = 1E-15;
           double c = Double.parseDouble(args[0]);
           double t = c;
              while (Math.abs(t - c/t) > t * EPS)
                      { t = (c/t + t) / 2.0; }
              System.out.println(t);
          }
      }

Above is one version of Newton Raphson i found in Java. I suppose it works but I'm having a hard time wrapping my head around how it actually works. { t = (c/t + t) / 2.0; } really confuses me.

I'm familiar with x_n+1 = x_n - f(x_n)/ f'(x_n) but not the one implemented in the code above..

I think it's a rewrite of `(c+t^2) / (2.0 * t)`, if that helps — OneCricketeer, Feb 28 '16 at 19:22

Ctx · Accepted Answer · 2016-02-28T19:40:53.837

4

To be able to apply the Newton-Raphson-method you have to find a function, for which the zero is the solution you are looking for. In the case of square-root this is:

f(x) = x^2 - c

Now if you find a x with f(x) = 0 you have found a square root of c.

f'(x) = 2*x

so

x - f(x)/f'(x) = x - ( (x^2 - c) / (2*x) ) = 1/2 * (x + c/x)

which is where your (c/t + t) / 2.0 comes from.

edited Feb 28 '16 at 19:40

answered Feb 28 '16 at 19:30

Ctx

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score 1 · Answer 2 · answered Feb 28 '16 at 19:29

1

Let t = sqrt(c)

=> t^2 = c   
=> 2t^2 - t^2 = c   
=> 2t^2 = t^2 + c

Let divide both side by 2t we get

=> t = (t^2 + c)/2t   
=> t = (t + c/t)/2   
=> t = (c/t + t)/2

answered Feb 28 '16 at 19:29

Ashraful Islam

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This only shows that the solution is a fixed point of the recursive equation – leonbloy Feb 28 '16 at 19:31

A version of Newton Raphson (Newton's method)

2 Answers2