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Newton Raphson's

  1. Nov 29, 2007 #1
    Uhm i mean, actually how do we predict the estimated roots before we implement this?

    Because i wonder if the estimated root is too big enough to predict and sometimes time-consuming, is there any way to predict the roots better and accurate before we implement the Newton Raphson's iterative methods? I am using plain subsitution into the function which sometimes have to check through all the numbers in order to get one estimated root beforehand.

    Thanks in advance.

  2. jcsd
  3. Nov 30, 2007 #2


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    We can't. One of the most famous "fractals" is the graph formed by looking at the three roots of z3= 1. These are, of course, 1, [itex]-(1/2)+ i\sqrt{3}/2[/itex], and [itex]-(1/2)-i\sqrt{3}/2[/itex]. Treat the x,y-plane as the complex plane, (x,y) corresponding to x+ iy. Taking each x+ iy as the "initial estimate" for Newton-Raphson, color the point "red", "blue", "green", or "black" according to whether the sequence converges to 1, [itex]-(1/2)+ i\sqrt{3}/2[/itex], and [itex]-(1/2)-i\sqrt{3}/2[/itex], or does not converge respectively. You will see large patches of "red", "blue", and "green" close to those respective roots but the boundary is extremely complex ("fractal"). In fact, every boundary point is a boundary point of all four sets simultaneously. It is possible that very tiny variations in choice of initial point will cause the iteration to converge to a different answer, or not converge at all.

    (By the way, when I first programmed a computer (with a "graphics" terminal) to do that, it took almost an hour. Now the same program runs in less that 10 seconds!)
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