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Thread: Numerical analysis- fixed point iteration method

  1. #1
    fpb
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    Numerical analysis- fixed point iteration method

    Q. Compute by the method of fixed point iteration a real root of the equation x^3 + x^2 - 1 =0 correct up to two decimal places.


    In the above problem if we take f(x)=0, then to solve it we can write f(x) in the form of x=g(x). Now g(x) can take three different forms.
    i) 1/sqrt(1+x)
    ii) 1/x+x^2
    iii) -1+ (1/x^2)

    Now my question is, for which form of g(x) I can be sure that the obtained root converges to the actual root?
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    Re: Numerical analysis- fixed point iteration method

    Quote Originally Posted by fpb View Post
    Q. Compute by the method of fixed point iteration a real root of the equation x^3 + x^2 - 1 =0 correct up to two decimal places.


    In the above problem if we take f(x)=0, then to solve it we can write f(x) in the form of x=g(x). Now g(x) can take three different forms.
    i) 1/sqrt(1+x)
    ii) 1/x+x^2
    iii) -1+ (1/x^2)

    Now my question is, for which form of g(x) I can be sure that the obtained root converges to the actual root?
    Looking at the Wiki on Fixed-point iteration I see we let

    $g(x) = x- \dfrac{f(x)}{f^\prime(x)}$

    in this case $g(x)=x-\dfrac{x^3+x^2-1}{3 x^2+2 x}$

    and let

    $x_{n+1}=g(x_n)$

    This recurrence will converge to some $x_f$ and $f(x_f)=0$

    I used $x_0=1$ and this recurrence converged very quickly to $x_f \approx 0.754878$

    ------------------------
    Playing with the 3 functions you specified only the first one converges and it does converge to the correct root.

    Looking a bit further at the Wiki page it looks like this is because the first function is the only one that is Lipschitz continuous.
    Last edited by romsek; Sep 5th 2016 at 09:20 PM.
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    fpb
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    Re: Numerical analysis- fixed point iteration method

    The method you explained above is called Newton- Raphson method which has 2nd order rate of convergence. But I dont think N-R method and fixed point iteration method are same.
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    Re: Numerical analysis- fixed point iteration method

    "Newton-Raphson" is a specific type of "fixed point iteration".

    But where did you get these "g" functions? I don't see that any of them are equivalent to this problem. Starting from x^3+ x^2- 1= 0 we can get x^2= 1- x^3 and then x= \frac{1- x^3}{x}= \frac{1}{x}- x^2 not \frac{1}{x}+ x^2. Or taking the square root of both sides of x^2= 1- x^3, x= \sqrt{1- x^3} but not [tex]\frac{1}{\sqrt{1+ x}}. Or we can write x^3= 1- x^2 and then x= \frac{1- x^2}{x^2}= \frac{1}{x^2}- 1 but not 1- \frac{1}{x^2}.

    In any case, how fast a fixed point iteration, x= g(x), converges depends on the derivatives of g. If p= g(p) and |f'(p)| lies between 0 and 1, we have linear convergence. If |f'(p)|= 0 and |f''(p)| lies between 0 and 1, we have quadratic convergence, etc.
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