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Thread: Convergent iteration

  1. #1
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    Convergent iteration

    Can someone help me with this problem?
    Consider the fixed point iteration
    $\displaystyle x_{k+1}=(m+1)x_k-{x_k}^2$, k=0,1,2,...where $\displaystyle 1\leq m \leq 2$
    a) Show that the iteration converges for any initial guess $\displaystyle x_0$ satisfying $\displaystyle m-1/5 \leq x_0 \leq m+1/5$
    b) Assume that the iteration converges, find m such that the method converges quadratically.

    I think I got part a) by letting $\displaystyle g(x)=(m+1)x-x^2$, then take the derivative $\displaystyle g'(x)=2x+m+1$. I then show that there exists $\displaystyle 0<k<1$ such that $\displaystyle \mid g'(x) \mid <k$, and by a theorem this implies the iteration converges for any initial guess in the specified interval.
    For part b), I need to find m such that $\displaystyle lim_{k\to \infty} \frac{|x_{k+1}-x|}{|x_k-x|^2}= lim_{k\to \infty} \frac{|(m+1)x_k-{x_k}^2-x|}{|x_k-x|^2}$ exists, but I don't get anywhere from here. Hope someone can give a hand.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by namelessguy View Post
    Can someone help me with this problem?
    Consider the fixed point iteration
    $\displaystyle x_{k+1}=(m+1)x_k-{x_k}^2$, k=0,1,2,...where $\displaystyle 1\leq m \leq 2$
    a) Show that the iteration converges for any initial guess $\displaystyle x_0$ satisfying $\displaystyle m-1/5 \leq x_0 \leq m+1/5$
    b) Assume that the iteration converges, find m such that the method converges quadratically.

    I think I got part a) by letting $\displaystyle g(x)=(m+1)x-x^2$, then take the derivative $\displaystyle g'(x)=2x+m+1$. I then show that there exists $\displaystyle 0<k<1$ such that $\displaystyle \mid g'(x) \mid <k$, and by a theorem this implies the iteration converges for any initial guess in the specified interval.
    For part b), I need to find m such that $\displaystyle lim_{k\to \infty} \frac{|x_{k+1}-x|}{|x_k-x|^2}= lim_{k\to \infty} \frac{|(m+1)x_k-{x_k}^2-x|}{|x_k-x|^2}$ exists, but I don't get anywhere from here. Hope someone can give a hand.
    Assuming that the itteration converges it is obvious it must converge to either $\displaystyle x=0$ or $\displaystyle x=m$, but it is easy to show that itteration will not converge to $\displaystyle x=0$.

    So suppose $\displaystyle x_n=m+\varepsilon$

    Then:

    $\displaystyle x_{n+1}=(m+1)(m+\varepsilon)-(m_\varepsilon)^2=m+\varepsilon(1-m)-\varepsilon^2$

    So if $\displaystyle m=1$ the itteration converges quadraticaly to $\displaystyle x=m=1$

    CB
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