Convergent iteration

**namelessguy** · December 17th 2008, 04:38 PM

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

I think I got part a) by letting $g(x)=(m+1)x-x^2$ , then take the derivative $g'(x)=2x+m+1$ . I then show that there exists $0<k<1$ such that $\mid g'(x) \mid <k$ , 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 $lim_{k\to \infty} \frac{|x_{k+1}-x|}{|x_k-x|^2}= lim_{k\to \infty} \frac{|(m+1)x_k-{x_k}^2-x|}{|x_k-x|^2}$ exists, but I don't get anywhere from here. Hope someone can give a hand.

**CaptainBlack** · December 17th 2008, 10:53 PM

Originally Posted by namelessguy

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

I think I got part a) by letting $g(x)=(m+1)x-x^2$ , then take the derivative $g'(x)=2x+m+1$ . I then show that there exists $0<k<1$ such that $\mid g'(x) \mid <k$ , 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 $lim_{k\to \infty} \frac{|x_{k+1}-x|}{|x_k-x|^2}= lim_{k\to \infty} \frac{|(m+1)x_k-{x_k}^2-x|}{|x_k-x|^2}$ 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 $x=0$ or $x=m$ , but it is easy to show that itteration will not converge to $x=0$ .

So suppose $x_n=m+\varepsilon$

Then:

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

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

CB

Thread: Convergent iteration

LinkBack

Thread Tools

Display

Convergent iteration

Similar Math Help Forum Discussions

How can you tell if a series is absolutely convergent or conditionally convergent?

Determine whether the series is absolutely convergent, conditionally convergent ....

absolutely convergent, conditionally convergent or divergent

determine if series is absolutely convergent conditionally convergent or divergent

Series: Absolutely convergent, con convergent or divergent

Search Tags