3
$\begingroup$

Consider the function $F(x) = x^2-2x+2$.

Find an interval in which the function is contractive and find the fixed point in this interval.

What is the convergence rate of the fixed point iteration: $x_{n+1} = F(x_n)$ in that interval?

I'm lost on fixed point iteration. I've watched a few youtube videos, but I'm still not understanding contractive intervals and methods for finding fixed points in functions.

$\endgroup$
1
$\begingroup$

You need an interval $[a,b]$ so that there's a $0\leq c<1$ such that $|f(x)-f(y)|\leq c|x-y|$ for all $x,y\in[a,b]$. In our case, we have $$ |f(x)-f(y)|=|x^2-y^2-(2x-2y)|\leq c|x-y|. $$ Notice that, in fact, $f(1)=1$, so you're going to want to choose an interval somewhere around $1$.

$\endgroup$
0
$\begingroup$

Try the form x(n+1)=(2x(n)-2)^(0.5) or x(n+1)=i(-2x(n)+2)^(0.5) of given equation. This converges to (1+i) which is one of the root of given equation. You note here that except infinities, any real or complex number as initial guess x(0) will converge x(n+1) to (1+i) .

$\endgroup$
-1
$\begingroup$

You need to consider the recurrence relation

$$ x_{n+1}= x_{n}^2-x_n+2 ,$$

which suggests considering the function

$$f(x) = x^2-x+2 \implies f'(x) = 2x-1. $$

Now, to find the interval, use the condition $|f'(x)|<1$. See a related problem.

$\endgroup$
  • $\begingroup$ Do you check $|f'(x)| < 1$ so that the derivative converges? In this case, I get that the interval is $x\in(0, 1)$. Once you have that interval, how do you find the fixed point? I tried iterating recurrence relation using 0, .5 and, 1, but got nothing concrete. $\endgroup$ – Neurax Mar 3 '14 at 17:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.