3

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.

Neurax
  • 1,005

3 Answers3

1

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 $1$.

Ian Coley
  • 6,000
0

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) .

Prasenjit D Wakode
  • 154
-1

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.

Community
  • 1
Mhenni Benghorbal
  • 47,431
  • 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. – Neurax Mar 03 '14 at 17:50