# Contractive Sequence

• Oct 24th 2006, 09:07 PM
OntarioStud
Contractive Sequence
Let x1 be greater than zero. x(n+1)=1/(2+xn) for all n greater than or equal to 2. Show that (xn) is a contractive sequence and find its limit.

Stuck on the proof. Pretty sure the limit would just be the roots of (x^2)+2x-1=0. Any help appreciated.
• Oct 25th 2006, 09:08 AM
OntarioStud
I just need a little help. I have that:
abs(x(n+1)-x(n))=abs(1/(2+xn)-1(2+x(n-1))).

I just don't know how to simplify this in terms of abs(xn-x(n-1)) and find what C in the contractive sequence is.

Could anyone just steer me in the right direction?
• Oct 25th 2006, 10:04 AM
ThePerfectHacker
We need to show,
$\displaystyle |x_{n+1}-x_n|\leq k|x_n-x_{n-1}|$
For some $\displaystyle k\in (0,1)$
Now,
Let $\displaystyle a=x_{n-1}$ then we have,
$\displaystyle x_{n+1}=\frac{1}{2+x_n}=\frac{1}{2+\frac{1}{2+a}}= \frac{a+2}{2a+5}$
$\displaystyle x_n=\frac{1}{a+2}$

Thus,
$\displaystyle \left| \frac{a+2}{2a+5}-\frac{1}{a+2} \right| \leq k \left| \frac{1}{a+2}-a\right|$
If und only if,
$\displaystyle \left| \frac{a^2+2a-1}{(2a+5)(a+2)}\right| \leq k \left|\frac{-a^2-2a+1}{2a+5} \right|$
If und only if,
$\displaystyle 0<\frac{1}{a+2}\leq k<1$
Thus, by choosing this konstant to be that value we can form a contractive sequence.