Results 1 to 3 of 3

Thread: Contractive Sequence

  1. #1
    Newbie
    Joined
    Jul 2006
    Posts
    12

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jul 2006
    Posts
    12
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. cauchy sequence, contractive sequence
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Mar 25th 2010, 06:25 AM
  2. question regarding contractive sequence
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Oct 9th 2009, 04:36 PM
  3. Contractive Map
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 18th 2008, 11:20 AM
  4. contractive sequence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jul 11th 2008, 09:40 AM
  5. Proof of contractive sequence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 2nd 2007, 07:50 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum