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.

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- Oct 24th 2006, 10:07 PMOntarioStudContractive 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, 10:08 AMOntarioStud
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, 11:04 AMThePerfectHacker
We need to show,

For some

Now,

Let then we have,

Thus,

If und only if,

If und only if,

Thus, by choosing this konstant to be that value we can form a contractive sequence.