cauchy sequence, contractive sequence
Hello,
Let X1 be in E. For n>=1, let Xn+1 = f(Xn)
show {Xn} is contractive
I have attempted several things:
1. |f(X2) - f(X1)|
= |X3-X2|
<= |X2-X1| + |X1 - X3| (by triangle inequality)
< |X2-X1|
By definition, a sequence is contractive if |Pn+1 -Pn|<= b|Pn-Pn-1| or |Pn+1-Pn|<= b^n-1|p2-p1|
So, I know |X3-X2| < |X2-X1|, but how do i find the b?
2. Show {Xn} is cauchy
to show that, I have to find an N s.t. |Pn -Pm|< epsilon for all n,m>=N.
So, |f(Xn-1)-f(Xm-1)|
= |Xn-Xm|
<= |Xn -p| + |p-Xm| (let |Xn - p| < epsilon/2 and |Xm -p| < epsilon/2)
< epsilon/2 + epsilon/2
=epsilon
But what is N?
Would either work?