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Math Help - cauchy sequence, contractive sequence

  1. #1
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    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?
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by inthequestofproofs View Post
    Let X1 be in E. For n>=1, let Xn+1 = f(Xn)
    show {Xn} is contractive
    You have not given enough information here. What is E? How is the function f defined?
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  3. #3
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    This is the whole question:
    Let E be a subset of R. If E is closed, and f: E -> E is contractive, prove there exists a unique point x0 in E s.t. f(x0) = x0 (a fixed point)

    On the back of the book, it is stated:
    "Let X1 be in E. For n>=1, let Xn+1 = f(Xn)
    show {Xn} is contractive"

    I don't understand how doing the latter proves the former.

    Thanks
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  4. #4
    Senior Member Tinyboss's Avatar
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    If f is contractive, then it's a quick step to show that the sequence f(x), f(f(x)), f(f(f(x))), etc. is Cauchy (hint: geometric series). Since E is closed and R is complete, that sequence has its limit L in E. Use the sequence definition of continuity to show that f(L)=L. To show uniqueness, suppose x and y are both fixed points. Then |f(x)-f(y)|=|x-y| is a contradiction unless |x-y|=0.
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