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Contraction Mapping Theorem

  1. Apr 5, 2009 #1
    If I take [tex]F(x)=\sqrt{1+x^2}[/tex], then the derivative is always less than one so this is a contraction mapping from R to R, right?

    But there is no fixed point where [tex]F(x)=x[/tex], where the contraction mapping theorem says there should be.

    So where have I gone wrong?

  2. jcsd
  3. Apr 5, 2009 #2


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    The derivative is less than 1, true. But it approaches 1 as x->infinity. So there is no q<1 such that f'(x)<q. It's NOT a contraction mapping. Look again at the definition of 'contraction mapping'.
  4. Apr 6, 2009 #3
    Thanks :-)
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