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Find interval for contraction map

  1. May 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Find an interval [a, b] for which the Contraction Mapping
    Theorem guarantees convergence to the positive fixed point or verify that there is no
    such interval.


    2. Relevant equations

    [itex] x = g(x) = \frac{14}{13} - \frac{x^{3}}{13}[/itex]

    3. The attempt at a solution

    I know the solution is slightly greater than 2. So, I assumed the upper bound on my interval would be 3 and g(3)=-1. So, I picked my lower bound to be -1 so that the function mapped from [-1,3] to [-1,3]. But, I'm having trouble showing that:

    [itex] abs(g'(x)) \leq γ [/itex] for some 0 < γ < 1.

    Any help would be massively appreciated.
     
  2. jcsd
  3. May 5, 2013 #2
    The latter inequality is what you should use to find the interval. Find g'(x), and see in what interval the inequality is satisfied.
     
  4. May 5, 2013 #3
    I've found that [itex] abs(g'(x)) < \frac{\sqrt{39}}{3} [/itex]

    But, this value is less than [itex]\sqrt[3]{14}[/itex] which is obviously the root. So is there no interval?
     
  5. May 5, 2013 #4
    Where does ##\frac {\sqrt {39} } {9}## even come from?

    Regardless, this is NOT what you want. You need to let ## |g'(x)| < 1 ## and find the interval satisfying that.
     
  6. May 5, 2013 #5
    But if we want ## |g'(x)| < γ ##
    And ## |g'(x)| = \frac{-3x^{2}}{13} ##
    So ## |\frac{-3x^{2}}{13}| < γ ##
    And then ## |x^{2}| < \frac{13γ}{3} ##
    Implies ## |x| < \sqrt{\frac{13γ}{3}} ##
    Then ## |x| < \frac{\sqrt{39}}{3} ## since ## γ < 1 ##

    No?
     
  7. May 5, 2013 #6
    You got ##|x| < r##, which is what you want. Compare that to #3.
     
  8. May 5, 2013 #7
    Sorry, I don't understand what you mean?
     
  9. May 5, 2013 #8
    In #5, you got the interval for x where g'(x) satisfies the contraction map criteria. In #3, you got some bogus inequality.
     
  10. May 5, 2013 #9
    So is my interval ## [-\frac{\sqrt{39}}{3},\frac{\sqrt{39}}{3}] ## ?

    Because that doesn't contain a root of the equation?

    Sorry, I'm very confused.
     
  11. May 5, 2013 #10
    What makes you think so?
     
  12. May 5, 2013 #11
    Well ## \frac{14}{13} - \frac{(\sqrt[3]{14})^{3}}{13}=\frac{14}{13}-\frac{14}{13} = 0 ##

    So that's the root?
     
  13. May 5, 2013 #12
    That's the zero on the RHS. You still have a non-zero LHS.
     
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