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Thread: [SOLVED] Contraction Mapping Principle

  1. #1
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    [SOLVED] Contraction Mapping Principle

    Problem:
    1.Show that none of the following mappings $\displaystyle f:X \rightarrow X $ have a fixed point and explain why the contraction mapping principle is not contradicted.
    a. $\displaystyle (0,1) \subseteq R$ and $\displaystyle f(x)=\frac{x}{2}$
    b. X = R and $\displaystyle f(x)=x+1 $ for x in X.
    c. $\displaystyle {(x,y) \in R^2 | x^2+y^2=1}$ and $\displaystyle f(x,y)=(-y,x)$ for (x,y) in X

    2. Define the function $\displaystyle f:[1, \infty) \rightarrow R$ by
    $\displaystyle f(x)=1+\sqrt{x}$ for $\displaystyle x \geq 1$
    Show that this function has exactly one fixed point.
    ================================================== ======================
    Attempt:
    Problem 1:
    By a definition a point x in X is a fixed point for the mapping $\displaystyle f:X \rightarrow X $ provided that $\displaystyle f(x) = x $
    For (a) and (b) is it sufficient to show that $\displaystyle x \neq \frac{x}{2} $ and $\displaystyle x \neq x+1 $?
    For (c), the only fixed point would be $\displaystyle x= \frac{\sqrt{2}}{2},y= \frac{\sqrt{2}}{2}$, but $\displaystyle f \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) \neq f \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$

    Problem 2:
    If this function has a fixed point, then $\displaystyle f(x)=x= 1 +\sqrt{x} $

    $\displaystyle \implies x - \sqrt{x} = 1 \implies -x^{\frac{1}{2}} + x - 1 =0$
    I should slap myself, because I can't figure a way to solve for x. First glance, I thought of using the quadratic formula, but I can't do that.

    Thank you for your time.
    Last edited by Paperwings; Jul 31st 2008 at 04:47 PM.
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    2. Define the function $\displaystyle f:[1, \infty) \rightarrow R$ by
    $\displaystyle f(x)=1+\sqrt{x}$ for $\displaystyle x \geq 1$

    Show that this function has exactly one fixed point.
    If this has a fixed point it is a solution of:

    $\displaystyle 1+\sqrt{x}=x$

    or:

    $\displaystyle \sqrt{x}=x-1$,

    now square:

    $\displaystyle x^2-3x-1=0$

    Which can be solved using the quadratic formula.

    However we can show that this has exactly one real root $\displaystyle >1$ using Descartes rule of signs after the change of variables $\displaystyle y=x-1$, which gives:

    $\displaystyle y^2-1-1=0$

    which by Descartes rule of signs has exactly one positive root.


    RoNL
    Last edited by CaptainBlack; Aug 1st 2008 at 07:48 AM.
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    Quote Originally Posted by CaptainBlack View Post
    If this has a fixed point it is a solution of:

    $\displaystyle 1+\sqrt{x}=x$

    or:

    $\displaystyle \sqrt{x}=x-1$,

    now square:

    $\displaystyle x^2-x-1=0$
    Hello Captain Black,
    I am unsure how you went from this step to the "now square step:"

    Since
    $\displaystyle \sqrt{x}=x-1$, then by squaring both sides

    I get

    $\displaystyle \left( \sqrt{x}\right) ^2 = \left( x-1 \right) ^2$

    $\displaystyle \implies x = x^2 - 2x + 1 \implies x^2-3x+1=0$

    By the quadratic formula, then $\displaystyle x= \frac{3 \pm \sqrt{5}}{2}$, which are two real roots.

    But since the function is defined as $\displaystyle x \geq 1 $, then the only fixed point is $\displaystyle \frac{3+\sqrt{5}}{2}$
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Paperwings View Post
    Hello Captain Black,
    I am unsure how you went from this step to the "now square step:"

    Since
    $\displaystyle \sqrt{x}=x-1$, then by squaring both sides

    I get

    $\displaystyle \left( \sqrt{x}\right) ^2 = \left( x-1 \right) ^2$

    $\displaystyle \implies x = x^2 - 2x + 1 \implies x^2-3x+1=0$

    By the quadratic formula, then $\displaystyle x= \frac{3 \pm \sqrt{5}}{2}$, which are two real roots.

    But since the function is defined as $\displaystyle x \geq 1 $, then the only fixed point is $\displaystyle \frac{3+\sqrt{5}}{2}$
    Typing trouble, I'm pretty sure at some point that is what I had, but the post was lost, and I had to retype. I guess the error crept in at that point. The Decartes rule of signs argument still appliers though.

    RonL
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  5. #5
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    Ah, ok. Thank you.
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