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Math Help - Quadratic map

  1. #1
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    Quadratic map

    Consider f_\lambda (x) = \lambda x(1 - x) for x, \lambda \in \mathbb{R}.

    1) Show that K_\lambda : = \{ x \in \mathbb{R}: the sequence x, f(x), f(f(x)), \ldots is bounded \} is always compact.

    2) For which values \lambda > 0 is K_\lambda connected?
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  2. #2
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    consider the intersection of the graphs of y=f(x) and y=x, which is a parabola and a line.
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  3. #3
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    I managed to show 1) myself, but what do I get from this intersection?
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  4. #4
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    Quadratic Map

    Quote Originally Posted by EinStone View Post
    Consider f_\lambda (x) = \lambda x(1 - x) for x, \lambda \in \mathbb{R}.

    1) Show that K_\lambda : = \{ x \in \mathbb{R}: the sequence x, f(x), f(f(x)), \ldots is bounded \} is always compact.

    2) For which values \lambda > 0 is K_\lambda connected?
    One thing that you get from the intersection of y=x and y=f_\lambda (x) is that for that value of x, call it x_0, f_\lambda (x_0)=x_0, in fact f_\lambda (f_\lambda (\dots f_\lambda (x_0)\dots ))=x_0

    The vertical line, x={1\over2} is the symmetry axis of the parabola y=f_\lambda (x), so f_\lambda(1-x_0)=f_\lambda(x_0)=x_0.

    Another thing to consider is that the vertex of this parabola is at (x,\, y)=\left({1\over2},\, {\lambda\over4}\right).

    What happens at x={1\over2} if \lambda>4.

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