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Thread: Quadratic map

  1. #1
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    Quadratic map

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

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

    2) For which values $\displaystyle \lambda > 0$ is $\displaystyle K_\lambda$ 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 $\displaystyle f_\lambda (x) = \lambda x(1 - x) $ for $\displaystyle x, \lambda \in \mathbb{R}.$

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

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

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

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

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

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