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I am trying to do this question using the Fixed Point Factor Theorem. I keep getting an answer $>0$ at the end of my long division of $f(x)-x$ into $f^2(x)-x$ therefore I must using the wrong divider. Can somebody help me please?

user4167
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Josephine
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    I don't know anything about dynamical systems, but it seems to me that in this context $f^2(x)$ should mean the iterated function $f(f(x))$. Did you accidentally calculate $f(x)^2=f(x)\cdot f(x)$? – Jyrki Lahtonen May 20 '13 at 10:02

2 Answers2

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Fixed points ( period 1 points): $f(x)=x$

so solve $x^2-4x+3=0$ (which leads to $x=3$ or $x=1$)

Period two points: $f(f(x))=x$

so

$f(f(x))= x^4-6x^3+12x^2-9x+3$, so to get only period 2 points we want to divide the fixed points out of $x^4-6x^3+12x^2-10x+3=(x-3)(x-1)(x^2-2x+1)$.

As it turns out, $x^2-2x+1$ has only $x=1$ as roots, so there are no proper period-2 points.

Adam
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Hagen von Eitzen
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Considering polynomials close to the given one, set $f(x)=x^2-3x+3+a$ to get \begin{align} f^2(x)-x &= f(x)^2-3f(x)+3+a-x=(f(x)-x)(f(x)+x-3)+x^2-4x+3+a \\ &=(f(x)-x)(x^2-2x+1+a) \end{align} For $a<0$ you find a 2-cycle from the roots of $x^2-2x+1+a=0$, for $a>0$ there are no real roots from that factor, thus at $a=0$, the original problem, you have a bifurcation point at $x=1$.

Lutz Lehmann
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