We have got the quadratic map $Q_a(x) = ax(1-x)$, with $a\in [0,4]$ and critical point $c = 1/2$ of period 3 ($c_k = Q^k_a(c)$). We know, that $I_1 = [c_2, c]$ and $I_2 = [c, c_1]$ form a Markov partition of $[c_2, c_1]$.

i.e.: $$ Q_a(I_1 = [c_2,c]) = [c,c_1] = I_2\\ Q_a(I_2 = [c,c_1]) = [c_1, c_2] = [c_2, c]\cup [c, c_1] = I_1 \cup I_2 $$ The transition matrix is defined by $A_{i,j} = 1$ if $Q_a(I_i) \supset I_j$, therefore

$$ A = \left(\begin{matrix} 0 & 1\\ 1 & 1 \end{matrix}\right). $$ Now we have to show, that the number of n-periodic points is equal to $tr(A^n)$.

Can anyone please help me with this proof. It should be very straightforward, but i have no idea how to write it down.

  • $\begingroup$ Of course, my bad, will correct it $\endgroup$ – iqopi May 18 '18 at 13:18

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