In web page http://mathworld.wolfram.com/LogisticMap.html it's written that for $$x_{1,2}=\dfrac{1}{2} \Bigg[(1+r^{-1}) \pm r^{-1} \sqrt{(r-3)(r+1)} \Bigg] \tag 1 $$ a 2-cycle occurs. Although don't we know theoritically that for $\geq 4$ logistic map is chaotic?
I searched for a 2-cycle for r=4 and $x_0=\dfrac{\sqrt{5}}{8}(\sqrt{5}+1)$ which occurs from (1) and I actually found that $x_0=x_2$. My estimations are:
$$\boldsymbol{x_1}=4x_0(1-x_0)$$ $$=4\dfrac{\sqrt{5}}{8}\big(\sqrt{5}+1 \big) \big(1-\dfrac{\sqrt{5}}{8}(\sqrt{5}+1) \big)$$ $$=\dfrac{\sqrt{5}}{2} \big(\sqrt{5}+1 \big) \big( 1-\dfrac{5}{8}-\dfrac{\sqrt{5}}{8} \big)=\dfrac{\sqrt{5}}{2} \big(\sqrt{5}+1 \big) \big(\dfrac{3}{8} -\dfrac{\sqrt{5}}{8} \big)$$ $$=\dfrac{\sqrt{5}}{16} \big(\sqrt{5}+1 \big)(3-\sqrt{5})=\dfrac{\sqrt{5}}{16}(2\sqrt{5}-2)$$ $$\boldsymbol{=\dfrac{\sqrt{5}}{8}( \sqrt{5}-1)}$$
$$\boldsymbol{x_2}=4x_1(1-x_1)$$ $$=4\dfrac{\sqrt{5}}{8} \big(\sqrt{5}-1 \big) \big( 1-\dfrac{\sqrt{5}}{8}(\sqrt{5}-1) \big) = \dfrac{\sqrt{5}}{2} \big(\sqrt{5}-1 \big) \big(1-\dfrac{5}{8}+\dfrac{\sqrt{5}}{8} \big) $$ $$=\dfrac{\sqrt{5}}{2} \big(\sqrt{5}-1 \big) \big( \dfrac{3}{8}+\dfrac{\sqrt{5}}{8} \big) =\dfrac{\sqrt{5}}{16}(\sqrt{5}-1)(3+\sqrt{5})$$ $$= \dfrac{\sqrt{5}}{16}(2 \sqrt{5}+2)\boldsymbol{=\dfrac{\sqrt{5}}{8}( \sqrt{5}+1)}$$
Can you please check my work and give me theoritical feedback if possible?
You have found the $2-$cycle ,namely $${\text {{$x_0$, $x_1$}}}$$ where, $$x_0=\dfrac{\sqrt{5}}{8}(\sqrt{5}+1)$$ and
$$\boldsymbol{x_1}=4x_0(1-x_0)\boldsymbol{=\dfrac{\sqrt{5}}{8}( \sqrt{5}-1)} $$ Note that $f(x_0) =x_1$ and $f(x_1)=x_0.$
Thus,
$$x_0 \to x_1\to x_0\to x_1 \to x_0\to......$$ indicate the existence of the $2-$cycle.
Note that the existence of a 2-cycle does not necessarily mean that your function is not chaotic. Changing the initial point may change the dynamics of the orbit significantly.