Transition from one difference equation to another

Question

I have initial equation: $y(n+1)=y^2(n)+C$, $C\leq\frac{1}4$

How I can show that if $C=\frac{a}2+\frac{a^2}4$ then by using replacement $y(n)=\alpha x(n)+\beta$

I will get the equation $x(n+1)=ax(n)(1-x(n))$

Answer 1

Using $C=a/2+a^2/4$ and $y=\alpha x+\beta$ in the equation leads to $$ \alpha x(n+1) + \beta = (\alpha^2 x^2(n) + 2\alpha\beta x(n) + \beta^2) + \frac{a}{2} + \frac{a^2}{4} $$ or $$ x(n+1) = x(n)(\alpha x(n) + 2\beta) + \frac{1}{\alpha} \left(\beta(\beta-1) + \frac{a}{2} + \frac{a^2}{4}\right). $$

The term in brackets vanishes if $$ \beta^2 - \beta + \frac{a}{2} + \frac{a^2}{4} = 0, $$ leading to $$ \beta = \frac{1 \pm i(a+1)}{2}. $$ If we choose $\alpha = -2\beta$, we have $$ x(n+1) = 2\beta x(n)(1 - x(n)), $$ which is the logistic equation, but with a complex coefficient. If we rather use $C=a/2-a^2/4$, we can choose $2\beta = a$ and $\alpha=-1$, leading then to $$ x(n+1) = a x(n)(1- x(n)). $$