Problem source

The numbers $x_n$,  where $n=0$, $1$, $2$, $\ldots$ , satisfy 
\[
x_{n+1} = kx_n(1-x_n) \;.
\]
\begin{questionparts}
\item Prove that, if $0 < k < 4$ and $0 < x_0 < 1$, then $0 < x_n < 1$ for all $n\,$.
\item Given that $x_0=x_1=x_2 = \cdots =a\,$, with  $a\ne0$ and $a\ne1$, find $k$ in terms of $a\,$.
\item Given instead that $x_0=x_2=x_4 = \cdots = a\,$, with   $a\ne0$ and $a\ne1$, show that $ab^3 -b^2 +(1-a)=0$, where $b=k(1-a)\,$. Given, in addition, that $x_1 \ne a$, find the possible values of $k$ in terms of $a\,$.
\end{questionparts}

Solution source

\begin{questionparts}
\item Consider $f(x) = x(1-x) = x - x^2 = \tfrac14 - (x - \tfrac12)^2$ which is clearly in $(0,\tfrac14)$ when $x \in (0,1)$, therefore if $0 < k < 4$ then $f(x) \in (0, 1)$ and so by induction $x_n \in (0,1)$.

\item Suppose $a = g(a)$ then $a = ka(1-a) \Rightarrow 1 = k(1-a) \Rightarrow k = \frac{1}{1-a}$ (since $a \neq 0, 1$)

\item If $g(g(a)) = a$ then
\begin{align*}
&& a &= kg(a)(1-g(a)) \\
&&&= k^2a(1-a)(1-ka(1-a)) \\
&&&= -k^3a^2(1-a)^2 + k^2a(1-a) \\
\Rightarrow && 1 &= -k^3a(1-a)^2 + k^2(1-a) \\
\Rightarrow && 1-a &=  -k^3a(1-a)^3+k^2(1-a)^2 \\
\Rightarrow && 1-a &= -ab^3+b^2 \\
\Rightarrow && 0 &= ab^3-b^2+(1-a)
\end{align*}

Note that \begin{align*}
&& 0 &= ab^3-b^2+(1-a) \\
&&&= (b-1)(ab^2-(1-a)b - (1-a))
\end{align*} and since $b \neq 1$ (otherwise $x_2 =0$ which is a contradiction) we must have $b = \frac{1-a \pm \sqrt{(1-a)^2+4a(1-a)}}{2a} = \frac{1-a\pm \sqrt{1+2a-3a^2}}{2a}$ and so

$k = \frac{b}{1-a} = \frac{1-a \pm \sqrt{1+2a-3a^2}}{2a(1-a)}$
\end{questionparts}