Problem source

\begin{questionparts}
  \item The sequence of numbers $u_0, u_1, \ldots $ is given by
    $u_0=u$ and, for $n\ge 0$,
    \begin{equation}
      u_{n+1} =4u_n(1- u_n)\,.
      \tag{$*$}
    \end{equation}
    In the case $u= \sin^2\theta$ for some given angle $\theta$, write down and simplify expressions for $u_1$ and $u_2$ in terms of $\theta$. Conjecture an expression for $u_n$ and prove your conjecture.
  \item The sequence of numbers $v_0, v_1, \ldots$ is given by $v_0=
    v \text{ and, for }n\ge 0$,
    \[
    v_{n+1} = -pv_n^2 +qv_n +r\,,
    \]
    where $p$, $q$ and $r$ are given numbers, with $p\ne0$. Show that a substitution of the form $v_n =\alpha u_n +\beta$, where $\alpha$ and $\beta$ are suitably chosen, results in the sequence $(*)$ provided that
    \[
    4pr = 8 +2q -q^2 \,.
    \]
    Hence obtain the sequence satisfying $v_0=1$ and, for $n\ge0$, $v_{n+1} = -v_n^2 +2 v_n +2 \,$.
  \end{questionparts}

Solution source

\begin{questionparts}

\item Suppose $u_0 = u = \sin^2 \theta$ then 
\begin{align*}
&& u_1 &= 4 u_0 (1-u_0) \\
&&&= 4 \sin^2 \theta ( 1- \sin^2 \theta) \\
&&&= 4 \sin^2 \theta \cos^2 \theta \\
&&&= (2 \sin \theta \cos \theta)^2 \\
&&&= (\sin 2 \theta)^2 = \sin^2 2 \theta \\
\\
&& u_2 & = 4u_1 (1-u_1) \\
&&&= 4 \sin^2 2\theta \cos^2 2 \theta \\
&&&= \sin^2 4 \theta
\end{align*}

Claim: $u_n = \sin^2 2^n \theta$.

Proof: (By Induction) Base case is clear, suppose it's true for $n=k$, then

\begin{align*}
&& u_{k+1} &= 4u_k(1-u_k) \\
&&&= 4 \sin^2 2^k \theta(1-\sin^2 2^k \theta) \\
&&&= (2 \sin 2^k \theta \cos 2^k \theta)^2 \\
&&&= (\sin 2^{k+1} \theta)^2 \\
&&&= \sin^2 2^{k+1} \theta
\end{align*}

Therefore since it is true for $n = 1$ and if it's true for $n = k$ it is true for $n=k+1$ it must be true for all $k$.

\item Suppose $v_n = \alpha u_n + \beta$ then

\begin{align*}
&& (\alpha u_{n+1}+\beta) &= -p(\alpha u_n + \beta)^2 + q(\alpha u_n + \beta) + r \\
&&&= -p\alpha^2u_n^2+\alpha(q-2p\beta) u_n -p \beta^2 +q \beta+r \\
\Rightarrow && u_{n+1} &= u_n(q-2p\beta -p \alpha u_n) -(p\beta^2-(q-1)\beta-r)
\end{align*}

So if $\alpha = \frac{4}{p}$ and $q-2p\beta = 4$ ie $\beta = \frac{q-4}{2p}$  then we also need  the constant term to vanish, ie

\begin{align*}
0 &&&= p\beta^2-(q-1)\beta+r \\
&&&= p \left (\frac{q-4}{2p} \right)^2 - (q-1) \frac{q-4}{2p} - r \\
\Rightarrow && 0 &= p(q-4)^2 -(q-1)(q-4)2p - 4p^2r \\
\Rightarrow && 0 &= (q-4)^2-2(q-1)(q-4)-4pr \\
&&&= q^2-8q+16-2q^2+10q-8-4pr \\
\Rightarrow && 4pr &= -q^2+2q+8
\end{align*}

Suppose $v_{n+1} = -v_n^2 + 2v_n +2$ then since $4\cdot 1 \cdot 2 = 8$ and $8 + 4 -4 = 8$ we can apply our method.

$v_n = 4u_n + \frac{-2}{2} = 4u_n -1 = 4\sin^2 (2^{n-1} \pi)-1$

\end{questionparts}