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{\bf Question}

By applying the Period-Doubling Theorem to $g_a^2=[ax(1-x)]^2$
show that a 4-cycle is created as a increases through $1+\sqrt 6$.
\medskip

{\bf Answer}
%Question 7 - note reference to question 6%
If the 2-cycle is $\{p,q\}$ as in question 6, then

\begin{eqnarray*} (g_a^4)' & = &
{g_a}'(g_a^3(x)){g_a}'(g_a^2(x)){g_a}'(g_a(x)){g_a}(x)\\ & = &
a^4(1-2g_a^3(x)(1-2g_a^2(x))(1-2g_a(x))(1-2x) \end{eqnarray*}

Now

$$\begin{array} {rcl} \ds\frac{\partial g_a}{\partial a}(p) & = &
p(1-p) = \ds\frac{1}{a}q;\\ \ds\frac{\partial {g_a}^2}{\partial
a}(p) & = & \ds\frac{1}{a}p+a(1-2q)\ds\frac{1}{a}q;\\
\ds\frac{\partial {g_a}^3}{\partial a}(p) & = &
\ds\frac{1}{a}q+a(1-2p)\left[\ds\frac{1}{a}p+a(1-2q)\ds\frac{1}{a}q\right].
\end{array}$$

Hence

$$\begin{array} {rcl} \ds\frac{\partial}{\partial a}(g_a^4)'(p) &
= & 4a^3(1-2p)^2(1-2q)^2\\ & - &
2a^4\left(\ds\frac{1}{a}q+a(1-2p)\left[\ds\frac{1}{a}p+a(1-2q)\ds\frac{1}{a}q\right]\right)(1-2p)^2(1-2q)\\
& - &
2a^4(1-2q)\left(\ds\frac{1}{a}p+a(1-2q)\ds\frac{1}{a}q\right)(1-2q)(1-2p)\\
& - & 2a^4(1-2q)(1-2p)\ds\frac{1}{a}q(1-2p) \end{array}$$

When \underline{$ a=1+\sqrt 6$} we have $(g_a^2)'(p)=\underline{
a^2(1-2p)(1-2q)=-1}$\ so we find the above simplifies to

$\begin{array} {rcl} \left.\ds\frac{\partial}{\partial a}
(g_a^4)'(p)\right|_{a=1+\sqrt 6} & = &
\ds\frac{4}{a}+2a^2p(1-2p)^2+2a^2(1-2q)\left(\ds\frac{1}{a}p+(1-2q)q\right)+2aq(1-2p)\\
& = &
\ds\frac{4}{a}+2a^2(p(1-2p)^2+q(1-2q)^2)+2a(p(1-2q)+q(1-2p)).
\end{array}$

Now using \underline {$p+q=\ds\frac{1}{a}+1$} (=$b$, say) and
\underline{$ pq=\ds\frac{b}{a}$} we find the above $\begin{array}
{cl} = &
\ds\frac{4}{a}+2a^2b\left(1-\ds\frac{8}{a^2}\right)+2ab\left(1-\ds\frac{4}{a}\right)\\
 = & \ds\frac{4}{a}+2b(a^2+a-12) >0 \end{array}$

because $a^2_a-12=(1+\sqrt 6)^2+(1+\sqrt 6)-12=3\sqrt 6 -4>0.$

Hence the bifurcation from a 2-cycle to a 4-cycle is
supercritical.
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