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{\bf Question}

Let $f_a9x)=a-x^2$.

Find:

\begin{description}
\item[(i)]
the value $a_1$ of $a$ such that $f_a$ has exactly one fixed
point,
\item[(ii)]
the largest value $a_2$ of $a$ for which $f-a$ has \underline {no}
2-cycle,
\item[(iii)]
the value $a_3$ of $a$ at which an attracting 2-cycle becomes
repelling.
\end{description}
Show that the conditions of the period-doubling theorem are
satisfied for $f_a^2$ and also $f_{a_3}^2$ (at the appropriate
points).
\medskip

{\bf Answer}
\begin{description}
\item[(i)]

Exactly one fixed point when the parabola $y-ax^2$ is tangent to
the line $y=x$, i.e. equation $a-x^2=x$ has repeated roots.
Condition ${\lq\lq} {b^2 -4ac}"$ here is $1=-4a$, i.e. \underline
{$a_1=-\ds\frac{1}{4}$}

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\item[(ii)]
$f_a^2(x)=a-(a-x^2)^2$, so fixed points of $f_a^2$ where
$x=a-(a-x^2)^2$, that is $x^4-2ax^2+x-(a-a^20=0$.  Left hand side
vanishes at fixed points of $f_a$, so has $(x_2+x-a)$ as a factor:
we find LHS$=(x^2+x-a)(x^2-x+(1-a))$.  Thus per-2 points are roots
of $x^2-x+(1-a)=0$; these are real if and only if
$a>\ds\frac{3}{4}={\rm max}\ a$ with no 2-cycle.


\item[(iii)]
If $\{p,q\}$ is a 2-cycle then $(f_a^2)'(p)=f_a'(q)f_a'(p)=4pq$
since ${f_a}'(x)=-2x$.  Thus 2-cycle repelling when $4q=-1$, i.e.
$pq=-\ds\frac{1}{4}$.  But $pq=$ product of roots of
$(x^2-x+(1-a))=0$, i.e. $pq=(1-a)$.  Therefore 2-cycle becomes
repelling where $-\ds\frac{1}{4}=(1-a)$, i.e. \underline{$
a_3=\ds\frac{5}{4}$}.
\end{description}
\medskip

We have $f_a'(x)=-2x$. When $a=a_2=\ds\frac{3}{4}$ the fixed
points are $x=\ds\frac{1}{2},\ -\ds\frac{3}{2}$ so \underline {$
f_{a_2}'\left(\ds\frac{1}{2}\right)=-1$}.

Now $(f_a^2)(x)=f_a'(f_a(x))\cdot f_a'(x)=-2(a-x^2) \cdot -2x$
giving $(f_a^2)'\left(\ds\frac{1}{2}\right)=2a-\ds\frac{1}{2}$;
then $\left.\underline {\ds\frac{\partial}{\partial a}
(f_a^2)'\left(\ds\frac{1}{2}\right)}\right|_{a=a_2} =2 \ne 0.$

[Since $2>0$ and $Sf_a<0$ the bifurcation is supercritical.]
\medskip

For the 2-cycle $\{p,q\}$ we have \underline
{$(f_{a_3}^2)'(p)=-1$} (that's how $a_3$ was found).  Now
$(f_a^4)'(x)=16f_a^3(x) \cdot f_a^2(x) \cdot f_a(x) \cdot x$
(Chain Rule).  We have $\ds\frac{\partial}{\partial a} f_a(x)=1,
\ds\frac{\partial}{\partial a} f_a^2(x)=1-2f_a(x),
\ds\frac{\partial}{\partial a} f_a^3(x)=1-2f_a^2(x)(1-2f_a(x))$
(using $f_a^2(x)=a-(f_a(x))^2, f_a^3(x)=a-(f_a^2(x))^2)$ and if
$\{p,q\}$ is a 2-cycle for $f_a$ these give \underline {$
\ds\frac{\partial}{\partial a} f_a(p)=1$},\ \underline {$
\ds\frac{\partial}{\partial a} f_a^2(p)=1-2q$},\ \underline {$
\ds\frac{\partial}{\partial a} f_a^3(p)=1-2p+4pq$}. We use these
to differentiate $(f_a^4)'(p)$ as a product:

$\ds\frac{\partial}{\partial a}
(f_a^4)'(p)=16[(1-2p+4pq)pq+q(1-2q)q+qp1]p$. When
$a=a_3=\ds\frac{3}{4}$, $p$ and $q$ are the roots of
$x^2-x-\ds\frac{1}{4}=0$ so $pq=-\ds\frac{1}{4},\ p+q=1$.  So
\underline {$\ds\frac{\partial}{\partial a}
(f_a^4)'(p)$}=$16\left[\ds\frac{1}{2}p^2-\ds\frac{1}{4}q(1-2q)-\ds\frac{1}{4}p\right]=8(p^2+q^2)-4=8(p+q)^2$\underline{$=8
\ne 0.$}

[Since $8>0$ and $Sf_a^2 <0$ (since $Sf_a<0$) the bifurcation from
a 2-cycle to a 4-cycle at $a=a_3$ is supercritical.]

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