= Solution
The exact second-iterate equation factors as
$$
F(F(x))-x
=x(\lambda x^2-\lambda-1)(\lambda x^2-\lambda+1)
(\lambda^2x^4-\lambda^2x^2+1).
$$
The positive nonfixed period-two points come from the final factor. If $y=x^2$, their squared values are
$$
y_\pm=\frac{1\pm\sqrt{1-4/\lambda^2}}2,
\qquad
y_++y_-=1,
\qquad
y_+y_-=\frac1{\lambda^2}.
$$
They are real and distinct for $\lambda>2$.
The <multiplier of a periodic orbit of an iteration> is
$$
F'(x_+)F'(x_-)
=\lambda^2(1-3y_+)(1-3y_-)
=\lambda^2\left(1-3(y_++y_-)+9y_+y_-\right)
=9-2\lambda^2.
$$
For $\lambda=2+\mu$ with $0<\mu\ll1$,
$$
9-2\lambda^2=1-8\mu+O(\mu^2),
$$
whose modulus is less than one. The new two-cycle is therefore locally asymptotically stable.
Back to article page
