 Solution

The nonzero fixed point of the logistic map is $x_{\ast }=(\mu -1)/\mu$, with multiplier $F^{\prime }(x_{\ast })=2-\mu$. It is stable for $1<\mu <3$, and its multiplier crosses $-1$ at $\mu =3$. To identify the emerging orbit explicitly, factorThe remaining two roots, interchanged by $F$, areThey coincide with $x_{\ast }$ at three and are distinct for $\mu >3$. Their cycle multiplier isIt lies between $-1$ and $1$ for $3<\mu <1+\sqrt{6}$. This establishes the first period-doubling bifurcation and the stable bifurcating two-cycle.

At ${\mu }_2=1+\sqrt{6}$, ${\Lambda }_2=-1$ and $d{\Lambda }_2/d\mu =-2\sqrt{6}\ne 0$. The second flip is nondegenerate: for $H=F^2$ at either cycle point, the cubic flip coefficient is $c=(H^{\prime \prime \prime }/6)+(H^{\prime \prime }/2{)}^2=-\mathcal{S}H/6$ when $H^{\prime }=-1$. Here $\mathcal{S}$ denotes the Schwarzian derivative. Since $\mathcal{S}F=-6/(1-2x{)}^2<0$ and $\mathcal{S}(F^2)=(\mathcal{S}F\circ F)(F^{\prime }{)}^2+\mathcal{S}F$, this coefficient is positive. Locally, after following the fixed point of $H$, its second iterate has the form $H^2(y)-y=2a(\mu -{\mu }_2)y-2c{y}^3+\cdots$, with $a=2\sqrt{6}>0$. Nonzero roots therefore appear with $y^2\sim a(\mu -{\mu }_2)/c$ on the increasing-parameter side. These are a four-cycle of $F$, proving the second period-doubling bifurcation occurs at $\mu =1+\sqrt{6}$.