 Solution

Let $x^{\ast }(\lambda )=\sqrt{1-1/\lambda }$ and center the map at this parameter-dependent fixed point:A Taylor polynomial in $X$, with the coefficients expanded at $(x,\lambda )=(1/\sqrt{2},2)$, givesunder the period-doubling scaling $X=O(\sqrt{\mu })$. The coefficients areHere we used $F_x(x^{\ast }(\lambda ),\lambda )=3-2\lambda$, $F_{xx}=-6\lambda x$, and $F_{xxx}=-6\lambda$.

Composing the local map with itself and retaining the terms of order $\mu X$ and $X^3$ givesBesides the fixed solution $X=0$, the two leading solutions therefore satisfyThus the period-two points born at $\lambda =2$ areto leading order. With the coefficients above, the radicand is positive for $\mu >0$.