Past exam of the topics in applied mathematics course of the University of Oxford / 2007 / A / 1 / iii / Solution

The fixed-point multiplier decreases continuously with $\gamma$ and first reaches the unit circle at $-1$. Solving $1-{\gamma }_{\ast }(1-\beta /\alpha )=-1$ givesIt is stable for $1<\gamma <{\gamma }_{\ast }$ and unstable for $\gamma >{\gamma }_{\ast }$. At the critical value linear stability is inconclusive. The crossing at $-1$ is the local signature of a period-doubling bifurcation, rather than a crossing through $+1$.