1. The problem statement, all variables and given/known data Consider the families of iterating functions Fλ(x) = λ(x3 - x). Fλ(x) undergoes a bifurcation at λ=1, about the fixed point x=0. Figure out what ilk of bifurcation is occurring for Fλ(x) and prove your assertion rigorously. 2. Relevant equations My book says this about period-doubling bifurcations, and I need to prove all four of these to prove the problem. Definition: A one-parameter family of functions Fλ undergoes a period-doubling bifurcation at the parameter value λ=λ0 if there is an open interval and an ε such that: 1. For each λ in the interval [λ0 - ε, λ0 + ε], there is a unique fixed point pλ for Fλ in I. 2.For λ0 - ε < λ < λ0, Fλ has no cycles of period 2 in I and pλ is attracting (resp. repelling). 3.For λ0 < λ < λ0 + ε, there is a unique 2-cycle q1λ,q2λ in I with Fλ(q1λ)=q2λ. This 2-cycle is attracting (resp. repelling). Meanwhile, the fixed point pλ is repelling (resp. attracting). 4.As λ -> λ0, we have qiλ -> pλ0 Also, these theorems are necessary (I think). Chain Rule Along A Cycle: Suppose x0, x2, ..., xn-1 lie on a cycle of period n for F with xi = Fi(x0). Then (Fn)'(x0) = F'(xn-1) * ... * F'(x1) * F'(x0). The corollary for this is: Suppose x0, x1, ..., xn-1 lie on an n-cycle for F. Then (Fn)'(x0) = (Fn)'(x1) = ... = (Fn)'(xn-1) 3. The attempt at a solution\ I have almost no idea where to start the proof. So far I have: The interval I can = (-1, 1). To answer 1., I know that there are fixed points at 0, +sqrt(2), -sqrt(2). So to have a unique fixed point pλ in the interval [λ0 - ε, λ0 + ε], 0<ε<sqrt(2). Can I just choose an arbitrary ε, like ε=1? To answer 2., if ε=1, λ0 - ε < λ < λ0, so 0 - 1 < λ < 0, so -1<λ<0. I used algebra and got 2-cycles for x = -1, -.7548777, 0, 1 1.4655712, but all but x=1.4655712 lie in the interval I. Could someone check this? I may have made a mistake in my algebra. I don't see why else I would get this. I really appreciate the help!!