I'm trying to find the first 3 period doubling bifurcations for this function
G(x) = acos(l*sin(2x))
A period doubling bifurcation will occur where G'(x) = -1.
I can take the derivative ...
G'(x) = [-1 / ((1 - (l*2cos(2x))^1/2]*l*2cos(2x)
but in order to find the next two bifurcations, I have to take the G' o G'... and so on. This gets hairy quick.
So I'm asking for some help in simplifying the above system.
I know that I can do this (as reviewed in another thread):
cos(G(x)) = l*sin(2x)
But, I'm not really seeing how that helps...
(yes. homework. sort of... this problem was on a test that I've already taken... but it bothered me that I couldn't figure it out in the allotted time.)
Thanks for the help.
