2周期点 x=f(f(x)=a2x(1-x)(1-ax(1-x) 得 x1=0,x2=(a-1)/a, x34=(1+a±√a2-2a-3)/2a,a23• 2 周期点 得 ( ( )) (1 )(1 (1 )) 2 x = f f x = a x − x − ax − x 0, ( 1)/ , x1 = x2 = a − a (1 2 3)/ 2 , 3 2 x3,4 = + a  a − a − a a 