Problem:
Let f(x)=4x−x2. Given x0​, consider the sequence defined by xn​=f(xn−1​) for all n≥1. For how many real numbers x0​ will the sequence x0​, x1​, x2​, … take on only a finite number of different values?
Answer Choices:
A. 0
B. 1 or 2
C. 3, 4, 5 or 6
D. more than 6 but finitely many
E. infinitely many
Solution:
Note that x0​=0 gives the constant sequence 0,0,…, since f(0)=4⋅0−02=0. Because f(4)=0,x0​=4 gives the sequence 4,0,0,… with two different values. Similarly, f(2)=4 so x0​=2 gives the sequence 2,4,0,0,… with three values. In general, if x0​=an​ gives the sequence an​,an−1​,…,a2​,a1​,a1​,… with n different values, and f(an+1​)=an​, then x0​=an+1​ gives a sequence with n+1 different values. (It could not happen that an+1​=ai​ for some i<n+1; why?) Thus, it follows by induction that there is a sequence with n distinct values for every positive integer n− as soon as we verify that there is always a real number an+1​ such that f(an+1​)=an​. This follows from the quadratic formula: First, the solutions to f(an+1​)=4an+1​−an+12​=an​ are an+1​=2±4−an​​. Second, if 0≤an​≤4, then an+1​ is real; in fact, 0≤an+1​≤4 (why?). Third, 0=a1​≤4. Thus, by induction, all terms satisfy 0≤an​≤4; in particular, they are all real.