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.