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Problem of the Month
Problem 0: September 2023

In this problem, f will always be a function defined by f(r)=ar+bcr+d where a, b, c, and d are integers. These integers will vary throughout the parts of the problem.

Given such a function f and a rational number r1, we can generate a sequence r1,r2,r3, by taking rn=f(rn1) for each n2. That is, r2=f(r1), r3=f(r2), r4=f(r3), and so on. Unless there is some point in the sequence where f(rn1) is undefined, a sequence of this form can be made arbitrarily long.

These sequences behave in different ways depending on the function f and the starting value r1. This problem explores some those behaviours.

  1. Suppose f(r)=2r1r+2.

    1. With r1=32, compute r2, r3, and r4.

    2. Find a rational number r1 with the property that r2 is defined, but r3 is not defined.

  2. Suppose f(r)=r+32r1.

    1. With r1=37, compute r2, r3, r4, and r5.

    2. Determine all rational values of r1 with the property that there is some integer n1 for which f(rn) is undefined. For all other values of r1, find simplified formulas for r2023 and r2024 in terms of r1.

  3. Suppose f(r)=r+2r+1.

    1. With r1=1, compute r2 through r9. Write down decimal approximations of r2 through r9 (after computing them exactly).

    2. Suppose r is a positive rational number. Prove that |f(r)2r2|=|12r+1|

    3. Suppose r1 is a positive rational number. Prove that |rn2|<12n1|r12| for each n2. Use this result to convince yourself that as n gets large, rn gets close to 2, regardless of the choice of the positive value r1. Can you modify f slightly so that the sequence always approaches 3?

  4. Explore the behaviour of the sequences generated by various values of r1 for each of the functions below. Detailed solutions will not be provided, but a brief discussion will. f(r)=r3r2,f(r)=r15r+3,f(r)=r1r+2,f(r)=2r+23r+3,f(r)=r+1r2