IMO 1981 SL 16

Origin: GBR

Problem

A sequence of real numbers u1, u2, u3, . . . is determined by u1 and the following recurrence relation for n \geq1: 4un+1 = 3\sqrt64un + 15. Describe, with proof, the behavior of un as n \to\infty.

Solution

The sequence {un} is bounded, whatever u1 is. Indeed, assume the opposite, and let um be the first member of the sequence such that |um| > max{2, |u1|}. Then |um−1| = |u3 m −15/64| > |um|, which is impos- sible. Next, let us see for what values of um, um+1 is greater, equal, or smaller, respectively. If um+1 = um, then um = u3 m+1 −15/64 = u3 m −15/64; i.e., um is a root of x3 −x −15/64 = 0. This equation factors as (x + 1/4)(x2 −x/4 − 15/16) = 0, and hence um is equal to x1 = (1 − \sqrt 61)/8, x2 = −1/4, or x3 = (1 + \sqrt 61)/8, and these are the only possible limits of the sequence. Each of um+1 > um, um+1 < um is equivalent to u3 m −um −15/64 < 0 and u3 m −um −15/64 > 0 respectively. Thus the former is satisfied for um in the interval I1 = (−\infty, x1) or I3 = (x2, x3), while the latter is satisfied for um in I2 = (x1, x2) or I4 = (x3, \infty). Moreover, since the function f(x) = x + 15/64 is strictly increasing with fixed points x1, x2, x3, it follows that um will never escape from the interval I1, I2, I3, or I4 to which it belongs initially. Therefore: (1) if u1 is one of x1, x2, x3, the sequence {um} is constant; (2) if u1 \inI1, then the sequence is strictly increasing and tends to x1; (3) if u1 \inI2, then the sequence is strictly decreasing and tends to x1; (4) if u1 \inI3, then the sequence is strictly increasing and tends to x3; (5) if u1 \inI4, then the sequence is strictly decreasing and tends to x3.