Competition problem — Algebra — Maths Olympiad Prep

Library / /389 of 462

Algebra Difficulty 6.9 National Olympiad Prove it Ireland

The maximum of two real numbers $a$ and $b$ is defined as follows:
$\max\{a, b\} = \begin{cases} a, & \text{if } a \ge b, \\ b, & \text{otherwise.} \end{cases}$
For any two positive real numbers $x_0 > 0$ , $x_1 > 0$ a sequence of real numbers $x_n$ is defined recursively as given below. Find $x_{2010}$ .
$x_{n+1} = \frac{4 \max\{x_n, 4\}}{x_{n-1}} \quad \text{for } n \ge 1.$

Solution

The recurrence is a version of what is sometimes called the Lyness max equation. All solutions are periodic with period 5 which can be established by computation. Hence $x_{2010} = x_0$ .
The change of variable $x_n = 4y_n$ puts the recurrence in standard form
$y_{n+1} = \frac{\max\{y_n, 1\}}{y_{n-1}}$
after which the periodicity can be established by considering the four cases indicated below:

	$y_0 \le 1$ , $y_1 \le 1$	$y_0 \le 1$ , $y_1 > 1$	$y_0 > 1$ , $y_1 \le 1$	$y_0 > 1$ , $y_1 > 1$
$y_2 =$	$1/y_0$	$y_1/y_0$	$1/y_0$	$y_1/y_0$
$y_3 =$	$1/(y_0 y_1)$	$1/y_0$	$1/y_1$	$\max\{1/y_0, 1/y_1\}$
$y_4 =$	$1/y_1$	$1/y_1$	$y_0/y_1$	$y_0/y_1$
$y_5 =$	$y_0$	$y_0$	$y_0$	$y_0$
$y_6 =$	$y_1$	$y_1$	$y_1$	$y_1$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.