HMMT 二月 2003 · 冲刺赛 · 第 9 题

HMMT February 2003 — Guts Round — Problem 9

[6] For x a real number, let f ( x ) = 0 if x < 1 and f ( x ) = 2 x − 2 if x ≥ 1. How many solutions are there to the equation f ( f ( f ( f ( x )))) = x ? 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

解析

For x a real number, let f ( x ) = 0 if x < 1 and f ( x ) = 2 x − 2 if x ≥ 1. How many solutions are there to the equation f ( f ( f ( f ( x )))) = x ? Solution: 2 Certainly 0 , 2 are fixed points of f and therefore solutions. On the other hand, there can be no solutions for x < 0, since f is nonnegative-valued; for 0 < x < 2, we have 0 ≤ f ( x ) < x < 2 (and f (0) = 0), so iteration only produces values below x , and for x > 2, f ( x ) > x , and iteration produces higher values. So there are no other solutions.