HMMT 二月 2010 · 代数 · 第 9 题

英文解析

9. [ 7 ] Let f ( x ) = cx ( x − 1), where c is a positive real number. We use f ( x ) to denote the polynomialnobtained by composing f with itself n times. For every positive integer n , all the roots of f ( x ) arereal. What is the smallest possible value of c ?
   n c
   Answer: 2 We first prove that all roots of f ( x ) are greater than or equal to − and less than orc n c − 1 1 c 1 c 4
   equal to 1 + . Suppose that r is a root of f ( x ). If r = − , f ( r ) = { } and − < < 1 + since
   4 4 2 4 2 4
   c is positive. Suppose r 6 = − ; by the quadratic formula, there exist two complex numbers r , r suchc
   1 2
   n 14
   that r + r = 1 and f ( r ) = f ( r ) = r . Thus all the roots of f ( x ) (except ) come in pairs that sum
   1 2 1 2
   c 2
   n n +1 − 1
   to 1. No root r of f ( x ) can be less than − , otherwise f ( x ) has an imaginary root, f ( r ). Also,
   c c 4
   no root r of f ( x ) can be greater than 1 + , otherwise its “conjugate” root will be less than − .n
   4 4
   √
   ( )
   1 4 x n
   Define g ( x ) = 1 + 1 + , the larger inverse of f ( x ). Note that g ( x ) is the largest element of
   2 c − n n cf ( x ) (which is a set). g (0) should be less than or equal to 1 + for all n . Let x be the nonzeroreal number such that g ( x ) = x ; then cx ( x − 1) = x = ⇒ x = 1 + . x < g ( x ) < x if x > x 140
   0 0 0 0 0 o 0 0
   and x < g ( x ) < x if x < x ; it can be proved that g converges to x . Hence we have the requirementnc
   0 0 0
   1 cthat x = 1 + ≤ 1 + = ⇒ c ≥ 2.
   c 40
   c 3
   We verify that c = 2 is possible. All the roots of f n ( x ) will be real if g (0) ≤ 1 + = . We know−
   4 2
   3 3 3 3
   2 n n +1
   that 0 < = ⇒ g (0) < , so g (0) < and g (0) < g (0) < for all n . Therefore all the roots of
   2 2 2 2
   f ( x ) are real.n