HMMT 二月 2010 · TEAM1 赛 · 第 9 题

HMMT February 2010 — TEAM1 Round — Problem 9

[ 30 ] Let p ( x ) = a x + a x + . . . + a be a polynomial with complex coefficients such that a 6 = 0 n n − 1 0 i a n − 1 i − 1 for all i . Prove that | r | ≤ 2 max | | for all roots r of all such polynomials p . Here we let | z | i =0 a i denote the absolute value of the complex number z .

解析

[ 30 ] Let p ( x ) = a x + a x + . . . + a be a polynomial with complex coefficients such that a 6 = 0 n n − 1 0 i a n − 1 i − 1 for all i . Prove that | r | ≤ 2 max | | for all roots r of all such polynomials p . Here we let | z | i =0 a i denote the absolute value of the complex number z . n n − 1 n Solution: If r is a root, then − a r = a r + . . . + a . By the Triangle Inequality, |− a r | ≤ n n − 1 0 n n − 1 n n − 1 | a r | + ... + | a | . Rearranging this inequality yields | a r | − | a r | − ... − | a | ≤ 0. n − 1 0 n n − 1 0 a i − 1 Now suppose | r | = k max | | . Applying this over values of i ranging from m + 1 to n (assuming a i n | a r | n m m + 1 ≤ n ), we get | a r | ≤ . This, along with the above equation, yields: m n − m k Team Round A ( ) 1 1 1 1 n | a r | · 1 − − − − . . . − = 0 n 2 3 n k k k k 1 1 This is only true when a = 0, r = 0, or (1 − − − . . . ) = 0. The first option is impossible by the n 2 k k constraints in the problem. The second option implies k = 0. The third option implies that k < 2; a n − 1 1 1 1 i − 1 otherwise (1 − − − . . . − ) would always remain positive. Either way, | r | ≤ 2 max | | . 2 n i =0 k k k a i