PUMaC 2012 · 加试 · 第 2 题

PUMaC 2012 — Power Round — Problem 2

It is not necessary to do the problems in order, although it is a good idea to read all the problems, so that you know what is permissible to assume when doing each problem. However, please collate the solutions in order in your solution packet. Each problem should start on a new page, and solutions should be written on one side of the paper only. Each page should also have on it the team name and problem number.

解析

It is possible to solve this problem by bashing out the computations. We give a rather slicker proof: First show D ( f f ) = f Df + f Df for all 1 2 1 2 2 1 f , f ∈ C [ X ] , by writing out both sides. Then 1 2 D ( f f ) Df Df 1 2 1 2 = + f f f f 1 2 1 2 3 Since ( Df ) /f = ( Dg ) /g and DA = DB = 0 , we apply the above lemma to the linear factors of f and g to obtain 1 1 1 1

. . . + = + . . . + X − a X − a X − b X − b 1 m 1 n Since m = n , it suffices to prove that the a and b are the same up to j j ordering. We know { a } and { b } are at least the same set of numbers, j j because both sides must blow up in absolute value when X gets very close to a root on one side. To show that the roots occur with the same multiplicity on both sides, cancel out all common linear factors from f and g to obtain new polynomials f and g , respectively, which do not 0 0 share any linear factors. Repeating the above argument for f , g shows 0 0 f = g = 1 , as needed. 0 0 3 Algebraic Numbers (32 points) 3.1 (7 points) Let α ∈ Q .