HMMT 二月 2019 · 团队赛 · 第 10 题
HMMT February 2019 — Team Round — Problem 10
题目详情
- [ 60 ] Prove that for all positive integers n , all complex roots r of the polynomial 2 n 2 n − 1 n +1 n n − 1 P ( x ) = (2 n ) x + (2 n − 1) x + · · · + ( n + 1) x + nx + ( n + 1) x + · · · + (2 n − 1) x + 2 n lie on the unit circle (i.e. | r | = 1).
解析
- [ 60 ] Prove that for all positive integers n , all complex roots r of the polynomial 2 n 2 n − 1 n +1 n n − 1 P ( x ) = (2 n ) x + (2 n − 1) x + · · · + ( n + 1) x + nx + ( n + 1) x + · · · + (2 n − 1) x + 2 n lie on the unit circle (i.e. | r | = 1). Proposed by: Faraz Masroor Note that neither 0 nor 1 are roots of the polynomial. Consider the function n n − n n − 1 − n +1 1 − 1 Q ( x ) = P ( x ) /x = (2 n ) x + (2 n ) x + (2 n − 1) x + (2 n − 1) x + · · · + ( n + 1) x + ( n + 1) x + n. All 2 n of the complex roots of P ( x ) will be roots of Q ( x ). iθ If | x | = 1, then x = e , and n − n n − 1 − n +1 − 1 Q ( x ) = (2 n )( x + x ) + (2 n − 1)( x + x ) + · · · + ( n + 1)( x + x ) + n inθ − inθ i ( n − 1) θ − i ( n − 1) θ iθ − iθ = (2 n )( e + e ) + (2 n − 1)( e + e ) + · · · + ( n + 1)( e + e ) + n = (2 n )(2 cos( nθ )) + (2 n − 1)(2 cos(( n − 1) θ )) + · · · + ( n + 1)(cos( θ )) + n, which is real. Thus on the unit circle, we have Q ( x ) is real, and we want to show it has 2 n roots there. Rewrite 2 n 2 n − 1 n +1 n n − 1 P ( x ) = (2 n ) x + (2 n − 1) x + · · · + ( n + 1) x + nx + ( n + 1) x + · · · + 2 n 2 n 2 n − 1 = (2 n )( x + x + · · · + 1) 2 n − 1 2 n − 2 n n − 1 n − 2 2 − ( x + 2 x + · · · + ( n − 1) x + nx + ( n − 1) x + · · · + 2 x + x ) 2 n +1 x − 1 2 n − 2 2 n − 3 n n − 1 n − 2 = 2 n − x ( x + 2 x + · · · + ( n − 1) x + nx + ( n − 1) x + · · · + 2 x + 1) x − 1 2 n +1 x − 1 n − 1 n − 2 2 = 2 n − x ( x + x + · · · + x + 1) x − 1 ( ) 2 2 n +1 n x − 1 x − 1 = 2 n − x , x − 1 x − 1 and thus ( ) 2 2 n +1 n 2 n x − 1 x x − 1 Q ( x ) = − . n n x x − 1 x x − 1 2 π i j 2 n Consider the roots of unity r = e , for j = 0 to 2 n − 1. There are 2 n such roots of unity: they all j 2 n n n have r = 1, and they alternate between those which satisfy r = 1 or r = − 1. At those x = r , if j j j j n r = 1 but x 6 = 1, then j ( ) 2 2 n +1 n 2 n x − 1 x x − 1 Q ( x ) = − n n x x − 1 x x − 1 ( ) 2 1 x − 1 1 − 1 = 2 n − x = 2 n > 0 . x − 1 x − 1 At x = 1, we can easily see Q (1) > 0. n If r = − 1, then j ( ) 2 2 n +1 n 2 n x − 1 x x − 1 Q ( x ) = − n n x x − 1 x x − 1 ( ) 2 1 x − 1 − 1 − 1 = − 2 n + x x − 1 x − 1 4 x = − 2 n + 2 ( x − 1) 4 = − 2 n + x − 2 + 1 /x 4 = − 2 n + < − 2 n − 4 < 0 2 π 2 cos( j ) − 2 2 n since the denominator of this second term is strictly negative ( j 6 = 0). Thus at each of the 2 n -roots of unity, Q ( x ) alternates in sign, and because Q ( x ) is real and continuous on the unit circle, it has at least one root between every pair of consecutive roots of unity. Since there are 2 n of these pairs, and we know that Q ( x ) has exactly 2 n roots (by the Fundamental Theorem of Algebra), we have found all of Q ’s roots, and therefore those of P .