HMMT 二月 2015 · 代数 · 第 7 题
HMMT February 2015 — Algebra — Problem 7
题目详情
- Suppose ( a , a , a , a ) is a 4-term sequence of real numbers satisfying the following two conditions: 1 2 3 4 • a = a + a and a = a + a ; 3 2 1 4 3 2 • there exist real numbers a, b, c such that 2 an + bn + c = cos( a ) n for all n ∈ { 1 , 2 , 3 , 4 } . Compute the maximum possible value of cos( a ) − cos( a ) 1 4 over all such sequences ( a , a , a , a ). 1 2 3 4 2
解析
- Suppose ( a , a , a , a ) is a 4-term sequence of real numbers satisfying the following two conditions: 1 2 3 4 • a = a + a and a = a + a ; 3 2 1 4 3 2 • there exist real numbers a, b, c such that 2 an + bn + c = cos( a ) n for all n ∈ { 1 , 2 , 3 , 4 } . Compute the maximum possible value of cos( a ) − cos( a ) 1 4 over all such sequences ( a , a , a , a ). 1 2 3 4 √ Answer: − 9 + 3 13 Let f ( n ) = cos a and m = 1. The second (“quadratic interpolation”) n condition on f ( m ), f ( m + 1), f ( m + 2), f ( m + 3) is equivalent to having a vanishing third finite difference f ( m + 3) − 3 f ( m + 2) + 3 f ( m + 1) − f ( m ) = 0 . 3 This is analogous to the “number theoretic” proof of the uniqueness of the base 2 expansion of a nonnegative integer. Algebra This is equivalent to f ( m + 3) − f ( m ) = 3 [ f ( m + 2) − f ( m + 1)] ⇐⇒ cos( a ) − cos( a ) = 3 (cos( a ) − cos( a )) m +3 m m +2 m +1 ( ) ( ) a + a a − a m +2 m +1 m +2 m +1 = − 6 sin sin 2 2 ( ) ( ) a a m +3 m = − 6 sin sin . 2 2 ( ) ( ) a a m +3 m Set x = sin and y = sin . Then the above rearranges to 2 2 2 2 2 2 (1 − 2 x ) − (1 − 2 y ) = − 6 xy ⇐⇒ x − y = 3 xy. √ − 3 ± 13 2 2 Solving gives y = x . The expression we are trying to maximize is 2( x − y ) = 6 xy , so we want 2 √ − 3+ 13 x , y to have the same sign; thus y = x . 2 Then | y | ≤ | x | , so since | x | , | y | ≤ 1, to maximize 6 xy we can simply set x = 1, for a maximal value of √ √ − 3+ 13 6 · = − 9 + 3 13. 2 2