HMMT 二月 2017 · ALGNT 赛 · 第 6 题

英文解析

6. A polynomial P of degree 2015 satisfies the equation P ( n ) = for n = 1 , 2 , . . . , 2016. Findn 2
   b 2017 P (2017) c .
   Proposed by: Alexander Katz
   Answer: − 9
   2 2
   Let Q ( x ) = x P ( x ) − 1. Then Q ( n ) = n P ( n ) − 1 = 0 for n = 1 , 2 , . . . , 2016, and Q has degree 2017.
   Thus we may write
   Q ( x ) = x P ( x ) − 1 = ( x − 1)( x − 2) . . . ( x − 2016) L ( x )2
   where L ( x ) is some linear polynomial. Then Q (0) = − 1 = ( − 1)( − 2) . . . ( − 2016) L (0), so L (0) = − .1
   2016!
   Now note that
   ′ 2 ′
   Q ( x ) = x P ( x ) + 2 x p ( x )
   2016
   ∑
   ′
   = ( x − 1) . . . ( x − ( i − 1))( x − ( i + 1)) . . . ( x − 2016) L ( x ) + ( x − 1)( x − 2) . . . ( x − 2016) L ( x )
   i =1
   Thus
   ( )
   2016! 2016! 2016!
   ′ ′
   Q (0) = 0 = L (0) + + . . . + + 2016! L (0) − 1 − 2 − 2016
   ( )
   1 1 1 H
   ′
   2016
   whence L (0) = L (0) + + . . . + = − , where H denotes the n th harmonic number.
   1 2 2016 2016!n
   H x +1
   2016
   As a result, we have L ( x ) = − . Then
   2016!
   ( )
   2017 H + 1
   2016
   Q (2017) = 2017 P (2017) − 1 = 2016! −2
   2016!
   which is − 2017 H − 1. Thus
   2016 − H
   2016
   P (2017) = .
   2017
   From which we get 2017 P (2017) = − H . It remains to approximate H . We alter the well known
   2016 2016
   approximation
   ∫
   1 n
   H ≈ dx = log xn
   1 xto
   ∫
   1 1 1 1 n
   H ≈ 1 + + dx = 1 + + log(2016) − log(3) ≈ log(2016) +
   2 x 2 2 nso that it suffices to lower bound log(2016). Note that e ≈ 20, which is close enough for our purposes.33
   6 7 3 5 0 . 6 7 . 6
   Then e ≈ 400 = ⇒ e ≈ 1080, and e ≈ 20 < 2 = ⇒ e << 2 = ⇒ e < 2016, so thatlog(2016) > 7 . 6. It follows that H ≈ log(2016) + 0 . 5 = 7 . 6 + 0 . 5 > 8 (of course these are loose
   2016
   estimates, but more than good enough for our purposes). Thus − 9 < 2017 P (2017) < − 8, making ouranswer − 9 .
   Alternatively, a well-read contestant might know that H ≈ log n + γ , where γ ≈ . 577 is the Eulern
   Mascheroni constant. The above solution essentially approximates γ as 0 . 5 which is good enough forour purposes.