HMMT 二月 2014 · 冲刺赛 · 第 28 题

HMMT February 2014 — Guts Round — Problem 28

[ 17 ] Let f ( n ) and g ( n ) be polynomials of degree 2014 such that f ( n ) + ( 1) g ( n ) = 2 for n = 2014 1 , 2 , . . . , 4030. Find the coe cient of x in g ( x ). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND Organization Team Team ID#

解析

[ 17 ] Let f ( n ) and g ( n ) be polynomials of degree 2014 such that f ( n ) + ( − 1) g ( n ) = 2 for n = 2014 1 , 2 , . . . , 4030. Find the coefficient of x in g ( x ). 2014 3 Answer: Define the polynomial functions h and h by h ( x ) = f (2 x ) + g (2 x ) and 2014 1 2 1 2 · 2014! 2 x 2 x − 1 h ( x ) = f (2 x − 1) − g (2 x − 1). Then, the problem conditions tell us that h ( x ) = 2 and h ( x ) = 2 2 1 2 for x = 1 , 2 , . . . , 2015. By the Lagrange interpolation formula, the polynomial h is given by 1 2015 2015 ∑ ∏ x − j 2 i h ( x ) = 2 . 1 i − j i =1 j =1 i 6 = j 2014 So the coefficient of x in h ( x ) is 1 ( ) 2015 2015 2015 2014 ∑ ∏ ∑ 1 1 2014 4 · 3 2 i 2 i 2015 − i 2 = 2 ( − 1) = i − j 2014! i − 1 2014! i =1 j =1 i =1 i 6 = j where the last equality follows from the binomial theorem. By a similar argument, the coefficient of 2014 2014 2 · 3 x in h ( x ) is . 2 2014! 1 2014 We can write g ( x ) = ( h ( x/ 2) − h (( x + 1) / 2)). So, the coefficient of x in g ( x ) is 1 2 2 ( ) 2014 2014 2014 1 4 · 3 2 · 3 3 − = . 2014 2014 2014 2 2 · 2014! 2 · 2014! 2 · 2014!