HMMT 二月 2011 · ALGCALC 赛 · 第 19 题
HMMT February 2011 — ALGCALC Round — Problem 19
题目详情
- Let 1 F ( x ) = , 5 2011 (2 − x − x ) ∞ ∑ n and note that F may be expanded as a power series so that F ( x ) = a x . Find an ordered pair of n n =0 a n positive real numbers ( c, d ) such that lim = c . d n →∞ n √ 2 2
解析
- Let 1 F ( x ) = , 5 2011 (2 − x − x ) ∞ ∑ n and note that F may be expanded as a power series so that F ( x ) = a x . Find an ordered pair of n n =0 a n positive real numbers ( c, d ) such that lim = c . d n →∞ n 1 5 Answer: ( , 2010) First notice that all the roots of 2 − x − x that are not 1 lie strictly 2011 6 2010! 5 outside the unit circle. As such, we may write 2 − x − x as 2(1 − x )(1 − r x )(1 − r x )(1 − r x )(1 − r x ) 1 2 3 4 (1 − x ) 1 b b b 0 1 4 where | r | < 1, and let = + + . . . + . We calculate b as lim = i 5 0 x → 1 5 (2 − x − x ) (1 − x ) (1 − r x ) (1 − r x ) (2 − x − x ) 1 4 ( − 1) 1 lim = . 4 x → 1 ( − 1 − 5 x ) 6 Now raise the equation above to the 2011th power. ( ) 2011 1 1 / 6 b b 1 4 = + + . . . + 5 2011 (2 − x − x ) (1 − x ) (1 − r x ) (1 − r x ) 1 4 Expand the right hand side using multinomial expansion and then apply partial fractions. The result − k − k will be a sum of the terms (1 − x ) and (1 − r x ) , where k ≤ 2011. i − k Since | r | < 1, the power series of (1 − r x ) will have exponentially decaying coefficients, so we only i i ( ) n + k − 1 − k n − k need to consider the (1 − x ) terms. The coefficient of x in the power series of (1 − x ) is , k − 1 which is a ( k − 1)th degree polynomial in variable n . So when we sum up all coefficients, only the − 2011 2010 power series of (1 − x ) will have impact on the leading term n . 1 − 2011 2011 The coefficient of the (1 − x ) term in the multinomial expansion is ( ) . The coefficient of 6 ( ) n +2010 n − 2011 1 2010 the x term in the power series of (1 − x ) is = n + . . . . Therefore, ( c, d ) = 2010 2010! 1 ( , 2010). 2011 6 2010! √ 2 2