PUMaC 2016 · 代数（A 组） · 第 8 题

PUMaC 2016 — Algebra (Division A) — Problem 8

Define the function f : R {− 1 , 1 } → R to be a b ∞ 2 3 ∑ x f ( x ) = . a +1 b +1 2 3 1 − x a,b =0 p 1 Suppose that f ( y ) − f ( ) = 2016. Then y can be written in simplest form as . Find p + q . y q ( R {− 1 , 1 } refers to the set of all real numbers excluding − 1 and 1.) 1

解析

For | x | < 1, we have: a b a b 2 3 − 2 3 ∑ x x − 1 f ( x ) − f ( x ) = + a +1 b +1 a +1 b +1 2 3 − 2 3 1 − x x − 1 a,b ≥ 0 a b a b 2 3 5 · 2 3 ∑ ∑ x + x = a +1 b +1 2 3 1 − x a ≥ 0 b ≥ 0 ( ) ∑ ∑ 2 u +1 a 2 = x a ≥ 0 u ≥ 0 ∑ v = x v> 0 x = 1 − x Similarly, for | x | > 1, we have: 1 − 1 f ( x ) = f ( x ) = . 1 − x 1 1 Since the range of over the domain | x | > 1 is ( −∞ , 0) ∪ (0 , ), it follows that | y | < 1. 1 − x 2 y 2016 Setting = 2016 yields y = , so p + q = 4033 . 1 − y 2017 Problem written by Bill Huang. If you believe that any of these answers is incorrect, or that a problem had multiple reason- able interpretations or was incorrectly stated, you may appeal at tinyurl.com/pumacappeals . All appeals must be in by 1 PM to be considered. 2