PUMaC 2013 · 代数（B 组） · 第 3 题
PUMaC 2013 — Algebra (Division B) — Problem 3
题目详情
- [ 4 ] Let x = 1 / 20, x = 1 / 13, and 1 2 2 x x ( x + x ) n n +1 n n +1 x = n +2 2 2 x + x n n +1 ∑ ∞ for all integers n ≥ 1. Evaluate (1 / ( x + x )). n n +1 n =1
解析
- If all 4 roots are real, then there are ( ) 3 + 4 − 1 = 15 (B.22) 4 possibilities for f , either by brute-force enumeration or the lemma below. Altogether there are 10 + 18 + 15 = 43 possibilities. Lemma. Let r ( n, N ) be the number of ways of choosing N things from n distinguished choices, allowing repetition in choice but not distinguishing order of choosing. Then ( ) n + N − 1 r ( n, N ) = . (B.23) N Proof. Imagine having N stars in a row and n − 1 vertical bars to place between them. We claim there is a bijection between the possible placements of the bars relative to the horizontal row of stars—where a single bar can be placed either left of the stars, right of the stars, or in between two adjacent stars, and multiple bars can occupy the same position—and the number of ways of choosing N things from n distinguished choices, allowing repetition but not distinguishing choice order. For, add an “immobile” bar to the right of the stars. Now match the n bars with the choices. Let the number of times a given choice was picked equal the number of stars between its bar and the bar immediately to the left. It is straightforward to verify that this is a bijection. Thinking of the bars, sans the immobile one, together with the stars as a row of n + N − 1 objects, the number of allowed placements of the free-moving bars is exactly the number of ways of choosing N things at all once from n + N − 1 choices, as needed. Remark. The proof above has led to this lemma being called “Stars and Bars.” That this is also the nickname of the flag of the defunct Confederate States of America is purely coincidental! References [Ka] R. Kanigel. The Man Who Knew Infinity: a Life of the Genius Ramanujan . New York: Charles Scribner’s Sons (1991). ISBN 0-684-19259-4. 6