HMMT 二月 2015 · 代数 · 第 3 题
HMMT February 2015 — Algebra — Problem 3
题目详情
- Let p be a real number and c 6 = 0 an integer such that ( ) 10 1 − (1 + x ) p c − 0 . 1 < x < c + 0 . 1 10 1 + (1 + x ) − 100 − 100 for all (positive) real numbers x with 0 < x < 10 . (The exact value 10 is not important. You could replace it with any “sufficiently small number”.) Find the ordered pair ( p, c ).
解析
- Let p be a real number and c 6 = 0 an integer such that ( ) 10 1 − (1 + x ) p c − 0 . 1 < x < c + 0 . 1 10 1 + (1 + x ) − 100 − 100 for all (positive) real numbers x with 0 < x < 10 . (The exact value 10 is not important. You could replace it with any “sufficiently small number”.) Find the ordered pair ( p, c ). 1 Note that this does actually have integer solutions by Bezout’s identity, as gcd(13 · 31 , 5 · 31 , 5 · 13) = 1. Algebra Answer: ( − 1 , − 5) This is essentially a problem about limits, but phrased concretely in terms of − 100 “small numbers” (like 0 . 1 and 10 ). 10 2 1 − (1+ x ) − 10 x + O ( x ) We are essentially studying the rational function f ( x ) := = , where the “big-O” 10 1+(1+ x ) 2+ O ( x ) 2 notation simply make precise the notion of “error terms”. − 10 x Intuitively, f ( x ) ≈ = − 5 x for “small nonzero x ”. (We could easily make this more precise if 2 we wanted to, by specifying the error terms more carefully, but it’s not so important.) So g ( x ) := p p +1 x f ( x ) ≈ − 5 x for “small nonzero x ”. • If p + 1 > 0, g will approach 0 (“get very small”) as x approaches 0 (often denoted x → 0), so there’s no way it can stay above the lower bound c − 0 . 1 for all small nonzero x . • If p + 1 < 0, g will approach −∞ (“get very large in the negative direction”) as x → 0, so there’s no way it can stay below the upper bound c + 0 . 1 for all small nonzero x . • If p + 1 = 0, g ≈ − 5 becomes approximately constant as x → 0. Since c is an integer , we must have c = − 5 (as − 5 is the only integer within 0 . 1 of − 5). − 100 Remark. Why does ( p, c ) = ( − 1 , − 5) actually satisfy the inequality? This is where the 10 kicks in: for such small values of x , the “error” | g ( x ) − ( − 5) | of the approximation g ≈ − 5 does actually lie within the permitted threshold of ± 0 . 1. (You can easily work out the details yourself, if you’re interested. It’s something you might want to work out once or twice in your life, but rational functions are “well-behaved” enough that we can usually rely on our intuition in these kinds of scenarios.)