HMMT 二月 2015 · 代数 · 第 3 题

英文解析

3. Let p be a real number and c 6 = 0 an integer such that
   ( )
   1 − (1 + x )10
   c − 0 . 1 < x < c + 0 . 1 p
   1 + (1 + x )10 − 100 − 100
   for all (positive) real numbers x with 0 < x < 10 . (The exact value 10 is not important. Youcould replace it with any “sufficiently small number”.)
   Find the ordered pair ( p, c ).
   Note that this does actually have integer solutions by Bezout’s identity, as gcd(13 · 31 , 5 · 31 , 5 · 13) = 1.1
   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”
   1+(1+ x ) 2+ O ( x )10
   notation simply make precise the notion of “error terms”.2 − 10 x
   Intuitively, f ( x ) ≈ = − 5 x for “small nonzero x ”. (We could easily make this more precise ifwe wanted to, by specifying the error terms more carefully, but it’s not so important.) So g ( x ) :=2
   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), sothere’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’sno 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 musthave 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 kicksin: for such small values of x , the “error” | g ( x ) − ( − 5) | of the approximation g ≈ − 5 does actuallylie within the permitted threshold of ± 0 . 1. (You can easily work out the details yourself, if you’reinterested. It’s something you might want to work out once or twice in your life, but rational functionsare “well-behaved” enough that we can usually rely on our intuition in these kinds of scenarios.)