Let k be the real cube root of 2. Find integers A, B, C, a, b, c such that | (Ax2 + Bx + C)/(ax2 + bx + c) - k | < | x - k | for all non-negative rational x.
Solution
Taking the limit, we must have (Ak2+ Bk + C) = k(ak2 + bk + c), so A = b, B = c, C = 2a. Now notice that (bx2 + cx + 2a) - k(ax2 + bx + c) = (b - ak)x2 + (c - bk)x + (2a - ck) = (x - k)( (b - ak)x + c - ak2). So we require | (b - ak)x + c - ak2| < | ax2 + bx + c | for all x ≥ 0.
There are many ways to satisfy this. For example, take a = 1, b = c = 2. Then (b - ak)x is always positive and less than bx for positive x, and c - ak2 is positive and less than c.
© John Scholes
jscholes@kalva.demon.co.uk
13 Jul 2002