19th APMC 1996 Problem 3 The polynomials p n (x) are defined by p 0 (x) = 0, p 1 (x) = x, p n+2 (x) = x p n+1 (x) + (1 - x) p n (x). Find the real roots of each p n (x).
0 Solution We show by induction that p n (x) = (x/(x-2)) ((x-1) n -1). For n = 1, rhs = x = lhs. For n = 2, rhs = (x/(x-2)) (x 2 -2x) = x 2 = lhs. So suppose the result is true for n+1 and n+2. Then p n+2 (x) = xp n+1 (x) + (1-x)p n (x) = xp n+1 (x) - (x-1)p n (x) = (x/(x-2)(x(x-1) n+1 -x) - (x/(x-2))((x-1) n+1 -x+1) = (x/(x-2)((x-1) n+2 -1), so the result is true for n+2. Hence it is true for all x. The only real roots of y n = 1 are 1 and possibly -1, so the only real roots of (x-1) n = 1 are 2 and possibly 0. But any root x = 2 is removed by the denominator, so the only real root is 0. 19th APMC 1996 © John Scholes jscholes@kalva.demon.co.uk 28 Nov 2003 Last corrected/updated 28 Nov 03