39th Putnam 1978 Problem B5 Find the real polynomial p(x) of degree 4 with largest possible coefficient of x 4 such that p( [-1, 1] ) ∈ [0, 1].
Let p(x) = a 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0 . Then a 4 x 4 - a 3 x 3 + a 2 x 2 - a 1 x + a 0 also satisfies the conditions and hence also a 4 x 4 + a 2 x 2 + a 0 . So we may take a 3 = a 1 = 0. Now p(0), p(1/2), p(1) belong to [0, 1], so 0 ≤ a 0 ≤ 1, 0 ≤ a 4 /4 + a 2 /2 + a 0 ≤ 1, 0 ≤ a 4 + a 2 + a 0 ≤ 1. Hence: -2 ≤ -2a 0 ≤ 0, 0 ≤ a 4 + 2a 2 + 4a 0 ≤ 4, -2 ≤ -2a 4 - 2a 2 - 2a 0 ≤ 0. Adding: -4 ≤ -a 4 < 4. So the maximum value of a 4 is at most 4. This can be achieved by 4(x 2 - 1/2) 2 = 4x 4 - 4x 2 + 1. 39th Putnam 1978 © John Scholes jscholes@kalva.demon.co.uk 30 Nov 1999