60th Putnam 1999 Problem A5 Find a constant k such that for any polynomial f(x) of degree 1999, we have |f(0)| ≤ k ∫ -1 1 |f(x)| dx.
The basic idea is simple. Divide the interval [-1, 1] into 2000 subintervals of length 1/1000, then there must be at least one subinterval K such that there is no root of f(x) whose real part falls in K. Take K' to be the central 1/3 of K. Then every point of K' is a distance at least 1/3000 from any root of f(x). Now |f(x)| = ∏|x - α| ≥ (1/3000) 1999 . Hence the integral is at least (1/3000) 2000 . That would work if we were looking for a lower bound for the ratio of the integral to the leading coefficient. But we were asked for the ratio of the integral to the modulus of constant term (which is ∏ | α| ), so we need to estimate ∏|x - α|/|α|. If f(0) = 0, then any constant k works, so we can assume that all the roots α ≠ 0. If |α| ≥ 1, then |x|  1/3000. If |α|  |x - α| > 1/3000. So we have established that |f(x)|/|f(0)| > (1/3000) 1999 on the interval K'. Hence the integral is at least (1/3000) 2000 |f(0)| and we have established the result with k = (1/3000) 2000 . 60th Putnam 1999 © John Scholes jscholes@kalva.demon.co.uk 14 Dec 1999