28th Putnam 1967

We are given a positive integer n and real numbers a_i such that |∑₁ⁿ a_k sin kx| ≤ |sin x| for all real x. Prove |∑₁ⁿ k a_k| ≤ 1.

Solution

Put f(x) = ∑₁ⁿ a_k sin kx. We note that ∑₁ⁿ k a_k = f '(0).

We also have f '(0) = lim ( f(x) - f(0) )/x = lim f(x)/x = lim f(x)/sin x lim (sin x)/x = lim f(x)/sin x. But |f(x)| ≤ |sin x|, so |f(x)/sin x| ≤ 1 and hence |f '(0)| = |lim f(x)/sin x| ≤ 1.

28th Putnam 1967

© John Scholes
jscholes@kalva.demon.co.uk
14 Jan 2002