26th Putnam 1965 Problem B1 X is the unit n-cube, [0, 1] n . Let k n = ∫ X cos 2 ( π(x 1 + x 2 + ... + x n )/(2n) ) dx 1 ... dx n . What is lim n→∞ k n ?
Let y i = 1 - x i and change variables from x i to y i . The sum (x 1 + x 2 + ... + x n )/(2n) becomes 1/2 - (y 1 + y 2 + ... + y n )/(2n), so the integrand becomes sin 2 ( π(y 1 + y 2 + ... + y n )/(2n) ). So the integral is identical to the original except that cos has been changed to sin. Thus we can add it to the original to get 2k n = ∫ X dx 1 dx 2 ... dx n = 1. So k n = 1/2 for all n. 26th Putnam 1965 © John Scholes jscholes@kalva.demon.co.uk 25 Jan 2002