28th Putnam 1967

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Problem A1

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

 

Solution

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