f : [0, 2π) → [-1, 1] satisfies f(x) = ∑0n (aj sin jx + bj cos jx) for some real constants aj, bj. Also |f(x)| = 1 for just 2n distinct values in the interval. Show that f(x) = cos(nx + k) for some k.
We call f(x) a trigonometric sum of degree n. Let x1, x2, ... , x2n be the points at which f(x) = 1 or -1. Two insights are needed. The first is that f '(xi) = 0. For if the value was non-zero, then f(x) would lie outside [-1, 1] for points sufficiently close (and on the correct side). [Note that there is a minor complication at 0. If f(0) = 1 and f '(0) < 0, then the aberrant points would be just less than 2π.]
The second is that if a trigonometric sum of degree n has 2n roots (counting multiplicities), then it is determined up to a multiplicative constant (and if it has more than 2n roots then it is identically zero). We will establish that later. But 1 - f(x)2 has the 2n roots xi. Also, they all have multiplicity at least 2 since f '(xi) = 0. So we have at least 4n roots (counting multiplicities). Using the familiar formulae cos(A + B) etc we can write 1 - f(x)2 in the same form as f(x) but with the sum from 0 to 2n.
Similarly, f '(x) is a trigonometric sum of degree n with the 2n roots xi. If we square it, each root is doubled, so f '(x)2 is a trigonometric sum of degree 2n with the 4n roots xi (each counted twice). Hence we must have f '(x)2 = multiple of (1 - f(x)2). Both are non-negative, so it must be a positive multiple. Hence we can take the square root to get f '(x) = N √(1 - f(x)2 ) for some fixed N. Solving, we get f(x) = cos(Nx + k). But for this to have just 2n values ±1 we must have N = n.
To prove the result about trigonometric polynomials put z = eix, then f(x) = z-n p(z), where p is a polynomial in z of degree 2n. Hence p has at most 2n roots and is determined by those roots up to a multiplicative constant. Note that if f '(x) = 0 and f(x) = 0, then p(z) = 0 and since 0 = f '(x) = combination of p(z) and p'(z), then p'(z) = 0 also, so double roots of f have corresponding roots of multiplicity at least 2 in p.
23rd Putnam 1962
© John Scholes
5 Feb 2002