f **:** [0, 2π) → [-1, 1] satisfies f(x) = ∑_{0}^{n} (a_{j} sin jx + b_{j} cos jx) for some real constants a_{j}, b_{j}. Also |f(x)| = 1 for just 2n distinct values in the interval. Show that f(x) = cos(nx + k) for some k.

**Solution**

We call f(x) a trigonometric sum of degree n. Let x_{1}, x_{2}, ... , x_{2n} be the points at which f(x) = 1 or -1. Two insights are needed. The first is that f '(x_{i}) = 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 x_{i}. Also, they all have multiplicity at least 2 since f '(x_{i}) = 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 x_{i}. If we square it, each root is doubled, so f '(x)^{2} is a trigonometric sum of degree 2n with the 4n roots x_{i} (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 = e^{ix}, 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.

© John Scholes

jscholes@kalva.demon.co.uk

5 Feb 2002