### 34th Putnam 1973

Problem B6

Show that sin2x sin 2x has two maxima in the interval [0, 2π], at π/3 and 4π/3. Let f(x) = the absolute value of sin2x sin34x sin38x ... sin32n-1x sin 2nx. Show that f(π/3) >= f(x). Let g(x) = sin2x sin24x sin28x ... sin22nx. Show that g(x) ≤ 3n/4n.

Solution

Let h(x) = sin2x sin 2x. Then h(x) = 2 sin3x cos x, and h'(x) = 2 sin2x (3 - 4 sin2x). So h'(x) = 0 at x = 0, π/3, 2π/3, 4π/3, 5π/3, 2π in the interval [0, 2π]. We easily see that h(x) has maxima at π/3 and 4π/3, minima at 2π/3 and 5π/3 and inflexions at 0, 2π. At the inflexions h(x) = 0, at the other stationary values, |h(x)| = (3√3)/8.

Notice that, for n > 1, 2nπ/3 = 2π/3 or 4π/3 (mod 2π). So at x = 2nπ/3, |h(x)| attains its maximum value.

Let fn(x) = sin2x sin34x sin38x ... sin32n-1x sin 2nx. Then f2(x) = h(x) h(2x) and in general fn(x) = fn-1(x) h(2n-1x), so that fn(x) = h(x) h(2x) h(4x) ... h(2n-1x). At x = π/3 each of |h(2ix)| is maximised separately and hence |fn(x)| is maximised.

We have in fact that |fn(x)| ≤ ( (√3)/2 )3n. Hence also |sin x| |fn(x)| |sin22nx| ≤ ( (√3)/2 )3n. But |sin x| |fn(x)| |sin22nx| = g(x)3/2, so |g(x)| ≤ ( (√3)/2 )2n = (3/4)n, and hence g(x) ≤ (3/4)n (g is certainly non-negative).

© John Scholes
jscholes@kalva.demon.co.uk
22 Aug 2001