Show that sin^{2}x sin 2x has two maxima in the interval [0, 2π], at π/3 and 4π/3. Let f(x) = the absolute value of sin^{2}x sin^{3}4x sin^{3}8x ... sin^{3}2^{n-1}x sin 2^{n}x. Show that f(π/3) >= f(x). Let g(x) = sin^{2}x sin^{2}4x sin^{2}8x ... sin^{2}2^{n}x. Show that g(x) ≤ 3^{n}/4^{n}.

**Solution**

Let h(x) = sin^{2}x sin 2x. Then h(x) = 2 sin^{3}x cos x, and h'(x) = 2 sin^{2}x (3 - 4 sin^{2}x). 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, 2^{n}π/3 = 2π/3 or 4π/3 (mod 2π). So at x = 2^{n}π/3, |h(x)| attains its maximum value.

Let f_{n}(x) = sin^{2}x sin^{3}4x sin^{3}8x ... sin^{3}2^{n-1}x sin 2^{n}x. Then f_{2}(x) = h(x) h(2x) and in general f_{n}(x) = f_{n-1}(x) h(2^{n-1}x), so that f_{n}(x) = h(x) h(2x) h(4x) ... h(2^{n-1}x). At x = π/3 each of |h(2^{i}x)| is maximised separately and hence |f_{n}(x)| is maximised.

We have in fact that |f_{n}(x)| ≤ ( (√3)/2 )^{3n}. Hence also |sin x| |f_{n}(x)| |sin^{2}2^{n}x| ≤ ( (√3)/2 )^{3n}. But |sin x| |f_{n}(x)| |sin^{2}2^{n}x| = 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