Let R be the reals. Define a_{n} by a_{1} = α ∈ R, a_{n+1} = cos a_{n}. Show that a_{n} converges to a limit independent of α.

**Solution**

|cos x| ≤ 1, so -1 ≤ a_{2} ≤ 1. Hence 0 < a_{3} < 1 and so 0 < a_{n} < 1 for n ≥ 3.

cos x is strictly monotonic decreasing over the range 0 to 1, so if a_{n} < a_{n+1} then a_{n+2} < a_{n+1}. Now the gradient of cos x is greater than -1 throughout the interval (0, 1), so if a_{n} < a_{n+1} then cos a_{n} - cos a_{n+1} < a_{n+1} - a_{n}. Hence a_{n} < a_{n+2} < a_{n+1}. Similarly, if a_{n} > a_{n+1}, then a_{n} > a_{n+2} > a_{n+1}. Thus either a_{2n} is an increasing sequence bounded above and a_{2n+1} is a decreasing sequence bounded below or vice versa. Hence both a_{2n} and a_{2n+1} converge.

By the mean value theorem |a_{n+1} - a_{n+2}| = |cos a_{n} - cos a_{n+1}| = sin k |a_{n} - a_{n+1}| for some k in (0, 1). But sin k < sin 1 < 0.9. So a_{2n} and a_{2n+1} must converge to the same limit. Suppose this limit is h. Then h = cos h. But cos x is strictly decreasing in (0, 1) and x is strictly increasing, so they have a single point of intersection in (0, 1). Thus h must be the unique point in (0, 1) at which h = cos h.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002