Let R be the reals. Define an by a1 = α ∈ R, an+1 = cos an. Show that an converges to a limit independent of α.
Solution
|cos x| ≤ 1, so -1 ≤ a2 ≤ 1. Hence 0 < a3 < 1 and so 0 < an < 1 for n ≥ 3.
cos x is strictly monotonic decreasing over the range 0 to 1, so if an < an+1 then an+2 < an+1. Now the gradient of cos x is greater than -1 throughout the interval (0, 1), so if an < an+1 then cos an - cos an+1 < an+1 - an. Hence an < an+2 < an+1. Similarly, if an > an+1, then an > an+2 > an+1. Thus either a2n is an increasing sequence bounded above and a2n+1 is a decreasing sequence bounded below or vice versa. Hence both a2n and a2n+1 converge.
By the mean value theorem |an+1 - an+2| = |cos an - cos an+1| = sin k |an - an+1| for some k in (0, 1). But sin k < sin 1 < 0.9. So a2n and a2n+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