-1 < a < 1. The sequence x1, x2, x3, ... is defined by x1 = a, xn+1 = ( √(3 - 3xn2) - xn)/2. Find necessary and sufficient conditions on a for all members of the sequence to be positive. Show that the sequence is always periodic.
Answer
0 < a < (√3)/2
period 2
Solution
Put x1 = sin k. Then x2 = (√3)/2 cos k - (1/2) sin k = sin 60o cos k - cos 60o sin k = sin(60o-k). Hence x3 = sin(60o - 60o + k) = x1.
For sin k and sin(60o-k) to be positive we need 0 < k < 60o and hence 0 < a < (√3)/2.
Thanks to Suat Namli
© John Scholes
jscholes@kalva.demon.co.uk
12 February 2004
Last corrected/updated 12 Feb 04