6.5.8 Example H The equation 2yk V1– y? (6.111) Yk+1 = can be transformed into a simpler form by letting yk = sin ak; this gives sin xk+1 = sin 2xk (6.112) %3D the solution of which is Cek+1 = (-1)"2xk + nA, (6.113) %3D where n is an integer. There are two cases to consider. Let n = 2m be an even integer. Therefore, ( Xk+1 – 2xk = 2m . (6.114) and = c2k – 2mr. (6.115) Consequently, Yk = sin(c2* – 2mn) = sin(c2*). (6.116) %3D | Let n = 2m +1 be an odd integer. In this instance, we have k+1 +2xk = (2m + 1)7, ) Ck = c(-2)* + /3(2m + 1)7) (6.117) (6.118) %3D and Yk = sin[c(-2)k + 3(2m + 1)7]. (6.119)