6.5.9 Example I Let N be a positive number and consider the following nonlinear difference equation: Yk+1 = /½(Yk – N/Yk). (6.120) Using the fact that cot2 0 (cot (20) = - 1 (6.121) %3D 2 cot 0 we can make the substitution yk VN cot xk, reducing equation (6.120) to the form cot xk+1 = cot (2ak). (6.122) A solution of this latter equation is Xk+1 2xk. (6.123) Therefore, Xk = c2k (6.124) and VN cot(c2*).) (6.125) Yk =

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