Some Notes on the Difference Equation xn+1=α+(xn−1/xnk)

Abstract

We investigate the behavior of the solutions of the recursive sequence xn+1=α+xn-1/xnk, n=0,1,…, whereα∈[0,∞),k∈(0,∞),and the initial conditionsx-1,x0are arbitrary positive numbers. Included are results that considerably improve those in the recently published paper by Hamza and Morsy (2009).

Some Discussions on the Difference Equationxn+1=α+(xn-1mBakery1 2015Journal of Function Spaces0000View full textAdd to dashboardCite

“…In Figure 3 we illustrate the result of Theorem 11 for = 3, = 2, and = 0.5, with the initial conditions −1 = 0.7, 0 = 0.54 and choosing = 0.2; in this case the subsequence ( 2 ) of the solution of (1) is bounded with 0.5 < 2 < 0.7 for all ≥ 0, while the subsequence ( 2 +1 ) is unbounded, so the terms of ( 2 ) appear approximately zero in comparison with the terms of ( 2 +1 ). Also Figure 4 explains the same result for = 0.3, = 2, and = 1.5; Gümüs andÖcalan in [5] showed that when > 1 and → ∞, every solution of (1) is unbounded. Figure 5 gives a counter example of this result for = 10 10 , = 1, and = 1.1.…”

“…they investigated the behavior of positive solutions of (15); they proved that when > 1, every positive solution of (15) is bounded and when > 1/ ≥ 1, the equilibrium point of (15) is globally asymptotically stable and they showed that (15) has periodic solutions without conditions on and . These results are improved in [5] by Gümüs andÖcalan; they investigated the boundedness character of positive solutions of (15); they proved that if 0 ≤ < 1, then there exist unbounded solutions of (15) in the cases when ∈ (0, ∞) and → 0 + or > 1 and → ∞; then every positive solution of (15) is unbounded. When 0 + , ∞, and > 1, then every positive solution of (15) is bounded.…”

“…The graph of the sequence x n when = 1.1, m = 1, and k = When > 1 and → ∞ the positive solution converges to , which gives a counter example of Theorem 2.1-(b) and the results that followed in[5]. Here = 1.1, = 1, and = 10 10 .…”

“…(ii) A necessary and sufficient condition for one of the roots of (5) In this case is unstable and is called a saddle point. (iii) A necessary and sufficient condition that a root of (5) has absolute value equal to 1 is…”