On the positive solutions of the difference equation xn+1=(xn−1)/(1+xnxn−1)

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Abstract

The positive solutions of the difference equation xn+1=(xn−1)/(1+xnxn−1), n=0,1,2,…, are investigated, where x−1 and x0 are the positive real numbers.

Introduction

Olwaidy [2] has studied the periodic cycle of xn+2=(an+bnxn)/xn−1. Owaidy [3] has also investigated the boundedness, persistence and asymptotic behaviour of positive solutions of the equation xn+1=A/xnp+B/xn−1q+C/xn−2s. Zhang [1] has obtained some global attractivity results for the rational recursive xn+1=(a+bxn2)/(1+xn−12). Li [5] has obtained sufficient conditions for the global asymptotic stability of the difference equation xn+1=f(xn)g(xnk). Devault [6] has investigated the global stability and periodic character of solution yn+1=(p+ynk)/(qyn+ynk). Cunningham [4] has investigated the global character of solutions of the nonlinear xn+1=(α+βxn)/(Bxn+Cxn−1).

Similar to the references, in this paper, we define new difference xn+1=(xn−1)/(1+xnxn−1), n=0,1,2,… and investigate the positive solutions of the difference equationxn+1=xn−11+xnxn−1,n=0,1,2,…where x−1 and x0 are the positive real numbers.

Section snippets

Main results

Theorem 2.1

Let x−1=k and x0=h be positive real numbers. Then all solution of Eq. (1) arexn=k∏i=0[(n+1)/2]−1(2hki+1)i=0[(n+1)/2]−1[(2i+1)hk+1]noddh∏i=0(n/2)−1[(2i+1)hk+1]i=0n/2(2ihk+1)neven

Proof

Assume that n is even. Then n−1 and n+1 are odd. If we substitute (2) in (1), then we have the following equals:xn−11+xn−1xn=ki=0(n/2)−1(2hki+1)i=0(n/2)−1[(2i+1)hk+1]1+ki=0(n/2)−1(2hki+1)i=0(n/2)−1[(2i+1)hk+1]·hi=0(n/2)−1[(2i+1)hk+1]i=0n/2(2ihk+1)=k∏i=0(n/2)−1(2hki+1)∏i=0n/2(2hki+1)i=0(n/2)−1[(2i+1)hki+1]∏i=0n/2

Numerical results

Example 3.1

Let xn+1=(xn−1)/(1+xnxn−1), n=0,1,2,…,999, x−1=100 and x0=30 we have the following table:

nxnnxn
10.3332222592e−17490.1525641154e−2
215.002499587500.8739491132
2490.2644249062e−29990.1321133966e−2
2501.51271478810000.7569252522

Example 3.2

Let xn+1=(xn−1)/(1+xnxn−1), n=0,1,2,…,999, x−1=0.001 and x0=0.001 we have the following table:

nxnnxn
10.9999990000e−37490.9996251940e−3
20.9999990000e−37500.9996251950e−3
2490.9998750075e−39990.9995003740e−3
2500.9998750075e−310000.9995003750e−3

Example 3.3

Let xn+1=(xn−1)/(1+xnxn−1), n

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