## On the recursive sequence $$x_{n+1}=x_{n-1}/g(x_n)$$.(English)Zbl 1019.39010

Several properties of the solutions of the recurrence relation $x_{n+1}= {x_{n-1} \over g(x_n)},\;n=1,2,\dots,$ subject to the initial conditions $$x_{-1} >0$$ and $$x_0>0$$ are obtained, where $$g\in C^1(R_+)$$, $$g(0)=1$$ and $$g'(x)>0$$ for $$x\in R_+$$. Then by continuity arguments, it is shown that for any $$u\in(0, \infty)$$, there is a solution $$\{x_n\}$$ satisfying $$x_{-1}=u$$ and $$x_0g(x_0) >u$$ such that $$x_0>x_1>x_2> \dots$$ and $$\lim_{n\to \infty}x_n=0$$.

### MSC:

 39A11 Stability of difference equations (MSC2000) 39B05 General theory of functional equations and inequalities
Full Text: