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ON CONVERGENCE OF A RECURSIVE SEQUENCE xn+1=f(xn−1,xn)x_{n+1} = f(x_{n-1}, x_n)

Sin-Ei Takahasi, Yasuhide Miura, Takeshi Miura

Source record

Source: Crossref

Published: Mar 1, 2006

DOI: 10.11650/twjm/1500403851

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Source abstract

C. H. Gibbons, M. R. S. Kulenovic and G. Ladas [1] have posed the following problem: Is there a solution of the difference equation: xn+1=βxn−1β+xn,x−1,x0>0,β>0(n=0,1,2,… ) x_{n+1} = \frac{\beta x_{n-1}}{\beta + x_n}, \quad x_{-1}, x_0 \gt 0, \beta \gt 0 \quad (n = 0, 1, 2, \dots) such that lim⁡n→∞xn=0\displaystyle \lim_{n \to \infty} x_n = 0? S. Stevic [2] gives an affirmative answer to this open problem and generalize this result to the equation of the form: xn+1=xn−1g(xn),x−1,x0>0(n=0,1,2,… ) x_{n+1} = \frac{x_{n-1}}{g(x_n)}, \quad x_{-1}, x_0 \gt 0 \quad (n = 0, 1, 2, \dots) by using his ingenious device. In this note, we generalize the result of Stevic to the equation of the form: xn+1=f(xn−1,xn),x−1,x0>0(n=0,1,2,… ). x_{n+1} = f(x_{n-1}, x_n), \quad x_{-1}, x_0 \gt 0 \quad (n = 0, 1, 2, \dots). However our proof is simple and short.

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