In case $\check{b}<0$ and ${\check{a}}^2<-4\check{b}$, we show that there are eventually periodic solutions when either $tan^{-1}\frac{\sqrt{-4\check{b}-{\check{a}}^2}}{\check{a}} \in]\dfrac{\pi}{2},\pi[$ (with $\check{a}<0$) is a rational multiple of $\pi$ or $tan^{-1}\frac{\sqrt{-4\check{b}-{\check{a}}^2}}{\check{a}} \in]0,\dfrac{\pi}{2}[$ (with $\check{a}>0$) as well.
📌 3 Maddede Bu Makale
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Amaç & Hipotez:In this paper, we study the admissible solutions of the nonlinear system of difference equations $$x_{n+1}=\dfrac{y_{n}}{x_{n}},\quad y_{n+1}=\dfrac{y_{n} }{\check{a}x_{n}+\check{b}y_{n}},\quad n=0,1,\ldots,$$ where $\check{a}$, $\check{b}$ are real numbers and the initial values $x_{0},y_{0}$ are nonzero real numbers.
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Temel Bulgular & Veri:In this paper, we study the admissible solutions of the nonlinear system of difference equations $$x_{n+1}=\dfrac{y_{n}}{x_{n}},\quad y_{n+1}=\dfrac{y_{n} }{\check{a}x_{n}+\check{b}y_{n}},\quad n=0,1,\ldots,$$ where $\check{a}$, $\check{b}$ are real numbers and the initial values $x_{0},y_{0}$ are nonzero real numbers.
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Sonuç & Katkı:In case $\check{b}<0$ and ${\check{a}}^2<-4\check{b}$, we show that there are eventually periodic solutions when either $tan^{-1}\frac{\sqrt{-4\check{b}-{\check{a}}^2}}{\check{a}} \in]\dfrac{\pi}{2},\pi[$ (with $\check{a}<0$) is a rational multiple of $\pi$ or $tan^{-1}\frac{\sqrt{-4\check{b}-{\check{a}}^2}}{\check{a}} \in]0,\dfrac{\pi}{2}[$ (with $\check{a}>0$) as well.
🔬Tür:Research Article
📍Kapsam:Genişletilmiş Akademik Örneklem
⚙️Değişkenler:System of difference equations · Forbidden set · Periodic solution
Abstract
In this paper, we study the admissible solutions of the nonlinear system of difference equations $$x_{n+1}=\dfrac{y_{n}}{x_{n}},\quad y_{n+1}=\dfrac{y_{n} }{\check{a}x_{n}+\check{b}y_{n}},\quad n=0,1,\ldots,$$ where $\check{a}$, $\check{b}$ are real numbers and the initial values $x_{0},y_{0}$ are nonzero real numbers. In case $\check{b}<0$ and ${\check{a}}^2<-4\check{b}$, we show that there are eventually periodic solutions when either $tan^{-1}\frac{\sqrt{-4\check{b}-{\check{a}}^2}}{\check{a}} \in]\dfrac{\pi}{2},\pi[$ (with $\check{a}<0$) is a rational multiple of $\pi$ or $tan^{-1}\frac{\sqrt{-4\check{b}-{\check{a}}^2}}{\check{a}} \in]0,\dfrac{\pi}{2}[$ (with $\check{a}>0$) as well.
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