The recursive sequence xn+1 = g(xn,xn−1)/(A + xn)

In this note, we investigate the periodic character of solutions of the nonlinear, second-order difference equation

where the parameter A and the initial conditions x₀ and x₁ are positive real numbers. We give sufficient conditions under which every positive solution of this equation converges to a period two solution.     

差分方程xn+1=xn+xαn/nβ解的渐近性

我们研究差分方程(1)的解的渐近性和全局吸引性.     

On the Behavior of Solutions of x n+1 = p+(x n−1/xn )

Our goal in this article is to complete the study of the behavior of solutions of the equation in the title when the parameter p is positive and the initial conditions are arbitrary positive numbers. Our main focus is the case 0 < p < 1. We will show that in this case, all solutions which do not monotonically converge to the equilibrium have a subsequence which converges to p and a subsequence which diverges to infinity. For the sake of completeness, we will also present the results (which were previously known) with alternative proofs for the case p = 1 and the case p > 1.     

差分方程xn+1=xanern(1-xn-k)正解的渐近性

本文研究时滞Logistic方程 xn+1=xanern(1-xn-k),n=0,1,2,… (*) 其中rn是非负实数列,a∈(0,1],k为非负整数,获得了保证方程(*)的每一正解趋于1的一些充分条件,推广和改进了文[1]的结果.     

关于Diophantine方程(n+1)+(n+2)y=nz

设n是正整数.本文证明了:方程(n+1)+(n+2)y=nz仅当n=3时有正整数解(y,z)=(1,2).     

More on the Difference Equation y n + 1 = ( p + y n −1 )/( qy n + y n −1 )

From previous studies of the equation in the title with positive parameters p and q and positive initial conditions we know that if q h 4 p + 1 then the equilibrium is a global attractor. We also know that if q > 4 p + 1 then every solution eventually enters and remains in the interval [ p / q , 1]. In this strip there exists a "unique" prime period two solution that is locally asymptotically stable. In this paper, we provide more insight as to the behavior of solutions of the equation in the title in the strip [ p / q , 1], where a one-dimensional stable manifold lives.     

对"条件x+y=1下1/xn+λ/yn(n≥2)的最小值"再探究

文[1]把解决问题(1):“已知z,Y∈R+且z+.y一1,求去+4j,的最小值’’的“]一z+y”的代换方法移植到问题(2):巳知z,Y∈R+且z+y一1,求壶+歹8的最小值’’时思此时z—i歹舞,y—i歹斋.路受阻后,提出在(≯1+多)( )中,括号内 。=应配上什么式子才能解出的疑问,由此利用文[1]中的(*)式和待定系数法探讨出了一般性的结论:“已知z,y E R+且z+y一1,若^>o,则当且仅当导一^南时,击+号("≥2)取得最小值为(1+A南)一十一. 笔者读后颇受启发,但在(去+io)( )中,括号内到底应配上什么式子呢?文[1]的一般性结论能否再推广?为此,本文再作些探究.1 大胆尝试,印…     

对"条件x+y=1下1/xn+λ/yn(n≥2)的最小值"再研究

1 问题回顾 文[1]由问题:"已知x,y∈R 且x y=1,求1/x 4/y的最小值"推广为下述定理:     

an+1=p(n)a2/n+f(n)an+r(p(n)≠0)型数列的求解方法

数列问题的背景新颖,能力要求高,内在联系密切,思维方法灵活,因此倍受命题者的青睐.解答数列问题要求熟练掌握数列基础知识,灵活运用基本数学思想方法,善于转化.an+1=p(n)@a2n+f(n)@an+r(p(n)≠0)型数列是数列和二次函数、不等式相结合的典范,难度较大.求解此类问题的思维模式是:观察-归纳-猜想-证明.求解的主要方法是:分析法,比较法,消去法,综合法,放缩法,数学归纳法.     

On the recursive sequencex n +1=α-(x n /x n −1)

We study the global asymptotic stability, global attractivity, boundedness character, and periodic nature of all positive solutions and all negative solutions of the difference equation $$x_{n + 1} = \alpha - \frac{{x_n }}{{x_{n - 1} }}, n = 0,1,...,$$ where α∈R is a real number, and the initial conditionsx?1,x ₀ are arbitrary real numbers.     

数列a_n=n(n+1)前n项和公式的三种推导方法

<正>在《数列》教学过程中,很多学生对数列a_n=n(n+1)前n项和公式S_n=n(n+1)(n+2)/3比较陌生,为了让学生在了解课本知识的基础上有所拓展,本文特总结了3种证明方法,以期为学生解决疑惑,起到举一反三的效果.证法1∵a_n=n(n+1)=n~2+n,     

重要极限llim from (n→∞)(1+n1/2)~n=e的注记

利用平均值不等式对重要极限limn→∞ 1 +1nn=e给出一个较简捷的证明方法。利用证明过程中所得到的不等式还可求得e的任意精度的近似值。