Boundedness properties of the difference equation x n+1=(α+βx n +δx n−2)/(x n−1)

Consider the third-order difference equation x _n+1 = (α+βx _n +δx _n ? 2)/(x _n ? 1) with α ∈ [0,∞) and β,δ ∈ (0,∞). It is shown that this difference equation has unbounded solutions if and only if δ>β.     

Global Behavior of Solutions of x n + 1 = max {x n , A}/x n x n – 1

We investigate the asymptotic behavior, the oscillatory character, and theperiodic nature of solutions of the difference equation

A Note on the Recursive Sequence x n + 1 = p k x n + p k − 1 x n − 1 +...+ p 1 x n − k + 1

We present some comments on the behavior of solutions of the difference equation where p _i 0, i = 1,..., k, k N, and x _–k ,..., x _–1 R.     

K 0 of hypersurfaces defined by x 1 2 + ... + x n 2 = ±1

Let k be a field of characteristic ≠ 2 and let Q _n,m (x ₁, ..., x _n , y ₁, ..., y _m ) = x ₁ ² +...+x _n ² ? (y ₁ ² +...+y _m ² ) be a quadratic form over k. Let R(Q _n,m ) = R _n,m = k[x ₁, ..., x _n , y ₁, ..., y _m ]/(Q _n,m ? 1). In this note we will calculate $\tilde K_0 \left( {R_{n,m} } \right)$ for every n,m ≥ 0. We will also calculate CH ₀(R _n,m ) and the Euler class group of R _n,m when k = ?.     

On the Diophantine Equations (2 n − 1)(6 n − 1) = x 2 and (a n − 1)(a kn − 1) = x 2

In this paper we prove that the equation (2 ⁿ – 1)(6 ⁿ – 1) = x ² has no solutions in positive integers n and x. Furthermore, the equation (a ⁿ – 1) (a ^kn – 1) = x ² in positive integers a > 1, n, k > 1 (kn > 2) and x is also considered. We show that this equation has the only solutions (a,n,k,x) = (2,3,2,21), (3,1,5,22) and (7,1,4,120).     

On Globally Periodic Solutions of the Difference Equation x n + 1 = f ( x n )/ x n − 1

We study the second-order difference equation x n +1 = f ( x n ) x n m 1 where f ] C 1 ([0, X ),[0, X )) and x n ] (0, X ) for all n ] Z . For the cases p h 5, we find necessary and sufficient conditions on f for all solutions to be periodic with period p . We answer some questions and conjectures of Kulenovi ' and Ladas.     

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.     

活用点（x1＋x2，x1x2）的轨迹方程解题

在学习直线与圆锥曲线的位置关系时,不少学生使用韦达定理具有一定的盲目性.特别是遇到较复杂的问题时,更是如此.对此,在教学中我们给学生装上“轨迹思想”的方向盘,使问题有了很好的解决.我们引入弦的端点坐标(x1,y1),(x2,y2),构造点(x2+x2,x1x2),或点(y1+y2,y1y2),先求出点的轨迹方程,再结合韦达定理求出该点的坐标,代入所求轨迹方程,或利用点的存在域x1x2≤14(x1+x2)2,然后求解.这样处理,思路清晰,许多问题迎刃而解.例1已知A,B为抛物线y2=2px(p>0)上两点,且OA⊥OB,原点O在AB上的射影为D(2,1),求此抛物线方程.解设A,B的坐标分别为(x1…     

话说不等式x1x2＜0

1 高中教师的赛题以下的这道赛题,本来是用来考高中教师的:赛题试求最大的常数λ,使得下列不等式对于满足条件x+y+z=0的实数x,y,z恒成立:1/5x2+6x+12+1/5y2+6y+12+1/5z2+6z+12≤λ趣事 一位初中学生看到了这道题目,他说式中的λ=1/4.     

话说不等式x1x2＜0

1 高中教师的赛题以下的这道赛题,本来是用来考高中教师的:赛题试求最大的常数λ,使得下列不等式对于满1足条件x+y+z=0的实数x,y,z恒成立:1/5x2+6x+12+1/5y2+6y+12+1/5z2+6z+12≤λ.     

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.     

A note on the diophantine equation (a n − 1)(b n − 1) = x 2

Let a, b be fixed positive integers such that a ≠ b, min(a, b) > 1, ν(a?1) and ν(b ? 1) have opposite parity, where ν(a ? 1) and ν(b ? 1) denote the highest powers of 2 dividing a ? 1 and b ? 1 respectively. In this paper, all positive integer solutions (x, n) of the equation (a ⁿ ? 1)(b ⁿ ? 1) = x ² are determined.     

On the diophantine equation x 2 − Dy 2 = n

We propose a method to determine the solvability of the diophantine equation x2-Dy2=n for the following two cases:(1) D = pq,where p,q ≡ 1 mod 4 are distinct primes with(q/p)=1 and(p/q)4(q/p)4=-1.(2) D=2p1p2 ··· pm,where pi ≡ 1 mod 8,1≤i≤m are distinct primes and D=r2+s2 with r,s ≡±3 mod 8.     

Zolotarev's first problem — the best approximation by polynomials of degree ≤n − 2 tox n − nσx n−1 in [−1, 1]

Chebyshev determined $$\mathop {\min }\limits_{(a)} \mathop {\max }\limits_{ - 1 \le x \le 1} |x^n + a_1 x^{n - 1} + \cdots + a_n |$$ as 2^1?n , which is attained when the polynomial is 2^1?n T _n(x), whereT _n(x) = cos(n arc cosx). Zolotarev's First Problem is to determine $$\mathop {\min }\limits_{(a)} \mathop {\max }\limits_{ - 1 \le x \le 1} |x^n - n\sigma x^{n - 1} + a_2 x^{n - 2} + \cdots + a_n |$$ as a function ofn and the parameter σ and to find the extremal polynomials. He solved this in 1878. Another discussion was given by Achieser in 1928, and another by Erdös and Szegö in 1942. The case when 0≤|σ|≤ tan²(π/2n) is quite simple, but that for |σ|> tan²(π/2n) is quite different and very complicated. We give two new versions of the proof and discuss the change in character of the solution. Both make use of the Equal Ripple Theorem.     

方程x1＋x2＋……＋xm=n的正整数解的个数及其应用

在近几年的高考试题中，出现了可化为求方程x1＋x2＋…＋xm=n（m，n∈N^＋，m≤n）的正整数解的个数的问题，下面就这个问题谈几点看法，供大家参考。     

浅谈求函数y=a1x2+b1x+c1/a2x2+b2x+c2值域的判别式法

函数y=a1x2+b1x+c1/a2x2+b2x+c2值域的求法,很多资料上给出方法是判别式(即△)法,而一旦自变量的范围给以限定,当△法失效时,还有其他方法吗?一般资料上就避而不谈了.要全面系统解决函数y=a1x2+b1x+c1/a2x2+b2x+c2值域的问题,本文以为需解决以下三个事情:①判别式法的过程和依据,②自变量有限制时还能用判别式法吗?③自变量有范围限制,问题可以归结为三类常见函数:反比例函数;y=t+c/t(c＞0);y=t+c/t(c＜0)的值域求法.     

关于公式1+2+…+n=(n(n+1))/2

<正>九年义务教育三年制初中《代数》第一册(上)第38页,B组第二题为:正整数从1开始,逐个相加,一直加到n,它们的和记作s,即s=1+2+3+…+n(n表示一个正整数),写出计算s的公式.这道题目中含有字母且设问富有思考性,解题方法体现了数学方法,更重要的是结果能作公式用,而且应用分层次用.为帮助同学理解这些特点,现对这道题进行解读,供同学们参考.     

关于函数方程f1n1+af1m1f2m2+f2n2=1

对于函数方程f1n1+af1m1f2m2+f2n2=1,其中a∈C/{0},n1,n2,m1,m2∈N,给出存在非常数亚纯函数解和整函数解的必要条件.