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.     

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.     

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.     

On the Diophantine equation x 2 − kxy + y 2 − 2 n = 0

In this study, we determine when the Diophantine equation x ²?kxy+y ²?2 ⁿ = 0 has an infinite number of positive integer solutions x and y for 0 ? n ? 10. Moreover, we give all positive integer solutions of the same equation for 0 ? n ? 10 in terms of generalized Fibonacci sequence. Lastly, we formulate a conjecture related to the Diophantine equation x ² ? kxy + y ² ? 2 ⁿ = 0.     

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).     

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.     

The diophantine equation x 2 + 2 a · 17 b = y n

Let ?, ? be the sets of all integers and positive integers, respectively. Let p be a fixed odd prime. Recently, there have been many papers concerned with solutions (x, y, n, a, b) of the equation x ² + 2 ^a p ^b = y ⁿ , x, y, n ε ?, gcd(x, y) = 1, n ? 3, a, b ε ?, a ? 0, b ? 0. And all solutions of it have been determined for the cases p = 3, p = 5, p = 11 and p = 13. In this paper, we mainly concentrate on the case p = 3, and using certain recent results on exponential diophantine equations including the famous Catalan equation, all solutions (x, y, n, a, b) of the equation x ²+2 ^a · 17 ^b = y ⁿ , x, y, n ε ?, gcd(x, y) = 1, n ? 3, a, b ∈ ?, a ? 0, b ? 0, are determined.     

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.     

无理函数y=(√a1x+b1)+(√a2x+b2)的最值求法

无理函数y=(√a1x+b1)+(√a2x+b2)(a1,a2,b1,b2均不为0)(1)的最值问题,是代数中较为典型的一类最值问题之一.当a1a2≥0时,函数(1)为单调函数,求出定义域后利用单词性很容易确定最大值和最小值.但当a1a2＜0时,函数(1)最值的求解具有一定的难度.     

运用均值不等式求y=m/ax+b+n/c-dx的最值

例1函数y=max b nc-dx(a,b,c,d,m,n均为常数,且ad>0)在定义域内恒有max b>0且nc-dx>0,求这个函数的最值.解∵(x ba) (cd-x)=d(ax b) a(c-dx)ad=bd acad,∴adac bd[(x ba) (cd-x)]=1,∴y=max b nc-dx=adac bd[(x ba) (cd-x)].(max b nc-dx)=adac bd[(x ba) (cd-x)].(m a x ba n     

The integral points on elliptic curves y 2 = x 3 + (36n 2 − 9)x − 2(36n 2 − 5)

Let n be a positive odd integer. In this paper, combining some properties of quadratic and quartic diophantine equations with elementary analysis, we prove that if n > 1 and both 6n ² ? 1 and 12n ² + 1 are odd primes, then the general elliptic curve y ² = x ³+(36n ²?9)x?2(36n ²?5) has only the integral point (x, y) = (2, 0). By this result we can get that the above elliptic curve has only the trivial integral point for n = 3, 13, 17 etc. Thus it can be seen that the elliptic curve y ² = x ³ + 27x ? 62 really is an unusual elliptic curve which has large integral points.     

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.     

The actions ofS n+1 andS n on the cohomology ring of a coxeter arrangement of typeA n−1

Let be the complexified Coxeter arrangement of hyperplanes of typeA _n−1. In this paper we construct anS _n+1 extension of the naturalS _n action on the complex cohomology ring of the complement ofA _n−1. Recurrence formulas connecting characters with respect to theS _n and theS _n+1 action are given.     

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.     

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.