关于Sk=1+2k+3k+…+nk的探究

1.递推公式由二项展开式得对i求和上式左边可化为(n+1)k+1-1+Sk+1,从而有(1)由(1)式得递推公式     

Hamiltonian[k,k+1]-因子

本文考虑n/2-临界图中Hamiltonian[k,k+1]-因子的存在性。Hamiltonian[k,k+1]-因子是指包含Hamiltonian圈的[k,k+1]-因子;给定阶数为n的简单图G,若δ(G)≥n/2而δ(G\e)相似文献   

Обобшенные проиэводные и граничные своиства проиэводных интеграла Пуассона для полупространства R + k+1 (k>1)

Реэуме. Вводятся понятия обобшеннои частнои проиэводнои и обобшенного оератора Лапласа для функции многих переменных. Докаэана теоремы типа теоремы Фату о граничных своиствах частхых проиэводных интегралал Пуассона для полупространства, когда плотностя интеграла имеет обобшеннуу частнуу проиэводнуу и обобшенныи оператор Лапласа.     

一类隐式的k+1阶线性k步法

本文首先给出了一类比Adams-Moulton方法的绝对稳定区间大的隐式k+1阶线性k步法基本公式.求出了3-9步新公式的分数形式的精确系数,阶数,局部截断误差主项系数和绝对稳定区间,然后构造了由4阶隐式新公式和同阶显式Nyström公式组合而成的预估-校正方法,比著名的Adams-Bashforth-Moulton和Nyström-Adams-Moulton预估校正方法的绝对稳定区间大,最后用对比数值试验对结果进行了验证.     

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.     

$$L^p$$ Christoffel–Minkowski problem: the case $$1< p<k+1$$1<p<k+1

We consider a fully nonlinear partial differential equation associated to the intermediate \(L^p\) Christoffel–Minkowski problem in the case \(1<p<k+1\). We establish the existence of convex body with prescribed k-th even p-area measure on \(\mathbb S^n\), under an appropriate assumption on the prescribed function. We construct examples to indicate certain geometric condition on the prescribed function is needed for the existence of smooth strictly convex body. We also obtain \(C^{1,1}\) regularity estimates for admissible solutions of the equation when \( p\ge \frac{k+1}{2}\).     

The extension of the D(?k 2)-pair {k 2,k 2 + 1}

Let n be a nonzero integer. A set of m distinct positive integers is called a D(n)-m-tuple if the product of any two of them increased by n is a perfect square. Let k be a positive integer. In this paper, we show that if {k ², k ²+1, c, d} is a D(?k ²)-quadruple with c < d, then c = 1 and d = 4k ²+1. This extends the work of the first author [20] and that of Dujella [4].     

随机图中[k,k+1]-因子的存在性

设G=G(n,p)是一个随机图,其顶点数为n,任两个顶点之间有边相关联的概率为p=p(n),k是一个正整数满足knp-2(nplogn)~(1/2).图G的—个支撑子图F称作是图G的—个[k,k+1卜因子,如果对任一个x∈V(G),都有k≤dF(x)≤k+1.我们证明任意满足p≥n~(-2/3)的随机图G(n,p)几乎一定包含[k,k+1]-因子.     

猜想M(2k,k+1)=3k-1+[(k-1)/2]的反例

Brualdi与Jung在[1]中研究了一类具有固定线和k的n×n矩阵上的最大跳跃数M(n,k),并提出猜想M(2k, k + 1) = 3k - 1 + [(k-1)/2].本文给出了这一猜想的两个反例．     

不等式k~(1/2)+1/(k+1)~(1/2)>(k+1)~(1/2)的几种证法

为什么要证明不等式k~(1/2)+1/(k+1)~(1/2)>(k+1)~(1/2)下面通过实例来说明,高中数学第三册P.147.3(4)题:求证1/1~(1/2)+1/2~(1/2)+…+1/n~(1/2)>n~(1/2)(n>1)。我们用数学归纳法来证明。 (1)当n=2时不等式左边=1/1~(1/2)+1/2~(1/2)=(2+2~(1/2))/2右边=2~(1/2)=(2~(1/2)+2~(1/2))/2,显然不等式成立。 (2)假设当n=k(k>1)时不等式成立,     

Evaluation of Series Involving the Product of the Tail of $${\zeta(k)}$$ and $${\zeta(k+1)}$$ζ(k+1)

We aim at evaluating the following class of series involving the product of the tail of two consecutive zeta function values

$$\sum\limits_{n=1}^{\infty}\left(\zeta(k)-1-\frac{1}{2^k}-\cdots-\frac{1}{n^k}\right)\left(\zeta(k+1)-1-\frac{1}{2^{k+1}}-\cdots-\frac{1}{n^{k+1}}\right),$$

where \({k\geq 2}\) is an integer. We show that the series can be expressed in terms of Riemann zeta function values and a special integral involving a polylogarithm function.

     

关于(4k+1)连通的(8k+6)维闭光滑流形的同胚分类

本文给出了光滑流形S^4k+2×S^4k+3(k≥1)上自微分同胚拟同痕的充分必要条件,并且计算了这些拟同痕类全体按复合作乘法所构成的群Π₀Diff#_m¹S^4k+2×S^4k+3,作为这些结果的一个直接应用,我们对一类(4k+1)连通的(8k+6)维闭光滑流形做了完全分类。     

New Rapidly Convergent Series Concerning ζ(2k+1)

Values of new series sum(((2n-1)!ζ(2n))/(2n + 2k)!)α2n from n=1 to ∞,sum(((2n-1)!ζ(2n))/(2n+2k +1)!)β2n from n=1 to ∞ are given concerning ζ(2k + 1),where k is a positive integer,α can be taken as 1,1/2,1/3,2/3,1/4,3/4,1/6,5/6 and β can be taken as 1,1/2.Some previous results are included as special cases in the present paper and new series converges more rapidly than those exsiting results for α = 1/3,or α = 1/4,or α = 1/6.     

具有鉴相特性为g(φ)=(1+k)sinφ/1+k cosφ的二阶锁相环路方程的定性分析

§1.引言 在文[1]中我们曾对具有正切鉴相特性的二阶锁相环路方程进行了定性分析。这样就对为什么具有正切鉴相特性的锁相环路没有失锁点及其快捕带是整个条形域:     

极小拟(k+1)连通图的最小度

给出了极小拟5连通图有围长大于或等于4的极小拟（k）＋1连通图的最小度。