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1.
方世跃 《数学通讯》2003,(11):19-20
设数列 {an}是等差数列 ,公差为d ,则an + 2 ·an-a2 n + 1=-d2 .此结论的证明不难 .an + 2 ·an-a2 n + 1=(an + 1+d) (an + 1-d) -a2 n + 1=a2 n + 1-d2 -a2 n + 1=-d2 .若从等差数列的特征去思考 ,它有an + 2+an=2an + 1这一递推关系式 ,那么此结论是否有其一般的规律呢 ?思考 1 在数列 {an}中 ,若an + 2 +an=pan + 1(n∈N ,p为非零常数 ) ,则an + 2 ·an-a2 n + 1=?探索 设bn=an + 2 ·an-a2 n + 1,则bn + 1-bn =an + 3 ·an + 1-a2 n + 2 -an + 2 ·an+a2 n + 1=an + 1(an + 3 +an + 1) -an + 2(an + 2 +an) =pan + 1·an + 2 - pan…  相似文献   

2.
试题研讨(24)     
题 1  ( 2 0 0 4年中山市高三联考试题 )数列 {an}满足 a1=0 ,且 an+ 1=3an + 1an + 3,试问 :m为何值时 ,总存在正整数 n使 an+ 1相似文献   

3.
一个数列的连续项之间的关系叫递归关系.由递归关系确定的数列叫递归数列.中学教材中的等差、等比数列,是最基本、应用最广的递归数列.递归数列在高考试题中时有所见.本文主要介绍一阶递归数列:an 1=f(an)(有一个初始条件)及二阶递归数列:an 1=f(an,an-1)(有两个初始条件)中的简单类型.1.第一类:a1=aan 1=can d 型若c=1,则an 1=an d,显然{an}是首项为a、公差为d的等差数列;若d=0,c≠0,则an 1=can,显然{an}是首项为a、公比为c的等比数列;下面就一般情形予以研究.例1 已知数列{an}中,a1=3,an 1=2an 4,求这数列的通项.解 由an 1=2an 4,得an=…  相似文献   

4.
逐差累加是法宝,出神入化显奇巧   总被引:1,自引:0,他引:1  
先给出一个简单的数列问题: 已知数列{an}的首项a1=1,an+1=an+1,求数列{an}的通项公式. 一方面,很明显an+1-an=1,故根据等差数列的定义可求得:an=n.  相似文献   

5.
数列是一种特殊的函数 ,所以数列中也必然存在着周期问题 ,有些数列题 ,表面上看与周期无关 ,但实际上隐含着周期性 ,一旦揭示了其周期 ,问题便迎刃而解 ,下面略举几例说明 .例 1 在数列 {an}中 ,a1=13,a2 =5 6 ,对所有的自然数n ,都有an + 1=an+an + 2 ,求a2 0 0 5.解 ∵an + 1=an +an + 2 ,∴an + 2 =an + 1+an+ 3,两式相加 ,整理得 an+ 3=-an,∴an + 6 =-an+ 3=an,∴数列是以 6为一个周期的周期数列 ,∴a2 0 0 5=a6× 334 + 1=a1=13.例 2 设数列a1,a2 ,… ,an,…满足a1=a2 =1,a3=2 ,且对任何自然数n都有anan + 1an + 2 ≠ 1,又anan +…  相似文献   

6.
通过给定的等比数列,可以构造出一组勾股数. 性质 如果等比数列{an}满足:(1)an>0,(2)an单调递增,(3)an为奇数(或偶数),那么an+1-an-1/2,an,an+1+an-1/2(n≥2)是一组勾股数. 证明 ∵(1)an>0,(2)an单调递增,(3)an为奇数(或偶数),  相似文献   

7.
题目 (2009,湖北理15)已知数列an满足:a1=m(m为正整数),an+1=(an)/(2),(当an为偶数时)3an+1,(当an为奇数时).若a6=1,则m所有可能的取值为____.……  相似文献   

8.
题197设a1=1,an 1=2an n 1.(Ⅰ)是否存在常数p,q使{an pn q}为等比数列?若存在,求出p,q的值,若不存在,说明理由:(Ⅱ)求{an}的通项公式;(Ⅲ)当n≥5时,证明:an>(n 2)2.解(Ⅰ)由an 1 p(n 1) q=2(an pn q)得an 1=2an pn (q-p).可见应有p=1,q-p=1,p=1,q=2,∴an 1 (n 1) 2=2(an n 2).  相似文献   

9.
试题研讨(24)     
题1 (2004年中山市高三联考试题)数列{an}满足a1=0,且an 1=3an 1/an 3,试问:m为何值时,总存在正整数n使an 1<m成立?  相似文献   

10.
题162在数列{an}中,已知an≥1,a1=1,且an 1-an=an 1 2an-1,n∈N*.1)记bn=(an-21)2,n∈N*,求证:数列{bn}是等差数列;2)求an的通项公式;3)对于任意的正整数k,是否存在m∈N*,使得am=k若存在,求出m的值;若不存在,请说明理由.解1)∵an 1-an=an 1 2an-1(n∈N*),∴an 12-an2-an 1 an=2  相似文献   

11.
在格蕴涵代数中提出了扩张滤子的概念,讨论了扩张滤子与滤子,扩张滤子与素滤子,扩张滤子与滤子的根,扩张滤子与准素滤子,扩张滤子与最大滤子之间的关系.得到了扩张滤子的一些性质.最后,证明了在格H蕴涵代数中,扩张滤子与扩张滤子的根相等.  相似文献   

12.
The main result of the paper is a structure theorem concerning the ideal extensions of archimedean ordered semigroups. We prove that an archimedean ordered semigroup which contains an idempotent is an ideal extension of a simple ordered semigroup containing an idempotent by a nil ordered semigroup. Conversely, if an ordered semigroup S is an ideal extension of a simple ordered semigroup by a nil ordered semigroup, then S is archimedean. As a consequence, an ordered semigroup is archimedean and contains an idempotent if and only if it is an ideal extension of a simple ordered semigroup containing an idempotent by a nil ordered semigroup.  相似文献   

13.
Izuru Mori 《代数通讯》2013,41(6):2071-2091
B-construction is a way of obtaining a graded algebra from the triple consisting of an additive category, an object, and an autoequivalence, while C-construction is a way of obtaining an algebra (without unity) from the pair consisting of an additive category and a set of objects. In this article, we study and compare three important classes of algebras in noncommutative algebraic geometry and representation theory of finite dimensional algebras, namely, quantum polynomial algebras, preprojetive algebras and trivial extensions, via these constructions.  相似文献   

14.
洪勇 《数学学报》2008,51(2):365-370
对带对称齐次核K(m,n)的级数算子T:T{an}=Σ_(n=1)~∞ K(m,n)an,{an}∈lω(n),l={{an)| an≥0,Σ_(n=1)~∞ω(n)an<+∞},本文研究了T的范数刻画,并讨论其应用.  相似文献   

15.
pth Power Lagrangian Method for Integer Programming   总被引:1,自引:0,他引:1  
When does there exist an optimal generating Lagrangian multiplier vector (that generates an optimal solution of an integer programming problem in a Lagrangian relaxation formulation), and in cases of nonexistence, can we produce the existence in some other equivalent representation space? Under what conditions does there exist an optimal primal-dual pair in integer programming? This paper considers both questions. A theoretical characterization of the perturbation function in integer programming yields a new insight on the existence of an optimal generating Lagrangian multiplier vector, the existence of an optimal primal-dual pair, and the duality gap. The proposed pth power Lagrangian method convexifies the perturbation function and guarantees the existence of an optimal generating Lagrangian multiplier vector. A condition for the existence of an optimal primal-dual pair is given for the Lagrangian relaxation method to be successful in identifying an optimal solution of the primal problem via the maximization of the Lagrangian dual. The existence of an optimal primal-dual pair is assured for cases with a single Lagrangian constraint, while adopting the pth power Lagrangian method. This paper then shows that an integer programming problem with multiple constraints can be always converted into an equivalent form with a single surrogate constraint. Therefore, success of a dual search is guaranteed for a general class of finite integer programming problems with a prominent feature of a one-dimensional dual search.  相似文献   

16.
建立[4]中定义的L-fuzzy拟序集的一种理想完备化,即证明了L-fuzzy拟序集上的所有L-fuzzy理想组成的集合,赋以合适的程度映射构成L-fuzzydomain,任意从L-fuzzy拟序集到L-fuzzydomain的L-fuzzy单调映射都可以扩张成为一个L-fuzzyScott连续映射。  相似文献   

17.
借助L-fuzzy关系在L-fuzzy中集中引入L-fuzzy偏序,自然地有了L-fuzzy偏序集,进一步借助水平截集刻画了L-fuzzy偏序集。  相似文献   

18.
设 p是任意奇素数 ,证明了任意角不能通过圆规直尺作图 p等分 .进而证明了任意角可以 m等分的充要条件是 m是 2的方幂  相似文献   

19.
Proper orthogonal decomposition (POD) finds an orthonormal basis yielding an optimal reconstruction of a given dataset. We consider an optimal data reconstruction problem for two general datasets related to balanced POD, which is an algorithm for balanced truncation model reduction for linear systems. We consider balanced POD outside of the linear systems framework, and prove that it solves the optimal data reconstruction problem. The theoretical result is illustrated with an example.  相似文献   

20.
二阶常微分方程初值问题C^0有限元的超收敛   总被引:2,自引:0,他引:2  
基于在一个单元上的改进的单元正交展开,得到二阶常微分方程C^0有限元Uh的误差表示式,以及节点和单元内部一些特征点上的超收敛结果。  相似文献   

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