新题征展(52)

A 题组新编 1.(1)满足条件{1,2}M{1,2,3,4,5}的集合M共有个;(2)满足条件M∪{a,b,c}={a,b,c,d,e}的集合M共有个;(3)M{1,2,3,4,5},且满足条件若a∈M,则6-a∈M,这样的非空集合M共有个;(4)A∪B {a,b}的集合A、B共有对;(5)A∪B={a,b,c}的集合A、B共有对.     

新题征展(52)

A　题组新编1.(1)满足条件 { 1,2 } M { 1,2 ,3,4 ,5 }的集合 M共有个 ;(2 )满足条件 M∪ { a,b,c} ={ a,b,c,d,e}的集合 M共有个 ;(3) M { 1,2 ,3,4 ,5 } ,且满足条件 :若 a∈ M,则 6 - a∈ M,这样的非空集合 M共有个 ;(4 ) A∪ B ={ a,b}的集合 A、B共有对 ;(5 ) A∪ B ={ a,b,c}的集合 A、B共有对 .2 .(1)若 f (x) =x1 x,则 f(1) f(2 ) f(3) … f(2 0 0 4 ) f(12 ) f(13) f(14 ) … f(12 0 0 4 ) =;(2 )若 f(x) =x21 x2 ,则 f (1) f(2 ) f(3) … f(2 0 0 4 ) f(12 ) f(13) f(14 ) … f(12 0 0 4 ) =;(3)若 f(x…     

新题征展(35)

A　题组新编1 .已知曲线 C:xy - 2 kx k2 =0与直线 l:x - y 8=0有唯一的公共点 ,而数列{an}的首项 a1=2 k,点 ( an- 1,an)恒在曲线上( n≥ 2 ) ,数列 {bn}满足关系 bn =1an - 2 .( 1 )问数列 {bn}是等差数列吗 ?( 2 )求数列 {an}的通项公式 .2 .已知二次函数 f ( x) =ax2 bx c有f ( 0 ) =3,且直线 y =5x 1与 f( x)的图像相切于点 ( 2 ,1 1 ) .( 1 )求函数 f ( x)的解析式 ;( 2 )若 f( n)为数列 {an}的前 n项和 ,求数列 {an}的通项公式 ;( 3)求limn→∞ ( 1a2 a3 1a3a4 1a4 a5 … 1an- 1an) .B　藏题新掘3.在边长为 1的正△ …     

HAUSDORFF DIMENSION OF CUTSET OF COMPLEX VALUED RADEMACHER SERIES

Let{a.}.>1beasequenceofrealnumberssatisfyingEla.l~coandtiman=0.n=1n-cocoThenRademacherseriesZa.(l--Ze.(x))takeseverypreassignedrealvalueN(cardinaln=1numberofthecontinuum)timesforx6(0,1],whereE.(x)isthen-thdigitofthe(unique)non--terminating2--adicexpansionofx6(0,l].W.A.Beyer[1]showsthatif{a.}.>1Efi,{a.}411,thenforanyaCR,dimH{xE(0,1];Za.(1--26.(x))~a}~1;ifn=1co{a.}.21411,nlLmcoan~0,,Zla.--a.--if相似文献   

Powers of the Catalan Generating Function and Lagrange''s 1770 Trinomial Equation Series

The Catalan numbers $1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862,\ldots$ are given by $C(n)=\frac{1}{n+1}\binom{2n}{n}$ for $n\geq 0$. They are named for Eugene Catalan who studied them as early as 1838. They were also found by Leonhard Euler (1758), Nicholas von Fuss (1795), and Andreas von Segner (1758). The Catalan numbers have the binomial generating function $$\mathbf{C}(z) = \sum_{n=0}^{\infty}C(n)z^n = \frac{1 - \sqrt{1-4z}}{2z}$$ It is known that powers of the generating function $\mathbf{C}(z)$ are given by $$\mathbf{C}^a(z) = \sum_{n=0}^{\infty}\frac{a}{a+2n}\binom{a+2n}{n}z^n.$$ The above formula is not as widely known as it should be. We observe that it is an immediate, simple consequence of expansions first studied by J. L. Lagrange. Such series were used later by Heinrich August Rothe in 1793 to find remarkable generalizations of the Vandermonde convolution. For the equation $x^3 - 3x + 1 =0$, the numbers $\frac{1}{2k+1}\binom{3k}{k}$ analogous to Catalan numbers occur of course. Here we discuss the history of these expansions. and formulas due to L. C. Hsu and the author.     

置换多项式$x^{1+\frac{q-1}{m}}+ax$

设$m$为正整数, $F_{q^r}$是特征为$p$的有限域. 本文证明了如果$p>m^2-m$且$q\equiv 1\pmod{m}$, 则多项式$x^{1+\frac{q-1}{m}}+ax~(a\neq0)$不是$F_{q^r}~(r\geq2)$上的置换多项式. 本文还证明了$q\equiv 1\pmod{7}$且$p\neq 2, 3$时, $x^{1+\frac{q-1}{7}}+ax~(a\neq0)$不是$F_{q^r}~(r\geq2)$上的置换多项式     

新题征展(125)

A 题组新编 1.(1)设数列{an}由a1=5,a2=23,an+2=5an=1-an(nΕN*)确定. ①求证{an+1-5-√21/2an}是等比数列; ②求数列{an}的通项公式. (2)若把无理数(5+√21/2)2011写成小数,求其个位数字及十分位、百分位上的数字.     

新题征展(116)

A 题组新编 1.(胡寅年)设{an}是首项为1的正项数列,且n+1a2n+1-na2n+an+1an=0. (1)求{an}的通项公式; (2)设bn=(ln(1+an))/(an),证明:①bn≤bn+1;②ln2≤bn相似文献   

On Products of Property b1

In this note,we present that:(1)Let X=σ{Xα:α∈A} be|A|-paracompact (resp.,hereditarily |A|-paracompact).If every finite subproduct of {Xα:α∈A} has property b1 (resp.,hereditarily property b1),then so is X.(2) Let X be a P-space and Y a metric space.Then,X×Y has property b1 iff X has property b1.(3) Let X be a strongly zero-dimensional and compact space.Then,X×Y has property b1 iff Y has property b1.     

具有性质$b_1$空间的乘积性

In this note,we present that:(1)Let X=σ{Xα:α∈A} be|A|-paracompact (resp.,hereditarily |A|-paracompact).If every finite subproduct of {Xα:α∈A} has property b1 (resp.,hereditarily property b1),then so is X.(2) Let X be a P-space and Y a metric space.Then,X×Y has property b1 iff X has property b1.(3) Let X be a strongly zero-dimensional and compact space.Then,X×Y has property b1 iff Y has property b1.     

新题征展(71)

A题组新编1.(1)已知数列{an}满足a1=m,a2=s(m≠s),且对任意不小于3的正整数n,均有an=an-1 an-22,求limn→∞an.(2)已知数列{an}满足a1=m,a2=s,a3=p,(m,s,p两两不等),且对任意不小于4的正整数n,均有an=an-1 an-2 an-33,求limn→∞an.2.(1)函数f(x)满足2x=m f(x)m-f(x)(m为大于零     

不等式

一、选择题(有且仅有一个答案正确) ,.。,>亡“的一个充分必要条件是(). (A)。>卜.(B)。>lb! (C)la}>b.(D)!a{>{b!, 2.当。>b>。{l于,一厂列不等式恒成立的是(). (A)ab>ao.(B)a}‘1>bl。】. _性)】a6})!。广1·戈D)(“一6)1‘一白!)0· 3.设a、b是满足。b相似文献   

放缩的方法探究

<正>考题(2014年新课标全国卷Ⅱ第17题)已知数列{an}满足a1=1,an+1=3an+1.(1)证明:{an+1/2}是等比数列,并求{an}的通项公式;(2)证明:1/a1+1/a2+...+1/an<3/2.不难证得(1)数列{an+1/2}是以3/2为首项,     

高一数学同步训练题

内容 :1.代数 :集合、映射与函数 ;　 2 .立体几何 :平面 ,空间两直线 .　　选择题1　已知全集Ｉ ={ 1,2 ,3 ,4,5 } ,Ａ∩Ｂ ={ 2 } ,Ａ∩Ｂ ={ 1,4} ,则Ｂ等于 (　　 )(Ａ) { 3} .　　　 (Ｂ) { 5 } .(Ｃ) { 1,2 ,4} .(Ｄ) { 3 ,5 } .2　设集合Ｍ ={ 1999,2 0 0 0 ,2 0 0 1} ,Ｎ ={ｘ|ｘ∈Ｍ } ,则Ｍ与Ｎ的关系是 (　　 )(Ａ)Ｍ =Ｎ .(Ｂ)Ｍ Ｎ .(Ｃ)Ｍ Ｎ .(Ｄ)Ｍ∩Ｎ = .3　三个平面最多可把空间分成 (　　 )(Ａ) 4个部分 .(Ｂ) 6个部分 .(Ｃ) 7个部分 .(Ｄ) 8个部分 .4　已知ａ ,ｂ是异面直线 ,直线ｃ平行于直线ａ ,那么ｃ与ｂ (　　 …     

新题征展(94)

A题组新编 　　1.(姜本超)(1)若a1=cos a/2,an/an-1=cos(2n-2·a),(0相似文献   

THE GROWTH OF RANDOM DIRICHLET SERIES (I)

We consider random Dirichlet serieswhere an C C, 0 5 A. T co, Zn(w) is a sequence of random variables defined in the probabilityspace (fi, F, P), s = a it(a, t E R).Conveniently we consider Dirichlet seriesThe convex regularized sequence of {-- In la. l} is noted as {-- In la; I}, set a.(w) = a.Zn(w),the convex regularized sequence of {-- In la.(w)l} is noted as {-- In la;(w) l} where a.(a.(w)) isthe abscissa of convergence about f(s)(f(s, w)).Lemma 1 (i) If Z.(w) satisfiesthen a.s.…     

名校基础知识自测

★高一年级 北京第十二中学(100071)李有毅一、选择题1.卜列四个关系式中正确的是(). (A)g任{a}(B)a星{a} ((、){a}任{a,b}(D)a〔{a,b}2.满足{l}里A里{1,2,3}的集合A的个数为().(A)l(B)2(C)3(D)43.已知尸一{、}二2一3二+2一0},T一{y{yS一定之一一5}.则尸nTUS一().(A)2)(B){1,2}(C){一2,2}(D){1}设全集u一{2,3,5},A={}a一5{,2},CoA一{5},贝日u的值为().(A)2(B)8(C)2或8(D)一2或8已知集合{‘·{一2了.>2了,·>。)一{工}了<一5或二>4},则,丫+n的值为().(A)一8(11)l()(C)8(D)80若集合A一{二i“厂一a二+1<。}一②,则实数“的值的集合…     

$K_{5}\times S_{n}$ 的交叉数

把完全图$K_{5}$的五个顶点与另外$n$个顶点都联边得到一类特殊的图$H_{n}$.文中证明了$H_{n}$的交叉数为$Z(5,n)+2n+\lfloor \frac{n}{2}\rfloor+1$,并在此基础上证明了$K_{5}$与星$K_{1,n}$的笛卡尔积的交叉数为$Z(5,n)+5n+\lfloor\frac{n}{2} \rfloor+1$.     

一道调考题的解法及推广

湖北省八市2007年高三三月调考数学试卷(理科)第10题:已知数列{an}为等差数列,从集合A={a1,a2,…,a20}中取出3个不同的数,使这3个数成等差数列,则不同的等差数列共有()(A)90个.(B)120个.(C)180个.(D)200个.[解法1]分类讨论设数列{an}的公差为d,由集合中元素的互异性知d≠0,所取出的公差为d的等差数列是由a1,a2,a3的各项分别加上0,d,2d,…,17d构成的,共有18个,同样可知公差为2d,3d,…,9d的等差数列分别有16,14,…,2个,又每个公差为kd(1≤k≤9,k∈N*)的等差数列都对应一个公差为-kd的等差数列,故不同的等差数列共有2(2 4 … 18)=180个.选(C…