关于n~(1/k)的两个不等式

文 [1 ]中给出了一个涉及n的不等式 :设正整数n >1 ,则2n + 23·n - 2 (n - 1 ) + 23·n - 1≤n <4n + 36 ·n - 4(n - 1 ) + 36 ·n - 1 ( 1 )由不等式 ( 1 ) ,可推出2n + 23·n - 2 - 13≤∑nk=1k≤4n + 36 ·n - 16 ( 2 )当且仅当n =1 ,2时 ,式中等号成立 .本文给出类似于不等式 ( 1 )的关于 kn的两个不等式 ,并提出一个猜想 .定理 1　设正整数n >1 ,则1 2n + 71 6 ·3n - 1 2 (n - 1 ) + 71 6 ·3n - 1 <3n <3n + 24 ·3n - 3(n - 1 ) + 24 ·3n - 1 ( 3)证　要证上限不等式 3n <3n + 24 ·3n- 3(n - 1 ) + 24 ·3n - 1 ,只要证( 3n - 2…     

上期课外练习解答

高一年级1.∵　f(2 ) =f(1)·f(1) =1,f(3 ) =f(1)·f(2 ) =1,f(4 ) =f(3 )·f(1) =1……由归纳得f(1) =f(2 ) =f(3 ) =… =f(2 0 0 3 ) =1.∴　原式 =1.2 .当x为非零实数 ,故　f(x + 1) =f(x)·f(1) 　f(x + 1)f(x) =f(1) =3 ,故　f(2 )f(1) + f(4 )f(3 ) +… + f(2n)f(2n -1) =3n .∴　n =667.3 .f(x) =a + 1-2ax + 2 欲使f(x)在 (-2 ,+∞ )上是增函数 ,只须使 1-2a <0 ,故a的取值范围是 (12 ,+∞ ) .高二年级1.记f(x) =x2 -2x +a ,g(x) =x2 -2bx + 5由函数图象易知A B f(1) =a -1≤ 0 ,f(3 ) =3 +a≤ 0 ,且 g(1) =6-2b≤ 0 ,g(3 ) =1…     

高階等差級数

1+1+1+…+1=n,1+2+3+…+n=(1/2)n(n+1),1+3+6+…+(1/2)n(n+1)==(1/1·2·3)n(n+1)(n+2)1+4+10+…+(1/1·2·3)n(n+1)(n+2)==(1/1·2·3·4)n(n+1)(n+2)(n+3),…………………要証明这些式子是不困难的。例如:因为     

关于n~(1/k)的不等式猜想的研讨

文 [1 ]给出了一个关于kn的不等式猜想 ,猜想的右侧不等式是 :正整数n ,k >1 ,则nk 2时 ,( 1 )式成立 .为证明上述结论 ,先给出两个引理引理 1　 [贝努利 (Bernoulli)不等式 ]若x >- 1且k是正整数 ,则 ( 1 +x) k≥ 1 +kx .等号当且仅当x =0时成立 .利用二项式定理易证引理 1 .引理 2 [2 ] 　若 - 1 相似文献   

两个关于三角形边角关系的结论

定理 1　设a、b、c为△ABC的三边 ,当an,bn,cn(n∈N+,n <5 )组成等差数列时∠B≤ 60°.证明　当n=1时 ,2b=a+c由cosB =a2 +c2 -b22ac=a2 +c2 - 14(a+c) 22ac =34× a2 +c22ac - 14≥12 　即B ≤ 60°当n =2时 ,2b2 =a2 +c2cosB =a2 +c2 -b22ac=a2 +c2 - 12 (a2 +c2 )2ac =12 ·a2 +c22ac ≥ 12 　即B≤ 60°当n =3时 ,12 (a3+c3)≥ ( a+c2 ) 3 (a3+c3) 3≥ ( a3+c32 ) 2 (a+c) 3 (a+c) 3(a2 +c2 -ac) 3≥ ( a3+c32 ) 2 (a+c) 3 (a2 +c2 -ac)≥ ( a3+c32 ) 2 (a2 +c2 -ac)≥ ( a3+c32 ) 23 a2 +c2 -ac≥b2 B ≤ 60°当n =4时 ,(a-c) 4 …     

用“判别式法”求极值辨析

求二次函数型的极值常可运用“判别式法”(以下简称“△法”)。但运用“△法”求极值可能产生增解或失解,学生在解题时常常忽略这个问题而出现一些错误,下面略举几例说明: 例1 求函数y=2-(4/x)-3x的极值(x>0) 错解函数可变形为3x~2+(y-2)x+4=0 (1) ∵x∈R ∴△=(y-2)~2-4·3·4≥0 解之得 y≤2-(4(3)~(1/2))或y≥2+4(4)3~(1/2)。简析:y极小=2+4(3)~(1/2)了就是用“△法”产生不符合题意的答案,事实上,当y=2+4(3)~(1/2)时,方程(1)化为3x~2+4(3)~(1/2)x+4=0(3~(1/2)x+2)~2x=-(2(3)~(1/2))/3<0。     

问题与解答

一本期问题 1 △ABC中,已知BC、CA、AB边上的高分别是h_a=6、h_b=4、h_C=3,试求△ABC的面积。 2 设以r为半径的圆内接正992边形P_1P_2…P_(992),P是圆周上的任意一点,求证PP_1~2+PP_2~2+…+PP_(992)~2=1984r~2。上海金山县中学生朱维欧提供 3 证明当n是自然数时,2~(1/2)·4~(1/4)·8~(1/8)…2~n(2~n)~(1/2)<4。 4 设x、y为正整数,且3x~2+2y~2=6x,问x取何值时,x~2+y~2达到最大值,并求出此最大值。巴东安居中学谭志新提供 5 求证 lg1+lg2+…+lgn相似文献   

上期课外练习解答

高一年级1.设ｆ(ｘ) =(ｘ - 1)ｌｏｇ23 ａ - 6ｘ·ｌｏｇ3 ａ +ｘ + 1=( 1+ｌｏｇ23 ａ - 6ｌｏｇ3 ａ)ｘ + 1-ｌｏｇ23 ａ ,∵ ｆ(ｘ)在 [0 ,1]上恒成立 ,由一次函数的单调性知 :ｆ( 0 ) >0 ,ｆ( 1) >0 ,　解得　 13 <ａ <33 .2 .设每期期初存入金额Ａ ,连存ｎ次 ,每期的利率为Ｐ ,那么到第ｎ期期末时 ,本金为ｎＡ ,则应得到的全部利息之和为 :Ｓｎ=ＡＰ +ＡＰ·2 +… +Ａ·ｐ·ｎ =ｎ(ｎ + 1)2 ＡＰ ,应纳税为　 ｎ(ｎ + 1)2 ＡＰ× 2 0 % =ｎ(ｎ + 1)10 ＡＰ ,实际取出　Ａ[ｎ + 2ｎ(ｎ + 1)5Ｐ] ,当Ａ =110 0 ,ｎ =12 ,Ｐ =0 .165%时 ,…     

关于补差集与Hadamard矩阵

本文着重讨论了H型补差集,主要结果是: (1) 证明了存在2~i·10~j 18~k·26~r·50~s·82~t阶H型2-补差集;其中i,j,k,r,s,t,为任意非负整数; (2) 给出了71阶和73的H型4-补差集; (3) 定义了v阶Abel群上的C划分, 给出了v=37和61时的C划分,指出了v∈S=S_2∪S_1∪S_3时存在C划分,其中 S_1={2k+1:O≤k≤16}∪{59} S_2={2~i·lO~j·26~k+1:i, j, k为任意非负整数}, S_3={37,61}: (4) 指出了当v′∈S,u∈W=W_1∪W_2∪W_3时,存在v′v阶H型4-补差集,其中 W_1={3~n:n≥1}, W_2={2k+1:0≤k≤14}∪{37,43}, W_3={n:2n-1≡1(mod4)是一素数的方幂}; (5) 利用C划分和[3]的一个结果证明了,当m∈S,n∈W_3时,存在2mn~r(n+1)阶H阵(r≥O); (6) 最后还证明,当在同一个u≡3(mod4)阶Abel群上存在{u;k;λz}差集和{u;1/2(u-1);1/4(u-3)}差集时,且存在v+l=u+1-4(k-λ)阶skew type H阵,则存在uv~r(v+1)阶H阵(r≥O).     

2004年第五期初中数学问题参考解答

一、求证 :f(n) =an + 2 +(a +1 ) 2n + 1被a2 +a +1整除 ,其中a是整数 ,n是自然数 .证明 :( 1 )当n =0时 ,f( 0 ) =a2 +(a +1 ) =a2 +a+1能被a2 +a +1整除 .( 2 )假设当n =k时 ,f(k) =ak+ 2 +(a +1 ) 2k+ 1能被a2 +a +1整除 .当n =k +1时 ,有f(k +1 ) =ak+ 3 +(a +1 ) 2 (k + 1) + 1=a·ak + 2 +(a+1 ) 2k+ 1·(a+1 ) 2=a·ak+ 2 +a2 ·(a +1 ) 2k + 1+2a·(a +1 ) 2k+ 1+(a+1 ) 2k + 1=[a·ak+ 2 +a·(a +1 ) 2k+ 1]+[a2 (a +1 ) 2k+ 1+a·(a +1 ) 2k + 1+(a+1 ) 2k+ 1]=a[ak + 2 +(a+1 ) 2k + 1]+(a +1 ) 2k + 1·(a2 +a +1 ) .∵a是整数…     

On a class of self-similar processes with stationary increments in higher order Wiener chaoses

We study a class of self-similar processes with stationary increments belonging to higher order Wiener chaoses which are similar to Hermite processes. We obtain an almost sure wavelet-like expansion of these processes. This allows us to compute the pointwise and local Hölder regularity of sample paths and to analyse their behaviour at infinity. We also provide some results on the Hausdorff dimension of the range and graphs of multidimensional anisotropic self-similar processes with stationary increments defined by multiple Wiener–Itô integrals.     

Qualitative Analysis of Bifurcating Solutions in the Lengyel-Epstein Model

One of the most fundamental problems in theoretical biology is to explain the mechanisms by which patterns and forms are created in the＇living world. In his seminal paper ＂The Chemical Basis of Morphogenesis＂, Turing showed that a system of coupled reaction-diffusion equations can be used to describe patterns and forms in biological systems. However, the first experimental evidence to the Turing patterns was observed by De Kepper and her associates（1990） on the CIMA reaction in an open unstirred reactor, almost 40 years after Turing＇s prediction. Lengyel and Epstein characterized this famous experiment using a system of reaction-diffusion equations. The Lengyel-Epstein model is in the form as follows     

A REPRESENTATION FORMULA RELATED TO SCHR(O)DINGER OPERATORS

Schr(o)dinger operator is a central subject in the mathematical study of quantum mechanics.Consider the Schrodinger operator H = -△ V on R, where △ = d2/dx2 and the potential function V is real valued. In Fourier analysis, it is well-known that a square integrable function admits an expansion with exponentials as eigenfunctions of -△. A natural conjecture is that an L2 function admits a similar expansion in terms of "eigenfunctions" of H, a perturbation of the Laplacian (see [7], Ch. Ⅺ and the notes), under certain condition on V.     

On a class of Riemann surfaces

It is considered the class of Riemann surfaces with dimT1 = 0, where T1 is a subclass of exact harmonic forms which is one of the factors in the orthogonal decomposition of the spaceΩH of harmonic forms of the surface, namely The surfaces in the class OHD and the class of planar surfaces satisfy dimT1 = 0. A.Pfluger posed the question whether there might exist other surfaces outside those two classes. Here it is shown that in the case of finite genus g, we should look for a surface S with dimT1 = 0 among the surfaces of the form Sg\K , where Sg is a closed surface of genus g and K a compact set of positive harmonic measure with perfect components and very irregular boundary.     

(0,1 ;0)-INTERPOLATION ON INFINITE INTERVAL (-∞, +∞)

In this paper, we study the explicit representation and convergence of (0, 1; 0)-interpolation on infinite interval, which means to determine a polynomial of degree ≤ 3n - 2 when the function values are prescribed at two set of points namely the zeros of Hn(x) and H′n(x) and the first derivatives at the zeros of H′n(x).     

CONTENTS OF VOLUME 30

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Instructions for contributions

正Applied Mathematics-A Journal of Chinese Universities,Series B(Appl.Math.J.Chinese Univ.,Ser.B)is a comprehensive applied mathematics journal jointly sponsored by Zhejiang University,China Society for Industrial and Applied Mathematics,and Springer-Verlag.It is a quarterly journal with     

Journal of Mathematical Research with Applications Guide for Authors

正Journal overview:Journal of Mathematical Research with Applications(JMRA),formerly Journal of Mathematical Research and Exposition(JMRE)created in 1981,one of the transactions of China Society for Industrial and Applied Mathematics,is a home for original research papers of the highest quality in all areas of mathematics with applications.The target audience comprises:pure and applied mathematicians,graduate students in broad fields of sciences and technology,scientists and engineers interested in mathematics.     

Staircase transportation problems with superadditive rewards and cumulative capacities

A cumulative-capacitated transportation problem is studied. The supply nodes and demand nodes are each chains. Shipments from a supply node to a demand node are possible only if the pair lies in a sublattice, or equivalently, in a staircase disjoint union of rectangles, of the product of the two chains. There are (lattice) superadditive upper bounds on the cumulative flows in all leading subrectangles of each rectangle. It is shown that there is a greatest cumulative flow formed by the natural generalization of the South-West Corner Rule that respects cumulative-flow capacities; it has maximum reward when the rewards are (lattice) superadditive; it is integer if the supplies, demands and capacities are integer; and it can be calculated myopically in linear time. The result is specialized to earlier work of Hoeffding (1940), Fréchet (1951), Lorentz (1953), Hoffman (1963) and Barnes and Hoffman (1985). Applications are given to extreme constrained bivariate distributions, optimal distribution with limited one-way product substitution and, generalizing results of Derman and Klein (1958), optimal sales with age-dependent rewards and capacities.To our friend, Philip Wolfe, with admiration and affection, on the occasion of his 65th birthday.Research was supported respectively by the IBM T.J. Watson and IBM Almaden Research Centers and is a minor revision of the IBM Research Report [6].