二重三角级数和函数的范数研究

对形如 ∞n=0 ∞m=0amncosmxcosny等二重三角级数的和函数进行了研究 ,并证明了其范数‖f(x ,y)‖ p =∫π-π∫π-π｜f(x ,y) ｜p1dxp2 /p1dy1 /p2 <∞所满足的几个不等式 .     

争鸣

问题问题109已知函数f(x)满足:f(x y) f(x-y)=2f(x)·f(y),且f(0)≠f(π2)=0,求f(π)及f(2π)的值.解法1令x=y=0,得f(0)=1.令x=y=π2,得f(π)=-1.令x=y=π,得f(2π)=1.解法2令x=y=0,得f(0)=1.令x=32π,y=π2,得f(2π)=-f(π).再令x=y=π,得f(2π) 1=2f2(π),∴2f2(π) f(π)-1=0.∴f(π)=12或f(π)=-1,从而f(2π)=-12或f(2π)=1.问题出在哪里?问题110人教版高一数学(上)P8,有下面一段话:容易知道,对于集体A,B,C,如果A B,B C,那么A C.事实上,设x是集合A的任意一个元素,因为A B,所以x∈B,又因为B C,所以x∈C,从而A C.这个证明严格吗?…     

从一道习题谈ξ(2n)的求法

一般的高等数学教科书或习题集在Fourier级数这一章都安排有类似以下的例题或习题:求x2/4-π|x|/2 π2/6在[-π,π]上的Fourier级数展开式,并计算∑∞n=11/n2的值.它的答案是x24-π2|x| π26=∑∞n=11n2cosnx,-π≤x≤π.(1)　在上式中令x=0得∑∞n=11n2=π26.仔细观察(1)式的右边会发现如果对它积分2次,再令x=0就会出现和式∑∞n=11/n4.一般地对(1)式右边不断积分重复2k-2次,再令x=0就会出现和式∑∞n=11/n2k.这就启示我们也许可以通过上述方式来求级数∑∞n=11/n2k的值.下面我们就来实现它.为符号简单起见,记ξ(2k)=∑∞n=11n2k,k≥1.把(…     

On a Refinement of Hardy-Hilbert's Inequality and Its Applications

§1. Introduction If p＞1, 1p+1q=1, an≥0, bn≥0, and 0＜∑∞n=1-λapn＜∞, 0＜∑∞n=1-λbqn＜∞ (λ=0,1), then∑∞m=1-λ∑∞n=1-λambnm+n+λ＜πsin(π/p)∑∞n=1-λapn1/p∑∞n=1-λbqn1/q,(1.1)where the constant π／sinπp is best possible for λ=0, or 1. For λ=0,1, (1.1) is named of HardyHilberts inequality, which is important in analysis and applications (see ［1］, Chapt. 9). On (1.1) for λ=0, by estimating a weight coefficient, Xu［2］ gave a refinement as∑∞m=1∑∞n=1ambnm+n＜∑∞n=…     

新题征展(36)

A　题组新编1 . ( 1 )函数 y =π - x2 -x2 -π(　　 ) ;( 2 )设 e为自然对数的底 ,则函数y = eπ - x2| 4- x| - 4(　　 ) ;( 3)函数 y =12 sin(πx) .( 1ax - 1 12 ) 3 3(　　 ) ;( 4 ) f ( x)不是常函数 ,且 f( x)满足f ( 8 x) =f( 8- x) ,f ( x 2 ) =f( x - 2 ) ,则 f ( x) (　　 ) .( A)是奇函数 ,不是偶函数( B)是偶函数 ,不是奇函数( C)是奇函数 ,也是偶函数( D)既不是奇函数 ,也不是偶函数2 .( 1 ) f( x)为奇函数是 f ( 0 ) =0的(　　 ) ;( 2 ) sinθ <0是θ在第三或第四象限的(　　 ) ;( 3) p为假或 q为假是 p为真且 q为真…     

一道三角函数题参考答案的探究

<正>近日做到这样一道题目:已知f(sinθ)=cos2θ+cosθ.(1)求y=f(cosx)解析式;(2)求(1)中函数在x∈[0,π/2]上的最大值和最小值.参考答案是:解(1)∵cosx=sin(π/2-x),∴y=f(cosx)=f[sin(π/2-x)]=cos[2(π/2-x)]+cos(π/2-x)=cos (π-2x)+sinx=-cos2+sinx=     

清华大学·微积分试题(A卷)1999.7.12

一、填空题(每小题4分,共40分)1.幂级数∑∞n=0(-1)n 1xn3n 2(n 4)的收敛半径是;收敛域是.2.函数f(x)在区间[0,1]上的表达式为2-x,f(x)在区间[0,1]上的正弦展开和余弦展开分别是S1(x)=∑∞n=1bnsinnπx和S2(x)=a02 ∑∞n=1ancosnπx,则S1(0)=,S2(0)=.3.设L是抛物线y=x2(-1≤x≤1),x增加方向为正向,则∫Lxdl=;∫Lxdy-ydx=.4.设S为半球面z=1-x2-y2,则S(x y z)dS=.5.设L是平面上一条逐段光滑的简单闭曲线,它包围的区域D的面积等于A,a1,a2,a3,b1,b2,b3是常数.则∮L(a1x a2y a3)dx (b1x b2y b3)dy=.6.设S为平面x y z=1在第一挂限的部分上侧…     

西北工业大学·高等数学(B卷)

一、填空题 (本题共 5小题 ,每小题 4分 ,满分 2 0分 )1 .设 z =e- ( yx xy) ,则 dz| ( 1,2 ) =2 .由曲面 z =4-12 (x2 y2 )与平面 z =2所围成的立体的体积等于3.设Σ是平面 x y z =6被圆柱 x2 y2 =1所载下的部分取上侧 ,则 Σzdxdy =4.设 f (x)是以 2π为周期的周期函数 ,在区间 (-π,π]上有 f (x) =1 -x,　 -π 相似文献   

2007年普通高等学校招生全国统一考试 山东数学(理科)

一、选择题:本大题共12小题,共60分1.若z=cosθ isinθ(i为虚数单位),则使z2=-1的θ值可能是A.6πB.4πC.3πD.2π2.已知集合M={-1,1},N={x|21<2x 1<4,x∈Z},则M∩N=A.{-1,1}B.{-1}C.{0}D.{-1,0}3.下列几何体各自的三视图中,有且仅有两个A视.图①相②同的是B.①③C.①④D.②④4.设α∈-1,1,21,3,则使函数y=xα的定义域为R且为奇函数的所有α值为A.1,3B.-1,1C.-1,3D.-1,1,35.函数y=sin2x π6 cos2x 3π的最小正周期和最大值分别为A.π,1B.π,2C.2π,1D.2π,26.给出下列三个等式:f(xy)=f(x) f(y),f(x y)=f(x)f(y),f(x y)=f(x) f(y)…     

2006年高考数学全国卷Ⅰ

一、选择题:共12小题,每小题5分,共60分.1.设集合M={x|x2-x<0},N={x||x|<2},则A.M∩N=B.M∩N=MC.M∪N=MD.M∪N=R2.已知函数y=ex的图像与函数y=f(x)的图像关于直线y=x对称,则A.f(2x)=e2x(x∈R)B.f(2x)=ln2·lnx(x>0)C.f(2x)=2ex(x∈R)D.f(2x)=lnx+ln2(x>0)3.双曲线mx2+y2=1的虚轴长是实轴长的2倍,则m=A.-41B.-4C.4D.414.如果复数(m2+i)(1+mi)是实数,则实数=A.1B.-1C.2D.-25.函数f(x)=tanx+4π的单调增区间为A.kπ-2π,kπ+2π,k∈ZB.(kπ,(k+1)π),k∈ZC.kπ-34π,kπ+4π,k∈ZD.kπ-4π,kπ+34π,k∈Z6.△ABC的内角A、B、…     

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.     

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.     

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].