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1.
二重三角级数和函数的范数研究   总被引:2,自引:1,他引:1  
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 .  相似文献   

2.
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?…  相似文献   

3.
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(…  相似文献   

4.
On a Refinement of Hardy-Hilbert''''s Inequality and Its Applications   总被引:1,自引:0,他引:1  
§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=…  相似文献   

5.
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…  相似文献   

6.
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…  相似文献   

7.
<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=  相似文献   

8.
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   

9.
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)…  相似文献   

10.
设Ω=[-πxπ,-πyπ],C(Ω)表示关于x,y均以2π为周期的连续函数空间.若f(x,y)∈C(Ω),取结点组为(xk,yl)=(2k+2n 1)π,(2l 2+m 1)πk=0,1,2,…,2n,l=0,1,2,…,2m,则我们获得一个二元三角插值多项式Cn,m(f;x,y)=M1N∑k=2n0∑l=2m0f(xk,yl).1+2∑nα=1cosα(x-xk)+2∑mβ=1cosβ(y-yl)+4∑nα=1∑mβ=1cosα(x-xk)cosβ(y-yl)其中M=2m+1,N=2n+1.为改进其收敛性,本文构造一个新的因子ρα,β,使得带有该因子ρα,β的二元三角插值多项式Ln,m(f;x,y)可以在全平面上一致地收敛到每个连续的f(x,y),且具有最佳逼近阶.  相似文献   

11.
As early as in 1990, Professor Sun Yongsheng, suggested his students at Beijing Normal University to consider research problems on the unit sphere. Under his guidance and encouragement his students started the research on spherical harmonic analysis and approximation. In this paper, we incompletely introduce the main achievements in this area obtained by our group and relative researchers during recent 5 years (2001-2005). The main topics are: convergence of Cesaro summability, a.e. and strong summability of Fourier-Laplace series; smoothness and K-functionals; Kolmogorov and linear widths.  相似文献   

12.
<正>Submission Authors must use LaTeX for typewriting,and visit our website www.actamath.com to submit your paper.Our address is Editorial Office of Acta Mathematica Sinica,Academy of Mathematics and Systems Science,Chinese Academy of Sciences,Beijing 100190,P.R.China.  相似文献   

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14.
正August 10-14,2015Beijin,China The International Congress on Industrial and Applied Mathematics(ICIAM)is the premier international congress in the field of applied mathematics held every four years under the auspices of the International Council for Industrial and Applied Mathematics.From August 10 to 14,2015,mathematicians,scientists  相似文献   

15.
In this paper, we study the commutators generalized by multipliers and a BMO function. Under some assumptions, we establish its boundedness properties from certain atomic Hardy space Hb^p(R^n) into the Lebesgue space L^p with p 〈 1.  相似文献   

16.
In this paper we study best local quasi-rational approximation and best local approximation from finite dimensional subspaces of vectorial functions of several variables. Our approach extends and unifies several problems concerning best local multi-point approximation in different norms.  相似文献   

17.
<正>May 26,2014,Beijing Science is a human enterprise in the pursuit of knowledge.The scientific revolution that occurred in the 17th Century initiated the advances of modern science.The scientific knowledge system created by  相似文献   

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19.
<正>August 10-14,2015Beijing,ChinaThe International Congress on Industrial and Applied Mathematics(ICIAM)is the premier international congress in the field of applied mathematics held every four years under the auspices of the International Council for Industrial and Applied Mathematics.From August 10 to 14,2015,mathematicians,scientists  相似文献   

20.
Let P(z)=∑↓j=0↑n ajx^j be a polynomial of degree n. In this paper we prove a more general result which interalia improves upon the bounds of a class of polynomials. We also prove a result which includes some extensions and generalizations of Enestrǒm-Kakeya theorem.  相似文献   

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