On the Best Degree of Approximation by Euler (E,q)-Means

Let f(x)∈L_(2π) and its Fourier series by f(x)～α_0/2+sum from n=1 to ∞(α_ncosnx+b_nsinx)≡sum from n=0 to ∞(A_n(x)). Denote by S_n (f,x) its partial sums and by E_n~q(f,x) its Euler (E, q)-means, i. e. E_n~q(f,x)=1/(1+q)~π sum from m=0 to n((?)q~(n-m)S_m(f,x)), with q≥0 (E_n~0≡S_n). In [1] Holland and Sahney proved the following theorem. THEOREM A Ifω(f,t) is the modulus of continuity of f∈C_(2π), then the degree of approximation of f by the (E,q)-means of f is givens by##特殊公式未编改     

关于凸函数的Hadamard不等式的拓广(英文)

设,是区间[a,b]上连续的凸函数。我们证明了Hadamard的不等式 f(a+b/2)≤1/b-a integral from a to b (f(x)dx)≤f(a)+f(b)/2可以拓广成对[a,b]中任意n+1个点x_0,…,x_n和正数组p_0,…,p_n都成立的下列不等式 f(sum from i=0 to n (p_ix_i)/sum from i=0 to n (p_i))≤|Ω|~(-1) integral from Ω (f(x(t))dt)≤sum from i=0 to n (p_if(x_i)/sum from i=0 to n (p_i),式中Ω是一个包含于n维单位立方体的n维长方体,其重心的第i个坐标为sum from i=i to n (p_i)/sum from i=i-1 (p_i),|Ω|为Ω的体积,对Ω中的任意点t=(t_1,…,t_n) ω(t)=x_0(1-t_1)+sum from i=1 to n-1 (x_i(1-t_(i+1))) multiply from i=1 to i (t_i+x_n) multiply from i=1 to n (t_i)。不等式中两个等号分别成立的情形亦已被分离出来。 此不等式是著名的Jensen不等式的精密化。     

甩瓦累-布然平均逼近连续函数

<正> §1.总说§1.1 设 f(x)∈C_(2π),f(x)～a_0/2+sum form n=1 to ∞ a_ncosnx+b_nsin nx≡sum form n=0 to ∞ A_n(x)记 S_n(f,x)=sum form v=0 to n A_v(x).称σ_(n,p)(f,x)=1/p+1 sum form v=n-p to n S_v(f,x)为 f(x)的瓦累-布然平均.记△_u~kf(x)=sum form v=0 to k (-1)~v(?)f[x+(k-2v)u].称函数ω_k(f,t)=(?)|△~u_kf(x)|为 f(x)的 k 阶连续模.简记ω(f,t)=ω_1(f,t).假如 f(x)的共轭函数     

复旦大学1984年硕士研究生入学考试试题

1.设x_0,x_1,…,x_n,x是n+2个相异点,证明 f(x_0,x_1,…,x_n,x)=sum from i=0 to n(f(x_j,x)/(multiply from (?) to n(x_j-x_1))) 其中f(xj,x)和f(x_o,x_1,…,x_n,x)分别表示函数f(x)的一阶和n+1阶差商。 2.设n阶线性方程组Ax=b中n×n矩阵A的顺序主子式det(A1)≠0(i=1,…n),令(n+1)×(n+1)矩阵B为     

关于线性算子的高阶饱和

设f(x)∈C_(2π)。而f(x)～sum from k=0 ( )A_k(f_1k)≡α_0/2 sum from k=1 ( )(α_kcoskx b_ksinkx)。 又设 U_n(f,x)=1/πintegral from -πto π(f(x t)u_n(t)dt,) 其中u_n(t)=1/2 sum from k=1ρ_k~(n)coskt满足条件: integral from 0 to k(|u_n(t)|dt=O(1),)ρ_k~(n)→1(n→∞;k=1,2,…,)。设m是正整数,ρ_0~(n)=1。记~mρ_k~(n)=sum form v=0 to ∞ ((-1)~(m~(-v))(m v)ρ_k v~(n) (k=0,1,…,)。)T.Nishishiraho考虑了在ρ_k~(n)=O(k>n)的情况下U_n(f,x)的饱和问题,证明了。 定理A 设{_n}是收敛于0的正数列,使得     

数列的循环求和法——探索一类数列和的表达式

数列求和的方法很多,己有许多杂志刊登了各种数列求和方法的文章,本文提及的循环求和法,其思想方法是通过式子变形,使所求和重复出现,造成循环,亦即构造出含有所求和S的方程S=f(s),然后解出S。问题:求 sum from k=1 to n (k·2~k)sum from k=1 to n (k·2~k)=sum from k=0 to (n-1) ((k+1)2~(k+1))=2 sum from k=0 to (n-1) k2~k+sum from k= to (n-1) (2(k+1))=2[sum from k=1 to n (k·2~k-n·2~n)]+sum from k=1 to n 2~k∴ sum from k=1 to n (k·2~k)=n·2~(n+1)-(2~(n+1)-2) 有许多同志会感兴趣于研究sum from k=1 to n (k~p 2~k)     

Fejèr-Korovkin算子及一种内插线性算子的渐近展开

设f(x)∈C_(2π)。本文讨论两种线性算子对f(x)的逼近,全文分两个部分。 在第一部分中,我们考虑在正卷积型三角多项式线性算子中占重要地位的Fejr-Korovkin算子K_n(f,x)=1/π integral from -x to π (f(x+t)k_n(t)dt),其中k_n(t)≡1/2+sum from k=1 to n (ρ_k~((n)) cos kt)=1/2+sum from k=1 to n (F_n(k/n+2)coskt),F_n(x)=(1-x)cosπX+1/n+2 cot π/n+2·sinπx.由于它满足Korovkin条件:所以有下述结果:设f(x)∈C_(2π),f″(x)∈C_(2π)。那么,当n→∞时,成立着     

关于积分∫f(x,{Nx})dx带准确余项的渐近展开式的一种新证明

本文对于积分from n=0 to 1 f(x,{Nx})dx带准确余项的渐近展开式from n=0 to 1 f(x,{Nx})dx=from n=0 to 1 from n=0 to 1f(x,y)dxdy+sum from k=1 to r 1/(k!) (1/N)~k from n=0 to 1[f~((k-1,0))(1,y)(?)_k(y-N)-f~((k-1,0))(O,y)B_k(y)]dy-1/(r|)(1/N)~r from n=0 to 1 from n=0 to 1 f~((r,O))(x,y)(?)_r(y-Nx)dxdy给出了一种简捷的推导,这种推导只需普通的分析知识,无需用到Euler-Maclaurin求和公式及Bernoulli多项式的Raabe乘积定理。     

退缩椭圆型方程的边值问题

<正> 在 R~n 的有界凸区域Ω上考虑椭圆型方程Lu≡sum from i,j=1 to n (a_(ij)(x)u_(xi)_(xj)+sum from i=1 to n b_i(x)u_i+c(x)u=f(x),(1)设对 x∈(?)及所有的实数组(ξ_1,ξ_2,…,ξ_n)sum from i,j=1 to n a_(ij)(x)ξ_iξ_j≥λ(x)sum from i=1 to n ξ_i~2≥0,a_(ji)(x)∈C(?),即算子 L(u)可能退缩而为退缩椭圆型算子。记(?)的边界为∑,∑上满足 sum from ij=1 to n a_(ij)n_in_j=0的点集为∑_0,(n_1,…,n_n)表示∑上的内单位法向量,∑_3=∑\∑_0,设其 n-1维测度非零,则对方程(1)可提如下的边值问题:     

OSCILLATORY AND ASYMPTOTIC BEHAVIORS OF FIRST ORDER FUNCTIONAL DIFFERENTIAL EQUATIONS

In this paper the author discusses the following first order functional differentialequations: x'(t) +integral from n=a to b p(t, ξ)x[g(t, ξ)]dσ(ξ)=0, (1) x'(t) +integral from n=a to b f(t, ξ, x[g(t, ξ)])dσ(ξ)=0. (2)Some suffcient conditions of oscillation and nonoseillafion are obtained, and two asymptolioproperties and their criteria are given. These criferia are better than those in [1, 2], and canbe used to the following equations: x'(t) + sum from i=1 to n p_i(t)x[g_i(t)] =0, (3) x'(t) + sum from i=1 to n f_i(t, x[g_i(t)] =0. (4)     

含根式不等式的处理方法

含根式不等式因技巧性较强,历年来颇受命题者喜爱,下面请欣赏几例. 一、三角代换例1 已知xi≥0,x0=0,sum from i=0 to nxi=1.求证:sum from i=1 to n xi/(1+x0+…+xi-1)(xi+…+xn)~(1/2)<π/2.证明 令x0+…+xi-1=sinθi-1,0=θ0≤θ1≤θ2≤…≤θn=π/2.则原式=sum from i=1 to n sinθi-sinθi-1/cosθi-1     

一类指数函数的高阶导数

在微积分学中,指数函数f(x)=e~(-x)~(-2)(x≠0)是一个非常简单而十分重要的初等偶函数,尤其是在函数的幂级数展开中,需要研究这个指数函数的有限形式的高阶导数及其性质.本文对此问题进行了研究,并得到如下结果:设f(x)=e~(-x)~(-2)(x≠0)的n阶导数为f_n(x)=fn(x)e~(-x)~(-2),则f_n(x)=sum from i=1 to n(-1)~(n+i)C_i(n)x~(-n-2i),其中C_1(n)=(n+1)!,C_i(n)=2sum from j=i to n(n+2i-1)!/(j+2i-1)!C_(i-1)(j-1),(1i≤n).     

一类非线性Volterra积分微分方程组及褶积型积分方程组的解之延展和界值

对非线性Volterra型积分微分方程组x'(t)=f(t,x(t))+sum from j=1 to m(integral from n=0 to t(A_j(t,s)g_j(s,x(s))ds)),t∈R_+ (1)以及褶积型积分方程组y(t)=F(t)+sum from j=1 to m(integral from n=0 to t(B_j(t-s)G_j(s,y(s))ds)),t∈R_+ (2)我们得到了如下结果:定理1 若方程组(1)满足下列条件1)f(t,η),g_j(t,η)∈c[R_+×R~n,R_n],A_j(t,s)∈c[R_+×R_+,R~(n×n)],它们使得(1)     

用希尔伯特空间理论讨论傅立叶级数

<正> 函数和它的傅立叶级数之间的关系,常见的有下列四种。命题1 (狄里赫勒定理)若f(x)∈C[-π,π),或在[-π,π]上只有有限个第一类间断点,并且可以把[-π,π]分为f(x)的有限个单调区间,则有f(x)=a_0/2+sum from i=1 to ∞(a_icosix+b_isinix)(1)其中x∈(-π,π)为f(x)的连续点,a_i,b_i为f(x)的傅立叶系数(以下同)。当x∈(-π,π)为f(x)的间断点时,则(1)式友端改为[f(x—0)+f(x+0)]/2。当x=±π时,则(1)式左端改为[f(-π+0)+f(π-0)]/2。命题2 若f(x)∈L_2[-π,π],则对任意确定的n,有||f(x)—a_0/2—sum from i=1 to n(a_1cosix+bsinix)||_2     

关于求sum from k=0 to n f(k)的一个计算公式

文[1]中讨论了利用差分多项式求sum from k=1 to n f(k)的一个方法。本文将给出直接求sum from k=0 to n f(k)的一个计算公式,作为特例,并给出求自然数方幂和的一个计算公式。设f(k)是K的m(m∈N)次多项式。定义P_m(x)=1/m! x(x-1)…(x-m+1),称为m阶差分多项式,P_0(x)=1称为零阶差分多项式。     

Fourier级数及其导级数的逼近

设p_m≥0↓,sum from k=0 to n(p_n)=P_m,n=0,l,…,p_0=P_0=1,P_n→∞(n→∞)若N_n=1/P_n sum from k=0 ton(p_(n,k)S_k→S(n t。0→∞)),则说{S_k}关于算子(N,p_n)收敛于S.设f(x)∈L_(?),S_n(f,x)为     

SOME GENERALIZATION OF GRONWALL-BIHARI INTEGRAL INEQUALITIES AND THEIR APPLICATIONS

The interest of this paper lies in the estimates of solutions of the three kinds of Gronwail-Bihari integral inequalities:(Ⅰ) y(x)≤f(x) sum from i=1 to n(g_i(x)integral from n=0 to x(h_i(d)y(s)ds)),(Ⅱ) y(x)≤f(x) g(x)φ(integral from n=0 to x(h(s)w(y(s))ds))(Ⅲ) y(x)≤f(x) sum from i=1 to n(g_i(x)integral from n=0 to a(h_i(s)y(s)ds g_(n 1)φ(integral from n=0 to x(h_(n 1)(s)w(y(t))ds)).The results include some modifications and generalizations of the results of D. Willett, U. D. Dhongade and Zhang Binggen. Furthermore, applying the conclusion on the above inequalities to a Volterra integral equation and a differential equation, the authors obtain some new better results.     

Kantorovitch算子的L^＊M逼近

1.符号与基本结果对对[0,1]上的可积函数f(x),Kantorovitch算子定义为: K_n(f,x)=(n+1)sum from k=0 to n(p_(n-K)(x)integral from ?(f(t)dt)其中p_(n-K)(x)=(n K)x~K(1-x)~(n-K),I_K=[K/(n+1),(K+1)/(n+1)]。记M(u)是N-函数,N(v)是其young意义下的余函数,用M(u)∈△_2表示,存在正数c,u_0满足     

关于一个Ляпунов函数的讨论

定义函数(?)是正数.s=1,2,…,n.令φ(x)=(φ_1(x_1),φ_2(x_2),…,φ_n(x_n))及V(x)=φ(x)·x=sum from (?)=1 to n φ_s(x_s)x_s,(1)则 V(x)为无限大定正函数,V(x)在 R~n 中满足 Lipshitz 条件.又定义(?)则有:命题1 任给 n 维常向量 x,f,则(?)1/h(V(x+hf)-V(x))=sum from s=1 to n φ_s(x_s+β(x_s)f_s)f_s.式中 x_s,f_s 表 x 及 f 的第 s 个分量.     

二次指派问题的一个新的限界方法

二次指派问题(QAP)的数学模型是:min{z(x)=sum from i=1 to n sum from =1 to n a_(ip)x_(ip)+sum from i=1 to n sum from p=1 to n sum from j=1 to n sum from q=1 to n c_(ipjq)x_(ip)x_(jq)|x∈},(1)这里∈(n~2维布尔集)是满足如下约束的集合:sum from i=1 to n x_(ip)=1,1≤p≤n,(2)sum from p=1 to n x_(ip)=1,1≤i≤n,(3)x_(ip)=0,1,1≤i,p≤n.(4)因为 x_(ip)~2=x_(ip)并且有约束(2)和(3),我们可以约定 c_(ipjq)=0,当 i=j 或 p=q.如果所有二次项的系数都可以写成