The Jackson Inequality for the Best L~2-Approximation of Functions on [0,1] with the Weight x

Let L^2（[0, 1], x） be the space of the real valued, measurable, square summable functions on [0, 1] with weight x, and let n be the subspace of L2（[0, 1], x） defined by a linear combination of Jo（μkX）, where Jo is the Bessel function of order 0 and {μk} is the strictly increasing sequence of all positive zeros of Jo. For f ∈ L^2（[0, 1], x）, let E（f, n） be the error of the best L2（[0, 1], x）, i.e., approximation of f by elements of n. The shift operator off at point x ∈[0, 1] with step t ∈[0, 1] is defined by T（t）f（x）=1/π∫0^π f（√x^2 ＋t^2-2xtcosO）dθ The differences （I- T（t））^r/2f = ∑j=0^∞（-1）^j（j^r/2）T^j（t）f of order r ∈ （0, ∞） and the L^2（[0, 1],x）- modulus of continuity ωr（f,τ） = sup{｜｜（I- T（t））^r/2f｜｜：0≤ t ≤τ] of order r are defined in the standard way, where T^0（t） = I is the identity operator. In this paper, we establish the sharp Jackson inequality between E（f, n） and ωr（f, τ） for some cases of r and τ. More precisely, we will find the smallest constant n（τ, r） which depends only on n, r, and % such that the inequality E（f, n）≤ n（τ, r）ωr（f, τ） is valid.     

第四章 反三角函数、三角方程和解三角形

§1 反三角函数要点 1.反三角函数的概念和性质; 2.常用恒等式,如 sin(arcsinx)=x,(x∈[-1,1]),arcsin(sinx)=x,(x∈[-π/2,π/2]),arcsinx arccosx=π/2,(x∈[-1,1])     

上期课外练习参考解答

1.令cosx=t,t∈[-1,1],则f(t)=t+3+1/(t+3)在[-1,1]内的值域为:f(-1)≤f(e)≤f(1),即5/2≤y≤17/4.2.把已知方程化为x2-4x+4=0(x>1),即x=2.由题意得B/A=2,sinC/sinA=2,于是有B=2A,sinC=2sinA,而A+B+C=π,∴C=π-3A,∴sinC=sin3A=2sinA,即3sinA-4sin3A=2sinA.     

选择题的编制和解法

一、选择题的编制如何编制数学选择题,根据初步实践,我们认为必须注意以下几个方面: (一)在确定命题的测试内容后,编制选择题要语言简练,尤要注意词义不应含糊,以免造成考生对考题难以判断。例1 函数y=cosx(x∈[-π/2、π/2])与函数y=arccosx(x∈[-1,1]都不是     

一类推广后的Feigenbaum函数方程的光滑解

考虑一类推广后的Feigenbaum函数方程｛g（0）=1,-1≤g（x）≤1,x∈[-1,1],h（g（x））=g（g（h（x）））其中h（x）是[-1,1]上的递减光滑奇函数且满足h（0）=0,-1〈h’（x）〈0,z∈[-1,1]．利用构造性方法讨论上述方程的光滑解的存在性及唯一性．     

一道西班牙数学竞赛题的探究

<正>第31届西班牙数学奥林匹克第2题为命题1如果(x+（x²+1）^1/2)(y+（y²+1）^1/2) =1,那么x+y=0.文[1]、[2]给出了命题1的三种证法,文[2]还给出了命题1的类似命题2如果x,y∈[1,+∞),或x,y∈(—∞,—1],且(x+（x²—1）^1/2)(y+（y²—1）^1/2)=1,那么x=y.     

How to use a definite integral to find area(2)

Another example,the sine and cosine curves intersect fromx=π/4to x=5π/4.However,we want to find the area of the region bounded by the sine and cosine curves fromx=0to x=2π.Analysis:Let f(x)=sinx,and g(x)=cosx.Let f(x)=g(x),and find xin sinx=cosxas x∈[0,2π]x=π/4 or x=5π/4.     

反三角函数定义域的教学作法点滴

关于反正弦函数y=arcsinx与反余弦函数y=arccosx的定义域为[-1,1]的教学,应十分注意。否则将影响到以后涉及到反正弦函数和反余弦函数的教学内容的学习。为了加强这一内容的教学,笔者认为应从以下几个方面加强练习,以巩固概念的学习。 1.求函数的定义域:进行求复合形式的反三角函数的定义域的练习.将有助于深化对定义域概念的掌握。例1 求下列函数的定义域: (1)y=arcsin(3x-2); (2)y=arceos(ctgx); (3)y=arcsin 2x/1 x~2 arceos 1-x~2/1 x~2. IM7“IWel“答案(1)x∈[1/3,1]; (2)x∈[kπ π/4,kπ 3π/4];     

一道错解例题引发的探究性教学

南方出版社出版的高中学业水平考试达标测评丛书《系统集成》（2014年湖南省专用）第64页有这么一道例题： 题目设函数f（x）=a·b,其中向量a=（cos2x＋1,1）,b=（1,3（1/2）sin 2x＋m）.（1）求f（x）的最小正周期;（2）当x∈[0,π6]时,-4〈f（x）〈4恒成立,求实数m的取值范围．     

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=     

从一道全国高中数学联赛预赛题想起

<正>题目(2014年山东赛区预赛第二题)已知函数f(x)=sinx+(1+cos²x)(1/2)(x∈R),则函数f(x)的取值范围____.解设t=sinx,则t∈[-1,1],原函数可化为g(t)=t+(2-t2x)(1/2)(x∈R),则函数f(x)的取值范围____.解设t=sinx,则t∈[-1,1],原函数可化为g(t)=t+(2-t²)(1/2),t∈[-1,1],即原题等价于求g(t)的值域问题,下面从不同角度来研究此函数的值域.一、解法探究解法1平方再开方.     

最值问题的六种解法

<正>题目(2014年浙江省高中数学竞赛试题)设实数x,y满足方程(x+2)²+y2+y²=1,则y/x的最大值为.解法1令x=-2+cosθ,y=sinθ,θ∈[0,2π),y/x=k.则y/x=sinθ/-2+cosθ=k,即kcosθ-sinθ=2k,     

Stability of a generalized quadratic functional equation in Schwartz distributions

We consider the Hyers-Ulam stability problem of the generalized quadratic functional equation
uoA＋voB-2woP1 - 2ko P2 =0,
which is a distributional version of the classical generalized quadratic functional equation
f（x＋y）＋g（x - y） - 2h（x） - 2k（y）=0     

对无理函数最值问题的多种解法的探讨

试题 已知函数y=√(3-x)+√x+1的最大值为M,最小值为m,则m/M的值为A.1/4 B.1/2 C.√2/2 D.√3/2此题作为一道选择题,我们容易得出答案为C,但此题同时也是一道典型的形如y=√(ax+b)+√(cx+d)(ac＜0)的求函数最值的题.它是高中数学的一个热点同时也是一个难点.本文研究此题的多种解法,与大家共勉.1 利用二次函数的性质求最值解法1显然y≥0,两边平方的y2 =4+2√(3+2x-x2),移项得y2-4=2√(3+2x-x2).因为x∈[-1,3],所以3+2x-x2 ∈[0,4],.即2√(3+2x-x2)∈[0,4],所以ymax=2√2,ymin=2.解法2由上面变形得到的y2-4=2√(3+2x-x2),两边再平方整理得4x2 -8x+y4-8y2 +4=0.(*)记f(x)=4x2 -8x+y4-8y2 +4,方程(*)在x∈[-1,3]有解.     

Comonotone approximation with interpolation at the ends on an interval

Let a function f ∈ C[-1, 1], changes its monotonisity at the finite collection Y := {y1,… ,ys} of s points yi ∈ (-1, 1). For each n ≥ N(Y), we construct an algebraic polynomial Pn, of degree ≤ n, which is comonotone with f, that is changes its monotonisity at the same points yi as f, and |f(x)-Pn(x)|≤c(s)ω2(f,(√1-x2)/n), x∈[-1,1],where N(Y) is a constant depending only on Y, c(s) is a constant depending only on s and ω2 (f, t) is the second modulus of smoothness of f.     

A Generalized Cubic Functional Equation

In this paper, we determine the general solution of the functional equation f1 （2x ＋ y） ＋ f2（2x - y） = f3（x ＋ y） ＋ f4（x - y） ＋ f5（x） without assuming any regularity condition on the unknown functions f1,f2,f3, f4, f5 ： R→R. The general solution of this equation is obtained by finding the general solution of the functional equations f（2x ＋ y） ＋ f（2x - y） = g（x ＋ y） ＋ g（x - y） ＋ h（x） and f（2x ＋ y） - f（2x - y） = g（x ＋ y） - g（x - y）. The method used for solving these functional equations is elementary but exploits an important result due to Hosszfi. The solution of this functional equation can also be determined in certain type of groups using two important results due to Szekelyhidi.     

一数学竞赛题变式推广的简证

命题 如果m〉0,x,y∈[m,＋∞）,或x,y∈（-∞,m],且（x＋√x^2＋m^2）（y＋√y^2＋m^2）=m^2,那么x=y．     

具有p-Laplacian算子的二阶微分方程Picard边值问题(英文)

Suffcient conditions for the existence of at least one solution of two-point boundary value problems for second order nonlinear differential equations {[φ（x（t））] ＋ kx（t） ＋ g（t,x（t）） = p（t）,t ∈（0,π） x（0） = x（π） = 0 are established,where [φ（x）] =（｜x ｜p-2x） with p 1.Our result is new even when [φ（x）] = x in above problem,i.e.p = 2.Examples are presented to illustrate the effciency of the theorem in this paper.     

抽象函数错题分析二例

问题1定义在R上的函数f（x）是偶函数,且f（x＋π/2）=-f（x）,当x∈[0,π/2]时,f（x）=sinx,求f（5π/3）的值． 《中学生数学》2007年1月（上）《一道错题的发现》指出问题1是一道错题．     

On the primitive divisors of the recurrent sequence u_(n+1)=(4cos~2(2π/7)-1)u_n-u_(n-1) with applications to group theory

Consider the sequence of algebraic integers un given by the starting values u0=0,u1=1 and the recurrence u_(n+1)=(4cos~2(2π/7)-1)u_n-u_(n-1).We prove that for any n ■{1,2,3,5,8,12,18,28,30}the n-th term of the sequence has a primitive divisor in Z[2 cos(2π/7)].As a consequence we deduce that for any sufficiently large n there exists a prime power q such that the groupcan be generated by a pair x,y with χ~2=y~3=(xy)~7=1 and the order of the commutator[x,y]is exactly n.The latter result answers in affirmative a question of Holt and Plesken.