Einige Eigenschaften der Lösungen vonw′=a(z)+b(z) w+c(z) w 2

Ohne ZusammenfassungHerrn A.Ostrowski zum sechzigsten Geburtstage gewidmet     

Compositions of analytic functions of the form Fn(z) = Fn−1(fn(z)), fn(z) → f(z)

Frequently, in applications, a function is iterated in order to determine its fixed point, which represents the solution of some problem. In the variation of iteration presented in this paper fixed points serve a different purpose. The sequence {F_n(z)} is studied, where F₁(z) = f₁(z) and F_n(z) = F_n−1(f_n(z)), with f_n → f. Many infinite arithmetic expansions exhibit this form, and the fixed point, α, of f may be used as a modifying factor (z = α) to influence the convergence behaviour of these expansions. Thus one employs, rather than seeks the fixed point of the function f.     

不等式k~(1/2)+1/(k+1)~(1/2)>(k+1)~(1/2)的几种证法

为什么要证明不等式k~(1/2)+1/(k+1)~(1/2)>(k+1)~(1/2)下面通过实例来说明,高中数学第三册P.147.3(4)题:求证1/1~(1/2)+1/2~(1/2)+…+1/n~(1/2)>n~(1/2)(n>1)。我们用数学归纳法来证明。 (1)当n=2时不等式左边=1/1~(1/2)+1/2~(1/2)=(2+2~(1/2))/2右边=2~(1/2)=(2~(1/2)+2~(1/2))/2,显然不等式成立。 (2)假设当n=k(k>1)时不等式成立,     

Quadratic differentials (A(z − a)(z − b)/(z − c)2) dz2 and algebraic Cauchy transform

We discuss the representability almost everywhere (a.e.) in C of an irreducible algebraic function as the Cauchy transform of a signed measure supported on a finite number of compact semi-analytic curves and a finite number of isolated points. This brings us to the study of trajectories of the particular family of quadratic differentials A(z ? a)(z ? b)×(z ? c)^?2 dz². More precisely, we give a necessary and sufficient condition on the complex numbers a and b for these quadratic differentials to have finite critical trajectories. We also discuss all possible configurations of critical graphs.     

由k~(1/2)+1/(k+1)~(1/2)>(k+1)~(1/2)的证明所想到的

本刊1984年第8期《不等式k~(1/2)+1/(k+1)~(1/2)>(k+1)~(1/2)的几种证法》一文(以下简称文〔1〕中,对不等式k~(1/2)+1/(k+1)~(1/2)>(k+1)~(1/2)作出了六种证法。其实,这个不等式还有一种更简单的证法,现补充如下,我们姑且叫它为〔1〕的第七个证法吧: g) 间接证法:     

Some inequalities for polynomials satisfying p(z)≡znp(1/z)

Let be a polynomial degreen and let . Then according to Bernstein’s inequality ‖p’‖≤n‖p‖. It is a well known open problem to obtain inequality analogous to Bernstein’s inequality for the class II_n of polynomials satisfying p(z)≡zⁿp(1/z). Here we obtain an inequality analogous to Bernstein’s inequality for a subclass of II_n. Our results include several of the known results as special cases.     

关于ξ(1/2+it)

<正> 曾经从事于求上极限Q使得(?)成立的这个问题有 Van der Corpur,Koksma,Walfisz,Tichmarsh,Phillips Titchmarsh 和闵词鹤,他们的结果是     

SOME INEQUALITIES FOR POLYNOMIALS SATISFYING p(z)=z~n p(1/z)

Let p(z)=be a polynomial degree n and let Then accord-ing to Bernstein's inequality ||p'||相似文献   

New Solutions for (1+1)-Dimensional and (2+1)-Dimensional Kaup–Kupershmidt Equations

In this paper, using the exp-function method we obtain some new exact solutions for (1+1)-dimensional and (2+1)-dimensional Kaup–Kupershmidt (KK) equations. We show figures of some of the new solutions obtained here. We conclude that the exp-function method presents a wider applicability for handling nonlinear partial differential equations.     

关于亚纯函数φ(z)f~n(z)f~((k))(z)的值分布

设n和k为任意的正整数,f(z)是复平面上超越亚纯函数,φ(z)为f(z)的不恒为零的小函数,讨论了亚纯函数φ(z)fn(z)f(k)(z)值分布,并提出一个新的定理,进行了较为详细的证明.     

关于asinα+bcosα=(a~2+b~2)~(1/2)sin(α+φ)中(φ)的确定

全日制十年制学校高中课本《数学》第一册160页例8“化asina+bcosa为一个角的一个函数的形式”。1979年2月第一版6月第一次印刷的本子中结论为     

(P~2+k)~(1/2)是无理数的证明

<正>定理若p为正整数,k为半偶数,则(p²+k)(1/2)是无理数.先给出半偶数的概念:能被2整除但不能被4整除的偶数称为半偶数.如2,6,10,14等.注意,定理中的半偶数的条件是必要的,否则定理不真.如(22+k)(1/2)是无理数.先给出半偶数的概念:能被2整除但不能被4整除的偶数称为半偶数.如2,6,10,14等.注意,定理中的半偶数的条件是必要的,否则定理不真.如(2²+12)(1/2)=16不是无理数,原因为12不是半偶数.下面用穷举反证法分两部分证明定理.     

The Value Regions of the Systems {f(z 1), f′(z 1)} and {f(z 1), f(z 2)} in the Class of Typically Real Functions

In the class T consisting of regular and typically real functions in the disk |x| < 1, the value regions of the system {f(z ₁), f(z ₁)} and {f(z ₁), f(z ₂)} are found for fixed z ₁ and z ₂. As an application, the value regions of f(z ₁) and f(z ₂) are found for f T with fixed value f(z ₁). Bibliography: 11 titles.     

关于公式1+2+…+n=(n(n+1))/2

<正>九年义务教育三年制初中《代数》第一册(上)第38页,B组第二题为:正整数从1开始,逐个相加,一直加到n,它们的和记作s,即s=1+2+3+…+n(n表示一个正整数),写出计算s的公式.这道题目中含有字母且设问富有思考性,解题方法体现了数学方法,更重要的是结果能作公式用,而且应用分层次用.为帮助同学理解这些特点,现对这道题进行解读,供同学们参考.     

函数y=((x±d)~2+a)~(1/2)+((c-x)~2+b)~(1/2)的极小值

本刊1983年增刊中曾发表过关于求函数y=(x~2+a)~(1/2)+((c-x)~2+b)~(1/2))的极小值的一篇文章,该文介绍了一种几何方法,的确比判别式法和导数法简捷。但该文的形式较特殊,在应用中受到一定的限制。为了得到一般的形式,本文在其基础上作些改进,使它的应用范围更加广泛。     

用 a+b/2≥a~(1/2)证明 a+b+c/3bc~(1/3)

高中代数教材在证明平均值不等式a+b/2≥ab~(1/2)和a+b+c/3≥(abc)~(1/3)时,各自采用了独立的证法。我们为强调基础知识的作用,采用二元平均不等式证明三元平均不等式的方法。设a,b,c∈R~+,求证a~3+b~3+c~3≥3abc.     

关于 X_(n+1)=(x_n(x_n~2+3))/(3x_n~2+1)的通项公式

这是八六年高考数学第八题:已知x_1>0,x_1≠1 且x_n+1=x_n(x_n~2+3)/3x_n~2+1(n=1,2,…)。试证:数列{x_n}或者对任意自然数都满足x_nx_(n+1)。此题证法很多,先求通项公式是一个类型的方法,下面给出一种求通项公式的简便方法。由已知