应用a~3 b~3 c~3=3abc解题

常见到三数和为零,求三数立方和的问题。这时应用若a b c=0,则a~3 b~3 c~3=3abc解题非常精炼。下举几例。例1 若a b c=0,a~3 b~3 C~3=0,求a~(1991) b~(1991) c~(1991)的值。解∵a b c=0, ∴a~3 b~3 c~3=3abc=0, ∴a、b、c中有一个为零,另两个互为相反     

a~3 b~3 c~3-3abc的综合应用

在因式分解当中,有一个很重要的公式：a~3 b~3 c~3-3abc=(a b c)(a~2 b~2 c~2- ab-bc-ca)．在做一些复杂题时,往往能因为它而化难为易,化复杂为简单．它的特殊之处在于两点：①当a b c=0时,a~3 b~3 c~3=3abc．②当a b c≠0,a~3 b~3 c~3=3abc时,     

a~3 b~3 c~3≥3abc的几种证法

一个人数学基础是否扎实,主要看他对一些基本定理的理解.然而,同学们往往忽视命题的证明过程,只记其结论,这未免有些不好.在自学中,我对a3 b3 c3≥3abc这一定理发掘出几种证法.     

不等式x~3 y~3 z~3≥3xyz的几何证法

我们先证ｘ２＋ｙ２≥２ｘｙ（ｘ、ｙ∈Ｒ＋,当ｘ＝ｙ时,等号成立）证明　如图１,设正方形ＡＢＣＤ的边长为ｘ,正方形ＢＥＦＪ的边长为ｙ,在ＡＢ上取ＡＨ＝ｙ,则ＨＢ＝ｘ－ｙ,故ＨＥ＝ＨＢ＋ＢＥ＝ｘ－ｙ＋ｙ＝ｘ,∴　Ｓ矩ＡＨＰＤ＝Ｓ矩ＨＥＦＫ＝ｘｙ．由图１显然有　Ｓ正ＡＢＣＤ＋Ｓ正ＢＥＦＪ≥Ｓ矩ＡＨＰＤ＋Ｓ矩ＨＥＦＫ,即　　　ｘ２＋ｙ２≥２ｘｙ（当且仅当ｘ＝ｙ时,等号成立）再证　ｘ３＋ｙ３＋ｚ３≥３ｘｙｚ（ｘ、ｙ、ｚ∈Ｒ＋,当且仅当ｘ＝ｙ＝ｚ时,等号成立）证明　如图２,设三个正方体ＶＡＢ、ＶＣＤ、ＶＥＦ…     

On the rank of 3×3×3-tensors

Let U, V and W be finite dimensional vector spaces over the same field. The rank of a tensor τ in U???V???W is the minimum dimension of a subspace of U???V???W containing τ and spanned by fundamental tensors, i.e. tensors of the form u???v???w for some u in U, v in V and w in W. We prove that if U, V and W have dimension three, then the rank of a tensor in U???V???W is at most six, and such a bound cannot be improved, in general. Moreover, we discuss how the techniques employed in the proof might be extended to prove upper bounds for the rank of a tensor in U???V???W when the dimensions of U, V and W are higher.     

环F_3+uF_3+vF_3+uvF_3上的循环码

研究了环R=F_3+uF_3+vF_3+uvF_3上循环码的结构(u~2=u,v~2=v,uu=uu),证明了该环上的循环码是主理想生成的,并给出了其上循环码的生成多项式.     

不等式a^3＋b^3＋c^3≥3abc的新证法

对于三元基本不等式“若ａ ,ｂ ,ｃ∈Ｒ+,则ａ3+ｂ3+ｃ3≥ 3ａｂｃ” ,老教材是利用因式分解的办法 ,将ａ3+ｂ3+ｃ3- 3ａｂｃ化为12 (ａ +ｂ +ｃ) [(ａ -ｂ) 2 +(ｂ -ｃ) 2 +(ｃ -ａ) 2 ]后 ,再判断其值的正负而获证的 ,新教材是利用构造的办法 ,联想构造不等式“若ａ ,ｂ∈Ｒ+,则ａ3+ｂ3≥ａ2 ｂ +ａｂ2 ” ,后利用二元基本不等式“若ａ ,ｂ∈Ｒ+,则ａ +ｂ≥ 2ａｂ”而证得的 .显然老教材中的证明对因式分解要求较高 ,学生较难掌握 ,故老教材中的证明被新教材中的证明取而代之了 ,但新教材中的构造证法技巧性亦较强 ,且构造的是一个一般性…     

3×3实矩阵的Volterra乘子

本文应用代数知识,详细讨论3×3实矩阵A存在Volterra乘子D的充分必要条件,以及A的Volterra乘子D唯一的充分必要条件,并且确定A的Volterra乘子中各元素的最小存在范围及具体表达式(当D存在唯一时)。     

导出1^3＋2^3＋3^3…＋n^3的公式的一种方法

笔者曾在《中学生数学》2002年第5期（上）发表了《导出1^2＋2^2＋…＋n^2的公式的一种方法》的文章,主要阐述用“全等三角形”数阵并进行“二维”变化进行叠加,得到n∑k=1k^2结论的方法．     

不定方程 x~3+y~3=z~2与 x~3+y~3=z~4

在 x,y 互素的条件下,本文给出不定方程 x~3+y~3=z~2所有的整数解,并证明不定方程 x~3+y~3=z~4无 xyz≠0之整数解.     

“a~3 b~3=3ab~2”的平几证法

BC=a,艺BAC=20。,求证a“ 吞习=3a乙’。 一般书籍多采用三角法证明,本文提供几种纯平几证法。 图1所示,在AC上取点D,使乙C刀D=20“。那么△BCD。△ABC,则有‘BD二。,Dc=髻,’D二吞“一aZ b这些关系式,以下直接引用。L证一:以A为圆心,b为半径作圆(图2),争延长BD、CA分别交圆于E     

导出1~3+2~3+3~3+…+n~3的公式的一种方法

笔者曾在《中学生数学》2002年第5期(上)发表了《导出12+22+…+n2的公式的一种方法》的文章,主要阐述用全等三角形数阵并进     

Equivariant Lagrangian Minimal S^3 in CP^3

The equivariant Lagrangian minimal immersion of Sa into C.Pa is studied. The complete classification and analytic expression for such kinds of immersions are given.     

Hopf bifurcation withD 3 ×D 3-symmetry

A generic Hopf bifurcation involving an eight dimensional center eigenspace is considered for systems possessing aD ₃ ×D ₃-symmetry. This kind of Hopf bifurcation can occur in systems of three interacting groups of oscillators, where each group itself is composed of three individual oscillators. The terminology micro and Macro is introduced here to denote symmetry operations acting on individual oscillators and on the whole groups, respectively. The normal form for the Hopf bifurcation admits 11 distinct periodic solutions with maximal isotropy subgroups. These are classified and their branching-types and stabilities are determined in terms of the cubic and relevant quintic coefficients of the normal form. The symmetry properties of these solutions when only certain Macro variables in the oscillator groups are observed are discussed in the context of the remaining symmetry. Furthermore, the relation of the normal form to the corresponding one for a singleD ₃-symmetry is established by restricting the system to four dimensional fixed point subspaces associated with submaximal isotropy subgroups. Based on this information the possibility of quasiperiodic solutions and of a particular class of heteroclinic cycles is discussed.Dedicated to Prof. Klaus Kirchgässner on the occasion of his 60th birthday     

Indefinite numerical range of 3 × 3 matrices

The point equation of the associated curve of the indefinite numerical range is derived, following Fiedler’s approach for definite inner product spaces. The classification of the associated curve is presented in the 3 × 3 indefinite case, using Newton’s classification of cubic curves. Illustrative examples of all the different possibilities are given. The results obtained extend to Krein spaces results of Kippenhahn on the classical numerical range. The work of this author was partially supported by the Portuguese foundation FCT, in the scope of program POCI 2010.     

不定方程x_1~3 x_2~3 x_3~3 x_4~3=0的整数解的一个必要条件

伍伟东同学这篇文章,是对不定方程“x_1~3 x_2~3 x_3~3 x_4~3=0的整数解”这一难而又有趣味的问题的一个探讨,指出了“此方程的每组整数解的四数之和必为6的倍数”。对于高一学生来说,能得到这一结果,说明他具有很强的运用所学知识解决问题的能力。     

A new central polynomial for 3 × 3 matrices

Abstract

Orthodox semigroups have been studied by many authors, in particular by Hall, Yamada and Petrich. In this paper, we give the standard representation of orthodox semigroups and investigate various e-varieties of orthodox semigroups which are determined by the standard representations.     

x~3+y~3+z~3=0无整数解的证明

<正> 文1证明了x~3+y~3+z~3=0无xyz≠0的整数解。其中重要的根据是;若s~3=a~2+3b~2,(a,b)=1,(a,b)的最大公约数,记为(a,b),则有s=n~2+3v~2,且a=n(n~2-9n~2),b=3v(n~2-v~2).例如91~3=836~2+3·135~2,求得上述的s=4~2+3·5~2,而不是4~2+3·5~2.     

公式1~3 2~3 … n~3=…的多证

问题求证:13 23 33 … n3=1/4n2·(n 1)2(正整数立方数列的前n项和公式)．证法一如图,作出函数y=2x(x≥0)的图像．在x轴上顺次截取线段OO1, O1O2,O2O3,…, On-1On使OO1=1, O1O2=2,O2O3=3,…,On-1On=n,过O1、O2、O3、…、On．分别作x轴的垂线,交半直线y=2x(x≥0)于P1、P2、P3、…、Pn,则P1O1=2×1=2,P2O2=