一个定理的再推广

文[1]对文[2]中的定理推广为:若方程x f(x)=m和x f-1(x)=m的根分别为a,b.则a b=m.经类比探讨,笔者得到如下结论.定理若方程x·f(x)=m和x·f-1(x)=m分别有唯一根a,b.则a·b=m.该定理的证明用到类似文[2]的引理:若函数y=f(x)与y=g(x)图象的交点为P(x0,y0),则点P′(y0,x0)一定是函     

BOUNDEDNESS AND CONVERGENCE FOR THE NON-LINARD TYPE DIFFERENTIAL EQUATION

In this article, the author studies the boundedness and convergence for the non-Lienard type differential equation (x|·)=a(y)-f(x) (y|·)=b(y)β(x)-g(x) e(t) where a(y),b(y),f(x),g(x),β(x) are real continuous functions in y∈R or x∈R,β(x)≥0 for all x and e(t) is a real continuous function on R = {t: t≥0} such that the equation has a unique solution for the initial value problem. The necessary and sufficient conditions are obtained and some of the results in the literatures are improved and extended.     

世界各地数学奥林匹克试题选登

1找到所有映射f:R→R,满足f(f(x) y)=f(x2-y) 4f(x)y,其中x,y∈R.解映射f(x)=0和f(x)=x2显然符合条件.下面证明不存在其它的映射符合要求.设映射f:R→R满足f(f(x) y)=f(x2-y) 4f(x)y(1)其中x,y∈R.令a=f(0).在(1)中取x=0则对任意y∈R,f(a y)=f(-y) 4ay(2)在(2)式中先取y=0,则有f(a)=a.取y=-a,则有a=a-4a2,即a=0.因此由(2)式知f是一个偶函数.在(1)式中令y=-f(x)及y=x2.比较其结果有4(f(x))2=4x2f(x).因而f(x)=0或f(x)=x2.现假设存在x0使得f(x0)≠0,则x0≠0及f(x0)=x02.因为f是偶函数.我们假设x0>0.令x为任意非零实数,在(1)式中令y=-x0,则…     

椭圆焦点弦中的新结论

1·引言文[1]介绍了椭圆x2a2 by22=1焦点三角形的若干性质,读后很受启发,笔者研究了焦点弦的若干性质·2·几个结论定理1设P是椭圆x2a2 by22=1上任意一点,F1、F2是两个焦点,弦PP1、PP2分别过焦点F1、F2,过P1、P2的切线交于P′,则P′点的轨迹方程为:x2a2 (ab22 y2c2)=1·证明设P(acosθ,bsinθ),F1(-c,0),F2(c,0),c2=a2-b2,P1(图x1,1y1),P2(x2,y2)·直线PP1方程为y=acbossiθnθ c(x c),b2x2 a2y2=a2b2,b2(acosθ c)2x2 a2b2sin2θ(x c)2=a2b2(acosθ c)2,x2项的系数为b2(a2sin2θ a2cos2θ 2accosθ c2)=b2(a2 c2 2accosθ)·x项的…     

陕西省第六次大学生高等数学竞赛本科组初赛试题

一、填空题:(6×4′=24′)1·设[x]表示不超过x的最大整数,则limx→0sinx|x|-2[x]=1.2·d4dx42 x1-x2x=0=48.3·设函数f(x,y)可微,f(0,0)=0,fx(0,0)=m,fy(0,0)=n,φ(t)=f[t,f(t,t)],则φ′(0)=m mn n2.4·设ddx∫2xf(2t)dt=x(x>0),则∫f(x)dx=-61x3 c.5·设f(x)在区间[-π,π]上连续,且满足f(x-π)=-f(x),则f(x)的傅立叶系数a2n=0.6·设质点在变力F=(3x 4y)i (7x-y)j的作用下,沿椭圆ax2 y2=4的逆时针方向运动一周所作的功等于6π,则a=4.二、选择题(8×4′=32′)7·当x→0时,下列无穷小量中最高阶的无穷小量是(D)A·∫0x1n(1 t3/2)dt;B·ta…     

四类基本初等函数性质的逆向探究

<正>学完高中数学必修一,我们知道,正比例函数f(x)=kx(k≠0)满足性质f(x+y)=f(x)+f(y),指数函数f(x)=a~x(a>0,a≠1)满足性质f(x+y)=f(x)f(y),对数函数f(x)=log_ax(a>0,a≠1)满足性质f(xy)=f(x)+f(y),幂函数f(x)=x~α(α∈R)满足性质f(xy)=f(x)f(y).然后,我们就会反过来想,满足这些性质的函数是唯一确定的吗?直到学习了导数,在     

抽象函数与解题模型

我们这里所说的“抽象函数”是指那些没有给出函数的具体解析式,只给出一些特殊条件或特征的函数(如函数递推式,函数的定义域、函数性质及特征、部分图象等)尽管这类函数问题高度抽象,但往往有它所对应的具体函数模型.例如:f(x y)=f(x)·f(y)对应的是指数函数ax y=ax·ay,f(xy)=f(x) f(y),对应的是对数函数loga(xy)=logax logay,f(x y)=f(x) f(y)对应的是正比例函数k(x y)=kx ky,f(x±y)=f(x)g(y)±g(x)f(y)正弦型的三角函数.f(x±y)=f(x)f(y)g(x)g(y)余弦型的三角函数等等.除此之外面对抽象函数数学题,我们的解题思路常常有:(1)合理赋值,化…     

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?…     

一类弹性梁方程的正解存在性

利用Schauder不动点定理,讨论了两端简单支撑的静态梁方程 y^(4)(x)=f(x,y(x)),y(0)=a,y(1)=b,y“(0)=c,y“(1)=d在非齐边界条件下正解的存在性结果。     

由一道习题谈函数图像关于直线对称问题

题目 有下列四个命题:①若函数f(x-a)=f(a-x),则f(x)的图像关于y轴对称;②函数f(x)满足f(a+x)=f(a-x),则f(x)的图像关于直线x=a对称;③函数y=f(a-x)与y=f(a+x)的图像关于y轴对称;④函数y=f(x-a)与y=f(a-x)的图像关于直线x=a对称.其中正确的命题是___.     

一类特殊分布纪录值之和的中心极限定理

摘要：设{X，Xn,n≥1)为独立同分布的服从某连续分布F的随机变量序列，X^(1)=X1，X^(2),X^(3),…为其纪录值序列．令ψ(u)=F^-1(1-e^-u)．其中F^-1是F的反函数．本文研究当ψ(u)=log^pu时Tn=∑k=1^nX^(k)=^dn∑k=1^nψ(Sn)的极限性质．解决了户为所有正整数时Tn的中心极限定理．     

ADDITIVE FAMILIES OF INVARIANTS OF FORMS OF DEGREE d

Abstract

Let d be a positive integer, and F be a field of characteristic zero. Suppose that for each positive integer n, I _n, is a GL _n,(F)- invariant of forms of degree d in x₁, …, x _n, over F. We call {I _n} an additive family of invariants if I _p+q (f ⊥ g) = I _p(f).I _q(g) whenever f; g are forms of degree d over F in x _l, …, x _p; …, x _q respectively, and where (f ⊥ g)(x _l, …, x _p+q) = f(x ₁, …, x _p,) + g (x _p+1, …, x _p+q). It is well-known that the family of discriminants of the quadratic forms is additive. We prove that in odd degree d each invariant in an additive family must be a constant. We also give an example in each even degree d of a nontrivial family of invariants of the forms of degree d. The proofs depend on the symbolic method for representing invariants of a form, which we review.     

一类pnm阶群的构造

本文研究了一类特殊的pnm阶有限群的构造.利用求解数论同余方程的方法和群的扩张理论,得到了具有m阶循环正规子群,其补子群为循环群的Pnm阶有限群的构造及相关的计数定理.     

COMPARISON OF PRODUCTS OF THE VARIETIES OF LIE ALGEBRAS

Abstract

In Fortes (2001), we introduced a notion of order for associative pairs and we obtained a Goldie-like characterization of left orders in a semiprime pair coinciding with its socle. In this paper, we take up again that notion of order to establish a Faith-Utumi theorem, which studies left orders in a prime pair coinciding with its socle.     

有限Abel群上子集等价类的计数

设G为一有限Abel群，|G|=v,D1,D2是G的两个子集，如果存在t∈Zv,(t,v)=1,s∈G使D1＝tD2 s，则称D1与D2是等价的。文中给出了G的k-子集等价类的计数公式，同时也给出了G的的所有子集等价类的计数公式。     

函数集的完全性

1.:敲g(x)篇〔一二,二]上之非降的有界缝差两数,业具有性鬓(K)s‘二一0,一。(:);f--:.,。g。尹(:)!d:一郁匕,(‘一”,”;dg)篇在〔一二,司上定羲业且满足修件:,一{户,(柳dg(·)}青<一,>l的可测蝮值函数族{f(幻}.封龄一徊乙“(一二,侧d刃中之子族凌B(幻},若由f(劣)(乙,(一二,二:dg),夕>1生+上夕q=1,及f--:ha”“’“““’一0纷{B(x)}之任何B(哟成立必滇致f(幻在〔一二,司上规乎虚虚等焚零则释{B(x)}在乙“(一二,侧dg)中完全. 函数族的完全性是舆函数横造的一些简题很有阴保的.徙【l]我们知道{e‘”}豁。是在乙,(一二,州dg),,>1,中完全的,…     

ESTIMATES OF N-FUNCTION AND m-FUNCTION OF MEROMORPHIC SOLUTIONS OF SYSTEMS OF COMPLEX DIFFERENCE EQUATIONS

We apply Nevanlinna theory of the value distribution of meromorphic functions to study the properties of Nevanlinna counting function and proximity function of meromorphic solutions of a type of systems of complex difference equations. Our results can give estimates on the proximity function and the counting function of solutions of systems of difference equations. This implies that solutions have a relatively large number of poles. It extend some result concerning difference equations to the systems of difference equations.