满足I(x,I(y,x))=1的IT,N蕴涵的解

研究了I(x,I(y,x))=1,其中I为由强否定N和连续三角模T生成的IT,N蕴涵,即IT,N(x,y)=N(T(x,N(y))),x,y∈[0,1],给出了满足I(x,I(y,x))=1,x,y∈[0,1]的充要条件;最后,研究了由强否定N和一类不连续的三角模min0生成的Imin0,N蕴涵,给出了满足I(x,I(y,x))=1,x,y∈[0,1]的充要条件.     

韦达定理之逆定理不成立

题:解方程组解:观察方程组的特征易看出左边相加有1 (1 y)(1 z),且右边相加为1,故有如下简捷解法: ① ②,整理得:(1 y)(1 z)=0, ∴1 y=0,1 z=0,即y=-1,z=-1 故原主程组的解为{y=-1,z=-1。} 由上述方程组及其解,我们有一个意外的收获——韦达定理之逆定理的一个反例: 原主程组实际为:{yz=4 y z=-5} 由韦达定理逆定理知满足此方程组即满足原方程组的y、z之(实数)值应为方程x~2 5x 4=0的两根; 从上述原方程组的解显见y=-1,z=-1,则有x~2 5x 4=0有二重根,应有△=0;     

双格半群在模糊逻辑中的典例

分析了n元模糊逻辑函数集合中的偏序结构,论证了该集合M-={f|f:[0,1]n→[0,1],(A)x∈[0,1]n,f(x)∈[0,1]}是一个双格半群.并且M-关于其上定义的等值关系构成的商集W={Cf|(A)g∈Cf(∈)M-,f(x)=g(x),f,g∈M,x∈[0,1]n}也构成一个双格半群.     

含有n个滞量的三维微分差分方程组周期解的存在性

本文研究含有n 个滞量的三维微分差分方程组x(t)=sum from i=1 to ∞(1/i)f[x(t),x(t-τ_i),y(t),y(t-τ_i),z(t),z(t-τ_i)]y(t)=sum from i=1 to ∞(1/i)g[x(t),x(t-τ_i),y(t),y(t-τ_i),z(t),z(t-τ_i)](τ_i>0)z(t)=sum from i=1 to ∞(1/i)h[x(t),x(t-τ_i),y(t),y(t-τ_i),z(t),z(t-τ_i)]周期解的存在性,给出了方程组周期解周期的取值范围.推广并改进了文[1]的结果.     

一个问题的分析与拓展

题:已知f(x)是定义在[0,1]上的非负函数,且f(1)=1,对任意x,y,x+y∈[0,1]都有:f(x+y)≥f(x)+f(y),求证:f(x)≤2x(x∈[0,1]).本题为2011年清华大学保送生考试题,难度     

数学问题解答

20 0 4年 1月号问题解答(解答由问题提供人给出 )1 471　求方程组 x+y =ztz+t =xy的非负整数解 .解　因为方程组中x与y ,z与t可以互换 ,所以可以先求满足 0 ≤x≤y ,0 ≤z≤t的整数解组 (x,y ,z,t) .( 1 )若x、z中有一个为零 ,不妨设x=0 ,则由原方程组消去t得 :y+z2 =0所以y =z=0 ,t= 0 .即 ( 0 ,0 ,0 ,0 )是原方程组求的一组解 .( 2 )若x ,z都不是 0 ,但是有一个为 1 ,设x=1 ,则由原方程组消去y得 :t+z=zt - 1所以 (z- 1 ) (t- 1 ) =2 ,因为z,t为正整数且z≤t,所以z - 1 =1t- 1 =2 得z=2 ,t =3,y=5即 ( 1 ,5 ,2 ,3)是原方程组的一组解 ,同…     

数学问题解答

1999年11月号数学问题解答(解答由问题提供人给出)1221.求方程组x y z=3x3 y3 z3=3的所有整数解.解　原方程组化为x y=3-z(1)x3 y3=3-z3(2)(1)3-(2),得3xy(x y)=24-27z 9z2(3)(1)代入(3),可得xy=8-9z 3z23-z(4)由(1)、(4)知x、y是以下二次方程的两个整数根:t2-(3-z)t 8-9z 3z23-z=0解得t1,2=3-z±(z-1)2·z 5z-32=3-z±(z-1)2(1 8z-3)2(5)由此知,x、y、z均为整数当且仅当z-1=0或z-3=1或z-3=-8,即z=1或z=4或z=-5.将其依次代入求根公式(5),得原方程组的所有整数解(共四组):x=1y=1z=1或x=-5y=4z=4或x=4y=-5z=4或x=4y=4z=-5注:(5)式中根号内的(z…     

一类非自治非线性微分方程周期解的存在性

本文讨论非自治非线性微分方程组■=ф(y)-f(x),■=-g(x)+e(t) (1)周期解的存在性.N.Levinson 曾给出■(y)≡y、g(x)≡x 时系统(1)存在周期解的条件,井竹君推广了文[1]的工作.本文给出方程组(1)存在周期解的一组充分条件,进一步推广了文[2]的结果.     

在皮亚诺算术模型上方括号划分的一个定理

本文解决文献[1]中提出的问题:如果I是半正则的,m,n∈ω,则I→[I]_m~n(?)I→[I]_(m 1)~(n 1),下文中所引用的术语和记号在[1]—[3]中均可找到。引理1 如果I是半正则的,I→[I]_m~n,则I→[I]_(m 1)~n。证.设F:[I]~n→(m 1)是规则划分函数,定义函数G如下:     

联立Pell方程组x~2-ay~2=1和y~2-bz~2=1的解数

设a,b是正整数.我们研究了联立Pell方程组x~2-ay~2=1,y~2-bz~2=1的正整数解(x,y,z)的个数.本文利用Bennett关于联立Padé逼近的一个结果,证明了该方程组至多只有两组正整数解(x,y,z),从而改进了Bennett(1998),袁平之(2004)等人的结论.     

Universal sums of three quadratic polynomials

Let a,b,c,d,e and f be integers with a≥ c≥ e> 0,b>-a and b≡a(mod 2),d>-c and d≡c(mod 2),f>-e and f≡e(mod 2).Suppose that b≥d if a=c,and d≥f if c=e.When b(a-b),d(c-d) and f(e-f) are not all zero,we prove that if each n∈N={0,1,2,...} can be written as x(ax+b)/2+y(cy+d)/2+z(ez+f)/2 with x,y,z∈N then the tuple(a,b,c,d,e,f) must be on our list of 473 candidates,and show that 56 of them meet our purpose.When b∈[0,a),d∈[0,c) and f∈[0,e),we investigate the universal tuples(a,b,c,d,e,f) over Z for which any n∈N can be written as x(ax+b)/2+y(cy+d)/2+z(ez+f)/2 with x,y,z∈Z,and show that there are totally 12,082 such candidates some of which are proved to be universal tuples over Z.For example,we show that any n∈N can be written as x(x+1)/2+y(3y+1)/2+z(5z+1)/2 with x,y,z∈Z,and conjecture that each n∈N can be written as x(x+1)/2+y(3y+1)/2+z(5z+1)/2 with x,y,z∈N.     

On Hyers-Ulam stability for a class of functional equations

Summary In this paper we prove some stability theorems for functional equations of the formg[F(x, y)]=H[g(x), g(y), x, y]. As special cases we obtain well known results for Cauchy and Jensen equations and for functional equations in a single variable. Work supported by M.U.R.S.T. Research funds (60%).     

On The Generalized Cocycle Equation of Ebanks and Ng

I show that in order to solve the functional equation $$F_{1}(x+y,z)+F_{2}(y+z,x)F_{3}(z+x,\ y)+F_{4}(x,y)+F_{5}(y,z)+F_{6}(z,x)=0$$ for six unknown functions (x,y,z are elements of an abelian monoid, and the codomain of each F _j is the same divisible abelian group) it is necessary and sufficient to solve each of the following equations in a single unknown function $$\matrix{\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad G(x+y,\ z)- G(x,z)- G(y,z)=G(y+z,x)- G(y,x)- G(z,x)\cr \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad H(x+y,\ z)- H(x,z)- H(y,x)+H(y+z,\ x)- H(y,x)- H(z,x)\cr +H(z+x,\ y)- H(z,y)- H(x,y)=0.}$$      

指数Diophantine方程(bn)~x+(2n)~y=((b+2)n)~z的例外解

设b是大于3的正奇数.运用初等方法讨论了方程(bn)^x+(2n)x+(2n)^y=((b+2)n)y=((b+2)n)^z适合(x,y,z)≠(1,1,1)的正整数解(x,y,z,n).证明了:i)对于任何给定的正整数N,存在无穷多个b可使该方程有满足min{x,y,z}≥N的正整数解(x,y,z,n);ii)对于任何给定的b,该方程仅有有限多组正整数解(x,y,z,n)满足y>z=x.     

在共振点附近的一类二阶泛函微分方程的解析解

在复域C内研究一类包含未知函数迭代的二阶微分方程x″(z)=G(z,x(z),x~2(z),…,x~m(z))解析解的存在性.通过Schr(?)der变换,即x(z)=y(αy~(-1)(z)),把这类方程转化为一种不含未知函数迭代的泛函微分方程α~2y″(αz)y″(z)-αy′(αz)y″(z)= (y′(z))~3G(y(z),y(αz),…,y(α~mz)),并给出它的局部可逆解析解.本文不仅讨论了双曲型情形0<|α|<1和共振的情形(α是一个单位根),而且还在Brjuno条件下讨论了共振点附近的情形(即单位根附近).     

The conditional distributivity of nullnorms over uninorms

Aequationes mathematicae - The aim of this work is to study the conditional distributivity given by the following functional equation: $$\begin{aligned} F(x,U(y,z))=U(F(x,y),F(x,z)),\ \ x,y,z\in...     

3-Variable Additive {\rho} -Functional Inequalities and Equations

In this paper, we introduce and investigate additive \({\rho}\) -functional inequalities associated with the following additive functional equations $$\begin{array}{lll} \,\,\,\,\,\,\, f(x+y+z) - f(x)-f(y)-f(z) \,\,\,\, = 0 \\ 2f \left(\frac{x+y}{2}+z \right) - f(x)-f(y)-2f(z) = 0 \\ \,\,2f \left(\frac{x+y+z}{2} \right) - f(x)-f(y)-f(z) = 0\end{array}$$ Furthermore, we prove the Hyers–Ulam stability of the additive \({\rho}\) -functional inequalities in complex Banach spaces and prove the Hyers–Ulam stability of additive \({\rho}\) -functional equations associated with the additive \({\rho}\) -functional inequalities in complex Banach spaces.     

The stability of additive $(\alpha,\beta)$-functional equations

In this paper, we investigate the following $(\alpha,\beta)$-functional equations $$ 2f(x)+2f(z)=f(x-y)+\alpha^{-1}f(\alpha (x+z))+\beta^{-1}f(\beta(y+z)),~~~~~~~~~(0.1) $$ $$ 2f(x)+2f(y)=f(x+y)+\alpha^{-1}f(\alpha(x+z)) +\beta^{-1}f(\beta(y-z)),~~~~~~~~~~~(0.2) $$ where $\alpha,\beta$ are fixed nonzero real numbers with $\alpha^{-1}+\beta^{-1}\neq 3$. Using the fixed point method and the direct method, we prove the Hyers-Ulam stability of the $(\alpha,\beta)$-functional equations $(0.1)$ and $(0.2)$ in non-Archimedean Banach spaces.     

关于满足I(x,n(x))=n(x)的D-蕴涵的刻画

研究I(x,n(x))=n(x),其中I为由连续三角模T、连续三角余模S和强否定n生成的D-蕴涵,即I(x,y)=S(T(n(x),n(y)),y),给出了满足I(x,n(x))=n(x)的充要条件。     

一个有理分式Diophantine方程

本文研究了方程(x2-1)/(y2-1)=(z2-1)2.利用四次和二次方程的性质,获得了它的全部正整数解(x,y,z).