数学问题解答

2007年1月号问题解答(解答由问题提供人给出)1651已知x,y,z∈R ,n∈N,求证:nx xy z x nyy z x yz nz≤n3 2.(陕西省绥德中学刘永春718000)证明当n=1时,不等式显然成立.当n≥2时,设nx y z=a,x ny z=b,x y nz=c,则x=(nn( n1) a1-)-b2-c,y=(nn( n1) b1-)-c2-a,z=(nn( n1) c1-)-a2-     

一道课本题的推广

新教材第二册上P30复习参考题第 8题 :已知a >b >c ,求证 :1a -b+ 1b -c+ 1c -a>0 .现对该题进行如下推广 .推广 1　若a >b >c ,m ,n均为正数 ,则 ma -b+ nb -c+ (m +n) 2c-a ≥ 0 .证　∵ (a -c ) ( ma -b + nb -c) =m(a -b +b -c)a -b + n(a -b +b -c)b -c =m +n + [m·b -ca -b+n·a -bb -c]≥m +n + 2mn =(m +n) 2 ,故 :ma -b+ nb -c+ (m +n) 2c -a ≥ 0 .推广 2　若a1>a2 >a3 >… >an >an + 1,则1a1-a2+ 1a2 -a3+… + 1an-an + 1+ n2an + 1-a1≥ 0证　利用柯西不等式 .∵ (an -an + 1) ( 1a1-a2+ 1a2 -a3+… +1an-an + 1) =[(a1-a2 ) …     

有趣的数形结合

例 1.求函数y =x - 3-x - 1的值域解 :y =x - 3-x - 1=- 4　 (x 3)2 - 2x　 (- 1 x 3)4　 (x - 1)得　y∈ - 4,4 (如图 )变式 :已知 :a 相似文献   

一类二阶泛函微分方程多个周期解的存在性

该文利用Mawhin重合度延拓定理研究了一类二阶泛函微分方程x″(t)+f(t,x(t),x(t-τ(t)))[x′(t)]~n+a(t)x~2(t)+b(t)x(t)=p(t)(n≥2)的多个周期解的问题,得到了这类方程至少存在两个周期解的结果.     

Fermat素数与Jemanowicz猜想

设r是正整数.本文运用初等数论方法证明了方程((2~(r+1)+1)n)~x+((2~(2r+1)+2~(r+1)n)~y=((2~(2r+1)+2~(r+1)+1)n)~2适合(x,y,z)≠(2,2,2)以及n1的正整数解(x,y,z,n)都满足xzy特别是当2~(r+1)是素数时,该方程仅有正整数解(x,y,z,n)=(2,2,2,t),其中t是任意正整数,即此时Jemanowicz猜想成立.     

On universal sums of polygonal numbers

For m = 3, 4,..., the polygonal numbers of order m are given by pm(n) =(m- 2) n2 + n(n= 0, 1, 2,...). For positive integers a, b, c and i, j, k 3 with max{i, j, k} 5, we call the triple(api, bpj, cpk)universal if for any n = 0, 1, 2,..., there are nonnegative integers x, y, z such that n = api(x) + bpj(y)+ cpk(z). We show that there are only 95 candidates for universal triples(two of which are(p4, p5, p6) and(p3, p4, p27)), and conjecture that they are indeed universal triples. For many triples(api, bpj, cpk)(including(p3, 4p4, p5),(p4, p5, p6) and(p4, p4, p5)), we prove that any nonnegative integer can be written in the form api(x) + bpj(y) + cpk(z) with x, y, z ∈ Z. We also show some related new results on ternary quadratic forms,one of which states that any nonnegative integer n ≡ 1(mod 6) can be written in the form x2+ 3y2+ 24z2 with x, y, z ∈ Z. In addition, we pose several related conjectures one of which states that for any m = 3, 4,...each natural number can be expressed as pm+1(x1) + pm+2(x2) + pm+3(x3) + r with x1, x2, x3 ∈ {0, 1, 2,...}and r ∈ {0,..., m- 3}.     

关于微分差分方程的边值问题

本文考虑含小参数ε>0且自变量具有固定时滞1的微分差分方程边值问题(?)其中L[y(x,ε)]=εy″(x,ε)-a(x,ε)y′(x,ε)-b(x,ε)y(x,ε),R[y(x,ε)]=A(x,ε)y′(x-1,ε)+B(x,ε)y(x-1,ε)+f(x,ε),T 是一正数,10下讨论了边值问题解的存在性、唯一性和区间-1≤x≤T 上当ε→0~+时解的一致有效估计.     

变幻莫测富韵味 演绎直观不了情——一类动圆切线方程求法的探究和启迪

1 引例 已知动圆x2+y2-2ax+2(a-2)y+2=0(a∈R,a≠1),求圆的切线方程. 通常的解法是将圆的一般方程改写为标准方程(x-a)2+[y-(2-a)]2=2(a-1)2,把切线l方程设为斜截式:y=k1x+m,利用圆心C(a,2-a)到l的距离等于半径2|a-1|,得到关于a的恒等式,求得k1=1,m=0,所以l的方程是y=x.     

求函数y=ax^2＋bx＋c／px^2＋qx＋r值域的方法探讨

求形如函数y =ax2 +bx +cpx2 +qx +r px2 +qx +r≠ 0的值域问题是众多求函数值域中的基本类型 ,其方法也是学生必须掌握的 ,分析探讨如下 :一、当函数的定义域D为全体实数时 ,即x∈R时 ,px2 +qx +r≠ 0 ;可以采用判别式法求函数的值域 ,即原函数可化为关于x的一元二次方程(yp -a)x2 + (yq -b)x +yr -c =0　 ( )(1)当yp -a =0时 ,即y =ap 时 ,若 ( )有解 ,则y可取到 ap ,若 ( )无解 ,则y不能取到 ap;(2 )当yp -a≠ 0时 ,一元二次方程 ( )有实根 ,所以其判别式△≥ 0 ,即 (yq -b) 2 - 4(yp -a) (yr -c)≥ 0 ,可解得y的取值范围 ;综全 (1) (2 …     

中立型Emden-Fowler微分方程的振动性

该文主要研究了二阶中立型Emden-Fowler微分方程(r(t)|z′(t)|~(α-1)z′(t))′+p(t)|z′(t)|~(α-1)z′(t)+q(t)|x(σ(t))|~(β-1)x(σ(t)=0的振动性,其中z(t)=x(t)+g(t)x(τ(t)).利用广义Riccati变换和积分平均技巧建立新的振动准则,推广和改进了一些文献中的结果.     

一类非线性复微分差分方程解的不存在性

该文研究了一类复微分差分方程[f(z)f'(z)]n + fm(z + r) = 1,[f(z)f'(z)]n + [f(z + r)-f(z)]m = 1,[f(z) f'(z)] 2 + P2(z) f2(z + η) = Q(z)eα(z) 的超越整函数解,其中P(z), Q(z)为非零多项式,α(z)为多项式,...     

非线性微分系统解的有界性和周期解的存在性

本文运用了比较新的手法,证明了非线性微分系统(dx)/(dt)=1/(a(x))[c(y)-b(x)];(dy)/(dt)=-a(x)[h(x)-e(t)](1)(其中a(x),b(x),h(x),c(y),e(t)为连续可微函数,x,y∈R,t∈[0,+∞),且a(x)>0)解的有界性及周期解的存在性,并应用该结论讨论了强迫振动方程:x+(f(x)+g(x)x)x+h(x)=e(t)(2)(其中f(x),g(x)为连续可微函数,x∈R,h(x),e(t)同上)解的有界性及周期解的存在性.     

一类二阶变系数常微分方程的可积条件

给出了一类二阶变系数常微分方程y″+[pu(x)-v(x)]y′+[qu2(x)+r(u(x)v(x)-u(′x))]y=f(x)及y″+[pu(x)-v(x)]y′+[qu2(x)+r(u(x)v(x)-u(′x))]y=f(x)[y-′ru(x)y]n可积的充分条件及其通解表达式,并举例说明它的应用.     

Hlawka’s functional inequality

The paper is devoted to the functional inequality (called by us Hlawka’s functional inequality) $$f(x+y)+f(y+z)+f(x+z)\leq f(x+y+z)+f(x)+f(y)+f(z)$$ for the unknown mapping f defined on an Abelian group, on a linear space or on the real line. The study of the foregoing inequality is motivated by Hlawka’s inequality: $$\|x+y\|+\|y+z\|+\|x+z\|\leq\|x+y+z\|+\|x\|+\|y\|+\|z\|,$$ which in particular holds true for all x, y, z from a real or complex inner product space.     

一类具分布时滞的微分方程的正周期解的存在性与多解性

讨论具分布时滞的微分方程x′(t)=-a(t,x)x(t)+∫-0τf(t,r,x(t+r))dr,x′(t)=a(t,x)x(t)-∫0-τf(t,r,x(t+r))drx′(t)=-g(t,x(t))+∫0-τf(t,r,x(t+r))dr,x′(t)=g(t,x(t))-∫0-τf(t,r,x(t+r))dr正周期解问题,利用锥不动点定理,获得了这类问题正解存在性和多重性的充分条件,推广了已有文献的相关结果.     

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.     

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).     

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 $$\matrix{\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad G(x+y,\ z)- G(x,z)- G(y,z)=G(y+z,x)- G(y,x)- G(z,x)\cr \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad H(x+y,\ z)- H(x,z)- H(y,x)+H(y+z,\ x)- H(y,x)- H(z,x)\cr +H(z+x,\ y)- H(z,y)- H(x,y)=0.}$$      

非线性两点边值问题的存在唯一性

对于非线性两点边值问题x″=f(t,x,x′),　x(0)=A,　x(1)=B,我们在f,fx,fx′,β(t),α′(t)都连续,且fx≥-β(t), -α(t)≤fx′≤M(1+|x′|),max β(t)-α2(t)+2α′(t)4t∈[0,1] ≤λ2<π24,α(1)≤2λcotλ,这些条件之下证明解之存在且唯一.     

一类二阶四点边值问题多个正解的存在性

研究了下面的二阶四点边值问题x″(t)+q(t)f(t,x(t),x′(t))=0,00.首先计算了相应齐次问题的Green函数,然后运用其Green函数的性质及Avery-Peterson不动点定理,我们得到了该边值问题至少存在三个正解.