构造二次函数巧证一道国际竞赛题

进入初三年级,我们学习了二次方程ax^2＋bx＋c=0根的判别式△=b^2-4ac,学习了二次函数f（x）=ax^2＋bx＋c与x轴有无交点的判别方法,将二次函数f（x）=ax^2＋bx＋c化简变形得到f（x）=a[（x＋b/2a）^2-△/4a^2],当a〉0,△=b^2-4ac≤0时,有f（x）≥0．     

The Hyers-Ulam stability of a functional equation deriving from quadratic and cubic functions in quasi-��-normed spaces

In this paper, we investigate the Hyers-Ulam stability of the following function equation 2f（2x ＋ y） ＋ 2f（2x - y） = 4f（x ＋ y） ＋ 4f（x - y） ＋ 4f（2x） ＋ f（2y） - Sf（x） - 8f（y） in quasi-β-normed spaces.     

A Generalized Cubic Functional Equation

In this paper, we determine the general solution of the functional equation f1 （2x ＋ y） ＋ f2（2x - y） = f3（x ＋ y） ＋ f4（x - y） ＋ f5（x） without assuming any regularity condition on the unknown functions f1,f2,f3, f4, f5 ： R→R. The general solution of this equation is obtained by finding the general solution of the functional equations f（2x ＋ y） ＋ f（2x - y） = g（x ＋ y） ＋ g（x - y） ＋ h（x） and f（2x ＋ y） - f（2x - y） = g（x ＋ y） - g（x - y）. The method used for solving these functional equations is elementary but exploits an important result due to Hosszfi. The solution of this functional equation can also be determined in certain type of groups using two important results due to Szekelyhidi.     

2009年浙江高考理科压轴题的别解

2009年浙江高考理科压轴题为：已知函数f（x）=x3-（k2-k＋1）x2＋5x-2,g（x）=k2x2＋kx＋1,其中k∈R．     

On the Boundary Behaviour,Including Second Order Effects,of Solutions to Singular Elliptic Problems

For γ≥1 we consider the solution u=u（x） of the Dirichlet boundary value problem Δu ＋ u^-γ=0 in Ω, u=0 on δΩ. For γ= 1 we find the estimate u（x）=p（δ（x））[1＋A（x）（log 1/δ（x）^-6], where p（r） ≈ r r√2 log（1/r） near r = 0,δ（x） denotes the distance from x to δΩ, 0 〈ε 〈 1/2, and A（x） is a bounded function. For 1 〈 γ 〈 3 we find u（x）=（γ＋1/√2（γ-1）δ（x））^2/γ＋[1＋A（x）（δ（x））2γ-1/γ＋1] For γ3= we prove that u（x）=（2δ（x））^1/2[1＋A（x）δ（x）log 1/δ（x）]     

一数学竞赛题变式推广的简证

命题 如果m〉0,x,y∈[m,＋∞）,或x,y∈（-∞,m],且（x＋√x^2＋m^2）（y＋√y^2＋m^2）=m^2,那么x=y．     

综合题新编选登

题130 设定义在R上的函数 f（x）=a0x^4＋a1x^3＋a2x^2＋a3x＋a4（a0,a1,a2,a3,a4∈R）， 当x=-1时,f（x）取极大值2/3，且函数y=f（x＋1）的图象关于点（-1,0）对称。     

Solution and stability of a cubic functional equation

In this paper, we investigate the general solution and the stability of a cubic functional equation f（x ＋ ny） ＋ f（x - ny） ＋ f（nx） = n^2 f（x ＋ y） ＋ n^2 f（x - y）＋ （n^3 - 2n^2 ＋ 2）f（x）,where n ≥ 2 is an integer. Furthermore, we prove the stability by the fixed point method.     

On the Non-Commutative Neutrix Product of the Distributions x＋^λ and x^＋μ

Let f and g be distributions and let gn = （g ＊ δn）（x）, where δn （x） is a certain converging to the Dirac delta function. The non-commutative neutrix product fog of f and g to be the limit of the sequence {fgn }, provided its limit h exists in the sense that sequence is defined N-lim n-∞（f（x）g,, （x）, φ（x）〉 = （h（x）, φ（x）},for all functions p in 2. It is proved that （x^λ＋1n^px＋）0（x^μ＋1n^qx＋）=x＋^λμ1n^p＋qx＋,（x^λ-1n^qx-）=x-^λ＋μ1n^p＋qx-,for λ＋μ〈-1; λ,μ, λ＋μ≠-1,-2…and p,q=0,1,2……     

Stability of a generalized quadratic functional equation in Schwartz distributions

We consider the Hyers-Ulam stability problem of the generalized quadratic functional equation
uoA＋voB-2woP1 - 2ko P2 =0,
which is a distributional version of the classical generalized quadratic functional equation
f（x＋y）＋g（x - y） - 2h（x） - 2k（y）=0     

两道考研试题的探讨

主要讨论了函数f（x）=ln（1＋x＋x^2＋…＋x^m）（-1≤x≤1,m为正整数）关于x的幂级数展开式在点x=1处的敛散性．     

两道考研试题的探讨

主要讨论了函数f（x）=ln（1＋x＋x^2＋…＋x^m）（-1≤x≤1，m为正整数）关于x的幂级数展开式在点x=1处的敛散性．     

RESEARCH ANNOUNCEMENTS——Dynamical Behavior for the Three-dimensional Generalized Hasegawa-Mima Equations

We consider the following generalized three-dimensional （3-D） dissipative Hasegawa-Mima equations： △ut - ut ＋ {u, △u} ＋ knuy - vz ＋ α△（u - △u） ＋ f（x, y, z） = 0, （1） vt ＋ {u, v} ＋ uz ＋ γv - β△v = g（x, y, z） （2） with initial datum v｜t=0=u0（x,y,z）,v｜t=0=v0（x,y,z）,（x,y,z）∈Ω∈R^3 （3）.     

西班牙数学竞赛一题推广的另证

第31届西班牙数学奥林匹克第2题为 命题1如果（x＋√x^2＋1）（y＋√y^2＋1）=1,则x＋y=0．文[1]给出下面推广： 命题2如果m〉0,x,y∈[m,＋∞）或x,y∈（-∞,＋m]且（x＋√x^2-m^2）（y＋√y^2-m^2）=m^2,那么x=Y． 文[1]采用换元法证明了命题2,仔细研读后笔者给出命题2的另一种简洁证法。     

一道预赛题的简解

2013年湖北省高中数学联赛高一、高二预赛题中均有以下一道试题：求函数y=x2＋x（x2-1）1/2的值域.命题组提供的参考解法如下：易求得函数的定义域为{x｜x≤-1或x≥1}.（1）易知函数y=x2＋x（x2-1）1/2是[1,＋∞）上的增函数,所以,当x≥1时,y≥1.（2）当x≤-1时,     

找回遗漏的解

题目1已知函数f（x）=｜x＋1｜＋｜x＋2｜＋…＋｜x＋2011｜＋｜x-1｜＋｜x-2｜＋…＋｜x-2011｜（x∈R）,且f（a^2-3a＋2）=f（a-1）,则满足条件的所有整数a的和是_____．     

剖析三个数学问题

问题1已知函数f（x）=ax^2＋bx的图像过点（-4n,0）,g（x）=2ax＋b,g（0）=2n（n∈N^＊）．     

Existence Results for a Class of Semilinear Elliptic Systems

In this paper, we study the existence of nontrivial solutions for the problem
{-△u=f（x,u,v）＋h1（x）in Ω
-△v=g（x,u,v）＋h2（x）inΩ
u=v=0 onδΩ
where Ω is bounded domain in R^N and h1,h2 ∈ L^2 （Ω）. The existence result is obtained by using the Leray-Schauder degree under the following condition on the nonlinearities f and g：
{lim s,｜t｜→＋∞f（x,s,t）/s=lim ｜s｜,t→＋∞g（x,s,t）/t=λ＋1 uniformly on Ω,
lim -s,｜t｜→＋∞f（x,s,t）/s=lim ｜s｜,-t→＋∞g（x,s,t）/t=λ-,uniformly on Ω,
where λ＋,λ-∈（0）∪σ（-△）,σ（-△）denote the spectrum of -△. The cases （i） where λ＋ = λ_ and （ii） where λ＋≠λ_ such that the closed interval with endpoints λ＋,λ_ contains at most one simple eigenvatue of -△ are considered.     

利用导数值域求割线斜率值域的探讨

问题 已知函数f（x）=-＋x3＋ax2＋b（a,b∈R）,若函数y=f（x）的图象上任意不同两点连线的斜率小于2,求a的取值范围．     

Liouvillian and analytic integrability of the quadratic vector fields having an invariant ellipse

We characterize the Liouvillian and analytic integrability of the quadratic polynomial vector fields in R2 having an invariant ellipse.More precisely,a quadratic system having an invariant ellipse can be written into the form x=x2＋y2-1＋y（ax＋by＋c）,y=x（ax＋by＋c）,and the ellipse becomes x2＋y2=1.We prove that（i） this quadratic system is analytic integrable if and only if a=0;（ii） if x2＋y2=1 is a periodic orbit,then this quadratic system is Liouvillian integrable if and only if x2＋y2=1 is not a limit cycle;and（iii） if x2＋y2=1 is not a periodic orbit,then this quadratic system is Liouvilian integrable if and only if a=0.