INJECTIVE MAPS ON PRIMITIVE SEQUENCES OVER Z/（p^e）

Let Z/(pe) be the integer residue ring modulo pe with p an odd prime and integer e ≥ 3. For a sequence (a) over Z/(pe), there is a unique p-adic decomposition (a) = (a)0 (a)1·p … (a)e-1 ·pe-1, where each (a)i can be regarded as a sequence over Z/(p), 0 ≤ i ≤ e - 1. Let f(x) be a primitive polynomial over Z/(pe) and G' (f(x), pe) the set of all primitive sequences generated by f(x) over Z/(pe). For μ(x) ∈ Z/(p)[x] with deg(μ(x)) ≥ 2 and gcd(1 deg(μ(x)),p- 1) = 1,set ψe-1 (x0, x1,…, xe-1) = xe-1·[ μ(xe-2) ηe-3 (x0, x1,…, xe-3)] ηe-2 (x0, x1,…, xe-2),which is a function of e variables over Z/(p). Then the compressing map ψe-1: G'(f(x),pe) → (Z/(p))∞,(a) (→)ψe-1((a)0, (a)1,… ,(a)e-1) is injective. That is, for (a), (b) ∈ G' (f(x), pe), (a) = (b) if and only if ψe - 1 ((a)0, (a)1,… , (a)e - 1) =ψe - 1 ((b)0,(b)1,… ,(b)e-1). As for the case of e = 2, similar result is also given. Furthermore, if functions ψe-1 and ψe-1 over Z/(p) are both of the above form and satisfy ψe-1((a)0,(a)1,… ,(a)e-1) = ψe-1((b)0,(b)1,… ,(b)e-1) for (a),(b) ∈ G'(f(x),pe), the relations between (a) and (b), ψe-1 and ψe-1 are discussed.     

Z/(2e)上本原序列不同压缩映射的导出序列

设 f( x)是 Z/ ( 2 e)上 n次强本原多项式 ,对形如 xe- 1 +η( x0 ,… ,xe- 2 )的二个 e元布尔函数 Φ( x0 ,… ,xe- 1 )和 Ψ( x0 ,… ,xe- 1 )及二条序列 a,b∈G( f( x) ) e,若Φ( a0 ,… ,ae- 1 ) =Ψ ( b0 ,… ,be- 1 ) ,给出了函数Φ ( x0 ,… ,xe- 1 )和Ψ ( x0 ,… ,xe- 1 )之间的关系与序列 a和 b之间的关系 .所给出的结论进一步说明了导出的二元序列具有良好的密码性质     

Z／（2∧e）上本原序列不同压缩映射的导出序列

设f(x）是Z/（2∧e）上n次强本原多项式,对形如xe-1 η（x0,…,xe-2）的二个e元布尔函数φ（xo,…,xe-1）和ψ（x0,…,xe-1）及二条序列a,b∈G（f（x)）e,若φ（a0,…,ae-1）=ψ（b0,…,be-1）,给出了函数φ（x0,…,xe-1）和ψ（x0,…,xe-1）之间的关系与序列a和b之间的关系,所给出的结论进一步说明了导出的二元序列具有良好的密码性质。     

On the Factorizations of （mg, mf）-graph

Let G be an (mg, mf)-graph, where g and f are integer-valued functions defined on V(G) and such that 0≤g(x)≤f(x) for each x ∈ V(G). It is proved that(1) If Z ≠ , both g and f may be not even, G has a (g, f)-factorization, where Z = {x ∈ V(G): mf(x)-dG(x)≤t(x) or dG(x)-mg(x)≤ t(x), t(x)= f(x)-g(x)>0}.(2) Let G be an m-regular graph with 2n vertices, m≥n. If (P1, P2,..., Pr) is a partition of m, P1 ≡ m (mod 2), Pi ≡ 0 (mod 2), i = 2,..., r, then the edge set E(G) of G can be parted into r parts E1 , E2,...,Er of E(G) such that G[Ei] is a Pi-factor of G.     

环Z/(pe)上本原权位序列0元素分布的保熵性(Ⅱ)

设Re=Z/(3e)为整数模3e剩余类环, e≥2.环风Re上序列a有唯一的权位分解 ,其中ai是{0,1,2}上序列.称ai为a的第i权位序列,ae-1为a的最高权位序列.它们可自然视为Z/(3)上序列.设f(x)是Re上本原多项式,a和b是Re上由f(x)生成的序列,a≠0(mod3e-1),本文证明了最高权位序列 的0元素分布包含原序列a的所有信息,即,对所有非负整数t,若ae-1(t)=0当且仅当be-1(t)=0,则a=b.并由此得到: (i)两条不同的本原权位序列是线性无关的; (ii)任给正整数k,函数 是保熵函数,即对由f(x)生成的序列a和b,a=b当且仅当 (mod3).     

Upper Bounds on Character Sums with Rational Function Entries

We obtain formulac and estimates for character sums of the type S(x,f,p^m)=∑x=1^p^mx(f(f)),where p^m is a prime power with m≥2,x is a multiplicative character(mod p^m),and f=f1/f2 is a rational function over Z.In particular,if p is odd,d=deg(f1) deg(f2) and d^*=max(deg(f1),deg(f2)) then we obtain |S(x,f,p^m)|≤(d-1)p^m(1-1/d^*) for any non-constant f (mod p) and primitive character x.For p=2 an extra factor of 2√2 is needed.     

剖析命题修正结论完善简解

文[1]给出了如下命题:命题如果x＞0时,f(x),g(x)连续可导,且limx→0f(x)=limx→0g(x),则当x≥0(或x＞0)时,若f(x)≥g(x)恒成立,那么f′(x)≥g′(x)恒成立.并利用该命题简解了一类高考压轴题:“对(A)x≥0,f(x)≥g(x)或f(x)≤g(x)恒成立,其中f(x)或g(x)含参数a,试确定参数a的取值范围.”简解的思路是:对(A)x≥0,只要对f(x)≥g(x)或f(x)≤g(x)两边取导数,再从f′(x)≥g′(x)或f′(x)≤g'(x)中分离出参数a,转化为最值问题.     

Z2k--线性负循环码

Wolfmann引进了Z4-负循环码.Zn4上的负移位γ是指Zn4上的满足γ(a0,a1,…,an-1)=(-an-1,a0,a1,…,an-2)的置换;长度为n的Z4-负循环是指Zn4的子集C满足γ(C)=C.他给出了在环Z4[x]/(x2 1)中多项式表示的Z4-负循环码;证明了Z4-线性负循环码Gray映射下的象是二元距离不变量循环码等.本文有两个目的:一是给出在环Z2k[x]/(x2 1)中的多项式表示的Z2k-负循环码及其对偶;二是由Z4-负循环码在Gray映射下的象构造出具有优良关连性质的二元周期序列族.     

2020年全国Ⅰ卷文科数学函数题的探究

<正>1问题呈现(2020全国Ⅰ卷文科数学第20题)已知函数f(x)=ex-a(x+2).(1)当a=1时,讨论f(x)的单调性;(2)若f(x)有两个零点,求a的取值范围.分析与解(1)当a=1时,f(x)=e^x-(x+2),∴f′(x)=ex-(x+2),∴f′(x)=e^x-1,令f′(x)=0,我们得到x=0,所以当x<0时,f′(x)<0;当x>0时,f′(x)>0;所以f(x)在(-∞,0)上单调递减,在(0,+∞)上单调递增.     

关于L′Hospital法则的一个注记

在微积分这门课程当中 ,常常碰到求 00 型、∞∞ 型的极限问题 ,解决这类问题的一种简单而有效的方法是 L′Hospital法则。过去 ,在不少有关的书籍中 [1,2 ,3] ,∞∞ 的 L′Hospital法则的表述是 :若 (1 )函数 f (x)和 g(x)在 (a,a δ)有定义 (δ>0 ) ,limx→ a 0 f (x) =∞ ,limx→ a 0 g(x) =∞(2 ) f(x)和 g(x)在 (a,a δ)都可导 ,g′(x)≠ 0 ,并且 limx→ a 0f′(x)g′(x) =A　 (包括 A=∞的情形 )则limx→ a 0f (x)g(x) =limx→ a 0f′(x)g′(x) =A　　最近 ,我们在 [4,5,6 ]中看到 ,去掉条件 (1 )中的 limx→ a 0 f (x) =∞…     

综合题新编选登

题153设函数f(x)=ax-(a 1)ln(x 1),其中a>0.1)求f(x)的单调区间;2)当x>0时,证明不等式:1 xx0),由f′(x)=0,解得x=1a.当x变化时,f′(x),f(x)的变化情况如下表:x(-1,1a)1a(1a, ∞)f′(x)-0 f(x)极小值由上表可知,当x∈(-1,1a)时,f′(x)<0,函数f(x)在(-1,1a)内单调递减;当x∈(1a, ∞)时,f′(x)>0,函数f(x)在(1a, ∞)内单调递增.所以,函数f(x)的单调减区…     

“三招齐下”破解含参数函数的导数应用的题

导数在高中数学中可以说是“叱咤风云”,具有深刻的内涵与丰富的外延,在应用中显示出独特的魅力和势不可挡的渗透力.导数是解决函数、方程、不等式、数列和曲线等问题的利器,是沟通初等数学与高等数学的桥梁.以函数为载体,以导数为工具,考查函数性质及导数应用为目标,是最近几年函数与导数交汇试题的显著特点和命题趋向.对导数应用的考查的广度和深度也在不断拓宽、加深.尤其是运用导数确定含参数函数的参数取值范围的问题,这类问题不仅综合性强、难度高,而且解题思路妙、方法巧,学生不容易掌握.例1(2010年全国Ⅱ理科)设函数f(x)=1-e(-s)(Ⅰ)证明:当x＞-1时f(x)≥者;(Ⅱ)设当x≥0时f(x)≤x/(ax+1),求a的取值范围.参考答案(Ⅰ)要证明当x＞-1时,f(x)≥x/(x+1),只需证明ex≥1+x.令g(x)=ex-x-1,则g′(x)=ex-1.当x≥0时,g′(x)≥0,g(x)在[0,+∞)是增函数;当x≤0时,g′(x)≤0,g(x)在,(-∞,0]是减函数.于是g(x)在x=0处达到最小值,因而当x∈ R时,g(x)≥g(0),即es≥1+x.所以当x＞-1时,f(x)≥x/(x+1).     

对一道微分问题的修正

文[1]习题3-1(P81)第3题(是非题)如下:设函数f(x),g(x)在[a,b]上连续,在(a,b)内可导,且在[a,b]上f′(x)≤g′(x),则有f(b)-f(a)≤g(b)-g(a).与文[1]配套的[2](P105)给出的解答是:答不对.虽然由拉格朗日定理得f(b)-f(a)b-a=f′(ξ),ξ∈(a,b)(1)g(b)-g(a)b-a=g′(ξ),ξ∈(a,b)(2)且有f′(x)≤g(x).但f′(ξ)不一定小于等于g′(ξ),因为(1)(2)式中的ξ不一定是相同的.我们认为上述解答是错的,也就是说,原命题是成立的.下面给出证明.证明令F(x)=f(x)-g(x),由题意,F(x)在[a,b]上连续,在(a,b)内可导,再由拉格朗日定理得F(b)-F(a)b-a=F′(ξ),…     

一些同余问题的例外集合

利用特征和与指数和的估计,研究了一些同余问题的例外集合.具体来说,设p为充分大的素数,集合Y■Zp,正整数N相似文献   

一个非等价转化问题的探究

某资料上有这样一个问题:问题|2x-a|+2/x≥1对任意x>0都成立,求a的取值范围.给出的解法是:原不等式等价于a≤2x+2/x-1或a≥2x-2/x+1,令f(x)=2x+2/x-1,g(x)=2x-2/x+1,则原不等式对任意的x>0都成立,等价于:对任意的x>0都有a≤f(x)或a≥g(x).由f′(x)=2-2/x~2,g′(x)=2+2/x~2可得:在(0,+∞)上,[f(x)]_(min)=f(1)=3,g(x)是增函数,值域为R,所以a≤f(x)对任意x>0都成立     

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…     

WEAK TRAVELLING WAVE FRONT SOLUTIONS OF GENERALIZED DIFFUSION EQUATIONS WITH REACTION

The author demonstrate that the two-point boundary value problem {p′(s)=f′(s)-λp^β(s)for s∈(0,1);β∈(0,1),p(0)=p(1)=0,p(s)&gt;0 if s∈(0,1),has a solution(λ^-，p^-（s）),where |λ^-| is the smallest parameter,under the minimal stringent restrictions on f(s), by applying the shooting and regularization methods. In a classic paper, Kohmogorov et.al.studied in 1937 a problem which can be converted into a special case of the above problem. The author also use the solution(λ^-,p^-(s)) to construct a weak travelling wave front solution u(x,t)=y(ξ),ξ=x-Ct,C=λ^-N/(N+1),of the generalized diffusion equation with reaction δ/δx（k（u）|δu/δx|^n-1 δu/δx)-δu/δt=g（u），where N&gt;0,k(s)&gt;0 a.e.on(0,1),and f(a):=n+1/N∫0ag(t)k^1/N(t)dt is absolutely continuous ou[0,1],while y(ξ) is increasing and absolutely continuous on (-∞,+∞) and (k(y(ξ))|y′(ξ)|^N)′=g(y(ξ))-Cy′(ξ)a.e.on(-∞，+∞）,y(-∞)=0,y(+∞)=1.     

L(d1, d2,..., dt)-Number λ(Cn; d1, d2,...,dt) of Cycles

An L(d1,d2,...,dt)-labeling of a graph G is a function f from its vertex set V(G) to the set {0, 1,..., k} for some positive integer k such that {f(x) - f(y)| ≥ di, if the distance between vertices x and y in G is equal to i for i = 1,2,...,t. The L(d1,d2,...,dt)-number λ(G;d1,d2,... ,dt) of G is the smallest integer number k such that G has an L(d1,d2,... ,dt)labeling with max{f(x)|x ∈ V(G)} = k. In this paper, we obtain the exact values for λ(Cn; 2, 2,1) and λ(Cn; 3, 2, 1), and present lower and upper bounds for λ(Cn; 2,..., 2,1,..., 1)     

另解两值联赛题

(一)2010年全国高中数学联赛一试试题(9)已知函数f(x)=ax~2+bx~2+cx+d(a≠0),当0≤x≤1,| f′(x)≤1 |,试求a的最大值.解由于f′(x)=3ax~2+2bx+c当a>0时,表示一条下凹的抛物线,从题设条件,可知0≤x≤1,-1≤f′(x)≤1.从图线中可以得到     

Determination of the limits for multivariate rational functions

The purpose of this paper is to solve the problem of determining the limits of multivariate rational functions.It is essential to decide whether or not limxˉ→0f g=0 for two non-zero polynomials f,g∈R[x1,...,xn]with f(0,...,0)=g(0,...,0)=0.For two such polynomials f and g,we establish two necessary and sufcient conditions for the rational functionf g to have its limit 0 at the origin.Based on these theoretic results,we present an algorithm for deciding whether or not lim(x1,...,xn)→(0,...,0)f g=0,where f,g∈R[x1,...,xn]are two non-zero polynomials.The design of our algorithm involves two existing algorithms:one for computing the rational univariate representations of a complete chain of polynomials,another for catching strictly critical points in a real algebraic variety.The two algorithms are based on the well-known Wu’s method.With the aid of the computer algebraic system Maple,our algorithm has been made into a general program.In the final section of this paper,several examples are given to illustrate the efectiveness of our algorithm.