Maximum Modulus of Polynomials

Let P(z) =n∑j=0 a_jz~j be a polynomial of degree n and let M(P, r) = max|z|=r|P(z)|. If P(z) ≠ 0 in |z| 1, then M( P, r) ≥ ((1 + r)/ (1 + ρ))~ n M( P, ρ).The result is best possible. In this paper we shall present a refinement of this result and some other related results.     

Some Sharpening and Generalizations of a Result of T. J. Rivlin

Let p(z)=a_0+a_1z+a_2z~2+a_3z~3+···+a_nz~n be a polynomial of degree n.Rivlin[12]proved that if p(z)≠0 in the unit disk,then for 0r≤1,max|z|=r|p(z)|≥((r+1)/2)~nmax|p(z)||z|=1.In this paper,we prove a sharpening and generalization of this result and show by means of examples that for some polynomials our result can significantly improve the bound obtained by the Rivlin’s Theorem.     

关于Diophantine方程a~x+b~y=c~z的Terai猜想

设m是正偶数.证明了(A)若b是奇素数,且a=m|m~6-21m~4+35m~2-7|,b=|7m~6-35m~4+21m~2-1|,c=m~2+1,则Diophantine方程G:a~x+b~y=c~z仅有正整数解(x,y,z)=(2,2,7);(B)若m2863,且a=m|m~8-36m~6+126m~4-84m~2+9|,b=|9m~8-84m~6+126m~4-36m~2+1|,c=m~2+1,则Diophantine方程G仅有正整数解(x,y,z)=(2,2,9);(C)若a,b,c适合a=m|∑_(i=0)~((r-1)/2)(-1)~i(_(2i)~r)m~(r-2i-1)|,b=|∑_(i=0)~((r-1)/2)(-1)~i(_(2i+1)~r)m~(r-2i-1)|,c=m~2+1,r≡1(mod4),2|x,2|y,且b为奇素数或m145r(log r),则方程G仅有解(x,y,z)=(2,2,r).     

Some Generalization for an Operator Which Preserving Inequalities Between Polynomials

For a polynomial p(z) of degree n which has no zeros in |z| 1, Dewan et al.,(K. K. Dewan and Sunil Hans, Generalization of certain well known polynomial inequalities, J. Math. Anal. Appl., 363(2010), 38–41) established zp′(z) +nβ2p(z) ≤n2{( β2 + 1+β2)max|z|=1|p(z)|-( 1+β2- β2)min|z|=1|p(z)|},for any complex number β with |β|≤ 1 and |z| = 1. In this paper we consider the operator B, which carries a polynomial p(z) into B[ p(z)] := λ0p(z) + λ1(nz2)p′(z)1!+ λ2(nz2)2 p′′(z)2!,where λ0, λ1, and λ2are such that all the zeros of u(z) = λ0+c(n,1)λ1z+c(n,2)λ2z2lie in the half plane |z| ≤ |z-n/2|. By using the operator B, we present a generalization of result of Dewan. Our result generalizes certain well-known polynomial inequalities.     

用向量证明不等式两例

由向量的内积:a·b=|a|·|b|·cosθ, 可得 因为 -1≤cosθ≤1, 所以有 这个结论在证明不等式时常常用到. 例1 已知口a2+b2+c2=1,x2+y2+z2= 1,其中a、b、c、x、y、z均为实数,求证: -1≤ax+by+cz≤1. 证明 设p=(a,b,c), q=(x,y,z), 则 ,即.     

L~p INEQUALITIES AND ADMISSIBLE OPERATOR FOR POLYNOMIALS

Let p(z) be a polynomial of degree at most n. In this paper we obtain some new results about the dependence of p(Rz)-βp(rz) + α (R+1/r+1)n-|β | p(rz) s on p(z) s for every α, β∈ C with |α|≤ 1, |β | ≤ 1, R > r 1, and s > 0. Our results not only generalize some well known inequalities, but also are variety of interesting results deduced from them by a fairly uniform procedure.     

带解析系数的二维奇异积分方程

本文我们研究二维奇异积分方程和它的共轭奇异积分方程这里G表示单位圆|z|<1,a(z),b(z),c(z)是G内的解析函数,我们不但建立了解的表达式,而且找出了这些方程可解的必要和充分条件。 我们研究奇异积分方程这里G是复平面z=x+iy上的单位圆:|z|<1.a(z),b(z),c(z)是G内的解析函数,属于C~1((?))。复值函数f和φ分别是L_p((?)),p>2中的已知和未知函数,同时还研究和它共轭的非齐次积分方程这里g和ψ分别是共轭空间L_q((?)),1/p+1/q=1中的已知和未知的复值函数,A和A~*由关系式Re(Aφ,ψ)=Re(φ,A~*ψ)相联系。     

高中代数复习(续)

二、复数复数这一章很多题都是用到任意复数z。z=a+bi(a,b∈R)或z=r(cosθ+isinθ)这个表示法来解或证的。例1.解方程|z|+z=8—4i求复数z。解:设z=a+bi(a,b∈R)|z|=(a~2+b~2)~(1/2)。由题设(a~2+b~2)~(1/2)+a+bi=8—4i由复数相等的条件得:     

Some Integral Mean Estimates for Polynomials with Restricted Zeros

Let P(z) be a polynomial of degree n having all its zeros in |z| ≤ k. Fork = 1,it is known that for each r 0 and |α|≥ 1,n(|α|- 1) {∫2π0|P(eiθ)|rdθ}1/r 0r≤ {∫2π0|1+ eiθ|rdθ}1/rmax|z|=|Dα P(z)|.In this paper, we shall first consider the case when k ≥ 1 and present certain generalizations of this inequality. Also for k ≤ 1, we shall prove an interesting result for Lacunary type of polynomials from which many results can be easily deduced.     

关于一类拟共形映照的偏差估计

1.引言 设B={t:|t-c|≤k},其中c是复数,k是非负实数,且|c|+k<1。设f(z)是C到(?)上的拟共形映照,且适合如下条件:在区域1<|z|<∞上它是单叶解析的,有展开式 f(z)=z+sum from n=1 to ∞(b_n/z~n),在区域|z|<1上,它的复伸张μ(z)=f_z/f_z几乎处处落在B中,(即|μ(z)-c|≤k a. e.).记这样的f(z)全体为Σ′(B).Schiffer, M. 和Schobor, G. 证明了Σ′(B)是紧族,并对系数b_1获得了估计     

On Some Inequalities Concerning Rate of Growth of Polynomials

In this paper we consider a class of polynomials P(z) = a0+∑n v=t a v z v, t ≥ 1not vanishing in |z|k, k≥1 and investigate the dependence of max|z|=1|P(Rz)-P(rz)on max|z|=1|P(z)|, where 1 ≤ r R. Our result generalizes and refines some know polynomial inequalities.     

Positivity of Fock Toeplitz Operators via the Berezin Transform

This paper deals with the relationship between the positivity of the Fock Toeplitz operators and their Berezin transforms. The author considers the special case of the bounded radial function φ(z) = a + be~(-α|z|~2)+ ce~(-β|z|~2), where a, b, c are real numbers and α, β are positive numbers. For this type of φ, one can choose these parameters such that the Berezin transform of φ is a nonnegative function on the complex plane, but the corresponding Toeplitz operator Tφ is not positive on the Fock space.     

A FUNDAMENTAL INEQUALITY AND ITS APPLICATION

Let f(z) be meromorphie in |z|k+4+[2/k].In this note,a fundamental inequality is established such that thecharacreristic function T(r,f)can be limibd by N(r,1/f)and _(τ-1)(r,1/(f~(k)-1).As anapplication,the following criterion for normality is also proved:Let be a family ofmeromorphic functions in a region D.If for every f(z)∈ ,f(z)≠0 and all the zeros off~(k)(z)-1 are of multiplicity >k+4+[2/k]in D,then is normal there.     

Inequalities Concerning The Maximum Modulus of Polynomials

Let P(z) be a polynomial of degree n having all its zeros in |z|≤k, k ≤1, then for every real or complex number β, with |β|≤ 1 and R ≥ 1, it was shown by A.Zireh et al. [7] that for |z|=1,min|z|=1|P(Rz)+β((R+k)/(1+k))~nP(z)|≥k~(-n)|R~n+β((R+k)/(1+k))~n|min|z|=k|P(z)|.In this paper, we shall present a refinement of the above inequality. Besides, we shall also generalize some well-known results.     

一道联考试题的讨论

湖北省部分重点中学 2 0 0 3届第一次联考数学试卷上有这样一道题 :已知 f(x) =ax2 +bx +c,如果x∈ [-1 ,1 ]时 ,均有 | f(x) |≤ 1 .1 )求证 :|c|≤ 1 ;2 )当x∈ [- 1 ,1 ]时 ,试求 g(x) =|cx2+bx +a|的最大值 ;3)试给出一个这样的 f(x) ,使 g(x)确实取到上述最大值 .命题者的解答如下 :解　∵x∈ [- 1 ,1 ]时 ,| f(x) |≤ 1恒成立 ,令x =0 ,得 |c|≤ 1 .2 )∵g(x) =|cx2 +bx +a|=|cx2 -c+c+bx +a|≤ |cx2 -c| + |c+bx +a|=|c| ( 1 -x2 ) + |c +bx +a|≤ |c| + |c+bx +a| ,由于函数 φ(x) =|c +bx +a|在 [- 1 ,1 ]的端点处取到最大值 .所以…     

A reverse Denjoy theorem Ⅲ

Suppose that C 1 and C 2 are two simple curves joining 0 to ∞, non-intersecting in the finite plane except at 0 and enclosing a domain D which is such that, for all large r, the set {θ : re iθ∈ D} has measure at most 2α, where 0 α π. Suppose also that u is a non-constant subharmonic function in the plane such that u(z) = Φ(|z|) for all large z ∈ C 1 ∪ C 2 ∪～D, where Φ(|z|) is a convex, non-decreasing function of |z| and ～D is the complement of D. Let A D (r, u) = inf{u(z) : z ∈ D and |z| = r}. It is shown that if A D (r, u) = O(1) then lim inf r→∞ B(r, u)/r π/(2α) 0.     

探究向量模的求法

题目（苏北2013年调研）已知平面向量a,b,c两两所成角为2π/3,并且|a|=1,|b|=2,|c|=3,求|a+b+c|的值.分析求向量的模,利用模长公式|a|=a⁽1/（?）=x²+y^21/2解决.解|a+b+c|= a+b+c^1/2=（?）=3^1/33.进一步思考变式1已知平面向量a,b,c两两所成角相等,并且|a|=1,|b|=2,|c|=3,求|a+b+c|的值.分析本题得了解对向量的夹角的定义,夹     

On extremal properties for the polar derivative of polynomials

If p(z) is a polynomial of degree n having all its zeros on |z| = k, k ≤ 1, then it is proved[5] that max |z|=1 |p′(z)| ≤ kn1n + kn m|z|=ax1 |p(z)|. In this paper, we generalize the above inequality by extending it to the polar derivative of a polynomial of the type p(z) = cnzn + ∑n j=μ cn jzn j, 1 ≤μ≤ n. We also obtain certain new inequalities concerning the maximum modulus of a polynomial with restricted zeros.     

Area Estimates of Images of Disks under Analytic Functions

Let D_r := {z = x + iy ∈ C : |z| r}, r ≤ 1. For a normalized analytic function f in the unit disk D := D1, estimating the Dirichlet integral Δ(r, f) =∫∫_(D_r)|f'(z)|~2 dxdy, z = x + iy, is an important classical problem in complex analysis. Geometrically, Δ(r, f) represents the area of the image of D_r under f counting multiplicities. In this paper, our main ob jective is to estimate areas of images of D_r under non-vanishing analytic functions of the form(z/f)~μ, μ 0, in principal powers,when f ranges over certain classes of analytic and univalent functions in D.     

数学问题解答

(1/2)(a+b+c),内切圆、外接圆半径分别为r、R。试证:△ABC为锐角、直角、钝角三角形的充要条件分别是p>2R+r、p 2R+r、p<2R|r。