共查询到20条相似文献,搜索用时 15 毫秒
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2009年全国高考数学1卷(理)20题(以下简称20题):
在数列{an}中a1=1,an+1=(1+1/n)an+n+1/2^n. 相似文献
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2005年高考重庆卷(理)压轴题为:数列{an}满足a1=1,且an+1=(1+1/n^2+n)an+1/2^n(n≥1). 相似文献
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Radon不等式设ai≥0,bi〉0(i=1,2,…,n),l∈N,则
^n∑i=1 ai^l+1/bi^l≥(^n∑i=1 ai)^l+1/(^n∑i=1 bi)^l
本文将(1)式推广如下: 相似文献
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本刊文[1]对文[2]中的第一个不等式给予推广,对第二个不等式的推广提出一个猜想:设xi〉0(i=1,2,3,…,n),n∑i=1xi=1.则n∏i=1(1/1-xi+xi)≥(n/n-1+1/n)^n. 相似文献
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文[1]中猜想:f(x)=a/sin^n x+b/cos^n x(0〈x〈π/2,a,b∈R^+,n∈N+),当且仅当x=arctan ^n+2√a/b时,取最小值(a 2/n+2+b 2/n+2)n+2/2。笔者发现不但此猜想是正确的,而且还得到它的一个推广,下面给出推广及证明(初等证明). 相似文献
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一种拟Grünwald插值算子的Lp收敛速度 总被引:3,自引:0,他引:3
1 引言
设f(x)为[-1,1]上的连续函数,则以第二类Chebyshev多项式Un(x)(Un(cosθ)=sin(n+1)θ/sinθ的全部零点{xk=cos k/n+1 π}^n k=1为插值结点组的f的Grunwald插值多项式为 相似文献
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P(z)=∑v=0^n cvz^v
be a polynomial of degree n and let M(f, r) = max|z|=r |f(z) | for an arbitrary entire function f(z). If P(z) has no zeros in |z| 〈 1 with M(P,1) = 1, then for |α| 〈 1, it is proved by Jain[Glasnik Matematicki, 32(52) (1997), 45-51] that
|P(Rz)+α(R+1/2)^nP(z)|≤1/2{|1+α(R+1/2)^n|+|R^n+α((R+1/2)^n|},R≥1,|z|=1.
In this paper, we shall first obtain a result concerning minimum modulus of polynomials and next improve the above inequality for polynomials with restricted zeros. Our result improves the well known inequality due to Ankeny and Rivlin and besides generalizes some well known polynomial inequalities proved by Aziz and Dawood. 相似文献
be a polynomial of degree n and let M(f, r) = max|z|=r |f(z) | for an arbitrary entire function f(z). If P(z) has no zeros in |z| 〈 1 with M(P,1) = 1, then for |α| 〈 1, it is proved by Jain[Glasnik Matematicki, 32(52) (1997), 45-51] that
|P(Rz)+α(R+1/2)^nP(z)|≤1/2{|1+α(R+1/2)^n|+|R^n+α((R+1/2)^n|},R≥1,|z|=1.
In this paper, we shall first obtain a result concerning minimum modulus of polynomials and next improve the above inequality for polynomials with restricted zeros. Our result improves the well known inequality due to Ankeny and Rivlin and besides generalizes some well known polynomial inequalities proved by Aziz and Dawood. 相似文献
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《数学教学》2012年第12期的数学问题874为:题目 已知 m,n∈N+,m,n≥2,xi∈R+(i=1,2,…,m),(^m∑i=1)xi=S,n∈N+,求证:(^m∑i=1)^n√xi/S-xi≥.看完此题,笔者不禁想起了文[1]中的不等式:题源1已知a,b,c为正数,求证:√a/(b+c)+√b/(c+a)+√c/(a+b)〉2。 相似文献
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一、圆环染色问题计算公式
如图1所示,把一个圆环(从圆环“中心”出发,以环“半径”为界)分成n(n≥2)个扇形区域A1A2…An,现有m(m≥2)种不同颜色为这n个区域染色,要求相邻两个区域An与An+1颜色不同,则共有an=(m-1)^n+(-1)^n(m-1)种不同的染色方法。 相似文献
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In this paper, by using superposition method, we aim to show that ∑^n i=1 (2/- 1)^2k-1 is the product of n2 and a rational polynomial in n2 with degree k- 1, and that ∑^ni=1 (2i - 1)^2k is the product of n(2n - 1)(2n + 1) and a rational polynomial in (2n - 1)(2n + 1) with degree k - 1. Moreover, recurrence formulas to compute the coefficients of the corresponding rational polynomials are also obtained. 相似文献
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利用数列{(1+1/n)^n}的单凋性和数学归纳法改进了Minc—sathre不等式的上下界,并由改进后的Minc—Sathre不等式得出n!的一个估计式. 相似文献
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