共查询到20条相似文献,搜索用时 31 毫秒
1.
2.
With the help of continued fractions, we plan to list all the elements of the set Q△ = {aX2 + bXY + cY2 : a,b, c ∈Z, b2 - 4ac = △ with 0 ≤ b 〈 √△}of quasi-reduced quadratic forms of fundamental discriminant △. As a matter of fact, we show that for each reduced quadratic form f = aX2 + bXY + cY2 = (a, b, c) of discriminant △〉0(and of sign σ(f) equal to the sign of a), the quadratic forms associated with f and defined by {〈a+bu+cu2,b+2cu.c〉},with 1≤σ(f)u≤b/2|c| (whenever they exist), 〈c,-b-2cu,a+bu+cu2〉 with b/2|c|≤σ(f)u≤[w(f)]=[b+√△/2|c|], are all different from one another and build a set I(f) whose cardinality is #I(f)={1+[ω(f)],when(2c)|b,[ω(f)],when (2c)|b. If f and g are two different reduced quadratic forms, we show that I(f) ∩ I(g) = Ф. Our main result is that the set Q△ is given by the disjoint union of all I(f) with f running through the set of reduced quadratic forms of discriminant △〉0. This allows us to deduce a formula for #(Q△) involving sums of partial quotients of certain continued fractions. 相似文献
3.
在学习了空间向量的法向量的求法后,我联想到高一曾学过在平面直角坐标系下直线ax+by+c=0(ab不全为零)的一个法向量是(a,b),那么(a,b,c)会不会就是空间平面ax+by+cz+d=0(a,b,c不全为零)的一个法向量呢? 相似文献
6.
1.人教A版选修2-1P98A组第11题已知向量a,b,c是空间的一个单位正交基底,向量a+b,a-b,c是空间的另一个基底,若向量p在基底口,b,c下的坐标为(1,2,3),求p在基底a+b,a-b,c下的坐标. 相似文献
7.
8.
瓦西列夫不等式:
设n,b,c〉0,n+b+c=1,则a^2+b/b+c+b^2+c/c+a+c^2+a/a+b≥2. 相似文献
10.
文[1]证明了这样一个不等式,若n,b,c为正实数,则.√a/b+c+√b/a+c+√c/a+b〉2. 相似文献
11.
瓦西列夫不等式的加强 总被引:2,自引:0,他引:2
本刊曾刊登了瓦西列夫提出的如下优美的不等式:设a,b,C〉0,a+b+c=1,则,^2a+b/b+c+b^2+c/c+a+c^2+a/+a+b≥2①笔者经过探索,得到了①的一个加强结果: 相似文献
12.
13.
这是第42届IMO第二题:对所有正实数a,b,c,证明:a/√a^2+8bc+b/√b^2+8ca+c/√c^2+8ab≥1.文[1]中宋庆老师将其加强为:若a,b,c,为正数,则a/√a^2+2(b+c)^2+b/√b^2+2(c+a)^2+c/√c^2+2(a+b)^2≥1. 相似文献
14.
我们都知道,勾股数组{a,b,c}的双参数表示是本文给出勾股数组的单参数表示也是求勾股数的一个具体方法。定理设{a,b,c}是个勾股数组,即 相似文献
15.
贵刊文[1]介绍了俄罗斯杂志《中学数学》刊登的一组不等式,其中之一是下面的瓦西列夫不等式:
设a,b,c〉0,且a+b+c=1,则
a^2+b/b+c+b^2+c/c+a+c^2+a/a+b≥2 (1) 相似文献
16.
17.
题目设n、b、c为正实数,证明:(2a+b+c)^2/2a^2+(b+c)^2+(2b+a+c)^2/ab^2+(a+c)^2+(2c+a+b)^2/2c^2+(a+b)^2这是第32届美国数学奥林匹克试题,文[1]给出了该问题的一种证明方法,本文再给出另一种证明方法,并把它加以推广. 相似文献
18.
Let C be a closed convex weakly Cauchy subset of a normed space X. Then we define a new {a,b,c} type nonexpansive and {a,b,c} type contraction mapping T from C into C. These types of mappings will be denoted respectively by {a,b,c}-ntype and {a,b,c}-ctype. We proved the following: 1. If T is {a,b,c}-ntype mapping, then inf{ || T(x)-x|| :x C C} =0, accordingly T has a unique fixed point. Moreover, any sequence {Xn}n∈NN in C with limn→∞||T(xn) - Xn|| = 0 has a subsequence strongly convergent to the unique fixed point of T. 2. If T is {a,b,c}-ctype mapping, then T has a unique fixed point. Moreover, for any x∈C the sequence of iterates {Tn (x)}n∈N has subsequence strongly convergent to the unique fixed point of T. This paper extends and generalizes some of the results given in [2,4, 7] and [13]. 相似文献
19.
题目设a,b,c∈R^+,且abc+1,求证:1/1+2a+1/1+2b+1/1+2c≥1.文[1]中给出了如下证法:首先我们证明: 相似文献
20.
题目 已知a,b,c是正实数,证明:
(2a+b+c)^2/2a^2+(b+c)^2+(2b+c+a)^2/2b^2+(c+a)^2+(2c+a+b)^2/2c^2+(a+b)^2≤8 ①
这是2003年美国数学奥林匹克竞赛第五题,文[1]及文[2]分别用不同的方法对该题目作出精彩的证明,本文利用“变量标准化”方法给出该竞赛题的别证. 相似文献