一个有极限的单调增加数列

证明了{n(16n^2+4n+3)/16n^2-4~n+3^(1/2) integral from 0 to π/2 sin^nxdx}为严格单调增加数列,且极限为π/2^(1/2),因而得π(16n^2+36n+23)/2(n+1)(16n^2+28n+15)^(1/2)相似文献   

一个有极限的单调减少数列

证明了{(n(4n+1)/4n-1)~(1/2)∫π/20 sin~nxdx}为严格单调减少数列,且极限为(π/2)~(1/2),因而得(π(4n-1)/2n(4n+1))~(1/2)∫π/20 sin~nxdx (π(4 n+5)/2(n+1)(4n+3))~(1/2).     

两个极限相等的有趣数列

证明了两个有趣数列{n~(1/2)∫_0~(π/2)sin~nxdx},{(n+1)~(1/2)∫_0~(π/2)sin~nxdx}的极限均为(π/2)~(1/2),且(π/2(n+1))~(1/2)∫_0~(π/2)sin~nxdx(π/2n)~(1/2).     

On the Approximation of Combination of Linear Operators

Let f(x)∈C_(2π).For Valle-Poussin integrals V_n(f,x)=(2n)!! 1(2n-1)!! 2πintegral grom -πto π(f(x 1)cos~(2n)t/2 dt), Z.Ditzian and G.Freud considered the approximation of their combination writingV_(n,1)(f,x)=2V_(2n-1)(f,x)-V_(n-1)(f,x),V_(n,2)(f,x)=8/3V_(4n-1)(f,x)-2V_(2n-1)(f,x) 1/3V_(n-1)(f,x), they proved that V_(n,1)(f,x)-f(x)=O(ω_4(f,1/n~(1/2))), V_(n,2)(f,x)-f(x)=O(ω_6(f,1/n(1/2))) In this paper, using the asymptotic expansions of linear operators with many terms,we generalize the above result to the case of eombination of m terms, where mis an arbtirary positive integer.     

关于积的三个三角恒等式的推广

一易证下列三个恒等式成立: (1)sinθsin(θ+π/ 3)sin(θ+2π/ 3) =sin3θ/4; (2)cosθcos(θ+π/3)cos(θ+2π/3) =-1/4cos3θ; (3)tgθtg(θ+π/3)tg(θ+2π/3) =-tg3θ。本文把上述三个恒等式予以推广,其一般形式为: (Ⅰ) multiply form j=1 to n sin(θ+(j-1)/nπ)=sinnθ/2~(n-1); (Ⅱ) multiply form j=1 to n cos(θ+(j-1)/nπ) =(-1)~(n-2) sinnθ/2~(n/1) (n为偶数), (-1)~(n-1)~2 cosnθ/2~(n-1)(n为奇数);     

On the generalized Chern conjecture for hypersurfaces with constant mean curvature in a sphere

Let M be a compact hypersurface with constant mean curvature in Denote by H and S the mean curvature and the squared norm of the second fundamental form of M,respectively.We verify that there exists a positive constant γ(n) depending only on n such that if |H| ≤γ(n) and β(n,H)≤S ≤β(n,H)+n/18,then S≡β(n,H) and M is a Clifford torus.Here,β(n,H)=n+n~3/2(n-1)H~2+n(n-2)/2(n-1)(1/2)n~2H~4+4(n-1)H~2.     

一个有趣数列的单调性

利用不等式(π4(n-1)/2n 4(n+))^(1/2)1<∫_0^(π/2)sin^nxdx<(π(4n+5)/2 (n+1)(4n+3))^(1/2),对有趣数列{(n+c)^(1/2)∫_0^(π/2)sin^nxdx}的单调性再次进行了分析论证,并对已有结论进行了改进.     

1956年北京、天津数学競賽試題及解答

(1) 証明:对任何正整数n n~3+(3/2)n~2+(1/2)n-1都是整数,並且用3除时余2, 証明1.n~3+(2/3)n~2-1=(n(n+1)/2)(2n+1)-1==(2n(2n+1)(2n+2)/8)-3+2。对任何整数n說來,(n(n+1))/2是整数,所以原式为整数。在相鄰的三个整数2n,2n+1,2n+2中至少有一个是3的倍数,因为3与8互質,8除得尽分子,分子提出3后还能被8除尽,所以(2n(2n+1)(2n+2)/8)-3是3的倍数,原式用3除时余2。 証明2.用f(n)代表原式,現在用数学归納法証明我們的問題。在n=1时,     

一个有趣的整除问题

本刊1983年第四期的问题征解上有这样一题求证(1+2+3+…+1983)|(1~5+2~5+3~5+…+1683~5)。在解此题时,我们从1~3+2~3+3~3+…+n~3=(1/4)n~2)n+1)~2=〔(1/2)n(n+1)〕~2=(1+2+3+…+n)~2,发现(1+2+3…+n)|(1~3++2~3+3~3+…+n~3)对任意自然数n皆成立。我们试问是否也有(1十2+3+…+n)|(1~5+2~5+3~5+…+n~5)对任意的自然数n皆成立呢!回答是肯定的。不难证明1~5+2~5+3~5+…+n~5=(?)(n+1)~2(2n~2+2n-1),因此,(1+2+3     

一道高三联考压轴题的多方法解析

正题目:已知f(x)=lnx,g(x)=mx-1.(1)若f(x)≤g(x)恒成立,求m的取值范围.(2)若n∈N*,n1,求证3/2~2+5/3~2+…+(2n-1)/n~22lnn.这道题表面是一道不等式题目,仔细深入挖     

关于Chowla的一个猜想

在1978年赫尔辛基的ICM会议上,Apry给出(3)=sum from n=1 to (?) (1/n~3)是无理数的证明。为此,Apry定义了一个迭代数列a_n: a_0=1,a_1=5,n~3a_n-(34n~3-51n~2+27n-5)a_(n-1)+(n-1)~3a_(n-2)=0,(1) 它满足 这里Chowla在[1]中讨论了Apry数a_n的同余性质,他证明了a_(5n+1)≡0(mod p),a_(5n+3)≡0(mod p)以及对于奇素数p恒成立a_p≡5(mod p~2)。在文章最后     

2011年清华金秋营数学试题及解答

1.求sinπnsin2πn…sin(n-1)πn的值.解设ε=cosπn+isinπn(i为虚数单位),则1,ε,ε2,…,ε2(n-1)为x2n-1=0的根,且sinkπn=εk-ε-k2i=ε2k-12iεk,所以sinπnsin2πn…sin(n-1)πn=(ε2-1)(ε4-1)…[ε2(n-1)-1]2n-1in-1ε12n(n-1)()n-1(2)(4)…[2(n-1)]     

常数项级数求和方法举例

一、利用级数和的定义求和例1 求级数sum from n=1 to ∞((2n 1)/n~2(n 1)~2)的和解:由于(2n 1)/n~2(n 1)~2=((n 1)~2-n~2)/(n~2(n 1)~2)=1/n~2-1/((n 1)~2)于是,S_n=(1-1/2~2) (1/2~2-1/3~2) … (1/n~2-1/(n 1)~2)=1-(1/(n 1)~2)所以,该级数的和S=(?)=1     

On simultaneous approximation by lagrange interpolating polynomials

This paper considers to replace △_m(x)=(1-x~2)~2(1/2)/n +1/n~2 in the following result for simultaneousLagrange interpolating approximation with (1-x~2)~2(1/2)/n: Let f∈C_(-1.1)~0 and r=[(q+2)/2],then|f~(k)(x)-P_~(k)(f,x)|=O(1)△_(n)~(a-k)(x)ω(f~(a),△(x))(‖L_n-‖+‖L_n‖),0≤k≤q,where P_n( f ,x)is the Lagrange interpolating polynomial of degree n+ 2r-1 of f on the nodes X_nU Y_n(see the definition of the text), and thus give a problem raised in [XiZh] a complete answer.     

推广的Wallis数列的单调性

讨论了推广的Wallis数列{(n+c)(1/2)∫π/20sin~nxdx}(n≥1,c为非负常数)的单调性.黄永忠等(2016)证明了当0≤c≤1/2该数列严格递增;当1/2c≤1该数列对于充分大的n严格递减.本文给出了此结论的一个新的简洁证明,并对相关问题做了讨论.进一步,证明了当且仅当c2π~2-16/16-π~2=0.609945…,推广的Wallis数列为严格递减数列.     

數學問题及解答欄

144,假如m/n是2~(1/3)的一個近似值,試証明(m~2+mn+2n~2)/(m~2+mn+n~2)是2~(1/3)的一個更好的近似值,並且(m~2+mn+2n~2)/(m~2+mn+n~2)-2~(1/3)與m/n-2~(1/3)異號。(選自蘇聯“數學教學”1954年第4期間題欄), 145.設P,Q,R,p,q,r那是整數,且p,q,r互素,證明:如 P/p+(Q/q)+(R/r)是一整數,那末P/p,Q/q,R/r,也是整數。 146.求不定方程x~2+y~2+z~2=3xyz的正整數解之適合於0相似文献   

一个常数竟有两个整数部分

题求1 1/2~(1/2) 1/3~(1/2) …1/100~(1/2)的整数部分解∵(n 1)~(1/2) n~(1/2)>2n~(1/2)>n~(1/2) (n-1)~(1/2)∴(n 1)~(1/2)-n~(1/2)<1/(2n~(1/2))相似文献   

Exact Solution to an Extremal Problem on Graphic Sequences with a Realization Containing Every 2-Tree on k Vertices

A simple graph G is a 2-tree if G=K_3,or G has a vertex v of degree 2,whose neighbors are adjacent,and G-v is a 2-tree.Clearly,if G is a 2-tree on n vertices,then |E(G)|=2 n-3.A non-increasing sequence π=(d_1,...,d_n) of nonnegative integers is a graphic sequence if it is realizable by a simple graph G on n vertices.[Acta Math.Sin.Engl.Ser.,25,795-802(2009)] proved that if k≥2,n≥9/2 k~2+19/2 k and π=(d_1,...,d_n) is a graphic sequence with∑_(i=1)~n di(k-2)n,then π has a realization containing every 1-tree(the usual tree) on k vertices.Moreover,the lower bound(k-2)_n is the best possible.This is a variation of a conjecture due to Erdos and Sos.In this paper,we investigate an analogue problem for 2-trees and prove that if k≥3 is an integer with k≡i(mod 3),n≥ 20[k/3] ~2+31[k/3]+12 and π=(d_1,...,d_n) is a graphic sequence with ∑_(i=1)~n d_imax{k-1)(n-1), 2 [2 k/3] n-2 n-[2 k/3] ~2+[2 k/3]+1-(-1)~i}, then π has a realization containing every 2-tree on k vertices.Moreover,the lower bound max{(k-1)(n-1), 2[2 k/3]n-2 n-[2 k/3] ~2+[2 k/3]+1-(-1)~i}is the best possible.This result implies a conjecture due to [Discrete Math.Theor.Comput.Sci.,17(3),315-326(2016)].     

常见累进数列n{(n+1)…(n+m))的求和方法

大家知道1·2+2·3+3·4+…n(n+1)的求和可利用通项公式来求,即: 1·2+2·3+3·4+…+n(n+1)=(1~2+2~2+3~2+…+n~2)+(1+2+3+… +n)=(1/6)n(n+1)(2n+1)+(1/2)n(1+n)-(1/3)n(n+1)(n+2) 但是用这种方法求和涉及到数列1~2,2~2,3~2…n~2的求和,如果给出累进数列的每项乘积因子则又涉及数列{n~3},{n~4},…的求和,所以利用通项求常见累进数列     

Wallis公式的精致化

证明了∫_0~(π/2)sin~αxdx(α→+∞)渐进展开式的存在性,并给出了展开式系数的递推公式,进而得到了((2n-1)!!)/((2n)!!)/((2n)!!)的渐进展开式和精致化的Wallis公式.在精致化的Wallis公式中取到1/n~m项,对π的逼近精度可达O(1/(n~(m+1))),比原来的Wallis公式的精度大为提高.