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1.
关于Littlewood的一个问题   总被引:1,自引:0,他引:1  
本文证明了: (1)如果{a_n}_n~N=1是非负不减序列,p>0,q>0,0≤r≤1,且p(q+r)≥q+p,则sum from n=1 to N(a_n~pA_n~q)(sum from m=n to N(a_n~(1+p/q)~r≤1·sum from n=1 to N(a_n~pA_n~q)~(1+p/q),其中A_n=sum from m=n to n (a_m).上述不等式在0≤r≤1时完全解决了H.Alzer~([4])在1996年提出的一个问题,且1是最佳常数; (2)如果{a_n}_n~N=1是非负序列,p,p≥1,r>0,r(p-1)≤2(q-1),令α=((p-1)(q+r)+p~2+1)/(p+1) β=(2p+2r+p-1)/(q+1),σ=(q+r-1)/(p+q+r)则sum from n=1 to N (a_n~p)sum from i=1 to n (a_i~qA_i~r)≤2~σsum from n=1 to N(a_n~αA_n~β)(0.2)(0.2)式改进了G.Be(?)et~([2,3])在1987年对Littlewood一个问题的结果,常数因子的3/2降为2~(3/2)=1.2598…  相似文献   

2.
Three Primes Theorem in a Short Interval (Ⅳ)   总被引:1,自引:0,他引:1  
In this paper, using sieve method and Iwaniec's mean value method, we prove that for sufficiently large odd number N, equationN=p1+p2+p3,N/3-N0.63651相似文献   

3.
本文证明了对5≤s≤8,几乎所有的满足某些同余条件的正整数N都可以表示为N=p31+···+p3s,|pi-(N/s)1/3|≤N1/3-θs,其中θ5=7261-2ε,θ6=5159-ε,θ7=11333-ε,θ8=19561-ε.  相似文献   

4.
It is proved that each sufficiently large integer N=5(mod24) can be written as N=p1^2+p2^2+p3^2+p4^2+p5^2 with|pj=√N/5|±、≤U=N^1/2-1/35+e,where pj ae primes.This result,which is obtained by an iterative method and a hybrid estimate for Dirichlet polynomial, improves the previous results in this direction.  相似文献   

5.
关于A+,A+MN的表达式及其应用   总被引:5,自引:0,他引:5  
For A ∈ Cm×nr, let M and N be Hermitian positive definite matrices oforder m and n respectively. We derived the representation of the Moore-PenroseA+MN in terms of maximal nonsin-inverse A+ and weighted Moore-Penrose inverse +gular submatrices of A. In our notation,A+ = | detA[plq]|2A+pq1/vol2(A) (p,q)∈N(A)AM+N = 1/vol2(A)(p,q)∈N(A) |detA[p|q]|2N-1/2A+pqM1/2where A=M1/2 AN -1/2. From this, we propose a new method to calculate A+A+MN. The results generalize that of Moore-Penrose inverse in [2][3].  相似文献   

6.
This paper deals with blowing up of solutions to the Cauchy problem for a class of general- ized Zakharov system with combined power-type nonlinearities in two and three space dimensions. On the one hand, for c0 = +∞ we obtain two finite time blow-up results of solutions to the aforementioned system. One is obtained under the condition α≥ 0 and 1 + 4/N ≤ p N +2/N-2 or α 0 and 1 p 1 + 4/N (N = 2, 3); the other is established under the condition N = 3, 1 p N +2/N-2 and α(p-3) ≥ 0. On the other hand, for c0 +∞ and α(p-3) ≥ 0, we prove a blow-up result for solutions with negative energy to the Zakharov system under study.  相似文献   

7.
该文利用伸缩变换结合重排不等式等技巧得到了修正Gross-Pitaevskii方程对应极小化问题极小元的存在性与非线性项指数p的依赖关系.当2 p 2+4/N时,对任意c 0,极小化问题存在极小元.若p=2+4/N且c≤‖φ‖_2或者c(3/2)~(N/4)‖φ‖_2(‖φ‖_2的定义见第一节)或p 2+4/N,问题不存在极小解.而对于p=2+4/N且‖φ‖_2 c(3/2)~(N/4)‖φ‖_2,不知道是否存在极小解.  相似文献   

8.
<正>2017年全国初中数学邀请赛第11题:已知二次函数y=x2+2mx-3m+1,自变量x及实数p、q满足4p2+2mx-3m+1,自变量x及实数p、q满足4p2+9q2+9q2=2,1/2x+3pq=1,且y的最小值为1.求m的值.解由1/2x+3pq=1可得x+6pq=2,即2p×3q=2-x.∵4p2=2,1/2x+3pq=1,且y的最小值为1.求m的值.解由1/2x+3pq=1可得x+6pq=2,即2p×3q=2-x.∵4p2+9q2+9q2=2,∴4p2=2,∴4p2+2×2p×3q+9q2+2×2p×3q+9q2=2+2×(2-x)=6-2x,即(2p+3q)2=2+2×(2-x)=6-2x,即(2p+3q)2=6-2x.  相似文献   

9.
设p是奇素数,N(p)是椭圆曲线E:y2=2px(x2+1)的正整数点(x,y)的个数.主要讨论了N(p)的性质,运用初等方法及四次Diophantine方程的性质,对某些特殊素数p,给出了N(p)的上界.证明了当p≡1(mod 8)且p=s2+32t,其中s,t是正整数时,N(p)≤3;当p≡1(mod 8)且p+s...  相似文献   

10.
设N为大偶数,以D(N)表示将N表成两个素数之和的表法个数,即 D(N)=sum from N=P_1+P_3 (1)。Hardy和Littlewood利用“圆法”证明了下面的结果 D(N)=(?)(N)N/log~2N+R (1)这里 (?)(N) 2 multiply from p>2((1-1/(p-1)~2) multiply from p\N P>2 (1+1/p-2),(2) R=(sum from q>Q(μ~2(q)/φ~2(q))C_q(-N))N/log~2N+integral from E (S~2(α,N)e~(-2πtαN)dα) (3) S(α,N)=sum from p≤N (e~(2πiαp)),C_q(-N)=sum from n=1 to q (e~(2πiNh/q))Q=log~(16)N,E表示在通常意义下的余区间,这就提出了下面的猜想 D(N)~(?)(N)N/log~2·(4)熟知Goldbach猜想的困难在于误差项R的处理,至今“圆法”是提出猜想(4)的唯一的方法,本文提出了另一种途径来研究猜想(4)。而且方法是初等的,看起来是更为直接的方法。令 (?)(N)=sum from d≤N(Λ(d)Λ(N-d))。 显然 D(N)=(?)(N)/log~2N[1+O(log log N/log N)]+O(N/log~3N).本文证明了下面两个定理: 定理1 设N为大偶数,这里证明定理1的方法是初等的,这就建议我们提出猜想(4)。 定理2 用Bombieri定理可以证明 R_1=R_2=O(Nlog~(-1)N)。从上面两个定理看出,研究Goldbach猜想的困难,在于处理余项R_3。  相似文献   

11.
Exponential sums over primes in short intervals   总被引:3,自引:0,他引:3  
In this paper we establish one new estimate on exponential sums over primes in short intervals. As an application of this result, we sharpen Hua's result by proving that each sufficiently large integer N congruent to 5 modulo 24 can be written as N = p12 p22 p32 p42 p52, with |pj-(N/5)~(1/2)|≤U = N1/2-1/20 ε, where pj are primes. This result is as good as what one can obtain from the generalized Riemann hypothesis.  相似文献   

12.
孟宪萌 《数学学报》2007,50(2):255-260
设P_k表示素因子个数不超过k的殆素数.本文证明了对几乎所有充分大的偶数n≠2(mod6),方程n=p_1+p_2有素数解p_1,p_2,且p_1+2=P_3;对任何充分大的奇数N≠1(mod6),方程N=p_1+p_2+p_3有素数解p_1,p_2,p_3,且p_2+2=P_3, p_3+2=P_2.  相似文献   

13.
Hua’s theorem with nine almost equal prime variables   总被引:2,自引:1,他引:1  
We sharpen Hua’s result by proving that each sufficiently large odd integer N can be written as
, where p j are primes. This result is as good as what was previously derived from the Generalized Riemann Hypothesis.   相似文献   

14.
In this paper the following result is proved: There is an absolute positive integer c such that for every large odd integer N the Diophantine equation with prime variables $N=p_1+p_2+p_3,N/3-U 相似文献   

15.
王明强 《数学学报》2004,47(4):695-702
H表示一个正整数N的集合,使对任意的正整数q,同余方程a+b~2≡N(mod q)在模q的既约剩余系中有解a;b.E(x)表示N≤x,N∈H,但不能表成p_1+p_2~2=N的数的个数,其中p_1,p_2个表示素数,则E(x)<相似文献   

16.
堆垒素数论的一些新结果   总被引:1,自引:0,他引:1  
潘承洞 《数学学报》1959,9(3):315-329
<正> (?)在1937年证明了所有充分大的奇数 N 皆可表成三素数之和,即有N=p_1+p_2+p_3,其中 p_i(i=1,2,3)为奇素数.而本文的目的在于限制 p_i(i=1,2,3)的变化范围.证明了下面三个定理:定理1.°设 N 为充分大的奇数,则必有 pi(i=1,2,3)满足  相似文献   

17.
On sums of a prime and four prime squares in short intervals   总被引:1,自引:1,他引:0  
In this paper, we prove that each sufficiently large integer N ≠1(mod 3) can be written as N=p+p1^2+p2^2+p3^2+p4^2, with
|p-N/5|≤U,|pj-√N/5|≤U,j=1,2,3,4,
where U=N^2/20+c and p,pj are primes.  相似文献   

18.
证明了一类n阶(n=P_1P_2…p_m,p_i(i=1,2,…,m)互异为素数)环是有限循环环,并讨论了他们的结构及相关性质,最后给出了这类n阶环有零因子或有子域的充要条件.主要结果:P_1P_2…P_m阶环共有2m个,它们是(p_(1m个,它们是(p_(1k_1) p_(2k_1) p_(2k_2)…p_(mk_2)…p_(mk_m)Z)/(p_(1k_m)Z)/(p_(1k_1+1)p_(2k_1+1)p_(2k_2+1)…p_(mk_2+1)…p_(mk_m+1)Z),其中k_i=0或1,1≤i≤m;阶是n=P_1P_2…p_m的环R可唯一分解为m个素数阶理想的直和,即R=〈α〉=(?);含pi(1≤i≤m)阶子域的P_1P_2…P_m阶环共有2k_m+1)Z),其中k_i=0或1,1≤i≤m;阶是n=P_1P_2…p_m的环R可唯一分解为m个素数阶理想的直和,即R=〈α〉=(?);含pi(1≤i≤m)阶子域的P_1P_2…P_m阶环共有2(m-1)个,它们是p_(1(m-1)个,它们是p_(1k_1) p_(2k_1) p_(2k_2)…p_(mk_2)…p_(mk_m)Z)/(p_(1k_m)Z)/(p_(1k_1+1)p_(2k_1+1)p_(2k_2+1)…p_(mk_2+1)…p_(mk_m+1)Z),其.中k_i=0,k_j=0或1,1≤j≤m,j≠i.  相似文献   

19.
In this note, we prove some results of Hua in short intervals. For example, each sufficiently large integer N satisfying some congruence conditions can be written as
$ \left\{ {\begin{array}{*{20}{c}} {N = p_1^2 + p_2^2 + p_3^2 + p_4^2 + {p^k}}, \hfill \\ {\left| {{p_j} - \sqrt {N/5} } \right| \leqslant U,\left| {p - {{\left( {N/5} \right)}^{\tfrac{1}{k}}}} \right|\leqslant UN - \tfrac{1}{2} + \tfrac{1}{k},j = 1,2,3,4,} \hfill \\ \end{array} } \right. $
where \( U = N\tfrac{1}{2} - \eta + \varepsilon \) with \( \eta = \frac{2}{{\kappa \left( {K + 1} \right)\left( {{K^2} + 2} \right)}} \) and \( K = {2^{k - 1}},k\geqslant 3. \)
  相似文献   

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