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标题: Goldbach Theorem [打印本页]

作者: 数学1+1    时间: 2014-6-14 11:42
标题: Goldbach Theorem
                                Goldbach Theorem6 d, R8 T) x+ b5 e8 e5 x
                         Union of set construction and analysis9 g. v  P. q" @5 O! L- C

/ Z/ G- R! b" R+ }  a       Loudi Xiaoguang mathematical research studio Su Xiaoguang
8 ?, x( P2 f) R6 ^6 E/ j# q  Abstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the:A={N │ N = (N-i) + i, N is a natural number, i belong to N} Clearly A is a countable set。C={N │ N = Pi + Pj, N is even number, Pi, Pj is  prime number,i, j is  natural number} Clearly, C is A subset。So C is a countable set.If x ={x|x no less than 6 and not more than N is an even number,x=P1+P2, P1, P2 is  prime number},card (x)=M (x).Clearly, x is C subset。So x is a countable set.We can get the M(x) range,If N ={N|N = P1 + P2, N is even number, P1, P2 is  prime number},card (N)=D(N),D(N)=M(N)-M(N-2)6 R, \, k7 @5 ^; \/ o# H
So when N> 800000 when, D(N)No less than1.8432(1-1/logN)N/log^2_(N-2);I
. E/ T( x$ [- P9 r8 t4 |Not more than5.0176[1+2/logN +o(1)]N/log[(N-2)/2]log(N-2)。+ Z: U& w7 D& S0 ?$ l$ P" v( a. Q
Key words: Germany,Goldbach,Union of set,even number, prime number, 7 |6 |- b1 H$ Z! S& W  ~, [8 ~
MR (2000) theme classification: 11 P32
( [8 C% Y" o# E& J3 t0 d, uEmail:suxiaoguong@foxmail. Com
作者: MichaeLonger    时间: 2014-6-17 12:44
英语不好。。。
作者: MichaeLonger    时间: 2014-6-17 12:44
看不明白啊。。
作者: MichaeLonger    时间: 2014-6-17 12:45
哎。。。。。
作者: MichaeLonger    时间: 2014-6-17 12:45
英语渣好成问题。。
作者: 数学1+1    时间: 2014-7-16 09:30
Goldbach formula:4 S" O5 t7 H: h( a
1.8432(1-1/log N)N/log(N-2)log(n-2)<D(N)<2.5088 S(N)N/[log(N-2)/2][log(N-2)/2]) |( x' E- X8 U1 o; Q
For N>800000,D(N) mean# U( F. F( q7 u. F9 l% s9 L) Q
       N=P_1+P_2
% O" {/ j% i/ Z& UThe number of elements。prime number P_1,P_2>2。
3 z4 m" _! `# B  O5 A      S(N)=1-(2 log 2 log 2)/[log(N-2)log(N-2)]-2{[ log(N-2)/2]log N(N-2)}/[2 log N log(N-2)log N/2]$ m) g: I% O2 [; r
               +[log N(N-2)/4]/[2 (log N/2)(log N/2)]1 c( o% \( |4 {% X+ B
               +{2 [log(N-2)/2] [log(N-2)/2]log N(N-2)}/[log N log N log(N-2)log(N-2)]




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