Goldbach Theorem

数学1+1

                                Goldbach Theorem
                         Union of set construction and analysis
6 ]. O2 V6 W+ \7 F5 h1 L
6 n, a0 c! f# S. P  c　　　　　　　Loudi Xiaoguang mathematical research studio Su Xiaoguang
　　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)
So when N> 800000 when, D(N)No less than1.8432(1-1/logN)N/log^2_(N-2);I
8 m( \  U5 Z& t4 ?8 ?Not more than5.0176[1+2/logN +o(1)]N/log[(N-2)/2]log(N-2)。
" Y8 [$ |4 {, r1 \; CKey words: Germany，Goldbach，Union of set,even number, prime number,
" [6 Y% C7 l% ^, ]" \! KMR (2000) theme classification: 11 P32
Email:suxiaoguong@foxmail. Com

MichaeLonger

英语不好。。。

MichaeLonger

看不明白啊。。

MichaeLonger

哎。。。。。

MichaeLonger

英语渣好成问题。。

数学1+1

Goldbach formula:
: A8 r1 m8 M) [ 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]
2 }6 b$ S2 _* o5 Q5 F1 Y" [0 l: f9 f, SFor N>800000,D(N) mean
       N=P_1+P_2
The number of elements。prime number P_1,P_2>2。
5 ?, u# j! v9 [& |      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]
; k% P$ ?+ k1 [8 Z               +[log N(N-2)/4]/[2 (log N/2)(log N/2)]
- C* I/ F# _1 Q$ n0 R7 x               +{2 [log(N-2)/2] [log(N-2)/2]log N(N-2)}/[log N log N log(N-2)log(N-2)]