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标题: Goldbach’s problem [打印本页]
作者: 数学1+1    时间: 2013-12-6 12:27
标题: Goldbach’s problem
Goldbach’s problem                    Su XiaoguangAbstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the：[code]<SPAN style="FONT-FAMILY: Arial; COLOR: #333333; FONT-SIZE: 12pt; mso-font-kerning: 0pt; mso-ansi-language: EN" lang=EN></SPAN>[/code]A= \bigcup_{i=0}^{\infty }A_{i},A_{i}=\left \{ i+0,i+1,i+2,\cdots \right \}\Rightarrow B(x,N)=\sum_{N\leqslant x,B(N)\neq 0}1,B(N)=\sum_{n+m=N,0\leq n,m\leq N}1.C=\bigcup_{i=0}^{\infty }C_{i},C_{i}=\left \{ p_{i}+p_{0},p_{i}+p_{1},p_{i} +p_{i},\cdots \right \}\wedge N> 800000\Rightarrow M(x)=\sum_{N\leq x,D(N)\neq 0}1DeducedD(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+log\frac{2}{N}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
, @% s* L$ @; W) P6 L Key words: Germany，Goldbach，even number, Odd number ，prime number, MR (2000) theme classification: 11 P32 Email:suxiaoguong@foxmail. com
0 m4 Z0 M/ |& x$ r' Y/ V- Q
作者: 数学1+1    时间: 2013-12-6 13:21
                  Goldbach’s problem (pdf)
                       Su Xiaoguang& b5 x% ]1 d0 f+ _, [$ W1 e
     
Abstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the：
A= \bigcup_{i=0}^{\infty }A_{i},A_{i}=\left \{ i+0,i+1,i+2,\cdots \right \}\Rightarrow B(x,N)=\sum_{N\leqslant x,B(N)\neq 0}1,B(N)=\sum_{n+m=N,0\leq n,m\leq N}1.
C=\bigcup_{i=0}^{\infty }C_{i},C_{i}=\left \{ p_{i}+p_{0},p_{i}+p_{1},p_{i} +p_{i},\cdots \right \}\wedge N> 800000\Rightarrow M(x)=\sum_{N\leq x,D(N)\neq 0}1
Deduced
D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge
1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+log\frac{2}{N}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
: L3 W% N5 v0 S) x$ N
Key words: Germany，Goldbach，even number, Odd number ，prime number,
MR (2000) theme classification: 11 P32
Email:suxiaoguong@foxmail. com
作者: 数学1+1    时间: 2013-12-9 10:43
                  Goldbach’s problem
% ~* D  [  P1 S0 `7 D                    Su Xiaoguang5 j9 T3 K4 d+ M" A  A5 v. q
Abstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the：
" k0 ?! x) Q3 O+ @0 X% |, ]$ ]; H; e& [. h7 `+ Q& K: r
A= \bigcup_{i=0}^{\infty }A_{i},A_{i}=\left \{ i+0,i+1,i+2,\cdots \right \}\Rightarrow B(x,N)=\sum_{N\leqslant x,B(N)\neq 0}1,B(N)=\sum_{n+m=N,0\leq n,m\leq N}1.2 F/ Q  g1 g* G
C=\bigcup_{i=0}^{\infty }C_{i},C_{i}=\left \{ p_{i}+p_{0},p_{i}+p_{1},p_{i} +p_{i},\cdots \right \}\wedge N> 800000\Rightarrow M(x)=\sum_{N\leq x,D(N)\neq 0}1" M7 A" c* D. J) O3 J$ W
Deduced
! f4 ?+ e, C& c& [+ BD(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge * k2 q4 D! H9 F/ U
1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+log\frac{2}{N}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}6 u' l2 B( W' m. }  S( ?

" [) D, @2 t: `8 a1 v0 Y2 ~Key words: Germany，Goldbach，even number, Odd number ，prime number,
  B6 D  C, ?# s; T: J9 yMR (2000) theme classification: 11 P32
' B0 U: }2 w  A2 i; s4 DEmail:suxiaoguong@foxmail. com) x) Z: k& S3 F+ B' J( v# k0 S' f
§ 1 Introduction
6 ~- X; [: ?  z          In 1742, the German mathematician Christian Goldbach (1690-1764)， Put forward two speculated about the relationship between positive integers and prime number，using analytical language expressed as:
9 Y! p( Y- s( I- ]% \（A）For even number N" P. Q4 g& ]) x7 D! D& w4 W; j

+ V, s5 d8 ~6 t* ]6 [; m5 RN\geq 6\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}Is a prime number}1>0" A7 i; M  j5 _4 U

- f0 G% ~4 I# @(B)  For odd number N0 X" }* f4 T1 D0 o/ w
6 k4 e! q- T/ u4 B- B( W
N\geq 9\Rightarrow T(N)=\sum_{p_{1}+p_{2}+p_{3}=N.p_{1},p_{2},p_{3}\geq 3}1>0' s2 E7 p- R3 K8 C
! u1 m8 Y' B4 `( ?. x: l* \
This is the famous GOldbach conjecture。If the proposition (A) true, then the proposition (B) True。So, as long as we prove Proposition (A), Launched immediately conjecture (B) is correct7 B& S7 p4 V1 u2 |
          : i: n( S* [$ e, ~8 ^8 \, a. _& c
§2 Correlation set constructor- `- [9 {/ ]$ J$ @  Z7 X
A_{0}=\left \{ 0+0,0+1,0+2,\cdots  \right \}! T3 p! J' z/ Z& ^# Y3 _/ q8 P
A_{1}=\left \{ 1+0,1+1,1+2,\cdots  \right \}
4 u, f* V) j/ d% {' e* u) u A_{2}=\left \{ 2+0,2+1,2+2,\cdots  \right \}8 z  `6 |1 Q4 B
\cdots
& P0 H$ n+ V1 T# |- q5 M" |7 SA=\bigcup_{i=0}^{\infty }A_{i}\Rightarrow B(N)=\sum_{n_{1}+n_{2}=N.0\leq n _{1},n_{2}\leq N}1,B(x,N)=\sum_{N\leq x,B(N)\neq 0}1      （1）
6 j8 p' S6 n, A3 l( u( L9 S) X; up_{0}=2,p_{1}=3,p_{2}=5,\cdots C_{0}=\left \{ p_{0} +p_{0},p_{0}+p_{1},p_{0}+p_{2},\cdots \right \} C_{1}=\left \{ p_{1} +p_{0},p_{1}+p_{1},p_{1}+p_{2},\cdots \right \} C_{2}=\left \{ p_{2} +p_{0},p_{2}+p_{1},p_{2}+p_{2},\cdots \right \} C=\bigcup_{i=0}^{\infty }C_{i}\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N.p_{1},p_{2}\geq 3}1\wedge M(x)=\sum_{N\leq x,D(N)\neq 0}  （2）      
* Q6 }2 q1 U! |! T# x* [, d  §3    Ready  Theorem4 `( Q0 L$ V! D
Theorem 1; W: U- u9 i1 t  x8 c
M_{i}=(x_{1}^{(i)},x_{2}^{(i)},\cdots ,x_{i}^{(i)},\cdots ),Is a countable set\Rightarrow M=\bigcup_{i=1}^{N}M_{i},Is a countable set! X# L: P; S  x5 @% i% S
  .Proof: Suppose M_{1},M_{2},\cdots ,M_{N},Is a countable set, M=\bigcup_{i=1}^{N}M_{i}
9 e2 a4 s& m9 s5 S\because M_{1}:x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},\cdots ,x_{i}^{(1)},\cdots2 n: T3 C7 ~# `
M_{2}:x_{1}^{(2)},x_{2}^{(2)},x_{3}^{(2)},\cdots ,x_{i}^{(2)},\cdots6 R' s1 H9 @& A, _7 w5 p# Q( O
M_{N}:x_{1}^{(N)},x_{2}^{(N)},x_{3}^{(N)},\cdots ,x_{i}^{(N)},\cdots
8 N. ^. J$ w" L2 j) x# q8 x4 J% Q\cdots0 x7 K3 x) p/ m* u
\therefore M:x_{1}^{(1)},x_{1}^{(2)},\cdots ,x_{1}^{(N)},x_{2}^{(1)},x_{2}^{(2)},\cdots ,x_{2}^{(N)},\cdots Countable4 l! j5 N6 k7 u2 U0 i! p  i
     Theorem 2 (Prime number theorem)
2 m) z# z. [/ v. \( @  |$ B8 ~8 J1 Z- q- B3 X1 S) V6 ?& M* a  m( [7 z
\pi (x)\sim \frac{x}{logx}^{\left [ 1 \right ]}
! n6 h# s! N8 `     Theorem 3  For even number x
6 y5 U& r: |: K% a5 Gx>800000\wedge M_{1}=minM(x)\Rightarrow M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x) -1\right ] . u5 K' q, t4 K
Proof: According to Theorem 1, (1)  
& h) [! s  S# D+ x! L  F  \because A_{i},A_{j} Countable,) y2 E3 X! j/ r, P( G; z" u8 |
\therefore A Countable\wedge B(x,N)=\frac{1}{2}(N+1)(N+2)
+ p' y/ z$ r6 x# i0 N- _Similarly, according to Theorem 1, (2), C countable
) y( p1 X$ O! [% q; R6 m( A Suppose3 o* n" a1 o7 U5 ]
      M_{1}(x)=minM(x)
! h8 R. f) z+ O# Uaccording to （2）, Then we have, e3 X" Z7 V4 O+ F% p1 X6 {' t5 n, }: y
. M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x)-1 \right ]
+ ~/ p$ ?5 p# o; O  X& g Theorem 4  For even number x/ L+ _+ y* Q! s% r+ u& ~
x>800000\wedge M_{2}(x)=maxM(x)\Rightarrow M_{2}(x)=4\pi (\frac{x}{2})\pi (x)-2\pi ^ {2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)                        （3）( d  u" t' M! Y) j
Proof: According to (2)，Then we have& k' H( p* S' ~1 b  g* r
M(x)=\sum_{N\leq x,D(N)\neq 0}1< \sum_{3\leq p_{1},p_{2}\leq \frac{x}{2}}1+\sum_{3\leq p_{1}\leq \frac{x}{2},\frac{x}{2}< p_{2}< x}+\sum_{\frac{x}{2}<p _{1}< x,3\leq p_{2}\leq \frac{x}{2}}0 s* l' ^- ~. W8 I$ b
     Suppose
3 K1 @! J6 L6 z; q+ R7 U1 I' K      M_{2}(x)=maxM(x)
9 @& D* X; e* b9 H+ q3 g\therefore M_{2}(x)
1 _5 y) g: s# {* F+ ]8 z$ I! N=\frac{1}{2}\cdot 2\pi (x)\left [ 2\pi (\frac{x}{2})-1 \right ]-\frac{1}{2}\left [ 2\pi (\frac{x}{2})-\pi (x) \right ]\left [ 2\pi (\frac{x}{2})-\pi (x) +1\right ]\cdot 2# l) L+ D  N( g: ^+ i
=4\pi (x)\pi (\frac{x}{2})-2\pi ^{2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)/ |* k# P' G6 V9 D; f; ^- z% e! L; _
§4 Goldbach's problem end/ _7 A5 H* e" V5 v
Theorem 5  For evem number N5 \* G# D" B5 i; Q# B- `
N> 800000\Rightarrow D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}8 c' Z7 J% r/ z
     Proof: According to Theorem 2
* G) n9 I* o1 eN> 800000\Rightarrow \alpha \frac{N}{logN}\leq \pi (N)\leq \beta \frac{N}{logN}      (4)7 b* d- ^. m3 S( w
Let   c_{1}=min(\alpha ,\beta ),
4 C8 N2 @- Z: x1 K# n$ `+ J) `) V4 ?5 ?According to Theorem 3,Then we have/ i5 p/ T" u* _" |* Y8 g0 v6 \7 Q
D_{1}(N)=M_{1}(N)-M_{1}(N-2)
7 q8 }' ~: j5 M/ qClear2 l+ ?2 s4 I4 e+ ?0 ]8 ~# ]
D(N)\geq D_{1}(N)8 E. }* D( N( o
\because log(1+x)=\int_{0}^{x}\frac{dt}{1+t}=x-\int_{0}^{x}\frac{t}{1+t}dt8 A( v( q. B* I9 C% _
\because x\geq-\frac{1}{2} \Rightarrow log(1+x)=x+o(x^{2})           (5)' ^- E; }" e* n( L* n  X( @, d, B
\therefore D_{1}(N)=2c_{1}^{2}(1-\frac{1}{logN})\frac{N}{log^{2}(n-2)}+o(1)% W2 `2 p" J3 w& `
N\rightarrow \infty ,o(1)\rightarrow 0\Rightarrow
& X8 h/ W- Q6 T6 J+ _D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
7 P2 P8 G0 Y4 Y# l8 b6 y2 L Theorem 6  For evem number N: A# @3 ?7 |& Z/ R/ L8 A: ^4 Q. O
N> 800000\Rightarrow D(N)\leq
. _/ Q/ l8 H( S7 x7 K8 W' M5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
: l" V  ?4 b4 d- cProof : According to (4)
* ~. g) Q) \! T3 j/ yLet  c_{2}=max(\alpha ,\beta )! j9 `6 N8 F5 s) e
According to Theorem 4,Then we have
2 w9 D8 y, n+ ~) i# PD_{2}(N)=M_{2}(N)-M_{2}(N-2)
/ K$ S: j. {; V% h\because D(N)\leq D_{2}(N)7 z% s* K; y' Q0 _  W: _
According to (5), Then we have. L+ p, @) {3 H. l; |
D(N)\leq 1 x1 k: F, ]# t& {7 l  [/ u0 h5 R
5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
1 N; V# v' P6 S* {Theorem 7 (Goldbach Theorem)  
) u; i. T* x$ v+ B* bFor evem number N, `6 \" D$ @6 H' d" D8 y: c1 k
N\geq 6\Rightarrow D(N)= \sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\geq 1/ c; u- ?+ N, g: v+ Q
Proof : According to Shen Mok Kong verification
0 q# V1 f5 v- X8 F6\leq N\leq 3.3\times 1000000^{\left [ 3 \right ]}\Rightarrow D(N)\geq 1
4 S* Q4 S) \& ]6 C& mAccording to Theorem 5, Theorem 6, Then we have5 e0 N5 P( x- Z3 @( L
N> 800000\Rightarrow 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
: M9 c5 O" T' q1 ~3 [' U\therefore N\geq 6\Rightarrow D(N)\geq 1
/ d5 F5 H# {9 P( n! u7 y4 Q7 sLemma 1 For odd number N; }' e! F. P# o5 l4 z4 h
N\geq 9\Rightarrow
! O( I* Y. }( O2 j1 [/ v9 J, rT(N)=\sum_{p_{1}+p_{2}+p_{3}=N,p_{1},p_{2},p_{3}\geq 3}1\geq 1# V/ F5 o( _- R: B, _: C
Proof et  n\geq 4+ ?7 }+ p* a( t4 s
\because 2n+1=2(n-1)+3: a) B% C4 _0 Q/ [/ i2 \( T
According to Theorem 7,  Then we have( F& ~. \0 N! ~5 O, @1 o
N\geq 9\Rightarrow T(N)\geq 16 x  l. o2 j9 W, ]$ A* A  n

+ ?$ o" k* N+ E5 q9 `$ P0 H0 P) e; l  q
    References" f( Z& C& C* r! |3 M' h- [
[1]  Wang yuan,TANTAN SUSHU,Shanghai, Shanghai Education Publishing House(1983),42.
( Q( h; \8 a2 Q! w5 d  h( j5 u[2]  U﹒Dudley,Elementary number theory, Shanghai, Shanghai Science and Technology Press,(1980),195.6 A1 ]3 R2 @" X) g+ p% S6 _
[3] Pan Chengdong,Pan Chengbiao,Goldbach conjecture,Beijing,Science Publishing house,(1984),1.# U9 L% \4 [" j: a# `, c

( W. O9 o; ~6 {2 L9 L7 r* m. J
作者: 数学1+1    时间: 2013-12-9 11:45
阅读本帖需具备阅读LATEX文件的知识，作者有一word文件上传，有兴趣的读者可下载阅读。
作者: 数学1+1    时间: 2013-12-9 13:04
                  Goldbach’s problem
                    Su Xiaoguang
摘要：哥德巴赫问题是解析数论的一个重要问题。作者研究了
A= \bigcup_{i=0}^{\infty }A_{i},A_{i}=\left \{ i+0,i+1,i+2,\cdots \right \}\Rightarrow B(x,N)=\sum_{N\leqslant x,B(N)\neq 0}1,B(N)=\sum_{n+m=N,0\leq n,m\leq N}1.
C=\bigcup_{i=0}^{\infty }C_{i},C_{i}=\left \{ p_{i}+p_{0},p_{i}+p_{1},p_{i} +p_{i},\cdots \right \}\wedge N> 800000\Rightarrow M(x)=\sum_{N\leq x,D(N)\neq 0}1
Deduced
D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge
1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+log\frac{2}{N}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
Key words: Germany，Goldbach，even number, Odd number ，prime number,
MR (2000) theme classification: 11 P32
Email:suxiaoguong@foxmail. com
§ 1  引言9 O/ ~, t  w+ {2 ]. u7 b
      1742年，德国数学家Christian Goldbach提出了关于正整数和素数之间关系的两个推测，用分析的语言表述为:
（A）对于偶数N
N\geq 6\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}Is a prime number}1>0
(B)  对于奇数N
N\geq 9\Rightarrow T(N)=\sum_{p_{1}+p_{2}+p_{3}=N.p_{1},p_{2},p_{3}\geq 3}1>0
        这就是著名的哥德巴赫猜想，如果命题（A）真，那么命题（B）真，所以，只要我们证明命题（A），立即推出猜想（B）是正确的
         
§2相关集的构造
+ e1 |* g& ^/ w7 g' x3 T# T1 |; F, \0 Z A_{0}=\left \{ 0+0,0+1,0+2,\cdots  \right \}
A_{1}=\left \{ 1+0,1+1,1+2,\cdots  \right \}
A_{2}=\left \{ 2+0,2+1,2+2,\cdots  \right \}
\cdots
A=\bigcup_{i=0}^{\infty }A_{i}\Rightarrow B(N)=\sum_{n_{1}+n_{2}=N.0\leq n _{1},n_{2}\leq N}1,B(x,N)=\sum_{N\leq x,B(N)\neq 0}1      （1）
p_{0}=2,p_{1}=3,p_{2}=5,\cdots C_{0}=\left \{ p_{0} +p_{0},p_{0}+p_{1},p_{0}+p_{2},\cdots \right \} C_{1}=\left \{ p_{1} +p_{0},p_{1}+p_{1},p_{1}+p_{2},\cdots \right \} C_{2}=\left \{ p_{2} +p_{0},p_{2}+p_{1},p_{2}+p_{2},\cdots \right \} C=\bigcup_{i=0}^{\infty }C_{i}\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N.p_{1},p_{2}\geq 3}1\wedge M(x)=\sum_{N\leq x,D(N)\neq 0}  （2）      
  §3    预备定理. s, j3 S" n! @% P+ I0 X7 X, \" f
定理 1
M_{i}=(x_{1}^{(i)},x_{2}^{(i)},\cdots ,x_{i}^{(i)},\cdots ),Is a countable set\Rightarrow M=\bigcup_{i=1}^{N}M_{i},Is a countable set
  .Proof: Suppose M_{1},M_{2},\cdots ,M_{N},Is a countable set, M=\bigcup_{i=1}^{N}M_{i}
\because M_{1}:x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},\cdots ,x_{i}^{(1)},\cdots
M_{2}:x_{1}^{(2)},x_{2}^{(2)},x_{3}^{(2)},\cdots ,x_{i}^{(2)},\cdots
M_{N}:x_{1}^{(N)},x_{2}^{(N)},x_{3}^{(N)},\cdots ,x_{i}^{(N)},\cdots
\cdots
\therefore M:x_{1}^{(1)},x_{1}^{(2)},\cdots ,x_{1}^{(N)},x_{2}^{(1)},x_{2}^{(2)},\cdots ,x_{2}^{(N)},\cdots Countable
     定理2 (素数定理)
\pi (x)\sim \frac{x}{logx}^{\left [ 1 \right ]}
      定理3  对于偶数x
x>800000\wedge M_{1}=minM(x)\Rightarrow M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x) -1\right ]
证明 根据定理1, (1)  
  \because A_{i},A_{j} Countable,
\therefore A Countable\wedge B(x,N)=\frac{1}{2}(N+1)(N+2) $ n2 u6 B# J& a
类似地，根据定理1, (2), C可数  3 e; V( s; S9 i$ E& z2 V
设      M_{1}(x)=minM(x)
根据（2），那么我们有.: H/ e) x3 U0 L& L6 |( g$ }% t
M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x)-1 \right ]
定理4  对于偶数x
x>800000\wedge M_{2}(x)=maxM(x)\Rightarrow M_{2}(x)=4\pi (\frac{x}{2})\pi (x)-2\pi ^ {2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)                        （3）
证明: 根据（2），那么我们有
5 [' U# Z% |8 A/ W$ t$ R  M(x)=\sum_{N\leq x,D(N)\neq 0}1< \sum_{3\leq p_{1},p_{2}\leq \frac{x}{2}}1+\sum_{3\leq p_{1}\leq \frac{x}{2},\frac{x}{2}< p_{2}< x}+\sum_{\frac{x}{2}<p _{1}< x,3\leq p_{2}\leq \frac{x}{2}}
     设   M_{2}(x)=maxM(x)
\therefore M_{2}(x)
=\frac{1}{2}\cdot 2\pi (x)\left [ 2\pi (\frac{x}{2})-1 \right ]-\frac{1}{2}\left [ 2\pi (\frac{x}{2})-\pi (x) \right ]\left [ 2\pi (\frac{x}{2})-\pi (x) +1\right ]\cdot 2
=4\pi (x)\pi (\frac{x}{2})-2\pi ^{2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)
§4 Goldbach's problem 终结
5 y1 z9 ?# I/ L/ P3 x- U& M0 }定理 5  对于偶数N
N> 800000\Rightarrow D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
    证明: 根据定理22 S) s0 Z  l' a: s& j
N> 800000\Rightarrow \alpha \frac{N}{logN}\leq \pi (N)\leq \beta \frac{N}{logN}      (4)
让  c_{1}=min(\alpha ,\beta ),
根据定理3，然后我们有; a- v. G5 t; ]7 E& T
      D_{1}(N)=M_{1}(N)-M_{1}(N-2)
显然: C6 H6 S* @) N8 ^3 |# J1 r" J; |$ V2 z
       D(N)\geq D_{1}(N)
\because log(1+x)=\int_{0}^{x}\frac{dt}{1+t}=x-\int_{0}^{x}\frac{t}{1+t}dt
\because x\geq-\frac{1}{2} \Rightarrow log(1+x)=x+o(x^{2})           (5)
\therefore D_{1}(N)=2c_{1}^{2}(1-\frac{1}{logN})\frac{N}{log^{2}(n-2)}+o(1)
N\rightarrow \infty ,o(1)\rightarrow 0\Rightarrow
D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
定理6  对于偶数N
N> 800000\Rightarrow D(N)\leq
5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
证明: 根据(4)
让  c_{2}=max(\alpha ,\beta )
根据定理4，然后我们有9 M& W2 r( v! N
       D_{2}(N)=M_{2}(N)-M_{2}(N-2)
\because D(N)\leq D_{2}(N)
根据(5)，那么我们有- p, ^- W+ Z* e: p% h
       D(N)\leq
5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
定理7 (Goldbach Theorem)  
对于偶数N
N\geq 6\Rightarrow D(N)= \sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\geq 1
证明: 根Shen Mok Kong 的验证6 H0 v) I! s, G6 a" F) r
      6\leq N\leq 3.3\times 1000000^{\left [ 3 \right ]}\Rightarrow D(N)\geq 1
根据定理5, 定理 6, 然后我们有
$ b0 ~+ E; w9 k0 {0 W: F& V/ a      N> 800000\Rightarrow 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
\therefore N\geq 6\Rightarrow D(N)\geq 1
引理1 对于奇数N
N\geq 9\Rightarrow
T(N)=\sum_{p_{1}+p_{2}+p_{3}=N,p_{1},p_{2},p_{3}\geq 3}1\geq 1
证明: 让 n\geq 4
\because 2n+1=2(n-1)+3
根据定理7，然后我们有! _4 u5 ^/ z3 U' e5 q
      N\geq 9\Rightarrow T(N)\geq 1
    References
[1]  Wang yuan,TANTAN SUSHU,Shanghai, Shanghai Education Publishing House(1983),42.
[2]  U﹒Dudley,Elementary number theory, Shanghai, Shanghai Science and Technology Press,(1980),195.
[3] Pan Chengdong,Pan Chengbiao,Goldbach conjecture,Beijing,Science Publishing house,(1984),1.
作者: 数学1+1    时间: 2013-12-9 13:56
我国数学家华罗庚，闵嗣鹤均对M(x)的下界做过研究，潘承洞，潘承彪对D(N)的上界做过研究，他们留下了遗憾，也留下了经验.
作者: 数学1+1    时间: 2013-12-13 15:32
D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge
7 k+ h4 U$ d8 J4 a2 t; }1.83150(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}\leq D(N)\leq 4.36166\left [ 1+log\frac{2}{N}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
: e) c% J* [! e# h* }+ G3 H, H
作者: 数学1+1    时间: 2013-12-14 11:37
若N＞800000，6 i) {5 U2 U0 Y! A2 o6 P  b* R# c  X6 s
则   1.83150（1－1／logN）[N／log^2(N-2)]≤D(N) ≤4.36166[1+2／logN +o(1)]×" u' L. `8 M, i
N／{log[(N-2)／2]log(N-2)}( ^6 p3 X: T7 d% X# F/ {+ U9 V
这就是哥德巴赫公式，有兴趣的读者不妨检测一下。
作者: 1300611016    时间: 2013-12-28 18:21
楼主的帖子怎么样？赶紧试试这里的快速回复给楼主点评论
作者: 1300611016    时间: 2013-12-28 18:31
本帖最后由 1300611016 于 2014-1-4 09:08 编辑
( ~  I0 H& }$ ?/ T9 r
: h# z2 ^) x$ [  R2 k% s太烦，可以用一个简明的形式，如·同偶质数对·形式展开详细见http://www.madio.net/thread-202136-1-1.html
7 b1 u. l# j' c" P. T1 e一般的用简明浅显的形式表述更容易推广，如能用初等数学表述这一问题，可以尝试一下。但不妨碍专业研究。
作者: 数学1+1    时间: 2013-12-29 21:30
1300611016:
+ [* a1 Q" o. u9 `( R      你如果能多读几本数论方面的著作，你就能理解哥德巴赫猜想，理解哥德巴赫猜想中D(N)表示什么？也就不感觉烦了。



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