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Goldbach’s problem

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    发表于 2013-12-6 12:27 |只看该作者 |倒序浏览
    |招呼Ta 关注Ta
    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)}
    % l$ M: y, _, j" ` Key words: Germany,Goldbach,even number, Odd number ,prime number, MR (2000) theme classification: 11 P32 Email:suxiaoguong@foxmail. com
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                      Goldbach’s problem (pdf)
                           Su Xiaoguang$ [% W6 a+ u) s3 x, ^
         
    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)}
    1 u' H0 B7 j0 U/ I& e% m/ }) d
    Key words: Germany,Goldbach,even number, Odd number ,prime number,
    MR (2000) theme classification: 11 P32
    Email:suxiaoguong@foxmail. com
    回复

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                      Goldbach’s problem
    - c/ J4 {. B( O/ W1 G8 [% u/ L5 v                    Su Xiaoguang
    % W& z! e9 Y+ k( ?( EAbstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the:
    $ h! T- q0 x2 D& L. C; s& p2 o1 [( v0 I0 y0 d0 v2 Y7 l; E' N, N
    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.
    & {9 s% E) O. [2 UC=\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. S6 X6 u* C/ X$ L3 x1 G
    Deduced
    4 U; S: J, C4 l# J& ]3 UD(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge
    * W) Z. ]( z( K1.8432(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)}
    3 D1 d5 D9 ]+ k! A: Z7 M  A4 C9 L' Y5 P/ h# P- \, A" c
    Key words: Germany,Goldbach,even number, Odd number ,prime number, % m9 o: ?& O* Z& D8 V' }- V/ @
    MR (2000) theme classification: 11 P32 * C: j* A/ o- G1 ?( m5 R4 [8 t
    Email:suxiaoguong@foxmail. com
    & A: v3 o" E  S' [8 S! \§ 1 Introduction5 g" ~# G0 }. N1 T% ]+ h
              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:
    % @7 j1 K8 k: q* C2 b* y" R# b& L(A)For even number N( ]3 X9 y: d* {' q

    6 P$ h% L/ O5 ~* QN\geq 6\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}Is a prime number}1>0
    / l5 w* f, f& t* w
    ; h1 a& H7 W3 @. ~( x' M(B)  For odd number N
    1 R( V( s+ u# y3 A) A$ W& d9 S1 T! b
    6 o& G5 F: x! b- a1 R; a" ]" |) tN\geq 9\Rightarrow T(N)=\sum_{p_{1}+p_{2}+p_{3}=N.p_{1},p_{2},p_{3}\geq 3}1>0( i6 g; Z3 R4 }" u/ u" j

    ' ^, x- [2 l3 o; r& IThis 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 correct
    1 @# m0 g( C2 i4 J8 J" J: i, e          % k. h5 d$ q* ]4 H# S6 o0 ~. _
    §2 Correlation set constructor
    7 r" y" U5 N, E$ G1 ?: d1 |$ b$ _6 MA_{0}=\left \{ 0+0,0+1,0+2,\cdots  \right \}2 b9 e  ]7 U0 n( ]
    A_{1}=\left \{ 1+0,1+1,1+2,\cdots  \right \}
    0 g1 J9 T* ]# Y" k A_{2}=\left \{ 2+0,2+1,2+2,\cdots  \right \}
    : |% b- c6 x- ]8 E6 B, p, Y \cdots
    . w' Z* q1 T! a( }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)
    " k( o4 a' _( H$ `- _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)      
    . V1 s0 I' x6 b, j  §3    Ready  Theorem
    9 I# q4 h8 j/ Y' u4 e, NTheorem 1
    ) I- A3 K; g  D# WM_{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
    2 B) @+ e. S% X  .Proof: Suppose M_{1},M_{2},\cdots ,M_{N},Is a countable set, M=\bigcup_{i=1}^{N}M_{i}
    7 p7 L$ J3 E) Y# S7 C7 g4 g4 X\because M_{1}:x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},\cdots ,x_{i}^{(1)},\cdots
    " J- h5 A8 S; g  s( }2 T: f8 \M_{2}:x_{1}^{(2)},x_{2}^{(2)},x_{3}^{(2)},\cdots ,x_{i}^{(2)},\cdots% @: S5 d) l; Y% G- Z# Q
    M_{N}:x_{1}^{(N)},x_{2}^{(N)},x_{3}^{(N)},\cdots ,x_{i}^{(N)},\cdots
    0 W3 E/ Q& N$ q( Q; _' F\cdots
    ! m' e$ }8 P; U& ]6 `\therefore M:x_{1}^{(1)},x_{1}^{(2)},\cdots ,x_{1}^{(N)},x_{2}^{(1)},x_{2}^{(2)},\cdots ,x_{2}^{(N)},\cdots Countable
    : v  l# R3 E4 e; a, Y& ^     Theorem 2 (Prime number theorem)+ J/ h5 \8 `+ d7 ]
    # k! g: Y) C% X, G6 _- N- `
    \pi (x)\sim \frac{x}{logx}^{\left [ 1 \right ]}- E& Q5 o3 N5 P. W
         Theorem 3  For even number x
    $ G6 l8 O. }( C, i* bx>800000\wedge M_{1}=minM(x)\Rightarrow M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x) -1\right ]
    9 k) `& u/ v9 e8 T* LProof: According to Theorem 1, (1)  : m( h6 T1 ?/ p5 _: G0 s( l
      \because A_{i},A_{j} Countable,
    7 L1 s  c3 E# P9 h. b9 ? \therefore A Countable\wedge B(x,N)=\frac{1}{2}(N+1)(N+2) + o3 Q3 W9 {! c) W
    Similarly, according to Theorem 1, (2), C countable
    $ D# o4 a) B$ O* \. Q Suppose1 k4 o' G9 ^# a# |  O8 E
          M_{1}(x)=minM(x)
    - W! j) [+ A' y9 h  Y5 _according to (2), Then we have) F1 H+ w8 ~8 M/ g
    . M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x)-1 \right ]
    7 Q9 P2 g. u  h0 d- n( | Theorem 4  For even number x' V/ ~8 F& u6 I" |' d4 Y: O
    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)
    : J% G: g2 x5 |$ s' A" Q( PProof: According to (2),Then we have. ]6 L. L: _9 m
    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}}! w; Y6 n1 ~- M$ E" \! P' F% z
         Suppose! K9 d1 i0 O- J, U7 l$ W: Z6 [
          M_{2}(x)=maxM(x): s- H3 j  v# M& o+ w7 M
    \therefore M_{2}(x)
    " ?7 O+ T7 {5 |% u+ `, 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
    + T' }2 Z- Z2 u=4\pi (x)\pi (\frac{x}{2})-2\pi ^{2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)
    ( |; G  Q: x# W3 J§4 Goldbach's problem end
    " K% V+ S) r1 |" WTheorem 5  For evem number N
    1 D- d3 d7 S* z& Y4 z; ON> 800000\Rightarrow D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
    ) d5 ^- L/ b4 l     Proof: According to Theorem 2
    ! o& u4 g& O" s4 y" @: c; WN> 800000\Rightarrow \alpha \frac{N}{logN}\leq \pi (N)\leq \beta \frac{N}{logN}      (4). ^( l% x7 J, o+ h( R  W/ z
    Let   c_{1}=min(\alpha ,\beta ),! ]  }/ Y1 o% v" Z; Y$ V) P
    According to Theorem 3,Then we have
    $ {# A, M7 m; ?" kD_{1}(N)=M_{1}(N)-M_{1}(N-2)$ F2 |8 [2 q& K% [
    Clear* g1 l# X3 F7 ]; l
    D(N)\geq D_{1}(N)* I! h4 v7 D4 c) q- e2 U
    \because log(1+x)=\int_{0}^{x}\frac{dt}{1+t}=x-\int_{0}^{x}\frac{t}{1+t}dt
    5 P! U! |) b0 v. L\because x\geq-\frac{1}{2} \Rightarrow log(1+x)=x+o(x^{2})           (5)+ i3 B! Q4 o' w. y! P9 T4 ]
    \therefore D_{1}(N)=2c_{1}^{2}(1-\frac{1}{logN})\frac{N}{log^{2}(n-2)}+o(1)
    * J6 l1 o5 V5 Y9 x, @+ ^" @N\rightarrow \infty ,o(1)\rightarrow 0\Rightarrow
    # C; o  M  k5 D% N3 @. cD(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
    - S( s/ i: E3 {: X Theorem 6  For evem number N
    8 W2 s  v1 {+ q- h+ nN> 800000\Rightarrow D(N)\leq " ?& w. Z! `. u9 q% ^- ?' y" i) ~
    5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}, S0 {5 X4 ^& V/ _- i" o, b
    Proof : According to (4)
    1 V# M# f1 k" l: j$ P7 J* {" pLet  c_{2}=max(\alpha ,\beta )
    9 Y5 R6 V7 q7 h" {6 q! B' X/ pAccording to Theorem 4,Then we have, u/ P- q. S% i
    D_{2}(N)=M_{2}(N)-M_{2}(N-2)
    8 C9 w' R4 f9 N/ r' S\because D(N)\leq D_{2}(N)% z. J4 `# l0 p7 e* I
    According to (5), Then we have
    ( l( t, b7 e" Q; ?/ `5 P% w6 ID(N)\leq
    1 ]) c* d  h8 p8 \8 J9 \5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
    % i; E1 w6 r. }. [7 t/ [3 L5 wTheorem 7 (Goldbach Theorem)  ) i* u5 j8 O9 \+ C, Q9 Z
    For evem number N7 {# [$ y7 ?) e& g, R: P
    N\geq 6\Rightarrow D(N)= \sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\geq 1
      v5 U3 f) h2 @- Y1 j, k* ~Proof : According to Shen Mok Kong verification
    % ~- e' e# ^0 C8 n8 y1 u, ^4 V6\leq N\leq 3.3\times 1000000^{\left [ 3 \right ]}\Rightarrow D(N)\geq 1+ U) X  ~9 b' m8 }
    According to Theorem 5, Theorem 6, Then we have
    & X$ i3 J1 d5 v1 R  aN> 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)}
    / h- S% r% _4 m7 }+ h  p9 O  ?\therefore N\geq 6\Rightarrow D(N)\geq 1
    & s: Q! d" K2 B6 BLemma 1 For odd number N
    ; r; d+ i; P0 Q5 J. t8 A1 [N\geq 9\Rightarrow 3 d) ]- p8 w0 S8 W
    T(N)=\sum_{p_{1}+p_{2}+p_{3}=N,p_{1},p_{2},p_{3}\geq 3}1\geq 1/ a2 _- k" w7 ]7 ^
    Proof et  n\geq 44 \: y2 {& q8 B
    \because 2n+1=2(n-1)+34 W+ }! c* B2 t1 G( r
    According to Theorem 7,  Then we have. ]3 r) g1 @3 P9 D1 f0 d0 I$ }
    N\geq 9\Rightarrow T(N)\geq 1
    + K5 t9 f! I% j6 O5 y
    & w/ A% b0 ~" ~- \& M: b( U0 X: O1 e5 a9 d0 @: W$ B+ k. h0 w
        References  m# o& a/ a2 L2 W% }+ h- n
    [1]  Wang yuan,TANTAN SUSHU,Shanghai, Shanghai Education Publishing House(1983),42., ~3 ^# }9 I, Y# U, O. W  v9 H
    [2]  U﹒Dudley,Elementary number theory, Shanghai, Shanghai Science and Technology Press,(1980),195.# \! H' a1 @/ F
    [3] Pan Chengdong,Pan Chengbiao,Goldbach conjecture,Beijing,Science Publishing house,(1984),1.+ {  f- m! b. T: n* m

    % ]; G5 ^6 Q# [; W( c+ x6 @6 k
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    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。
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                      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
    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
    5 }2 g+ o2 H, Q  e* H* F      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
    % r9 V5 w- \, Y; b1 ^: H
    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* q3 e4 s8 C) N
    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) ; ~/ E$ G7 n, p( d: ^( r! K0 K7 ]4 @
    类似地,根据定理1,
    (2), C可数  
    5 r* f* g" K5 u  ^% |* e% x  {# H
    设      M_{1}(x)=minM(x)
    根据(2),那么我们有.: {* q  G* v+ t$ ^7 i* T' m
    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
    8 [4 s* K) d3 F5 D; z8 c, 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
    , I8 e  l) H$ L; q! x$ B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2) @! O4 L* o3 z& m) [
    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
      }: n9 B) b6 R( e9 J# ~% _      D_{1}(N)=M_{1}(N)-M_{1}(N-2)
    显然
    * S" J/ A6 x5 `% ?       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
    2 {5 c8 b4 p! O2 ]" U8 Z       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
    ; K& w3 x- {( E+ X( p8 U       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4 G1 O9 W: w0 B% v7 E' a2 o
          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" ]! [3 |8 _8 i: \. f
          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2 y' {' g7 ^9 F5 T4 L2 s0 e( 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.
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    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.
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    D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge 7 W- S6 e. f8 Y" w+ `
    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)}
    % G& l9 ]6 m* ?# U5 F
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    若N>800000,
    7 I! R, u- V/ X3 `8 ]" E. p. H7 U则   1.83150(1-1/logN)[N/log^2(N-2)]≤D(N) ≤4.36166[1+2/logN +o(1)]×) }* D/ w' A7 @: L- _9 }
    N/{log[(N-2)/2]log(N-2)}0 j- u3 \. J& V) u1 I
    这就是哥德巴赫公式,有兴趣的读者不妨检测一下。
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    本帖最后由 1300611016 于 2014-1-4 09:08 编辑
    ' H5 l' X6 J( `) Y" `4 N3 R2 Z! L; ]( i) c
    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
    * ], A7 U9 ]/ _5 ?! X4 S: P! a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。
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