QQ登录

只需要一步,快速开始

 注册地址  找回密码
查看: 2848|回复: 10
打印 上一主题 下一主题

Goldbach’s problem

[复制链接]
字体大小: 正常 放大
数学1+1        

23

主题

14

听众

2532

积分

升级  17.73%

  • TA的每日心情
    开心
    2025-4-20 11:19
  • 签到天数: 842 天

    [LV.10]以坛为家III

    新人进步奖

    跳转到指定楼层
    1#
    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)}
      [2 N) V8 k  b Key words: Germany,Goldbach,even number, Odd number ,prime number, MR (2000) theme classification: 11 P32 Email:suxiaoguong@foxmail. com+ O" u% I" M" d2 P- W, w. T
    zan
    转播转播0 分享淘帖0 分享分享0 收藏收藏0 支持支持0 反对反对0 微信微信
    数学1+1        

    23

    主题

    14

    听众

    2532

    积分

    升级  17.73%

  • TA的每日心情
    开心
    2025-4-20 11:19
  • 签到天数: 842 天

    [LV.10]以坛为家III

    新人进步奖

                      Goldbach’s problem (pdf)
                           Su Xiaoguang* d/ k, Q/ I- p7 y
         
    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)}
    - l$ }* a+ f+ l
    Key words: Germany,Goldbach,even number, Odd number ,prime number,
    MR (2000) theme classification: 11 P32
    Email:suxiaoguong@foxmail. com
    回复

    使用道具 举报

    数学1+1        

    23

    主题

    14

    听众

    2532

    积分

    升级  17.73%

  • TA的每日心情
    开心
    2025-4-20 11:19
  • 签到天数: 842 天

    [LV.10]以坛为家III

    新人进步奖

                      Goldbach’s problem
    * l, z% u+ f" A& Y+ r7 P                    Su Xiaoguang
    , m# `! P; [3 @) C4 Y" |$ bAbstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the:
    5 W! X$ f& `0 ]' Y! u) R+ @
    0 v' s7 M7 J9 k7 u6 PA= \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 k; Z; r+ K2 m# i7 q7 HC=\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! q& U9 z. U8 l7 ]7 z- C. X: W
    Deduced, n) W0 [( a# Q6 L; D1 `" v) q3 u
    D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge 5 n1 N# C; b; x& A  _
    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)}9 x& D& Y, c6 @9 c" y+ S
      J$ B6 Y" z& L- B& N+ I9 |
    Key words: Germany,Goldbach,even number, Odd number ,prime number, % h% K- K' p, Y1 H0 G
    MR (2000) theme classification: 11 P32
    9 O( z" o% l3 r2 O- mEmail:suxiaoguong@foxmail. com
    ( h, W9 W9 g' I& Y% _; T2 w§ 1 Introduction
    " `- F9 ?' c3 X$ h& l: Z1 \0 C* 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:
    , s) k6 H( S2 U: m: E(A)For even number N) d6 E; ?0 F/ G- N) X
    2 Q- P( o+ z- g9 q+ B* p/ @# i
    N\geq 6\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}Is a prime number}1>0
    1 ^/ h0 p1 ~( n9 j
    & _3 a, w7 j  N- ](B)  For odd number N, R0 B" T+ `( f  B2 N  H

    1 t+ R6 ~9 M. k9 `7 \) mN\geq 9\Rightarrow T(N)=\sum_{p_{1}+p_{2}+p_{3}=N.p_{1},p_{2},p_{3}\geq 3}1>06 K" J5 ~+ M/ ~$ H9 w) l9 e8 p
    % H% }3 W) v* A( z8 H2 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 correct9 p$ W! F- [$ H
              6 [7 J4 Y' g8 q. @, B9 _( Z$ s
    §2 Correlation set constructor. `( h% u" R2 `+ R& J
    A_{0}=\left \{ 0+0,0+1,0+2,\cdots  \right \}
    ) Z$ A" J, P/ j A_{1}=\left \{ 1+0,1+1,1+2,\cdots  \right \}0 @& o' U& c. `6 m9 P
    A_{2}=\left \{ 2+0,2+1,2+2,\cdots  \right \}0 w5 n; Y$ `( p! {2 z
    \cdots' K& c$ b& C7 O$ L' M: @
    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)8 x5 l1 C: h; k6 A$ `
    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 U0 ?, Q$ R0 W$ I# l0 e9 Y
      §3    Ready  Theorem
    * [% C# j. S; c$ f% H2 WTheorem 1
    . e: s' z1 P. o5 _: SM_{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
    - a, _+ ^6 @) _! M. ?  .Proof: Suppose M_{1},M_{2},\cdots ,M_{N},Is a countable set, M=\bigcup_{i=1}^{N}M_{i}$ n4 r/ ]) ?9 k- w9 j6 ^0 {
    \because M_{1}:x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},\cdots ,x_{i}^{(1)},\cdots( m* `- N5 O7 s5 _  @& \+ Y
    M_{2}:x_{1}^{(2)},x_{2}^{(2)},x_{3}^{(2)},\cdots ,x_{i}^{(2)},\cdots4 D) Q! \9 o- Y. O
    M_{N}:x_{1}^{(N)},x_{2}^{(N)},x_{3}^{(N)},\cdots ,x_{i}^{(N)},\cdots
    ; w) v1 \7 z* X: J\cdots* g9 l/ y; `- w6 u& K$ G9 f
    \therefore M:x_{1}^{(1)},x_{1}^{(2)},\cdots ,x_{1}^{(N)},x_{2}^{(1)},x_{2}^{(2)},\cdots ,x_{2}^{(N)},\cdots Countable
    + C* D1 `8 [+ W) D     Theorem 2 (Prime number theorem)1 T* r' c5 s) J

    1 V5 `& v5 @' X' i\pi (x)\sim \frac{x}{logx}^{\left [ 1 \right ]}, n0 B7 s9 E; s- O' L
         Theorem 3  For even number x9 H5 @  z* Z/ O2 e( j
    x>800000\wedge M_{1}=minM(x)\Rightarrow M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x) -1\right ]
    % ^! F) l. j9 b- C7 ^( F! a3 X4 B9 aProof: According to Theorem 1, (1)  + {$ X$ c( o" q' ?, H# p' [
      \because A_{i},A_{j} Countable,
    ! Q& C5 P2 e/ y: y5 i# L \therefore A Countable\wedge B(x,N)=\frac{1}{2}(N+1)(N+2)
    9 ]% p. k$ N$ i' R& N5 ?$ BSimilarly, according to Theorem 1, (2), C countable- s0 v# p; q6 ?/ D/ L
    Suppose
    ' \1 d0 V$ y0 Q$ s4 i      M_{1}(x)=minM(x)9 i4 X, n# H& e' N, C$ n7 S1 f
    according to (2), Then we have; m( L2 }6 z; |1 R' I
    . M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x)-1 \right ]7 m- W* s4 [7 q" C; c* ~  l) `
    Theorem 4  For even number x# c& {# h( f0 N2 ^. M
    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)
    . s2 e0 P% l0 z# J* X5 @Proof: According to (2),Then we have. I+ ?) n$ A( t0 Z- G
    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}}5 t) N5 q: u3 w2 @& R7 n% M
         Suppose
    7 K; }& j6 Y. Z- {/ U      M_{2}(x)=maxM(x)
    : H5 Z; S; T" m- r  d\therefore M_{2}(x)
    - K6 \" y, ?" D0 S/ Q=\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& x0 M; f5 l# J- s- ?
    =4\pi (x)\pi (\frac{x}{2})-2\pi ^{2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)
    1 P6 ]+ y6 \# d" d7 P§4 Goldbach's problem end
    8 ^# A, K1 N$ o0 NTheorem 5  For evem number N& y' h, Z& p7 i, N0 ~! U
    N> 800000\Rightarrow D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
      W; m9 \7 a, l8 V0 ]+ F     Proof: According to Theorem 2
    + C+ F1 k% d" \7 E# m$ k$ PN> 800000\Rightarrow \alpha \frac{N}{logN}\leq \pi (N)\leq \beta \frac{N}{logN}      (4)' Y3 j, C" q( A0 N- [, J& p
    Let   c_{1}=min(\alpha ,\beta ),6 n4 N8 P3 q. a; p8 v
    According to Theorem 3,Then we have4 b  t1 k& ]: f% q* y0 q% X& A
    D_{1}(N)=M_{1}(N)-M_{1}(N-2)
    & F5 M8 c. D+ GClear+ L  S: t8 f% Z2 m$ i6 U1 _
    D(N)\geq D_{1}(N)
    3 e2 o( R9 x2 a( u+ ~) Y4 K\because log(1+x)=\int_{0}^{x}\frac{dt}{1+t}=x-\int_{0}^{x}\frac{t}{1+t}dt+ }5 B0 W, \, c. f  I4 \3 N
    \because x\geq-\frac{1}{2} \Rightarrow log(1+x)=x+o(x^{2})           (5)
      G, E# O4 K% ~4 o\therefore D_{1}(N)=2c_{1}^{2}(1-\frac{1}{logN})\frac{N}{log^{2}(n-2)}+o(1)
    9 D0 L2 _  I; I4 V8 ]N\rightarrow \infty ,o(1)\rightarrow 0\Rightarrow
      ]0 t6 {* \8 R$ V: aD(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
    , {1 @, S$ C1 `5 B# q Theorem 6  For evem number N
    " B- f$ v& P( b1 yN> 800000\Rightarrow D(N)\leq
    , s4 G% ]' A4 S+ y0 v, \0 x5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}% ]1 r8 m3 K8 U1 o
    Proof : According to (4)" q1 X0 J/ J! H6 V$ t  m
    Let  c_{2}=max(\alpha ,\beta )
    3 O6 a- i( d2 t' q8 X/ ZAccording to Theorem 4,Then we have) H1 @  O. [5 M( ]8 ?. a/ \
    D_{2}(N)=M_{2}(N)-M_{2}(N-2)- k, h! M  W; P9 n8 N- g0 t
    \because D(N)\leq D_{2}(N)
    + k6 y, n- U) K* d8 XAccording to (5), Then we have. X5 h( `# v( T  G
    D(N)\leq
    2 K2 M& i, W4 n" p5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}
    # m" c* ~5 X) w% Z; U! o( }Theorem 7 (Goldbach Theorem)  : e* F. |4 ]$ W
    For evem number N
    ' t1 |' x$ c. u, j" i/ sN\geq 6\Rightarrow D(N)= \sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\geq 1
    # q' `0 |0 s$ XProof : According to Shen Mok Kong verification
      @. \/ h" T! [3 B& ^3 F6\leq N\leq 3.3\times 1000000^{\left [ 3 \right ]}\Rightarrow D(N)\geq 1
    & x( V+ x" e  V% S( {According to Theorem 5, Theorem 6, Then we have
    * p7 f1 _5 L$ i# c1 N9 j2 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)}  N/ ]. T5 I; {# T; h
    \therefore N\geq 6\Rightarrow D(N)\geq 1; R- _% ?; F+ N2 H) j
    Lemma 1 For odd number N& d- ?7 [, q$ W2 r! {, f) u
    N\geq 9\Rightarrow
    1 o6 i; U) Y- DT(N)=\sum_{p_{1}+p_{2}+p_{3}=N,p_{1},p_{2},p_{3}\geq 3}1\geq 16 a' P7 P/ w2 ?# ~
    Proof et  n\geq 4
    & @. M1 R# K2 _# t3 S" r\because 2n+1=2(n-1)+3
    # I/ }6 A5 ?" W$ qAccording to Theorem 7,  Then we have9 K0 t4 s' H) Z5 R8 y9 Y
    N\geq 9\Rightarrow T(N)\geq 1
    & t( F/ `# \& h5 t3 W! o5 a' F$ U! O2 F% Y; [! z/ g

    ; A* \- A0 ]1 k; v    References
    0 h8 }- H) ]8 p5 s: a* w4 G3 P7 Z! L[1]  Wang yuan,TANTAN SUSHU,Shanghai, Shanghai Education Publishing House(1983),42.$ X! l) K6 B* @1 @' L" T
    [2]  U﹒Dudley,Elementary number theory, Shanghai, Shanghai Science and Technology Press,(1980),195.' O7 F$ y$ y, s8 C, y
    [3] Pan Chengdong,Pan Chengbiao,Goldbach conjecture,Beijing,Science Publishing house,(1984),1.0 `* v& l6 c4 ~' h' M2 s1 z3 C
    1 M+ h% s1 S) M* ~7 s; r- H( B8 s
    回复

    使用道具 举报

    数学1+1        

    23

    主题

    14

    听众

    2532

    积分

    升级  17.73%

  • TA的每日心情
    开心
    2025-4-20 11:19
  • 签到天数: 842 天

    [LV.10]以坛为家III

    新人进步奖

    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。
    回复

    使用道具 举报

    数学1+1        

    23

    主题

    14

    听众

    2532

    积分

    升级  17.73%

  • TA的每日心情
    开心
    2025-4-20 11:19
  • 签到天数: 842 天

    [LV.10]以坛为家III

    新人进步奖

                      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
    ; X+ G) z7 M5 m# d      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
    2 k# v/ _; I$ X" h1 |
    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, A5 V0 w  q) O3 F; x& _
    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) + V8 v+ D# g& y0 Y) V( N' p. o4 `
    类似地,根据定理1,
    (2), C可数  8 |4 Y: Y6 S) o# f) y
    设      M_{1}(x)=minM(x)
    根据(2),那么我们有.
    3 j% s/ I0 J  o% E- L' ^7 N7 i 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2 l8 o8 i* L$ N8 b9 f: }- U; K
      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9 ~% f; s: S2 m4 E1 V( G
    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
    ' m9 F6 @- s& t% H& M7 G' ? 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
    7 b, L6 j6 r6 B" m' i: h: p      D_{1}(N)=M_{1}(N)-M_{1}(N-2)
    显然  s3 \5 u# C3 p, G6 x5 ~. i
           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& a( W+ l* I0 [/ ?
           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; @0 v! c: w" f
           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+ b3 N8 I5 M! W
          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8 f# O9 R2 M9 e
          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. j. X) z# {. y4 A* V9 }- ]& }
          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.
    回复

    使用道具 举报

    数学1+1        

    23

    主题

    14

    听众

    2532

    积分

    升级  17.73%

  • TA的每日心情
    开心
    2025-4-20 11:19
  • 签到天数: 842 天

    [LV.10]以坛为家III

    新人进步奖

    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.
    回复

    使用道具 举报

    数学1+1        

    23

    主题

    14

    听众

    2532

    积分

    升级  17.73%

  • TA的每日心情
    开心
    2025-4-20 11:19
  • 签到天数: 842 天

    [LV.10]以坛为家III

    新人进步奖

    D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge
    , Q8 Z) {3 @$ T& {/ S+ G1.83150(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)}
    ) {  ]- H% K( A8 E7 L* w4 ?7 r
    回复

    使用道具 举报

    数学1+1        

    23

    主题

    14

    听众

    2532

    积分

    升级  17.73%

  • TA的每日心情
    开心
    2025-4-20 11:19
  • 签到天数: 842 天

    [LV.10]以坛为家III

    新人进步奖

    若N>800000,
    0 p! j: W6 P8 \0 p7 \7 A% ?! |  ^7 ^则   1.83150(1-1/logN)[N/log^2(N-2)]≤D(N) ≤4.36166[1+2/logN +o(1)]×
    - p4 S( d0 n4 T/ n+ R7 tN/{log[(N-2)/2]log(N-2)}. F- y0 F& |8 j0 x/ M" u- i
    这就是哥德巴赫公式,有兴趣的读者不妨检测一下。
    回复

    使用道具 举报

    11

    主题

    12

    听众

    1704

    积分

    升级  70.4%

  • TA的每日心情
    开心
    2016-6-3 20:54
  • 签到天数: 300 天

    [LV.8]以坛为家I

    自我介绍
    菩提本无树,明镜亦非台。本来无一物,何处惹尘埃。

    社区QQ达人

    群组数学建模培训课堂1

    群组数模思想方法大全

    回复

    使用道具 举报

    11

    主题

    12

    听众

    1704

    积分

    升级  70.4%

  • TA的每日心情
    开心
    2016-6-3 20:54
  • 签到天数: 300 天

    [LV.8]以坛为家I

    自我介绍
    菩提本无树,明镜亦非台。本来无一物,何处惹尘埃。

    社区QQ达人

    群组数学建模培训课堂1

    群组数模思想方法大全

    本帖最后由 1300611016 于 2014-1-4 09:08 编辑 9 G3 G; M- Q- d2 O/ i
    6 G* q7 L) D# Z$ U# b$ Y
    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
    : O1 J' J- c" I1 W, R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。
    回复

    使用道具 举报

    您需要登录后才可以回帖 登录 | 注册地址

    qq
    收缩
    • 电话咨询

    • 04714969085
    fastpost

    关于我们| 联系我们| 诚征英才| 对外合作| 产品服务| QQ

    手机版|Archiver| |繁體中文 手机客户端  

    蒙公网安备 15010502000194号

    Powered by Discuz! X2.5   © 2001-2013 数学建模网-数学中国 ( 蒙ICP备14002410号-3 蒙BBS备-0002号 )     论坛法律顾问:王兆丰

    GMT+8, 2025-5-30 11:38 , Processed in 1.964290 second(s), 100 queries .

    回顶部