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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)}' C" y# K# @6 R4 |+ `
Key words: Germany，Goldbach，even number, Odd number ，prime number, MR (2000) theme classification: 11 P32 Email:suxiaoguong@foxmail. com
( l* [4 P" S  ?6 `( N
作者: 数学1+1    时间: 2013-12-6 13:21
                  Goldbach’s problem (pdf)
                       Su Xiaoguang) G8 @; h) z# h5 r* z
     
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)}

* O+ U3 _% Y" R7 _, F* O
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
+ {1 r- A; M  Q% ^' W4 M6 C                    Su Xiaoguang
8 q2 s* ^: |$ U. f! zAbstract: In the analytic number theory Goldbach problem is an important issue. The authors studied the：5 g8 y4 f& Q4 s( J

5 f$ v1 L4 X; n4 G( z: d0 bA= \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./ T& l+ z; L  z5 y& _
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
5 Z( d- I/ ~2 e6 h4 c/ qDeduced
& m! e8 U) g! H- e- k# tD(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\wedge 6 c0 U0 k9 G$ s8 K5 V+ g9 C
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)}7 k9 n0 `' ?; }
  F$ M, F& \( p) P2 T+ J9 d& p) n
Key words: Germany，Goldbach，even number, Odd number ，prime number,
: u- W6 M0 K) A% m3 O# X/ }/ p; RMR (2000) theme classification: 11 P32
% B% w: t, e  s4 `8 tEmail:suxiaoguong@foxmail. com$ M" s* r% |1 w1 j$ |8 x
§ 1 Introduction
# i# F, z9 N5 i# q          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:. y  ]1 `! n' R) D
（A）For even number N! M! Y; C& }! V' e) t

! W# U' T: F5 \5 c& W  o# iN\geq 6\Rightarrow D(N)=\sum_{p_{1}+p_{2}=N,p_{1},p_{2}Is a prime number}1>0: ^5 P) @: f. m- T

: f- `% @: P4 r8 ]( y. H! o(B)  For odd number N
% a! n  D1 Y/ _% ]. j$ n3 F9 q6 J0 l, I
N\geq 9\Rightarrow T(N)=\sum_{p_{1}+p_{2}+p_{3}=N.p_{1},p_{2},p_{3}\geq 3}1>0$ o+ C3 g1 f! o$ C
' B  s! W9 H. }, r
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 correct
) _* e+ W9 r/ K/ G          ; Q1 n8 Q2 r8 o; W* ]# {
§2 Correlation set constructor
6 `7 ?! x$ v$ a4 o- F7 Q( ]. CA_{0}=\left \{ 0+0,0+1,0+2,\cdots  \right \}* f! o' Z) E' o' t! G! Q  `
A_{1}=\left \{ 1+0,1+1,1+2,\cdots  \right \}
0 b- r9 s9 e+ Q7 \# ~; ~- j A_{2}=\left \{ 2+0,2+1,2+2,\cdots  \right \}4 F6 Y( d& m9 y" b/ A
\cdots
/ c; P/ D$ a' V0 M3 I6 nA=\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）4 E0 e6 U0 \+ h! w% U, u. E) 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）      & Y: Y2 @9 b5 t1 J
  §3    Ready  Theorem% A8 i: w& T3 W$ N. w9 S
Theorem 1
, w8 m; p5 v( D' T' 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
/ Z. K5 V4 B2 D' H# v0 D  .Proof: Suppose M_{1},M_{2},\cdots ,M_{N},Is a countable set, M=\bigcup_{i=1}^{N}M_{i}+ x6 t8 j0 c" K& _8 y  i
\because M_{1}:x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},\cdots ,x_{i}^{(1)},\cdots
+ B) h) F5 _- W" @7 SM_{2}:x_{1}^{(2)},x_{2}^{(2)},x_{3}^{(2)},\cdots ,x_{i}^{(2)},\cdots
! l6 _( o' a5 g, x! lM_{N}:x_{1}^{(N)},x_{2}^{(N)},x_{3}^{(N)},\cdots ,x_{i}^{(N)},\cdots
/ _# D, ^; y! N2 W* B* m$ w\cdots
  {" G" i, ?# p4 k3 |\therefore M:x_{1}^{(1)},x_{1}^{(2)},\cdots ,x_{1}^{(N)},x_{2}^{(1)},x_{2}^{(2)},\cdots ,x_{2}^{(N)},\cdots Countable3 E( t. j9 j$ ~! d. B2 W. k$ ?  b
     Theorem 2 (Prime number theorem), ~1 i7 i$ N. H, d" C

0 \6 E1 Q9 t7 A, \8 o\pi (x)\sim \frac{x}{logx}^{\left [ 1 \right ]}
. g: ?: N% ?, `+ b) |. Q     Theorem 3  For even number x
9 A3 c* f3 U  S3 {x>800000\wedge M_{1}=minM(x)\Rightarrow M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x) -1\right ] 1 W0 [" Y5 q: p4 R, l% B
Proof: According to Theorem 1, (1)  
# [% G. a! r, B3 _4 C  I7 J' m9 F1 Q  \because A_{i},A_{j} Countable,! n. s, u% z- ^, a$ M; Q
\therefore A Countable\wedge B(x,N)=\frac{1}{2}(N+1)(N+2) + c- _6 m; t! x, y  P
Similarly, according to Theorem 1, (2), C countable
# S! m+ O0 P+ Z& T Suppose4 f. u/ x7 D: f  C" l2 y
      M_{1}(x)=minM(x)4 c0 r1 E4 b) m7 q2 w
according to （2）, Then we have+ @3 }9 Y' g& ]$ {7 O- a; S
. M_{1}(x)=\frac{1}{2}\pi (x)\left [ \pi (x)-1 \right ]1 ]+ E* E* a9 |: I4 \
Theorem 4  For even number x" ~! Q* Z& R: b
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）; r! C4 H* o! Q9 g
Proof: According to (2)，Then we have
6 _4 ]  Z. S1 a) P) j9 k6 cM(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 t: `0 X) T5 j* E     Suppose
' I  P9 \/ g# I1 V! C      M_{2}(x)=maxM(x)7 |: }  L. O; a$ B* U; C
\therefore M_{2}(x)
! L; M- M8 d1 X/ m( w=\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, V- U3 i3 d1 O6 h
=4\pi (x)\pi (\frac{x}{2})-2\pi ^{2}(\frac{x}{2})-3\pi (\frac{x}{2})-\pi ^{2}(x)+\pi (x)
6 Y* E: T! K4 w2 y  \§4 Goldbach's problem end
$ t4 }+ ?( C$ J- D9 E- NTheorem 5  For evem number N8 k& N2 }' ^4 I. k+ \9 w8 P
N> 800000\Rightarrow D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
# T+ P' O' {! K+ z     Proof: According to Theorem 2
9 o* a% A6 z$ Y- C* |4 @3 ]N> 800000\Rightarrow \alpha \frac{N}{logN}\leq \pi (N)\leq \beta \frac{N}{logN}      (4)- @+ p9 A* @& n; v6 Y# l5 C: T( p
Let   c_{1}=min(\alpha ,\beta ),
& ?& T6 `& o1 e3 O4 JAccording to Theorem 3,Then we have
* _5 ?, s$ f( q; H7 GD_{1}(N)=M_{1}(N)-M_{1}(N-2)
* j: g& |9 L: H- R7 rClear
$ _. g# J. Y' s$ X7 FD(N)\geq D_{1}(N)) R1 ~3 K4 o( D' l. K6 {
\because log(1+x)=\int_{0}^{x}\frac{dt}{1+t}=x-\int_{0}^{x}\frac{t}{1+t}dt
0 j( m6 o( o" o  n" ^4 m\because x\geq-\frac{1}{2} \Rightarrow log(1+x)=x+o(x^{2})           (5)2 e% `$ W4 O  u* [
\therefore D_{1}(N)=2c_{1}^{2}(1-\frac{1}{logN})\frac{N}{log^{2}(n-2)}+o(1)
7 g* c! M( X" }& ^- g1 IN\rightarrow \infty ,o(1)\rightarrow 0\Rightarrow 7 S7 `9 ?: G5 K) [% c
D(N)\geq 1.8432(1-\frac{1}{logN})\frac{N}{log^{2}(N-2)}
8 w: [; Y/ ~9 s( H. x0 U5 t5 P# ^2 R/ _ Theorem 6  For evem number N3 y; A: X6 |: X6 n- t; ?
N> 800000\Rightarrow D(N)\leq & B+ d- D* j! M- N) H- J4 M
5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}$ X  N' }4 G& P6 g1 e
Proof : According to (4)+ {4 V8 P$ f. {& f) J6 z8 |: B0 y
Let  c_{2}=max(\alpha ,\beta )+ v8 Y& s' d( Q
According to Theorem 4,Then we have
- a; ^* l8 o/ I, a" m+ ?# t- BD_{2}(N)=M_{2}(N)-M_{2}(N-2)
; P# m5 R& y% `) N+ x7 O! L! n\because D(N)\leq D_{2}(N)
# V; W' c& W$ {' \6 j7 SAccording to (5), Then we have
! A) {; Y! y: z+ n) zD(N)\leq   A: K. H  n! E1 {4 E  D
5.0176\left [ 1+\frac{2}{logN}+o(1) \right ]\frac{N}{log\frac{N-2}{2}log(N-2)}; I( `3 l8 b8 g
Theorem 7 (Goldbach Theorem)  
% r3 ^; O0 g! o( q6 O5 M( a# ]+ RFor evem number N8 |! V0 E! Y4 [6 b0 @% j$ n/ @2 P
N\geq 6\Rightarrow D(N)= \sum_{p_{1}+p_{2}=N,p_{1},p_{2}\geq 3}1\geq 1
7 @) x9 w3 o: A" I( x# zProof : According to Shen Mok Kong verification
. P! }* I3 j* J1 K( {" L4 A- a, _6\leq N\leq 3.3\times 1000000^{\left [ 3 \right ]}\Rightarrow D(N)\geq 1
$ ~0 ]* Y1 g6 C# MAccording to Theorem 5, Theorem 6, Then we have' p9 ?) q- y) U
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)}3 v& l1 X/ o0 ^$ ~
\therefore N\geq 6\Rightarrow D(N)\geq 1
7 Q1 n. ]7 ^/ _3 O; {: yLemma 1 For odd number N4 f* I3 t, d7 i. _; ~9 ?! [
N\geq 9\Rightarrow $ h, Y* m8 _; w& d# @5 I0 F
T(N)=\sum_{p_{1}+p_{2}+p_{3}=N,p_{1},p_{2},p_{3}\geq 3}1\geq 12 T) ?& _' T( Z4 X" \
Proof et  n\geq 4
- {7 @" r3 k$ O' x& _/ D\because 2n+1=2(n-1)+3
6 z! B( [  u- D/ z" c9 }* AAccording to Theorem 7,  Then we have. w5 [! Y# r$ l' Z7 i
N\geq 9\Rightarrow T(N)\geq 1
6 ?3 u& Z, [1 F' X1 ~$ G' d3 s) ?1 Y. G
2 ?+ |% Y- Y4 @, ?4 c; M
    References
# L" I1 C) C) f: a[1]  Wang yuan,TANTAN SUSHU,Shanghai, Shanghai Education Publishing House(1983),42.% L, q, Q; I! V
[2]  U﹒Dudley,Elementary number theory, Shanghai, Shanghai Science and Technology Press,(1980),195.
9 c3 [& ^0 a6 y/ q# B! S[3] Pan Chengdong,Pan Chengbiao,Goldbach conjecture,Beijing,Science Publishing house,(1984),1.
6 h0 L; R6 K/ H0 P
6 I$ W: @! K1 B( o) U, D
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
  P) o; [$ Y% ]- R# O6 H      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
+ b  Y+ a- `9 X0 N2 r( m 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( f' ?  n; w* G8 h" 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)
) Y+ }; O6 X! K) t类似地，根据定理1, (2), C可数  # ^- V: H& Z  D: U
设      M_{1}(x)=minM(x)
根据（2），那么我们有." m' F) l  T! r: Q2 _
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
: v1 q2 E5 z* Z  V  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; v+ e: x! n1 }' H( a+ c5 m
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( @( |2 o$ \- q; x
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( m$ T3 P6 I3 k5 r+ f, s: y: n# a
      D_{1}(N)=M_{1}(N)-M_{1}(N-2)
显然& g8 h% ?$ r5 R$ q8 C
       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 e% b6 P) I# l& m. T
       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
$ |- ?- o- s2 y* h2 Q       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: Q1 D/ s, R9 H8 m( \$ }; [
      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# b1 i9 E/ H( 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)}
\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7 R0 L9 {5 X. B5 C" B
      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 ; w% d' D1 y' m) _
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)}
. _6 R2 V1 A0 N
作者: 数学1+1    时间: 2013-12-14 11:37
若N＞800000，) G6 z: I5 o( {' X' w* n- Y
则   1.83150（1－1／logN）[N／log^2(N-2)]≤D(N) ≤4.36166[1+2／logN +o(1)]×2 W6 j3 P1 M5 Q" h- s3 Y4 _* g
N／{log[(N-2)／2]log(N-2)}* l- `$ T; {  F4 m6 t) I+ C$ t
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
- P( Q; f. o. e( ^" c. P3 s" Y: Z0 M( p6 G& s8 T$ B
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
( ~2 Z& S) [; r/ c+ D' ]: l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:( X) p$ U2 L5 C0 B/ ~4 _
      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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