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
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.