User:Rii'jeg'fkep'c

See my classical Chinese page文言 also my simple English page Simple English

結構

「What I mean just technically speaking, it is incorrect to say Baidu blocks it. OK, I know how the Party and Baidu company evil, and I am f*cked by the holy Firewall almost everyday, but as a technical expert, I can still access Wikipedia, here, so that's all, and that's enough. So in an evil society, let us all be evil, that would be the best choice. I don't care about those non-technical things, as the ancient Chinese classics {左傳} says, 肉食者謀之，又何間焉。 Yao Ziyuan 21:47, 14 May 2006 (UTC)」

User's Language

Mardarin	Classical Chinese	English	ｎｉｈｏｎｇｏ
zh該用戶的母語是中文。 该用户的母语是中文。	lzh-4此士精於文言。	en-3This user can contribute with an advanced level of English.	ja-2この利用者は中級の日本語ができます。
korean	FRANÇAIS	LATINA	Ελληνική
ko-0이 사용자는 한국어를 모르거나, 이해하는 데 어려움이 있습니다.	fr-0Cet utilisateur ne comprend pas le français ou seulement avec des difficultés notables.	la-0Hic usor linguam latinam non intellegit aut eam intellegit aegre.	el-0Αυτός ο χρήστης δεν καταλαβαίνει ελληνικά.

1. FRANÇAIS
2. LATINA
3. Ελληνική

Baidu baike in Classical Chinese Wiki

百度百科亦有排版不暢之疾，尤以數學公式。常以b/a代${\frac {b}{a}}$，以(1011)₂代$1011_{(2)}$,以e^(-x^2)代$e^{-x^{2}}$,使人難閲之。

亦有排版不暢之疾，尤以數學公式。常代${\frac {b}{a}}$之以b/a，代$1011_{(2)}$之以(1011)_2，代$e^{-x^{2}}$之以e^(-x^2)，使人難閲之。

無窮性之證

孿生質數無窮性之證

由質數分佈定理可知，質數之距遞增。

設質數之距d,由定理可知

$d_{(p_{n},p_{n+m})}<d_{(p_{n+k},p_{n+m+k})}$

暫令二者等距，且令$(p_{n}+k)\in \mathbb {P} \ ,p_{n}=r,$則

$\Delta {p_{n}}^{2}=(p_{n}+k)^{2}-{p_{n}}^{2}$

$=r^{2}+2kr+k^{2}-{p_{n}}^{2}$

$2kr+k^{2}$

又質數平方內半質數（質數分佈缺陷）pp'有如下之分佈

${p_{3}}^{2}$ 5² $\varnothing$

${p_{4}}^{2}$ 7² $5\cdot 7$ $C_{2}^{2}$

${p_{5}}^{2}$ 11² $5\cdot 7\quad 5\cdot 11\quad 7\cdot 11$ ${\ce {C_3^2}}$

${P_{6}}^{2}$ 13² $5\cdot 7\quad 5\cdot 11\quad 5\cdot 13\quad 7\cdot 11\quad 7\cdot 13\quad 11\cdot 13$ ${\ce {C_4^2}}$

etc

${P_{n}}^{2}$ $C_{n-2}^{2}$

$C_{n-2}^{2}={\frac {1}{2}}(n-2)(n-3)={\frac {1}{2}}[n^{2}-5n+6]$

則$(p_{n}+k)^{2}$滿足

$n_{pp'}<{\frac {1}{2}}(r+k-2)(r+k-3)$

$={\frac {1}{2}}[r^{2}+2kr+k^{2}-5r-5k+6]$

則

$\Delta n_{pp'}<{\frac {1}{2}}[r^{2}+2kr+k^{2}-5k-5r+6]-{\frac {1}{2}}[r^{2}-5r+6]$

$={\frac {1}{2}}[k^{2}+2kr-5k]$

半質數之密度

${\frac {\Delta n_{pp'}}{\Delta {p_{n}}^{2}}}<{\frac {1}{2}}\cdot {\frac {[k^{2}+2kr-5k]}{k^{2}+2kr}}$

$={\frac {1}{2}}[1-{\frac {5k}{k^{2}+2kr}}]$

$\lim _{k\to \infty }\lim _{r\to \infty }{\frac {\Delta n_{pp'}}{\Delta {p_{n}}^{2}}}<{\frac {1}{2}}$

又質數之距遞增，故

$\lim _{r\to \infty }{\frac {\Delta n_{pp'}}{\Delta {p_{n}}^{2}}}=0$

即半質數之比趋於零，故孿生質數{6n-1,6n+1}無窮

6n-1,6n+1

senary

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