User:Rii'jeg'fkep'c

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

「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)」

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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    $$5\cdot 7 \quad 5 \cdot 11 \quad 7\cdot 11$$

$${P_6}^2$$   132    $$5\cdot 7 \quad 5 \cdot 11 \quad 5\cdot 13 \quad 7 \cdot 11 \quad 7 \cdot 13 \quad 11 \cdot 13$$

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

duodecimal