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神奇的圆周率

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APPSYZY 发表于 2019-4-19 21:34 |阅读模式
Pic.jpg
不知该如何证明。

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色k 发表于 2019-4-20 00:02
看不懂,码一下啊顺便给翻译下

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orzweb111 发表于 2019-4-20 04:05
回复 1# APPSYZY
You were asking the Probability of coprimality. Here is an answer en.wikipedia.org/wiki/Coprime_integers

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$\style{scale:11;fill:#eff}꩜$

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hbghlyj 发表于 2024-11-26 00:51
APPSYZY 发表于 2019-4-19 13:34
Given any pair of whole numbers chosen from a large, random collection of numbers, the probability that the two numbers have no common factor other than one (1) is
$$
\frac{6}{\pi^2}
$$
For example, using the small collection of numbers: $2,3,4,5,6$; there are 10 pairs that can be formed: $(2,3),(2,4)$, etc. Six of the 10 pairs: $(2,3),(2,5),(3,4),(3,5),(4,5)$ and $(5,6)$ have no common factor other than one. Using the ratio of the counts as the probability we have:
$$
\begin{aligned}
\frac{6}{\pi^2} & \approx \frac{6}{10} \\
\pi & \approx 3.162
\end{aligned}
$$

en.wikipedia.org/wiki/Coprime_integers#Probability_of_coprimality

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