为何987654321_123456789的值是8.0000000729

title	author	date	CreateTime	categories
为何 987654321/123456789 的值是 8.0000000729	lindexi	2018-05-23 16:11:54 +0800	2018-2-13 17:23:3 +0800

有时候，发现 987654321/123456789=8.0000000729 为什么后面还有几个数字？ 本文告诉大家，后面几个是如何算的

实际的值是

    8.0000000729000006633900060368490549353263999114702391943791‌​76668850507686539619‌​94751054152234592785‌​33479434654662855357‌​43198375263105214894‌​25745553774284539345‌​98930804850270324137‌​45994965088554182305‌​84305898317183674686‌​37143964598010077841‌​89170836121454608705‌​23693921765614688067‌​09366141055231883602‌​61014078375228113214‌​57583025264005529902‌​45032211229793122191‌​11741193916844864643‌​28826825392324111070‌​14941073835963771907‌​27032435615995164105‌​55599336055953958109‌​18101879354727102128‌​01662936495132722105‌​70777116194071757366‌​05299203108222748284‌​82700939192578546652‌​46477453742944829060‌​79794445326129452467

把 987654321/123456789 记为 p/q

那么 p 可以表示为

$$ p = n^{n-1} - \frac{n^{n-1}-1}{(n-1)^2}$$

$$ q = \frac{n(n^{n-1}-1)}{(n-1)^2}-1.$$

于是可以看到

$ p = (n-2)q + n-1 $

$$ \frac{p}{q} = n-2 + \frac{(n-1)^3}{n^n} \frac{1}{1 - \frac{n^2-n+1}{n^n}} = n-2 + \frac{(n-1)^3}{n^n} \sum_{k=0}^{\infty} \left(\frac{n^2-n+1}{n^n}\right)^k. $$

设n=10

$$\frac{987654321}{123456789} = 8 + \frac{729}{10^{10}}\sum_{k=0}^{\infty}\left(\frac{91}{10^{10}}\right)^k $$

https://math.stackexchange.com/questions/396135/why-is-frac987654321123456789-8-0000000729/396179

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标签：10,frac,8.0000000729,23,987654321,123456789
来源： https://blog.51cto.com/u_11283245/2952068