2018-5-23-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
8.0000000729000006633900060368490549353263999114702391943791766688505076865396199475105415223459278533479434654662855357431983752631052148942574555377428453934598930804850270324137459949650885541823058430589831718367468637143964598010077841891708361214546087052369392176561468806709366141055231883602610140783752281132145758302526400552990245032211229793122191117411939168448646432882682539232411107014941073835963771907270324356159951641055559933605595395810918101879354727102128016629364951327221057077711619407175736605299203108222748284827009391925785466524647745374294482906079794445326129452467
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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