2018-10-10 22:03:03 +00:00
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---
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id: 5900f3ad1000cf542c50fec0
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2020-12-16 07:37:30 +00:00
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title: 问题65:e的收敛
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2018-10-10 22:03:03 +00:00
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challengeType: 5
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videoUrl: ''
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---
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2020-12-16 07:37:30 +00:00
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# --description--
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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2的平方根可以写成无限连续分数。
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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√2= 1 + 1
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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2 + 1
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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2 + 1
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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2 + 1
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2018-10-10 22:03:03 +00:00
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2 + ......
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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可以写出无限连续分数,√2= \[1;(2)],(2)表示2无限重复。以类似的方式,√23= \[4;(1,3,1,8)]。事实证明,平方根的连续分数的部分值序列提供了最佳的有理近似。让我们考虑√2的收敛。
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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1 + 1 = 3/2
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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2
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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1 + 1 = 7/5
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2 + 1
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2
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1 + 1 = 17/12
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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2 + 1
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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2 + 1
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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2
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2018-10-10 22:03:03 +00:00
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1 + 1 = 41/29
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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2 + 1
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2 + 1
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2 + 1
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2
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因此,√2的前十个收敛的序列是:1,3 / 2,7 / 5,17 / 12,41 / 29,99 / 70,239 / 169,577 / 408,1333 / 985,3333 / 2378 ,...最令人惊讶的是重要的数学常数,e = \[2; 1,2,1,1,4,1,1,6,1,......,1,2k,1,......]。 e的会聚序列中的前十个项是:2,3,8 / 3,11 / 4,19 / 7,87 / 32,106 / 39,193 / 71,1264 / 465,1457 / 536,.... ..第10个收敛的分子中的数字之和为1 + 4 + 5 + 7 = 17。求e的连续分数的第100个收敛的分子中的位数之和。
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# --hints--
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`euler65()`应该返回272。
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2018-10-10 22:03:03 +00:00
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```js
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assert.strictEqual(euler65(), 272);
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2018-10-10 22:03:03 +00:00
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```
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2020-08-13 15:24:35 +00:00
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2020-12-16 07:37:30 +00:00
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# --solutions--
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