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Problem 65


Problem 65


Convergents of e

The square root of 2 can be written as an infinite continued fraction.

$$\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+…}}}}$$

The infinite continued fraction can be written, √2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, √23 = [4;(1,3,1,8)].

It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for √2.

$$1+\frac{1}{2}=\frac{3}{2}$$
$$1+\frac{1}{2+\frac{1}{2}}=\frac{7}{5}$$
$$1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}=\frac{17}{12}$$
$$1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}=\frac{41}{29}$$

Hence the sequence of the first ten convergents for √2 are:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, …

What is most surprising is that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1 , … , 1,2k,1, …].

The first ten terms in the sequence of convergents for e are:

2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, …

The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.


e的有理逼近
2的算术平方根可以写成无限连分数的形式。

$$\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+…}}}}$$

这个无限连分数可以简记为√2 = [1;(2)],其中(2)表示2无限重复。同样的,我们可以记√23 = [4;(1,3,1,8)]。

可以证明,截取算术平方根连分数表示的一部分所组成的序列,给出了一系列最佳有理逼近值。让我们来考虑√2的逼近值:

$$1+\frac{1}{2}=\frac{3}{2}$$
$$1+\frac{1}{2+\frac{1}{2}}=\frac{7}{5}$$
$$1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}=\frac{17}{12}$$
$$1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}=\frac{41}{29}$$

因此√2的前十个逼近值为:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, …

最令人惊讶的莫过于重要的数学常数e有如下连分数表示
e = [2; 1,2,1, 1,4,1, 1,6,1 , … , 1,2k,1, …]。

e的前十个逼近值为:

2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, …

第10个逼近值的分子各位数字之和为1+4+5+7=17。

求e的第100个逼近值的分子各位数字之和。