32 lines
1.1 KiB
Markdown
32 lines
1.1 KiB
Markdown
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The infinite continued fraction can be written, `sqrt(2) = [1;(2)]`, `(2)` indicates that `2` repeats ad infinitum.
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In a similar way, `sqrt(23) = [4;(1,3,1,8)]`.
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It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations.
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Let us consider the convergents for `sqrt(2)`.
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~~~
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1 + 1/2 = 3/2
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1 + 1/(2+ 1/2) = 7/5
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1 + 1/(2+ 1/(2+ 1/2)) = 17/12
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1 + 1/(2+ 1/(2+ 1/(2+ 1/2))) = 41/29
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~~~
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Hence the sequence of the first ten convergents for `sqrt(2)` are:
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~~~
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1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...
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~~~
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What is most surprising is that the important mathematical constant,
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~~~
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e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...]
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~~~
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The first ten terms in the sequence of convergents for `e` are:
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~~~
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2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...
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~~~
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The sum of digits in the numerator of the 10th convergent is `1+4+5+7=17`.
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Find the sum of digits in the numerator of the 100th convergent of the continued fraction for `e`.
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