1.7 KiB
id | challengeType | title |
---|---|---|
5900f3ad1000cf542c50fec0 | 5 | Problem 65: Convergents of e |
Description
√2 = 1 + 1
2 + 1
2 + 1
2 + 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 + 1 = 3/2
2
1 + 1 = 7/5
2 + 1
2
1 + 1 = 17/12
2 + 1
2 + 1
2
1 + 1 = 41/29
2 + 1
2 + 1
2 + 1
2
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.
Instructions
Tests
tests:
- text: <code>euler65()</code> should return 272.
testString: assert.strictEqual(euler65(), 272, '<code>euler65()</code> should return 272.');
Challenge Seed
function euler65() {
// Good luck!
return true;
}
euler65();
Solution
// solution required