2.0 KiB

Raw Blame History

id	title	challengeType	forumTopicId	dashedName
5900f3ad1000cf542c50fec0	Problem 65: Convergents of e	5	302177	problem-65-convergents-of-e

--description--

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

\\sqrt{2} = 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + ...}}}}

The infinite continued fraction can be written, \\sqrt{2} = \[1; (2)] indicates that 2 repeats ad infinitum. In a similar way, \\sqrt{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 \\sqrt{2}.

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

Hence the sequence of the first ten convergents for \\sqrt{2} are:

1, \\dfrac{3}{2}, \\dfrac{7}{5}, \\dfrac{17}{12}, \\dfrac{41}{29}, \\dfrac{99}{70}, \\dfrac{239}{169}, \\dfrac{577}{408}, \\dfrac{1393}{985}, \\dfrac{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, \\dfrac{8}{3}, \\dfrac{11}{4}, \\dfrac{19}{7}, \\dfrac{87}{32}, \\dfrac{106}{39}, \\dfrac{193}{71}, \\dfrac{1264}{465}, \\dfrac{1457}{536}, ...

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

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

--hints--

convergentsOfE() should return a number.

assert(typeof convergentsOfE() === 'number');

convergentsOfE() should return 272.

assert.strictEqual(convergentsOfE(), 272);

--seed--

--seed-contents--

function convergentsOfE() {

  return true;
}

convergentsOfE();

--solutions--

// solution required

2.0 KiB Raw Blame History