3.3 KiB
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 10th convergent is 1 + 4 + 5 + 7 = 17
.
Find the sum of digits in the numerator of the n
th convergent of the continued fraction for e
.
--hints--
convergentsOfE(10)
should return a number.
assert(typeof convergentsOfE(10) === 'number');
convergentsOfE(10)
should return 17
.
assert.strictEqual(convergentsOfE(10), 17);
convergentsOfE(30)
should return 53
.
assert.strictEqual(convergentsOfE(30), 53);
convergentsOfE(50)
should return 91
.
assert.strictEqual(convergentsOfE(50), 91);
convergentsOfE(70)
should return 169
.
assert.strictEqual(convergentsOfE(70), 169);
convergentsOfE(100)
should return 272
.
assert.strictEqual(convergentsOfE(100), 272);
--seed--
--seed-contents--
function convergentsOfE(n) {
return true;
}
convergentsOfE(10);
--solutions--
function convergentsOfE(n) {
function sumDigits(num) {
let sum = 0n;
while (num > 0) {
sum += num % 10n;
num = num / 10n;
}
return parseInt(sum);
}
// BigInt is needed for high convergents
let convergents = [
[2n, 1n],
[3n, 1n]
];
const multipliers = [1n, 1n, 2n];
for (let i = 2; i < n; i++) {
const [secondLastConvergent, lastConvergent] = convergents;
const [secondLastNumerator, secondLastDenominator] = secondLastConvergent;
const [lastNumerator, lastDenominator] = lastConvergent;
const curMultiplier = multipliers[i % 3];
const numerator = secondLastNumerator + curMultiplier * lastNumerator;
const denominator = secondLastDenominator + curMultiplier * lastDenominator;
convergents = [lastConvergent, [numerator, denominator]]
if (i % 3 === 2) {
multipliers[2] += 2n;
}
}
return sumDigits(convergents[1][0]);
}