118 lines
3.3 KiB
Markdown
118 lines
3.3 KiB
Markdown
---
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id: 5900f3ad1000cf542c50fec0
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title: 'Problem 65: Convergents of e'
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challengeType: 5
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forumTopicId: 302177
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dashedName: problem-65-convergents-of-e
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---
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# --description--
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The square root of 2 can be written as an infinite continued fraction.
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$\\sqrt{2} = 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + ...}}}}$
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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 $\\sqrt{2}$.
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$1 + \\dfrac{1}{2} = \\dfrac{3}{2}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2}} = \\dfrac{7}{5}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2}}} = \\dfrac{17}{12}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2}}}} = \\dfrac{41}{29}$
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Hence the sequence of the first ten convergents for $\\sqrt{2}$ are:
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$1, \\dfrac{3}{2}, \\dfrac{7}{5}, \\dfrac{17}{12}, \\dfrac{41}{29}, \\dfrac{99}{70}, \\dfrac{239}{169}, \\dfrac{577}{408}, \\dfrac{1393}{985}, \\dfrac{3363}{2378}, ...$
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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:
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$2, 3, \\dfrac{8}{3}, \\dfrac{11}{4}, \\dfrac{19}{7}, \\dfrac{87}{32}, \\dfrac{106}{39}, \\dfrac{193}{71}, \\dfrac{1264}{465}, \\dfrac{1457}{536}, ...$
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The sum of digits in the numerator of the 10<sup>th</sup> convergent is $1 + 4 + 5 + 7 = 17$.
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Find the sum of digits in the numerator of the `n`<sup>th</sup> convergent of the continued fraction for `e`.
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# --hints--
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`convergentsOfE(10)` should return a number.
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```js
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assert(typeof convergentsOfE(10) === 'number');
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```
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`convergentsOfE(10)` should return `17`.
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```js
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assert.strictEqual(convergentsOfE(10), 17);
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```
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`convergentsOfE(30)` should return `53`.
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```js
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assert.strictEqual(convergentsOfE(30), 53);
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```
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`convergentsOfE(50)` should return `91`.
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```js
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assert.strictEqual(convergentsOfE(50), 91);
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```
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`convergentsOfE(70)` should return `169`.
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```js
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assert.strictEqual(convergentsOfE(70), 169);
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```
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`convergentsOfE(100)` should return `272`.
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```js
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assert.strictEqual(convergentsOfE(100), 272);
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```
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# --seed--
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## --seed-contents--
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```js
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function convergentsOfE(n) {
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return true;
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}
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convergentsOfE(10);
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```
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# --solutions--
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```js
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function convergentsOfE(n) {
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function sumDigits(num) {
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let sum = 0n;
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while (num > 0) {
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sum += num % 10n;
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num = num / 10n;
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}
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return parseInt(sum);
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}
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// BigInt is needed for high convergents
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let convergents = [
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[2n, 1n],
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[3n, 1n]
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];
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const multipliers = [1n, 1n, 2n];
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for (let i = 2; i < n; i++) {
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const [secondLastConvergent, lastConvergent] = convergents;
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const [secondLastNumerator, secondLastDenominator] = secondLastConvergent;
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const [lastNumerator, lastDenominator] = lastConvergent;
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const curMultiplier = multipliers[i % 3];
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const numerator = secondLastNumerator + curMultiplier * lastNumerator;
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const denominator = secondLastDenominator + curMultiplier * lastDenominator;
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convergents = [lastConvergent, [numerator, denominator]]
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if (i % 3 === 2) {
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multipliers[2] += 2n;
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}
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}
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return sumDigits(convergents[1][0]);
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}
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```
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