1080 lines
28 KiB
HTML
1080 lines
28 KiB
HTML
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<!doctype html>
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<html lang="en-US">
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<head>
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<meta charset="utf-8" />
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<meta name="viewport" content="width=device-width,initial-scale=1" />
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<link rel="stylesheet" href="/styles/default.css" />
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<link rel="stylesheet" href="/styles/junk.css" />
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<link rel="stylesheet" href="/styles/faulhaber.css" />
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<link rel="icon" href="/images/favicon.ico" />
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<title>Sums of Powers</title>
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</head>
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<body>
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<header>
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<a id="homelink" href="/">dairydemon.net</a>
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</header>
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<main>
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<article>
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<hgroup>
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<h1>Sums of Powers</h1>
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<p>Posted <time datetime="2026-04-15">April 15, 2026</time></p>
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</hgroup>
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<p>This page is all about how we can efficiently compute large sums of
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powers:</p>
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<div class="math-block">
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<math display="block">
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<mrow>
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<msup>
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<mn>1</mn>
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<mi>p</mi>
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</msup>
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<mo>+</mo>
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<msup>
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<mn>2</mn>
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<mi>p</mi>
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</msup>
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<mo>+</mo>
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<msup>
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<mn>3</mn>
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<mi>p</mi>
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</msup>
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<mo>+</mo>
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<mo>⋯</mo>
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<mo>+</mo>
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<msup>
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<mi>n</mi>
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<mi>p</mi>
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</msup>
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</mrow>
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</math>
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</div>
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<section>
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<h2>Demo</h2>
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<p>Note that large exponents will take longer to calculate, but your
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upper bound can be quite large - even thousands of digits!</p>
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<div id="demo-inputs">
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<label id="limit-label" for="limit"><math><mi>n</mi></math> (upper bound)</label>
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<input inputmode="numeric" pattern="\d+" id="limit" />
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<label id="exponent-label" for="exponent"><math><mi>p</mi></math> (exponent)</label>
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<input type="number" id="exponent" />
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<input type="button" id="enter" value="Enter" />
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</div>
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<output id="result" for="limit exponent"></output>
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</section>
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<section>
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<h2>How it works</h2>
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<p>A straightforward approach to computing a sum like this would be
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something like</p>
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<figure>
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<pre><code>function sumOfPowers(n, p) {
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let total = 0;
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for (let k = 1; k <= n; k++) {
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total += Math.pow(k, p);
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}
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return total;
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}</code></pre>
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<figcaption>A simple algorithm for computing sums of powers</figcaption>
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</figure>
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<p>This works fine for small values of <math><mi>n</mi></math>, but
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this algorithm requires we perform <math><mi>n</mi></math>
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exponentiations and <math><mi>n</mi></math> additions. In other
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words, the time to run grows as <math><mi>n</mi></math> increases,
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which makes computing sums with a high upper bound take a long time.
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This is <em>not</em> the algorithm the demo uses.</p>
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<p>In fact, the algorithm used in the demo looks quite different.
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It's not that many lines of code, but it's relying on some nifty (and
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slightly tricky) mathematics. If all you do is read the code, it's
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hard to believe it works, so rather than simply presenting it right
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now, I want to introduce it piece by piece and explain it as we
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go.</p>
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<section>
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<h3>Special cases</h3>
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<p>If we want to compute a sum where each term's exponent is 1, it
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turns out there's a simple formula we can use
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(<a href="https://en.wikipedia.org/wiki/Triangular_number">read more about why it works here</a>):</p>
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<div class="math-block">
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<math display="block">
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<mrow>
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<mn>1</mn>
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<mo>+</mo>
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<mn>2</mn>
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<mo>+</mo>
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<mn>3</mn>
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<mo>+</mo>
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<mo>⋯</mo>
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<mo>+</mo>
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<mi>n</mi>
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<mo>=</mo>
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<mfrac>
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<mrow>
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<mi>n</mi>
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<mo>(</mo>
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<mi>n</mi>
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<mo>+</mo>
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<mn>1</mn>
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<mo>)</mo>
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</mrow>
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<mn>2</mn>
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</mfrac>
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</mrow>
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</math>
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</div>
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<p>The nice thing about this formula is the number of operations we
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perform is no longer dependent on our upper bound - we'll always
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perform one addition, one multiplication, and one division. That
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means we can easily determine the sum of the numbers from 1 to some
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extreme limit like 1,000,000,000,000,000 much faster than with the
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naive algorithm.</p>
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<p>There are similar formulas for computing sums when our exponent
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is 2 or 3:</p>
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<div class="math-block">
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<math display="block">
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<mrow>
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<msup>
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<mn>1</mn>
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<mn>2</mn>
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</msup>
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<mo>+</mo>
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<msup>
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<mn>2</mn>
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<mn>2</mn>
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</msup>
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<mo>+</mo>
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<msup>
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<mn>3</mn>
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<mn>2</mn>
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</msup>
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<mo>+</mo>
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<mo>⋯</mo>
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<mo>+</mo>
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<msup>
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<mi>n</mi>
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<mn>2</mn>
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</msup>
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<mo>=</mo>
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<mfrac>
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<mrow>
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<mi>n</mi>
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<mo>(</mo>
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<mi>n</mi>
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<mo>+</mo>
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<mn>1</mn>
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<mo>)</mo>
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<mo>(</mo>
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<mn>2</mn>
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<mi>n</mi>
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<mo>+</mo>
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<mn>1</mn>
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<mo>)</mo>
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</mrow>
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<mn>6</mn>
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</mfrac>
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</mrow>
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</math>
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<math display="block">
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<mrow>
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<msup>
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<mn>1</mn>
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<mn>3</mn>
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</msup>
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<mo>+</mo>
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<msup>
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<mn>2</mn>
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<mn>3</mn>
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</msup>
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<mo>+</mo>
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<msup>
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<mn>3</mn>
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<mn>3</mn>
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</msup>
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<mo>+</mo>
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<mo>⋯</mo>
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<mo>+</mo>
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<msup>
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<mi>n</mi>
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<mn>3</mn>
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</msup>
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<mo>=</mo>
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<msup>
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<mrow>
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<mo>(</mo>
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<mfrac>
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<mrow>
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<mi>n</mi>
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<mo>(</mo>
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<mi>n</mi>
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<mo>+</mo>
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<mn>1</mn>
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<mo>)</mo>
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</mrow>
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<mn>2</mn>
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</mfrac>
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<mo>)</mo>
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</mrow>
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<mn>2</mn>
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</msup>
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</mrow>
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</math>
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</div>
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<p>It turns out there's a pattern that lets us construct
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polynomials like these for <em>any</em> exponent, but the pattern
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isn't super obvious.</p>
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</section>
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<section>
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<h3>Faulhaber's formula</h3>
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<p>
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<a href="https://en.wikipedia.org/wiki/Faulhaber%27s_formula">Faulhaber's formula</a>
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gives the general pattern these polynomials follow. The most
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commonly given version of the formula involves a special sequence
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called the
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<a href="https://en.wikipedia.org/wiki/Bernoulli_number">Bernoulli numbers</a>.
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</p>
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<p>Of course, there are algorithms to compute the Bernoulli
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numbers, but since they're rational numbers, there are tradeoffs to
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consider. If we use floats to store the sequence, we need to be
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mindful of rounding errors; additionally, after the 258th term, the
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Bernoulli numbers become too large to store in a 64 bit float. This
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means to compute sums of large powers precisely, we'll want a
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rational data type, which not every language has handy.</p>
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<p>I suppose if you need the coefficients of the polynomial for
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some reason, this is the formula for you, but if all you want is
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the sum of powers for some <math><mi>n</mi></math> and
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<math><mi>p</mi></math>, we can sidestep computing the Bernoulli
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numbers by taking another approach.</p>
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</section>
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<section>
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<h3>Stirling numbers</h3>
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<p>To begin explaining this alternate approach, we need to
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introduce the
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<a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind">Stirling numbers of the second kind</a>
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(as you might expect, there are Stirling numbers of the
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<em>first</em> kind too, but they aren't super relevant here).
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These numbers give the number of ways to partition a set of
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<math><mi>n</mi></math> elements into <math><mi>k</mi></math>
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non-empty subsets.</p>
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<p>For instance, there's 6 ways to partition the letters a, b, c,
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and d into 3 non-empty subsets:</p>
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<ul>
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<li>{{a,b},{c},{d}}</li>
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<li>{{a,c},{b},{d}}</li>
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<li>{{a},{b,c},{d}}</li>
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<li>{{a,d},{b},{c}}</li>
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<li>{{a},{b,d},{c}}</li>
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<li>{{a},{b},{c,d}}</li>
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</ul>
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<p>The Stirling numbers of the second kind (or just "Stirling
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numbers" from here on) happen to have a nice recursive formula that
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we can apply to compute them.</p>
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<div class="math-block">
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<math display="block">
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<mrow>
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<mrow>
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<mo>{</mo>
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<mfrac linethickness="0px">
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<mi>n</mi>
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<mi>k</mi>
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</mfrac>
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<mo>}</mo>
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</mrow>
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<mo>=</mo>
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<mi>k</mi>
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<mrow>
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<mo>{</mo>
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<mfrac linethickness="0px">
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<mrow>
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<mi>n</mi>
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<mo>−</mo>
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<mn>1</mn>
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</mrow>
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<mi>k</mi>
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</mfrac>
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<mo>}</mo>
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</mrow>
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<mo>+</mo>
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<mrow>
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<mo>{</mo>
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<mfrac linethickness="0px">
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<mrow>
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<mi>n</mi>
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<mo>−</mo>
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<mn>1</mn>
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</mrow>
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<mrow>
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<mi>k</mi>
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<mo>−</mo>
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<mn>1</mn>
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</mrow>
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</mfrac>
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<mo>}</mo>
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</mrow>
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</mrow>
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</math>
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</div>
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<p>To compute sums of powers for a given exponent
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<math><mi>p</mi></math>, we'll need to know</p>
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<div class="math-block">
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<math display="block">
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<mrow>
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<mrow>
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<mo fence="true" form="prefix">{</mo>
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<mfrac linethickness="0px">
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<mi>p</mi>
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<mn>0</mn>
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</mfrac>
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<mo fence="true" form="postfix">}</mo>
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</mrow>
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<mo separator="true">,</mo>
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<mrow>
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<mo fence="true" form="prefix">{</mo>
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<mfrac linethickness="0px">
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<mi>p</mi>
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<mn>1</mn>
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</mfrac>
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<mo fence="true" form="postfix">}</mo>
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</mrow>
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<mo separator="true">,</mo>
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<mrow>
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<mo fence="true" form="prefix">{</mo>
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<mfrac linethickness="0px">
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|
|
<mi>p</mi>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo fence="true" form="postfix">}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<mo separator="true">,</mo>
|
|||
|
|
<mo>…</mo>
|
|||
|
|
<mo separator="true">,</mo>
|
|||
|
|
<mrow>
|
|||
|
|
<mo fence="true" form="prefix">{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo fence="true" form="postfix">}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
</mrow>
|
|||
|
|
</math>
|
|||
|
|
</div>
|
|||
|
|
|
|||
|
|
<p>(We'll learn why in a bit.) Here's a function to get all those
|
|||
|
|
values.</p>
|
|||
|
|
|
|||
|
|
<figure>
|
|||
|
|
<pre><code>function stirling2(p) {
|
|||
|
|
if (p === 0) {
|
|||
|
|
return [1n];
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
let s = [0n];
|
|||
|
|
for (let i = 1; i < p + 1; i++) {
|
|||
|
|
s.push(1n);
|
|||
|
|
for (let j = i - 1; j > 0; j--) {
|
|||
|
|
s[j] = BigInt(j) * s[j] + s[j-1];
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
return s;
|
|||
|
|
}</code></pre>
|
|||
|
|
<figcaption>An algorithm for computing Stirling numbers of the second kind</figcaption>
|
|||
|
|
</figure>
|
|||
|
|
|
|||
|
|
<p>This algorithm is quadratic in the value of
|
|||
|
|
<math><mi>p</mi></math>, which isn't great, but it's simple to
|
|||
|
|
implement (and I don't know of any faster way).</p>
|
|||
|
|
|
|||
|
|
<section>
|
|||
|
|
<h4>Stirling numbers calculator</h4>
|
|||
|
|
<p>Be warned that these numbers can grow quite large, even for
|
|||
|
|
small values of <math><mi>p</mi></math>.</p>
|
|||
|
|
|
|||
|
|
<div id="stirling-input">
|
|||
|
|
<label id="stirling-label" for="stirling-limit"><math><mi>p</mi></math></label>
|
|||
|
|
<input type="number" id="stirling-limit" />
|
|||
|
|
|
|||
|
|
<input type="button" id="stirling-enter" value="Enter" />
|
|||
|
|
</div>
|
|||
|
|
|
|||
|
|
<output id="stirling-result" for="stirling-limit"></output>
|
|||
|
|
</section>
|
|||
|
|
</section>
|
|||
|
|
|
|||
|
|
<section>
|
|||
|
|
<h3>Connection to exponents</h3>
|
|||
|
|
<p>To understand what the Stirling numbers have to do with our
|
|||
|
|
problem, it helps to first introduce the notation for
|
|||
|
|
<a href="https://en.wikipedia.org/wiki/Falling_and_rising_factorials">falling factorials</a>.</p>
|
|||
|
|
<div class="math-block">
|
|||
|
|
<math display="block">
|
|||
|
|
<mrow>
|
|||
|
|
<msup>
|
|||
|
|
<mi>x</mi>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mi>n</mi>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>=</mo>
|
|||
|
|
<mi>x</mi>
|
|||
|
|
<mo>(</mo>
|
|||
|
|
<mi>x</mi>
|
|||
|
|
<mo>−</mo>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
<mo>(</mo>
|
|||
|
|
<mi>x</mi>
|
|||
|
|
<mo>−</mo>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
<mo>⋯</mo>
|
|||
|
|
<mo>(</mo>
|
|||
|
|
<mi>x</mi>
|
|||
|
|
<mo>−</mo>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
</mrow>
|
|||
|
|
</math>
|
|||
|
|
</div>
|
|||
|
|
|
|||
|
|
<p>It turns out there is a very elegant identity relating
|
|||
|
|
exponents, Stirling numbers, and falling
|
|||
|
|
factorials:<sup>1</sup></p>
|
|||
|
|
<div class="math-block">
|
|||
|
|
<math display="block">
|
|||
|
|
<mrow>
|
|||
|
|
<msup>
|
|||
|
|
<mi>x</mi>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
</msup>
|
|||
|
|
<mo>=</mo>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mn>0</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<msup>
|
|||
|
|
<mi>x</mi>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>0</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<msup>
|
|||
|
|
<mi>x</mi>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<msup>
|
|||
|
|
<mi>x</mi>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>2</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mo>⋯</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<msup>
|
|||
|
|
<mi>x</mi>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mi>n</mi>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
</mrow>
|
|||
|
|
</math>
|
|||
|
|
</div>
|
|||
|
|
|
|||
|
|
<p>We can apply this formula to every term of our power sum and
|
|||
|
|
rearrange to get</p>
|
|||
|
|
<div class="math-block">
|
|||
|
|
<math display="block">
|
|||
|
|
<mtable displaystyle="true">
|
|||
|
|
<mtr>
|
|||
|
|
<mtd style="padding-left:0em;padding-right:0em;">
|
|||
|
|
<mrow>
|
|||
|
|
<msup>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>3</mn>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mo>⋯</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</msup>
|
|||
|
|
<mo>=</mo>
|
|||
|
|
</mrow>
|
|||
|
|
</mtd>
|
|||
|
|
<mtd style="padding-left:0em;padding-right:0em;">
|
|||
|
|
<mrow>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mn>0</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<mo>(</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>0</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>0</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>3</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>0</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mo>⋯</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>0</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
</mrow>
|
|||
|
|
</mtd>
|
|||
|
|
</mtr>
|
|||
|
|
<mtr>
|
|||
|
|
<mtd style="padding-left:0em;padding-right:0em;">
|
|||
|
|
<mrow></mrow>
|
|||
|
|
</mtd>
|
|||
|
|
<mtd style="padding-left:0em;padding-right:0em;">
|
|||
|
|
<mrow>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<mo>(</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>3</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mo>⋯</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
</mrow>
|
|||
|
|
</mtd>
|
|||
|
|
</mtr>
|
|||
|
|
<mtr>
|
|||
|
|
<mtd style="padding-left:0em;padding-right:0em;">
|
|||
|
|
<mrow></mrow>
|
|||
|
|
</mtd>
|
|||
|
|
<mtd style="padding-left:0em;padding-right:0em;">
|
|||
|
|
<mrow>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<mo>(</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>2</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>2</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>3</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>2</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mo>⋯</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>2</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
</mrow>
|
|||
|
|
</mtd>
|
|||
|
|
</mtr>
|
|||
|
|
<mtr>
|
|||
|
|
<mtd style="padding-left:0em;padding-right:0em;">
|
|||
|
|
<mrow></mrow>
|
|||
|
|
</mtd>
|
|||
|
|
<mtd style="padding-left:0em;padding-right:0em;">
|
|||
|
|
<mrow>
|
|||
|
|
<mo>⋯</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
</mrow>
|
|||
|
|
</mtd>
|
|||
|
|
</mtr>
|
|||
|
|
<mtr>
|
|||
|
|
<mtd style="padding-left:0em;padding-right:0em;">
|
|||
|
|
<mrow></mrow>
|
|||
|
|
</mtd>
|
|||
|
|
<mtd style="padding-left:0em;padding-right:0em;">
|
|||
|
|
<mrow>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<mo>(</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>3</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mo>⋯</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
</mrow>
|
|||
|
|
</mtd>
|
|||
|
|
</mtr>
|
|||
|
|
</mtable>
|
|||
|
|
</math>
|
|||
|
|
</div>
|
|||
|
|
|
|||
|
|
<p>This might seem more complicated at first, but despite the
|
|||
|
|
similar notation to normal exponentiation, it's much easier to
|
|||
|
|
compute sums of falling factorials. For
|
|||
|
|
<math><mi>p</mi><mo>></mo><mn>0</mn></math>,</p>
|
|||
|
|
<div class="math-block">
|
|||
|
|
<math display="block">
|
|||
|
|
<mrow>
|
|||
|
|
<msup>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>3</mn>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mo>⋯</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
<mo>=</mo>
|
|||
|
|
<mfrac>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>(</mo>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<msup>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mrow>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</mrow>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
</mrow>
|
|||
|
|
<mrow>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</mrow>
|
|||
|
|
</mfrac>
|
|||
|
|
</mrow>
|
|||
|
|
</math>
|
|||
|
|
</div>
|
|||
|
|
<p>(For <math><mi>p</mi><mo>=</mo><mn>0</mn></math>, the sum is
|
|||
|
|
just <math><mi>n</mi></math>.)<sup>2</sup> This lets us
|
|||
|
|
simplify.<sup>3</sup></p>
|
|||
|
|
|
|||
|
|
<div class="math-block">
|
|||
|
|
<math display="block">
|
|||
|
|
<mrow>
|
|||
|
|
<msup>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mn>3</mn>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</msup>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mo>⋯</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<msup>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</msup>
|
|||
|
|
<mo>=</mo>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mn>0</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<mfrac>
|
|||
|
|
<mrow>
|
|||
|
|
<mo form="prefix" stretchy="false" lspace="0em" rspace="0em">(</mo>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<msup>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>2</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
</mrow>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mn>2</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<mfrac>
|
|||
|
|
<mrow>
|
|||
|
|
<mo form="prefix" stretchy="false" lspace="0em" rspace="0em">(</mo>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<msup>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>3</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
</mrow>
|
|||
|
|
<mn>3</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mn>3</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<mfrac>
|
|||
|
|
<mrow>
|
|||
|
|
<mo form="prefix" stretchy="false" lspace="0em" rspace="0em">(</mo>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<msup>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mn>4</mn>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
</mrow>
|
|||
|
|
<mn>4</mn>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mo>⋯</mo>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mrow>
|
|||
|
|
<mo>{</mo>
|
|||
|
|
<mfrac linethickness="0px">
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
</mfrac>
|
|||
|
|
<mo>}</mo>
|
|||
|
|
</mrow>
|
|||
|
|
<mfrac>
|
|||
|
|
<mrow>
|
|||
|
|
<mo form="prefix" stretchy="false" lspace="0em" rspace="0em">(</mo>
|
|||
|
|
<mi>n</mi>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
<msup>
|
|||
|
|
<mo>)</mo>
|
|||
|
|
<menclose notation="bottom">
|
|||
|
|
<mrow>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</mrow>
|
|||
|
|
</menclose>
|
|||
|
|
</msup>
|
|||
|
|
</mrow>
|
|||
|
|
<mrow>
|
|||
|
|
<mi>p</mi>
|
|||
|
|
<mo>+</mo>
|
|||
|
|
<mn>1</mn>
|
|||
|
|
</mrow>
|
|||
|
|
</mfrac>
|
|||
|
|
</mrow>
|
|||
|
|
</math>
|
|||
|
|
</div>
|
|||
|
|
|
|||
|
|
<p>The first term is 0 unless
|
|||
|
|
<math><mi>p</mi><mo>=</mo><mn>0</mn></math>, which is a trivial
|
|||
|
|
case, so we really only need to consider the remaining terms.</p>
|
|||
|
|
</section>
|
|||
|
|
|
|||
|
|
<section>
|
|||
|
|
<h3>Implementation</h3>
|
|||
|
|
<p>This formula might still seem a little complicated, but you may
|
|||
|
|
be surprised by how simple it is to implement. We already have a
|
|||
|
|
method to compute the Stirling numbers we need. With those values
|
|||
|
|
available, we can write a simple loop to compute each term and add
|
|||
|
|
to our total.</p>
|
|||
|
|
|
|||
|
|
<figure>
|
|||
|
|
<pre><code>function faulhaber(n, s) {
|
|||
|
|
const limit = BigInt(s.length);
|
|||
|
|
if (limit === 1n) {
|
|||
|
|
return n;
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
let q = n + 1n;
|
|||
|
|
let total = 0n;
|
|||
|
|
for (let k = 1n; k < limit; k++) {
|
|||
|
|
q *= n + 1n - k;
|
|||
|
|
total += s[k] * q / (k + 1n);
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
return total;
|
|||
|
|
}</code></pre>
|
|||
|
|
<figcaption>A function that takes an upper bound and Stirling
|
|||
|
|
numbers to compute a sum of powers</figcaption>
|
|||
|
|
</figure>
|
|||
|
|
</section>
|
|||
|
|
|
|||
|
|
<p>Note that the loop is dependent on <math><mi>p</mi></math> and not
|
|||
|
|
<math><mi>n</mi></math>, which allows us to compute sums with
|
|||
|
|
extremely large upper bounds very quickly.</p>
|
|||
|
|
|
|||
|
|
<section>
|
|||
|
|
<h3>Final thoughts</h3>
|
|||
|
|
<p>This approach spends most of its time computing the Stirling
|
|||
|
|
numbers. Additionally, it seems like the performance is largely
|
|||
|
|
hampered by how large the Stirling numbers get - computing them
|
|||
|
|
with a modulus is considerably quicker. The demo is a little
|
|||
|
|
clever, in that it only recomputes the Stirling numbers if the
|
|||
|
|
value of the exponent changes.</p>
|
|||
|
|
|
|||
|
|
<p>If you would rather use the more commonly known Faulhaber
|
|||
|
|
formula, you'll need to compute the Bernoulli numbers.
|
|||
|
|
<a href="https://arxiv.org/abs/1108.0286v3">Brent and Harvey</a>
|
|||
|
|
give a simple algorithm that is as efficient as how we're
|
|||
|
|
calculating the Stirling numbers. The needed binomial coefficients
|
|||
|
|
can also be calculated easily.</p>
|
|||
|
|
|
|||
|
|
<p>Furthermore, Brent and Harvey give another algorithm for the
|
|||
|
|
Bernoulli numbers that is even more optimized (although it is
|
|||
|
|
considerably more complicated). I believe using that algorithm in
|
|||
|
|
conjunction with the more traditional formula would be more
|
|||
|
|
efficient than the Stirling number approach.</p>
|
|||
|
|
|
|||
|
|
<p>However, in other ways, I like this formula better. For one
|
|||
|
|
thing, whichever algorithm you use, Bernoulli numbers are a bit
|
|||
|
|
harder to calculate, and require more lines of code. You'll have to
|
|||
|
|
compute the binomial coefficients, as well.</p>
|
|||
|
|
|
|||
|
|
<p>One nice thing about the Stirling approach is the terms are
|
|||
|
|
integers throughout. I think that makes this algorithm a little
|
|||
|
|
better for modular calculations.</p>
|
|||
|
|
</section>
|
|||
|
|
</section>
|
|||
|
|
|
|||
|
|
<section>
|
|||
|
|
<h2>Footnotes</h2>
|
|||
|
|
<ol id="footnote-list">
|
|||
|
|
<li><sup>1</sup> For proofs of this identity, see chapter 1.9 of
|
|||
|
|
<cite>Enumerative Combinatorics</cite> by Stanley and chapter 6.1
|
|||
|
|
of <cite>Concrete Mathematics</cite> by Graham, Knuth, and
|
|||
|
|
Patashnik.</li>
|
|||
|
|
<li><sup>2</sup> See chapter 2.6 of <cite>Concrete
|
|||
|
|
Mathematics</cite> for an explanation of this identity.</li>
|
|||
|
|
<li><sup>3</sup> A version of this formula is also mentioned in
|
|||
|
|
chapter 6.5 of <cite>Concrete Mathematics</cite>.</li>
|
|||
|
|
</ol>
|
|||
|
|
</section>
|
|||
|
|
</article>
|
|||
|
|
</main>
|
|||
|
|
|
|||
|
|
<footer>
|
|||
|
|
<p id="copyright">© 2026 filifa. This page is licensed under <a href="https://creativecommons.org/licenses/by-sa/4.0/">CC BY-SA 4.0</a> and the <a href="https://spdx.org/licenses/0BSD.html">BSD Zero Clause license</a>.</p>
|
|||
|
|
</footer>
|
|||
|
|
|
|||
|
|
<script src="faulhaber.js"></script>
|
|||
|
|
</body>
|
|||
|
|
</html>
|