Algorithms - JavaScript - 基礎概念 - 数学的な基礎(総和、公式の発見、帰納法による証明、階差数列)

The Art of Computer Programming
Volume 1
原著

開発環境

• macOS Sierra - Apple (OS)
• Emacs (Text Editor)
• JavaScript (プログラミング言語)
• Node.js, Safari(JavaScript エンジン)
• Learning JavaScript [邦訳](参考書籍)

The Art of Computer Programming　Volume 1 Fundamental Algorithms Third Edition 日本語版(Donald E. Knuth (著)、青木 孝 (著)、筧 一彦 (著)、鈴木 健一 (著)、長尾 高弘 (著)、有澤 誠 (その他)、和田 英一 (その他)、ドワンゴ)の第1章(基礎概念)、1.2(数学的な基礎)、演習問題11を取り組んでみる。

11. 
    $\begin{array}{l}n=0,\frac{1}{5}\\ n=1,\frac{1}{5}-\frac{27}{85}=\frac{17-27}{85}=-\frac{10}{85}=-\frac{2}{17}\\ n=2,-\frac{2}{17}+\frac{125}{629}=\frac{-74+125}{629}=\frac{51}{629}=\frac{3}{37}\\ n=3,\frac{3}{37}-\frac{343}{2405}=\frac{195-343}{2405}=-\frac{148}{2405}=-\frac{4}{65}\end{array}$

    分母に注目。

    $\begin{array}{l}5,17,37,65,···\\ 12,20,28,···\\ b_n=8n+12\\ n\ge 1\\ a_n=5+\frac{n\left(12+\left(8\left(n-1\right)+12\right)\right)}{2}\\ =5+\frac{n\left(12+8n+4\right)}{2}\\ =5+4n\left(n+2\right)\\ =4n^2+8n+5\\ =4{\left(n+1\right)}^2+1\end{array}$
    $\begin{array}{l}{S}_n=\frac{{\left(-1\right)}^n\left(n+1\right)}{4{\left(n+1\right)}^2+1}\\ \\ S_0=\frac{1}{5}\\ \\ S_n=S_{n-1}+\frac{{\left(-1\right)}^n{\left(2n+1\right)}^3}{{\left(2n+1\right)}^4+4}\\ =\frac{{\left(-1\right)}^{n-1}n}{4n^2+1}+\frac{{\left(-1\right)}^n{\left(2n+1\right)}^3}{{\left(2n+1\right)}^4+4}\\ =\frac{-{\left(-1\right)}^{n}n\left({\left(2n+1\right)}^4+4\right)+\left(4n^2+1\right){\left(-1\right)}^n{\left(2n+1\right)}^3}{\left(4n^2+1\right)\left({\left(2n+1\right)}^4+4\right)}\\ ={\left(-1\right)}^n\frac{\left(4n^2+1\right)\left(8n^3+12n^2+6n+1\right)-\left(16n^5+32n^4+24n^3+8n^2+5n\right)}{\left(4n^2+1\right)\left({\left(2n+1\right)}^4+4\right)}\\ ={\left(-1\right)}^n\frac{\left(32n^5+48n^4+32n^3+16n^2+6n+1\right)-\left(16n^5+32n^4+24n^3+8n^2+5n\right)}{\left(4n^2+1\right)\left({\left(2n+1\right)}^4+4\right)}\\ ={\left(-1\right)}^n\frac{16n^5+16n^4+8n^3+8n^2+n+1}{64n^6+128n^5+112n^4+64n^3+44n^2+8n+5}\\ =\frac{{\left(-1\right)}^n\left(n+1\right)\left(16n^4+8n^2+1\right)}{\left(4n^2+1\right)\left({\left(2n+1\right)}^4+4\right)}\\ =\frac{{\left(-1\right)}^n\left(n+1\right){\left(4n^2+1\right)}^2}{\left(4n^2+1\right)\left({\left(2n+1\right)}^4+4\right)}\\ =\frac{{\left(-1\right)}^n\left(n+1\right)\left(4n^2+1\right)}{\left({\left(2n+1\right)}^4+4\right)}\\ =\frac{{\left(-1\right)}^n\left(n+1\right)\left(4n^2+1\right)}{16n^4+32n^3+24n^2+8n+5}\\ =\frac{{\left(-1\right)}^n\left(n+1\right)}{4n^2+8n+5}\\ =\frac{{\left(-1\right)}^n\left(n+1\right)}{4{\left(n+1\right)}^2+1}\end{array}$

コード(Emacs)

HTML5

<pre id="output0"></pre>
n = <input id="n0" type="number" min="0" step="1" value="10">
<button id="run0">run</button>
<button id="clear0">clear</button>

<script src="sample11.js"></script>

JavaScript

let input0 = document.querySelector('#n0'),
    pre0 = document.querySelector('#output0'),
    btn0 = document.querySelector('#run0'),
    btn1 = document.querySelector('#clear0'),
    p = (x) => pre0.textContent += x + '\n',
    range = (start, end, step=1) => {
        let result = [];

        for (let i = start; i < end; i += step) {
            result.push(i);
        }
        return result;
    };

let term = (n) =>
    Math.pow(-1, n) * Math.pow(2 * n + 1, 3) / (Math.pow(2 * n + 1, 4) + 4),
    f = (n) => range(0, n + 1)
    .map((i) => term(i))
    .reduce((x, y) => x + y),
    s = (n) => Math.pow(-1, n) * (n + 1) / (4 * Math.pow(n + 1, 2) + 1);
        
    
let output = () => {
    let n = parseInt(input0.value, 10),
        a1 = f(n),
        a2 = s(n),
        d = Math.abs(a1 - a2);

    p(f(n));
    p(s(n));
    p(`誤差: ${d}`)
};

input0.onchange = output;
btn0.onclick = output;
btn1.onclick = () => pre0.textContent = '';

output();

n =