数学 – JavaScript - 整数 - 数学的帰納法と除法の定理(二項定理の証明)

代数系入門

学習環境/開発環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Google Chrome...)用JavaScript Library: MathJax
• OS X El Capitan - Apple (OS)
• Emacs (Text Editor)
• JavaScript (プログラミング言語)
• JavaScript Library
  • kjs-array
• Safari(Web browser)
• 参考書籍
  • JavaScript 第6版 (David Flanagan(著)、村上 列(翻訳)、オライリージャパン)
  • JavaScriptリファレンス 第6版(David Flanagan(著)、木下 哲也(翻訳)、オライリージャパン)

代数系入門 (松坂 和夫(著)、岩波書店)の第1章(整数)、2(数学的帰納法と除法の定理)、問題5.を取り組んでみる。

問題5.

$\begin{array}{l}n=0\\ {\left(x+y\right)}^n=1\\ {\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n-r}{y}^r}=1\\ \\ {\left(x+y\right)}^n={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n-r}{y}^r}\\ {\left(x+y\right)}^{n+1}\\ =\left(x+y\right){\left(x+y\right)}^n\\ =\left(x+y\right){\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n-r}{y}^r}\\ =x{\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n-r}{y}^r}+y{\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n-r}{y}^r}\\ ={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n+1-r}{y}^r}+{\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n-r}{y}^{r+1}}\\ ={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n+1-r}{y}^r}+{\displaystyle \sum _{r=1}^{n+1}\left(\begin{array}{c}n\\ r-1\end{array}\right)x^{n-\left(r-1\right)}{y}^{\left(r-1\right)+1}}\\ ={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n+1-r}{y}^r}+{\displaystyle \sum _{r=1}^{n+1}\left(\begin{array}{c}n\\ r-1\end{array}\right)x^{\left(n+1\right)-r}{y}^r}\\ =\left(\begin{array}{c}n\\ 0\end{array}\right)x^{n+1-0}{y}^0+{\displaystyle \sum _{r=1}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n+1-r}{y}^r}+{\displaystyle \sum _{r=1}^n\left(\begin{array}{c}n\\ r-1\end{array}\right)x^{\left(n+1\right)-r}{y}^r}+\left(\begin{array}{c}n\\ n+1-1\end{array}\right)x^{\left(n+1\right)-\left(n+1\right)}{y}^{n+1}\\ =\left(\begin{array}{c}n\\ 0\end{array}\right)x^{n+1-0}{y}^0+{\displaystyle \sum _{r=1}^n\left(\left(\begin{array}{c}n\\ r\end{array}\right)+\left(\begin{array}{c}n\\ r-1\end{array}\right)\right)x^{n+1-r}{y}^r}+\left(\begin{array}{c}n\\ n+1-1\end{array}\right)x^{\left(n+1\right)-\left(n+1\right)}{y}^{n+1}\\ =\left(\begin{array}{c}n+1\\ 0\end{array}\right)x^{n+1-0}{y}^0+{\displaystyle \sum _{r=1}^n\left(\begin{array}{c}n+1\\ r\end{array}\right)x^{n+1-r}{y}^r}+\left(\begin{array}{c}n+1\\ n+1\end{array}\right)x^{\left(n+1\right)-\left(n+1\right)}{y}^{n+1}\\ ={\displaystyle \sum _{r=0}^{n+1}\left(\begin{array}{c}n+1\\ r\end{array}\right)x^{\left(n+1\right)-r}{y}^r}\end{array}$

JavaScript で確認。

HTML5

<label for="x0">x = </label>
<input id="x0" type="number"  value="5">
<label for="y0">y = </label>
<input id="y0" type="number"  value="10">
<label for="n0">n = </label>
<input id="n0" type="number" min="0" step="1"  value="2">

<div id="output0"></div>

<script src="array.js"></script>
<script src="sample5.js"></script>

JavaScript

コード(Emacs)

(function () {
    var input_x = document.querySelector('#x0'),
        input_y = document.querySelector('#y0'),
        input_n = document.querySelector('#n0'),
        inputs = [input_n, input_x, input_y],
        div_output = document.querySelector('#output0'),
        nl = '<br>',
        factorial,
        combination,
        output;

    console.log(inputs);
    factorial = function (n) {
        return Array.range(1, n + 1)
            .reduce(function (x, y) {
                return x * y;
            }, 1);
    };
    combination = function (n, r) {
        return factorial(n) / (factorial(r) * factorial(n - r));
    };

    output = function () {
        var left,
            right,
            x = parseFloat(input_x.value),
            y = parseFloat(input_y.value),        
            n = parseInt(input_n.value, 10);

        left = Math.pow(x + y, n);
        right = Array.range(n + 1)
            .map(function(r) {
                return combination(n, r) *
                    Math.pow(x, n - r) *
                    Math.pow(y, r);
            })
            .sum();
        div_output.innerHTML =
            '左辺 = ' + left + nl +
            '右辺 = ' + right + nl +
            '左辺 = 右辺 ' + (left === right) + nl +
            '|左辺 - 右辺| = ' + Math.abs(left - right) + nl;
    };

    inputs.forEach(function (input) {
        input.onchange = output;
    });

    output();
}());

x = y = n =