数学 – JavaScript - 整数 - 数学的帰納法と除法の定理(帰納法による等式の証明)

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
  • kjs-math-number
• 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3.を取り組んでみる。

問題3.

1. 
   $\begin{array}{l}1+2+···+n+\left(n+1\right)\\ =\frac{n\left(n+1\right)}{2}+\left(n+1\right)\\ =\frac{\left(n+1\right)\left(\left(n+1\right)+1\right)}{2}\end{array}$
2. 
   $\begin{array}{l}{1}^2+2^2+···+n^2+{\left(n+1\right)}^2\\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6}+{\left(n+1\right)}^2\\ =\frac{\left(n+1\right)\left(n\left(2n+1\right)+6\left(n+1\right)\right)}{6}\\ =\frac{\left(n+1\right)\left(2n^2+7n+6\right)}{6}\\ =\frac{\left(n+1\right)\left(n+2\right)\left(2n+3\right)}{6}\\ =\frac{\left(n+1\right)\left(\left(n+1\right)+1\right)\left(2\left(n+1\right)+1\right)}{6}\end{array}$
3. 
   $\begin{array}{l}{1}^3+2^3+···+n^3+{\left(n+1\right)}^3\\ =\frac{n^2{\left(n+1\right)}^2}{4}+{\left(n+1\right)}^3\\ =\frac{{\left(n+1\right)}^2\left(n^2+4\left(n+1\right)\right)}{4}\\ =\frac{{\left(n+1\right)}^2\left(n^2+4n+4\right)}{4}\\ =\frac{{\left(n+1\right)}^2{\left(n+2\right)}^2}{4}\\ =\frac{{\left(n+1\right)}^2{\left(\left(n+1\right)+1\right)}^2}{4}\end{array}$
4. 
   $\begin{array}{l}1·2+2·3+···+n\left(n+1\right)+\left(n+1\right)\left(\left(n+1\right)+1\right)\\ =\frac{1}{3}n\left(n+1\right)\left(n+2\right)+\left(n+1\right)\left(\left(n+1\right)+1\right)\\ =\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)}{3}\\ =\frac{\left(n+1\right)\left(\left(n+1\right)+1\right)\left(\left(n+1\right)+2\right)}{3}\end{array}$
5. 
   $\begin{array}{l}1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+···+\frac{1}{2n-1}-\frac{1}{2n}+\frac{1}{2\left(n+1\right)-1}-\frac{1}{2\left(n+1\right)}\\ =\frac{1}{n+1}+\frac{1}{n+2}+···+\frac{1}{2n}+\frac{1}{2\left(n+1\right)-1}-\frac{1}{2\left(n+1\right)}\\ =\frac{1}{n+2}+···+\frac{1}{2n}+\frac{1}{2n+1}-\frac{1}{2\left(n+1\right)}+\frac{1}{n+1}\\ =\frac{1}{n+2}+\frac{1}{n+3}+···+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2\left(n+1\right)}\\ =\frac{1}{\left(n+1\right)+1}+\frac{1}{\left(n+1\right)+2}+···+\frac{1}{2\left(n+1\right)}\end{array}$
6. 
   $\begin{array}{l}\left(n+1\right)\left(n+2\right)\left(n+3\right)···\left(2n\right)\frac{\left(2n+1\right)\left(2\left(n+1\right)\right)}{n+1}\\ =2^n·1·3·5·····\left(2n-1\right)\frac{\left(2n+1\right)\left(2\left(n+1\right)\right)}{n+1}\\ =2^{n+1}·1·3·5·····\left(2n-1\right)\left(2n+1\right)\end{array}$
7. 
   $\begin{array}{l}1+2x+3x^2+···+n{x}^{n-1}+\left(n+1\right)x^n\\ =\frac{1-x^n}{{\left(1-x\right)}^2}-\frac{n{x}^n}{1-x}+\left(n+1\right)x^n\\ =\frac{\left(1-x^n\right)-n\left(1-x\right)x^n+{\left(1-x\right)}^2\left(n+1\right)x^n}{{\left(1-x\right)}^2}\\ =\frac{1-x^n-n{x}^n+n{x}^{n+1}+n{x}^n+x^n-2n{x}^{n+1}-2x^{n+1}+n{x}^{n+2}+x^{n+2}}{{\left(1-x\right)}^2}\\ =\frac{1-n{x}^{n+1}-2x^{n+1}+n{x}^{n+2}+x^{n+2}}{{\left(1-x\right)}^2}\\ =\frac{1-x^{n+1}}{{\left(1-x\right)}^2}-\frac{x^{n+1}+n{x}^{n+1}-n{x}^{n+2}-x^{n+2}}{{\left(1-x\right)}^2}\\ =\frac{1-x^{n+1}}{{\left(1-x\right)}^2}-\frac{-\left(n+1\right)x^{n+2}+\left(n+1\right)x^{n+1}}{{\left(1-x\right)}^2}\\ =\frac{1-x^{n+1}}{{\left(1-x\right)}^2}-\frac{\left(n+1\right)x^{n+1}\left(1-x\right)}{{\left(1-x\right)}^2}\\ =\frac{1-x^{n+1}}{{\left(1-x\right)}^2}-\frac{\left(n+1\right)x^{n+1}}{1-x}\end{array}$

JavaScript で確認。

HTML5

<label for="n0">n = </label>
<input id="n0" type="number" min="1" step="1" value="100">
<label for="x0">n = </label>
<input id="x0" type="number" value="10">
<div id="output0"></div>

<script src="array.js"></script>
<script src="number.js"></script>
<script src="sample3.js"></script>

JavaScript

コード(Emacs)

var input_n = document.querySelector('#n0'),
    input_x = document.querySelector('#x0'),
    inputs = [input_n, input_x],
    div_output = document.querySelector('#output0'),
    nl = '<br>',
    accumulate,
    sum,
    mul,
    funcs,
    output;

accumulate = function (op, func, initial, a, b) {
    return Array.range(a, b + 1)
        .map(function (n) {
            return func(n);
        })
        .reduce(function (x, y) {
            return op(x, y);
        }, initial);
};

sum = function (func, n) {
    return accumulate(
        function (x, y) {
            return x + y;
        }, func, 0, 1, n);
};
mul = function (func, n) {
    return accumulate(
        function (x, y) {
            return x * y;
        }, func, 1, 1, n);
};
funcs = [
    [
        function (n) {
            return sum(function (n) {
                return n;
            }, n);
        },
        function (n) {
            return n * (n + 1) / 2;
        }
    ],
    [
        function (n) {
            return sum(function (n) {
                return Math.pow(n, 2);
            }, n);
        },
        function (n) {
            return n * (n + 1) * (2 * n + 1) / 6;
        }
    ],
    [
        function (n) {
            return sum(function (n) {
                return Math.pow(n, 3);
            }, n);
        },
        function (n) {
            return Math.pow(n, 2) * Math.pow(n + 1, 2) / 4;
        }
    ],
    [
        function (n) {
            return sum(function (n) {
                return n * (n + 1);
            }, n);
        },
        function (n) {
            return 1 / 3 * n * (n + 1) * (n + 2);
        }
    ],
    [
        function (n) {
            return sum(function (n) {
                return 1 / (2 * n - 1) - 1 / (2 * n);
            }, n);
        },
        function (n) {
            return sum(function (i) {
                return 1 / (i + n);
            }, n);
        }
    ],
    [
        function (n) {
            return mul(function (n) {
                return 2 * n;
            }, n);
        },
        function (n) {
            return Math.pow(2, n) * mul(function (n) {
                return 2 * n - 1;
            }, n);
        }
    ],
    [
        function (n, x) {
            return sum(function (n) {
                return n * Math.pow(x, n -1);
            }, n);
        },
        function (x, n) {
            return (1 - Math.pow(x, n)) / Math.pow(1 - x, 2) -
                n * Math.pow(x, n) / (1 - x);
        }
    ]
];

output = function () {
    var n = parseInt(input_n.value, 10),
        x = parseFloat(input_x.value),
        code = 'a'.charCodeAt(0);

    div_output.innerHTML = '';
    Array.range(7).map(
        function (i) {
            return String.fromCharCode(code + i);
        })
        .forEach(function (ch, i) {
            var left = funcs[i][0](n, x),
                right = funcs[i][1](n, x);
            
            div_output.innerHTML +=
                '(' + ch + ')' + nl +
                ' 左辺: ' + left + ', 右辺: ' + right + nl +
                '左辺 = 右辺: ' + (left === right) + nl +
                '|左辺 - 右辺| = ' + (left - right).abs() + nl + nl;
        });
};

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

output();

n = n =