数学 - JavaScript - 「場合の数」 を数える - 順列・組合せ – 二項定理

数学読本〈4〉

学習環境

• 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

数学読本〈4〉数列の極限，順列/順列・組合せ/確率/関数の極限と微分法(松坂 和夫(著)、岩波書店)の第15章(「場合の数」 を数える - 順列・組合せ)、15.3(二項定理)、二項定理、問36、37.を取り組んでみる。

36. 1. 
       $\begin{array}{l}4a^4+4a^{3}2b+6a^2{\left(2b\right)}^2+4a{\left(2b\right)}^3+{\left(2b\right)}^4\\ =4a^4+8a^{3}b+24a^2{b}^2+32a{b}^3+16b^4\end{array}$
    2. 
       $\begin{array}{l}{x}^5+5x^4\left(-y\right)+10x^3{\left(-y\right)}^2+10x^2{\left(-y\right)}^3+5x{\left(-y\right)}^4+{\left(-y\right)}^5\\ =x^5-5x^{4}y+10x^3{y}^2-10x^2{y}^3-5x^{4}y-y^5\end{array}$
    3. 
       $\begin{array}{l}{2}^6+6·2^{5}x+15·2^4{x}^2+20·2^3{x}^3+15·2^2{x}^4+6·2x^5+x^6\\ =64+192x+240x^2+160x^3+60x^4+x^6\end{array}$
    4. 
       $\begin{array}{l}{\left(2x\right)}^7+7{\left(2x\right)}^6\left(-3y\right)+21{\left(2x\right)}^5{\left(-3y\right)}^2+35{\left(2x\right)}^4{\left(-3y\right)}^3+35{\left(2x\right)}^3{\left(-3y\right)}^4+21{\left(2x\right)}^2{\left(-3y\right)}^5+7\left(2x\right){\left(-3y\right)}^6+{\left(-3y\right)}^7\\ =128x^7-1344x^{6}y+6048x^5{y}^2-15120x^4{y}^3+22680x^3{y}^4-20412x^2{y}^5+10206x{y}^6-2187y^7\end{array}$
37. 1. 
       ${\displaystyle \sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)a^n{b}^{n-i}}$
    2. 
       ${\displaystyle \sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right){\left(-1\right)}^{n-i}{x}^{n-i}}$

コード(Emacs)

HTML5

<button id="run0">run</button>
<button id="clear0">clear</button>
<pre id="output0"></pre>
<script src="sample36.js"></script>

JavaScript

let input0 = document.querySelector('#n0'),
    btn0 = document.querySelector('#run0'),
    btn1 = document.querySelector('#clear0'),
    pre0 = document.querySelector('#output0'),
    p = (x) => pre0.textContent += x + '\n';

let range = (start, end, step=1) => {
    let iter = (i, result) => {
        return i >= end ? result : iter(i + step, result.concat([i]));
    }
    return iter(start, []);
};
let factorial = (n) => {
    return n <= 1 ? 1 : n * factorial(n - 1);
};

let combination = (n, r) => {
    return factorial(n) / (factorial(r) * factorial(n - r));
};
let term = (n, i, a, b) => {
    return combination(n, i) * Math.pow(a, n - i) * Math.pow(b, i);
};
let f = (n, a1, b1, a2, b2) => {
    return range(0, n + 1)
        .map((i) => `${term(n, i, a1, b1)}${a2}^${n - i}${b2}^${i}`)
        .join(' + ');
};
let output = () => {
    p(f(4, 1, 2, 'a', 'b'));
    p(f(5, 1, -1, 'x', 'y'));
    p(f(6, 2, 1, '', 'x'));
    p(f(7, 2, -3, 'x', 'y'));
};

btn0.onclick = output;
btn1.onclick = () => {
    pre0.textContent = '';
};

output();