数学 - 解析学 - JavaScript - 数列と級数 - 数列の収束条件(下に有界な単調減少数列、極限値、平方根、近似(ニュートン法))

解析入門〈1〉

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
• Emacs (Text Editor)
• JavaScript (プログラミング言語)
• D3.js (JavaScript Library)
• 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)
  • インタラクティブ・データビジュアライゼーション(Scott Murray (著)、長尾 高弘 (翻訳)、オライリージャパン)

解析入門〈1〉(松坂 和夫(著)、岩波書店)の第2章(数列と級数)、2.2(数列の収束条件)、問題2.2、3.を取り組んでみる。

問3

1. 
   $\begin{array}{l}{a}_n-a_{n+1}\\ =a_n-\frac{1}{2}\left(a_n+\frac{\alpha }{a_n}\right)\\ =\frac{1}{2}\left(a_n-\frac{\alpha }{a_n}\right)\\ =\frac{1}{2}\left(\frac{a_n{}^2-\alpha }{a_n}\right)\\ \\ a_n{}^2>\alpha \\ と仮定する。\\ a_{n+1}^2-\alpha \\ ={\left(\frac{1}{2}\left(a_n+\frac{\alpha }{a_n}\right)\right)}^2-\alpha \\ =\frac{{\left(a_n{}^2+\alpha \right)}^2}{4a_n{}^2}-\alpha \\ =\frac{{\left(a_n^2+\alpha \right)}^2-4\alpha a_n{}^2}{4a_n{}^2}\\ =\frac{a_n{}^4+2\alpha a_n{}^2+{\alpha }^2-4\alpha a_n{}^2}{4a_n{}^2}\\ =\frac{a_n{}^4-2\alpha a_n{}^2+{\alpha }^2}{4a_n{}^2}\\ =\frac{{\left(a_n{}^2-\alpha \right)}^2}{4a_n{}^2}>0\\ a_{n+1}{}^2>\alpha \\ \\ a_n-a_{n+1}>0\\ a_n>a_{n+1}\\ \\ a_n>0\\ \left(a_n\right)は単調減少で下に有界なので収束する。\\ \beta =\frac{1}{2}\left(\beta +\frac{\alpha }{\beta }\right)\\ 2\beta =\beta +\frac{\alpha }{\beta }\\ {\beta }^2=\alpha \\ \beta =\sqrt{\alpha }\\ \underset{n\to \infty }{\lim }{a}_n=\sqrt{\alpha }\end{array}$
2. 
   $\begin{array}{l}{\epsilon }_{n+1}=a_{n+1}-\sqrt{\alpha }\\ =\frac{1}{2}\left(a_n+\frac{\alpha }{a_n}\right)-\sqrt{\alpha }\\ =\frac{a_n{}^2+\alpha -2a_n\sqrt{\alpha }}{2a_n}\\ =\frac{{\left(a_n-\sqrt{\alpha }\right)}^2}{2a_n}\\ =\frac{{\epsilon }_n{}^2}{2a_n}\\ <\frac{{\epsilon }_n{}^{2·}}{2\sqrt{\alpha }}\\ =\frac{{\epsilon }_n{}^2}{\beta }\\ {\epsilon }_{n+1}<\frac{{\epsilon }_n{}^2}{\beta }\\ \frac{{\epsilon }_{n+1}}{\beta }<\frac{{\epsilon }_n{}^2}{{\beta }^2}={\left(\frac{{\epsilon }_n}{\beta }\right)}^2<{\left(\frac{{\epsilon }_{n-1}}{\beta }\right)}^{2·2^1}<···<{\left(\frac{{\epsilon }_1}{\beta }\right)}^{2·2^{n-1}}={\left(\frac{\epsilon }{\beta }\right)}^{2n}\\ \frac{{\epsilon }_{n+1}}{\beta }<{\left(\frac{\epsilon }{\beta }\right)}^{2n}\end{array}$

HTML5

<div id="graph0"></div>
<div id="output0"></div>
<label for="alpha0">α = </label>
<input id="alpha0" type="number" min="2" value="2">
<label for="a1">a1 = </label>
<input id="a1" type="number" min="2" value="2">
<label for="n0">n = </label>
<input id="n0" type="number" min="1" step="1" value="10">


<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/d3/4.2.6/d3.min.js" integrity="sha256-5idA201uSwHAROtCops7codXJ0vja+6wbBrZdQ6ETQc=" crossorigin="anonymous"></script>
<script src="sample3.js"></script>

JavaScript

{
    let div_graph = document.querySelector('#graph0'),
        div_output = document.querySelector('#output0'),
        input_alpha = document.querySelector('#alpha0'),
        input_a = document.querySelector('#a1'),
        input_n = document.querySelector('#n0'),
        inputs = [input_alpha, input_a, input_n],
        width = 600,
        height = 600,
        padding = 50;

    let f = (alpha, prev) => {
        return 1 / 2 * (prev + alpha / prev);
    };
    let getPoints = (alpha, a1, n) => {
        let prev = a1,
            result = [[1, prev]];

        for (let i = 2; i <= n; i += 1) {
            let next = f(alpha, prev);
            result.push([i, next]);
            prev = next;
        }
        return result;
    }
    let plot = () => {
        let alpha = parseFloat(input_alpha.value),
            a1 = parseFloat(input_a.value);

        if (a1 > Math.sqrt(alpha)) {
            let n = parseInt(input_n.value, 10),
                points = getPoints(alpha, a1, n);

            console.log(points);
            let xscale = d3.scaleLinear()
                .domain([1, n])
                .range([padding, width - padding])
            let yscale = d3.scaleLinear()
                .domain([points[points.length - 1][1] * 0.9, a1])
                .range([height - padding, padding]);

            console.log(points.map((x) => x[1]));
            div_graph.innerHTML = '';
            let svg = d3.select('#graph0')
                .append('svg')
                .attr('width', width)
                .attr('height', height);

            svg.selectAll('circle')
                .data(points)
                .enter()
                .append('circle')
                .attr('cx', (d) => xscale(d[0]))
                .attr('cy', (d) => yscale(d[1]))
                .attr('r', 2)
                .attr('color', 'green');

            let xaxis = d3.axisBottom().scale(xscale);
            let yaxis = d3.axisLeft().scale(yscale);

            svg.append('line')
                .attr('x1', xscale(1))
                .attr('y1', yscale(Math.sqrt(alpha)))
                .attr('x2', xscale(n))
                .attr('y2', yscale(Math.sqrt(alpha)))
                .attr('stroke', 'red');
            
            svg.append('g')
                .attr('transform', `translate(0, ${height - padding})`)
                .call(xaxis);
            svg.append('g')
                .attr('transform', `translate(${padding}, 0)`)
                .call(yaxis);

            div_output.innerHTML =
                points.map((x) => `${x[0]}: ${x[1]}`).join('<br>') +
                `<br>Math.sqrt(${alpha}): ${Math.sqrt(alpha)}`;
            
        } else {
            div_output.innerHTML = '注意: a1は√α より大きい数';
        }
    };

    inputs.forEach((input) => input.onchange = plot);
    plot();

}

1: 2
2: 1.5
3: 1.4166666666666665
4: 1.4142156862745097
5: 1.4142135623746899
6: 1.414213562373095
7: 1.414213562373095
8: 1.414213562373095
9: 1.414213562373095
10: 1.414213562373095
Math.sqrt(2): 1.4142135623730951

α = a1 = n =