数学 - Python - JavaScript - 解析学 - 関数の近似、テイラーの定理 - テイラーの定理(方程式の解の近似値を求める方法、ニュートン(Newton)の方法)

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
参考書籍

解析入門〈2〉(松坂和夫(著)、岩波書店)の第6章(関数の近似、テイラーの定理)、6.1(テイラーの定理)、問題4.を取り組んでみる。

1. $\begin{array}{l} f (a) < 0, f (b) > 0 で、 f は [a, b] で連続なので、中間値の定理により、 \\ \exists ξ \in [a, b] [f (ξ) = 0] \\ また、 f (x) > 0 より、 f は狭義単調増加より、 \\ \exists 1 ξ \in [a, b] [f (ξ) = 0] \end{array}$
2. $点 (b_{n}, f (b_{n})) における接線と x 軸との交点の x 座標が b_{n + 1} 。$
3. $\begin{array}{l} f' (x), f'' (x) > 0 \\ ξ < b_{n + 1} < b_{n} \\ \lim_{n \to \infty} b_{n} = α \\ α = α - \frac{f (α)}{f' (α)} \\ f (α) = 0 \\ α = ξ \\ \lim_{n \to \infty} b_{n} = ξ \end{array}$
4. $\begin{array}{l} ξ < c_{n} < b_{n} \\ f (ξ) = f (b_{n}) + f' (b_{n}) (ξ - b_{n}) + \frac{f'' (c_{n})}{2!} {(ξ - b_{n})}^{2} \\ 0 = f (b_{n}) + f' (b_{n}) (ξ - b_{n}) + \frac{f'' (c_{n})}{2} {(ξ - b_{n})}^{2} \\ 0 = \frac{f (b_{n})}{f' (b_{n})} + ξ - b_{n} + \frac{f'' (c_{n})}{2 f' (b_{n})} {(ξ - b_{n})}^{2} \\ b_{n} - \frac{f (b_{n})}{f' (b_{n})} - ξ = \frac{f'' (c_{n})}{2 f' (b_{n})} {(b_{n} - ξ)}^{2} \\ b_{n + 1} - ξ = \frac{f'' (c_{n})}{2 f' (b_{n})} {(b_{n} - ξ)}^{2} \end{array}$
5. $\begin{array}{l} 0 < b_{n + 1} - ξ = \frac{f'' (c_{n})}{2 f' (b_{n})} {(b_{n} - ξ)}^{2} \\ \leq \frac{δ_{2}}{2 δ_{1}} {(b_{n} - ξ)}^{2} \\ = A {(b_{n} - ξ)}^{2} \\ A (b_{n + 1} - ξ) \leq {(A (b_{n} - ξ))}^{2} \\ A (b_{n + 1} - ξ) \leq {(A (b_{1} - ξ))}^{2 n} \\ b_{n + 1} - ξ \leq \frac{1}{A} {(A (b_{1} - ξ))}^{2 n} \end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

from sympy import pprint, symbols, Derivative, plot, solve

print('(a)')
x = symbols('x', positive=True)
a = 1
b = 2
f = x ** 2 - 2
f1 = Derivative(f, x, 1).doit()
f2 = Derivative(f, x, 2).doit()

delta1 = 1
delta2 = 3
p = plot(f, f1, f2, delta1, delta2, (x, a, b), show=False, legend=True)
p.save('sample4.svg')

s = solve(f, x)
pprint(s)
x0 = s[0]

print('(b)')


def bn(n):
    if n == 1:
        return b
    b0 = bn(n - 1)
    return b0 - f.subs({x: b0}) / f1.subs({x: b0})

for i in range(1, 10):
    print('n = {0}: {1}, {2}, {3}'.format(
        i, float(bn(i)), float(bn(i) - x0), bn(i + 1) < bn(i)))

入出力結果(Terminal, IPython)

$ ./sample4.py
(a)
[√2]
(b)
n = 1: 2.0, 0.585786437626905, True
n = 2: 1.5, 0.08578643762690495, True
n = 3: 1.4166666666666667, 0.0024531042935716178, True
n = 4: 1.4142156862745099, 2.12390141475512e-06, True
n = 5: 1.4142135623746899, 1.5948618246068547e-12, True
n = 6: 1.4142135623730951, 8.992928321650453e-25, True
n = 7: 1.4142135623730951, 2.859283843333951e-49, True
n = 8: 1.4142135623730951, 2.8904771932153646e-98, True
n = 9: 1.4142135623730951, -1.512731216738015e-123, True
$

HTML5

<div id="graph0"></div>
<pre id="output0"></pre>
<label for="r0">r = </label>
<input id="r0" type="number" min="0" value="0.5">
<label for="dx">dx = </label>
<input id="dx" type="number" min="0" step="0.0001" value="0.001">
<br>
<label for="x1">x1 = </label>
<input id="x1" type="number" value="-2">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="5">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="-5">
<label for="y2">y2 = </label>
<input id="y2" type="number" value="10">
<br>
<label for="n0">n = </label>
<input id="n0" type="number" min="1" step="1" value="1">

<button id="draw0">draw</button>
<button id="clear0">clear</button>

<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="sample4.js"></script>

JavaScript

let div0 = document.querySelector('#graph0'),
    pre0 = document.querySelector('#output0'),
    width = 600,
    height = 600,
    padding = 50,
    btn0 = document.querySelector('#draw0'),
    btn1 = document.querySelector('#clear0'),
    input_r = document.querySelector('#r0'),
    input_dx = document.querySelector('#dx'),
    input_x1 = document.querySelector('#x1'),
    input_x2 = document.querySelector('#x2'),
    input_y1 = document.querySelector('#y1'),
    input_y2 = document.querySelector('#y2'),
    input_n = document.querySelector('#n0'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
              input_n],
    p = (x) => pre0.textContent += x + '\n',
    range = (start, end, step=1) => {
        let res = [];
        for (let i = start; i < end; i += step) {
            res.push(i);
        }
        return res;
    };

let f = (x) => x ** 2 - 2,
    f1 = (x) => 2 * x,
    f2 = (x) => 2,
    a = 1,
    b = 4,
    bn = (n) => {
        if (n === 1) {
            return b;
        }
        let b0 = bn(n - 1);
        return b0 - f(b0) / f1(b0);
    },
    g = (f, f1, x0) => (x) => f1(x0) * (x - x0) + f(x0);

let draw = () => {
    pre0.textContent = '';

    let r = parseFloat(input_r.value),
        dx = parseFloat(input_dx.value),
        x1 = parseFloat(input_x1.value),
        x2 = parseFloat(input_x2.value),
        y1 = parseFloat(input_y1.value),
        y2 = parseFloat(input_y2.value),
        n = parseInt(input_n.value, 10);
    
    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2 || n < 1) {
        return;
    }

    let points = [],
        fg = g(f, f1, bn(n));

    for (let x = x1; x <= x2; x += dx) {
        let y = f(x);
        
        if (Math.abs(x) < Infinity) {
            points.push([x, y]);
        }
    }
    let t = points.length;
    for (let x = x1; x <= x2; x += dx) {
        let y = fg(x);
        
        if (Math.abs(x) < Infinity) {
            points.push([x, y]);
        }
    }
    
    let xscale = d3.scaleLinear()
        .domain([x1, x2])
        .range([padding, width - padding]);

    let yscale = d3.scaleLinear()
        .domain([y1, y2])
        .range([height - padding, padding]);

    let xaxis = d3.axisBottom().scale(xscale);
    let yaxis = d3.axisLeft().scale(yscale);
    div0.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', r)
        .attr('fill', (d, i) =>
              i < t ? 'green' : 'blue');

    svg.selectAll('line')
        .data([[x1, 0, x2, 0], [0, y1, 0, y2]])
        .enter()
        .append('line')
        .attr('x1', (d) => xscale(d[0]))
        .attr('y1', (d) => yscale(d[1]))
        .attr('x2', (d) => xscale(d[2]))
        .attr('y2', (d) => yscale(d[3]))
        .attr('stroke', 'black');
    
    svg.append('g')
        .attr('transform', `translate(0, ${height - padding})`)
        .call(xaxis);

    svg.append('g')
        .attr('transform', `translate(${padding}, 0)`)
        .call(yaxis);

    p(`b_${n} = ${bn(n)}`);
    p(`√2  = ${Math.sqrt(2)}`);
};

inputs.forEach((input) => input.onchange = draw);
btn0.onclick = draw;
btn1.onclick = () => pre0.textContent = '';
draw();

r = dx =
x1 = x2 =
y1 = y2 =
n =

Kamimura's blog

2017年6月8日木曜日

数学 - Python - JavaScript - 解析学 - 関数の近似、テイラーの定理 - テイラーの定理(方程式の解の近似値を求める方法、ニュートン(Newton)の方法)

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