数学 - Python - JavaScript - 細分による加法 - 積分法 – 定積分の性質と計算 - 簡単な例(絶対値、累乗(べき乗)、三角関数(正弦、2倍角)、指数関数)

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, Microsoft Edge, Google Chrome...)用JavaScript Library: MathJax
参考書籍

数学読本〈5〉微分法の応用/積分法/積分法の応用/行列と行列式(松坂和夫(著)、岩波書店)の第19章(細分による加法 - 積分法)、19.3(定積分の性質と計算)、簡単な例、問22.を取り組んでみる。

1. $\begin{array}{l} x^{2} - 1 = 0 \\ x = \pm 1 \\ \int_{0}^{1} (1 - x^{2}) d x + \int_{1}^{3} (x^{2} - 1) d x \\ = {[x - \frac{1}{3} x^{3}]}_{0}^{1} + {[\frac{1}{3} x^{3} - x]}_{1}^{3} \\ = 1 - \frac{1}{3} + 9 - 3 - (\frac{1}{3} - 1) \\ = 8 - \frac{2}{3} \\ = \frac{22}{3} \end{array}$
2. $\begin{array}{l} \int_{- 1}^{1} | x | d x - \int_{- 1}^{1} | x - 1 | d x \\ = 2 \int_{0}^{1} x d x + \int_{- 1}^{1} (x - 1) d x \\ = 2 {[\frac{1}{2} x^{2}]}_{0}^{1} + \int_{- 1}^{1} x d x - \int_{- 1}^{1} 1 d x \\ = 2 \cdot \frac{1}{2} + 0 - 2 \\ = - 1 \end{array}$
3. $\begin{array}{l} x - x^{2} \\ = - x (x - 1) \\ - \int_{- 1}^{0} (x - x^{2}) d x + \int_{0}^{1} (x - x^{2}) d x - \int_{1}^{2} (x - x^{2}) d x \\ = - {[\frac{1}{2} x^{2} - \frac{1}{3} x^{3}]}_{- 1}^{0} + {[\frac{1}{2} x^{2} - \frac{1}{3} x^{3}]}_{0}^{1} - {[\frac{1}{2} x^{2} - \frac{1}{3} x^{3}]}_{1}^{2} \\ = \frac{1}{2} + \frac{1}{3} + \frac{1}{2} - \frac{1}{3} - (2 - \frac{8}{3} - (\frac{1}{2} - \frac{1}{3})) \\ = - 1 + \frac{7}{3} + \frac{1}{2} \\ = \frac{11}{6} \end{array}$
4. $\begin{array}{l} \sqrt{x} - 2 = 0 \\ x = 4 \\ - \int_{1}^{4} (\sqrt{x} - 2) d x + \int_{4}^{9} (\sqrt{x} - 2) d x \\ = - {[\frac{2}{3} x^{\frac{3}{2}} - 2 x]}_{1}^{4} + {[\frac{2}{3} x^{\frac{3}{2}} - 2 x]}_{4}^{9} \\ = - ((\frac{2}{3} \cdot 2^{3} - 8) - (\frac{2}{3} - 2)) + ((\frac{2}{3} \cdot 3^{3} - 18) - (\frac{2}{3} \cdot 2^{3} - 8)) \\ = - (\frac{16}{3} - 8 + \frac{4}{3}) - (\frac{16}{3} - 8) \\ = - \frac{36}{3} + 16 \\ = - 12 + 16 \\ = 4 \end{array}$
5. $\begin{array}{l} 0 \leq x \leq π \\ 0 \leq 2 x \leq 2 π \\ 2 x = π \\ x = \frac{π}{2} \\ \int_{0}^{\frac{π}{2}} \sin 2 x d x - \int_{\frac{π}{2}}^{π} \sin 2 x d x \\ = - {[\frac{1}{2} \cos 2 x]}_{0}^{\frac{π}{2}} + {[\frac{1}{2} \cos 2 x]}_{\frac{π}{2}}^{π} \\ = - \frac{1}{2} (\cos π - \cos 0) + \frac{1}{2} (\cos 2 π - \cos π) \\ = - \frac{1}{2} (- 1 - 1) + \frac{1}{2} (1 + 1) \\ = 1 + 1 \\ = 2 \end{array}$
6. $\begin{array}{l} e^{x} = 2 \\ \log e^{x} = \log 2 \\ x \log e = \log 2 \\ x = \log 2 \\ 0 < \log 2 < 1 \\ \int_{0}^{\log 2} (2 - e^{x}) d x + \int_{\log 2}^{1} (e^{x} - 2) d x \\ = {[2 x - e^{x}]}_{0}^{\log 2} + {[e^{x} - 2 x]}_{\log 2}^{1} \\ = 2 \log 2 - e^{\log 2} - (- 1) + e - 2 - (e^{\log 2} - 2 \log 2) \\ = 4 \log 2 - 2 + 1 + e - 2 - 2 \\ = 4 \log 2 + e - 5 \end{array}$

コード(Emacs)

Python 3

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

from sympy import pprint, symbols, Integral, plot, Abs, sqrt, sin, pi, exp

print('22.')
x = symbols('x')
# (4) は計算が終了しなかったから省略
fs = [(Abs(x ** 2 - 1), (0, 3)),
      (Abs(x) - Abs(x - 1), (-1, 1)),
      (Abs(x - x ** 2), (-1, 2)),
      # (Abs(sqrt(x) - 2), (1, 9)),
      (Abs(sin(2 * x)), (0, pi)),
      (Abs(exp(x) - 2), (0, 1))]

# 積分の値を求められないのは、SymPy の問題か、やり方が違うからか分からず。
for i, (f, (x1, x2)) in enumerate(fs, 1):
    print(f'({i})')
    I = Integral(f, (x, x1, x2))
    for o in [I, I.doit()]:
        pprint(o)
    try:
        p = plot(f, show=False, legend=True)
        p.save(f'sample22_{i}.svg')
    except Exception as err:
        print(type(err), err)
    print()

入出力結果(Terminal, IPython)

$ ./sample22.py
22.
(1)
3            
⌠            
⎮ │ 2    │   
⎮ │x  - 1│ dx
⌡            
0            
3            
⌠            
⎮ │ 2    │   
⎮ │x  - 1│ dx
⌡            
0            

(2)
1                    
⌠                    
⎮  (│x│ - │x - 1│) dx
⌡                    
-1                   
1                    
⌠                    
⎮  (│x│ - │x - 1│) dx
⌡                    
-1                   

(3)
2             
⌠             
⎮  │ 2    │   
⎮  │x  - x│ dx
⌡             
-1            
2             
⌠             
⎮  │ 2    │   
⎮  │x  - x│ dx
⌡             
-1            

(4)
π              
⌠              
⎮ │sin(2⋅x)│ dx
⌡              
0              
π              
⌠              
⎮ │sin(2⋅x)│ dx
⌡              
0              

(5)
1            
⌠            
⎮ │ x    │   
⎮ │ℯ  - 2│ dx
⌡            
0            
1            
⌠            
⎮ │ x    │   
⎮ │ℯ  - 2│ dx
⌡            
0            

$

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="-10">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="10">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="-10">
<label for="y2">y2 = </label>
<input id="y2" type="number" value="10">

<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="sample22.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'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2],
    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 f4 = (x) => Math.abs(Math.sqrt(x) - 2),
    f41 = (x) => Math.sqrt(x) - 2;

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);

    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
        return;
    }
    
    let points = [],
        lines = [[1, y1, 1, y2, 'red'],
                 [9, y1, 9, y2, 'red']],
        fns = [[f41, 'blue'],
               [f4, 'green']],
        fns1 = [],
        fns2 = [];

    fns.forEach((o) => {
        let [fn, color] = o;
        for (let x = x1; x <= x2; x += dx) {
            let y = fn(x);

            if (Math.abs(y) < Infinity) {
                points.push([x, y, color]);
            }
        }
    });
    fns1.forEach((o) => {
        let [fn, color] = o;
        
        lines.push([x1, fn(x1), x2, fn(x2), color]);
    });
    fns2.forEach((o) => {
        let [fn, color] = o;

        for (let x = x1; x <= x2; x += dx0) {
            let g = fn(x);
            
            lines.push([x1, g(x1), x2, g(x2), color]);
        }        
    });
    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('line')
        .data([[x1, 0, x2, 0], [0, y1, 0, y2]].concat(lines))
        .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', (d) => d[4] || 'black');
    
    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) => d[2] || 'green');
    
    svg.append('g')
        .attr('transform', `translate(0, ${height - padding})`)
        .call(xaxis);

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

    [fns, fns1, fns2].forEach((fs) => p(fs.join('\n')));
};

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

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

Kamimura's blog

2017年8月27日日曜日

数学 - Python - JavaScript - 細分による加法 - 積分法 – 定積分の性質と計算 - 簡単な例(絶対値、累乗(べき乗)、三角関数(正弦、2倍角)、指数関数)

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