数学 - Python - JavaScript - 解析学 - 積分法 - 定積分の計算(三角関数(正弦、余弦、逆正弦、逆正接、2倍角))

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

解析入門〈2〉(松坂和夫(著)、岩波書店)の第8章(積分の計算)、8.2(定積分の計算)、問題1-1、2、3.を取り組んでみる。

1. $\begin{array}{l} - a \leq x \leq a \\ - 1 \leq \sin t \leq 1 \\ - a \leq a \sin t \leq a \\ x = a \sin t \\ 0 \leq x \leq a \\ 0 \leq a \sin t \leq a \\ 0 \leq \sin t \leq 1 \\ 0 \leq t \leq \frac{π}{2} \\ d x = a \cos t d t \\ \sqrt{a^{2} - x^{2}} = a \sqrt{1 - \sin^{2} t} \\ = a \cos t \\ \int_{0}^{\frac{π}{2}} a (\cos t) a (\cos t) d t \\ = a^{2} \int_{0}^{\frac{π}{2}} \cos^{2} t d t \\ = a^{2} \int_{0}^{\frac{π}{2}} \frac{\cos 2 t + \cos 0}{2} d t \\ = a^{2} \int_{0}^{\frac{π}{2}} \frac{\cos 2 t + 1}{2} d t \\ = \frac{a^{2}}{2} {[\frac{1}{2} \sin 2 t + t]}_{0}^{\frac{π}{2}} \\ = \frac{a^{2}}{2} ((\frac{1}{2} \sin π + \frac{π}{2}) - (\frac{1}{2} \sin 0 + 0)) \\ = \frac{a^{2}}{2} (\frac{π}{2}) \\ = \frac{π a^{2}}{4} \end{array}$
2. $\begin{array}{l} {[\arcsin \frac{x}{a}]}_{0}^{a} \\ = \arcsin 1 - \arcsin 0 \\ = \frac{π}{2} \end{array}$
3. $\begin{array}{l} \int_{0}^{1} \frac{1}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} d x \\ \int_{0}^{1} \frac{1}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x \\ t = x + \frac{1}{2} \\ x = 0, t = \frac{1}{2} \\ x = 1, t = \frac{3}{2} \\ x = t - \frac{1}{2} \\ d x = d t \\ \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{t^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x \\ = \frac{1}{\frac{\sqrt{3}}{2}} {[\arctan \frac{x}{\frac{\sqrt{3}}{2}}]}_{\frac{1}{2}}^{\frac{3}{2}} \\ = \frac{2}{\sqrt{3}} (\arctan \frac{\frac{3}{2}}{\frac{\sqrt{3}}{2}} - \arctan \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}) \\ = \frac{2}{\sqrt{3}} (\arctan \sqrt{3} - \arctan \frac{1}{\sqrt{3}}) \\ = \frac{2}{\sqrt{3}} (\frac{π}{3} - \frac{π}{6}) \\ = \frac{2 π}{6 \sqrt{3}} \\ = \frac{π}{3 \sqrt{3}} \end{array}$

コード(Emacs)

Python 3

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

from sympy import pprint, symbols, Integral, sqrt, Pow

print('1.')
x = symbols('x', real=True)
a = symbols('a', positive=True)
fs = [(sqrt(Pow(a, 2) - Pow(x, 2)), (0, a)),
      (1 / sqrt(Pow(a, 2) - Pow(x, 2)), (0, a)),
      (1 / (1 + x + x ** 2), (0, 1))]

for i, (f, (x1, x2)) in enumerate(fs, 1):
    print(f'({i})')
    I = Integral(f, (x, x1, x2))
    pprint(I)
    I = I.doit()
    pprint(I)
    print('factor:')
    pprint(I.factor())
    print('expand:')
    pprint(I.expand())

入出力結果(Terminal, IPython)

$ ./sample1.py
1.
(1)
a                
⌠                
⎮    _________   
⎮   ╱  2    2    
⎮ ╲╱  a  - x   dx
⌡                
0                
   2
π⋅a 
────
 4  
factor:
   2
π⋅a 
────
 4  
expand:
   2
π⋅a 
────
 4  
(2)
a                
⌠                
⎮      1         
⎮ ──────────── dx
⎮    _________   
⎮   ╱  2    2    
⎮ ╲╱  a  - x     
⌡                
0                
π
─
2
factor:
π
─
2
expand:
π
─
2
(3)
1              
⌠              
⎮     1        
⎮ ────────── dx
⎮  2           
⎮ x  + x + 1   
⌡              
0              
√3⋅π
────
 9  
factor:
√3⋅π
────
 9  
expand:
√3⋅π
────
 9  
$

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="-5">
<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="5">

<br>
<label for="a0">a = </label>
<input id="a0" type="number" min="0" value="2">

<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="sample1.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_a0 = document.querySelector('#a0'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
             input_a0],
    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 f3 = (x) => 1 / (1 + x + 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),
        a0 = parseFloat(input_a0.value);
    
    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
        return;
    }

    let points = [],
        lines = [[a0, y1, a0, y2, 'red'],
                 [1, y1, 1, y2, 'blue']],
        f1 = (x) => Math.sqrt(a0 ** 2 - x ** 2),
        f2 = (x) => 1 / Math.sqrt(a0 ** 2 - x ** 2),
        fns = [[f1, 'green'],
               [f2, 'brown'],
               [f3, 'orange']];

    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]);
                }
            }
        });
    
    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) => d[2] || 'green');

    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.append('g')
        .attr('transform', `translate(0, ${height - padding})`)
        .call(xaxis);

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

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

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

Kamimura's blog

2017年7月23日日曜日

数学 - Python - JavaScript - 解析学 - 積分法 - 定積分の計算(三角関数(正弦、余弦、逆正弦、逆正接、2倍角))

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