数学 - Python - JavaScript - 大小関係を見る - 不等式 – 不等式の解法 – 2次不等式(2つの2次不等式、共通部分)

学習環境

数学読本〈1〉数・式の計算/方程式/不等式 (松坂和夫(著)、岩波書店)の第4章(大小関係を見る - 不等式)、4.2(不等式の解法)、2次不等式の問10の解答を求めてみる。

1. $\begin{array}{l} (x - 3) (x + 2) > 0 \\ x < - 2, 3 < x \\ (x - 1) (2 x + 5) < 0 \\ - \frac{5}{2} < x < 1 \end{array}$
  
  よって、求める x の値の範囲は、
  
  $- \frac{5}{2} < x < - 2$
2. $\begin{array}{l} x^{2} - 1 > 1 \\ x < - 1, 1 < x \\ (x - 4) (x + 2) \leq 0 \\ - 2 \leq x \leq 4 \\ - 2 \leq x < - 1, 1 < x \leq 4 \end{array}$
3. $\begin{array}{l} (x - 2) (x - 5) \geq 0 \\ x \leq 2, 5 \leq x \\ (x - 3) (2 x + 1) > 0 \\ x < - \frac{1}{2}, 3 < x \\ x < - \frac{1}{2}, 5 \leq x \end{array}$
4. $\begin{array}{l} (x + 2) (3 x - 2) < 0 \\ - 2 < x < \frac{2}{3} \\ (x + 5) (2 x - 1) \leq 0 \\ - 5 \leq x \leq \frac{1}{2} \\ - 2 < x \leq \frac{1}{2} \end{array}$
5. $\begin{array}{l} (x - 5) (x + 7) \geq 0 \\ x \leq - 7, 5 \leq x \\ - 10 < x - 2 < 10 \\ - 8 < x < 12 \\ - 8 < x \leq - 7, 5 \leq x < 12 \end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, plot, factor
from sympy.solvers.inequalities import reduce_inequalities

print('10.')

x = symbols('x')
ts = [(x ** 2 - x - 6 > 0, 2 * x ** 2 + 3 * x - 5 < 0),
      (x ** 2 > 1, x ** 2 - 2 * x - 8 <= 0),
      (x ** 2 - 7 * x + 10 >= 0, 2 * x ** 2 - 5 * x - 3 > 0),
      (3 * x ** 2 + 4 * x - 4 < 0, 2 * x ** 2 + 9 * x - 5 <= 0),
      (x ** 2 + 2 * x - 35 >= 0, abs(x - 2) < 10)]

for i, (a, b) in enumerate(ts, 1):
    print(f'({i})')
    pprint(reduce_inequalities(a) & reduce_inequalities(b))
    print()

p = plot(x ** 2 + 2 * x - 35, abs(x - 2), 10,
         (x, -9, 13), legend=True, show=False)
colors = ['red', 'green', 'blue']
for i, _ in enumerate(p):
    p[i].line_color = colors[i]

p.save('sample10.png')

入出力結果(Terminal、cmd(コマンドプロンプト)、Jupyter(IPython))

$ python3 sample10.py
10.
(1)
((-∞ < x ∧ x < -2) ∨ (3 < x ∧ x < ∞)) ∧ -5/2 < x ∧ x < 1

(2)
-2 ≤ x ∧ x ≤ 4 ∧ ((-∞ < x ∧ x < -1) ∨ (1 < x ∧ x < ∞))

(3)
((5 ≤ x ∧ x < ∞) ∨ (x ≤ 2 ∧ -∞ < x)) ∧ ((-∞ < x ∧ x < -1/2) ∨ (3 < x ∧ x < ∞))

(4)
-5 ≤ x ∧ x ≤ 1/2 ∧ -2 < x ∧ x < 2/3

(5)
((5 ≤ x ∧ x < ∞) ∨ (x ≤ -7 ∧ -∞ < x)) ∧ -8 < x ∧ x < 12

$

HTML5

<div id="graph0"></div>
<pre id="output0"></pre>
<label for="r0">r = </label>
<input id="r0" type="number" min="0" value="1">
<label for="dx">dx = </label>
<input id="dx" type="number" min="0" step="0.001" value="0.005">
<br>
<label for="x1">x1 = </label>
<input id="x1" type="number" value="-10">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="15">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="-10">
<label for="y2">y2 = </label>
<input id="y2" type="number" value="15">

<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="sample10.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';

let fns = [[x => x ** 2 + 2 * x - 35, 'red'],
           [x => Math.abs(x - 2), 'green'],
           [x => 10, 'blue']];
    
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 = [],
        x0 = 10,
        y0 = 12,
        lines = [[-8, y1, -8, y2, 'orange'],
                 [-7, y1, -7, y2, 'brown'],
                 [5, y1, 5, y2, 'purple'],
                 [12, y1, 12, y2, 'pink']];
        
    fns
        .forEach((o) => {
            let [f, color] = o;
            for (let x = x1; x <= x2; x += dx) {
                let y = f(x);

                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('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].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 =

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2019年2月3日日曜日

数学 - Python - JavaScript - 大小関係を見る - 不等式 – 不等式の解法 – 2次不等式(2つの2次不等式、共通部分)

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