数学 – Python - JavaScript - 数学はここから始まる - 数 – 方程式とその解法 – 2次方程式の解法(一般化、解の公式)

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3章(数学の威力を発揮する - 方程式)、3.1(方程式とその解法)、2次方程式の解法の問4.を取り組んでみる。

1. $\begin{array}{l} x = \frac{- 5 \pm \sqrt{25 + 96}}{8} \\ = \frac{- 5 \pm \sqrt{121}}{8} \\ = \frac{- 5 \pm 11}{8} \\ = - 2, \frac{3}{4} \end{array}$
2. $\begin{array}{l} x = \frac{2 \pm \sqrt{4 + 16}}{2} \\ = \frac{2 (1 \pm \sqrt{1 + 4})}{2} \\ = 1 \pm \sqrt{5} \end{array}$
3. $\begin{array}{l} x = \frac{- 8 \pm \sqrt{8^{2} + 4 \cdot 16}}{2} \\ = \frac{2 (- 4 \pm 2 \sqrt{4 + 16})}{2} \\ = - 4 \pm 2 \sqrt{5} \end{array}$
4. $\begin{array}{l} x = \frac{11 \pm \sqrt{121 - 76}}{2} \\ = \frac{11 \pm \sqrt{45}}{2} \\ = \frac{11 \pm 3 \sqrt{5}}{2} \end{array}$
5. $\begin{array}{l} x = \frac{- 17 \pm \sqrt{1 7^{2} - 24 \cdot 12}}{12} \\ = \frac{- 17 \pm \sqrt{289 - 288}}{12} \\ = - \frac{3}{2}, - \frac{4}{3} \end{array}$
6. $\begin{array}{l} x = \frac{4 \pm \sqrt{4^{2} - 4 \cdot 3 \cdot (- 2)}}{2 \cdot 3} \\ = \frac{2 (2 \pm \sqrt{4 + 6})}{2 - 3} \\ = \frac{2 \pm \sqrt{10}}{3} \end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, solve, plot

print('4.')

x = symbols('x', real=True)

eqs = [(4, 5, -6),
       (1, -2, -4),
       (1, 8, -16),
       (1, -11, 19),
       (6, 17, 12),
       (3, -4, -2)]

for i, (a, b, c) in enumerate(eqs, 1):
    print(f'({i})')
    pprint(solve(a * x ** 2 + b * x + c, dict=True))
    print()

p = plot(*[a * x ** 2 + b * x + c for a, b, c in eqs], show=False, legend=True)
colors = ['red', 'green', 'blue', 'orange', 'brown', 'purple']
for i, color in enumerate(colors):
    p[i].line_color = color
p.save('sample4.svg')

入出力結果(Terminal, Jupyter(IPython))

$ ./sample4.py
4.
(1)
[{x: -2}, {x: 3/4}]

(2)
[{x: 1 + √5}, {x: -√5 + 1}]

(3)
[{x: -4 + 4⋅√2}, {x: -4⋅√2 - 4}]

(4)
⎡⎧     3⋅√5   11⎫  ⎧   3⋅√5   11⎫⎤
⎢⎨x: - ──── + ──⎬, ⎨x: ──── + ──⎬⎥
⎣⎩      2     2 ⎭  ⎩    2     2 ⎭⎦

(5)
[{x: -3/2}, {x: -4/3}]

(6)
⎡⎧   2   √10⎫  ⎧     √10   2⎫⎤
⎢⎨x: ─ + ───⎬, ⎨x: - ─── + ─⎬⎥
⎣⎩   3    3 ⎭  ⎩      3    3⎭⎦

$

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="0.001" 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">

<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'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2],
    p = (x) => pre0.textContent += x + '\n';

let fns = [[x => 4 * x ** 2 + 5 * x - 6, 'red']];
    
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 = [[-2, y1, -2, y2, 'green'],
                 [3 / 4, y1, 3 / 4, y2, 'blue']];

    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 =

Kamimura's blog

2018年10月27日土曜日

数学 – Python - JavaScript - 数学はここから始まる - 数 – 方程式とその解法 – 2次方程式の解法(一般化、解の公式)

0 コメント:

コメントを投稿