数学 – Python - JavaScript - 数学はここから始まる - 数 – 2次方程式と複素数 – 2次式の因数分解(複素数の範囲)

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)、3.2(2次方程式と複素数)、2次式の因数分解の問11.を取り組んでみる。

1. $\begin{array}{l} \frac{- 89 \pm \sqrt{8 9^{2} - 4 \cdot 56 \cdot 35}}{112} \\ = \frac{- 89 \pm \sqrt{7921 - 7840}}{112} \\ = \frac{- 89 \pm \sqrt{81}}{112} \\ = \frac{- 89 \pm 9}{112} \\ = \frac{- 98}{112}, \frac{- 80}{112} \\ = - \frac{7}{8}, - \frac{5}{7} \end{array}$
2. $\frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$
3. $\begin{array}{l} x^{2} = - \frac{25}{9} \\ x = \pm \frac{5}{3} i \end{array}$
4. $\begin{array}{l} \frac{4 \pm \sqrt{4^{2} - 4 \cdot 3^{2}}}{2 \cdot 3} \\ = \frac{4 \pm 2 \sqrt{4 - 9}}{2 \cdot 3} \\ = \frac{2 \pm \sqrt{5} i}{3} \end{array}$
5. $\begin{array}{l} \frac{- 14 \sqrt{2} \pm \sqrt{1 4^{2} \cdot 2 - 4 \cdot 2 \cdot 13}}{2 \cdot 2} \\ = \frac{- 14 \sqrt{2} \pm 2 \sqrt{7 \cdot 22 - 2 \cdot 13}}{2 \cdot 2} \\ = \frac{- 7 \sqrt{2} \pm \sqrt{98 - 26}}{2} \\ = \frac{- 7 \sqrt{2} \pm \sqrt{72}}{2} \\ = \frac{- 7 \sqrt{2} \pm 6 \sqrt{2}}{2} \\ = - \frac{13 \sqrt{2}}{2}, - \frac{\sqrt{2}}{2} \end{array}$

コード(Emacs)

Python 3

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

print('11.')

x = symbols('x')
coefficients = [(56, 89, 35),
                (1, -1, -1),
                (9, 0, 25),
                (3, -4, 3),
                (2, 14 * sqrt(2), 13)]
fs = [a * x ** 2 + b * x + c for a, b, c in coefficients]

for i, f in enumerate(fs, 1):
    print(f'({i})')
    pprint(solve(f, x, dict=True))
    print()

p = plot(*fs, show=False, legend=True)
colors = ['red', 'green', 'blue', 'orange', 'brown']
for i, color in enumerate(colors):
    p[i].line_color = color
p.save('sample11.svg')

入出力結果(Terminal, Jupyter(IPython))

$ ./sample11.py
11.
(1)
[{x: -7/8}, {x: -5/7}]

(2)
⎡⎧   1   √5⎫  ⎧     √5   1⎫⎤
⎢⎨x: ─ + ──⎬, ⎨x: - ── + ─⎬⎥
⎣⎩   2   2 ⎭  ⎩     2    2⎭⎦

(3)
⎡⎧   -5⋅ⅈ ⎫  ⎧   5⋅ⅈ⎫⎤
⎢⎨x: ─────⎬, ⎨x: ───⎬⎥
⎣⎩     3  ⎭  ⎩    3 ⎭⎦

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

(5)
⎡⎧   -13⋅√2 ⎫  ⎧   -√2 ⎫⎤
⎢⎨x: ───────⎬, ⎨x: ────⎬⎥
⎣⎩      2   ⎭  ⎩    2  ⎭⎦

$

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="0.001">
<br>
<label for="x1">x1 = </label>
<input id="x1" type="number" value="-1.5">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="1.5">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="-1.5">
<label for="y2">y2 = </label>
<input id="y2" type="number" value="1.5">

<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="sample11.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 => 56 * x ** 2 + 89 * x + 35, '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 = [[-7/8, y1, -7/8, y2, 'green'],
                 [-5/7, y1, -5/7, 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11月11日日曜日

数学 – Python - JavaScript - 数学はここから始まる - 数 – 2次方程式と複素数 – 2次式の因数分解(複素数の範囲)

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