数学 - Python - JavaScript - 図形の変換の方法 - 線形写像・1次変換 – 平面の1次変換 - 原点のまわりの回転(回転を表す行列、逆変換、直線、楕円、曲線)

数学読本〈6〉

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)
• Nebo(Windows アプリ)
• iPad Pro + Apple Pencil
• MyScript Nebo(iPad アプリ)
• 参考書籍
  • Pythonからはじめる数学入門
  • JavaScriptリファレンス 第6版
  • インタラクティブ・データビジュアライゼーション

数学読本〈6〉線形写像・1次変換/数論へのプレリュード/集合論へのプレリュード/εとδ/落ち穂拾い など(松坂 和夫(著)、岩波書店)の第22章(図形の変換の方法 - 線形写像・1次変換)、22.2(平面の1次変換)、原点のまわりの回転、問11.を取り組んでみる。

11. 1. 
       

       問題の直線上の点(u, v)を原点のまわりに60度だけ回転した点を(x, y)とすれば、

       $(\begin{array}{cc}\cos \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}\right)& -\sin \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}\right)\\ \sin \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}\right)& \cos \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}\right)\end{array}\left)(\begin{array}{c}x\\ y\end{array}\right)=(\begin{array}{c}u\\ v\end{array})$
       $(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\left)(\begin{array}{c}x\\ y\end{array}\right)=(\begin{array}{c}u\\ v\end{array})$
       $\begin{array}{}u=\frac{1}{2}x+\frac{\sqrt{3}}{2}y\\ v=-\frac{\sqrt{3}}{2}x+\frac{1}{2}y\end{array}$

       また、

       $u+v=1$

       が成り立つので、

       $\begin{array}{}\frac{1}{2}x+\frac{\sqrt{3}}{2}y-\frac{\sqrt{3}}{2}x+\frac{1}{2}y=1\\ \left(1-\sqrt{3}\right)x+\left(\sqrt{3}+1\right)y=2\end{array}$
    2. 
       
       $(\begin{array}{cc}\cos \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{6}\right)& -\sin \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{6}\right)\\ \sin \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{6}\right)& \cos \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{6}\right)\end{array}\left)(\begin{array}{c}x\\ y\end{array}\right)=(\begin{array}{c}u\\ v\end{array})$
       $(\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\left)(\begin{array}{c}x\\ y\end{array}\right)=(\begin{array}{c}u\\ v\end{array})$
       $\begin{array}{}\frac{\sqrt{3}}{2}x+\frac{1}{2}y=u\\ -\frac{1}{2}x+\frac{\sqrt{3}}{2}y=v\end{array}$
       $\begin{array}{}\frac{u^2}{4}+v^2=1\\ \frac{1}{4}\left(\frac{3}{4}{x}^2+\frac{1}{4}{y}^2+\frac{\sqrt{3}}{2}xy\right)+\left(\frac{1}{4}{x}^2+\frac{3}{4}{y}^2-\frac{\sqrt{3}}{2}xy\right)=1\\ 3x^2+y^2+2\sqrt{3}xy+4x^2+12y^2-8\sqrt{3}xy=16\\ 7x^2+13y^2-6\sqrt{3}xy-16=0\end{array}$
    3. 
       
       $(\begin{array}{cc}\cos \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}\right)& -\sin \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}\right)\\ \sin \left(-\frac{x}{4}\right)& los\left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{4}\right)\end{array}\left)(\begin{array}{c}x\\ y\end{array}\right)=(\begin{array}{c}u\\ v\end{array})$
       $(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\left)(\begin{array}{c}x\\ y\end{array}\right)=(\begin{array}{c}u\\ v\end{array})$
       $\begin{array}{}u=\frac{1}{\sqrt{2}}\left(x+y\right)\\ v=\frac{1}{\sqrt{2}}\left(-x+y\right)\end{array}$
       $\begin{array}{}\sqrt{\frac{1}{\sqrt{2}}\left(x+y\right)}+\sqrt{\frac{1}{\sqrt{2}}\left(-x+y\right)}=1\\ \sqrt{x+y}+\sqrt{-x+y}=\sqrt{2}\end{array}$

コード(Emacs)

Python 3

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

x, y = symbols('x, y')
a = [
    (
        x + y - 1,
        (1 - sqrt(3)) * x + (sqrt(3) + 1) * y - 2
    ),

    (
        x ** 2 / 4 + y ** 2 - 1,
        7 * x ** 2 + 13 * y ** 2 - 6 * sqrt(3) * x * y - 16
    ),
    (
        sqrt(x) + sqrt(y) - 1,
        sqrt(x + y) + sqrt(-x + y) - sqrt(2)
    )
]

fs = []
for i, (f, g) in enumerate(a, 1):
    print(f'({i})')
    for h in [f, g]:
        fs += solve(h, y)
        pprint(solve(h, y))
        print()
    print()

p = plot(*fs, show=False, legend=True)
p.save('sample11.svg')

入出力結果(Terminal, Jupyter(IPython))

$ ./sample11.py
(1)
[-x + 1]

[-√3⋅x + 2⋅x - 1 + √3]


(2)
⎡    __________      __________⎤
⎢   ╱    2          ╱    2     ⎥
⎢-╲╱  - x  + 4    ╲╱  - x  + 4 ⎥
⎢───────────────, ─────────────⎥
⎣       2               2      ⎦

⎡              _____________                _____________⎤
⎢             ╱      2                     ╱      2      ⎥
⎢3⋅√3⋅x   4⋅╲╱  - 4⋅x  + 13   3⋅√3⋅x   4⋅╲╱  - 4⋅x  + 13 ⎥
⎢────── - ──────────────────, ────── + ──────────────────⎥
⎣  13             13            13             13        ⎦


(3)
⎡        2⎤
⎣(√x - 1) ⎦

⎡ 2    ⎤
⎢x    1⎥
⎢── + ─⎥
⎣2    2⎦



$

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="0.005">
<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="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',
    range = (start, end, step=1) => {
        let res = [];
        for (let i = start; i < end; i += step) {
            res.push(i);
        }
        return res;
    };

let f1 = (x) => -x + 1,
    f2 = (x) => -Math.sqrt(3) * x + 2 * x - 1 + Math.sqrt(3),
    g11 = (x) => -Math.sqrt(- (x ** 2) + 4) / 2,
    g12 = (x) => -g11(x),
    g21 = (x) => (3 * Math.sqrt(3) * x - 4 * Math.sqrt(-4 * x ** 2 + 13)) / 13,
    g22 = (x) => (3 * Math.sqrt(3) * x + 4 * Math.sqrt(-4 * x ** 2 + 13)) / 13,
    h1 = (x) => (Math.sqrt(x) - 1) ** 2,
    h2 = (x) => 1 / 2 * (x ** 2 + 1);

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 = [],
        fns = [[f1, 'red'],
               [f2, 'blue'],
               [g11, 'green'],
               [g12, 'green'],
               [g21, 'orange'],
               [g22, 'orange'],
               [h1, 'brown'],
               [h2, 'purple']],
        fns1 = [],
        fns2 = [];

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

                points.push([x, y, color]);
            }
        });

    fns1
        .forEach((o) => {
            let [f, color] = o;
            
            lines.push([x1, f(x1), x2, f(x2), color]);
        });
    
    fns2
        .forEach((o) => {
            let [f, color] = o;
            for (let x = x1; x <= x2; x += dx0) {
                let g = f(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();

(x) => -x + 1,red
(x) => -Math.sqrt(3) * x + 2 * x - 1 + Math.sqrt(3),blue
(x) => -Math.sqrt(- (x ** 2) + 4) / 2,green
(x) => -g11(x),green
(x) => (3 * Math.sqrt(3) * x - 4 * Math.sqrt(-4 * x ** 2 + 13)) / 13,orange
(x) => (3 * Math.sqrt(3) * x + 4 * Math.sqrt(-4 * x ** 2 + 13)) / 13,orange
(x) => (Math.sqrt(x) - 1) ** 2,brown
(x) => 1 / 2 * (x ** 2 + 1),purple

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