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、像、単位ベクトル、三角関数(正弦、余弦))

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10.を取り組んでみる。

10. 
    
    $(\begin{array}{cc}\cos \frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}& -\sin \frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}\\ \sin \frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}& \cos \frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}\end{array})=(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array})$
    $(\begin{array}{cc}\cos \frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{2}& -\sin \frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{2}\\ \sin \frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{2}& \cos \frac{-n}{2}\end{array})=(\begin{array}{cc}0& -1\\ 1& 0\end{array})$
    $(\begin{array}{cc}\cos \frac{2}{3}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}& -\sin \frac{2}{3}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}\\ \sin \frac{2}{3}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}& \cos \frac{2}{3}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}\end{array})=(\begin{array}{cc}-\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array})$
    $(\begin{array}{cc}\cos \frac{3}{4}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}& -\sin \frac{3}{4}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}\\ \sin \frac{3}{4}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}& \cos \frac{3}{4}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}\end{array})=(\begin{array}{cc}-\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array})$
    $(\begin{array}{cc}\cos \mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}& -\sin \mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}\\ \sin \mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}& \cos \mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}\end{array})=(\begin{array}{cc}-1& 0\\ 0& -1\end{array})$
    $(\begin{array}{cc}\cos \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}\right)& -\sin \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}\right)\\ \sin \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}\right)& \cos \left(-\frac{\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}}{3}\right)\end{array})=(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& \frac{1}{2}\end{array})$
    $(\begin{array}{cc}\cos \left(-\frac{3}{4}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}\right)& -\sin \left(-\frac{3}{4}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}\right)\\ \sin \left(-\frac{3}{4}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}\right)& \cos \left(-\frac{3}{4}\mathrm{\pi \; <!--\; greek\; small\; letter\; pi\; -->}\right)\end{array})=(\begin{array}{cc}-\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array})$

    これらの回転による点（2，1）の像。

    $\left(1-\frac{\sqrt{3}}{2},\sqrt{3}+\frac{1}{2}\right)$
    $\left(-1,2\right)$
    $\left(-1-\frac{\sqrt{3}}{2},\sqrt{3}-\frac{1}{2}\right)$
    $\left(-\frac{2+1}{\sqrt{2}},\frac{2-1}{\sqrt{2}}\right)=\left(-\frac{3}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$
    $\left(-2,-1\right)$
    $\left(1+\frac{\sqrt{3}}{2},-\sqrt{3}+\frac{1}{2}\right)$
    $\left(\frac{-2+1}{\sqrt{2}},-\frac{2+1}{\sqrt{2}}\right)=\left(-\frac{1}{\sqrt{2}},-\frac{3}{\sqrt{2}}\right)$

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, sin, cos, pi, Matrix

θ = symbols('θ')
A = Matrix([[cos(θ), -sin(θ)],
            [sin(θ), cos(θ)]])

X = Matrix([[2],
            [1]])

for θ0 in [pi / 3, pi / 2, 2 * pi / 3, 3 * pi / 4, pi, -pi / 3, -3 * pi / 4]:
    B = A.subs({θ: θ0})
    for t in [θ0, B, (B * X).T]:
        pprint(t)
        print()
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample10.py
π
─
3

⎡     -√3 ⎤
⎢1/2  ────⎥
⎢      2  ⎥
⎢         ⎥
⎢√3       ⎥
⎢──   1/2 ⎥
⎣2        ⎦

⎡  √3              ⎤
⎢- ── + 1  1/2 + √3⎥
⎣  2               ⎦


π
─
2

⎡0  -1⎤
⎢     ⎥
⎣1  0 ⎦

[-1  2]


2⋅π
───
 3 

⎡      -√3 ⎤
⎢-1/2  ────⎥
⎢       2  ⎥
⎢          ⎥
⎢ √3       ⎥
⎢ ──   -1/2⎥
⎣ 2        ⎦

⎡     √3           ⎤
⎢-1 - ──  -1/2 + √3⎥
⎣     2            ⎦


3⋅π
───
 4 

⎡-√2   -√2 ⎤
⎢────  ────⎥
⎢ 2     2  ⎥
⎢          ⎥
⎢ √2   -√2 ⎥
⎢ ──   ────⎥
⎣ 2     2  ⎦

⎡-3⋅√2   √2⎤
⎢──────  ──⎥
⎣  2     2 ⎦


π

⎡-1  0 ⎤
⎢      ⎥
⎣0   -1⎦

[-2  -1]


-π 
───
 3 

⎡      √3 ⎤
⎢1/2   ── ⎥
⎢      2  ⎥
⎢         ⎥
⎢-√3      ⎥
⎢────  1/2⎥
⎣ 2       ⎦

⎡√3               ⎤
⎢── + 1  -√3 + 1/2⎥
⎣2                ⎦


-3⋅π 
─────
  4  

⎡-√2    √2 ⎤
⎢────   ── ⎥
⎢ 2     2  ⎥
⎢          ⎥
⎢-√2   -√2 ⎥
⎢────  ────⎥
⎣ 2     2  ⎦

⎡-√2   -3⋅√2 ⎤
⎢────  ──────⎥
⎣ 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">
<br>
<label for="θ0">θ0 = </label>
<input id="m0" type="number" step="0.01" value="3.14">

<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'),
    input_θ0 = document.querySelector('#m0'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
              input_θ0],
    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 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),
        θ0 = parseFloat(input_θ0.value);

    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
        return;
    }    

    let points = [],
        lines = [[0, 0, 2, 1, 'red'],
                 [0, 0,
                  2 * Math.cos(θ0) - Math.sin(θ0),
                  2 * Math.sin(θ0) + Math.cos(θ0),
                 'green']],
        fns = [[(x) => m0 * x, 'green']]
        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();

r = dx =
x1 = x2 =
y1 = y2 =
θ0 =