数学 - Python - JavaScript - 線形代数学 - R^n におけるベクトル – 複素数(三角関数(正弦、余弦)、加法定理、指数関数、累乗根、正多角形)

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)
数式入力ソフト(TeX, MathML): MathType
MathML対応ブラウザ: Firefox、Safari
MathML非対応ブラウザ(Internet Explorer, Microsoft Edge, Google Chrome...)用JavaScript Library: MathJax
参考書籍

ラング線形代数学(上)(S.ラング (著)、芹沢正三 (翻訳)、ちくま学芸文庫)の1章(R^n におけるベクトル)、6(複素数)、練習問題7.を取り組んでみる。

1. $\begin{array}{l} e^{i (θ_{1} + θ_{2})} \\ = \cos (θ_{1} + θ_{2}) + i \sin (θ_{1} + θ_{2}) \\ = \cos θ_{1} \cos θ_{2} + i (\sin θ_{1} \cos θ_{2} + \cos θ_{1} \sin θ_{2}) \\ = (\cos θ_{1} + i \sin θ_{1}) (\cos θ_{2} + i \sin θ_{2}) \\ = e^{i θ_{1}} \cdot e^{i θ_{2}} \end{array}$
2. 任意の複素数a + bi(a、bは実数)を考える。
  $\begin{array}{l} a + b i \\ = \sqrt{a^{2} + b^{2}} (\frac{a}{\sqrt{a^{2} + b^{2}}} + \frac{b}{\sqrt{a^{2} + b^{2}}} i) \\ \cos θ = \frac{a}{\sqrt{a^{2} + b^{2}}}, \sin θ = \frac{b}{\sqrt{a^{2} + b^{2}}} \\ \sqrt{a^{2} + b^{2}} (\frac{a}{\sqrt{a^{2} + b^{2}}} + \frac{b}{\sqrt{a^{2} + b^{2}}} i) \\ = \sqrt{a^{2} + b^{2}} (\cos θ + i \sin θ) \\ = \sqrt{a^{2} + b^{2}} e^{i θ} \\ r = \sqrt{a^{2} + b^{2}} \\ a + b i = r (\cos θ + i \sin θ) \end{array}$
3. (a)より。
  $\begin{array}{l} z_{1} z_{2} \\ = r_{1} e^{i θ_{1}} r_{2} e^{i θ_{2}} \\ = r_{1} r_{2} e^{i θ_{1}} e^{i θ_{2}} \\ = r_{1} r_{2} e^{i (θ_{1} + θ_{2})} \end{array}$
4. $\begin{array}{l} z = r e^{i θ} \\ z^{\frac{1}{n}} = r^{\frac{1}{n}} e^{i \frac{θ}{n}} \\ = r^{\frac{1}{n}} (\cos \frac{θ}{n} + i \sin \frac{θ}{n}) \\ w = r^{\frac{1}{n}} (\cos \frac{θ}{n} + i \sin \frac{θ}{n}) \\ w^{n} = z \\ k \in ℤ \\ z = r e^{i θ} \\ = r (\cos θ + i \sin θ) \\ = r (\cos (θ + 2 k π) + i \sin (θ + 2 k π)) \\ = r e^{θ + 2 k π} \\ z^{\frac{1}{n}} = r^{\frac{1}{n}} e^{\frac{θ + 2 k π}{n}} \\ = r^{\frac{1}{n}} (\cos \frac{θ + 2 k π}{n} + i \sin \frac{θ + 2 k π}{n}) \\ = r^{\frac{1}{n}} (\cos (θ + \frac{2 k π}{n}) + i \sin (θ + \frac{2 k π}{n})) \\ k = 0, 1, \dots, n - 1 \end{array}$
  よって、問題の方程式の解を満たすような複素数wはちょうどn個存在する。

コード(Emacs)

Python 3

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

print('7.')
w, z = symbols('w z', imag=True)
n = symbols('n', integer=True, positive=True)
pprint(solve(w ** n - z, w))

入出力結果(Terminal, Jupyter(IPython))

$ ./sample7.py
7.
⎡n ___⎤
⎣╲╱ z ⎦
$

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="0.0001" 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">
<br>
<label for="r1">r1 = </label>
<input id="r1" type="number" min="0" value="5">
<label for="θ0">θ = </label>
<input id="θ0" type="number" step="0.1" value="1">
<label for="n0">n = </label>
<input id="n0" type="number" min="1" step="1" 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="sample7.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_r1 = document.querySelector('#r1'),
    input_θ0 = document.querySelector('#θ0'),
    input_n0 = document.querySelector('#n0'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
              input_r1, input_θ0, input_n0],
    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),
        r1 = parseFloat(input_r1.value),
        θ0 = parseFloat(input_θ0.value),
        n0 = parseInt(input_n0.value, 10);

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

    let points = [],
        lines = [],
        f1 = (x) => Math.sqrt(r1 ** 2 - x ** 2),
        f2 = (x) => -Math.sqrt(r1 ** 2 - x ** 2),
        gx = (θ) => r1 * Math.cos(θ),
        gy = (θ) => r1 * Math.sin(θ),
        θs = range(0, n0).map((k) => θ0 + 2 * k * Math.PI / n0),
        fns = [[f1, 'green'],
               [f2, 'green']],
        fns1 = [],
        fns2 = [];

    θs.forEach((θ) => lines.push([0, 0, gx(θ), gy(θ), '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]);
            }
        });
    
    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 =
r1 = θ = n =

Kamimura's blog

2017年10月24日火曜日

数学 - Python - JavaScript - 線形代数学 - R^n におけるベクトル – 複素数(三角関数(正弦、余弦)、加法定理、指数関数、累乗根、正多角形)

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