数学 - Python - JavaScript - 線形代数学 - R^n におけるベクトル – ベクトルのノルム(射影、スカラー積(内積)の幾何学的性質)

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)、3(ベクトルのノルム)、練習問題3、4.を取り組んでみる。

1. $\begin{array}{l} \frac{- 2 - 1}{1 + 1} (- 1, 1) \\ = - \frac{3}{2} (- 1, 1) \\ = (\frac{3}{2}, - \frac{3}{2}) \end{array}$
2. $\begin{array}{l} \frac{12}{0 + 16} (0, 4) \\ = \frac{3}{4} (0, 4) \\ = (0, 3) \end{array}$
3. $\begin{array}{l} \frac{- 2 - 1 + 5}{1 + 1 + 1} (- 1, 1, 1) \\ = \frac{2}{3} (- 1, 1, 1) \\ = (- \frac{2}{3}, \frac{2}{3}, \frac{2}{3}) \end{array}$
4. $\begin{array}{l} \frac{1 - 6 - 12}{1 + 9 + 16} (- 1, 3, - 4) \\ = - \frac{17}{26} (- 1, 3, - 4) \\ = (\frac{17}{26}, - \frac{51}{26}, \frac{34}{13}) \end{array}$
5. $\begin{array}{l} \frac{2 π^{2} - 9 - 7}{4 π^{2} + 9 + 49} (2 π, - 3, 7) \\ = \frac{2 π^{2} - 16}{4 π^{2} + 58} (2 π, - 3, 7) \\ = \frac{π^{2} - 8}{2 π^{2} + 29} (2 π, - 3, 7) \\ = (\frac{2 π^{3} - 16 π}{2 π^{2} + 29}, \frac{- 3 π + 24}{2 π^{2} + 29}, \frac{7 π^{2} - 56}{2 π^{2} + 29}) \end{array}$
6. $\begin{array}{l} \frac{15 π - 6 - 4}{π^{2} + 9 + 1} (π, 3, - 1) \\ = \frac{15 π - 10}{π^{2} + 10} (π, 3, - 1) \\ = (\frac{15 π^{2} - 10 π}{π^{2} + 10}, \frac{45 π - 30}{π^{2} + 10}, \frac{- 15 π + 10}{π^{2} + 10}) \end{array}$
1. $\frac{- 2 - 1}{4 + 1} A = - \frac{3}{5} A$
2. $\frac{0 + 12}{1 + 9} A = \frac{6}{5} A$
3. $\frac{- 2 - 1 + 5}{4 + 1 + 25} A = \frac{2}{30} A = \frac{1}{15} A$
4. $\frac{1 - 6 - 12}{1 + 4 + 9} A = - \frac{17}{14} A$
5. $\frac{2 π^{2} - 9 - 7}{π^{2} + 9 + 1} A = \frac{2 π - 16}{π + 10} A$
6. $\frac{15 π - 6 - 4}{225 + 4 + 16} A = \frac{15 π - 10}{245} A = \frac{3 π - 2}{49} A$

コード(Emacs)

Python 3

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

from sympy import pprint, symbols, Matrix, pi

print('3.')
A = [(2, -1),
     (-1, 3),
     (2, -1, 5),
     (-1, -2, 3),
     (pi, 3, -1), (15, -2, 4)]

B = [(-1, 1),
     (0, 4),
     (-1, 1, 1),
     (-1, 3, -4),
     (2 * pi, -3, 7),
     (pi, 3, -1)]

for i, (A0, B0) in enumerate(zip(A, B), 1):
    print(f'{i}.')
    MA = Matrix(A0)
    MB = Matrix(B0)
    for X in [MA, MB, (MA.dot(MB) / MB.dot(MB)) * MB]:
        pprint(X.T)
        print()
    print()

print('4.')
for i, (A0, B0) in enumerate(zip(A, B), 1):
    print(f'{i}.')
    MA = Matrix(A0)
    MB = Matrix(B0)
    pprint(MA.dot(MB) / MA.dot(MA))
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample3.py
3.
1.
[2  -1]

[-1  1]

[3/2  -3/2]


2.
[-1  3]

[0  4]

[0  3]


3.
[2  -1  5]

[-1  1  1]

[-2/3  2/3  2/3]


4.
[-1  -2  3]

[-1  3  -4]

⎡17  -51   34⎤
⎢──  ────  ──⎥
⎣26   26   13⎦


5.
[π  3  -1]

[2⋅π  -3  7]

⎡    ⎛         2⎞     ⎛         2⎞     ⎛         2⎞⎤
⎢2⋅π⋅⎝-16 + 2⋅π ⎠  -3⋅⎝-16 + 2⋅π ⎠   7⋅⎝-16 + 2⋅π ⎠⎥
⎢────────────────  ────────────────  ──────────────⎥
⎢      2                 2                2        ⎥
⎣   4⋅π  + 58         4⋅π  + 58        4⋅π  + 58   ⎦


6.
[15  -2  4]

[π  3  -1]

⎡π⋅(-10 + 15⋅π)  3⋅(-10 + 15⋅π)  -(-10 + 15⋅π) ⎤
⎢──────────────  ──────────────  ──────────────⎥
⎢    2               2               2         ⎥
⎣   π  + 10         π  + 10         π  + 10    ⎦


4.
1.
-3/5

2.
6/5

3.
1/15

4.
-17 
────
 14 

5.
         2
-16 + 2⋅π 
──────────
  2       
 π  + 10  

6.
  2    3⋅π
- ── + ───
  49    49

$

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">

<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="sample3.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 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 = [[0, 0, 2, -1, 'red'],
                 [0, 0, -1, 1, 'green'],
                 [0, 0, -1, 3, 'blue'],
                 [0, 0, 0, 4, 'orange']],
        fns = [],
        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]);
            }
        });
    
    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 =

Kamimura's blog

2017年9月22日金曜日

数学 - Python - JavaScript - 線形代数学 - R^n におけるベクトル – ベクトルのノルム(射影、スカラー積(内積)の幾何学的性質)

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