数学 - Python - JavaScript - 線形代数学 - ベクトル空間 – 基底(垂直、1次独立、連続実関数、内積(スカラー積)、三角関数(正弦、余弦)、部分積分法)

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, 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2章(ベクトル空間)、3(基底)、練習問題9、10.を取り組んでみる。

$\begin{array}{l} x_{1} A_{1} + x_{2} A_{2} + \cdot \cdot \cdot + x_{r} A_{r} = O \\ A_{i} \cdot A_{j} = 0 (i \neq j) \\ x_{1} A_{1} \cdot A_{k} + \cdot \cdot \cdot + + \cdot \cdot \cdot + x_{r} A_{r} \cdot A_{k} = O \cdot A_{k} \\ x_{k} A_{k} \cdot A_{k} = 0 \\ x_{k} {| A_{k} |}^{2} = 0 \\ {| A_{k} |}^{2} \neq 0 \\ x_{k} = 0 \\ 1 次独立。 \end{array}$
$\begin{array}{l} 〈 \sin i t, \sin j t 〉 \\ = \int_{- π}^{π} \sin i t \cdot \sin j t d t \\ = \int_{- π}^{π} \sin i t \cdot (- \frac{1}{j} \cos j t)' d t \\ = {[\sin i t \cdot (- \frac{1}{j} \cos j t)]}_{- π}^{π} - \int_{- π}^{π} i \cos i t \cdot (- \frac{1}{j} \cos j t) t d t \\ = \frac{i}{j} \int_{- π}^{π} \cos i t \cdot \cos j t d t \\ = \frac{i}{j} \int_{- π}^{π} (\cos (i t + j t) + \sin i t \sin j t) d t \\ = \frac{i}{j} (\frac{1}{i + j} {[\sin (i + j) t]}_{- π}^{π} - \int_{- π}^{π} \sin i t \sin j t d t) \\ = - \frac{i}{j} \int_{- π}^{π} \sin i t \sin j t d t \\ (1 + \frac{i}{j}) \int_{- π}^{π} \sin i t \cdot \sin j t d t = 0 \\ 1 + \frac{i}{j} \neq 0 \\ \int_{- π}^{π} \sin i t \cdot \sin j t d t = 0 \end{array}$

コード(Emacs)

Python 3

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

from sympy import Symbol, symbols, solve, sin, Integral, pi, pprint

t = Symbol('t')
i, j = symbols('i j', integer=True, positive=True)

expr = Integral(sin(i * t) * sin(j * t), (t, -pi, pi))

pprint(expr.doit())

入出力結果(Terminal, IPython)

$ ./sample9.py
⎧π  for i = j
⎨            
⎩0  otherwise
$

HTML5

<div id="graph0"></div>
<pre id="output0"></pre>
<label for="dt">dt = </label>
<input id="dt" type="number" min="0" value="0.001">
<label for="x1">x1 = </label>
<input id="x1" type="number" value="-3.15">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="3.15">
<label for="r0">r = </label>
<input id="r0" type="number" min="0" value="0.5">
<br>
<label for="i0">i = </label>
<input id="i0" type="number" min="1" step="1" value="1">
<label for="j0">j = </label>
<input id="j0" type="number" min="1" step="1" value="2">
<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="sample9.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_dt = document.querySelector('#dt'),
    input_x1 = document.querySelector('#x1'),
    input_x2 = document.querySelector('#x2'),
    input_r = document.querySelector('#r0'),
    input_i = document.querySelector('#i0'),
    input_j = document.querySelector('#j0'),
    inputs = [input_dt, input_x1, input_x2, input_r, input_i, input_j],
    p = (x) => pre0.textContent += x + '\n';

let draw = () => {
    pre0.textContent = '';

    let dt = parseFloat(input_dt.value),
        x1 = parseFloat(input_x1.value),
        x2 = parseFloat(input_x2.value),
        r = parseFloat(input_r.value),
        i = parseInt(input_i.value, 10),
        j = parseInt(input_j.value, 10);
    
    if (dt === 0 || x1 >= x2|| r <= 0 || i < 1 || i > j) {
        return;
    }
    let points = [],
        f = (t) => Math.sin(i * t) * Math.sin(j * t);

    for (let t = x1; t <= x2; t += dt) {
        points.push([t, Math.sin(i * t)]);
    }
    let t1 = points.length;

    for (let t = x1; t <= x2; t += dt) {
        points.push([t, Math.sin(j * t)]);
    }
    let t2 = points.length;
    let area = 0;
    for (let t = x1; t <= x2; t += dt) {
        let k = f(t);

        area += dt * k;
        points.push([t, k]);
    }

    let xscale = d3.scaleLinear()
        .domain([x1, x2])
        .range([padding, width - padding]);
    let yscale = d3.scaleLinear()
        .domain([x1, x2])
        .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('circle')
        .data(points)
        .enter()
        .append('circle')
        .attr('cx', (d) => xscale(d[0]))
        .attr('cy', (d) => yscale(d[1]))
        .attr('r', r)
        .attr('fill', (d, i) =>
              i < t1 ? 'red' :
              i < t2 ? 'green' : 'blue');
    
    svg.append('g')
        .attr('transform', `translate(0, ${yscale(0)})`)
        .call(xaxis);

    svg.append('g')
        .attr('transform', `translate(${padding}, 0)`)
        .call(yaxis);

    p(area);
}

inputs.forEach((input) => input.onchange = draw);
btn0.onclick = draw;
btn1.onclick = () => pre0.textContent = '';
draw();

dt = x1 = x2 = r =
i = j =

Kamimura's blog

2017年5月16日火曜日

数学 - Python - JavaScript - 線形代数学 - ベクトル空間 – 基底(垂直、1次独立、連続実関数、内積(スカラー積)、三角関数(正弦、余弦)、部分積分法)

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