数学 - Python - JavaScript - 解析学 - 積分法 - 定積分の計算(三角関数(正弦、余弦)、和と積、加法定理)

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

解析入門〈2〉(松坂和夫(著)、岩波書店)の第8章(積分の計算)、8.2(定積分の計算)、問題3.を取り組んでみる。

1. $\begin{array}{l} \cos (m x + n x) = \cos m x \cos n x - \sin m x \sin n x \\ \cos (m x - n x) = \cos m x \cos n x + \sin m x \sin n x \\ \sin m x \sin n x = \frac{1}{2} (\cos (m - n) x - \cos (m + n) x) \\ 2 \int_{0}^{π} \frac{1}{2} (\cos (m - n) x - \cos (m + n) x) d x \\ = \int_{0}^{π} (\cos (m - n) x - \cos (m + n) x) d x \\ m \neq n \\ {[\frac{\sin (m - n) x}{m - n} - \frac{\sin (m + n) x}{m + n}]}_{0}^{π} \\ = 0 \\ m = n \neq 0 \\ \int_{0}^{π} (\cos (m - n) x - \cos (m + n) x) d x \\ \int_{0}^{π} (1 - \cos 2 m x) d x \\ = {[x - \frac{\sin 2 m x}{2 m}]}_{0}^{π} \\ = π \\ m = n = 0 \\ \int_{0}^{π} (\cos (m - n) x - \cos (m + n) x) d x \\ = \int_{0}^{π} (1 - 1) d x \\ = 0 \end{array}$
2. $\begin{array}{l} \sin (m x + n x) = \sin m x \cos n x + \cos m x \sin n x \\ \sin (m x - n x) = \sin m x \cos n x - \cos m x \sin n x \\ \sin m x \cos n x = \frac{1}{2} (\sin (m + n) x + \sin (m - n) x) \\ \int_{- π}^{π} \frac{1}{2} (\sin (m + n) x + \sin (m - n) x) d x \\ = 0 \end{array}$
3. $\begin{array}{l} \cos (m x + n x) = \cos m x \cos n x - \sin m x \sin n x \\ \cos (m x - n x) = \cos m x \cos n x + \sin m x \sin n x \\ \cos m x \cos n x = \frac{1}{2} (\cos (m + n) x + \cos (m - n) x) \\ \int_{- π}^{π} \frac{1}{2} (\cos (m + n) x + \cos (m - n) x) d x \\ = \frac{1}{2} \cdot 2 \int_{0}^{π} (\cos (m + n) x + \cos (m - n) x) d x \\ = \int_{0}^{π} (\cos (m + n) x + \cos (m - n) x) d x \\ m \neq n \\ {[\frac{\sin (m + n) x}{m + n} + \frac{\sin (m - n) x}{m - n}]}_{0}^{π} \\ = 0 \\ m = n \neq 0 \\ \int_{0}^{π} (\cos (m + n) x + \cos (m - n) x) d x \\ = \int_{0}^{π} (\cos (m + n) x + 1) d x \\ = {[\frac{\sin (m + n) x}{m + n} + x]}_{0}^{π} \\ = π \\ m = n = 0 \\ \int_{0}^{π} (\cos (m + n) x + \cos (m - n) x) d x \\ = \int_{0}^{π} 2 d x \\ = {[2 x]}_{0}^{π} \\ = 2 π \end{array}$

コード(Emacs)

Python 3

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

from sympy import pprint, symbols, Integral, sin, cos, pi

print('2.')
x = symbols('x')
m, n = symbols('m n', integer=True, nonnegative=True)
fs = [sin(m * x) * sin(n * x),
      sin(m * x) * cos(n * x),
      cos(m * x) * cos(n * x)]

for i, f in enumerate(fs, 1):
    print(f'({i})')
    I = Integral(f, (x, -pi, pi))
    pprint(I)
    pprint(I.doit())
    print()

入出力結果(Terminal, IPython)

$ ./sample3.py
2.
(1)
π                      
⌠                      
⎮  sin(m⋅x)⋅sin(n⋅x) dx
⌡                      
-π                     
⎧0   for m = 0 ∧ n = 0
⎪                     
⎪-π     for m = -n    
⎨                     
⎪π       for m = n    
⎪                     
⎩0       otherwise    

(2)
π                      
⌠                      
⎮  sin(m⋅x)⋅cos(n⋅x) dx
⌡                      
-π                     
0

(3)
π                      
⌠                      
⎮  cos(m⋅x)⋅cos(n⋅x) dx
⌡                      
-π                     
⎧2⋅π  for m = 0 ∧ n = 0 
⎪                       
⎨ π   for m = -n ∨ m = n
⎪                       
⎩ 0       otherwise     

$

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="m0">m = </label>
<input id="m0" type="number" min="0" step="1" value="2">
<label for="n0">n = </label>
<input id="n0" type="number" min="0" step="1" value="3">

<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'),
    input_m0 = document.querySelector('#m0'),
    input_n0 = document.querySelector('#n0'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
              input_m0, 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),
        m0 = parseFloat(input_m0.value),
        n0 = parseFloat(input_n0.value);
        
        
    
    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
        return;
    }

    let points = [],
        f1 = (x) => Math.sin(m0 * x) * Math.sin(n0 * x),
        f2 = (x) => Math.sin(m0 * x) * Math.cos(n0 * x),
        f3 = (x) => Math.cos(m0 * x) * Math.cos(n0 * x),
        lines = [[-Math.PI, y1, -Math.PI, y2, 'brown'],
                 [Math.PI, y1, Math.PI, y2, 'brown']],
        fns = [[f1, 'red'],
               [f2, 'green'],
               [f3, 'blue']];

    fns
        .forEach((o) => {
            let [fn, color] = o;
            
            for (let x = x1; x <= x2; x += dx) {
                let y = fn(x);
                
                if (Math.abs(y) < Infinity) {
                    points.push([x, y, 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('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.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.append('g')
        .attr('transform', `translate(0, ${height - padding})`)
        .call(xaxis);

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

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

r = dx =
x1 = x2 =
y1 = y2 =
m = n =

Kamimura's blog

2017年7月28日金曜日

数学 - Python - JavaScript - 解析学 - 積分法 - 定積分の計算(三角関数(正弦、余弦)、和と積、加法定理)

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