## 2018年6月16日土曜日

### 数学 - Python - JavaScript - 解析学 - 積分 - 積分の計算 - 三角関数の積分(連続関数、フーリエ係数、余弦、絶対値、加法定理、2倍角)

1. $\begin{array}{}{c}_{0}=\frac{1}{2\pi }·4\underset{0}{\overset{\frac{\pi }{2}}{\int }}\mathrm{cos}x\mathrm{dx}\\ =\frac{2}{\pi }{\left[\mathrm{sin}x\right]}_{0}^{\frac{\pi }{2}}\\ =\frac{2}{\pi }\end{array}$
$n=1$

のとき。

$\begin{array}{}-\underset{-\pi }{\overset{\left(-\frac{\pi }{2}\right)}{\int }}{\mathrm{cos}}^{2}x\mathrm{dx}\\ =-\underset{-\pi }{\overset{\left(-\frac{\pi }{2}\right)}{\int }}\frac{\mathrm{cos}\left(2x\right)+1}{2}\mathrm{dx}\\ =-\frac{1}{2}{\left[\frac{\mathrm{sin}\left(2x\right)}{2}+x\right]}_{\left(-\pi \right)}^{\left(-\frac{\pi }{2}\right)}\\ =-\frac{1}{2}\left(-\frac{\pi }{2}+\pi \right)\\ =-\frac{\pi }{4}\end{array}$
$\begin{array}{}\frac{1}{2}{\left[\frac{\mathrm{sin}\left(2x\right)}{2}+x\right]}_{\left(-\frac{\pi }{2}\right)}^{0}=\frac{\pi }{4}\\ \frac{1}{2}{\left[\frac{\mathrm{sin}\left(2x\right)}{2}+x\right]}_{0}^{\frac{\pi }{2}}=\frac{\pi }{4}\\ -\frac{1}{2}{\left[\frac{\mathrm{sin}\left(2x\right)}{2}+x\right]}_{\frac{\pi }{2}}^{\pi }=-\frac{\pi }{4}\end{array}$
${a}_{n}=0$
$n\ne 1$

のとき。

$\begin{array}{}-\underset{-\pi }{\overset{-\frac{\pi }{2}}{\int }}\mathrm{cos}x\mathrm{cos}\left(nx\right)\mathrm{dx}\\ =-\underset{-\pi }{\overset{-\frac{\pi }{2}}{\int }}\frac{\mathrm{cos}\left(1+n\right)x+\mathrm{cos}\left(1-n\right)x}{2}\mathrm{dx}\\ =-\frac{1}{2}{\left[\frac{\mathrm{sin}\left(1+n\right)x}{1+n}+\frac{\mathrm{sin}\left(1-n\right)x}{1-n}\right]}_{-\pi }^{\left(-\frac{\pi }{2}\right)}\\ =-\frac{1}{2}\left(\frac{-\mathrm{sin}\left(1+n\right)·\frac{\pi }{2}}{1+n}+\frac{-\mathrm{sin}\left(1-n\right)\frac{\pi }{2}}{1-n}\right)\\ =\frac{1}{2}\left(\frac{\mathrm{sin}\left(1+n\right)\frac{\pi }{2}}{1+n}+\frac{\mathrm{sin}\left(1-n\right)\frac{\pi }{2}}{1-n}\right)\end{array}$
$\frac{1}{2}{\left[\frac{\mathrm{sin}\left(1+n\right)x}{1+n}+\frac{\mathrm{sin}\left(1-n\right)x}{1-n}\right]}_{\left(-\frac{\pi }{2}\right)}^{0}=\frac{1}{2}\left(\frac{\mathrm{sin}\left(1+n\right)\frac{\pi }{2}}{1+n}+\frac{\mathrm{sin}\left(1-n\right)\frac{\pi }{2}}{1-n}\right)$
$\frac{1}{2}{\left[\frac{\mathrm{sin}\left(1+n\right)x}{1+n}+\frac{\mathrm{sin}\left(1-n\right)x}{1-n}\right]}_{0}^{\frac{\pi }{2}}=\frac{1}{2}\left(\frac{\mathrm{sin}\left(1+n\right)\frac{\pi }{2}}{1+n}+\frac{\mathrm{sin}\left(1-n\right)\frac{\pi }{2}}{1-n}\right)$
$\begin{array}{}{a}_{n}=\frac{2}{\pi }\left(\frac{\mathrm{sin}\left(1+n\right)\frac{\pi }{2}}{1+n}+\frac{\mathrm{sin}\left(1-n\right)\frac{\pi }{2}}{1-n}\right)\\ =\frac{2}{\pi \left(1-{n}^{2}\right)}\left(\mathrm{sin}\left(1+n\right)\frac{\pi }{2}-n\mathrm{sin}\left(1+n\right)\frac{\pi }{2}+\mathrm{sin}\left(1-n\right)\frac{\pi }{2}+n\mathrm{sin}\left(1-n\right)\frac{\pi }{2}\right)\\ =\frac{2}{\pi \left(1-{n}^{2}\right)}\left(2\mathrm{cos}\frac{n\pi }{2}\right)\\ =\frac{4}{\pi \left(1-{n}^{2}\right)}\mathrm{cos}\frac{n\pi }{2}\end{array}$
$\begin{array}{}n=1\\ -\underset{-\pi }{\overset{-\frac{\pi }{2}}{\int }}\mathrm{cos}x\mathrm{sin}x\mathrm{dx}\\ =-\underset{-\pi }{\overset{-\frac{\pi }{2}}{\int }}\frac{\mathrm{sin}2x}{2}\mathrm{dx}\\ =\frac{1}{4}{\left[\mathrm{cos}\left(2x\right)\right]}_{-\pi }^{-\frac{\pi }{2}}\\ =-\frac{1}{2}\\ -\frac{1}{4}{\left[\mathrm{cos}\left(2x\right)\right]}_{\left(-\frac{\pi }{2}\right)}^{0}=0\\ -\frac{1}{4}{\left[\mathrm{cos}\left(2x\right)\right]}_{0}^{\frac{\pi }{2}}=0\\ \frac{1}{4}{\left[\mathrm{cos}\left(2x\right)\right]}_{\frac{\pi }{2}}^{\pi }=\frac{1}{2}\\ {b}_{n}=0\end{array}$
$\begin{array}{}n\ne 1\\ -\underset{-\pi }{\overset{-\frac{\pi }{2}}{\int }}\mathrm{cos}x\mathrm{sin}\left(nx\right)\mathrm{dx}\\ =-\underset{-\pi }{\overset{-\frac{\pi }{2}}{\int }}\frac{\mathrm{sin}\left(1+n\right)x+\mathrm{sin}\left(1-n\right)x}{2}\mathrm{dx}\\ =\frac{1}{2}{\left[\frac{\mathrm{cos}\left(1+n\right)x}{1+n}+\frac{\mathrm{cos}\left(1-n\right)x}{1-n}\right]}_{-\pi }^{-\frac{\pi }{2}}\\ =\frac{1}{2}{\left[\frac{\mathrm{cos}\left(1+n\right)x}{1+n}+\frac{\mathrm{cos}\left(1-n\right)x}{1-n}\right]}_{\pi }^{\frac{\pi }{2}}\end{array}$
$\begin{array}{}-\frac{1}{2}{\left[\frac{\mathrm{cos}\left(1+n\right)x}{1+n}+\frac{\mathrm{cos}\left(1-n\right)x}{1-n}\right]}_{\left(-\frac{\pi }{2}\right)}^{0}\\ -\frac{1}{2}{\left[\frac{\mathrm{cos}\left(1+n\right)x}{1+n}+\frac{\mathrm{cos}\left(1-n\right)x}{1-n}\right]}_{0}^{\frac{\pi }{2}}\\ \frac{1}{2}{\left[\frac{\mathrm{cos}\left(1+n\right)x}{1+n}+\frac{\mathrm{cos}\left(1-n\right)x}{1-n}\right]}_{\left(\frac{\pi }{2}\right)}^{\pi }\\ {b}_{n}=0\end{array}$

コード(Emacs)

Python 3

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

x = symbols('x')
n = symbols('n', positive=True, integer=True)

f = Function('f')(x)
c0 = 1 / (2 * pi) * (Integral(-f, (x, -pi, -pi / 2)) + Integral(f,  (x, -
pi / 2, 0)) + Integral(f, (x, 0, pi / 2)) + Integral(-f, (x, pi / 2, pi)))
an = 1 / pi * (Integral(-f * cos(n * x), (x, -pi, -pi / 2)) +
Integral(f * cos(n * x), (x, -pi / 2, 0)) +
Integral(f * cos(n * x), (x, 0, pi / 2)) +
Integral(-f * cos(n * x), (x, pi / 2, pi)))
bn = 1 / pi * (Integral(-f * sin(n * x), (x, -pi, -pi / 2)) +
Integral(f * sin(n * x), (x, -pi / 2, 0)) +
Integral(f * sin(n * x), (x, 0, pi / 2)) +
Integral(-f * sin(n * x), (x, pi / 2, pi)))

fs = [c0, an, bn]

for t in fs:
g = t.subs({f: cos(x)})
for s in [g, g.doit(), g.doit().simplify()]:
pprint(s)
print()
print()

p = plot(abs(cos(x)), abs(cos(x)) * cos(2 * x),
abs(cos(x)) * sin(2 * x), legend=True, show=False)
for i, color in enumerate(['red', 'green', 'blue']):
p[i].line_color = color
p.save('sample23.svg')


$./sample22.py 0 π ⌠ ⌠ ⎮ -sin(x) dx + ⎮ sin(x) dx ⌡ ⌡ -π 0 ─────────────────────────── 2⋅π 2 ─ π 2 ─ π 0 π ⌠ ⌠ ⎮ -sin(x)⋅cos(n⋅x) dx + ⎮ sin(x)⋅cos(n⋅x) dx ⌡ ⌡ -π 0 ───────────────────────────────────────────── π ⎛⎧ 0 for n = 1⎞ ⎛⎧ 0 for n = 1⎞ ⎛⎧ 0 for n = 1⎞ ⎜⎪ ⎟ ⎜⎪ ⎟ ⎜⎪ ⎟ ⎜⎪ n ⎟ ⎜⎪ 1 ⎟ ⎜⎪ n ⎟ ⎜⎨-(-1) ⎟ - ⎜⎨────── otherwise⎟ + ⎜⎨ (-1) 1 ⎟ ⎜⎪─────── otherwise⎟ ⎜⎪ 2 ⎟ ⎜⎪- ────── - ────── otherwise⎟ ⎜⎪ 2 ⎟ ⎜⎪n - 1 ⎟ ⎜⎪ 2 2 ⎟ ⎝⎩ n - 1 ⎠ ⎝⎩ ⎠ ⎝⎩ n - 1 n - 1 ⎠ ────────────────────────────────────────────────────────────────────────────── π ⎧ 0 for n = 1 ⎪ ⎪ ⎛ n ⎞ ⎨-⎝2⋅(-1) + 2⎠ ⎪─────────────── otherwise ⎪ ⎛ 2 ⎞ ⎩ π⋅⎝n - 1⎠ 0 π ⌠ ⌠ ⎮ -sin(x)⋅sin(n⋅x) dx + ⎮ sin(x)⋅sin(n⋅x) dx ⌡ ⌡ -π 0 ───────────────────────────────────────────── π ⎛⎧-π ⎞ ⎛⎧π ⎞ ⎜⎪─── for n = 1⎟ ⎜⎪─ for n = 1⎟ ⎜⎨ 2 ⎟ + ⎜⎨2 ⎟ ⎜⎪ ⎟ ⎜⎪ ⎟ ⎝⎩ 0 otherwise⎠ ⎝⎩0 otherwise⎠ ─────────────────────────────────── π 0$


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="0.001">
<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="n0">n = </label>
<input id="n0" type="number" min="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="sample23.js"></script>


JavaScript

let div0 = document.querySelector('#graph0'),
pre0 = document.querySelector('#output0'),
width = 600,
height = 600,
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_n0 = document.querySelector('#n0'),
inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
input_n0],
p = (x) => pre0.textContent += x + '\n';

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),
n0 = parseFloat(input_n0.value);

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

let points = [],
lines = [[-Math.PI, y1, -Math.PI, y2, 'orange'],
[Math.PI, y1, Math.PI, y2, 'brown']],
fns = [[(x) => Math.abs(Math.cos(x)), 'red'],
[(x) => Math.abs(Math.cos(x)) * Math.cos(n0 * x), 'green'],
[(x) => Math.abs(Math.cos(x)) * Math.sin(n0 * x), '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]);
}
});

let xscale = d3.scaleLinear()
.domain([x1, x2])
let yscale = d3.scaleLinear()
.domain([y1, y2])

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].forEach((fs) => p(fs.join('\n')));
};

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