2017年12月13日水曜日

数学 - Python - JavaScript - 解析学 - 距離空間の位相 - n次元実数空間における曲線(対数関数、積分、置換積分法、曲線の長さ、パラメーター表示に変換)

1. 曲線のパラメーター表示を

$\left(t,\mathrm{log}t\right)$

と考える。

$\begin{array}{}\gamma \left(t\right)=\left(t,\mathrm{log}t\right)\\ \gamma \text{'}\left(t\right)=\left(1,\frac{1}{t}\right)\\ \left|\gamma \text{'}\left(t\right)\right|=\sqrt{1+\frac{1}{{t}^{2}}}\end{array}$

もとめる曲線の長さ。

$\begin{array}{}{\int }_{1}^{3}\left|\gamma \text{'}\left(t\right)\right|\mathrm{dt}\\ ={\int }_{1}^{3}\sqrt{1+\frac{1}{{t}^{2}}}\mathrm{dt}\\ ={\int }_{1}^{3}\sqrt{\frac{{t}^{2}+1}{{t}^{2}}}\mathrm{dt}\\ ={\int }_{1}^{3}\frac{\sqrt{{t}^{2}+1}}{t}\mathrm{dt}\end{array}$
$\begin{array}{}u=\sqrt{{t}^{2}+1}\\ \frac{du}{\mathrm{dt}}=\frac{2t}{2\sqrt{{t}^{2}+1}}\\ \frac{du}{\mathrm{dt}}=\frac{t}{\sqrt{{t}^{2}+1}}\\ \mathrm{dt}=\frac{\sqrt{{t}^{2}+1}}{t}du\\ 1\le t\le 3\\ \sqrt{2}\le u\le \sqrt{10}\end{array}$
$\begin{array}{}L\left(t\right)=\underset{\sqrt{2}}{\overset{\sqrt{10}}{\int }}\frac{\sqrt{{t}^{2}+1}}{t}·\frac{\sqrt{{t}^{2}+1}}{t}du\\ =\underset{\sqrt{2}}{\overset{\sqrt{10}}{\int }}\frac{{t}^{2}+1}{{t}^{2}}du\\ ={\int }_{2}^{\sqrt{10}}\left(1+\frac{1}{{t}^{2}}\right)du\\ =\underset{\sqrt{2}}{\overset{\sqrt{10}}{\int }}1du+{\int }_{\sqrt{2}}^{\sqrt{10}}\frac{1}{{t}^{2}}du\end{array}$
$\begin{array}{}{u}^{2}={t}^{2}+1\\ {t}^{2}={u}^{2}-1\end{array}$
$\begin{array}{}L\left(t\right)=\sqrt{10}-\sqrt{2}+\underset{\sqrt{2}}{\overset{\sqrt{10}}{\int }}\frac{1}{{u}^{2}-1}du\\ \frac{1}{{u}^{2}-1}=\frac{1}{u+1}·\frac{1}{u-1}\\ \frac{A}{u+1}+\frac{B}{u-1}=\frac{\left(A+B\right)u-A+B}{{u}^{2}-1}\\ A+B=0\\ -A+B=1\\ B=\frac{1}{2},A=-\frac{1}{2}\\ \frac{1}{{u}^{2}-1}=-\frac{1}{2\left(u+1\right)}+\frac{1}{2\left(u-1\right)}\end{array}$
$\begin{array}{}L\left(t\right)\\ =\sqrt{10}-\sqrt{2}+\frac{1}{2}\underset{\sqrt{2}}{\overset{\sqrt{10}}{\int }}\left(-\frac{1}{u+1}+\frac{1}{u-1}\right)du\\ =\sqrt{10}-\sqrt{2}+\frac{1}{2}{\left[-\mathrm{log}\left(u+1\right)+\mathrm{log}\left(u-1\right)\right]}_{\sqrt{2}}^{\sqrt{10}}\\ =\sqrt{10}-\sqrt{2}+\frac{1}{2}\left(\mathrm{log}\frac{\sqrt{10}-1}{\sqrt{10}+1}-\mathrm{log}\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\\ =\sqrt{10}-\sqrt{2}+\frac{1}{2}\mathrm{log}\frac{\left(\sqrt{10}-1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{10}+1\right)\left(\sqrt{2}-1\right)}\end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, sqrt, Integral

t = symbols('t')
f = sqrt(1 + 1 / t ** 2)
I = Integral(f, (t, 1, 3))
for o in [I, I.doit()]:
pprint(o)
print()


$./sample5.py 3 ⌠ ⎮ ________ ⎮ ╱ 1 ⎮ ╱ 1 + ── dt ⎮ ╱ 2 ⎮ ╲╱ t ⌡ 1 -√2 - asinh(1/3) + log(1 + √2) + √10$


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="sample5.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'),
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 f = (x) => Math.log(x);

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 = [[1, y1, 1, y2, 'red'],
[3, y1, 3, y2, 'blue']],
fns = [[f, 'green']];

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])

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('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();