数学 - Python - JavaScript - 解析学 - 微分と基本的な関数 - 平均値の定理 - 増加・減少関数(光の速度、空気中、水中、最短時間の経路、試行、事象、確率、尤度関数(likelihood function)、軸と直線による三角形の面積)

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

解析入門原書第3版 (S.ラング(著)、松坂和夫(翻訳)、片山孝次(翻訳)、岩波書店)の第2部(微分と基本的な関数)、第5章(平均値の定理)、3(増加・減少関数)、補充問題23、24、25.を取り組んでみる。

$\begin{array}{l} P_{1} Q = \sqrt{y_{1}^{2} + x^{2}} \\ Q P_{2} = \sqrt{y_{2}^{2} + {(a - x)}^{2}} \\ f (x) = \frac{P_{1} Q}{v_{1}} + \frac{P_{2} Q}{v_{2}} \\ = \frac{\sqrt{y_{1}^{2} + x^{2}}}{v_{1}} + \frac{\sqrt{y_{2}^{2} + {(a - x)}^{2}}}{v_{2}} \\ f' (x) = \frac{{(y_{1}^{2} + x^{2})}^{- \frac{1}{2}} 2 x}{2 v_{1}} - \frac{{(y_{2}^{2} + {(a - x)}^{2})}^{- \frac{1}{2}} 2 (a - x)}{2 v_{2}} \\ \frac{{(y_{1}^{2} + x^{2})}^{- \frac{1}{2}} 2 x}{2 v_{1}} - \frac{{(y_{2}^{2} + {(a - x)}^{2})}^{- \frac{1}{2}} 2 (a - x)}{2 v_{2}} = 0 \\ \frac{1}{v_{1}} \cdot \frac{x}{\sqrt{y_{1}^{2} + x^{2}}} = \frac{1}{v_{2}} \cdot \frac{a - x}{\sqrt{y_{2}^{2} + {(a - x)}^{2}}} \\ \frac{\sin θ_{1}}{v_{1}} = \frac{\sin θ_{2}}{v_{2}} \end{array}$
$\begin{array}{l} L' (p) = s p^{s - 1} {(1 - p)}^{n - s} - p^{s} (n - s) {(1 - p)}^{n - s - 1} \\ s p^{s - 1} {(1 - p)}^{n - s} - p^{s} (n - s) {(1 - p)}^{n - s - 1} = 0 \\ p^{s - 1} {(1 - p)}^{n - s - 1} (s (1 - p) - p (n - s)) = 0 \\ s (1 - p) - p (n - s) = 0 \\ p = \frac{s}{n} \end{array}$
点(3, 2)を通り、傾き(m)負の直線の方程式。
$\begin{array}{l} m < 0 \\ y = m (x - 3) + 2 \end{array}$
この直線とx軸との交点の座標を(a, 0)とする。
$\begin{array}{l} 0 = m (a - 3) + 2 \\ m = - \frac{2}{a - 3} \\ y = - \frac{2}{a - 3} (x - 3) + 2 \end{array}$
直線とy軸との交点の座標を求める。
$\begin{array}{l} y = - \frac{2}{a - 3} (- 3) + 2 \\ = \frac{6}{a - 3} + 2 \\ = \frac{2 a}{a - 3} \\ (0, \frac{2 a}{a - 3}) \end{array}$
三角形の面積が最小になるようなaの値を求める。
$\begin{array}{l} f (a) = \frac{1}{2} a \frac{2 a}{a - 3} \\ = \frac{a^{2}}{a - 3} \\ f' (a) = \frac{2 a (a - 3) - a^{2}}{{(a - 3)}^{2}} \\ = \frac{a (a - 6)}{{(a - 3)}^{2}} \\ \frac{a (a - 6)}{{(a - 3)}^{2}} = 0 \\ a = 6 \end{array}$
求める直線の方程式。
$\begin{array}{l} y = - \frac{2}{3} (x - 3) + 2 \\ 3 y = - 2 x + 6 + 6 \\ 2 x + 3 y - 12 = 0 \end{array}$

コード(Emacs)

Python 3

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

from sympy import pprint, symbols, Derivative, solve

print('24.')
p, n, s = symbols('p n s', nonnegative=True)
L = p ** s * (1 - p) ** (n - s)
d = Derivative(L, p, 1)
pprint(d)
L1 = d.doit()
pprint(L1)
s = solve(L1, p)
pprint(s)

print('25.')
x = symbols('x')
a = symbols('a', positive=True)
g = -2 / (a - 3) * (x - 3) + 2
f = a ** 2 / (a - 3)
d = Derivative(f, a, 1)
f1 = d.doit()
pprint(d)
pprint(f1)

s = solve(f1)
pprint(s)

for a0 in s:
    g0 = g.subs({a: a0})
    pprint(g0)

入出力結果(Terminal, IPython)

$ ./sample23.py
24.
∂ ⎛ s         n - s⎞
──⎝p ⋅(-p + 1)     ⎠
∂p                  
 s                  n - s    s           n - s
p ⋅(-n + s)⋅(-p + 1)        p ⋅s⋅(-p + 1)     
───────────────────────── + ──────────────────
          -p + 1                    p         
⎡s⎤
⎢─⎥
⎣n⎦
25.
  ⎛   2 ⎞
d ⎜  a  ⎟
──⎜─────⎟
da⎝a - 3⎠
      2           
     a        2⋅a 
- ──────── + ─────
         2   a - 3
  (a - 3)         
[6]
  2⋅x    
- ─── + 4
   3
$

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="0">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="1">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="0">
<label for="y2">y2 = </label>
<input id="y2" type="number" value="0.002">
<br>
<label for="dx0">dx0 = </label>
<input id="dx0" type="number" min="0" value="0.01">
<label for="n0">n = </label>
<input id="n0" type="number" min="0"  value="10">
<label for="s0">s = </label>
<input id="s0" type="number" min="0"  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="sample23.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_dx0 = document.querySelector('#dx0'),
    input_n0 = document.querySelector('#n0'),
    input_s0 = document.querySelector('#s0');
    
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
              input_dx0, input_n0, input_s0],
    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),
        dx0 = parseFloat(input_dx0.value),
        n0 = parseFloat(input_n0.value),
        s0 = parseFloat(input_s0.value);

    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2 || s0 > n0) {
        return;
    }    

    let points = [],
        x3 = s0 / n0,
        f = (x) => x ** s0 * (1 - x) ** (n0 - s0),
        f1 = (x) => s0 * x ** (s0 - 1) * (1 - x) ** (n0 - s0) - x ** s0 * (n0 - s0) * (1 - x) ** (n0 - s0 - 1),
        g = (x0) => (x) => f1(x0) * (x - x0) + f(x0),
        lines = [[x3, y1, x3, y2, 'red']],
        fns = [[f, 'green']],
        fns1 = [],
        fns2 = [[g, 'orange']];

    fns
        .forEach((o) => {
            let [f, color] = o;
            for (let x = x1; x <= x2; x += dx) {
                let y = f(x);

                if (Math.abs(y) < Infinity) {
                    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 =
dx0 = n = s =

Kamimura's blog

2017年7月4日火曜日

数学 - Python - JavaScript - 解析学 - 微分と基本的な関数 - 平均値の定理 - 増加・減少関数(光の速度、空気中、水中、最短時間の経路、試行、事象、確率、尤度関数(likelihood function)、軸と直線による三角形の面積)

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