数学 - Python - JavaScript - 解析学 - 積分 - 積分法 - 上方和および下方和(累乗(べき乗、2乘)、関数、区間、分割)

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)
Nebo(Windows アプリ)
iPad Pro + Apple Pencil
MyScript Nebo(iPad アプリ)
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3部(積分)、第9章(積分法)、5(上方和および下方和)、練習問題1.を取り組んでみる。

1. 上方和。
  
  $\begin{array}{l} \frac{1}{2} ({(\frac{3}{2})}^{2} + 2^{2}) \\ = \frac{1}{2} (\frac{9}{4} + 4) \\ = \frac{25}{8} \end{array}$
  
  下方和。
  
  $\begin{array}{l} \frac{1}{2} (1^{2} + {(\frac{3}{2})}^{2}) \\ = \frac{1}{2} (1 + \frac{9}{4}) \\ = \frac{13}{8} \end{array}$
2. 上方和。
  
  $\begin{array}{l} \frac{1}{3} ({(\frac{4}{3})}^{2} + {(\frac{5}{3})}^{2} + 2^{2}) \\ = \frac{1}{3} \cdot \frac{16 + 25 + 36}{9} \\ = \frac{1}{3} \frac{77}{9} \\ = \frac{77}{27} \end{array}$
  
  下方和。
  
  $\begin{array}{l} \frac{1}{3} ({(\frac{3}{3})}^{2} + {(\frac{4}{3})}^{2} + {(\frac{5}{3})}^{2}) \\ = \frac{1}{3} \cdot \frac{9 + 16 + 25}{9} \\ = \frac{50}{27} \end{array}$
3. 上方和。
  
  $\begin{array}{l} \frac{1}{4} ({(\frac{5}{4})}^{2} + {(\frac{6}{4})}^{2} + {(\frac{7}{4})}^{2} + {(\frac{8}{4})}^{2}) \\ = \frac{1}{4} \cdot \frac{1}{16} \frac{1}{6} \cdot (8 (8 + 1) (16 + 1) - 4 (4 + 1) (8 + 1)) \\ = \frac{1}{16 \cdot 6} (2 \cdot 9 \cdot 17 - 5 \cdot 9) \\ = \frac{1}{16} \cdot \frac{1}{2} (2 \cdot 3 \cdot 17 - 15) \\ = \frac{102 - 15}{32} \\ = \frac{87}{32} \end{array}$
  
  下方和。
  
  $\begin{array}{l} \frac{1}{4} ({(\frac{4}{4})}^{2} + {(\frac{5}{4})}^{2} + {(\frac{6}{4})}^{2} + {(\frac{7}{4})}^{2}) \\ = \frac{1}{4} \cdot \frac{1}{16} \cdot \frac{1}{6} (7 \cdot 8 \cdot 15 - 3 \cdot 4 \cdot 7) \\ = \frac{1}{16} \cdot \frac{1}{6} (7 \cdot 2 \cdot 15 - 3 \cdot 7) \\ = \frac{1}{16} \cdot \frac{7}{2} (2 \cdot 5 - 7) \\ = \frac{7}{32} \cdot 3 \\ = \frac{63}{32} \end{array}$
4. 上方和。
  
  $\begin{array}{l} \frac{1}{n} ({(1 + \frac{1}{n})}^{2} + {(1 + \frac{2}{n})}^{2} + \dots + {(1 + \frac{n}{n})}^{2}) \\ = \frac{1}{n} ({(\frac{n + 1}{n})}^{2} + {(\frac{n + 2}{n})}^{2} + \dots + {(\frac{n + n}{n})}^{2}) \\ = \frac{1}{n^{3}} \cdot \frac{1}{6} ((2 n) (2 n + 1) (4 n + 1) - n (n + 1) (2 n + 1)) \\ = \frac{1}{6 n^{2}} \cdot (2 n + 1) (2 (4 n + 1) - (n + 1)) \\ = \frac{(2 n + 1) (7 n + 1)}{6 n^{2}} \end{array}$
  
  下方和。
  
  $\begin{array}{l} \frac{1}{n} (1^{2} + {(1 + \frac{1}{n})}^{2} + \dots + {(1 + \frac{n - 1}{n})}^{2}) 2 \\ = \frac{1}{n} (\frac{n^{2}}{n^{2}} + \frac{{(n + 1)}^{2}}{n^{2}} + \dots + \frac{(2 n - 1)}{n^{2}}) \\ = \frac{1}{n^{3}} \frac{1}{6} ((2 n - 1) (2 n) (4 n - 2 + 1) - (n - 1) n (2 n - 2 + 1)) \\ = \frac{1}{6 n^{3}} ((2 n - 1) (2 n) (4 n - 1) - (n - 1) n (2 n - 1)) \\ = \frac{1}{6 n^{2}} (2 n - 1) (8 n - 2 - n + 1) \\ = \frac{(2 n - 1) (7 n - 1)}{6 n^{2}} \end{array}$

コード(Emacs)

Python 3

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

i, n = symbols('i, n', integer=True)
x = symbols('x')
f = x ** 2
u = 1 / n * summation(f.subs({x: 1 + i / n}), (i, 1, n))
l = 1 / n * summation(f.subs({x: 1 + i / n}), (i, 0, n - 1))

for n0 in [2, 3, 4]:
    d = {n: n0}
    for g in [u, l]:
        pprint(g.subs(d))
        print()
    print()

for t in [u, l]:
    pprint(t.factor())
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample1.py
25/8

13/8


77
──
27

50
──
27


87
──
32

63
──
32


(2⋅n + 1)⋅(7⋅n + 1)
───────────────────
           2       
        6⋅n        

(2⋅n - 1)⋅(7⋅n - 1)
───────────────────
           2       
        6⋅n        

$

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="0">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="5">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="0">
<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" 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="sample1.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_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',
    range = (start, end, step=1) => {
        let res = [];
        for (let i = start; i < end; i += step) {
            res.push(i);
        }
        return res;
    };

let f = (x) => x ** 2;

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 = parseInt(input_n0.value, 10);

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

    let points = [],
        lines = [],
        fns = [[f, 'green']],
        fns1 = [],
        fns2 = [];

    for (let i = 1; i <= 2; i += 1 / n0) {
        lines.push([i, y1, i, y2, 'red']);
    }
    fns
        .forEach((o) => {
            let [f, color] = o;
            for (let x = x1; x <= x2; x += dx) {
                let y = f(x);

                points.push([x, y, color]);
            }
        });

    fns1
        .forEach((o) => {
            let [f, color] = o;
            
            lines.push([x1, f(x1), x2, f(x2), 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();

(x) => x ** 2,green

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

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2018年1月14日日曜日

数学 - Python - JavaScript - 解析学 - 積分 - 積分法 - 上方和および下方和(累乗(べき乗、2乘)、関数、区間、分割)

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