数学 - Python - 解析学 - ベクトル - ベクトル積 - 性質、等式、内積

続解析入門 (原書第2版)
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原著: Calculus of Several Variables

学習環境

続解析入門 (原書第2版) (S.ラング(著)、松坂和夫(翻訳)、片山孝次(翻訳)、岩波書店)の第1章(ベクトル)、7(ベクトル積)の練習問題8の解答を求めてみる。

$\begin{array}{l} A = (a_{1}, a_{2}, a_{3}) \\ B = (b_{1}, b_{2}, b_{3}) \\ C = (c_{1}, c_{2}, c_{3}) \end{array}$

とおく。

CP 1について。

$A \times B$

$= (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1})$

$= - (b_{2} a_{3} - b_{3} a_{2}, b_{3} a_{1} - b_{1} a_{3}, b_{1} a_{2 -} b_{2} a_{2})$

$= - (B \times A)$

CP 2 について。

$A \times (B + C)$

$= (a_{2} (b_{3} + c_{3}) - a_{3} (b_{2} + c_{2}), a_{3} (b_{1} + c_{1}) - a_{1} (b_{3} + c_{3}), a_{1} (b_{2} + c_{2}) - a_{2} (b_{1} + c_{1}))$

$\begin{array}{l} = (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1}) \\ + (a_{2} c_{3} - a_{3} c_{2}, a_{3} c_{1} - a_{1} c_{31} a_{1} c_{2} - a_{2} c_{1}) \end{array}$

$= (A \times B) + (A \times C)$

$(B + C) \times A$

$= ((b_{2} + c_{2}) a_{3} - (b_{3} + c_{3}) a_{2}, (b_{3} + c_{3}) a_{1} - (b_{1} + c_{2}) a_{3}, (b_{1} + c_{1}) a_{2} - (b_{2} + c_{2}) a_{1})$

$\begin{array}{l} = (b_{2} a_{3} - b_{3} a_{2}, b_{3} a_{1} - b_{1} a_{3}, b_{1} a_{2} - b_{2} a_{1}) \\ + (c_{2} a_{3} - c_{3} a_{2}, c_{3} a_{1} - a_{3} c_{1}, c_{1} a_{2} - c_{2} a_{1}) \end{array}$

$= (B \times A) + (C \times A)$

CP 3について。

任意の a に対して

$(a A) \times B$

$= (a a_{2} b_{3} - a a_{3} b_{2}, a a_{3} b_{1} - a a_{1} b_{3}, a a_{1} b_{2} - a a_{2} b_{1})$

$= a (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1})$

$= a (A \times B)$

$a (A \times B)$

$= (a_{2} a b_{3} - a_{3} a b_{2}, a_{3} a b_{1} - a_{1} a b_{3,} a_{1} a b_{2} - a_{2} a b_{1})$

$= A \times (a B)$

CP 4について。

$(A \times B) \times C$

$= (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1}) \times (c_{1}, c_{2}, c_{3})$

$\begin{array}{l} = (a_{3} b_{1} c_{3} - a_{1} b_{3} c_{3} - a_{1} b_{2} c_{2} + a_{2} b_{1} c_{2}, \\ a_{1} b_{2} c_{1} - a_{2} b_{1} c_{1} - a_{2} b_{3} c_{3} + a_{3} b_{2} c_{3}, \\ a_{2} b_{3} c_{2} - a_{3} b_{2} c_{2} - a_{3} b_{1} c_{1} + a_{1} b_{3} c_{1}) \end{array}$

$(A \cdot C) B - (B \cdot C) A$

$= (a_{1} c_{1} + a_{2} c_{2} + a_{3} c_{3}) (b_{1}, b_{2}, b_{3}) - (b_{1} c_{1} + b_{2} c_{2} + b_{3} c_{3}) (a_{1}, a_{2}, a_{3})$

$\begin{array}{l} = (a_{2} b_{1} c_{2} + a_{3} b_{1} c_{3} - a_{1} b_{2} c_{2} - a_{1} b_{3} c_{3}, \\ a_{1} b_{2} c_{1} + a_{3} b_{2} c_{3} - a_{2} b_{1} c_{1} - a_{2} b_{3} c_{3}, \\ a_{1} b_{3} c_{1} + a_{2} b_{3} c_{2} - a_{3} b_{1} c_{1} - a_{3} b_{2} c_{2}) \end{array}$

よって、

$(A \times B) \times C = (A \cdot C) B - (B \cdot C) A$

（証明終）

コード

#!/usr/bin/env python3
from unittest import TestCase, main
from sympy import Matrix, symbols
from sympy.abc import f

print('8.')

a = Matrix(symbols('a:3'))
b = Matrix(symbols('b:3'))
c = Matrix(symbols('c:3'))


def eq(x, y):
    return all([x0.expand() == y0.expand() for x0, y0 in zip(x, y)])


class TestCrossProduct(TestCase):
    def test1(self):
        self.assertEqual(a.cross(b), -b.cross(a))

    def test2(self):
        self.assertTrue(eq(a.cross(b + c), a.cross(b) + a.cross(c)))
        self.assertTrue(eq((b + c).cross(a), b.cross(a) + c.cross(a)))

    def test3(self):
        self.assertTrue(eq((f * a).cross(b), f * a.cross(b)))
        self.assertTrue(f * a.cross(b), a.cross(f * b))

    def test4(self):
        self.assertTrue(eq(a.cross(b).cross(c), a.dot(c) * b - b.dot(c) * a))


if __name__ == "__main__":
    main()

入出力結果(Zsh、PowerShell、Terminal、Jupyter(IPython))

% ./sample8.py -v
8.
test1 (__main__.TestCrossProduct) ... ok
test2 (__main__.TestCrossProduct) ... ok
test3 (__main__.TestCrossProduct) ... ok
test4 (__main__.TestCrossProduct) ... ok

----------------------------------------------------------------------
Ran 4 tests in 0.042s

OK
%

Kamimura's blog

2020年7月13日月曜日

数学 - Python - 解析学 - ベクトル - ベクトル積 - 性質、等式、内積

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