数学 - Pytyhon - 解析学 - ベクトル - ベクトルのノルム - 射影、ノルム、内積、スカラー倍、2次元、3次元

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)、4(ベクトルのノルム)の練習問題3、4の解答を求めてみる。

1. $\begin{array}{l} \frac{A \cdot B}{B \cdot B} B \\ = \frac{- 2 - 1}{1 + 1} (- 1, 1) \\ = - \frac{3}{2} (- 1, 1) \\ = (\frac{3}{2}, - \frac{3}{2}) \end{array}$
2. $\frac{12}{16} (0, 4) = (0, 3)$
3. $\frac{- 2 - 1 + 5}{1 + 1 + 1} (- 1, 1, 1) = (- \frac{2}{3}, \frac{2}{3}, \frac{2}{3})$
4. $\begin{array}{l} \frac{1 - 6 - 12}{1 + 9 + 16} (- 1, 3, - 4) \\ = - \frac{17}{26} (- 1, 3, - 4) \\ = (\frac{17}{26}, - \frac{51}{26}, \frac{34}{13}) \end{array}$
5. $\begin{array}{l} \frac{2 π^{2} - 9 - 7}{4 π^{2} + 9 + 49} (2 π, - 3, 7) \\ = \frac{2 π^{2} - 16}{4 π^{2} + 58} (2 π, - 3, 7) \\ = \frac{π^{2} - 8}{2 π^{2} + 29} (2 π, - 3, 7) \\ = (\frac{2 π^{3} - 16 π}{2 π^{2} + 29}, \frac{- 3 π^{2} + 24}{2 π^{2} + 29}, \frac{7 π^{2} - 56}{2 π^{2} + 29}) \end{array}$
6. $\frac{15 π - 10}{π^{2} + 10} (π, 3, - 1) = (\frac{15 π^{2} - 10 π}{π^{2} + 10}, \frac{45 π - 30}{π^{2} + 10}, \frac{- 15 π + 10}{π^{2} + 10})$
1. $- \frac{3}{5} (2, - 1) = (- \frac{6}{5}, \frac{3}{5})$
2. $\frac{12}{10} (- 1, 3) = (- \frac{6}{5}, \frac{18}{5})$
3. $\frac{2}{30} (2, - 1, 5) = (\frac{2}{15}, - \frac{1}{15}, \frac{1}{3})$
4. $- \frac{17}{14} (- 1, - 2, 3) = (\frac{17}{14}, \frac{17}{7}, - \frac{51}{14})$
5. $\frac{2 π^{2} - 16}{π^{2} + 10} (π, 3, - 1) = (\frac{2 π^{3} - 16 π}{π^{2} + 10}, \frac{6 π^{2} - 48}{π^{2} + 10}, \frac{- 2 π^{2} + 16}{π^{2} + 10})$
6. $\begin{array}{l} \frac{15 π - 10}{225 + 4 + 16} (15, - 2, 4) \\ = \frac{3 π - 2}{49} (15, - 2, 4) \\ = (\frac{45 π - 30}{49}, \frac{- 6 π + 4}{49}, \frac{12 π - 8}{49}) \end{array}$

コード

#!/usr/bin/env python3
from unittest import TestCase, main
from sympy import symbols, Matrix, pi, Rational

print('3, 4.')


def eq(a, b):
    for a0, b0 in zip(a, b):
        if a0.simplify() != b0.simplify():
            return False
    return True


As = [Matrix(t) for t in [(2, -1), (-1, 3), (2, -1, 5),
                          (-1, -2, 3), (pi, 3, -1), (15, -2, 4)]]
Bs = [Matrix(t) for t in [(-1, 1), (0, 4), (-1, 1, 1),
                          (-1, 3, -4), (2 * pi, -3, 7), (pi, 3, -1)]]


class TestProjection(TestCase):
    def test_b(self):
        projs = [Matrix(t)
                 for t in [(Rational(3, 2),
                            -Rational(3, 2)),
                           (0, 3),
                           (-Rational(2, 3), Rational(2, 3), Rational(2, 3)),
                           (Rational(17, 26), -Rational(51, 26), Rational(34, 13)),
                           ((2 * pi ** 3 - 16 * pi) / (2 * pi ** 2 + 29),
                            (-3 * pi ** 2 + 24) / (2 * pi ** 2 + 29),
                            (7 * pi ** 2 - 56) / (2 * pi ** 2 + 29)),
                           ((15 * pi ** 2 - 10 * pi) / (pi ** 2 + 10),
                            (45 * pi - 30) / (pi ** 2 + 10),
                            (-15 * pi + 10) / (pi ** 2 + 10))]]
        for i, (A, B, proj) in enumerate(zip(As, Bs, projs)):
            print(f'({chr(ord("a") + i)})')
            self.assertTrue(eq(A.dot(B) / B.dot(B) * B, proj))

    def test_a(self):
        projs = [Matrix(t) for t in [
            (-Rational(6, 5), Rational(3, 5)),
            (-Rational(6, 5), Rational(18, 5)),
            (Rational(2, 15), -Rational(1, 15), Rational(1, 3)),
            (Rational(17, 14), Rational(17, 7), -Rational(51, 14)),
            ((2 * pi ** 3 - 16 * pi) / (pi ** 2 + 10),
             (6 * pi ** 2 - 48) / (pi ** 2 + 10),
             (-2 * pi ** 2 + 16) / (pi ** 2 + 10)),
            ((45 * pi - 30) / 49, (-6 * pi + 4) / 49, (12 * pi - 8) / 49)]]
        for i, (A, B, proj) in enumerate(zip(As, Bs, projs)):
            print(f'({chr(ord("a") + i)})')
            self.assertTrue(eq(B.dot(A) / A.dot(A) * A, proj))


if __name__ == "__main__":
    main()

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

% ./sample3.py -v
3, 4.
test_a (__main__.TestProjection) ... (a)
(b)
(c)
(d)
(e)
(f)
ok
test_b (__main__.TestProjection) ... (a)
(b)
(c)
(d)
(e)
(f)
ok

----------------------------------------------------------------------
Ran 2 tests in 1.044s

OK
%

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2020年6月8日月曜日

数学 - Pytyhon - 解析学 - ベクトル - ベクトルのノルム - 射影、ノルム、内積、スカラー倍、2次元、3次元

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