数学 - Python - 代数学 - 連立方程式と高次方程式 - 展開、因数分解、因数定理、解、複素数の範囲

学習環境

代数への出発 (新装版数学入門シリーズ) (松坂和夫(著)、岩波書店)の第5章(連立方程式と高次方程式) 、問8の解答を求めてみる。

1. $\begin{array}{l} 16 - 16 - 6 + 6 = 0 \\ (x - 2) (2 x^{2} - 3) = 0 \\ (x - 2) (\sqrt{2} x - \sqrt{3}) (\sqrt{2} x + \sqrt{3}) = 0 \\ x = 2, \pm \frac{\sqrt{6}}{2} \end{array}$
2. $\begin{array}{l} (x^{2} - 3 x + 2) (x - 3) = 24 \\ x^{3} - 6 x^{2} + 11 x - 30 = 0 \\ 125 - 150 + 55 - 30 = 0 \\ (x - 5) (x^{2} - x + 6) = 0 \\ x = \frac{1 \pm \sqrt{1 - 24}}{2} = \frac{1 \pm \sqrt{23} i}{2} \\ x = 5, \frac{1 \pm \sqrt{23} i}{2} \end{array}$
3. $\begin{array}{l} 5 - 12 + 1 + 6 = 0 \\ (x - 1) (5 x^{2} - 7 x - 6) = 0 \\ (x - 1) (x - 2) (5 x + 3) = 0 \\ x = - \frac{3}{5}, 1, 2 \end{array}$
4. $\begin{array}{l} 1 + 8 + 9 - 8 - 10 = 0 \\ (x - 1) (x^{3} + 9 x^{2} + 18 x + 10) = 0 \\ - 1 + 9 - 18 + 10 = 0 \\ (x - 1) (x + 1) (x^{2} + 8 x + 10) = 0 \\ x = - 4 \pm \sqrt{16 - 10} = - 4 \pm \sqrt{6} \\ x = \pm 1, - 4 \pm \sqrt{6} \end{array}$
5. $\begin{array}{l} (4 x^{2} - 1) (x^{2} + 3) = 0 \\ x = \pm \frac{1}{2}, \pm \sqrt{3} i \end{array}$
6. $\begin{array}{l} 16 - 40 + 20 - 6 + 18 \neq 0 \\ 81 - 135 + 45 - 9 + 18 = 0 \\ (x - 3) (x^{3} - 2 x^{2} - x - 6) = 0 \\ 27 - 18 - 3 - 6 = 0 \\ {(x - 3)}^{2} (x^{2} + x + 2) = 0 \\ x = \frac{- 1 \pm \sqrt{1 - 8}}{2} = \frac{- 1 \pm \sqrt{7} i}{2} \\ x = 3, \frac{- 1 \pm \sqrt{7} i}{2} \end{array}$
7. $\begin{array}{l} {(x^{2} - 5 x)}^{2} - 6 (x^{2} - 5 x) - 55 = 0 \\ (x^{2} - 5 x - 11) (x^{2} - 5 x + 5) = 0 \\ x = \frac{5 \pm \sqrt{25 + 44}}{2}, \frac{5 \pm \sqrt{25 - 20}}{2} \\ = \frac{5 \pm \sqrt{69}}{2}, \frac{5 \pm \sqrt{5}}{2} \end{array}$
8. $\begin{array}{l} x^{3} + x^{3} + 3 x^{2} + 3 x + 1 + x^{3} + 6 x^{2} + 12 x + 8 = x^{3} + 9 x^{2} + 27 x + 27 \\ 2 x^{3} - 12 x - 18 = 0 \\ x^{3} - 6 x - 9 = 0 \\ 27 - 18 - 9 = 0 \\ (x - 3) (x^{2} + 3 x + 3) = 0 \\ x = \frac{- 3 \pm \sqrt{9 - 12}}{2} = \frac{- 3 \pm \sqrt{3} i}{2} \\ x = 3, \frac{- 3 \pm \sqrt{3} i}{2} \end{array}$

#!/usr/bin/env python3 from unittest import TestCase, main from sympy import symbols, solveset, I, Rational, sqrt print('8.') class TestEquations(TestCase): def test(self): x = symbols('x') eqs = [2 * x ** 3 - 4 * x ** 2 - 3 * x + 6, (x - 1) * (x - 2) * (x - 3) - 24, 5 * x ** 3 - 12 * x ** 2 + x + 6, x ** 4 + 8 * x ** 3 + 9 * x ** 2 - 8 * x - 10, 4 * x ** 4 + 11 * x ** 2 - 3, x ** 4 - 5 * x ** 3 + 5 * x ** 2 - 3 * x + 18, (x ** 2 - 5 * x - 2) * (x ** 2 - 5 * x - 4) - 63, x ** 3 + (x + 1) ** 3 + (x + 2) ** 3 - (x + 3) ** 3] xss = [(2, -sqrt(6) / 2, sqrt(6) / 2), (5, (1 - sqrt(23) * I) / 2, (1 + sqrt(23) * I) / 2), (-Rational(3, 5), 1, 2), (-1, 1, -4 - sqrt(6), -4 + sqrt(6)), (-Rational(1, 2), Rational(1, 2), -sqrt(3) * I, sqrt(3) * I), (3, (-1 - sqrt(7) * I) / 2, (-1 + sqrt(7) * I) / 2), ((5 - sqrt(5)) / 2, (5 + sqrt(5)) / 2, (5 - sqrt(69)) / 2, (5 + sqrt(69)) / 2), (3, (-3 - sqrt(3) * I) / 2, (-3 + sqrt(3) * I) / 2)] for i, (eq, xs) in enumerate(zip(eqs, xss), 1): print(f'({i})') self.assertEqual(solveset(eq), set(xs)) if __name__ == "__main__": main()

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2020年5月12日火曜日

数学 - Python - 代数学 - 連立方程式と高次方程式 - 展開、因数分解、因数定理、解、複素数の範囲

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