高等数学

Wang Haihua

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方程组(System of Equations)

\begin{align} 3x+6y+2z&=-13\\ x+2y+z&=-5\\ -5x-10y-2z&=19 \end{align}

增广矩阵(Augumented Matrix)

$$ \left[\begin{matrix}3 & 6 & 2 & -13\\1 & 2 & 1 & -5\\-5 & -10 & -2 & 19\end{matrix}\right] $$$$ \left( \left[\begin{matrix}1 & 2 & 0 & -3\\0 & 0 & 1 & -2\\0 & 0 & 0 & 0\end{matrix}\right], \ \left( 0, \ 2\right)\right) $$

简化阶梯型矩阵（Reduced Row Echelon Form）

例子

考虑这个问题:已知点$(1,3)$,$(2, -2)$ ,$(3, -5)$, and $(4, 0)$，如何找到一个经过这些点的三次多项式？

三次多项式的形式为

\begin{align} y=a_0+a_1x+a_2x^2+a_3x^3 \end{align}

我们将点代入

\begin{align} (x,y)&=(1,3)\qquad\longrightarrow\qquad \ 2=a_0+3a_1+9a_2 +27a_3 \\ (x,y)&=(2,-2)\qquad\longrightarrow\qquad 3=a_0+a_1+a_2+a_3\\ (x,y)&=(3,-5)\qquad\longrightarrow\qquad 2=a_0-4a_1+16a_2-64a_3\\ (x,y)&=(4,0)\qquad\longrightarrow\qquad -2=a_0+2a_1+4a_2+8a_3 \end{align}

增广矩阵是 $$ \left[\begin{matrix}1 & 1 & 1 & 1 & 3\\1 & 2 & 4 & 8 & -2\\1 & 3 & 9 & 27 & -5\\1 & 4 & 16 & 64 & 0\end{matrix}\right] $$

解得最简阶梯形矩阵为

$$ \left[\begin{matrix}1 & 0 & 0 & 0 & 4\\0 & 1 & 0 & 0 & 3\\0 & 0 & 1 & 0 & -5\\0 & 0 & 0 & 1 & 1\end{matrix}\right] $$

最后一列是解，即三次多项式的系数。所求三次多项式为 \begin{align} y = 4 + 3x - 5x^2 + x^3 \end{align}

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import sympy as sy
def pl(m):
    x= '$$'+sy.latex(m)+'$$'
    print(x)
import numpy as np
import matplotlib.pyplot as plt
plt.rcParams['font.family']=['SimHei']
plt.rcParams['axes.unicode_minus'] = False
%matplotlib inline

A = sy.Matrix([[3, 6, 2, -13], [1, 2, 1, -5], [-5, -10, -2, 19]]);A

$\displaystyle \left[\begin{matrix}3 & 6 & 2 & -13\\1 & 2 & 1 & -5\\-5 & -10 & -2 & 19\end{matrix}\right]$

print(sy.latex(A))

\left[\begin{matrix}3 & 6 & 2 & -13\\1 & 2 & 1 & -5\\-5 & -10 & -2 & 19\end{matrix}\right]

A_rref = A.rref(); A_rref

(Matrix([
 [1, 2, 0, -3],
 [0, 0, 1, -2],
 [0, 0, 0,  0]]),
 (0, 2))

a, b, c = sy.symbols('a, b, c', real = True)
A = sy.Matrix([[1, 2, -3, a], [4, -1, 8, b], [2, -6, -4, c]]); A

$\displaystyle \left[\begin{matrix}1 & 2 & -3 & a\\4 & -1 & 8 & b\\2 & -6 & -4 & c\end{matrix}\right]$

A_rref = A.rref(); A_rref

(Matrix([
 [1, 0, 0,          2*a/7 + b/7 + c/14],
 [0, 1, 0,    16*a/91 + b/91 - 10*c/91],
 [0, 0, 1, -11*a/91 + 5*b/91 - 9*c/182]]),
 (0, 1, 2))

A_rref[0]

$\displaystyle \left[\begin{matrix}1 & 0 & 0 & \frac{2 a}{7} + \frac{b}{7} + \frac{c}{14}\\0 & 1 & 0 & \frac{16 a}{91} + \frac{b}{91} - \frac{10 c}{91}\\0 & 0 & 1 & - \frac{11 a}{91} + \frac{5 b}{91} - \frac{9 c}{182}\end{matrix}\right]$

A = sy.Matrix([[1, 1, 1, 1, 3], [1, 2, 4, 8, -2], [1, 3, 9, 27, -5], [1, 4, 16, 64, 0]]); A

$\displaystyle \left[\begin{matrix}1 & 1 & 1 & 1 & 3\\1 & 2 & 4 & 8 & -2\\1 & 3 & 9 & 27 & -5\\1 & 4 & 16 & 64 & 0\end{matrix}\right]$

A_rref = A.rref(); A_rref[0]

$\displaystyle \left[\begin{matrix}1 & 0 & 0 & 0 & 4\\0 & 1 & 0 & 0 & 3\\0 & 0 & 1 & 0 & -5\\0 & 0 & 0 & 1 & 1\end{matrix}\right]$

A = sy.Matrix([[1, 1, 1, 1, 3], [1, 2, 4, 8, -2], [1, 3, 9, 27, -5], [1, 4, 16, 64, 0]])
A_rref = A.rref()
A_rref = np.array(A_rref[0])
poly_coef = A_rref.astype(float)[:,-1]
x = np.linspace(-5, 5, 40)
y = poly_coef[0] + poly_coef[1]*x + poly_coef[2]*x**2 + poly_coef[3]*x**3.5
fig, ax = plt.subplots(figsize = (7, 5))
ax.plot(x, y, lw = 3, color ='red')
ax.scatter([1, 2, 3, 4], [3, -2, -5, 0], s = 100, color = 'blue', zorder = 3)
ax.grid()
ax.set_xlim([0, 5])
ax.set_ylim([-10, 10])

ax.text(1, 3.5, '$(1, 3)$', fontsize = 15)
ax.text(1.5, -2.5, '$(2, -2)$', fontsize = 15)
ax.text(2.7, -4, '$(3, -5.5)$', fontsize = 15)
ax.text(4.1, 0, '$(4, .5)$', fontsize = 15)
plt.savefig('images/lin0202.png')

<ipython-input-47-db7151f4f30a>:6: RuntimeWarning: invalid value encountered in power
  y = poly_coef[0] + poly_coef[1]*x + poly_coef[2]*x**2 + poly_coef[3]*x**3.5

$$\left[\begin{matrix}1 & 0 & 0 & 0 & 4\\0 & 1 & 0 & 0 & 3\\0 & 0 & 1 & 0 & -5\\0 & 0 & 0 & 1 & 1\end{matrix}\right]$$