例题 5.3

例题 5.3.

求一个三次多项式 $f (x)$ ,使得

$f (1) = 6$

$f (2) = 20$

$f (- 1) = 8$

$f (- 3) = 10$ .

\begin{proof} 设三次多项式为 $f (x) = a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$ . 由条件有

$f (1) = a_{3} + a_{2} + a_{1} + a_{0} = 6,$

$f (2) = 8 a_{3} + 4 a_{2} + 2 a_{1} + a_{0} = 20,$

$f (- 1) = - a_{3} + a_{2} - a_{1} + a_{0} = 8,$

$f (- 3) = - 27 a_{3} + 9 a_{2} - 3 a_{1} + a_{0} = 10 .$

将 $a_{3}, a_{2}, a_{1}, a_{0}$ 看作未知量,可得线性方程组

$⎩ ⎨ ⎧ a_{3} + a_{2} + a_{1} + a_{0} 8 a_{3} + 4 a_{2} + 2 a_{1} + a_{0} - a_{3} + a_{2} - a_{1} + a_{0} - 27 a_{3} + 9 a_{2} - 3 a_{1} + a_{0} = 6, = 20, = 8, = 10 .$

其系数行列式为

$D = 18 - 1 - 27 1419 12 - 1 - 3 1111 = - 240 .$

计算可得

$D_{1} = 6208101419 12 - 1 - 3 1111 = - 240, D_{2} = 18 - 1 - 27 620810 12 - 1 - 3 1111 = - 720,$

$D_{3} = 18 - 1 - 27 14196208101111 = 480, D_{4} = 18 - 1 - 27 1419 12 - 1 - 3 620810 = - 960,$

所以有

$a_{3} = \frac{D _{1}}{D} = 1, a_{2} = \frac{D _{2}}{D} = 3, a_{1} = \frac{D _{3}}{D} = - 2, a_{0} = \frac{D _{4}}{D} = 4.$

故而所求三次多项式为 $f (x) = x^{3} + 3 x^{2} - 2 x + 4$ . \end{proof}

Youliang Zhong

例题 5.3

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