题目
当 x=1 ,-1, 2时, f(x)=0 ,-3, 4,求f(x)的二次插值多项式.-|||-(1)用单项式基底.(2)用拉格朗日插值基底.(3)用牛顿基底·-|||-证明三种方法得到的多项式是相同的.
题目解答
答案
解析
步骤 1：单项式基底
设 ${p}_{2}(x)={a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$ .代入插值条件. $\left \{ \begin{matrix} {a}_{0}+{a}_{1}+{a}_{2}=0\\ {a}_{0}-{a}_{1}+{a}_{2}=-3\\ {a}_{0}+2{a}_{1}+4{a}_{2}=4\end{matrix} \right.$
步骤 2：解线性方程组
解上述方程组，得到 ${a}_{0}=-\dfrac {7}{3}$, ${a}_{1}=\dfrac {3}{2}$, ${a}_{2}=\dfrac {5}{6}$.
步骤 3：拉格朗日插值基底
${I}_{g}(x)=f({x}_{1}){l}_{h}(x)+f({x}_{1}){l}_{1}(x)+f({x}_{2})lg(x)$
$=0+(-3)\dfrac {(x-1)(x-2)}{(-1-1)(-1-2)}+4\dfrac {(x-1)(x+1)}{(2-1)(2+1)}$
步骤 4：牛顿插值基底
建立差商表 -1 -3 1 0 -$\dfrac {3}{2}$ 5/6 2 4 4
则 ${M}_{2}(x)=-3+\dfrac {3}{2}(x+1)+\dfrac {5}{6}(x+1)(x-1)$
设 ${p}_{2}(x)={a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$ .代入插值条件. $\left \{ \begin{matrix} {a}_{0}+{a}_{1}+{a}_{2}=0\\ {a}_{0}-{a}_{1}+{a}_{2}=-3\\ {a}_{0}+2{a}_{1}+4{a}_{2}=4\end{matrix} \right.$
步骤 2：解线性方程组
解上述方程组，得到 ${a}_{0}=-\dfrac {7}{3}$, ${a}_{1}=\dfrac {3}{2}$, ${a}_{2}=\dfrac {5}{6}$.
步骤 3：拉格朗日插值基底
${I}_{g}(x)=f({x}_{1}){l}_{h}(x)+f({x}_{1}){l}_{1}(x)+f({x}_{2})lg(x)$
$=0+(-3)\dfrac {(x-1)(x-2)}{(-1-1)(-1-2)}+4\dfrac {(x-1)(x+1)}{(2-1)(2+1)}$
步骤 4：牛顿插值基底
建立差商表 -1 -3 1 0 -$\dfrac {3}{2}$ 5/6 2 4 4
则 ${M}_{2}(x)=-3+\dfrac {3}{2}(x+1)+\dfrac {5}{6}(x+1)(x-1)$