## 利用 Lagrange 內插多項式推導 Vandermonde 矩陣的逆矩陣

$V=\begin{bmatrix} 1&1&\cdots&1\\ x_1&x_2&\cdots&x_n\\ x_1^2&x_2^2&\cdots&x_n^2\\ \vdots&\vdots&\vdots&\vdots\\ x_1^{n-1}&x_2^{n-1}&\cdots&x_n^{n-1} \end{bmatrix}$

$n\times n$ 階矩陣 $Z=[z_{ij}]$ 代表 Vandermonde 矩陣 $V=[v_{ij}]$ 的逆矩陣，即 $ZV=I$。對於任意 $1\le i,k\le n$，就有

$\displaystyle \sum_{j=1}^nz_{ij}v_{jk}=\sum_{j=1}^nz_{ij}x_k^{j-1}=\delta_{ik}$

\displaystyle\begin{aligned} L_i(t)&=\frac{(t-x_1)\cdots(t-x_{i-1})(t-x_{i+1})\cdots(t-x_n)}{(x_i-x_1)\cdots(x_i-x_{i-1})(x_i-x_{i+1})\cdots(x_i-x_n)}\\ &=\prod_{1\le p\le n\atop p\neq i}(t-x_p)\bigg/\prod_{1\le p\le n\atop p\neq i}(x_i-x_p).\end{aligned}

$t\in\{x_1,\ldots,x_n\}$，就有

$\displaystyle L_i(t)=\sum_{j=1}^nz_{ij}t^{j-1}$

$z_{ij}$ 可唯一決定 (因為 $\{x_1,\ldots,x_n\}$ 互異，共有 $n$ 個條件方程)。剩下的工作是從上式解出 $z_{ij}$

\displaystyle\begin{aligned} \prod_{1\le p\le n\atop p\neq i}(t-x_p)&=S_0t^{n-1}-S_1t^{n-2}+\cdots+(-1)^{n-1}S_{n-1}\\ &=\sum_{q=1}^n(-1)^{q-1}S_{q-1}t^{n-q}\\ &=\sum_{j=1}^n(-1)^{n-j}S_{n-j}t^{j-1},\end{aligned}

$\displaystyle S_r(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_{n})=\sum_{1\le p_1<\cdots

$\displaystyle L_i(t)=\sum_{j=1}^n(-1)^{n-j}S_{n-j}(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n)t^{j-1}\bigg/\prod_{1\le p\le n\atop p\neq i}(x_i-x_p)$

\displaystyle\begin{aligned} z_{ij}&=(-1)^{n-j}S_{n-j}(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n)\bigg/\prod_{1\le p\le n\atop p\neq i}(x_i-x_p)\\ &=(-1)^{j+1}S_{n-j}(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n)\bigg/\prod_{1\le p\le n\atop p\neq i}(x_p-x_i)\\ &=(-1)^{j+1}{\sum_{1\le p_1<\cdots

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