# Linear algebra and Lagrange interpolation formula

$\displaystyle{ A = \begin{bmatrix} \alpha_0^0 & \alpha_0^1 & \cdots & \alpha_0^n \\ \alpha_1^0 & \alpha_1^1 & \cdots & \alpha_1^n \\ \vdots & \vdots & & \vdots \\ \alpha_n^0 & \alpha_n^1 & \cdots & \alpha_n^n \end{bmatrix} }$,

$\displaystyle{ e^j e_i = [i = j] }$,

$\displaystyle{ [x^i](\sum_{i=0}^n \beta_i x^i) = \beta_i }$.

$\displaystyle{ [\alpha_j](f(x)) = [\alpha_j](\sum_{i=0}^n \beta_i x^i) = \sum_{i=0}^n \beta_i \alpha_j^i = f(\alpha_j) }$.

$\displaystyle{ L_i(x) = \prod_{j=0 \atop j \neq i}^n \frac{x-\alpha_j}{\alpha_i-\alpha_j} }$.

$\displaystyle{ [\alpha_j]L_i(x) = L_i(\alpha_j) = [i = j] }$

$\displaystyle{ f = \sum_{i=0}^n (e^i f) e_i }$.

$\displaystyle{ x^j = \sum_{i=0}^n ([\alpha_i]x^j) L_i(x) = \sum_{i=0}^n \alpha_i^j L_i(x) }$,

$\displaystyle{ A = \begin{bmatrix} \alpha_0^0 & \alpha_0^1 & \cdots & \alpha_0^n \\ \alpha_1^0 & \alpha_1^1 & \cdots & \alpha_1^n \\ \vdots & \vdots & & \vdots \\ \alpha_n^0 & \alpha_n^1 & \cdots & \alpha_n^n \end{bmatrix} }$

$\displaystyle{ f(x) = \sum_{i=0}^n \beta_i L_i(x) }$

### Author: hcsoso

Ph.D. student in the Department of Computer Science, University of Illinois at Urbana-Champaign.

### 2 thoughts on “Linear algebra and Lagrange interpolation formula”

1. 倒数段落里的“Lagrange差值公式”是否应为“Lagrange插值公式”呢？

• 的確! 謝謝你的細心, 歡迎常來看看! (As long as 我常常在更新…)