插值算法

线性插值

已知点$(x_0, y_0)$和$(x_1, y_1)$计算$[x_0, x_1]$区间内某一位置$x$在直线上的$y$值

$\cfrac{y-y_0}{x-x_0}=\cfrac{y_1-y_0}{x_1-x_0} \quad \to \quad y=\cfrac{x_1-x}{x_1-x_0}y_0+\cfrac{x-x_0}{x_1-x_0}y_1$

双线性插值

avater

$f(R_1) \approx \cfrac{x_2-x}{x_2-x_1}f(Q_{11})+\cfrac{x-x_1}{x_2-x_1}f(Q_{21})$ $f(R_1) \approx \cfrac{x_2-x}{x_2-x_1}f(Q_{12})+\cfrac{x-x_1}{x_2-x_1}f(Q_{22})$ $f(P) \approx \cfrac{y_2-y}{y_2-y_1}f(R_1)+\cfrac{y-y_1}{y_2-y_1}f(R_2)$ $\begin{align}\to \quad f(P)=\cfrac{1}{(y_2-y_1)(x_2-x_1)}(y_2-y)(x_2-x)f(Q_{11})+\cfrac{1}{(y_2-y_1)(x_2-x_1)}(y_2-y)(x-x_1)f(Q_{21})\\ +\cfrac{1}{(y_2-y_1)(x_2-x_1)}(y-y_1)(x_2-x)f(Q_{12})+\cfrac{1}{(y_2-y_1)(x_2-x_1)}(y_2-y)(x-x_1)f(Q_{22})\end{align}$ MathJax TeX Test Page

插值算法

插值算法

线性插值

双线性插值

CATALOG

FEATURED TAGS

FRIENDS