B样条函数

$\newcommand{\nj}[1]{{\left\langle#1\right\rangle}}\newcommand{\kh}[1]{{\left(#1\right)}}\newcommand{\zkh}[1]{{\left[#1\right]}}\newcommand{\dkh}[1]{{\left\{{#1}\right \}}}\newcommand{\norm}[1]{\left|#1\right|}\newcommand{\normm}[1]{\left\Vert#1\right\Vert}\newcommand{\calA}{{\mathcal{A}}}\newcommand{\calC}{{\mathcal{C}}}\newcommand{\calH}{{\mathcal{H}}}\newcommand{\calN}{{\mathcal{N}}}\newcommand{\calQ}{{\mathcal{Q}}}\newcommand{\calR}{{\mathcal{R}}}\newcommand{\calS}{{\mathcal{S}}}\newcommand{\calW}{{\mathcal{W}}}\newcommand{\calX}{{\mathcal{X}}}\newcommand{\calZ}{{\mathcal{Z}}}\newcommand{\bbC}{\mathbb{C}}\newcommand{\bbN}{\mathbb{N}}\newcommand{\bbR}{\mathbb{R}}\newcommand{\bbZ}{\mathbb{Z}}\newcommand{\sin}{{\rm sin}}\newcommand{\cos}{{\rm cos}}\newcommand{\Id}{{\rm Id}}$

样条函数空间

　　给定节点$\left\{\xi_i\right\}_{i=0}^n$，记其上分片$k$次多项式、$k-1$次连续可导函数构成的空间为$\calS\left\{k;\xi_0,\cdots\xi_n\right\}$。显然任意$s\in\calS$应有

$$s\kh{x}=\sum_{i=0}^kc_ix^i+\frac{1}{k!}\sum_{i=1}^{n-1}d_i\zkh{x-\xi_i}^k_+,$$

其中$\zkh{\cdot}_+=\max\left\{0,\cdot\right\}$。从此可看出dimS=