# Basic of $B$-splines

##### Posted on Sep 09, 20190 Comments
Tags: B-spline

This note is based on de Boor, C. (1978). A Practical Guide to Splines, Springer, New York.

Let $\tau:=(\tau_i)_1^n$ be a sequence of sites not necessarily distinct. We say that the function $p$ agrees with the function $g$ at $\tau$ provided that, for every site $\zeta$ that occurs $m$ times in the sequence $\tau_1,\ldots,\tau_n$, $p$ and $g$ agree $m$-fold at $\zeta$, that is, $$p^{(i-1)}(\zeta) = g^{(i-1)}(\zeta)\qquad \text{for }i=1,\ldots,m\,.$$
The $k$-th divided difference of a function $g$ at the sites $\tau_i,\ldots,\tau_{i+k}$ is the leading coefficient, i.e., the coefficient of $x^k$, of the order $k+1$ that agrees with $g$ at the sequence $(\tau_i,\ldots,\tau_{i+k})$, which is denoted by $$[\tau_i,\ldots,\tau_{i+k}]g\,.$$
For computations, $$[\tau_i,\ldots,\tau_{i+k}]g=\frac{g^{(k)}(\tau_i)}{k!}\quad \text{if }\tau_i=\cdots=\tau_{i+k}, g\in C^{(k)}\,,$$ while $$[\tau_i,\ldots,\tau_{i+k}]g = \frac{[\tau_i,\ldots,\tau_{r-1},\tau_{r+1},\ldots,\tau_{i+k}]g - [\tau_i,\ldots,\tau_{s-1},\tau_{s+1},\ldots,\tau_{i+k}]g}{\tau_s-\tau_r}$$ if $\tau_r,\tau_s$ are any two distinct sites in the sequence $\tau_i,\ldots,\tau_{i+k}$.
Let $t:=(t_j)$ be a nondecreasing sequence. The $j$-th normalized $B$-spline of order $k$ for the knot sequence $t$ os denoted by $B_{j,k,t}$ and is defined by the rule $$B_{j,k,t}(x):=(t_{j+k}-t_j)[t_j,\ldots,t_{j+k}](\cdot -x)_+^{k-1}\,, \quad\text{all }x\in\IR\,.$$ That is, $$B_{j,k,t}(x)=[t_{j+1},\ldots,t_{j+k}](\cdot -x)_+^{k-1} - [t_j,\ldots,t_{j+k-1}](\cdot -x)_+^{k-1}\,.$$
B-spline properties: (i) For $k > 1$, $$B_{jk} = \omega_{jk}B_{j,k-1} + (1-w_{j+1,k})B_{j+1,k-1}\,,$$ where $$\omega_{jk}(x)=\frac{x-t_j}{t_{j+k-1}-t_j}\,.$$ (ii) Support and positivity: The $B$-spline $B_{j,k,t}$ is pp of order $k$ with breaks $t_j,\ldots,t_{j+k}$, hence made up of at most $k$ nontrivial polynomial pieces, vanishes outside the interval $[t_j..t_{j+k})$, and is positive on the interior of that interval, that is, $B_{j,k,t}(x) > 0, t_j< x < t_{j+k}$ while $t_j=t_{j+k}$ implies $B_{jk}=0$.

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