# centered B-spline

• May 4th 2010, 03:13 PM
surjective
centered B-spline
Hello,

I'm trying to show that the centered B-spline $B_{2}$ is given as:

$B_{2}(x)= \left\lbrace \begin{matrix}1+x, \hspace{0,3cm} x\in[-1,0]\\ 1-x , \hspace{0,3cm} x\in[0,1]\\0, \hspace{0,3cm} otherwise \end{matrix} \right.$

I know that the centered B-spline is given as:

$B_{m}(x)=N_{m}\left(x+\frac{m}{2}\right)$

The spline $N_{m}$ is given as:

$N_{m}(x)=\frac{1}{(m-1)!} \sum_{j=0}^{m}(-1)^{j} \left(\begin{matrix} m \\ j \end{matrix} \right) (x-j)_{+}^{(m-1)}$

Furthermore by definition:

$f(x)_{+}=\max(0,f(x))$

So far I have computed the centered B-spline $(m=2)$:

$B_{2}(x)=N_{1}(x+1)=(x+1)_{+}-2(x)_{+}+(x-1)_{+}$

Could someone assist me getting from here to the function mentioned at the beginning?

Thanks very much indeed.
• May 4th 2010, 06:30 PM
Jose27
Quote:

Originally Posted by surjective
Hello,

I'm trying to show that the centered B-spline $B_{2}$ is given as:

$B_{2}(x)= \left\lbrace \begin{matrix}1+x, \hspace{0,3cm} x\in[-1,0]\\ 1-x , \hspace{0,3cm} x\in[0,1]\\0, \hspace{0,3cm} otherwise \end{matrix} \right.$

I know that the centered B-spline is given as:

$B_{m}(x)=N_{m}\left(x+\frac{m}{2}\right)$

The spline $N_{m}$ is given as:

$N_{m}(x)=\frac{1}{(m-1)!} \sum_{j=0}^{m}(-1)^{j} \left(\begin{matrix} m \\ j \end{matrix} \right) (x-j)_{+}^{(m-1)}$

Furthermore by definition:

$f(x)_{+}=\max(0,f(x))$

So far I have computed the centered B-spline $(m=2)$:

$B_{2}(x)=N_{1}(x+1)=(x+1)_{+}-2(x)_{+}+(x-1)_{+}$

Could someone assist me getting from here to the function mentioned at the beginning?

Thanks very much indeed.

Well, this is easy, just consider the different cases suggested:

For $x\geq 1$ we have $B_2(x)= x+1-2x+x-1=0$

For $0\leq x \leq 1$ we get $B_2(x)= x+1-2x= 1-x$

For $-1\leq x \leq 0$ we get $B_2(x)=x+1$

and finally if $x\leq -1$ then $B_2(x)=0$

where all this should follow by the definition of $f_+$