# Math Help - centered B-spline

1. ## 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.

2. 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 $f_+$