# centered B-spline

• May 4th 2010, 02:13 PM
surjective
centered B-spline
Hello,

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

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

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

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

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

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

Furthermore by definition:

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

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

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

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

Thanks very much indeed.
• May 4th 2010, 05:30 PM
Jose27
Quote:

Originally Posted by surjective
Hello,

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

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

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

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

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

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

Furthermore by definition:

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

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

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

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

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

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

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

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