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Math Help - centered B-spline

  1. #1
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    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.
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  2. #2
    Super Member
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    Quote Originally Posted by surjective View Post
    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_+
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