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- Apr 14, 2013

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Let $N_j$, $j=-k,\ldots , m-1$ the normalized B-splines of the set of nodes $x_0, \ldots , x_m$ of degree $k$.

Show that $$\sum_{j=-k}^{m-1}N_j(x)=1 \ \text{ for all } x\in [x_0, x_m]$$

A formula with divided differences is

\begin{align*}&N_j(x)=(x_{j+k+1}-x_j)B_j(x) \\& \text{ with } \ B_j=(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k+1}] =\frac{(\cdot -x)_+^k[x_{j+1}x_{j+2}\ldots x_{j+k+1}]-(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k}]}{x_{j+k+1}-x_j}\end{align*} where $(x)_+^k=\begin{cases}x^k & \text{ if } x\geq 0 \\ 0 & \text{ if } x<0\end{cases}$.

It holds that $B_j=(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k+1}]=f_x[x_jx_{j+1}x_{j+2}\ldots x_{j+k+1}]$ with $f_x(y)=(y-x)_+^k$, so is this the function for the divided difference?

Then we have \begin{align*}\sum_{j=-k}^{m-1}N_j(x)&=\sum_{j=-k}^{m-1}(x_{j+k+1}-x_j)B_j(x)\\ & =\sum_{j=-k}^{m-1}(x_{j+k+1}-x_j)\frac{(\cdot -x)_+^k[x_{j+1}x_{j+2}\ldots x_{j+k+1}]-(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k}]}{x_{j+k+1}-x_j}\\ & =\sum_{j=-k}^{m-1}\left ((\cdot -x)_+^k[x_{j+1}x_{j+2}\ldots x_{j+k+1}]-(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k}]\right ) \\ & = (\cdot -x)_+^k[x_{m}x_{m+1}\ldots x_{m+k}]-(\cdot -x)_+^k[x_{-k}x_{-k+1}x_{-k+2}\ldots x_{0}] \\ & = (\cdot -x)_+^k[x_{m}x_{m+1}\ldots x_{m+k}]-0[x_{-k}x_{-k+1}x_{-k+2}\ldots x_{0}] \\ & =1-0 \\ & =1\end{align*}

I haven't understood the part from the 4th equality.

Why is the sum equal to the last term minus the first term?

How do we get the zero and how do we get the one?