- #1

mathmari

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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? :unsure:

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? :unsure:

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