# Sum of normalized B-splines

#### mathmari

##### Well-known member
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Hey!! 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? • Klaas van Aarsen

#### Klaas van Aarsen

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Why is the sum equal to the last term minus the first term?
Hey mathmari !!

It's called a telescoping series.
Consider $$\sum_{i=0}^{N-1} (a_{i+1}-a_i) = (a_1-a_0) + (a_2-a_1)+\ldots +(a_N - a_{N-1}) = a_N-a_0$$ How do we get the zero and how do we get the one?
What does $(\cdot -x)_+^k[x_{m}x_{m+1}\ldots x_{m+k}]$ mean? As for the zero, I can already guess that we get something like $(x_0-x)_+^k$ and for all $x$ in $[x_0,x_m]$ we have that $x_0-x\le 0$, so that $(x_0-x)_+^k=0$. • mathmari

#### mathmari

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MHB Site Helper
It's called a telescoping series.
Consider $$\sum_{i=0}^{N-1} (a_{i+1}-a_i) = (a_1-a_0) + (a_2-a_1)+\ldots +(a_N - a_{N-1}) = a_N-a_0$$ We have that
\begin{align*} \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_{-k+1}x_{-k+2}\ldots x_{1}]-(\cdot -x)_+^k[x_{-k}x_{-k+1}x_{-k+2}\ldots x_{0}]+(\cdot -x)_+^k[x_{-k+2}x_{-k+3}\ldots x_{2}]-(\cdot -x)_+^k[x_{-k+1}x_{-k+2}x_{-k+3}\ldots x_{1}] +\ldots \\ & +\ldots
(\cdot -x)_+^k[x_{m-1}x_{m}\ldots x_{m+k-1}]-(\cdot -x)_+^k[x_{m-2}x_{m-1}x_{m}\ldots x_{m+k-2}]
+(\cdot -x)_+^k[x_{m}x_{m+1}\ldots x_{m+k}]-(\cdot -x)_+^k[x_{m-1}x_{m}x_{m+1}\ldots x_{m+k-1}] \\ & = (\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}]\end{align*} What does $(\cdot -x)_+^k[x_{m}x_{m+1}\ldots x_{m+k}]$ mean? As for the zero, I can already guess that we get something like $(x_0-x)_+^k$ and for all $x$ in $[x_0,x_m]$ we have that $x_0-x\le 0$, so that $(x_0-x)_+^k=0$. Since we have 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 $f_x(y)=(y-x)_+^k$ it must be the difference you are reffering to.

So do we have that $$(\cdot -x)_+^k[x_{-k}x_{-k+1}x_{-k+2}\ldots x_{0}]=([x_{-k}x_{-k+1}x_{-k+2}\ldots x_{0}]-x)_+^k$$ or how do we write this?

I understand what you say that it is zero because it is negative, but why is the other one equal to $1$ ? #### Klaas van Aarsen

##### MHB Seeker
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So do we have that $$(\cdot -x)_+^k[x_{-k}x_{-k+1}x_{-k+2}\ldots x_{0}]=([x_{-k}x_{-k+1}x_{-k+2}\ldots x_{0}]-x)_+^k$$ or how do we write this?
That does not look right to me.
It does not make sense that we would subtract $x$ from a product of $x_i$'s. The "dimensions" don't match.
So I think something else is intended. Something where we process the $x_i$ one by one or something like that. Once we know what it is, we can probably understand how it evaluates to $1$.

• mathmari

#### mathmari

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That does not look right to me.
It does not make sense that we would subtract $x$ from a product of $x_i$'s. The "dimensions" don't match.
So I think something else is intended. Something where we process the $x_i$ one by one or something like that. Once we know what it is, we can probably understand how it evaluates to $1$.
Here in Wikipedia it says that it is the n-th divided difference, but I haven't really understood that. #### Klaas van Aarsen

##### MHB Seeker
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Here in Wikipedia it says that it is the n-th divided difference, but I haven't really understood that. The mean value theorem for divided differences says:

For any $n + 1$ pairwise distinct points $x_0$, ..., $x_n$ in the domain of an $n$-times differentiable function $f$ there exists an interior point
$$\xi \in (\min\{x_0,\dots,x_n\},\max\{x_0,\dots,x_n\})$$
where the $n$th derivative of $f$ equals $n!$ times the $n$th divided difference at these points:
$$f[x_0,\dots,x_n] = \frac{f^{(n)}(\xi)}{n!}.$$ In our case we have $x\in[x_0,x_m]$ and:
$(\cdot -x)_+^k[x_{m}x_{m+1}\ldots x_{m+k}] =\frac{\left((\cdot -x)_+^k\right)^{(k)}(\xi)}{k!} =\begin{cases}\frac{\left((\cdot -x)^k\right)^{(k)}(\xi)}{k!} &\text{if }\xi-x>0\\ \frac{\left(0\right)^{(k)}(\xi)}{k!} &\text{if }\xi-x< 0 \end{cases} =\begin{cases}1 &\text{if }\xi-x>0\\ 0 &\text{if }\xi-x< 0 \end{cases}$
Since $\xi$ must be between $x_{m},x_{m+1},\ldots, x_{m+k}$, it follows that $\xi-x>0$. For the zero case, $\xi$ must be betweeen $x_{-k},\ldots,x_0$, so $\xi-x<0$. Last edited:
• mathmari

#### mathmari

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MHB Site Helper
In our case we have $x\in[x_0,x_m]$ and:
$(\cdot -x)_+^k[x_{m}x_{m+1}\ldots x_{m+k}] =\frac{\left((\cdot -x)_+^k\right)^{(k)}(\xi)}{k!} =\begin{cases}\frac{\left((\cdot -x)^k\right)^{(k)}(\xi)}{k!} &\text{if }\xi-x>0\\ \frac{\left(0\right)^{(k)}(\xi)}{k!} &\text{if }\xi-x< 0 \end{cases} =\begin{cases}1 &\text{if }\xi-x>0\\ 0 &\text{if }\xi-x< 0 \end{cases}$
Since $\xi$ must be between $x_{m},x_{m+1},\ldots, x_{m+k}$, it follows that $\xi-x>0$. For the zero case, $\xi$ must be betweeen $x_{-k},\ldots,x_0$, so $\xi-x<0$. I haven't really understood how we get the $1$ in the case $\xi-x>0$ #### Klaas van Aarsen

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I haven't really understood how we get the $1$ in the case $\xi-x>0$
Consider the function $f: \cdot \mapsto (\cdot - x)^k$.
Note that this is not a function of $x$. Instead it's a function of the unspecified $\cdot$.
So we have for instance $f(y) = (y-x)^k$, which is a function of $y$ and we treat $x$ as a constant. If we take the derivative of $f$, we get $f'(y)=k(y-x)^{k-1}$.
Repeat for a total of $k$ times, and we get $f^{(k)}(y)=k! (y-x)^0=k!$.
Now we substitute $\xi$ for $y$, but there is no $y$ anymore, so nothing happens. We actually have $(y-x)_+^k$ in the formula, which we evaluate for $y-x>0$, so our result is for the case that $\xi -x>0$.
Finally we divide $f^{(k)}(\xi)=k!$ by $k!$ to arrive at $1$. Last edited: