Marsden's Identity states that for every $\tau$ in $\mathbb{R }$:

$$(\cdot -\tau)^{k-1}=\sum_j\Psi_{j,k}(\tau)B_{j,k,t} \, ,$$

with $\Psi_{j,k}=(t_j-\tau)\times...\times(t_{j+k-1}-\tau)$.

Following de Boor's notation we have that $B_{j,k,t}$ stands for the $j-th$ B-spline of order $k$ defined over the knot vector $t$, i.e., $(t_{j+k}-t_j)\cdot [t_j,...,t_{j+k}](\cdot - x)_+^{k-1}$.

Also, define the space spanned by the B-splines as:

$\$_{k,t}:=\{\sum\alpha_jB_{j,k,t} : \alpha_j\in\mathbb{R}\}$

Technically, using Marsden's Identity I should be able to show that $\mathbb{P}_k$, the space of all polynomials of degree $<k$, is contained in $\$_{k,t}$, by putting $\alpha_j=\Psi_{j,k}$. But when I do that, I don't really see how this expression describes all possible polynomials of degree $k-1$.

Doesn't $(\cdot -\tau)^{k-1}$ represents all polynomials of degree $k-1$ where $\tau$ is a root with multiplicity $k-1$?

Am I missing something here?

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  • What are $B_{j,k,t}$? Same as the $B_{i,n,t}$ in en.wikipedia.org/wiki/B-spline#Cardinal_B-spline ? – darij grinberg Dec 7 at 16:17
  • I am using de Boor notation, $B_{j,k,t}$ stands for the $j-th$ B-spline of order $k$ defined over the knot vector $t$. The cardinal spline in the link is actually formed using a different normalization, in this case the definition is :$(t_{j+k}-t_j)\cdot [t_j,...,t_{j+k}](\cdot - x)_+^{k-1}$ – Ramiro Scorolli Dec 7 at 16:22

I found the solution after some research, hence I'll post it here in case anyone have curiosity:

Marsden's Identity states that for all $\tau$ in $\mathbb{R}$ it holds that: $$(\cdot -\tau)^{k-1}=\sum_j\Psi_{j,k}(\tau)B_{j,k} \, ,$$

It's straightforward to show that $((\cdot-\tau_j)^{k-1}:j=1,...,k)$ , $\tau_1<...<\tau_k$, is a basis for the space $\Pi_{<k}$ (the space of polynomials of degree smaller than $k$).

Hence any polynomials of degree $<k$ can be written as a linear combination of the elements in the basis, i.e.:

$$\beta_1(\cdot-\tau_1)^{k-1}+...+\beta_k(\cdot-\tau_k)^{k-1},$$

By Marsden's Identity we have that the latter equals: $$\sum_j\beta_1\Psi_{j,k}(\tau_1)B_{j,k} +...+\sum_j\beta_k\Psi_{j,k}(\tau_k)B_{j,k}$$

Reordering it yelds to: $$\sum_j(\beta_1\Psi_{j,k}(\tau_1)+...+\beta_k\Psi_{j,k}(\tau_k))B_{j,k}$$

Letting $\alpha_j=(\beta_1\Psi_{j,k}(\tau_1)+...+\beta_k\Psi_{j,k}(\tau_k))$ we have shown that any polynomial in $\Pi_{<k}$ is contained in $\$_{j,k}$.

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Ramiro Scorolli is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

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