# Marsden's Identity and B-splines

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 $$, is contained in $$\_{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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Ramiro Scorolli is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• 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 $$\Pi_{ (the space of polynomials of degree smaller than $$k$$).

Hence any polynomials of degree $$ 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 $$\Pi_{ is contained in $$\_{j,k}$$.

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