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?



Marsden's Identity and B-splines
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 $\$\_{j,k}$. 


