Divided differences and polar forms are two elegant tools for the representation of b-splines. I observed the following connection between both tools. But, since I didn't find it in the literature yet I believe that it is flawed. It would be nice if someone could point me to some literature where I can find this observation or show me its flaw.
Let $(x_l)_{l=0,\ldots,n}$ be a knot sequence for a spline of degree $d$ ($x_l\leq x_{l+1}$ , $x_l<x_{l+d+1}$ ). We assume furthermore $x_l<x_{l+1}$ for $l=d,\ldots,n-d-2$ .
In the following we employ the usual notation
\begin{align*} (x)_+ :=\begin{cases} 0 &@ x\leq 0\\ x &@ x > 0 \end{cases} \end{align*}
and the divided difference
\begin{align*} [x_0,\ldots,x_d]_x f(x), \end{align*}
i.e., the highest coefficient of the osculating interpolation polynomial of $f(x)$ with degree $d$ .
The diagonal $M(x,\ldots,x)$ of the multi-variate function
\begin{align*} M_l(y_1,\ldots,y_d) := (x_{l+d+1}-x_l)[x_l,\ldots,x_{l+d+1}]_{\bar x} \prod_{m=1}^d (\bar x - y_m)_+ \end{align*}
is the b-spline
\begin{align*} b_l(x) &= (x_{l+d+1}-x_l)[x_l,\ldots,x_{l+d+1}]_{\bar x}(\bar x-x)_+^d = M_l(x,\ldots,x). \end{align*}
Therefore, the diagonal $s(x):=S(x,\ldots,x)$ of
\begin{align*} S(y_1,\ldots,y_d) &= \sum_{l=0}^{n-d-1} q_l M_l(y_1,\ldots,y_d) \end{align*}
gives a spline with $q_l\in\mathbb{R}$ ($l=0,\ldots,n-d-1$ ) as degrees of freedom.
We define
\begin{align*} (f(x))_{x>y} &:= \begin{cases} 0 &@ x < y\\ f(x) &@ x \geq y \end{cases}. \end{align*}
The values of the polar form
\begin{align*} P_{l,k}(y_1,\ldots,y_d) := (x_{l+d+1}-x_l)[x_l,\ldots,x_{l+d+1}]_{\bar x} \prod_{m=1}^d (\bar x - y_m)_{\bar x > x_k} \end{align*}
are equal to $M_l(y_1,\ldots,y_d)$ for arguments $y_i\in [x_k,x_{k+1})$ ($i=1,\ldots,d$ ).
The diagonal $B_k(x,\ldots,x)$ of the polar form
\begin{align*} B_k(y_1,\ldots,y_d) &= \sum_{l=0}^{n-d-1} q_l P_{l,k}(y_1,\ldots,y_d) \end{align*}
equals the spline $s(x)$ for $x\in [x_k,x_{k+1})$ .
Since polar forms are uniquely defined through their diagonal the $B_k$ should be the same polar forms as described by Lyle Ramshaw in his report "Blossoms Are Polar Forms". (Is that right?)
Therefore, the control values $q_k$ should have the representation
\begin{align*} q_k = B_k(x_{k+1},\ldots,x_{k+d}). \end{align*}