Floater–Hormann weights equal the local-polynomial blend weights

Put $P_0(x)={\prod }_{j=0}^n(x-x_j)$ and $m=n-d$. Counting the $m$ factors to the right of a window gives ${\mu }_i(x)=(-1{)}^m{P}_0(x){\lambda }_i(x)\mathrm{.}$ Indeed, after cancellation each right factor $x_k-x$ contributes one minus sign when rewritten as $x-x_k$, and the remaining denominator factors are exactly the window nodes. Thus the cleared products agree as polynomials, including at nodes; summing gives the stated cleared-denominator identity. Away from nodes it also shows that the blend quotient with ${\lambda }_i$ equals the quotient with ${\mu }_i$ and $s$.

Expand each $p_i$ using [F2], and interchange the two finite sums. The coefficient of $f_j\mathrm{/}(x-x_j)$ in ${\sum }_i{\lambda }_i{p}_i$ is ${\tilde{w}}_j={\sum }_{i=\max (0,j-d)}^{\min (j,n-d)}(-1{)}^i{\prod }_{\begin{array}{c}i\le k\le i+d\\ k\mathrm{\ne }j\end{array}}\frac{1}{x_j-x_k}\mathrm{.}$ Within a window exactly $i+d-j$ factors $x_j-x_k$ are negative, so each summand has sign $(-1{)}^{2i+d-j}=(-1{)}^{j-d}$ and absolute value ${\prod }_{k\mathrm{\ne }j}\mathrm{\mid }{x}_j-x_k{\mathrm{\mid }}^{-1}$. Thus ${\tilde{w}}_j=w_j$ as in [F1].

The same expansion for constant data $f_k=1$ has each window polynomial equal to $1$ by [F2]. Therefore the coefficient of $1\mathrm{/}(x-x_j)$ in ${\sum }_i{\lambda }_i$ is also ${\tilde{w}}_j=w_j$. Dividing the expanded numerator by the expanded denominator proves that the blend equals the barycentric quotient in [F3] away from support nodes.

At a node $x_{\alpha }$, ${\mu }_i(x_{\alpha })$ is zero unless $\max (0,\alpha -d)\le i\le \min (\alpha ,m)$; for these indices all its factors are positive. This index set is nonempty, and every corresponding window contains $x_{\alpha }$. By [F2], $p_i(x_{\alpha })=f_{\alpha }$, so the blend has numerator $f_{\alpha }s(x_{\alpha })$ and $s(x_{\alpha })>0$. Its value is therefore $f_{\alpha }$, agreeing with the support convention in [F3].