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Set $r=k-1$, with $k\ge 2$ for the final linear-reproduction assertion, and use the normalized knot polynomial ${\psi }_i={\omega }_i/r!$. Differentiate the Marsden identity $r-j$ times with respect to the auxiliary variable $x$, for $0\le j\le r$. This givesThe Taylor polynomial expansion of an arbitrary polynomial $p$ of degree at most $r$ about $x$ is exact:Substitute the differentiated identity into each term and collect the B-spline coefficients. The result isThis is the normalized Marsden dual functional. There is no extra factorial outside the sum, because it is already included in ${\psi }_i$.

To prove independence of $x$ directly, differentiate the finite sum. The contribution from differentiating ${\psi }_i$ in term $j$ cancels the contribution from differentiating $p$ in term $j-1$. Explicitly,The two uncancelled endpoint terms vanish because $p^{(r+1)}={\psi }_i^{(r+1)}=0$. Thus the coefficient functional is constant in the auxiliary variable, as required.

The leading two terms of the knot polynomial giveFor a linear polynomial $p(t)=A+Bt$, only $j=0,1$ survive, whenceThe coefficient points $t_i^{\ast }$ are the Greville abscissae. Inserting these coefficients proves exact linear reproduction:All identities hold on the basic knot interval $[t_k,t_{n+1}]$, including the endpoints with the standard continuous endpoint convention. Repeated knots cause no difficulty in these polynomial differentiations.

Put $r=k-1$. The linear-reproduction request requires $k\ge 2$, which we use for that final step. Since ${\psi }_i={\omega }_i/r!$, the useful normalized Marsden dual functional isThere is no further factor $1/r!$ outside this sum: it has already been included in ${\psi }_i$.

To derive the formula directly from Marsden's identity, Taylor's theorem for the polynomial $p$ around the arbitrary point $x$ givesOn the other hand,Differentiate Marsden's identity $r-j$ times in $x$, multiply by $(-1{)}^j{p}^{(j)}(x)/r!$, and sum over $j$. The left side becomes $p(t)$, while the right side becomesOnly finite sums and derivatives of polynomials are involved.

Independence of the auxiliary point does not require an assumption about uniqueness of an expansion. Differentiate the formula for ${\lambda }_i$ itself:The first summand at $j=r$ and the second at $j=0$ vanish because both polynomials have degree at most $r$. All remaining terms cancel after shifting the index by one. Therefore ${\lambda }_i(p)$ is a constant in $x$, and it is visibly a linear functional of $p$.

For $p(t)=a+bt$, only $j=0,1$ remain. The leading two coefficients of the monic knot polynomial giveIt follows thatSubstituting into the expansion proveson the basic knot interval. Thus the Greville abscissae are exactly the sampling coefficients that reproduce linear polynomials; the constant case also gives the subpartition of unity for B-splines with equality on this interval.