= Solution
Write $j^{\underline m}=j(j-1)\cdots(j-m+1)$ for the <falling factorial>. At the sampling points, $f_{nm}(j/n)=n^{-m}j^{\underline m}$, which vanishes for $j<m$. For $j\ge m$, cancellation of the factorials gives
$$
j^{\underline m}\binom nj=\frac{n!}{(n-m)!}\binom{n-m}{j-m}=n^{\underline m}\binom{n-m}{j-m}.
$$
Insert this into the <Bernstein polynomial> and set $\ell=j-m$:
$$
\begin{aligned}
B_n(f_{nm},x)&=\frac{n^{\underline m}}{n^m}x^m\sum_{\ell=0}^{n-m}\binom{n-m}{\ell}x^\ell(1-x)^{n-m-\ell}\\
&=\frac{n^{\underline m}}{n^m}x^m=f_{nm}(1)x^m.
\end{aligned}
$$
The <binomial theorem> evaluates the remaining sum. For $m=0$, both sides equal one. This proves the \b[<Bernstein falling-factorial identity> $B_n(f_{nm},x)=f_{nm}(1)x^m$].
Back to article page
