= Solution
For $n\geq1$, the <Bernstein polynomial> is
$$
B_n(f;x)=\sum_{j=0}^nf(j/n)\binom njx^j(1-x)^{n-j}.
$$
The weights sum to one by the <binomial theorem>. Equivalently, if $K$ has the <binomial distribution> with parameters $n,x$, then $B_n(f;x)=\mathbb E f(K/n)$. This interpretation makes the cancellation of apparent high-degree terms transparent.
For the <falling factorial> $u^{\underline r}=u(u-1)\cdots(u-r+1)$, with $u^{\underline0}=1$, direct cancellation in the binomial weights gives
$$
\begin{aligned}
\mathbb E K^{\underline r}
&=\sum_{j=r}^nj^{\underline r}\binom njx^j(1-x)^{n-j}\\
&=n^{\underline r}x^r\sum_{j=r}^n\binom{n-r}{j-r}x^{j-r}(1-x)^{n-j}
=n^{\underline r}x^r.
\end{aligned}
$$
The <polynomials> $u^{\underline0},\ldots,u^{\underline m}$ form a monic triangular <basis> of the degree-at-most-$m$ <polynomials>. Hence $u^m=u^{\underline m}+\sum_{r<m}c_r u^{\underline r}$, and
$$
B_n(x^m;x)=\frac{n^{\underline m}}{n^m}x^m+\sum_{r<m}c_r\frac{n^{\underline r}}{n^m}x^r.
$$
For $m\leq n$, its <leading coefficient> is $n^{\underline m}/n^m\ne0$. If $p$ has actual degree $m$ and <leading coefficient> $a_m\ne0$, linearity shows that the <leading coefficient> of $B_np$ is $a_mn^{\underline m}/n^m$; lower-degree terms cannot cancel it. Thus the <Bernstein polynomial degree preservation> property gives
$$
\boxed{\deg B_np=m\quad\text{whenever }\deg p=m\leq n.}
$$
This includes the requested $m<n$, and also explains why the property cannot simply be extended to $m>n$.
The same moments give $\mathbb EK=nx$ and $\mathbb EK^2=n(n-1)x^2+nx$. Therefore
$$
\boxed{B_n1=1,\qquad B_nx=x,\qquad B_n(x^2)=x^2+\frac{x(1-x)}n.}
$$
The first two approximation errors vanish identically, while
$$
\boxed{\|B_n(x^2)-x^2\|_\infty=\frac1{4n}\longrightarrow0.}
$$
All three convergences are consequently <uniform convergence> on $[0,1]$.
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