Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 339 / 3 / a

First address the positivity request in part (a), which has no separate Solution heading in the stub. For $x\in [0,1]$, each $x^k(1-x{)}^{n-k}$ and each coefficient $c_k$ is nonnegative. Their sum is therefore nonnegative, including at the endpoints:This is the elementary direction of the positive Bernstein coefficients for a strictly positive polynomial criterion; the converse developed later needs strict positivity.

For completeness, the convergence fact listed as (i) also has a short proof. If $K\sim \mathrm{Bin}(n,x)$, the Bernstein polynomial is $B_n(f)(x)=\mathbb{E}[f(K/n)]$. Its mean argument is $x$ and $\mathrm{Var}(K/n)=x(1-x)/n\le 1/(4n)$. Given $ϵ>0$, uniform continuity of $f$ on the compact interval supplies $\delta >0$ such that $|f(s)-f(t)|<ϵ$ whenever $|s-t|\le \delta$. By Chebyshev's inequality, uniformly in $x$,Letting $n\to \infty$ and then $ϵ\downarrow 0$ proves $\boxed{\Vert B_n(f)-f{\Vert }_{\infty }\to 0}$. The two numbered items in the source are assumptions supplied for later parts, not additional exam questions; their stub sections contain these supporting derivations.

Consider $B_n$ on the finite-dimensional vector space $E=\mathbb{R}[x{]}_{\le d}$ and use the monomial basis. For $j\ge 1$, a binomial distribution and the falling factorial moment formula givewhere $S(j,\ell )$ is the Stirling number of the second kind and $(n{)}_{\ell }=n(n-1)\cdots (n-\ell +1)$. This follows by expanding $K^j={\sum }_{\ell }S(j,\ell )(K{)}_{\ell }$ and using $\mathbb{E}(K{)}_{\ell }=(n{)}_{\ell }{x}^{\ell }$. Also $B_n(1)=1$.

Each monomial maps to a polynomial of no larger degree. The resulting matrix is triangular with diagonal entries $1$ and $(n{)}_j/n^j$, $1\le j\le d$. When $n\ge d$ and $n\ge 1$, all entries are positive. ThereforeFor constant polynomials the conclusion is immediate. Taking positive indices avoids the undefined sample expression $k/n$ at $n=0$.