{c}
{title2=$b_{k,n}(x)=\binom nkx^k(1-x)^{n-k}$}

These $n+1$ nonnegative polynomials form a basis of polynomials of degree at most $n$, and sum to one on $[0,1]$. For $0\le j\le n$,
$$
x^j=\sum_{k=0}^n\frac{(k)_j}{(n)_j}b_{k,n}(x),
$$
where ratios for $j=0$ equal one and terms with $k<j$ vanish. The identity follows from $\binom nk(k)_j/(n)_j=\binom{n-j}{k-j}$ and the <binomial theorem>. It gives explicit basis coefficients using the <falling factorial>.

Bernstein basis

Consider $B_n$ on the <finite-dimensional vector space> $E=\mathbb R[x]_{\le d}$ and use the monomial basis. For $j\ge1$, a <binomial distribution> and the <falling factorial> moment formula give
$$
B_n(x^j)=\frac1{n^j}\sum_{\ell=0}^j S(j,\ell)(n)_\ell x^\ell,
$$
where $S(j,\ell)$ 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 $n\ge1$, all entries are positive. Therefore
$$
\boxed{B_n(E)\subseteq E,\qquad B_n:E\to E\text{ is invertible for }n\ge\max(1,d).}
$$
For constant polynomials the conclusion is immediate. Taking positive indices avoids the undefined sample expression $k/n$ at $n=0$.

Solution

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

{title2=$(a)_n=a(a+1)\cdots(a+n-1)$}

= Pochhammer symbol
{c}
{synonym}

For a nonnegative integer $n$, the rising factorial is the product of $n$ consecutive increments starting at $a$, with $(a)_0=1$. Another notation is $a^{\overline n}$; the <falling factorial> uses descending factors. It compactly expresses the coefficient recurrence of a <confluent hypergeometric function of the first kind>.

Rising factorial

{title2=$(t)_j=t(t-1)\cdots(t-j+1)$}

The falling factorial is the product of $j$ successive descending factors, with $(t)_0=1$. Another notation is $t^{\underline j}$; this article uses $(t)_j$ for descending factors, whereas the <rising factorial> uses the same notation for ascending factors. For a nonnegative integer $t$, it counts ordered selections of $j$ distinct objects and vanishes when $j>t$. Powers expand as $t^j=\sum_{\ell=0}^j S(j,\ell)(t)_\ell$, where $S(j,\ell)$ is a <Stirling number of the second kind>.

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 Incoming links: Falling factorial