For $n\in \mathbb{N}$ let $B_n$ be the linear operator taking a function $f$ on the unit interval $I=[0,1]$ to its $n$-th Bernstein polynomial $B_nf$, $$ B_nf(x):=\sum_{k=0}^n\binom{n}{k} f\Big(\frac{k}{n}\Big)x^k(1-x)^{n-k}\label{1}\tag{1}$$ The polynomial $B_nf(x)$ has a natural probabilistic interpretation, namely, it is the expected value of $f(\xi)$, where $\xi=\frac{1}{n}\sum_{j=1}^n \omega_j$ is the average of $n$ independent random variables with identical Bernoulli distribution of parameter $x$, that is, $\mathbb{P}(\omega_j=1)=x$. In fact, this is the starting point in the beautiful Bernstein's proof of the Weierstrass' density theorem via the WLLN. However, this question is about an alternative definition of the sequence $(B_n)_{n\ge0}$.

Let $D:C^1(I)\to C^0(I)$ be the derivative operator, and for all $n\ge1$, let $D_n:C^0(I)\to C^0(I)$ be the approximate discrete derivative given by the incremental ratio $$D_nf(x):=\frac{f\big( \frac{n-1}{n} x+\frac{1}{n}\big)-f\big( \frac{n-1}{n} x\big)}{\frac{1}{n}}, $$ (which is well-defined for $f\in C^0(I)$ and $x\in I$).

It is easy to check that definition \eqref{1} implies
$$DB_n=B_{n-1}D_n\label{2}\tag{2}$$
together with:

$$B_0f(x)=B_nf(0)=f(0)\label{3}\tag{3}$$

Conversely these two imply formula \eqref{1}, as it follows immediately by induction, at least, if we already have it (quite a common situation of formulas proven by induction). Thus, since \eqref{2} and \eqref{3} characterize $(B_n)_n$, we may even take them as an inductive definition of $(B_n)_n$. Note that replacing $D_n$ with $D$ in \eqref{2} gives the analogous inductive definition for the Taylor polynomials in $0$. (Incidentally, formula \eqref{2} is relevant in the approximation theory, in that it implies that for $f\in C^k(I)$ one has $B_nf\to f$ in $C^k$: this by induction from the case $k=0$, since $D_n$ converges strongly to $D$. Also, it says that if some derivative $f^{(k)}$ is non-negative on $I$, so is $(B_nf)^{(k)}$.)

**Question:** How can we deduce *naturally* formula \eqref{1} (i.e., assuming we don't know it, and we do not have a crystal ball to guess it) from \eqref{2} and \eqref{3}**?**