Let's say I have a (finite) sequence $(a_k)_{k=0}^{n}$, with $i=j\iff a_i=a_j$ (all $a_i$ are distinct). Let now $p_i(x)$ be the polynomial with $p_i(a_j)=0$ if $i\neq j$, and $p_i(a_i)=1$ (this polynomial has degree $n$). We see that
$$p_i(x)=\prod_{\substack{0\leq k\leq n\\k\neq i}}\frac{x-a_k}{a_i-a_k}$$
Now to approximate a function $f$, we could do
$$P(x)= \sum_{k=0}^nf(a_k)p_k(x)$$
So that $P(a_i)=f(a_i)$ for all $i$. If we would take the limit of $n$ to infinity, we'd get a power series; is this necessarily equal to $f$ (whenever $f$ is, analytic? Entire? I'm not sure what the requirement is here). Does it matter what $(a_n)$ is? Should it, for example, be bounded?
I tried to create such polynomial for $f(x)=a^x$ with $(a_k)_{k=0}^n$ defined as $a_k=\frac{k}{n}$ (so we're making a pretty good approximation on $[0,1]$) and worked out the constant and linear term of the power series that arises when we take the limit, and yep, the constant term is $1$ and the linear term is $\log(a)$.
Also, is this way of creating lower-order polynomial approximations to functions a thing that is being done already? It seems to be a lot better than simply cutting off the Taylor series, especially if you're using lower-order polynomials and if you're only interested in a smaller interval (for example, the second-degree polynomial on $[0,\pi]$ using $a_0=0$, $a_1=\frac{\pi}2$, $a_2=\pi$ is a lot better than the second degree taylor series, that is, $x$, since there is no quadratic term).