In a first calculus course it is often that one learns about the Taylor polynomials
$$f(x) \approx \sum_{k=0}^{N}\frac{f^{(k)}(x_0)(x-x_0)^k}{k!}$$
Which provide increasingly better approximation to a function which needs to be differentiable at the point as many times as the highest exponent. ($f^{(n)}(x_0)$ must exist).
Now to my question. What happens if we replace the sum by a product? Is there some concept we can use to create a multiplicative refinement as contrary to an additive one?
$$f(x) \approx \prod_{k=0}^N\mathcal F \{f,k,x_0\}(x)$$
Can we derive what this $\mathcal F$ thing (functional?) could be for this to make sense, and what must we demand of $f$ to hold on or around $x_0$?
Expanding a comment to an answer to highlight an interesting application. The Weierstrass Factorization Theorem is a way of representing an entire holomorphic function as an infinite product in terms of its growth rate at infinity and its zeroes.
I don't quite remember the chain of events, but Euler was the first to write down the product formula
$$ \frac{\sin(\pi x)}{\pi x} = \prod_{n = 1}^\infty \bigg( 1 - \frac{x^2}{n^2} \bigg) \, . $$
The Taylor series of the right-hand side expands as $1 - \frac{(\pi x)^2}{3!} + O(x^4)$, whereas the "monomial expansion" (glossing over how to make sense of that rigorously) of the right-hand side is of the form $1 - \sum_{n = 1}^\infty \frac{x^2}{n^2} + O(x^4)$. Equating the $x^2$ terms yields Euler's (first) solution to the Basel problem, $$ \sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $$