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If we look at the definition of "function", we can see it as a "relation" between two sets, or "mapping" of a given element to another element, my question is: why we can represent this relation as infinite summation of a simple polynomial? In particular, why "infinite"? I see it like a strange thing.

Post Nerdism
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    You can't represent an arbitrary function as a power series. Functions that can be represented as a power series are rather special - they're called "analytic". In particular, they must be differentiable to all orders. – Jair Taylor Nov 23 '19 at 20:06
  • @JairTaylor I know that. And I'm asking for this also, what is the restriction? I read about analytic function definition, but the definition says "it can be represented by power series...", which says nothing for me at least. – Post Nerdism Nov 23 '19 at 20:09
  • Not sure if it's too helpful, but see wikipedia for characterizations of analyticity. For complex-valued functions it is rather simpler: they are analytic if and only if they are complex-differentiable on a domain. – Jair Taylor Nov 23 '19 at 20:15
  • My intuition is that polynomials are not simple, but surprisingly complicated. For example, you can write down a polynomial going through any finite number of points (Lagrange interpolation), and you can approximate any continuous function on a bounded interval using polynomials (Stone-Weierstrass). You have more and more parameters to set as the degree of a polynomial increasing, so the set of polynomials becomes more and more expressive. – Jair Taylor Nov 23 '19 at 20:20

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You cannot, in general, write a function as a power series: e.g. $\chi_{\mathbb{Q}}(x)=\begin{cases} 0\ x\in \mathbb{R}-\mathbb{Q}\\1\ x\in \mathbb{Q}\end{cases}$ (actually, this is true for "most" of the functions, where most can be precised in a quite technical manner). You can, however, do it for a quite useful set of functions (namely, the analytic functions. A necessary condition for a function to be analytic js for it to be infinitely differentiable in $\mathbb{R}$, while in $\mathbb{C}$ being differentiable is a necessary and sufficient condition), which are defined as the functions that can be written as a power series.

So, not all of the functions can be represented as a power series. Between the analytic functions, however, why must we use an infinite sum? Well, that is simply because finite sums of polynomials are not enough: for a start, they cannot have an infinite set of zeros which is not the whole domain, while many interesting functions do ($\sin(x),\cos(x)...$). And for this reason we include an infinite numbers of addendums.

Since we have covered why an infinite sum, I am tempted to go on and analyze why a sum of polynomials.

A first answer is that these kind of sum comes naturally as a limit process from the Taylor expansion. But this is not the only reason. As stated in the comments by JamesTaylor, the set of polynomials satisfy a number of properties (they form a sub-algebra, with a constant, that separates points) that makes them fit to approximate the continuous function on an interval (thanks to the Stone-Weierstrass theorem).

As a final note, we do not use only infinite sum of polynomials to describe functions. We use a lot of functions: as an example, Fourier series are infinite sums of sines and cosines, and with those we are able to describe a larger class of functions than the class of analytics functions. In general, in an infinite dimensional Hilbert space we are able to write its elements as the infinite sum of some special elements, called a Hilbert basis of the space.