Can any continuous function be represented as an infinite polynomial?
Motivation: the antiderivative
$ \int^\ e^{-x^2}dx\ $
can be expressed as an infinite polynomial(write Taylor series for integrand function and integrate) but this antiderivative has no closed/elementary form expression according to Liouville's theorem but is clearly continuous. So are the rest of the non-elementary functions expressible as infinite polynomials? Fascinating.Any insights on how to proceed????
On
No!
The functions that are given by a convergent power series are really quite rare in the whole scheme of things. They are called analytic functions.
There are a whole class of functions that are called flat functions. These have all of their derivatives zero at a given point and so, as far as Taylor series can tell, are identically zero. The classical example of a flat function is $x \mapsto \operatorname{e}^{-1/x^2}$. Where $0 \mapsto 0$. In this case all of the derivatives are zero at zero (you have to take limits) and so, as far as Taylor series are concerned, this is the zero function.
In addition, some Taylor series only hold in cetain regions. For example, the Taylor series of $(1-x)^{-1}$ is given by $1+x+x^2+x^3+\cdots+x^k+\cdots$. This is fine for all $-1 < x < 1$, but when $|x|>1$ we have serious trouble.
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And there is something more weird such as the trajectory of one-dimensional Brownian motion, which is a continuous function but nowhere differentiable. Since power series are differentiable on interval of covergence(except for the endpoints), these nowhere differentiable continuous functions can not be represented as power series
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Not all continuous functions have a useful Taylor expansion (TE), such as $e^\frac{1}{x}$. (Although, you could say that its TE simply has a radius of convergence equal to zero.)
To answer your other point, the TE could be useful for integration, but not really. The problem is you still end up with an infinite sum, which is not elementary, and difficult for computers to calculate. I'll use your example with $e^{-x^2}$.
The TE for $e^{-x^2}$ is $\sum_{n=0}^{\infty}\frac{1}{n!}(-x^2)^n=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^{2n}$. Integrating, we get $\int\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^{2n}dx=\sum_{n=0}^{\infty}\int\frac{(-1)^n}{n!}x^{2n}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!(2n+1)}x^{2n+1}$.
Great! Now we have a definite way of finding values of erf, the integral of $e^{-x^2}$. But, as you said there is no way to express this as not an infinite series.
Finally, I am pretty sure that as long as you choose a point on the curve where the limit exists, you can write a Maclaurin series (general Taylor series) that represents that function to a certain radius of convergence. The problem is, that series stops being effective outside that radius. So, you could write a Maclaurin series for the Gamma function, but it doesn't tell you a whole lot about the function.
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Very late reply but might be somehow helpful for future understanding. As pointed out in the comments by @Tyler, the answer of your question would be it depends on your settings.
In the theory of functional analysis, the set of polynomials are dense in the vector space of continuous functions with the infinity norm. This follows directly from the Stone Weierstrass theorem. Thus, if you consider any compact interval (assumed it to be path connected) you can conclude there exists such an approximation. However, in my knowledge, the method to compute this might not be unique.
This idea can be extended to general functions taking functions in compact sets. Check wikipedia as pointed by the user in the comments
The following function is not only continuous, but has continuous derivatives of all orders. However, it is not equal to any Taylor series. $$f(x)=\begin{cases} e^{-1/x^2} & x>0\\ 0 & x\le 0\end{cases}$$