Can any continuous function be represented as an infinite polynomial?

Solution 1:

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.

Solution 2:

The following function is not only continuous, but has continuous derivatives of all orders. However, it is not equal to any Taylor series. $$f(x)=\begin{cases} e^{-1/x^2} & x>0\\ 0 & x\le 0\end{cases}$$

Solution 3:

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