Every continuous function $f:ℝ\to ℝ$ can be presented as a sum of a certain power series.

Question

I've encountered a statement that I am unable to prove or disprove:

Every continuous function $f:ℝ\to ℝ$ can be presented as a sum of a certain power series.

Is this statement true? If yes, can somebody prove it?

Answer 1

This is not true in all cases. Power series are differentiable almost everywhere but there exist continuous functions which are differentiable nowhere. Now consider the continuous smooth function $f(x)=\exp(-1/x^2)$ with $f(0)=0$. What power series would you like this to be equal to? You should find that the only power series it could be is: $0 + 0x + 0x^2 +\cdots$ but this function is not zero.

Answer 2

This isn't even true for smooth functions. The classic example: try to expand $$ f(x)=\begin{cases}e^{-1/x^2}&x\ne 0\\ 0&x=0\end{cases} $$ in a power series at $x=0$. Convince yourself that $f^{n}(0)$ exists for any $n$ and is zero (these are the coefficients of a hypothetical power series). But the function certainly isn't zero in any positive radius of the origin.