# Power series of a function

I am wondering if there are any functions $f(x)$ such that it cannot be expressed as a power series of $x$? This might turn out to be a silly question, but I can't think of one at the moment.

Thanks!

-

Yes, there are many. One dramatic example is $$f(x) = \begin{cases} 0 & x\in\mathbb Q \\ 1 & x\in\mathbb R\setminus\mathbb Q\end{cases}$$

but there are also more subtle ones, such as $$g(x) = \begin{cases} 0 & x=0 \\ e^{-1/x^2} & x\ne 0 \end{cases}$$ which is even differentiable arbitrarily many times everywhere but cannot be expressed as a power series in any interval that includes $0$.

You can get a $\mathcal C^\infty(\mathbb R)$ function that has a power series nowhere by letting $(q_n)_{n\in\mathbb N}$ enumerate the rational numbers and forming the infinite sum $$h(x) = \sum_{n=1}^{\infty} \frac{g(x-q_n)}{2^n}$$

-
And while you're at it, one might add that in a precise sense almost all functions, and even almost all $\mathcal C^\infty(\mathbb R)$ functions, are nowhere given by a power series. – Marc van Leeuwen Oct 5 '12 at 7:21

For example, $f(x)=e^{-\frac{1}{x^{2}}}$ for $x\neq0$ and $f(0)=0$, you can see that $f(x)$ is continuous and differentiable of infinite types, but $f(x)$ cannot be expressed as a power series of $x$ at $0$.

-

If a function $f(x)$ can be expressed as a power series of $x$,it is called analytic function.You can read more about them by searching.

There are many examples of non-analytic functions,one you may consider is given by http://en.wikipedia.org/wiki/Non-analytic_smooth_function.

Edit:Sorry,I did not see the answers posted during the time I was typing..

-