Taylor series not converging, other example than $\exp(-1/x^2)$?

The usual example for non-converging Taylor series is $g(x) = \exp(-1/x^2) \; \forall x \neq 0, g(0) = 0$: the Taylor series around $x=0$ is zero, but $g$ isn't zero for any $x \neq 0$.

What's not so nice about this example are the derivatives: \begin{align*}g'(x) &= \frac{2}{x^3}\exp(-1/x^2), \\ g''(x) &= \left(-\frac{6}{x^4}+\frac{4}{x^6}\right)\exp(-1/x^2), \\ \ldots \end{align*}

So obviously we have to calculate the derivative of $g$ in $x = 0$ by using the definition of the derivative. It is not possible to get it by "applying the rules" (e.g. $2\exp(-1/x^2)/x^3$ isn't defined for $x=0$).

The question is: Can there be a function $f\colon \mathbb{R}\rightarrow \mathbb{R}$ which has a Taylor series with radius of convergence $0$, but whose derivatives in the Taylor series can be calculated easily by just using the "usual" rules (derivatives of polynomials, product rule, chain rule, derivatives of $\exp, \sin, \ldots$)?

Solution 1:

Here is an example: $$f(x)=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-t^2-x^2t^4}dt,\qquad x\in\mathbb{R}.$$ The function $f(x)$ as well as all of its derivatives are well-defined at $x=0$. Moreover, one can compute them explicitly: $$f^{(2k)}(0)=(-1)^k\frac{(4k)!}{4^{2k}k!},\qquad f^{(2k+1)}(0)=0.$$ However the radius of convergence of the Taylor series is $0$.