Finding an infinitely differentiable function on $\mathbb{R}$ with a taylor series that only converges to $f(x)$ for $x\leq 0$

Question

I am having trouble with this question because to my knowledge there is no taylor series that converges for $x\leq 0 $ because the interval of convergence of a power series is (c-R,c+R) for some c. So therefore there can not be a taylor series that converges to $f(x)$ for $x\leq 0$. The problem that I am having with this is that I am unsure if my reasoning is correct and if that is enough to explain why no such function with this taylor series exists. Any suggestion would be greatly appreciated. Cheers

Answer 1

Consider :

$$f:\mathbb{R}\to\mathbb{R},x\mapsto\cases{0\qquad\textrm{if }x\le0\cr\exp\left(-\frac1x\right)\quad\textrm{otherwise}}$$

Its Taylor series (at $0$) is the null series.

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Proof sketch ...

There exists a sequence $(P_n)_{n\ge0}$ of polynomials such that :

$$\forall n\in\mathbb{N},\,\forall x>0,\,f^{(n)}(x)=P_n\left(\frac1x\right)\exp\left(-\frac1x\right)$$

This can be used to show that :

$$\forall n\in\mathbb{N},\,\lim_{x\to0^+}f^{(n)}(x)=0$$

And now, a last induction will show that, for all $n\ge 0$, $f^{(n)}$ has a right-derivative at $x=0$, which is zero.

Finally (since left derivatives at zero are obviously all zero), this proves that $\forall n\in\mathbb{N},\,f^{(n)}(0)=0$ which in turn shows the claim.