2

Suppose $f$ is $C^\infty$ on an open set $U$, and $x \in U$. Does $Tf$, the Taylor series centered at $x$, converge in a neighborhood of $x$ to some function(not necessarily the original function $f$)? To be more precise, if $B$ is an open ball centered at $x$ contained in $U$, do we have $Tf$ converges in $B$?

I’m not asking about Taylor series not equal to the original function. It’s a totally different question.

José Carlos Santos
  • 427,504
user655213
  • 53
  • 5
  • 2
    If by "smooth" you mean analytic, then yes; the power series has positive radius of convergence equal to distance of $x$ to the nearest singularity. But if smooth just means infinitely differentiable, then the answer is no. – hardmath Feb 01 '21 at 17:55
  • https://math.stackexchange.com/posts/comments/8273506?noredirect=1 does not answer my question. It just give an example not a proof – user655213 Feb 01 '21 at 18:13
  • That example’s Taylor series converge, but is it true for all smooth function? If yes, I need a proof – user655213 Feb 01 '21 at 18:18
  • I have voted to reopen. Note that I'm not the one who voted to close as duplicate. My Comment initially was asking you to clarify your Question. – hardmath Feb 01 '21 at 18:19
  • Ok, thanks. I will clarify my question – user655213 Feb 01 '21 at 18:23
  • The answer for my second question is no. Consider $\frac{1}{1-x}$, which is real analytic except at $1$. Modify this function near $x=1$ to be a smooth function on $\mathbb{R}$. Then the Taylor series of new function centered at $0$ still only converges on $|x| <1$ rather than converges on $\mathbb{R}$. – user655213 Feb 01 '21 at 18:45

1 Answers1

5

Yes, such functions exist. Take a power series centered at $0$ such that its radius of convergence is $0$, such as, say,$$\sum_{n=0}^\infty n^nx^n.\tag1$$Then, according to Borel's lemma, there is a smooth function $f\colon\Bbb R\longrightarrow\Bbb R$ such that$$(\forall n\in\Bbb N):f^{(n)}(0)=n!n^n.$$So, the Taylor series of $f$ centered at $0$ is the power series $(1)$.

José Carlos Santos
  • 427,504