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Just to be sure. I want to apply Taylor's Theorem (expand in a Taylor Series $f$) to an infinitely differentiable function $f$ in order to prove something. So as to do so, it would be enough if I state the theorem like this:

Let $f: D \to \Bbb{R}$ be an infinitely differentiable function, such that $x_0 \in D \subseteq \Bbb{R}. \,\,\cdots$

$\cdots$

Since $f$ is infinitely differentiable in $D$, we can apply Taylor's Theorem as follows

$f(x) = f(x_0) + (x-x_0)f'(x_0) + \cdots$

PinkyWay
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Mr. N
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    It doesn't seem quite correct to me. Just because $f$ is $C^{\infty}$ doesn't mean that the Taylor series will converge to $f$. – Célio Augusto Apr 30 '20 at 21:49
  • First, thanks for correcting the question. What did you use instead of "\mathds"? Secondly, about convergence: $f^{(n)}(x_0)$ for natural $n$ are defined. Isn't that enough to ensure convergence? Or if we evaluate at any point that belongs to $D$, would that be enough? – Mr. N Apr 30 '20 at 23:11
  • I used \mathbb. And this is not true for real-valued functions. Take, for example, $f(x)=\exp(-1/x^2)$, $f(0)=0$. All its derivatives vanishes at the origin, so its Taylor series does not converge to $f$. Take a look on the definition of analytic functions. – Célio Augusto May 01 '20 at 01:22
  • Thanks, I'll use it from now on. However, still on the question,, this function is not defined for $x=0$ (despite the limit exist), therefore, according to my statement, it would be impossible to apply Taylor's Theorem. This would be the same if we take $f(x) = x\cdot \ln(x) -x$. Zero is not defined for $f$, however, the $\displaystyle\lim_{x \to 0} f(x)= 0$. Both cases we should define $f$ as a piecewise function. Or should I rewrite the statement as follows: $f^{(k)}: D \to \mathbb{R}, ,, k \in \mathbb{N}$? – Mr. N May 01 '20 at 13:20
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    Of course it's defined on $0$! $f(0)=0$, I explicitely defined it. It doesn't matter if the fuction is defined piecewise or not; after all, every function can be defined piecewise. Take, for example, $g(x)=x/x$, $g(0)=1$. It's obviously the constant function $1$, its Taylor series converge. Would you say there is a problem with defining $g$, or taking its derivatives? Of course not. – Célio Augusto May 01 '20 at 17:14
  • Sorry, now I realized you have defined it. I thought you computed $f$ at $x=0$. On the matter of analytic functions, wouldn't that depend on the definition of Taylor's Theorem? Take a look at the edit, now I corrected the problem, I think... – Mr. N May 01 '20 at 20:42
  • It's still not correct. Taylor's Theorem is valid for finite-order polynomial approximations. You can't extrapole it to an infinite series, unless the function is analytic. – Célio Augusto May 01 '20 at 20:54
  • You can also use \Bbb instead of \mathbb. (: – PinkyWay May 01 '20 at 21:26
  • You might find this useful. – PinkyWay May 01 '20 at 21:40
  • Thanks again, @CélioAugusto for your answer and patience. We can conclude I need to read some Analysis book, could you recommend me? Thanks, Verk, for linking my question to this another one, however, that's the opposite to what I want to know. Anyway, thanks. – Mr. N May 01 '20 at 22:57
  • I wrote an anwser, summarizing my previous comments and giving recerences. If it's useful to you, you coan upvote and accept. You're welcome :) – Célio Augusto May 01 '20 at 23:23

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The proof is not correct. If a function if $C^{\infty}$ at $x_0$, that doesn't mean that its Taylor series converges fo $f$ in any neighborhood of $x_0$. For example, take

$$f(x)=\begin{cases}e^{-1/x^2}, & \text{if } \ x \neq 0; \\ 0,& \text{if } \ x=0\end{cases}.$$

It can be proven that $f^{k}(0)=0$ for every $k \in \Bbb{N}$. Therefore, the Taylor series of $f$ centered at $0$ converges in all real line, but of course, converges to the constan $0$, and not to $f$.

If the Taylor series of a $C^{\infty}$ function $f$ about $x_0$ converges to $f$ in some neighborhood fo $x_0$ the function $f$ is said to be analytic at $x_0$. Thus all analytic funtcions are $C^{\infty}$, but the converse is not true.

About the books: Rudin's Principles of Mathematical Analysis is the standard reference. There is also Royden's Real Analysis and Tao's Analysis I. If you know a little bit of Portuguese (or even Spanish would do) I strongly recommend Elon Lages' Curso de Análise, vol. I.

Célio Augusto
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  • Thanks again! I'll get these books. – Mr. N May 02 '20 at 13:46
  • Hello, again. I also found this on wikipedia: https://en.wikipedia.org/wiki/Taylor%27s_theorem. Surely, it does not substitute the books, but it may be useful for now. – Mr. N May 05 '20 at 14:28