Does the Taylor series of an infinitely differentiable function converge; and if yes, does it converge to the function?

Question

I have googled it, but I am not satisfied with those.

So my questions are:

1. Let $D$ be an open set in $\mathbb{R}$.

Let $f:D\rightarrow \mathbb{R}$ be a infinitely differentiable function.

Fix $x_0\in D$

Then, does $\sum_{n=0}^\infty \frac{f^n(x_0)}{n!}(x-x_0)^n$ converge on some neightborhood of $x_0$?

Secondly,

1. Let $D$ and $f$ be the domain and function illustrated as above.

Fix $x_0 \in D$

Assume, $\sum_{n=0}^\infty \frac{f^n(x_0)}{n!}(x-x_0)^n$ converge on $W\cap D$ where $W$ is some neightborhood $W$ of $x_0$.

Then, does this series coincide with $f$ on $W\cap D$?

Answer 1

Not every infinitely differentiable function possess a power series. For instance the function $ e^{-\frac{1}{x^2}} $ does not possess a power series centered at the point $x=0$.

Answer 2

Taylor's theorem says that every function $f(x)$ that has $n$ derivatives in some neighborhood of point $x_0$, is equal to the series (for every $x$ in that neighborhood)

$$f(x) = \sum_{n=0}^{n-1} \frac{f^n(x_0)}{n!}(x-x_0)^n + R_n(x)$$

Where $R_n$ is the remainder, $$ R_n(x) = \frac{(x-x_0)^{n}}{n!}f^{(n)}(\xi), $$

Where $\xi$ is some number between $x_0$ and $x$. Note that this remainder term is the just the generalization of the Lagrange mean value theorem, i.e. for n=1, $$ f(x) = f(x_0) + (x-x_0)f'(\xi) $$

Now if you let $n \to \infty$, the function will be equal to the infinite series (in a neighborhood of $x_0$) $$ \sum_{n=0}^{\infty} \frac{f^n(x_0)}{n!}(x-x_0)^n $$

Only if a neighborhood of point $x_0$ exists, such that for all x in that neighborhood the remainder term tends to zero

$$ \lim_{n \to \infty}R_n(x) = 0 $$

If this condition is satisfied the function is analytic at $x_0$.

You can easily prove that the remainder tends to zero if you can bound every derivative of your function, this is true for the sine function for example, for all $n$

$$ |\sin (x)^{(n)}| \leq 1 \Rightarrow \lim_{n \to \infty}\frac{(x-x_0)^{n}}{n!}|\sin (\xi)^{(n)}| \leq \lim_{n \to \infty}\frac{(x-x_0)^{n}}{n!} = 0 $$

Thus $\sin$ is an analytic function.

Answer 3

Borel's theorem (You can read about it here) provides strong counterexamples to your question.