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

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 $W\cap D$?

marked as duplicate by Andrés E. Caicedo, Davide Giraudo real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 9 '14 at 18:51

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 $x=0$.

• Specifically, this does not answer the first question, but does provide a counterexample for the second question. That function has a Taylor series at $x=0$ with radius of convergence $\infty$, but it doesn't converge to the function except at $0$. – Jonas Meyer Sep 9 '14 at 4:22
• @JonasMeyer: Leave the OP take his time to read my answer. Thanks for your comment. – Mhenni Benghorbal Sep 9 '14 at 4:24

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.