# Function - Analytic or not? (A General Approach)

I am doing a beginner's course in real analysis, so I am fairly new to it.

I've been told that a smooth function is infinitely differentiable, that is, all its derivatives exist. Moreover, a smooth function is called analytic at a point $$a$$ if the Taylor series of the function converges to the function in some neighbourhood of $$a$$.

My problem is, that these are merely definitions. I am not able to understand how I can show whether or not a given function is analytic at a given point.

$$ln(1+x)$$ for example, at $$x = 0$$. I don't see a method to check the convergence of that infinite series, other than the ratio test - which assures convergence in $$(-1,1)$$. So is the function analytic only in that interval? Is it possible that the Taylor series converges in a particular neighbourhood, but not to the value of the function at that point? Please help me prove or disprove this, and also let me know the ways to determine if a function is analytic or not, in general! Thanks :)

Let $$f:(x_0-\epsilon, x_0+\epsilon)\to \mathbb R$$ and suppose:

• $$f \in C^{\infty}((x_0-\epsilon, x_0+\epsilon))$$

• there exists $$M\in \mathbb R$$ sucht that $$|f^{(n)}(x)|\le M \frac {n!}{\epsilon^n}$$ $$\forall n \in \mathbb N, \forall x\in (x_0-\epsilon, x_0+\epsilon)$$

Then $$f$$ is analityc in $$(x_0-\epsilon, x_0+\epsilon)$$.

Try to demonstrate this proposition.

• I don't understand your notation, in the line right after "suppose:". What does the C stand for? – cogito_ai Sep 1 '19 at 12:44
• Infinite differentiable in that set. – Marco Lecci Sep 1 '19 at 12:46
• OK, thanks! Where can I find such notation? (Text reference if possible) – cogito_ai Sep 1 '19 at 12:48
• en.wikipedia.org/wiki/Smoothness – Marco Lecci Sep 1 '19 at 12:50
• More generally, $C^k(\Omega)$ is the set of $k$-times differentiable functions $f$ on $\Omega$ such that $f^{(k)}$ is continuous everywhere in $\Omega$. By default, $f$ takes it values in $\Bbb R$. If you want to talk about maps into some other space, then you include it in the notation: $C^3((0,1), \Bbb C)$ is the set of all functions from the unit interval $(0,1)$ into the complex numbers $\Bbb C$ which have a continuous third derivative. If there is no exponent, or if the exponent is $0$, it means the set of all continuous functions: $C(\Omega) = C^0(\Omega)$. – Paul Sinclair Sep 1 '19 at 21:28