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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 :)

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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.

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  • $\begingroup$ I don't understand your notation, in the line right after "suppose:". What does the C stand for? $\endgroup$ Sep 1, 2019 at 12:44
  • $\begingroup$ Infinite differentiable in that set. $\endgroup$ Sep 1, 2019 at 12:46
  • $\begingroup$ OK, thanks! Where can I find such notation? (Text reference if possible) $\endgroup$ Sep 1, 2019 at 12:48
  • $\begingroup$ en.wikipedia.org/wiki/Smoothness $\endgroup$ Sep 1, 2019 at 12:50
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    $\begingroup$ 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: $C(\Omega) = C^0(\Omega)$. $\endgroup$ Sep 1, 2019 at 21:28

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