1
$\begingroup$

A function $f : \Bbb R \rightarrow \Bbb R$ is said to be real analytic at a point $x$ if there exists a neighbourhood of $x$, in which it has a power series expansion.

Suppose $f$ is a real analytic at each $ x \in \Bbb R$,then does there exist a single power series expansion of $f$ ?

We know that this holds in the complex setting, but does there exist a similar result on $\Bbb R$

Does the result hold on compact subsets of $\Bbb R$ ?

$\endgroup$
  • 1
    $\begingroup$ Power series expansion takes different form for different points even for complex analytic functions. That is, the expansion $f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z-z_0)^n$ depends as well on $z_0$. What do you mean by 'a single power series expansion'? $\endgroup$ – Sangchul Lee Jan 20 '17 at 11:10
  • $\begingroup$ If a complex analytic function is analytic everywhere, the there exists a single power series expansion, which means that the radius of convergence of a power series( around any point) is infinity. I am asking if, this holds in the real case. $\endgroup$ – ssk Jan 20 '17 at 12:09
  • $\begingroup$ @ssk real analytic functions may not be analytic over $\mathbb C$ everywhere and thus may not have a power series with infinite radius of convergence. $\endgroup$ – Simply Beautiful Art Jan 20 '17 at 12:33
4
$\begingroup$

Real analytic functions are just special cases of complex analytic functions, so we can use the full power of complex analysis to create a counter-example. A quintessential example is

$$ f(x) = \frac{1}{1+x^2}. $$

This function is real-analytic. However, for any $x_0 \in \Bbb{R}$, the radius of convergence of the power series

$$ \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n \tag{*} $$

is exactly $R = \sqrt{\smash[b]{x_0^2} + 1}$ since the nearest singularities are $\pm i$. This tells that there exists no $x_0$ with which $\text{(*)}$ converges for all $\Bbb{R}$. (In fact, if such $x_0$ exists then all $x_0 \in \Bbb{R}$ should work as well, thus it is enough to check the failure at $x_0 = 0$.)

We can even explicitly compute that

$$ \frac{f^{(n)}(x_0)}{n!} = \frac{(-1)^n}{R^{n+1}} \sin((n+1)\operatorname{arccos}(x_0/R)),$$

from which we also confirm that the radius of convergence of $\text{(*)}$ is $R$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.