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 $\Bbb R$ ?

  • 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'? $\endgroup$ Jan 20, 2017 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, 2017 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. $\endgroup$ Jan 20, 2017 at 12:33

1 Answer 1


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 $R$.


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