Proving that a function is real-analytic

Question

I try to solve the following exercise:

Let $f:\mathbb{R}\to\mathbb{R}$ with $f(x):=\frac{1}{1+x^4}$. Prove that $f(x)$ is real analytic and compute the radius of convergence of it's Taylor series at any arbitrary point aroud $x_0\in\mathbb{R}$.

I know that an analytic function is locally equal to a power series. So I thought that the Taylor series is $f(x):=\frac{1}{1+x^4}=\sum\limits_{n=0}^\infty (-x^4)^n$ and it should converges for $|x|<1$, but I don't know if it is correct or how else I should do this exercise, since we have a real function so I can not use Cauchy-Riemann Equation or any other complex theorem.

Thanks for help

Answer 1

Analytic continuation to the rescue!

Extend your function to $\mathbb C\setminus\{\frac{\pm 1\pm i}{\sqrt2}\}\to \mathbb C$ by defining $f(z) = \frac1{1+z^4}$.

The extended function is complex differentiable everywhere in its domain and therefore equals its Taylor series around the center of any disc that fits in the domain.

For any given center point, the radius of convergence of the Taylor series is the shortest distance to any (non-removable) complex singularity of the extended function. This is still true when the chosen center happens to be real, in which case the Taylor coefficients are real too.