How to show real analyticity without extending to complex plane

Question

Suppose we have some $f \in C^\infty(\mathbb{R},\mathbb{R}).$ For example, $$f(x)=(1+x^2)^{-1}.$$ Using complex analysis, we can easily show $f$ is real analytic. Is there an easy, general method to show this which doesn't use complex analysis?

Answer 1

Sums, products, quotients, and compositions of real analytic functions are real analytic. So in your example, $1+x^2$ is real-analytic, and so is $1/x,$ hence so is $1/(1+x^2).$

Answer 2

The most popular general method is to calculate the general term $f^{(n)}(a)$, and if that's possible, and for every $a$, find an interval $[a-h,a+h]$, such that if $$ M_n=\max_{x\in[a-h,a+h]}\lvert \,f^{(n)}(x)\rvert, $$ then $$ \limsup_{n\to\infty} \left(\frac{k_n}{n!}\right)^{\!1/n}=L<\infty. $$ The $M_n$ can be approximated using Taylor expansion remainder.