Is the reciprocal of an analytic function analytic? Let $f(x)=\sum_{n=0}^{\infty}a_nx^n$ be a function on the real line with a positive(or infinite) radius of convergence about $x=0$, and $a_0\neq0$. Then is it true that the function $1/f(x)$ can also be Taylor expanded about the origin? The question is essentially whether the formal inverse series has a positive or infinite radius of convergence.
 A: ashpool: Yes, $1/f(x)$ is analytic, and in fact it is analytic in the largest disk centered at the origin where $f(x)$ is defined and non-zero. This disk exists, since $f(0)=a_0\ne0$.
One way to check this is to use that having a convergent power series expansion is equivalent to being complex differentiable. And one can check directly that $1/f$ is differentiable if $f$ is.

(Though I prefer the complex analytic approach, one can argue directly. The idea is simple, though I don't have time right now to give full details. Note that $f(x)=a_0(1+\epsilon(x))$ where $\epsilon(x)$ is analytic (so, continuous) and small when $x$ is small. Then $1/f(x)$ can be expanded as a geometric series in terms of $\epsilon(x)$, and you can expand $\epsilon$ and its powers term by term. The fact that it is small can be quantified to see that the coefficients resulting from this expansion give a series that converges in a small interval. In a sense, you are just comparing the formal series term by term with a numerical series that you know converges (Fix $0<\delta<1$ and pick $r$ so that $|\epsilon(x)|<\delta$ for all $x$ with $|x|<r$. Then work inside the interval of radius $r$). It is a bit messy but straightforward.)
A: The answer is yes: the given function has a Taylor expansion about the origin with positive radius of convergence (note that most consider $\infty$ to be a positive radius of convergence).
Probably the easiest way to see this is to appeal to complex analysis: the function $f$ equally well defines a complex analytic function on a neighborhood of $0$.  Recall that a function is complex analytic iff it is differentiable.  Moreover, the usual calculus proof adapts to this context to show that if $f$ is complex differentiable and $f(0) \neq 0$, then $\frac{1}{f}$ is differentiable at $0$, thus complex analytic.
It is also possible to prove this result by direct power series manipulations, but I confess I don't remember the argument at the moment: I believe it's a little tricky.  But you could consult a book on real analytic function theory, e.g. this one by Krantz and Parks.  (Added: the argument is given on pages 7 and 8 of this text.  I am about to sign off for a little while, but upon request I could paraphrase it here a little later on.)
A: Yes. I think the following is a way to do this without complex analysis (although I think it would be the easiest way).
For simplicity, let's assume $a_0 = 1$ (otherwise you just divide by $a_0$). And take a radius $\rho > 0$ such that $\sum_{n=1}^{\infty} |a_n| |x|^n < 1$ for all $|x| < \rho$ (there exists one since $x \mapsto \sum_{n=1}^{\infty} |a_n| |x|^n$ is continuous, and zero at the origin). Let's denote $b_n$ for the coefficient of the formal inverse and show by recurrence that $|b_n| \le \rho^{-n}$ (which will prove it converges within  the radius $\rho$).
For $n = 0$ it is obvious since $b_0 = 1$. Then for $n \ge 1$, you have
$$b_n = - \sum_{k=1}^{n} a_k b_{n-k}$$
And
$$|b_n| \le \sum_{k=1}^{n} |a_k| \rho^{k-n}$$
So
$$|b_n| \rho^n \le \sum_{k=1}^{n} |a_k| \rho^{k} \le \sum_{k=1}^{\infty} |a_k| \rho^{k} < 1$$
Which concludes.
