Existence of power series expansion for implicitly defined function

Question

I am trying to solve the following implicit equation for $g(x)$.

$F[ g(x) ] = y(x)$

For simplicity assume that $F$, $g$ and $y$ all map $\mathbb{R} \to \mathbb{R}$. It is known that, for every $x$ there exists a unique (I added "unique") number $g(x)$ such that $F[g(x)] = y(x)$. So the equation is well-posed.

The function $y(x)$ has a power series expansion in $x$ valid for all real $x$. The function $F[ g ]$ has a power series expansion in $g$ valid for all $g$. I would like to argue that $g(x)$ must therefore have a power series expansion in $x$ valid for all real $x$.

This seems like sound logic. But, before I put this in a paper, I would like to be sure that this is correct. As always, thanks in advance for any advice you can offer.

EDIT: I should have phrased things slightly differently. Instead of saying "F is invertible" I should have said, for every $x$ there exists a number $g(x)$ such that $F[g(x)] = y(x)$. This is obviously not the same thing as saying $F$ is invertible. I made this edit above.

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I greatly appreciate all of the comments that people have left. Certainly, my understanding of what can go wrong with my initial assumption has been greatly clarified. In an effort to see if my solution to a specific problem is valid, I will write it below. As always, many thanks for your comments. And, thank you for being patient who does not have a rigorous mathematical background.

I would like to find $\sigma^\epsilon$ that solves $v(\sigma^\epsilon) = u^\epsilon$ where $$u^\epsilon = \int d \lambda e^{\phi_0 (\lambda) + \epsilon \phi_1 (\lambda)} h(\lambda,k) \qquad h(\lambda,k) = \frac{-e^{k-i k \lambda }}{\sqrt{2 \pi } \left(i \lambda +\lambda ^2\right)}$$ and $$v(\sigma^\epsilon) = \int d\lambda e^{ \phi(\lambda;\sigma^\epsilon)} h(\lambda,k) , \qquad \phi(\lambda;\sigma^\epsilon) = \frac{1}{2}(\sigma^\epsilon)^2(-\lambda^2 - i \lambda)$$

All functions of $\lambda$ are analytic on the set {Im$(\lambda)<1$}. Integration is over a line parallel to the real axis in the stip of analyticity.

The way that I "solved" this was by expanding both sides in powers of $\epsilon$ ASSUMING that $\sigma^\epsilon$ has a power series expansion $$\sigma^\epsilon = \sigma_0 + \epsilon \sigma_1 + \cdots$$. I then matched terms of like powers of $\epsilon$ to find the cofficients {$\sigma_n$}.

My "solution" is indistinguishable for values of $k$ near $0$ (for basically any size $\epsilon$. But, as $k$ moves away from $x$ the convergence is quite bad. I don't know if this is a problem with my series solution or my numerical intergration scheme. So, if I somehow knew that expanding $\sigma^\epsilon$ in powers of $\epsilon$ were valid, then I would know that it the numerical integration scheme that is causing problems, and not my series solution.

Answer 1

If $g$ and $y$ are entire functions and there is some complex number $w$ such that $g'(w) = 0$ but $y'(w) \ne 0$ (or both $g'$ and $y'$ have zeros at $w$ but that of $g'$ has higher multiplicity), then $F$ can't be differentiable at $g(w)$. In many cases, the radius of convergence of a Taylor series for an implicit function $F$ is determined by this fact.

Answer 2

I found an acceptable answer to my particular question (not the general one).

1. The inverse of an invertible analytic function whose derivative is nowhere zero is analytic.
2. The composition of two analytic functions is analytic.

$F[ g(x) ] = y(x)$

$F[g]$ satisfies 1 in my case. And $y(x)$ is analytic. Therefore

$g(x) = F^{-1}[ y (x)]$ is analytic.