# Existence of power series expansion for implicitly defined function

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.

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This is false. Take $F[g] = g^2, y = x$. – Qiaochu Yuan Jun 28 '12 at 0:33
Look for versions of the "implicit function theorem" to see what is possible. – Gerald Edgar Jun 28 '12 at 0:35
(Well, this $F$ is not invertible, so take $F[g] = g^3$.) – Qiaochu Yuan Jun 28 '12 at 0:36
Specifically, take $F[g(x)]=g(x)^3$ and $y(x)=x$, so they obey your regularity and invertibility conditions. The unique solution is $g(x)=x^{1/3}$, all of whose Taylor expansions around whatever point you choose have a finite radius of convergence, due to a branch point at $x=0$. – Emilio Pisanty Jun 28 '12 at 1:09
Ah...I see. That is unfortunate. Any suggestions on how I might find conditions under which g(x) DOES have a power series expansion? I phrased my question rather generally. But, in fact, my question refers to a very specific problem. So, I still have hope that my logic, which is generally incorrect, may still be correct for my particular problem. – psyduck Jun 28 '12 at 1:19

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.

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If $F$ and $y$ are given and analytic, as in the OP's hypotheses, then your argument simply says that $y'(w)\neq0$ implies $g'(w)\neq0$, which is a condition on the unknown $g$ and thus probably acceptable. – Emilio Pisanty Jun 28 '12 at 1:06
Thanks Emilio (and Robert). This is quite helpful. Might either of you have a reference for Robert's statement? Citing "MathOverflow" probably won't be acceptable to whichever journal I eventually submit to. – psyduck Jun 28 '12 at 1:45
It's just the chain rule: $F \circ g = y$ with everything differentiable implies $F'(g(w)) g'(w) = y'(w)$, which is impossible if $g'(w) = 0$ and $y'(w) \ne 0$. – Robert Israel Jun 28 '12 at 4:40
Sorry, I seem to have taken $y$ and $g$ to be "known" and $F$ unknown. Of course it can work in the other direction too: if $F'(g(w)) = 0$ and $y'(w) \ne 0$ then $g$ can't be differentiable at $w$. – Robert Israel Jun 28 '12 at 4:52
For example, with $F(g) = g^3 + g$ and $y(z) = z$, the Maclaurin series of $g(z)$ (for the root with $g(0)=0$) has radius of convergence $2/(3 \sqrt{3})$, corresponding to the fact that $g(2 i/(3 \sqrt{3})) = i/3$ and $F'(i/3) = 0$. – Robert Israel Jun 28 '12 at 5:17

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.

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Consider $F[g]=g^3+g$ and $y(x)=x$. Then, $F$ is an invertible analytic function with positive derivative, so $g(x)$ is analytic. But, the expansion of $g$ as a power series does not converge for all $x$. If it did, then $F$ would also be invertible as a map on the complex plane, but it is not one-to-one on $\mathbb{C}$. – George Lowther Jun 28 '12 at 23:01
actually, I see that Robert Isreal mentioned this example. – George Lowther Jun 28 '12 at 23:03
It may also be worth mentioning that the only invertible entire functions are $F(z) = az+b$. – Robert Israel Jun 29 '12 at 1:39
I think what often happens when I post on this board is that people give me exactly the answer that I need, but it takes me a long time to realize it. In any case, I do learn a lot by posting my research question here. And there is no doubt that my papers are more rigorous for the help the MathOverflow community provides. – psyduck Jun 29 '12 at 1:45