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$\DeclareMathOperator\dom{dom}$Sorry to bother the community again with these type of questions about power series, I am ready to delete the question if it is not suitable.

Definition: I say a function $h$ is $(m,n)$-representable by power series iff $0\in \dom(h)$, $h(0)=0$, and $h$ is a restriction (to a smaller domain) of some power series (centered at $0$) that takes as input $m$ real variables and outputs a vector in $\mathbb{R}^n$. In other words, $0\in \dom(h)\subseteq B_r(0)$ (where $r$ is the radius of convergence of the power series), and for every $x\in \dom(h)$ we have that $h(x)$ equals the evaluation of the power series at the point $x$.

Question: Let $f$ be $(a,b)$-representable by power series. Let $g$ be $(b,c)$-representable by power series. It is given that $\dom(g\circ f)$ (i.e. $\{x\in \dom(f)\mid f(x)\in \dom(g)\}$) is path connected. It is also given that $\dom(f),\dom(g)$ are path connected. Must $g\circ f$ be $(a,c)$ representable by power series?

I edited my question to include the hypothesis of path connectivity of domains of $f,g$

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    $\begingroup$ Does the fact its multi variable make this considerably harder to prove than the usual result “composition of analytic functions is analytic”? I would try to look that result up and generalize it: math.stackexchange.com/questions/1345778/… $\endgroup$
    – Sidharth Ghoshal
    Commented Jun 30, 2023 at 1:40
  • $\begingroup$ @SidharthGhoshal The challenge is not about the multivariable part, but rather that I start from stronger hypotheses and want to reach stronger conclusions. Representability by power series is a stronger condition than just analyticity $\endgroup$
    – Amr
    Commented Jun 30, 2023 at 1:42
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    $\begingroup$ Doesn't substitution of power series correspond to composition of representable functions? $\endgroup$
    – LSpice
    Commented Jun 30, 2023 at 2:16
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    $\begingroup$ @M.G. I don't think your question is equivalent to my question. For example, if $f$ is identity, and $g$ is some power series with finite radius of convergence. Then we will have that $g\circ f$ is representalbe by power series , even though the radius of convergence of $g\circ f$ is not as big as the radius of convergence of $f$ $\endgroup$
    – Amr
    Commented Jun 30, 2023 at 2:25
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    $\begingroup$ @M.G. I would state what you stated slightly different . If $N$ is small enough so that $f^{-1}(N)$ is connected, does that necessarily mean that $f^{-1}(N)$ is small enough to fit inside the domain of convergence of the power series of $g\circ f$ ? $\endgroup$
    – Amr
    Commented Jun 30, 2023 at 3:11

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Maybe I misunderstood the question, is this a counterexample?

Let $c>1$, put $f(x) = \sin x$ and $g(x) = \tfrac{x}{c^2+x^2}$. Then $f$ is representable by a power series on $\mathbb{R}$, $g$ is representable by a power series on $(-c,c)$ and $f(\mathbb{R}) \subset (-c,c)$.

However, the meromorphic function $g \circ f$ has a pole at $i \sinh^{-1}(c)$, so the power series of $g \circ f$ is only convergent on $(- \sinh^{-1}(c), \sinh^{-1}(c))$, not on all of $\mathbb{R}$.

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  • $\begingroup$ Yes this is a counterexample. Thank you very much for your time. $\endgroup$
    – Amr
    Commented Jul 6, 2023 at 10:54
  • $\begingroup$ Dear Prof. David, Will the situation changed if I had changed the hypothesis and required my power series(es) to be be power series in a complex variable instead of a power series in a real variable ? $\endgroup$
    – Amr
    Commented Jul 12, 2023 at 17:34

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