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It is well known (see here for example) that a function over $\mathbb{R}$ is representable by a power series iff its analytic continuation to $\mathbb{C}$ is holomorphic on some open subset of $\mathbb{C}$ in the standard topology.

Does our ability to create new representations for a function over $\mathbb{R}$ become more robust if we allow ourselves to use Hahn series instead of only power/Laurent/Puiseux series?

In the case of Laurent series (a subset of Hahn series) our expressive power does indeed increase, as the simple poles of a complex function amount to negative terms in its Laurent series expansion -- accordingly, complex functions with simple poles can't be expressed as power series, but can be expressed as Laurent series. More concisely, $\mathbb{C}((X^\mathbb{Z}))$ has more representations for functions over $\mathbb{C}$ than $\mathbb{C}[[X]]$ does.

As an initial sub-question, is it known whether we can express functions with even more idiosyncratic behavior by using the full Puiseux series field $\mathbb{C}((X^{\mathbb{Q}}))$ as a set of possible representations? (Emil gives an obvious positive answer to this below) More generally does the Hahn series field $\mathbb{C}((X^{N_0}))$ admit more representations, where $N_0$ denotes the Surreal numbers?

I somewhat doubt that flat functions with essential singularities like $e^{-\frac{1}{x^2}}$ will be representable in this form, but it seems intuitive that we should be able to express more functions using the above Puiseux/Hahn series field than we can using the Laurent series field $\mathbb{C}((X^\mathbb{Z}))$.

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  • $\begingroup$ Yes, e.g., X^{1/2}. $\endgroup$ – Emil Jeřábek Dec 15 '17 at 7:40
  • $\begingroup$ @EmilJeřábek This is a positive answer to the sub question, but that representation is in the Puiseux series field. Is the Hahn extension essentially more powerful in a nontrivial way? $\endgroup$ – Alec Rhea Dec 15 '17 at 7:42
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    $\begingroup$ @EmilJeřábek Also, unless I'm mistaken we can even represent $\sqrt{x}$ in $\mathbb{R}[[X]]$ as $\sum_{n<\omega}\frac{\prod_{m<n}\frac{1-2m}{2}}{n!}(x-1)^n$ plus some combinatorial business to eliminate the $-1$ in the parenthesis, so I possibly want to refine the question for functions that don't admit representations as infinite members of simpler series fields. $\endgroup$ – Alec Rhea Dec 15 '17 at 10:21
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There are many papers, and even books, written on use of transseries to represent functions. And yes, $e^{-1/x^2}$ is one example.

There is also the theory of resurgence proposed by Écalle ... transseries that represent a function, but not by convergence. But instead by a generaized summability, like Borel summability.

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  • $\begingroup$ This is exactly along the lines of what I was looking for, thank you. $\endgroup$ – Alec Rhea Dec 15 '17 at 11:31
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    $\begingroup$ This lack of self-promotion is remarkable, but I think Transseries for beginners will prove to be a better introduction to transseries that this rather disappointing Wikipedia entry "look at this transserie: $\ee^{\sqrt{\ln \ln x}} + \ln \ln x + \frac{\ee^x}{1+x}$"! From what I've heard, Ecalle's theory is very effective, but difficult to the point that few algebraists have dared to look into it. Do you know good sources that deal (albeit partially) with the problem of summation in a manner accessible to the layman mathematician? $\endgroup$ – nombre Dec 15 '17 at 12:17
  • $\begingroup$ @nombre This is a very rough and intuitive first impression, but it seems that these generalized resummation methods amount to 'crushing' the coefficients as we sum them using the multiplicative inverse of some appropriately fast-growing function, indexed over the positions in the ordered group. Taylor series throwing in a $n!$ underneath each term is a simple example of this, and (from the wikipedia for Borel summation) it seems like this uses faster functions than $n!$ to crush the coefficients down sufficiently and make them converge. $\endgroup$ – Alec Rhea Dec 15 '17 at 15:08

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