Redundancy in transseries representation of functions?

"Transseries" are a kind of generalized power series that allow things like fractional exponents and exponentials (with another transseries as the exponent). I know very little about them but I have looked through e.g. Edgar's Transseries for beginners. The transseries considered there are "expansions around infinity", so that the powers of the variable $x$ must decrease to the right; thus for instance the usual power series $\sum_{n=0}^\infty \frac{x^n}{n!}$ for $e^x$ is not a transseries (at $\infty$), but we do I think have a transseries

$$\sum_{n=0}^\infty \frac{1}{n! x^n}$$

that is a sort of power series at $\infty$ for $e^{1/x}$.

On the other hand, unless I am confused, $e^{1/x}$ is also itself a transseries. So it seems that in the world of transseries we have two different ways to represent the same function $f(x)=e^{1/x}$, as "$e^{1/x}$" and as the above series "$\sum_{n=0}^\infty \frac{1}{n! x^n}$".

My question is, assuming this is correct, how should I think about this redundancy? Is it an unavoidable disadvantage of the extra expressivity of transseries over power series? Or is there some positive reason why we want to distinguish "$e^{1/x}$" and "$\sum_{n=0}^\infty \frac{1}{n! x^n}$"?

• @GerryMyerson As I said, a transseries is allowed to contain exponentials with transseries exponent, as well as powers of x. The basic "monomial" appearing in a transseries is of the form $x^n e^T$, where $T$ is another transseries. – Mike Shulman Jan 11 '16 at 4:51
• Oops! $T$ is not just any transseries, it has to be "purely large", which rules out $1/x$. – Mike Shulman Jan 11 '16 at 5:00
• You can know the rules which allow to avoid redundancies by combining "Orders of infinity" (in the sense of Hardy) and Neumann-Malcev series. – Duchamp Gérard H. E. Nov 28 '17 at 5:46

1 Answer

How embarassing; I puzzled over this for several days, only to figure out the answer an hour after posting the question.

A transseries is only allowed to contain exponentials $e^T$ when $T$ is a "purely large" transseries: one all of whose terms "go to infinity as $x$ does". That rules out $1/x$ as an allowable exponent. I hope/suppose that this restriction actually suffices to rule out any other apparent redundancies of this sort.