Are hyperreal numbers isomorphic to formal power series? I wonder whether hyperreal numbers isomorphic with formal Laurent series?
It seems that any hyperreal number can be represented in the form of Laurent series over $\omega$. For instance,
$e^{\omega}=\frac{\omega^0}{0!}+\frac {\omega^1}{1!}+\frac{\omega^2}{2!}+...+\frac{\omega^n}{n!}+...$
If so, it follows that any analytic function corresponds to a hyperreal number.
$f(x)=x$ corresponds to $\omega$
$f(x)=x^2$ corresponds to $\omega^2$
etc.
Thus operators on analytic functions are isomorphic to functions of hyperreals. Am I correct?
 A: One problem is that the set of formal Laurent series is not a real closed field (an ordered field where every positive element has a square root, and every polynomial of odd degree has a root). That particular problem would be fixed if one considered instead the real closure of the field of Laurent series, which is the field of Puiseux series (Laurent series involving powers of $x^{1/n}$ where $n$ is allowed to vary). So the question would become: is the field of hypperreals isomorphic to the field of Puiseux series? 
The answer is "certainly not"; the hyperreals are much "bigger", or it would be better to say "much more saturated" than Puiseux series. As you probably know, the hyperreals do not form a complete ordered field (only the standard real numbers do that), but they do have properties that approximate completeness. Namely, there is the $\eta_1$ property: given countable sets $L$ and $U$ such that every element of $L$ is less than every element of $U$, there exists an element that is an upper bound of all elements of $L$ and a lower bound of all elements of $U$. So for example: in the field of Puiseux series (and taking $x$ to be an infinitesimal element), there is a pretty big gap between all the rational constants (forming an $L$) and the $x^{-1/n}$ (forming a $U$). The hyperreals fill in such gaps. 
(Added on request.) The hyperreals thus form an ordered field extension of the field of Puiseux series, and admits elements such as $\sin \varepsilon$ where $\varepsilon$ is an infinitesimal. 
A: I would like to point out that it is not true that every every
hyperreal can be represented by a Laurent series in the way you
describe.
(Let me assume that by the term "hyperreals", you mean a
nonstandard elementary extension of the reals, with the transfer
property. It is not really correct to speak of "the" hyperreals,
since it is possible to have non-isomorphic fields with these
properties.)
In particular, what I claim is that it is never the case for a nonzero
hyperreal number $x$ that the hyperreal analogue of $e^x$ obeys
the equation $$e^x=
\frac{x^0}{0!}+\frac {x^1}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots,$$
if what is meant is that $n$ ranges only over the standard natural
numbers, which is how I take your example. In particular, the sequence of finite partial sums of
this series does not converge in the hyperreals, unless $x$ is
zero. So the series is simply not meaningful in this way in the
hyperreals.
Rather, the right way to handle series in the hyperreals is that
they should have nonstandard terms, including also the terms for
nonstandard natural numbers $n$. In particular, the transfer
principle implies that the meaning of $e^x$ in the hyperreals is
given by the nonstandard series representation
$$e^x=\sum_{n=0}^\infty
\frac{x^n}{n!},$$ where $n$ here ranges over all the nonstandard
natural numbers, not just the standard finite natural numbers.
The difference between the two series is precisely the nonstandard
part of the series, and this difference cannot be visible to the
hyperreal field, since from it you could define the standard cut,
which would violate the transfer principle.
