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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?

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    $\begingroup$ This is also asked at math.stackexchange.com/questions/1117477/… $\endgroup$
    – Ross Millikan
    Jan 24, 2015 at 14:55
  • $\begingroup$ The product of two Laurent series need not be a Laurent series. If you want a non-archimedean field that has well-defined multiplication and division, a good candidate is the Levi-Civita field, but the LC field doesn't have $e^\omega$ where $\omega$ is infinite. It would also be odd if a structure as "natural" as the Laurent series were isomorphic to the hyperreals, since the hyperreals are non-unique in ZFC. $\endgroup$
    – user21349
    Jan 24, 2015 at 20:04

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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.

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    $\begingroup$ (1) No, you can form a hypperreal $\sin \omega$. (2) Yes, one can regard the field of formal power series as an ordered subfield of the hyperreals. $\endgroup$
    – Todd Trimble ♦
    Jan 24, 2015 at 12:48
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    $\begingroup$ Oh, I see (and sorry I was confusing). Right, the way that Puiseux series are usually set up, the $x$ will be infinitesimal [I edited a mistake here in my answer; I often get confused which way it should go], and there, there is no such series for that would express an infinite element such as $e^y$ where $y = x^{-1}$. $\endgroup$
    – Todd Trimble ♦
    Jan 24, 2015 at 13:19
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    $\begingroup$ Todd, one may refer to $\sin(\omega)$ or $e^\omega$ for any hyperreal $\omega$, including infinite elements, simply by appealing to the transfer principle---every function on the reals has a nonstandard analogue, which satisfies all the same first-order expressible properties in the hyperreals that the original function did in the standard reals. $\endgroup$
    – Joel David Hamkins
    Jan 24, 2015 at 14:47
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    $\begingroup$ Anixx, based on your most recent comment I think you might be confusing the hyperreals with the surreals, which are two very different things. In particular, there's no "special" element of 'the' hyperreals which we deem "$\omega$." $\endgroup$
    – Noah Schweber
    Jan 24, 2015 at 16:34
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    $\begingroup$ Again, "$No$" is from the surreals, not the hyperreals. $\endgroup$
    – Noah Schweber
    Jan 25, 2015 at 2:48
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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.

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  • $\begingroup$ The number 3 is a nonzero hyperreal number. Are you saying in the hyperreals that $e^3$ is not $1 + 3 + 3^2/2! + ... + 3^n/n! + ...$ where $n$ runs over the nonnegative integers? $\endgroup$
    – KConrad
    Apr 9, 2017 at 21:14
  • $\begingroup$ In the hyper-real field, that sum does not converge if you use only standard $n$. You have to include the terms with nonstandard $n$, and in that case, you'll get $e^3$. Perhaps it would help you to think about the fact that in the hyperreals, the number $e^3$ is surrounded by a tiny interval of hyperreals infinitesimally close to $e^3$, and the standard part of the series is only getting you up to that whole interval and not to the center of it, which is what the rest of the series does. $\endgroup$
    – Joel David Hamkins
    Apr 9, 2017 at 21:19
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    $\begingroup$ At first your response really surprised me, but now I think I see the problem: convergence in the hyperreals uses its order topology and an infinitesimal neighborhood of the real number $e^3 = 20.085...$ is isolated from all those classical partial sums I wrote down above, and thus they don't converge to $e^3$. Is that all, or is there more to it than that? For the same reason, I guess, no classical power series centered at $0$ converges in the hyperreals except at 0, e.g., if $0 < |x| < 1$ in the reals then the usual geom. series $1 + x + x^2 + x^3 + ...$ does not converge to $1/(1-x)$. $\endgroup$
    – KConrad
    Apr 9, 2017 at 21:28
  • $\begingroup$ Yes, you've got it now. $\endgroup$
    – Joel David Hamkins
    Apr 9, 2017 at 21:30

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