14
$\begingroup$

For a sequence $(x_{\alpha})$ of surreal numbers indexed by the set of all ordinal numbers, we say that $\lim x_{\alpha}=l$ ($l$ is a surreal number) if for each surreal $\epsilon>0$, there exists an ordinal $\beta$ such that $|x_{\alpha}-l|<\epsilon$ for each ordinal $\alpha>\beta$.

Consider the sequence $x_{\alpha}=(1+1/\alpha)^\alpha$, where for each surreal $x,y$, $x^y$ is defined by $\exp(y\log x)$.

Q1: What is the normal form of $x_{\omega}$? Is its real part equals to $e$?

Q2: What is $\lim x_{\alpha}$?

$\endgroup$
5
  • 7
    $\begingroup$ What, $e$ is not surreal enough for you as it is? $\endgroup$
    – Asaf Karagila
    May 30, 2017 at 16:07
  • 7
    $\begingroup$ The set of all ordinal numbers? $\endgroup$
    – Nik Weaver
    May 30, 2017 at 16:34
  • $\begingroup$ I think you mean "standard part" rather than "real part". $\endgroup$ May 30, 2017 at 19:46
  • $\begingroup$ Of course many sequences indexed by the ordinal numbers do not converge at all in the surreals. That is, the left and/or right "set" has to be a proper class. Do you think this sequence is one? $\endgroup$ May 30, 2017 at 21:46
  • $\begingroup$ @JoelDavidHamkins : Any surreal $x$ can be written in a unique way as the sum $x_{\prec 1} + x_{\asymp 1} + x_{\succ 1}$ where $x_{\prec 1}$, the infinitesimal part, is infinitesimal, $x_{\asymp 1}$, the real part, is real, and $x_{\succ 1}$, the purely infinite part, has only strictly positive exponents in its normal form, or equivalently is a logarithm of some $\omega^y$. $\endgroup$
    – nombre
    May 30, 2017 at 22:01

1 Answer 1

7
$\begingroup$

$\DeclareMathOperator{\ee}{e}$If $\varepsilon$ is an infinitesimal surreal, the quantity $\log(1+\varepsilon)$ is actually equal to the formal sum à la Hahn series $\sum \limits_{n \in \mathbb{N}} \frac{(-1)^n\varepsilon^{n+1}}{n+1}$, and $\log(1+\varepsilon) - \varepsilon$ is negligeable with respect to $\varepsilon$. So the proof for real numbers can be applied here to show that the sequence converges to $\ee$.

As for $x_{\omega}$, this is $\exp(1 - \frac{1}{2\omega} + \frac{1}{3\omega^2} - ...)$ where $\exp(- \frac{1}{2\omega} + \frac{1}{3\omega^2} - ...)$ is infinitesimally close to $1$ because $a:= -\frac{1}{2\omega} + \frac{1}{3\omega^2} - ...$ is an infinitesimal. So the real part of $x_{\omega}$ is $\ee$.


For now I don't see what its normal form because some combinatorial cleverness seems to be required in unfolding $\sum \limits_{n \in \mathbb{N}} \frac{a^n}{n!}$. I'll edit this answer if I find something. In the meantime you can try to find it too: we know that exponents of $\omega$ in the $a^n$ and $\exp(a)$ will be negative integers. It might be easier to compute the exponential sum directly if you can find relations between the coeffifients $q_{n,k}$ of $\omega^{-k}$ in $a^n$ for different values of $n$.

$\endgroup$
6
  • 1
    $\begingroup$ @WillSawin That moment you forgot a dollar sign and the editing period timed out. $\endgroup$
    – user78249
    May 30, 2017 at 22:23
  • $\begingroup$ Doesn't this argument show that the limit is actually e? It seems to show that $(1+\epsilon)^{1/\epsilon}$ is equal to $e$ to within $O(\epsilon)$, so if $\epsilon$ gets smaller than the inverse of any ordinal, it should be equal ot $e$ within the inverse of every ordinal. $\endgroup$
    – Will Sawin
    May 30, 2017 at 22:46
  • $\begingroup$ @james.nixon Fixed. $\endgroup$
    – Will Sawin
    May 30, 2017 at 22:46
  • 2
    $\begingroup$ @WillSawin : $\DeclareMathOperator{\ee}{e}$$\DeclareMathOperator{\Ord}{Ord}$Yes, this shows that the limit is $e$, and for that matter, that $\exp$ can be defined as $x \mapsto \lim \limits_{\alpha \in \Ord} (1 + \frac{x}{\alpha})^{\alpha}$. What's missing in the anwser is the normal form of $x_{\omega}$. $\endgroup$
    – nombre
    May 31, 2017 at 7:11
  • 1
    $\begingroup$ @PaoloLipparini : I don't know if there is an easier way to define $s^{\alpha}$ for $\alpha \in \mathbf{Ord}$. There doesn't seem to be a strong connection between the standard surreal exponential function and ordinals. $\endgroup$
    – nombre
    Feb 14, 2020 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.