Can this number $\ln \omega$ be written in $\{L|R\}$ form? What's its birthday?
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6$\begingroup$ All surreals admit a left/right set representation, and further a unique one of smallest birthday length; I suspect Nombre will be able to provide an explicit description. $\endgroup$– Alec RheaCommented Jan 26, 2022 at 11:56
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2$\begingroup$ @AlecRhea This is not a bad guess! $\endgroup$– nombreCommented Jan 26, 2022 at 13:52
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$\begingroup$ For more information regarding surreal numbers, transseries, and their structures of fields with log and exp, I suggest you look at Alessandro Berarducci's survey paper Surreal numbers, exponentiation and derivations which is a nice read. $\endgroup$– nombreCommented Jan 26, 2022 at 19:40
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$\begingroup$ @nombre from that paper it looks like $\exp w\ne e^w$ in surreals, yes? $\endgroup$– AnixxCommented Jan 26, 2022 at 22:26
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$\begingroup$ @Anixx What do you mean by $e^\omega$ if not $exp(\omega)$ for some exponential function $exp$? $\endgroup$– Alec RheaCommented Jan 27, 2022 at 0:24
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1 Answer
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In general this is taken to mean the value of Gonshor's logarithm at $\omega$. This was defined in the tenth chapter of his 1986 book An introduction to the theory of surreal numbers where you can find a justification for my answer.
In an informal way, the function $\ln$ is the "simplest" function that is eventually smaller than each power function $x\mapsto x^r$ for $r \in \mathbb{R}^{>0}$ but eventually greater than any constant function. So if $\ln(\omega)$ could be the simplest number that is greater than each real number but smaller than each power $\omega^r$ for $r \in \mathbb{R}^{>0}$, that would be nice.
Indeed $\ln(\omega)=\{ \mathbb{R} \ | \ \omega^{r}: r \in \mathbb{R}^{>0}\}$. In Conway normal form, this is a monomial $\omega^{\omega^{-1}}$. You can also write $\ln(\omega)=\{ \mathbb{N} \ | \ \omega^{2^{-n}}: n \in \mathbb{N}\}$, then the difference is that the elements in brackets are simpler than $\ln(\omega)$ in the sense of the simplicity relation on surreal numbers.
re-edit: my past answer for the birth day was wrong. In fact each $\omega^{2^{-n}}$ has birth day $\omega+\omega^2.n$, so the birth day of $\ln(\omega)$ is actually $\omega^3$.
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2$\begingroup$ No, it's infinitely greater than 1. $\endgroup$ Commented Jan 26, 2022 at 16:24
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2$\begingroup$ @Anixx Being a generalization is perhaps a little too vague a phrazing to deduce mathematical statements. In the Levi-Civita field, one can extend the logarithm to finite (i.e. real + infinitesimal) elements, using a general method of extending analytic functions in Hahn series fields. $\endgroup$– nombreCommented Jan 26, 2022 at 17:29
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2$\begingroup$ @Anixx A logarithm on the Levi-Civita field would not behave as one expects. If one wants a logarithm that retains some first-order properties of the real logarithm, then $\ln(f)$ should be smaller than $f^q$ for all positive rationals $q$ whenever $f$ is positive infinite. This can never be the case with Levi-Civita series since the set $\{f^q:q>0\}$ is always coinitial among positive infinite elements when $f$ is positive infinite. $\endgroup$– nombreCommented Jan 26, 2022 at 17:30
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1$\begingroup$ @Anixx Yes it would. But in fact, analytic functions can only be extended to finite Hahn series of the form $r+\epsilon$ where the real number $r$ lies in the domain (within $\mathbb{R}$) of the corresponding function, and $\epsilon$ is an infinitesimal series. So $\ln$ remains undefined in that case at infinitesimals. In $\mathbf{No}$, it is defined at infinitesimals but using some more structure on $\mathbf{No}$ that you don't have on the Levi-Civita field (but that you have on fields of transseries). $\endgroup$– nombreCommented Jan 26, 2022 at 18:13
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2$\begingroup$ @Anixx As for your second question, I cannot answer, since there is no $\omega$ in the Levi-Civita field, nor is there a way that I know of to define $x^{\frac{1}{x}}$. $\endgroup$– nombreCommented Jan 26, 2022 at 18:14