10

Can this number $\ln \omega$ be written in $\{L|R\}$ form? What's its birthday?

Anixx
  • 9,302
  • 5
    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. – Alec Rhea Jan 26 '22 at 11:56
  • 2
    @AlecRhea This is not a bad guess! – nombre Jan 26 '22 at 13:52
  • 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. – nombre Jan 26 '22 at 19:40
  • @nombre from that paper it looks like $\exp w\ne e^w$ in surreals, yes? – Anixx Jan 26 '22 at 22:26
  • @Anixx What do you mean by $e^\omega$ if not $exp(\omega)$ for some exponential function $exp$? – Alec Rhea Jan 27 '22 at 00:24

1 Answers1

15

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 $\omega^3$.

nombre
  • 2,337
  • Is not $\omega^{1/\omega}$ infinitesimally close to $1$? – Anixx Jan 26 '22 at 15:45
  • 2
    No, it's infinitely greater than 1. – Philip Ehrlich Jan 26 '22 at 16:24
  • @PhilipEhrlich Wikipedia says that surreal numbers is a generalization of Levi-Civita field. Does this mean that in Levi-Civita also $\ln \omega=\omega^{1/\omega}$?... On the other hand, Hardy fields are also a generalization of Levi-Civita, and this is not true there for sure (taking $\omega$ as a germ of $f(x)=x$ at infinity)... – Anixx Jan 26 '22 at 17:16
  • 2
    @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. – nombre Jan 26 '22 at 17:29
  • 2
    @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. – nombre Jan 26 '22 at 17:30
  • @nombre in Levi-Civita if logarithm can be defined for $\varepsilon$, then it can be defined for $\omega=1/\varepsilon$ as well: it would be the same, but with opposite sign, would not it? – Anixx Jan 26 '22 at 17:42
  • @nombre also, is not in Levi-Civita $\omega^{1/\omega}$ infinitesimally close to 1? – Anixx Jan 26 '22 at 17:43
  • 1
    @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). – nombre Jan 26 '22 at 18:13
  • 2
    @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 $x^{\frac{1}{x}}$. – nombre Jan 26 '22 at 18:14
  • @nombre well, $\omega$ is short for $1/\varepsilon$ in Levi-Civita. I wonder, how $\ln \omega$ works in transseries... – Anixx Jan 26 '22 at 18:22