Timeline for Growth of the hyperexponential
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 20 at 9:37 | comment | added | nombre | Yes it is. In fact the hyperexponential function induces a bijection between positive numbers with purely infinite support and so-called $\kappa$-numbers, which are the simplest representatives for the wider equivalence classes $a \asymp_K b$ if $\exists n \in \mathbb[N},\log_n(a) \leqslant b \leqslant \exp_n(a)$. If you impose only empty infinitesimal support, then you remain $\log$-atomic, and more precisely you are a real-iterate of $\exp$ applied at a $\kappa$-number. You can find a list of relevant identities in Section 6.1 of the pre-print "The hyperserial field of surreal numbers". | |
| Sep 20 at 9:24 | comment | added | user23467 | I notice that the bounds you gave for b are infinitesimally different from a (because log_n(a) is infinite). So you're saying that exp_ω maps this infinitesimal neighborhood to an exponential level. Does that mean that if the infinitesimal support of a is empty, exp_ω(a) would be the log-atomic representative of that exponential level? | |
| Sep 20 at 7:44 | comment | added | nombre | The image of all such $b$ under $\exp_{\omega}$ is the whole exponential level of $\exp_{\omega}(a)$. | |
| Sep 20 at 6:03 | comment | added | user23467 | I don't understand. How do you go from the single number a to an entire class? Are you saying that exp_ω(b) is the same for every b in that class? | |
| Sep 19 at 17:09 | history | answered | nombre | CC BY-SA 4.0 |