Timeline for Are hyperreal numbers isomorphic to formal power series?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 9, 2017 at 21:30 | comment | added | Joel David Hamkins | Yes, you've got it now. | |
| Apr 9, 2017 at 21:28 | comment | added | KConrad | 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 $1/(1-x)$. | |
| Apr 9, 2017 at 21:19 | comment | added | Joel David Hamkins | 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. | |
| Apr 9, 2017 at 21:14 | comment | added | KConrad | 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? | |
| Jan 24, 2015 at 14:43 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
deleted 185 characters in body
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| Jan 24, 2015 at 14:29 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |