Timeline for Is there a name for this fast growing function?
Current License: CC BY-SA 3.0
18 events
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| S Jun 14, 2016 at 19:07 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
Corrected typo in title, improved overall spelling, minor stylistic changes
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| S Jun 14, 2016 at 19:07 | history | suggested | jeq | CC BY-SA 3.0 |
Typo in title.
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| Jun 14, 2016 at 18:54 | review | Suggested edits | |||
| S Jun 14, 2016 at 19:07 | |||||
| Jun 14, 2016 at 17:47 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 15, 2016 at 17:18 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 16, 2016 at 0:06 | comment | added | Steven Stadnicki | (Also, pedantry requires me to belatedly point out that by mathematical standards these functions aren't particularly 'fast-growing' at all - they're dominated by functions at level 3 of the usual fast-growing hierarchies. :-) See en.wikipedia.org/wiki/Fast-growing_hierarchy for details) | |
| Nov 17, 2015 at 11:21 | answer | added | András Salamon | timeline score: 1 | |
| Nov 16, 2015 at 19:38 | comment | added | Douglas Zare | See math.stackexchange.com/questions/135787/…. | |
| Nov 16, 2015 at 19:34 | comment | added | Steven Stadnicki | The log of this product is just $\sum_{j=1}^n j^i\log j$ and this sum is easily approximable by all the usual methods (e.g., Euler-Maclaurin); in particular, the lead term is $\Theta(n^{i+1}\log n)$ and the constant is easily found. | |
| Nov 16, 2015 at 19:10 | comment | added | András Salamon | For the revised question, there appears to be a straightforward bound $2^{n^{i+1}/\log(i+1)} \le F(n,i) \le 2^{n^{i+1+1/(e\ln 2)}}$ (modulo errors on my part) but getting it tighter than that seems to require a lot of work. | |
| Nov 16, 2015 at 18:38 | history | edited | user76479 | CC BY-SA 3.0 |
added 38 characters in body
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| Nov 16, 2015 at 15:24 | comment | added | Emil Jeřábek | $n^{n^{i+1}}$ is too much, it should be rather something like $\exp((1+o(1))n^{i+1}(\log n)/(i+1))$. Well, the $\Theta(n^{n^{i+1}})$ bound is already contradicted by the $i=0$ case. | |
| Nov 16, 2015 at 14:57 | comment | added | Douglas Zare | It's not reasonable to guess $\Theta$-level precision of rapidly growing functions. Try expressing $\Theta(2^{n!})$ in a nontrivial fashion. Stirling's formula $n! \sim \sqrt{2\pi n} (n/e)^n$ does not narrow down $n!$ enough to approximate $2^{n!}$ up to a constant or anywhere close. | |
| Nov 16, 2015 at 12:30 | comment | added | András Salamon | $\log F(n,i)=\sum_{j=1}^n j^{i+\log\log j/\log j}$. | |
| S Nov 16, 2015 at 11:32 | history | suggested | Tadashi |
Added relevant tag
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| Nov 16, 2015 at 11:04 | review | Suggested edits | |||
| S Nov 16, 2015 at 11:32 | |||||
| Nov 16, 2015 at 10:45 | history | edited | user76479 | CC BY-SA 3.0 |
edited title
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| Nov 16, 2015 at 10:26 | history | asked | user76479 | CC BY-SA 3.0 |