Timeline for Two (strictly related) proofs by induction of inequalities
Current License: CC BY-SA 3.0
24 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| S Jan 2, 2015 at 17:44 | history | bounty ended | CommunityBot | ||
| S Jan 2, 2015 at 17:44 | history | notice removed | CommunityBot | ||
| Jan 1, 2015 at 21:24 | history | edited | Vincenzo Oliva | CC BY-SA 3.0 |
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| Jan 1, 2015 at 20:30 | comment | added | Vincent | A bit late to the party here, but what is $a_1$? | |
| Jan 1, 2015 at 13:25 | comment | added | Vincenzo Oliva | *fedja's comments , I should not write at 4 am. @Pietro Thank you, I have modified something and fixed a typo, I hope now things are clearer. Let me know if I can improve the exposition further. | |
| Jan 1, 2015 at 13:22 | history | edited | Vincenzo Oliva | CC BY-SA 3.0 |
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| Jan 1, 2015 at 13:00 | comment | added | Pietro Majer | So $m=m(n)$; still there is a certain ambiguity in the formula $a_n=p_{m+1}a_{n+1}$: is $m+1$ there $m(n)+1$ or $m(n+1)+1$? And $q$? Is it given, or has it to be found; does it depend on $n$ or not? And the formula $a_n=p_q a_{n+1}$ remains really mysterious, as $a_n$ and $a_{n+1}$ have already been defined. I fear everybody still remains in a state of uneasiness described in fedja's initial comment. | |
| Jan 1, 2015 at 2:12 | history | edited | Vincenzo Oliva | CC BY-SA 3.0 |
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| Jan 1, 2015 at 2:09 | comment | added | Vincenzo Oliva | @PietroMajer It was previously $\lim_{n\to\infty}$, indeed because $m$ depends on $n$, I changed it after the comments of fedja. Yes, I think it was better before, I'm changing it back. | |
| Dec 31, 2014 at 23:31 | comment | added | Pietro Majer | The definition of $a_n$ suggests that $m$ depends on $n$, but then the $\lim_{n,m\to\infty}$ suggests that $m$ does not depend on $n$, and that $(n,m)$ is just a pair of non-negative independent integers. Was it $a_{n,m}$ instead of $a_n$? Not clear. | |
| Dec 28, 2014 at 1:04 | history | edited | Vincenzo Oliva | CC BY-SA 3.0 |
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| Dec 28, 2014 at 1:03 | comment | added | Vincenzo Oliva | @fedja As for $\lim_{n→∞}\frac{\log a_n}{p_m}$, I considered $p_m$ a function of $n$, as it is the largest prime factor of $a_n$. But I guess I can say $n,m→∞$. $p_{m+1}$ is the prime that follows $p_m$. So for example if $a_n=2^{10}⋅3^6⋅5^4, p_{m+1}=7$. | |
| Dec 28, 2014 at 0:42 | history | edited | Vincenzo Oliva | CC BY-SA 3.0 |
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| Dec 27, 2014 at 23:22 | comment | added | fedja | Now it looks a bit better :-). However I'm still a bit perplexed by the notation $a_n$ and the phrase $n\to\infty$ because there is no running $n$ anywhere else. Also, what on Earth is $p_{m+1}$ in this case? | |
| Dec 27, 2014 at 17:38 | history | edited | Vincenzo Oliva | CC BY-SA 3.0 |
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| Dec 27, 2014 at 17:22 | comment | added | Vincenzo Oliva | @fedja I'm sorry for that. Though I'm not totally sure how I could improve the exposition. Does stating $a_n=\prod_{i=1}^m p_i^{b_i}$ help? Anyway, perhaps you might be interested in this question, whose inequality implies $(1)$ here, and I guess looks a bit nicer (I hope I'll manage to keep your attention on my problem :D) | |
| Dec 27, 2014 at 16:50 | comment | added | fedja | Let's get the notation straight before talking about anything else: the opening phrase "Let $p_m$ be the largest prime factor of $a_n$" puts me in such a beautiful state of blissful ignorance about the indexation in the following three story formulae, that the proverbial ram looking at the new gate is an example of perfect comprehension and deep understanding compared to me looking at them ;-) | |
| S Dec 25, 2014 at 16:18 | history | bounty started | Vincenzo Oliva | ||
| S Dec 25, 2014 at 16:18 | history | notice added | Vincenzo Oliva | Draw attention | |
| Dec 23, 2014 at 10:55 | history | edited | Vincenzo Oliva | CC BY-SA 3.0 |
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| Dec 22, 2014 at 16:32 | history | edited | Vincenzo Oliva | CC BY-SA 3.0 |
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| Dec 22, 2014 at 10:51 | review | First posts | |||
| Dec 22, 2014 at 11:24 | |||||
| Dec 22, 2014 at 10:40 | history | asked | Vincenzo Oliva | CC BY-SA 3.0 |